CH EMISTRY & Chemical Reactivity John C. Kotz SU NY Disti nguished Teachi ng Pr ofessor State University o rNew York Co llege of O n eonta
Paul M. Treichel Profe ssor o f C he m istry Univers ity of Wisconsin-Madi son
John R. Townsend Associa te Pr ofesso r We st C heste r Un iversi ty of Pennsylvani a
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Isotopes 15 3 Iso tope Abu n dan ce I 54 Dete rm in ing Atom ic Mass an d Iso to pe Abun dance I 54
2.4
1.6
Physical and Ch em ical Changes I 17
Atomic 'Vei ght I 55 CUJe Stud)': Catc/tinK Cheaters with !sotoj)('S 158
2.5
The Periodic Table I 58 Devel o ping the Pe rio d ic Table 158 H istori cal Perspeaiues: T he Stot)' of the Periodic Table 159 Fe atures of t he Pe r iodi c Tab le 160 A Brief O verview of the Pe riod ic Tabl e an d the C he m ical Ele m e n ts 162
2.6
M ol ecules, Compounds, an d Fo rmulas 16 7 Fo rm u las 168 Mo lecu la r Mod el s 169
Case Study Ancient and Modem f lair Coloring 1 /8 CI IAPTER (;OALS REVISIT ED 120 KEY EQUA TI ONS 120
•
ST n )Y Q L' ESTI ONS 120
Let's Review: The Tools of Quantitative Chemistry I 24 Cop per 124 U n its of Measurement t 25 Tem pe rat ure Scale s 126 Le ngth , Vol um e , a nd Ma ss 127
v
Ionic Co m pou nd s : Formula'>, Nam es , and Properties I 70 Io ns I 71 Fo rm ulas of Io ni c Com pou nds I 74 Na m es o f Ion s I 76 Na mes o f Ion ic Com pou nd s I 77 Propertie s o f Ioni c Com po u n d s I 78
2.7
2.8
Molecular Compounds: Fo r m ulas and Names I 80
2.9
Atoms, M olecules, and the Mole I82 H istorical Perspectives: A mnleo Avogadro and HL~ Number I 83 Atoms a n d ~l o 1a r r"'f ass I 83 Mol e cules , Co m po u nds , a n d Mo la r Mass I 85
2.10 Describing Compound Fo r m ulas 188 Percen t Co m pos itio n I 88 Em p ir ica l an d Mo le cula r Form u las fro m Pe rcent Co m positio n I 90 Dete rmi ni ng a For m u la fro m Mass Dat a 193 A Closer look: Mass Sj)('clrom.f>fl)', Molar Mass,
3.8
Gas-Forming R eactions 1139
3.9
Oxidation-Reduction Reactions 1141 O xida tion -Reducti o n Re acti o ns a n d Elec tron Tr a nsfer 1 142 O xid at ion Nu m bers 1 144 A Closer Look: A n> Oxidation Numbers "Rm l" ? 1 144 Re cog ni zing O xid ati o n-Re d uc tio n Re acti o n s 1 146 Case Study: Killing Bacteria wilh Siluer 1 148
3.10 Classifying Re actions in Aqueous Solution 1 149 CHAPTER (;OALS REVI SIT ED 1 13 1 STU D Y Q UE ST IO NS 1152
Stoichiometry: Quantitative Information II About Chemical Reactions I 158 The Chemistry of a Sparkler 1 158
4.1
Mass R el ationships in Chemical R eactions: Stoichiometry 1 159
4.2
Reactions in \Vhich One Re actant Is Present in Limited Supply 1 163 A Stoichio rru-u-y Calc ula tio n with a Limi ting Re acta nt 1 163
CI IAI'TER GO ALS REVISIT ED I ss
4.3
Percent Yield I 168
KEYEQU ATJO NS 1 100
4.4
Chemical Equations and Ch emi cal Analysis 1169 Qu an tita tive Analysis o f a Mixtu re 1 169 Determ ini n g th e Fo rm u la of a Co m po u nd hy Com b ustio n 1 171
4.5
Measuring Concentrations of Compounds in Solution 1174 Solution Co ncen tra tio n: Mol a r ity I 174 Prepari n g Sol u tio ns of Known Co n ce n tra tio n I 177
4.6
pH, a Concentration Scale for Acids and Bases 1179 A ClOSN Look: Seri al Dilu tions 1 180
4. 7
Stoi chi ometry of R eactions in Aqueous Solution I 182 Sol ut ion Sto ich io met ry 1 18 2 Titra tio n : A iv1et hod of Ch e mi ca l A na lysis 1 183 C(1.\'(' Study: Hrno AJuch 5ialt IJ Ther(' in Smwater? 1 186 Standard izin g an Acid or Base 1 186 Det e rm ini n g Mol ar Mass b yTitra tio n 1 187 Ti tra tions Usin g O xid a tio n-Red u ct ion Re actio n s I 188 Case Stu dy: Forensi c Chem istry: Titnu ions fin d Food
4.8
Spectrophotometry, Anoth er Method o f Analysis 1189 Tran sm itt an ce , Ab sorbanc e , a n d th e Beer-Lambe rt Law 1 190 Spectro p ho tometric Ana lysis 1 19 2
and Isotopes I 95
2.11 Hydrated Compounds I 96 Case Study:
IVJUl I :~
in Those French Fries ? I 96
STUDY Q UE STIONS 1100
II Chemical Reactions I 112 Black Smokers 1112 3.1
Introduction to Ch emical Equations 1113 H istorical Perspectives: A nto ine Lau rentLanoisin; I i4 317 94 111 4
3. 2
Balancing Chemical Equations J 116
3.3
Introduction to Ch emical Equilibrium 1118
3.4
Chemical Reactions in Aqueous Solution 1121
3.5
Ions and Molecules in Aqueous Solution 1122 Solubili ty o f Io n ic Co mpo u nds in Wa ter 1 125
3. 6
Precipitation Reactions 1127 Predictin g th e Ou tcom e o f a Pre cipitatio n Reactio n 1 127 Net Ionic Eq ua tio ns 1 129
3. 7
Acids and Bases 1131 Aci ds a nd Rase s: Th e Arrh en iu s Defin ition 1 132 Acids a nd Rase s: Th e Bre n stcd -Lowr v Defini tio n 1 133 A Closer Look: T he Hydro ni u m IOll- T he H r Ion ill lla ler 1 13 4 Chemical Perspertiues: Su lfu ric A dd 1 13 5 Re acti o ns of Aci d s a nd Rases 1 136 O x id e s o f No n m e tals a n d Metals 1 138
vi
I
Contents
Tampn i ng
CIIAI'TER
1
188
(~OA LS
REVISITED I 194
KEY EQUATION S 1 195 STU J)" QU EST I ONS 1195
Principles of Chemical Reactivity: 11 Energy and Chemical Reactions I 208
O ther Fo ssil Fuel So u rces I 259 En viro nme n ta l Impa cts of Fos sil Fu el Usc 1260 En ergy in the Fu tu re: Choices and Alternatives I 262
A Hot Air Balloon 1208
5. 1
Fuel Cells I262
Energy: Some Basic Prin cipl es I 209 Co nserva tio n o f Energy 12 11 Temperat u re a nd He at 12 1 1 Syste ms and Su r rou nd in gs I 2 12 Di rcc tion al itv and Exten t o f Tran sfer o f H eat : Th e rm al Eq uili b rium j 2 12 A Closer Look: 1l7/fl1 t, H ml ? 1 213 Energy U ni ts 12 14 Chemical Perspectiv es: Food and Calories 1 215
A Hyd ro ge n Eco n o my I 26 3 Bioso u rccs of Ene rgy I 264 So lar Enc rg y I 26 5 What Does the Future H old for Energy? 1266 Sl ' GGESTE D READl N(;S I 266
STC I>Y Q L'EST IO:\S 1266
5.2
Specifi c Heat Capacity: Hea ting and Cooling 12 15 Q ua n tita tive As pec ts of En ergy Tr a n sferred a s l lea t l2 17
PART 2 ATOMS AND MOLECULES
5.3
Energy and Changes of State 1219 Case Studr Abba \' He/ rig fm tor 1 222
II The Structure of Atoms I 268
5.4
T he First Law of Thermodynamics 1222 En thalpy I 225 A Closer Look: P- F ll 'tn k 1 225 St a te Fu nc tio n s 1226
6.1
Electromagn e tic Radiation I 269
6. 2
Quantization: Planck , Einste in, Energy, and Photons 12 71 Pla n ck's Equati on I 271 Einstein ami th e Ph oto el e ct ric Effe ct I 27 3 Energy a nd C h e m istry: Us ing Plan c k's Eq uat io n I 27:1
6.3
Atomic Line Spectra and Ni els Bohr 1275 T he Bo h r Model of th e H ydroge n Atom I 2 76 Th e Boh r Th e ory and the Spec tra of Exc ite d Ato m s 1278 Case Studr It7w t M akes tlie Colon in Firro1orks ? 1 28 1
6.4
P article-Wave D uality: Prelude to Quantum M echanics 1282
Product- o r Reactant-Favored Reactions and T he r m o dyn am ics I 239 Case Study T he Furl ConlrovPrsy A lcohol raul Gasoline I 2 40
6.5
The Modern View of Electronic Structure: " rave o r Quantum Mech anics 128 3 Quantum Nu m be rs a nd Orb ital s 12KI) Sh e lls an d Subshc lls 12R6
CI IAPT ER G OA LS REV IS ITE D 124 I
6. 6
The Shapes o f Atomic O rbitals 1287 s Orb ita ls 1287 A CIOJa Look: f-f A tom Orbital SJwlJps-Wav f Fu n ctions mul No dI'S 1 28 9 IJOrb itals 1290 d Orb itals 129 1 / Orb itals 129 1
6.7
One More Electron Proper ty: Electron Spin I 291 T h e Electron Spi n Qua ntu m Nu m ber, 1TI,; 129 1 A Closer Look: Pa ramag netism an d Ferromagnetism 1 292 Dia m ag ne tism and Pa ram ag n etism 129:l ChemicalPerspea iues: QUflll lizn i SI)i1ls and AlR[ 1 294
Aurora Borealis 1268
5. 5
En thalpy Changes for Ch em ical Reactions 1227
5.6
Calo rimetry 1229 Co ns ta n t Pre ssure Calorime try, Mea su ring ~ H 1 2 29 Con stan t Volume Calori metry, Measu ring ~ U 12 3l
5.7
5.8
En thalpy Calculations 1233 He ss's Law 1233 Energy Lewl Diag ram s 123 4 Sta nd a rd En thalpie s of Fo rma tion 1 ~ 36 En th al py C han ge for a Re actio n 123 7 A Closer Look: II p,5.\ \ [ ,(I'll' and Equation. 5,6 1238
KEY EQUAT IONS 12 4 1 ST L' DY Ql:ESTIONS 12 42
INTE RCH APTE R
The Chemistry of Fuels and Energy Resources Supply a nd Dem and: The Bal an ce Shee t on Ene rgy I 255 Ene rgy Usag e 125 5 Energy Re so u rces 1256 Fossil Fuels 1257 Coal 1258 Natu ra l G as 1258 Pet ro le u m 1 ~ 59
I 254
CIIA PTER (;O A(.S REVISITE D 1295 KEY EQUAT I ONS 12Ytj ST L' l>Y Ql'ES Tl O~S
129 7
Contents
I
vii
The Structure of Atoms and Periodic II Trends I 304
8.3
At om Form al Charges in Cova le n t Mole cul es and Ions I 359 A Closer Look: Comj)(uiug Formal Charw and Oxidat ion ..V umhn 1 360
The Chromium-Bearing Mineral Crocoite, PbCr04 I 304
7. 1
Th e Pauli Ex cl us io n P rinciple I 305
8.4
Resonance 1361
7.2
Atomic Subshell En ergies and Ele ctron Assignments I 306 O rd e r' o f Subshcll En e rg ies an d Assign m en ts I :'l 0 7 Effective N uclear Charge, Z* I 308
8.5
7.3
Ele ctron Configurations of At oms I3 09 Ele ctro n Co n fig uration s o f th e Mai n Croup Elemen ts I 309 Elements o f Pe riod 313 13 Ele ct ro n Co nfi g urat ion s of th e Tra n sition Elem en ts 13 15
Exc e p tio n s to th e Octet Rule 1364 Com pou nd s in Which an A to m Has Fewer T han Eig h t Valence Electron s 1364 Co m po un ds in Whic h an A to m Has Mo re Th an Eig ht Vale nce Elect ro ns 1364 Mo lecules with a n Odd Nu m ber of Elec tro ns 1 ~ 6 6 Ca,H' Study: T he l mportun re (1an Odd-Electron Mole cu le, s o I 36i
8.6
Molecular Shapes 1367 Cen tral A tom s Surrou n de d O n ly by Si n g le-Bond Pai rs 1368 Ce n tral A toms with Sin g le- Bo nd Pa irs an d Lo n e Pa ir s 1:170 Mul tip le Bo n d s a n d Mo lecu lar Geom e try 1 37:1
8.7
Bond P ol arity and Electronegativity 1375 Ch a rg e Distr ib ution : Co m b in ing Formal Ch a rge an d Elc c tro ncga tiviry 1 377 A Closer look: f,'[ect IVn f'g alivil)' 1 J78
8.8
Bond and Molecular Polarity 1380 A Closer look: Visualizing Chargp Distribu tions (11a[ Molecu lar Polar ity-e-Electrosuuic Potential Surfaces ant! Pa rtial Cluu gt' 1382
8.9
Bond P roperties : Orde r, Length, Energy 1386 Bond O rd e r 1 ~S 6 Bond Le n gth 1 ~ 87 Bo nd Disso ciat ion Enthalp y 1388 Histori cal Perspectives: /) N A - lVa tsoll, Crick, ant! Fra nk lin 13 92
7.4
Electron Co n figura tio ns of Ions I 3 16 A Closer Look: Qu estions A bou t Tra nsition Element Electron Configurations I 3 J 7
7.5
At omic Prope rties an d P eriodic Trends I 3 19 Atomic Size I 3 19 Io n izati on Energy 1321 Ek-cu-on Affinit y I 32 4 Tre nds in Io n Size s I 326 Case Study: Met als in Biorhn nistl )' awl M edicine I 327
7.6
Periodic Trends and Chemical Properties I 328 CIIA PT ER (;OAI,5 REVISIT ED I :1:11 STU DY QUESTI ONS I 3:12
INTERCHAPTER
Milestones in the Development of Chemistry and the Modern View of Atoms and Molecules
I 338
G re e k Philosophers a nd Medieval Alchemists I339
8.10 DNA , Revisited 1392 CHA PT ER (; OAl.S REViSIT ED I :193
Chemists o f the 18th-19th Centu ries 1340
KEY EQUAT I ONS I 3'::1:')
Atomic Structure-Rema rkable Discoveries-1890s and Beyond I342 H istorical Perspectives: 20th,Cf'nlwy Gia n ts of Scinto' I 346
STU ny QUESTIONS I 3!tj
Bondi ng and Molecular Structure: Orbital Hybridization and Molecular Orbitals I 404
The Nature o f th e Ch em ica l Bond 1347 SlT(;(;ESTE J) REAJ)fI'\ ( ;S I 34 7
The Chemistry o f th e N o b le Gases 1404
STU In'QU ESTI ONS 1347
II Bonding and Molecular Structure I 348
9.1
Orbitals an d Theori es of Chemi cal Bonding 140 5
9.2
Valence Bond Theory I 406 T he Orbita l O verlap Mode l o f Bo n di n g 14()fi H yb rid izat ion of Atomic Orb ita ls 1408 Multi ple Bo nds 14 16 Ben zen e : A Special Case o f 1T Bo n din g 14 2 1
9.3
Molecular Orbital Theor y 14 22 Prin ci ple s o f Mo le cu la r Orbital T h eor y I 422 A Closer Look: Mo lecu lar Orbitalsfor Compounds Formed from p-Blodc Elements 1 42 9
Chem ical Bonding in DNA I 34 8 8.1
Chemical Bond Fo r m a tio n 1349
8.2
Covale n t Bonding and Lewis Structures I 350 Vale nce Ele ctro ns an d Lewis Sym bol s for A to m s I :tl)O Lewis Ele ct ron Do t Structures a nd th e O O eL Rule I 352 Drawing Lewis Electro n Dot Structures I 353 Predicting Lewis Structure s 1:\55
viii
I
Cont ents
Elec tro n Co n figurat ion s fo r H ct c ronu clca r Diat omic Mo lecules I 429 CaJ(' Stud)'." Two Chemical Bonding My steries I 4J () Re son a nce an d MO T he o ry 143 1 CHAPTER GO ALS REViS ITE D 143:1 KEY EQUA TI ONS 1433
ST L:I)'V QCESTIONS 14:14
Carbon: More Than Just Another Element 442
I
Cam p h o r, an "Aromatic" Molecule 1442
10.1
Why Carbon? 14 43 Struc tu ral Dive rsity I 443 Isom ers I 44 4 A Closer Look: Writing Formulas and Drawing St ru ctures I 445 Stab ility o f C arbo n Com po u nds I 446
Sickle Cd l Ane m ia I 500 En zyme s, Ac tive Sites, an d Lyso zym e 150 1 Nuclei c Acids I 503 N ucl eic Acid Structu re I 503 Protein Syn thes is I 50 4 T he Ri\'A \Vo rld a n d th e O rigin of Life I 506 Lipids and Cell Membranes I 507 Chemical Perspectives: AIDS awl Renerse Tra ns crip tase I 507 Metabolism 15 10 En c rgva n d A'I'P 1510 Oxidatio n-Re duct ion a n d NAn n 15 1 1 Resp ira t ion a nd Ph otosyn th e sis I 51 1 Concluding Remarks I 512 SL'(; GEST ED REAIH N(;S 1S l~ ST C DYQl'ESTl O;-';S 13 12
Chemical Perspea ioes: Chi mlit)' and FlejJ!lants I 447
PART 3
10.2 Hydrocarbons 1447 Alkanes I 44 8
STATES OF MATTER
A Closer Look: Flex ible Mo lecu les I 45 J
Alken e s an d Alkvn cs 145 3 Aroma tic Com pounds 1 458 A Closer Look: Petroleum Chemistry I 46/ 10.3 Alcohols, Eth ers, and Amines I 461 Alco hols a n d Eth ers 146 2 Propertie s of Alco h ol s an d Et hers 1464 Amines I 466 H istorical Pa .\j JPd iv PJ: M auori ne I 467 10.4 Co m p o u n d s with a Carb o nyl Group I 468 Alde hydes an d Keton es I 469 C arbox ylic Acid s I 47 1 Este rs 1 47 2 A Closer Look: C iu cose ami Sugan I 4 7J Ami de s 1475 A Closer Look: Fats an d Oils I 476 10.5 Polymers 1478 Classifying Pol ym ers I 478
m
Gases and Their Properties
I 514
The Atmosphere and Altitude Sickness I 5 14
11.1 G a.s P ressure 15 16 A Closer Look: Meas u ring Gas Pressu re I 5 J 7
11.2 G as Laws : The Experim ental Basis I 5 17 Bo yle 's Law: Th e Com p re ssibi lity of C ases I 5 17 Th e Effe c t of Tem pe ra tu re o n C as Volume: C harles's t aw 151 9 C ombin in g Boyle 's an d C ha rles's Laws: T he Ge ne ral Gas Law 152 1 Avo garlro 's Hyp oth esis 1.",22 11.3 The Ideal Gas Law I524 T he De nsity of Gase s 1 52 5 Calc ula ting the Mol a r Ma ss of a Ga s fro m P, F, a nd TDa ta I 526
1104 Gas Laws and Chemical Reactions I527
CaseSl uri)'." Bi odiesf'l-A Fuel / or the Futuret I 4 79
11 .5 Gas Mixtures and Partial Pressures I 530
Ad d itio n Po lym ers I 48 0 C on de nsa tio n Pol ym e rs 1484
11,6 The Kin etic-Mol ecular Th eory of G a.s es I5 32 H istorical Perspediues: Stu dies on Gases: R obert Boy le fUuiJarquf'J Clw rlf'J I 5 3J
Chemical Perspea iues: ,)'uIJer Dial JPrJ I 487
Mo lecul a r Spe ed an d Kin eti c Enc rj.,. '} I 533 Chemical Perspectives: T !lf' Ea rth:\ AtnwsIJ!lf'rl' I 534 Kin e tic-Mo le cu lar T heory a nd th e Gas Laws I 5 37
CHAPTER GO ALS REViS ITE D 14t\H ST n )\'QL:ESTlO;-';S 14HH
11.7 Diffusion and Effusio n I 5 38
INTE RCHAPTE R The Chemistry of Life-Biochemistry
I 496
Proteins I 497 Am in o Acids Are the Bu ildin g Blo cks of Protei n s 149 8 Protein Struc tu re a n d He m o gl o bi n 1 49 ~ )
11.8
Some Appli cations o f th e Gas Laws and Kin eti cMolecular Theory I 540 Se pa ra ting Isotope s I 54 0 De e p Sea Divin g I 5 40 Case Stud)'." You Stin k! I 54 J
Co ntents
I
ix
11.9 Nonide al Beh avio r: Real G as es 1542 CI IAI'TER GOALS REVISIT ED 1544 KEY EQUAT IONS 1:344
ST l 1DYQUESTIONS 154 6
ED Intermolecular Forces and Liquids I 554 Antarctica Scen e- Icebe rgs, Penguins, Snow, Ic c, and Fog 15 54
12.1 States o f Matter and In termol ecular Forces 1555 12.2 Intermolecular Forces Involving Polar Molecules 1557 Interacti on s Between Ion s a nd Mol ec u les wit h a Pe rm an e nt Dipole 155 7 Interactio ns Between Mo lec u les with Perm an e nt Dip o le s 1 55 8 A Closer Look: Hyd rated Salts 1559 H yd roge n Bo nd in g 1 56 1 Hydrogen Bo nding and th e U n usual Prope rtie s of Water 156:l A Closer Look: ll)'d mw'n Bonding in Biochemistry 1 565
12.3 Intermolecular Forces Involving Nonpolar Molecules I 565
Dipol cyl nduccd Dip o le Forces 1565 Lon do n Dispersio n Fo rces: Ind u ced Dipole /I nd uced Dipo le Forces 1 566 A Closer Look: M etha ne Hydr ates: An A nswn to ll-'orld Fuel Supplif's? 156 7 Su m m a ry o f Inte r molecula r Forces 1 56H
12.4 Properties o f Liquids 15 70 Vap orizatio n a nd Co n de nsa tio n 1570 Vapor Pre ssure 157:l Vapor Pre ssure , En tha lpy o fVa po rization , an d th e Clau sius-Clapeyro n Equa tion 15 75 Roi lin g Point 1 576 Cri tica l Te m pera ture and Pre ssure 1577 Surface Ten sio n , Ca pi llary Ac tio n , a nd Viscosity 15 78 Case Study: TJw .Mystn J of the l>i,m ptwaring Fin gn1Jrints 1 5 79 CI IAI'TER (;OAI,S REVISIT ED 15HO KEY EQUAT ION 158 1
STl1DYQUESTIONS I :'I HI
Ell The Chemistry of Solids I 588 Graphite to Graphene 1588
13.1 Crystal Lattices and U n it Cells 1589 A Closer Look: Packing Oranges 1 595 13.2 Structures and Formulas of Ionic Solids 1596 13.3 Bonding in Ionic Compounds: Lattice Energy 1599 Latt ice Energy 1599 Calc ula ting a Latt ice En tha lpy fro m Therm odyn ami c Data 1600
x
I
Contents
13.4 The Solid St at e : Othe r Kinds of Solid Materials I 602 Mo le cu la r Sol ids 1602 Network So lid s 1 602 Am orp ho us Soli d s 1603
13.5 Phase Changes Involving Solids 1604 Meltin g: Co nversion o f Sol id in to Liquid l 604 Su blima tion : Conversion o f Sol id into Vapor 1606
13.6 Phase Diagrams 1606 Water 160 6 Case Study: T hf' l Vorld \' I, i{;h k~ t Solid 1 607 Ph ase Diag ram s an d T he r mo dynamics 1 60R Ca rbo n Dio xide 160H CHA PTER (; OAI,5 REVISIT ED 16 10 STU DYQUEST IONS 16 10
II Solutions and Their Behavior I 616 Safe Flying I 616
14.1 Units of Concentration I 618 14.2 The Solution Pro cess 1620 A Closer Look: Supersaturated Solu tions
1 620 Liquid s Dissol vin g in Liquid s 1621 So lids Dissolvin g in Wat er 1622 Enthalpy of So lu tio n 1 623 En th alpy of So lu tio n : T herm o d ynam ic Data 1 625
14.3 Factors Affecting Solubility: Pressure and Temperature 1626 Di ssolvin g Gases in Liquid s: H e n ry's Law 1 626 Tem perature Effec ts on So lu hi litv: l.c C ha tcl icr's Prin cipl e 1 627
14.4 Colligativc Properties 16 28 Cha nges in Vapo r Pr essu re : Ra oult 's Law 1 629 Chemical Pt>rsjJl'(:tilJPJ: I lm l)' \ lruo and the Killer Lakes of Cmnm JO !I 1 630 Roilin g Po in L Elevati on 1632 Free zin g Point Depression 1 634 O smotic Pre ssu re 1 635 Co lliga t ive Prope rties an d Mo lar Mass Determ ination 16:l7 Colligative Prope rt ies of So lu tio ns Con ta ini ng Io ns 163 9 A Closer look: Osm osis roul Medicine 1 6]9 Cast> Study: H enry 's I,aw in a Soda Bottle 1 641
14.5 Colloids 1642 Types of Co lloids 1643 Su rfactants 1645 CHA PTER (; OALS REVISIT ED 1640 KEY EQUAT IONS 16 47
STUl n'Ql:ESTIONS I 0·18
INTE RCH APTE R The Chemistry of Modern Materials I 656 Me tals 1657 Bo nding in Me tal s I 65 7 Alloys: Mixtu res of Met a ls I 659 Semiconductors I 660 Bo nding in Se mi co nductors: Th e Ba nd C ap I 66 0 Applications of Se m ico n d uc tors: Diodes, LEDs. an d Transisto rs I 662
Ce ramics I66 3 Glass: A Diso rdere d Ceramic I 66:1 Fired Ceram ics fo r Special Pu rpose s: C em e n ts, C lays, a n d Refractor ie s I 665 Mod e rn Ce ra mics with Exce pt iona l Properti es I 666 Biomaterials: Learning fro m Nature I 667 The Future of Materials I 668 SUGGEST ED
READl N(~ S
I 669
ST C D YQUES TIO~S 1 6 6~)
PART 4 CONTROL OF REACTIONS
ED
15.5 A Microscopic View of Reaction Rates 1692 Conc e n tra tio n, Re acti o n Rate , an d Co llisio n T he or y 1692 Te mp e ra tu re , Reac tio n Rate, a nd Activation Energy 169:{ Activatio n En er gy 1694 Effect of a Te mpe ra tu re In cr e ase 1695 Effe ct of Mol e cu la r Orie n ta tion o n Rea ct io n Ra te 1695 Th e Arrhe n iu s Eq ua tion 1696 A Closer L ook: Reaction Coonlinnte Dingrrnns 1 697
Effe ct of Ca ta lysts on Re actio n Rat e 1699
15.6 Reaction Mechanisms 1701 Case Study: Enzyw,,,: Nature's Catalysts I i02 Molccu lari ty of Elem en ta ry Steps I 703 Rat e Eq ua tion s fo r Elemen ta ry Ste ps 1704 Mol ec u la rit v an d Reacti o n O rd e r 1704 Re action Mech a ni sms and Rat e Equ at ion s 170 5 CHA PT ER GOALS REVIS ITE D I ilO KEY EQ LTATI O NS I 7 1I STU DY QL'EST I O~S
17 12
Principles of Reactivity: Chemical Em Equilibria I 724 Dyn amic and Reversible! 1724
16.1 Chemical Equilibrium: A Review I 725
Chemical Kinetics: The Rates of Chemical Reactions I 670 Where Did the Indicator Go? 16 70
15.1 Rat es of Chem ical Reactions 1671 15.2 Reaction Conditions and R at e 16 76 15.3 Effec t of Concentration on Reaction Ra te 1677 Rat e Eq ua tion s 1678 Th e O rd er of a Rea ct ion I 679 T he Rat e Constan t, k 16 79 Det e rmin ing a Rat e Eq uatio n 1680
15.4 Co ncen tr a tio n- T ime Relationships: Integrated Rate Laws 168 3 Firs t-O rde r Rea ct io ns 1683 A Closer Look: Rate Latos, Rate Constants, an d Reaction Stoichi ometry 1 684
Sec on d-Order Reacti o n s 168 6 Ze ro-Ord er Rea c tion s 1687 Grap hi cal Method s fo r Det e rmi n in g Re acti o n O rde r a nrl the Ra te Consta n t 1687 I Ialf-Lifc and First-O rder Re actio n s 1690
16.2 The Equilibrium Co ns tan t a nd Re acti on Quotient I 726 Writin g Equi libr ium Co nsta n t Exp re ssio ns 1728 A Closf'r Looh: Equilibrium Cons tant Expressions for G(HI'SK- and s; 1 i30 T h e Mea n in g of the Equili briu m Co n stan t, KI 730 Th e Re acti on Q uotie n t, QI 732 16.3 Determining an Equilibrium Constant 1 734 16.4 Using Equilibrium Constants in Calculations 173 7 Ca lculat ion s Where th e Solu tio n Invol ves a Quad ra tic Ex p re ssion 17:{8 16.5 More About Bal anced Equations and Equilibrium Constants 174 1 16.6 Disturbing a Chemical Equilibrium 1744 Effe ct of th e Additio n o r Re mo val of a Reac tan t o r Prod u ct I 745 Effec t of Volu me Cha nges on Gas-Phase Eq u ilibri a 1746 Effect of Tem peratu re Cha nges on Equi libriu m Com posi tion 1748 Case Stu d)'.'Apj)lJi1lJfEqu ilibri u m C01/(1)t.5: T he H aberUmr:h ProU,H 1 749 CIIAI'TE R ( ;OAI.S REVISITE ll I 7~JO KEY EQLTATIONS I i51 STC \)YQl' ES TI O~S
1752
Co ntents
I
xi
m
The Chemistry of Acids and Bases
I 760
Aspirin Is Over 100 Ye ars Old! 1760 17.1
of Reactivity: Other Aspects lEI ofPrinciples Aqueous Equilibria I 810 Minerals and Gems-The Importance of Solubility I8 10
Acids an d Bases: A Review 176 1
17.2 The Brensted -Lowry Co n ce p t of Acids and Bases Ex te nde d I 762 Conj ugate Aci d-Base Pairs I 76 4
17.3 "Vater and th e pH Scal e I 76 5 Water Au to ion iza tion an d th e Wa ter Ion iza tion Co nstant , K.... I 765 Th e pi ' Scale I 767 Ca lcula ting pll I 7GH
17.4 Equilibrium Const an ts for Acids an d Bas es I 768 K Values fo r Polyp ro tic Acids I 77 '2 Aq ueous Solu tion s of SaILs I 773 A Lo ga rith m ic Scal e of Rela tive Acid Str e ngt h , p K" 1775 Rel a tin g t he Io n izat io n C onsta n ts fo r a n Acid a nd Its Conj ugate Base I 77.rs d
17.5 P redi cting the Direction o f Acid-Rase Re a ctions I 776
17.6 Typ es o f Acid-Base R ea cti ons I 778 T he React ion of a Str on g Acid with a Strong Rase I 779 T he React ion of a We ak Acid with a St rong Ba se I 7 79 Th e React io n of a Stro ng Acid with a \Ve a k Ra se I 7 79 Th e React io n of a ' Ve ak Ac id with a We ak Rase I 780 17.7 Calculations with Equilibrium Co ns tan ts I 780 Determ in in g Kfrom Init ial Conce n tra tions a n d Me asu red p H I 78 0 What Is the pH of a n Aq u eous So lu tio n of a We a k Ac id o r Rase ? 1782
18.1
The Common Ion Effect 18 11
18.2 Con trolling pH : Buffe r Solutions 1814 Gen eral Exp ressio ns fo r Ruffe r So lu tio n s I 81 f Prepari ng Bu ffer So luti on s I 8 18 H ow Do c s a Buffe r Ma in ta in p H? I 820 18.3 Acid-Base Titrations 18 21 Ca.H' Stu dy: Take A DprjJ Brfrlth.' I 822 Titrati on of a Stro ng Acid with a Strong Rase 1822 Ti tra tio n of a We ak Acid with a Stro ng Ra se 1824 T itration o f Wea k Pol yprotic Acids 1827 T itration o f a Weak Base wit h a Strong Acirl 1828 pH Indi cat o rs I 83 0 18.4 Sol ubility o f Salts 183 2 T he So lubi lity Pro du ct Cons ta nt, «; 1832 Rel a ting So lu b ility and «; 183 4 A Closer Look : Solubility Cal culations I 837 Solub ility a nd th e Co m mo n Ion Effect 1838 T he Effect o f Basic An io ns O il Sa lt Sol ubi lity I 840 18.5 Precipitation R eactions 184 2 K p a nd th e Reactio n Q u otie n t, (21843 K. p , the Rea ctio n Qu otien t, an d Pr e cip ita tio n Reac tion s I 844 18.6 Eq uilib ria Involving Co m p lex Ions 184 6 18.7 Solubility a nd Complex Ions 1847 CHA PTE R (; OALS REVI SIT ED I H49
17.8 Polyprotic Acids and Bases I 787
KE Y EQUAT I ONS 1850
17.9 The Lewis Concept of Aci d s and Bases I 789 Cast' Stu dy Uric A d d, Gou t, an d Bird Dmf1Jing\ I 789 Catio n ic Lewis Acids I 790 Mo lec ul a r Lewis Acids I 79 1 Molecu la r Lewis Rase s I 79 3
17.10 Molecular Structure, Bonding, a nd Acid-Base Behavior 1793 Acid Str e ng th of th e Hyd ro ge n H a lid e s, H X I 79 3 Co mpari ng Ox oacids, H N n ;,! a nd II N0 3 I 794 A Closer Look: A cid Strengt hs and Mo lecula r Structure I 795
Wh y Are Ca rboxylic Ac ids Brenstcd Acids? I 796 \Vh y Arc I Iyd ra ted Me tal Ca tio ns Bren su-d Acid s? I 797 Wh y Are Ani o ns Bron sted Base s? I 798 WI1\' Are Ammo n ia a nd Its De riva tive s Bro n stcd and Le\~·is Ba ses? I 79 8 CI IA PTER
( ~O A LS
REVI SITl:]) I i 99
KEY EQUATIO N S I HOO STU DY QUEST IONS I HOI
xii
I
Co ntents
STU DY QUESTIONS I K')O
Principles of Reactivity: Entropy and Free Energy
I 860
Can Ethanol Co n trib ute to Ener gy a nd Environmental Goals ? 18 60
19.1
Spontaneity and Energy Transfer as H eat I 862
19.2 Dispersal o f Energy: Entro py 1863 A Closer L ook: Rnsersible and l rreoersihle P r OU ,\ .H'S I 864 19.3 En tro p y: A Mi croscopic U nd ers tan d ing I 864 Dispe rsal of Energy 1864 Dispe rsal of Ma tte r : Dispe rsal of Energy Revisited I 866 A Su m mary: En tropy, En tro py C ha ng e , and En e rgy Dispe rsal I 868 19.4 Entropy Me asurement and Value s I 868 Sta ndard Entropy Valu es, So I 868 Determ in ing En tro py C ha nges in Ph ysica l a nd Che m ica l Pro ce sse s I 8 70
19.5 Entropy Changes and Spontaneity I 871 III Summ ar y: Spo n tan eous o r N o t? 187 4 19.6 Gibbs Fre e Energy I876 The Ch a nge in th e Gibbs Free Enc rr.,. ']', d GI 876 Gib bs Free En ergy, S po n ta nei ty, and Chem ical Equil ib riu m I 87 7 A Summa ry: G·ibhs Free Energy ( ~r Gand f}, rGO), th e Re act ion Qu otient «(2) and Eq u ilibri u m Consta n t ( K), a n d Re action Favo rability I 879 Wh at Is "Free" En e rgy? 1879 19.7 Calculating and Using Free Energy I 879 Sta nda rd Fre e Ene rgy o f Form at ion 187 9 Calc ula ting f}"G o, th e Fre e Energy Change fo r a Re act ion Un der Sta ndard Con d itio ns I 88 0 Free Ene rgy a nd Temp e ratu re l Sg l Case Stud)': Thrn nodynrunics and liuing Things I 884 Using th e Re lationship Betwe en f},r(;O an d KI885 f: I IAi'T ER {;O ALS REVISIT ED I SH6 KE Y EQUAT I ONS I tls7
ST U >Y QtJEST IONS 18 Hi
Principles of Reactivity: Electron Transfer Reactions I 896 Don't Hold onto That Money! 1896
20 .1 Oxidation-Reduction Reactions 1898 Balan ci n g O xidat ion-Redu cti on Equa tio ns 1 89 9 20 .2 Simple Voltaic Cells I 905 Vo ltaic Cel ls with Inert Elec tro des 1908 Electroc hem ica l o- n No tations 1 909 20 .3 Commercial Voltaic Cells I 909 H istorical Perspectiues: Frogs and Voltaic Piles 1 9 10 Prima ry Batteries: Dry Ce lls an d Alkaline Batteries 19 1 1 Secon da ry or Rechargea ble Batte rie s 19 12 Fu el Ce lls an d I Ivb rid Ca rs 19 I 4 20.4 Standard Electrochemical Potentials 19 15 Electromo tive Fo rce 19 15 Measu ring S ta nda rd Poten tials 1 9 16 Sta nda rd Reduct io n Potentials 19 17 A Closer Look: f:MF, Cell Potmtial, and Volfagp 1 9/8 Tabl e s ofSta ndard Redu ction Potentials 191 8 Using Tab les of Stan dard Reduct ion Potentials I 921 Rel at ive Str en gth s of Oxidiz ing and Reduci ng Age nts I 923 Chemical Perspeaiues: An Fledrochemicol Toothache! 1 925 20 .5 Electrochemical Cells Under Nonstandard Conditions I 925 Th e Ncrns t Eq uat ion 1 925
20 .7 Electrolysis: Ch emical Change Using El ectrical Energy 193 1 CaW'Study Ala ng fllu'sf' in tlre Oce ans 1 932 Ele ctrolysis of Molten Sa lts 1932 Electrolysis of Aq u eo us Solutio n s 193:1 20 .8 Counting Electrons 19 37 H istorical I'erspectiues: E/ntrochnllistl)' and Mi clm el Famdas 1 93 7 CIIA PT ER C OA I.S REVISITE D I 939 KEY EQ LTATIO NS 1940
STl: DY Q L'ESTIO\IS 1940
INTE RCH APTER
The Chemistry of the Environment I 948 The Atmosphere 1949 Nitrogen and Nitroge n O xid es 1 950 O xygen 1~).I') 1 O zon e 1952 C h loro fluorocarbo ns (Cres) an d Ozone 195 2 Carbo n Dioxide 195 ~ Climate Change I 954 G re en h o use G ases 1954
The Aqna Sphere (Water) I 955 T he O cean s 1 955 Drin king Wat c r! 956 Wat er Pollution 19 5 7 Chemical Perspea iues: Chlorituuion of Watn Su!,!,iips I 958
Green Chemistry 1959 Chemical Perspectives: Particulates awl Air Pollution
1
960
Sl 'GGESTED READl N( ; S I ~J6 1
STL' I>YQ L'ES TI O:\,S 196 1
PART 5 CHEMISTRY OF THE ELEMENTS Chemistry of the Main Group II The Elements I 962 Carbon and Silicon 1962
2I.I Element Abundances 1963 21.2 The Periodic Table: A Guide to the Elements 1964 Valence Electrons 19 64 Io ni c Co m po u n ds o f Ma in G ro u p Ele me n ts I 9 65 Mo le cul a r Com po un ds of Main Grou p Elem ents 196 6 A Closer Look: lly l mgOl, Heliu m, and Balloons I 968
20.6 Electrochemistry and Thermodynamics 1928 \ Vork a nd Free En ergy 19 28 Eo an d th e Eq u ilib rium Co ns ta n t I 9 29
Co ntents
I
xiii
21.3
H ydroge n 1968 Chemi cal a nd Physical Pr operti es of Hyd roge n I 968 Pr eparatio n of Hyd rogen I 969
21.4
T he Alkali Met als, Gro np 1A 197 1 Prepa ration of Sod ium and Potassium 197 1 Pr op e rties of So di um an d Po tassium 1972 A Closer look: T he Redu cing A bilil)' oIthe Alkali Alfl(J L~ I 973
Im po rta n t Lith iu m , So d iu m, and Potassium Co mpoun ds I 9 74 2 1.5 The Alkaline Earth Elements, Group 2A I 975 Prop erties of Calc iu m an d Magn esium 19 7(; Metallurgy of Magn esium 1976 Chemical Perspectives: Alkaline Har th Me tals
21. 10 T h e Halogens, Group 7A 11005 Preparat ion of th e Ele ments 1 1005 Fluo rin e Co m pou n ds 1 100 7 Ch lori n e Co m poun ds 1 1008 CHAPT ER GOALS REVISIT ED 110 10 STU D"QU EST IONS 110 11
Chemistry of the Transition ED The Elements I 1018 Memo ry Me tal I 1018
22 . 1 Properties of th e Transitio n Elements 1 1019 Elec tro n Co n figura tions 1 102 1 O xid at ion an d Reduct io n 1 1021
and BiololD' I 977 Calci um Min e ral s an d Th ei r App licat ions I 978 Chemical P"nj m ;ti l)(,S: 0IRo1TlrU/s, limestone, and Clulmj)(lgup I 978
21.6 Boron, Aluminum , an d the Group 3A Elemen ts 19 79 Che m istry of the C·Wllp 3A Elem ents I 97 9 Case Study: lIard Wain I 980 Boron Mine ral s an d Produ ction of the Elem cn tlDx l Me tallic Alu mi n u m an d Its Pro d u ctio n 19H1 Boron Co m po u n ds 198 3 Alum inum Co m po u n ds 1m~ 5 21.7 Silico n a nd th e Group 4A Ele ments 1986 Silic o n I 986 Silicon Dio xide 198 7 Silicate Mine rals with Ch ain an d Ribbon Structures I 988 Silicates with Sh eet Stru ctu res an d Alu minosili cat e s 198 9 Silicon e Po lyme rs 1990 Case Study L ead, Beethr m en, and a .I.\ fp t f'r)' Solvnl l 991
21.8 Nitrogen, Phosph orus, and the Gro up 5A Elements 1991 Propertie s of N itroge n an d Ph osphorus 1992 Nitroge n Co m pou n ds 1992 Case Study: ;\ Ilm lthy Saltuvuer Aquarium and the Nitrogen Cyrlp I 994 A Closer Look: M akin/; Phospho ru s 1 997
Hydroge n Com po unds of Pho sp horus a nd Oth e r C roup 5A Eleme n ts 1997 Ph osp horus O xides and Sulfide s 1997 Ph osphorus O xoacids a nd T h e ir Sa lts 19 99 21.9 Oxyge n, Su lfur, and th e G roup 6A Elements 110 0 1 Pr eparat ion and Properties of till' Elem e nts 1 1001 Sulfu r Co m po un ds 1 1003 A Closer Look: S not-tites and Su lfu r Chemistry 1 1004
xiv
I
Cont ent s
Chemical Perspraiues: Corrosion of Iron
1
1023
Perio d ic Tr ends in th e d-Block: Size, Den sity, Melting Po int 1 1024
22.2
Metallnrgy I 1025 Pvrom ctallurg y: Iro n Pro d u ctio n 1 102 6 Hvdrom ctal lu rgy: Co p per Production 1 l 02B
22.3
Coordination Co mpo un ds I 1029 Com p lexes a nd Liga nd s 11029 Fo rmu las of Coord in a tio n Com po un ds 1 1032 A Closer I.ook: H emoglobin
1
1033
Na mi ng Coord ination Co m po un ds I 10:14 22 .4 Structures of Co ordinatio n Co mpounds I 1036 Co m m o n Coord ination Ge o met ries 11036 Isom e rism 1 1036 22 .5 Bond in g in Co ordinatio n Co m p o unds 1 1040 T he d Orbitals: Ligand Fiel d Th eo ry 1 104 0 Ele ctron Co n fig urations an d Magneti c Propert ie s 1 1041 22.6 Co lo rs of Coordinat io n Co mpounds I 1045 Co lor 1 1045 T he Spec troc he mical Ser ies I 1046
22.7
O rga no m et alli c Chemistry: Th e Chemistry of Lo w-Valen t Me tal- O rgan ic Co mplexes 1 1048 Case Study: Accidental [)isw v f'r)' of a Chnno/hemp)'
AJ;e1I/ I 1049 Ca rbo n Monoxid e Com plexes of Metals I 1049 T he Effec tive At omic Nu m ber Rule a nd Hondi ng in O rganom et all ic Co mpound s 11()50 Ligands in O rgano metalli c Co m po un ds 11051 Case Study: Ferrocn U'- The Beginning ofa Chemical HI'1JUlu li01/ 1 1052 CHA PTE R (; OALS REVISIT ED I HFJ4
STU IJYQl TESTl ONS I I O;'}1
m
Nuclear Chemistry
I 1060
A Primordial Nuclear Reactor 11060
II Appendices I A-1 A
Usin g Logarithms a nd th e Quadrati c Equation I A-2
23 .1 Natural Radioactivity 1106 1
B
Some Important Physical Concepts I A-7
23.2 N uclear Reactions and Radioactive Decay 1106 2 Eq u a tio ns fo r Nu clea r Re action s I 1062 Radi oactive De cay Se ries I 1063 O th er Types of Radi oac tive De cay I 1066
C
Abbreviations and Use f ul Co nversi o n Factors IA-I O
o
Physi cal Co nsta nts I A·14
23.3 Stability of Ato m ic Nuclei 11067 Th e Band of Stabili ry an d Radi oacti ve De cay I 1068 Nuclear Bind in g Energy I 1069
E
A Brief Guide to Naming Organic Compounds I A-I7
F
Values for th e Ionization Energies and Ele ctron Affinities o f th e Ele m e nts I A-2 1
G
Vapo r Pressure of Water at Various Temperatures I A-22
23 .4 Rates of N uclear Decay 11072 Hal f-Life I 1072 Kineti cs of Nucl ea r De cay I 1073 Radiocar bon Dat ing I 1075
J
23.5 Artificial Nuclear Re actions I 107 7 A CloserLook: The Search forNem Elements v 1079
Solubility P roduct Co ns tan ts fo r Some Inorganic Com p o unds at 25 °C I A-26
K
Formation Co nsta nts fo r Some Co m p lex Ions in Aqueous Solution I A-28
L
Selected Thermodynamic Values I A-29
M
Standard Reduction Potentials in Aqueous Solution a t 25'C I A-36
N
Answ ers to Exe rcises I A-40
23 .6 N uclear Fission 11080 23.7 N uclear Fusio n 1108 1 23.8 Radiation Health and Safety 11082 Un its for Me asu ri n g Rad iat io n 110 82 Radiat ion : Dose s a nd Effec ts I 1083 A CloserLooh: lWw l !s a SajP Expo,w rr? 1 / 084 23.9 Ap p lica tio ns o f Nuclear Chem istry I 1084 Nu clear Medicin e: Me d ical Im agi ng 11085 Nuclear Med ici ne : Radiation T he ra py I IOSfi An alytical Me th o d s: Th e U se o f Rad ioac tive Iso tope s as Tra cers I iosn Analyt ical Me th ods: Iso to pe Dilu tio n I 1086 1\ C/OSfr L ook: Tn:!mpliuH/499m I I08 ? Space Sc ie nc e: Ne u tro n Activa tio n Ana lysis an d th e M o o n Ro cks I lOBS Food Scie n ce : Food Irra di a tio n I lOSS Case Stu d)'.' Nuclear Me di cine and Hyprrtll)'roidi.n ll I 108 9
H
Ionization Constan ts for Weak Acids at 25 °C I A-23 Ionization Co nsta nts fo r w eak Bas es at 25 °C I A-2 5
o
Answers to Selected Study Questions I A-62
P
Answe rs to Sel ect ed Inte rchapte r Study Q uestio ns I A-II8
Q
Answers to Ch apter Opening P uzzler and Cas e Study Questions I A-I 22
Index/Glossary
I 1-1
CIIA PT ER COALS REViSIT ED I 1090 KEY EQ UAT IONS 1 1090
ST L' DYQlJF.ST IO NS
1 1O ~Jl
Conten t s
I
xv
Go Chemistry Modules
<s:==-"'-".
The new Go Chem istry m o du le s arc min i video le ctu re s in clude d in ChcmistryNow that arc designed for po rtable usc
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Chapter 1
Basic Concepts of Chemistry
Mod ule 1
Exploring the Periodic Table
Chapter 2 Atoms, Molecules, and Ions
Mod ule 2
Ion Charges
Mod ule 3
Naming: Names to Formu las of Ionic Compounds
Module 4
The Mole
Mod ule 5
Solubility of Ionic Compounds
Mod ule 6
Net Ionic Equations
Mod ule 7
Simp le Stoichiometry
Chapter 3 Chemical Reactions
Chapter 4 Stoichiometry: Quantitative Information About Chemical Reactions
Module 8a Li miting Reactants - part 1 Mod ule 8b Li miting Reactants - part 2 Mod ule 9a pH- partl Mod ule 9b pH-part 2
Chapter 5 Principles of Chemica l Reactivity: Energy and Chemical Reactions
Module 10 Using Hess's Law
Chapte r 7 The Structure of Atoms and Periodic Trends
Module 11 Periodic Trends
Chapte r 8
Module 12 Lewis Elect ron Dot Structures
Bonding and Mo lecular Structure
Module 13 Molecular Polarity Chapter 9
Bon ding and Mo lecular Structure: Orbita l Hybridization and Molecular Orbitals
Module 14 Hybrid Orbitals
Chapter 10 Carbon: More Than Just Another Element
Module 15 Naming Organic Compounds
Chapter 11 Gases and Their Properties
Module 16 The Gas Laws and Kinetic Molecular Theory
Chapter 12 Intermolecular Forces and Liquids
Module 17 Identifyi ng Intermolecular Forces
Chapter 13 The Chemistry of Solids
Module 18 Unit Cells and Compound Formulas
Chapter 14 Solutions and Their Behavior
Module 19 Colligative Properties
Chapter 15 Chemical Kinetics: The Rates of Chemica l Reactions
Modu le 20 Half-Li fe and the Integrated First Order Equation
Chapter 16 Princip les of Reactivity: Chemical Equilibria
Mod ule 21 Solving an Eq uilibrium Problem
Chapter 17 The Chemistry of Acids and Bases
Mod ule 22 Eq uilibrium - pH of a Weak Acid
Chapter 18 Principles of Reactivity: Ot her As pects of Aqueous Equilibria
Module 23 Understanding Buffers
Chapter 19 Pri nciples of Reactivity: Entropy and Free Energy
Module 24 Free Energy and Equilibrium
Chapte r 20 Pri nciples of Reactivity: Electron Transfer Reactions
Module 25 Ba lancing Redox Equations
xvi
Preface he authof.S of thi ~ book have more, th an 100 years of ex perience teaching general che mis try and othe r areas o f che mis try at the co llege level. Altho ug h we have
T
be en a t diffe re nt in stitutions du rin g our ca ree rs, we
share several goals in co mmo n. One is to provide a bro ad overview of the principles o f che mis try, th e re activity of th e chemical ele me nts an d thei r co mpounds, and the ap plica tio ns o f che mistr y. To reac h th at goal with our stu-
de nts, we have tr ied to show th e close relatio n bet ween th e observat ions chem ists mak e of che m ical and physical changes in th e lab orat o ry and in nature and the way these cha nges are viewe d a t th e at omic a n d m ol e cular level.
Ano ther of o ur goals has bee n 10 cOllvey a se nse o f chem istry as a fie ld th at not o n ly has a lively history but also o ne that is curre n tly d ynam ic, with important new d evelo pm ents occ u rring evc ry yca r. Furthermore , we wan t to provid e so me ins igh t in to th e chemical aspec ts of th e wo rld aro u nd us. Indeed , a major obj ective o f thi s bo ok is to provide th e tools needed fo r you to functio n as a che mically literat e citize n . Learn ing so me thing o f the che m ica l world is just as important as understanding so me basic mathemati cs an d biol ogy an d as im po rtan t as having an appreciatio n fo r history, music, and literature. For exam ple, you shou ld know wha t mat e rial s are important to o u r ccono my, so me o f the reactio ns in plan ts and ani ma ls and in o u r e nviro n ment, and t he role tha t che m ists pl ay in protect ing th e enviro n me n t. These goals and our ap proach have bee n translated in to Chemistry & Chemical Reactioity, a bo ok th at has been used by mo re th an I million stu de n ts in its first six cd itions. \\'e are clearly gratifie d by th is success. But, at the same time , we know that the d et ails of o ur p resen tation an d organization can always be improved . In addition , th ere are signi fica n t advances in th e technology of co mmunicati ng information , an d we wan t to ta ke ad van tage of those ncw app ro ac hes. These have been th e impetus beh ind the p reparat ion of thi s new ed itio n, whic h incorporates a ne w o rganizat io n of m aterial , new ways to descr ibe co n te mpo rary uses of che m istry, new te chnologies, an d im proved int egra tion with ex isting techno log ies.
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Audience for Chemistry & Chemical Reactivity and OWL T he textbo ok and O\VL arc d esign cd for introductory co urses in che m istry for studen ts interested in fu rt her stu dy in scie nce, whe -the r th at scie nce is che m istry, biology, 'engince ring, gc o logy, physics, or related su bjects. O ur assu mp tion is th at stude n ts beginning thi s co u rse have had so me preparation in a lgebra and in gene ra l x vii
What's New in This Edition 1.
2.
3.
New chap ter introductions o n topics suc h as a ltitude sickness (page 5 14) a n d th e contribu tio n of ethan ol to e nv iro n me n ta l goals (page 860) . Ea ch of these c ha p te r-o pe n ing topics h as a question or two th at is a nswe re d in Ap pe n d ix Q.
On e o r m ore Case Studies arc p rese n te d in e ac h cha pter. These coyer p ract ical che m istry a n d pose que sti ons th at can be an swered u sing th e co nce p ts o ft hat cha p te r. Case Stu d ies cover suc h to p ics as silver in was h ing m ach in es ( page 148 ) , u sin g isotope s to ca tc h c hea te rs (page 58) , aqua riu m c he m istry (p ag e 99 2) , why garl ic stin ks (page ,r:j4 1). what is in th ose Fren ch fries (page 96) , why Beethoven died at an ea rly age (page 989) , and m an y o th e rs.
x vi ii
A bri ef di scu ssion of m odern organom etalli c c he m istry h as bee n ad d ed to Chap ter 22 .
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Ad d itio n a l ch all engin g que stion s h ave been ad de d to eac h chap te r.
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Ad d itio nal Chemical Perspea ioes and Case Studies bo xes h ave been au th o re d by J effrey Kacffab cr (U n iversity o f Flo rida) and Eri c Sce rr i (UC LA) .
5.
V T he OWL (O n line Web-based Learning) syste m ha s been u sed by ove r 100 ,000 stu d c n ts. The con ten ts of O\VI... are th e co n tents an d orga n iza tio n o f Chemistry & Chemical Rear!iv i!),. Fo r th e sixth edition , a bo u t 20 end-of-chapter que s~ tion s we re assignab le in O\VI.... Th at number h as been a p p rox im a tely d oubled fo r th e seve n th ed ition. In ad d itio n , the assets of Che m istryNowExercises, Tutorials, a nd Sim u latio ns t ha t allow stu de n ts to practice chemistry- are n ow full y in co rp o ra ted in O\VL.
6.
The new e-Book in OWL is a complet e electronic ve rs io n of th e te xt , fu lly assignable a n d linked to O\VI... homework. The e-book ca n be purch ased with th e printed bo ok o r as a n independent te xt replacement.
7.
... ~ . . __ ----!.S::.>- Go Chemistry modu les. T he re a re 27 mini-lectures that ca n be pl ayed on an iPod o r o th er ~lP3 pl aye r or on a com p u te r. The modules feature n arrated exam p les o f the most im pOl-tant material from each c h a p te r and focus on areas in wh ich we know from ex perie nce th at stu de n ts may n eed e xtra help.
New a n d co m p le tely rev ised l nterchup ters. John Ems ley, a no ted scie nce writer, rev ised th e int ercha p ter 011 the enviro n me n t (page 949) an d wro te a n ew intcrch aptcr o n th e hi story of chem istry (page 338).
4.
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"pH , a Concentration Scale
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pH is a way to express t he ocidity of a solution - t he concent ration of H,O· in solution.
Reorganization/addition of material:
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The first fou r c hapters in p articula r h ave been revised a n d co n de nse d .
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T he "mo les of reaction " concept is u sed in th e rmodyn a m ics.
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The m at e rial o n in te rmo lecular fo rce s (C hap te r 12) h as been se pa ra te d from so lids (C hap ter 13 ) .
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T he c h a p te r o n e n tro py a n d free e ne rgy (Ch a p te r 19 ) h as been thoroughly rev ise d.
Preface
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ISBN-IO: 0-49!i-38709-6; ISBN-13: 978-0-49!i-38709- 1
Ch e misrrvlvow's onlin e self-assessmen t tools give you th e ch oices a nd resou rces yo u need to study smarter. You can ex plo re a variety of tutori als, exercises, a nd simu lations (cr oss-re fere nced throughout th e text by margin a n no tatio ns) , view Active Figure interactive versions of key pieces o f a rt. from th e text, or tak e ch ap ter-spec ific Pre-Tests and ge t a Pe rsonalized Study Plan that di rects you to specific interactive mat eri als th at ca n h elp yo u maste r th e areas in wh ich you need add itio na l work. Includes access to . n-one tutoring an d Go Che mistry mini vide o lecon e.. o tu res. Access to Che mistryNow for two semesters may be included with each new te xtb ook or ca n be purchased at www.ichapters.com usin g ISBN 0-495-394 3 1-9.
A printed test bank of mo re th a n 1250 question s in a ran ge of diffi cu lty an d variety are correlate d d irectly to the ch ap te r sec tio ns found in the main text . Nu merical, ope n-ended, or co ncep tual p roblem s a re writte n in mu ltiple choice , fill-in-th e-blank, or short-answe r for mats. Both sing le- and multiple-step problems ar c presented fix eac h chap ter. Elec tro n ic files of th e Test Bank are included on the PowerLecture C n - R O~l \VebCT a nd Blackboard versions of th e test bank a re available on the in structor's co m pa n io n site a t academic.cengage.com /kotz.
Tr ansparen cies ISBN-lO: 0-490-38 714-2; ISBN-13: 9 78-0-49!i-387 I 4-5 A co llectio n of 150 full -eolor tr an sp arencies of key images sel ected from the text by th e a u th o rs. The Powe r Lecture C D- R O ~l includes all te xt a rt a n d man y photos to aid in preparin g tran sp aren cies for mat e rial not p rese n t in thi s set. xxii
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Go Chemistry for General Chemistry _
~"T'~'»
27-Mo d u le Set ISBN-IO: 0-490-3822 8-0; ISBN-13: 9 78-0490 -38228-7 These n ew mini video lectures, playable on video il'ods, iPhon es, and ~"IP3 p laye rs as well as on l'Tune s. include
anima tions and p roblem s for a qui ck su m mary of key co nc e p ts. In selected Go Che m istry modules, e-Flashca rds b riefly in troduce a key concep t an d the n test student u nde rstan d ing of the basics with a series of questions. Modules are also available se pa rately. Go Che mis try is in cluded in Chem istr yNow. To purch ase , enter ISBN 0-4 9 ~-3 82 2 8-0 a t www.ichapters.com.
v
OWL for General Chemistry See th e above d escription in th e in structor su p port materi als sec tio n .
Essential Math for Chemistry Students, Second Edition by David \V. Ball , Cleve land Stat e U nivers ity ISBN-IO: 0-495-01 327- 7; ISBN-13:
9 78-0-49~-0 132 7-3
This sho rt bo ok is inte nd ed for stude nts who lack confid en ce and /or co mpe te ncy in th e esse n tia l mathematical skills necessary to su rvive in general che m istry. Eac h cha p te r fo cu ses o n a specific type o f skill an d has wo rked -ou t ex amples to show how these skills tr an slate to che mical p roble m so lving.
Survival Guide for General Chemistry with Math Review, Se cond Edition b y Charles II . Atwoo d , Un ivers ity of Ge o rg ia ISBN-IO:
0-49~-3 87~1-7;
ISBN-13: 978-0-495-38 75 1-0
Intended to help you practic e for exa ms, thi s "su rvival gu ide" shows yo u how to so lve difficult probl em s by dissecting th em into m anageabl e ch u n ks. T he guid e includes three levels of profi ciency questions-A, B, and m inimal-to qu ickly build co nfide nce as yo u master the kn owled ge you nee d to succeed in yo u r co u rse.
For the Laboratory Brooks/Cole Lab Manuals Bro oks/ Cole offe rs a varie ty of print ed manuals to meet all general chemistr y labo ratory needs. Visit th e ch emistry site at academic.cengage.com/chemistry for a full listing and d escrip tion of th ese lab o ratory manuals and lab oratory not ebooks. All Bro ok s/ Cole lab man uals can be cu stom ized for yo u r specific needs. Signature Labs . . . for the customized laboratory Signa tu re Labs (www.signaturelabs.com ) co m bines th e res o urces o f Bro oks/ Col e , CE R, an d Oute rNet Publishing to provid e you un parall el ed se rv ice in ereati ng your id eal cus to mized lab p rog ram . Se lect the ex pe r iments and artw o r k yo u need fr om o u r co llection o f co n te n t and im ag ery to find th e perfect lab s
to m at ch yo u r co urse . Visit www.signaturelabs.com o r con tact yo u r Br oo ks/C o le r eprese nta tive for mo r e in formation .
Acknowledgments Becau se significa n t changes have been mad e fro m th e sixth editio n, preparing th is ne w ed itio n of ChemistrJ & Chemical Reactiv ity to o k almost 3 years o f co n tin uo us effo rt. However, as in o ur wor k o n th e first six ed itio ns, we have had th e su p po rt a nd en couragement o f o ur families an d o f so m e wonder fu l fri ends, co lleagues, and studen ts.
CENGAGE LEARNING Brooks/Cole Th e sixth ed ition of thi s bo ok was publish ed by Thomso n Brooks/ Col e . As o ften happens in th e modern publish ing in du stry, that co m pa ny was rece ntl y acq uired by an o ther group and the new nam e is Cengage Learning Brooks/Co le. In spite of th ese changes in own ership, we co n tin ue with the same excellent team we have had in place for th e previou s several years. The sixt h ed ition o f th e bo ok was very successful, in large part owing to the work of David H arri s, o ur publish er. David aga in saw us through much o f th e d evel opment of th is new ed itio n , but Lisa Lockwood recently assu med hi s du ties as o ur acquis itio ns edito r; she has considerable ex pe rience in textbook publishi ng and was also resp on sibl e fo r th e success of th e sixth ed itio n. will miss David but are lo ok in g fo rward to a clo se assoc iatio n with Lisa. Pet e r !\.IcGahey ha s been our th e Developmen t Ed ito r for the fifth an d sixt h editions and ag ain fo r this ed ition . Pe ter is blessed with e nergy, creativity, en th usiasm , intelligen ce , an d good humor. He is a trusted fri end and co nfidant and cheerfu lly answe rs our man y q uestions d u ring alm os t-daily phone ca lls. No book ca n be successful witho u t proper marketing. Amee Mosley was a great help in marketin g the sixth ed ition and she is back in th at ro le fo r thi s ed ition. She is kn owled geabl e ab out th e market a nd has worked tir elessly to br ing the boo k to eve ryo ne 's atten tio n. O ur team at Bro ok sZCo le is co m p leted with Tere sa Treg o , Prod uc tion Manager, an d Lisa Weber, Tec h n o logy Proj ect Ma nage r, Sche d ules are ve ry deman d ing in textbo ok publishin g, an d Teresa has helped to keep us on sche d u le . \Ve ce rta in ly ap precia te her organi zational skills . Lisa \Vebe r h as d irecte d the d evelopment of th e Go Chemistry m o dul e s a nd our ex pan de d use o f OWL. Peopl e outsid e of publishing often d o not reali ze th e nu m ber of people involved in p roducing a textbook .
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Anne William s of Grap h ic \\Torld Inc. gu ide d th e book th ro ug h its almo st year- lo ng p rodu ction . Mar cy Lunetta was the photo research er for th e book and was successful in fillin g ou r so me times o ffbeat requests for a particu lar photo .
Photography, Art, and Design Most o f th e co lo r photogr aphs for thi s edition were again be autifully created by Charles D. winte rs. He p rodu ced severa l dozen new images for thi s book, always with a creative eye. C ha rlie's wo rk ge ts better and better with eac h ed itio n. \I\'e have wo rked with Charlie for more th an 20 years and have be come clo se friends. \Ve listen to his j okes, both ne w and o ld-and always forge t them. When we fini sh th e book, we loo k forward to a kayakin g trip. Wh en th e fifth ed itio n was being plann ed, we br o ught in Patrick Harm an as a member o f th e team . Pat de sign ed th e first edition of th e General ChemisttyNow CD-RO l\'f , and we bel ieve its succ ess is in no small way co nnected to his d esign skill. For th e fifth ed itio n o f th e book, Pat wen t over almost every figu re, and almost ever y wo rd, to bri ng a fresh persp ective to ways to co m mu nicate che mistry and he did th e same for th e sixth edition. Once aga in he has wo rke d on designing and pr oducin g ne w illustrat ion s for the sevent h ed ition, and his creativity is o bvious in their clarity and beauty. Finally, Pat also d esigne d and produced th e Go Che m istry modules. As we have worked together so clo sely for so m any years, Pat has be come a good friend, as wel l, and we sha re int erests not o nly in beautiful books but in interestin g mu sic.
Other Collaborators \\re ha ve been fortunate to have a number o f other colleagues who have played valuable ro les in th is project. • Bill Vin ing (Sta te Un iversity of New York, On eonta ) , was the lead au thor of th e General ChemistryNow C [)- RO ~ f and of th e m edia asse ts in O \VL. lIe has bee n a fri end for man y years and re ce n tly too k th e place o f o ne o f th e au thors at SUN' Y~O ne o n ta. Bill has again appl ied h is co ns iderable e nergy and creativity in preparin g man y more O w l, questio n s with tutori als. • Su san \ bung (Ha rtwick Co llege) ha s been a goo d fri end and co llaborator through five editio ns and has again prepared the l nstru ctor's Resource Manual. She ha s a lways been helpful in p ro o fread ing , in answe ring qu estion s o n co n te nt, a nd in g iving us good advice . • Alto n Banks (N o rt h Carolina State Un ivers ity) has also been in volved for a number of ed itions p reparing the Studen t Solutions M anual: Roth Susan xxiv
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and Alto n have been very helpful in e nsuri ng th e acc u racy of th e Study Question answe rs in th e book, as well as in th eir res pec tive manu als. • Mich ael Mo ran (West Cheste r Un iversity) has upd ated and revised th e Study Guide th at was writte n byJohn To wnsend for the sixth ed itio n . T his book has h ad a histo ry o f excellen t study gu ides, and thi s m anual foll ows th at tr adition. • \Ve also wish to acknowled g e th e su ppo rt o f George Pu rvis and Fuj itsu fo r use of th e CAChe Scien tific so ftware for molecul ar modeling. All th e mol ecul ar mod els and th e el ectro stati c potential su rfaces in the bo ok were prepared using CAChe software . • J ay Freedman o nce again did a masterful job co mpiling the index/ glo ssary fo r thi s ed itio n. A major task is p ro ofreadi ng th e ho ok afte r it has been se t ill type. The bo ok is read in its en tirety by th e au thors and accuracy reviewe rs. Afte r mak ing co rrect jn ns , the bo ok is read a sec o nd tim e . Any errors rem aining at thi s point are ce rta in ly the resp o nsibility of th e au thors, a nd stude n ts and inst ru ctors sho uld co n tac t tbe au tho rs by e mail to o ffe r their suggestions. If thi s is d o ne in a timel y m an ne r, co rrec tions ca n be mad e when th e book is reprinted. \Ve wan t to th ank th e following acc u racy reviewers fo r th ei r invaluabl e assista nce. The book is immeasurably improved by th e ir work. • William Bro d erick, Monta na Stat e Unive rsity • Stephen Z. Go ld berg, Ade lp hi Un ivers ity • J effrey Alan Mac k, Ca lifo rn ia State Un iversity, Sacrame n to • Clyde Metz , Co llege of Charles to n • David Shin n, U n iversity of H awaii , Manoa • Scott R. White, Southern Arkansas University, Magnolia
Reviewers for the Seventh Edition • Ge ra ld M. Kore no wski. Renssel aer Pol ytechn ic In stitute • Robe rt L. La Duca, Michigan Sta te University • J effrey Alan Mack , Ca liforn ia Stat e Un iversity, Sacra me n to • Arman do M. Rivera-Figuero a, East Los Angel es Co llege • Dani el.J. Williams, Ken nesaw Sta te Un ivers ity • Steven G. \l\To od, Brigh am Young Univers ity • Roge r A. lIin richs, Weil! Corne ll Me dical Co llege in Q atar (reviewed th e Energy in te rc hap te r) • Leo nard Fine, Co lum bia Un iversity (reviewed the Mat eri als interch apter)
Advisory Board for the Seventh Edition As the new ed itio n was be ing plann ed , th is board listened 10 so me o f o ur id eas a n d m ade o the r suggestio ns . \\'e ho pe to co ntin ue our asso ciation wit h t hese e ne rgetic a n d c reative c hem ica l ed ucators. • Do nnie Hyers,Jo h nson Cou n ty Co mmun ity C.o llcgc • Sharo n Fe tze r G islaso n , U niversity of Illi no is, C hicago • Adrian George , Un iversity of Ne braska • George Gra n t, T id ewat er Community College, Virgin ia Beach Campus • Mich ael H am p to n. Uni versi ty of Ce ntra l Flo rid a • Milto n J ohnsto n, U niversi ty o f So uth Flo rida • .Jeffrey Ala n Mack, California Sta te Un iversity. Sacra men to
• William Bro d e rick, Mo n tana State Un iversity • Shane Street, Un iversity of Alabama • Xlart in Valla , University of Florida
About the Authors JOHN C. KOTZ, a State U n ive rs ity of New York Dis tingu ished Teac h in g Professo r, Em eri tu s, at the Co llege a t O neo n ta , was ed uca ted a t Washi ngton a nd Lee U n ive rsity and Cornell Unive rsity H e hel d Na tio na l In stitu tes o f Il callh postd o ctoral a p poin tme n ts at th e Un iversity o f Man ch ester In stitute fo r Scie nce a nd Te chnol ogy in Eng land and at Indian a U niver sity. I Ie has coa u tho re d three textb oo ks in severa l cd ino ns (/norgauir Cnemistrv; Chemistry & Chemical RractinitJ. and 'flip Chemical Ho rM) and th e General ChnnisI')'Nmo en-RO,\(. His research in inorga nic che mistr y an d electroc he m istry a lso h as been publish ed. li e was a Fulb rig h t Le cture r a nd Research Scholar in Portugal in )979 and a Visiting Pro fesso r there in 1 9 9 ~. li e wa s also a Visiting Pro fesso r a t the Institu te fo r C he mical Ed ucatio n (Un iversity of Wisconsin , 199 11992 ) , a t Auc kla nd Unive rsity in Xe w Zealand ( 1999 ). an d a t Pot ch efstroom University in Sou th Africa in 2006. li e has been a n invite d spe aker on che m ical educa tio n a t co nfere nces in South Afric a , Ne w Zealan d . and Brazil. H e also serve d 3 yea rs as a men to r fo r th e U.S. Nat ion a l Chemistry Ol ymp iad Team . l ie has rece ived several awa rds , amon g them a Sta te Un iversi ty o f Xe w Yo r k Chan ce llor's Award ( 19i 9). a Na tio nal Cat alyst Award fo r Exce lle nce in Teachin g ( 1992 ). th e Estcc Lc ct urership at th e Un ive rsity o f So u th Dako ta (l99R) , th e Visiting Scientist Award fro m the \\'este rn Co n nec ticu t Section o f the Ame rican Chem ica l Soc ie ty ( 1999), the Distin gu ished Ed uca tio n Award fro m th e Bingham to n (NY) Secti o n o f the Am erica n C he mical So ciet y (200 I ) , the SUNY Award fo r Re-
Left to right: Paul Treichel. John Townsend, and John Katz.
sea rch and Scholars hi p (200,:) , and th e Sq uibb Lectu reship in C he mistry at the U n ive rsity of Xorth Carolina-Ashe ville (200i). li e lIIay be co n tac te d bv email at ko tlj c@o neonta.edu .
PAUL M. TREICHEL received his B.S. d egree fro m th e Un iversity o f \\'isco nsin in 19:'jR and a Ph .D. from H arvard University in 1962. Afte r a yea r o f postdocto ral study in Londo n , he assu me d a faculty po sitio n at th e University o f Wisco nsin-Madi so n. li e served as d ep art me nt chair fro m 1986 th rough 199,:) and was awa rded a lI e lfaer Professors hip in 1996. He has held visiting facu lty po sitions in South Africa ( 1975 ) and in J ap an ( 1995). Retiring afte r 44 years as a fac u lty mem be r in 2007, he is cu rr en tly Eme ritus Professor o f Che mistry. During his faculty career he ta ug h t co urses ill genera l chem istry, inorgan ic chemistry, o rgano me ta llic che mistry, an d scie nti fic ethics. Pro fesso r Trei chel's research in organometallic a nd metal cluster che mistry and in mass spectr o me try, aided by 75 grad uate and u nd ergraduat e stud e n ts, has led to mo re th an 1iO pap ers in scie n tific j ournals. He may be co n tac te d by ema il at trc ichelescbem.wisc.edu. JOHN R. TOWNSEND, Associa te Pro fesso r o f Che m istry at \fest Ches ter Un ive rsity o f l'en nsylvau ia , co m plete d his B.A. in Che m istry as well as th e Approved Prog ra m for Teach er Ce rt ification in Che mi stry at th e University of De laware . Afte r a career teach ing h igh school science a nd math e m atics, he earned h is ~ I.S . an d Ph .D. in biophysic a l chemistry at Cornell Un ivers ity. At Cornell he also pe rfor med expe rime nts in the origins of life fiel d a nd received the DuP o n t Teaching Award . Aft e r teachi ng at Blo o ms bu rg Un ive rs ity, Dr. Tow nsend joined t he faculty at \\~e s t Ches ter University, whe re he coo rdi na tes the c hem istry e d uca tion program fo r pro spective high scho ol te ach e rs and the ge nera l che mistry lectu re p rog ram for scie nce majo rs. Hi s re search intere sts a rc in th e fiel d s of che mical ed uca tio n a nd bioc he m istry. li e may be co n ta cted by e mail a t jtown sc nd fcwcupa.ed u . Preface
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Contributors Wh en we d esign ed th is ed itio n, we d ecid ed to see k c he m ists outside o f our team to a u th o r so me o f th e su p p le me n ta l c h a p te rs a nd other m aterials. John Ernsley, University of Cambridge M ilestones in the Deoelopment oj Chemistrv and the Modern View of Atoms and M olecules a n d The Chemistry of the Entsinmment
Afte r 22 yea rs as a chem istry lecturer a t King 's Co llege London ,John Ems lcy be cam e a full -time sc ie nce write r in 1990. As the Scie nce Wri ter in Resid ence at Imperial Co llege Londo n fro m 1990 to 1997 , he wro te the "Mo lccu lc o f the Mon th " colu m n for The Independent n e wspa pe r. Emsley's main ac tivity is '''Tiling po pular scie nce bo ok s that feature che m istry a n d its rol e in everyd ay life. Recent publication s include J11e Consumer's Good Chemical Guide, whic h won the Scie nce Bo ok Pr ize of 199 5 ; M olecu les at an Exhibition; \-las it Something rouAte?; Nature 's Bu ilding Blocks; Th e Shocking H istory oj PhosphonlS; Hm it)', Vitality & Virilit)'; a n d The Elements of M urder: Hi s most recent book, published in 2007, is Better Looking, Belin Living, Belin Loving. Jeffrey J . Heaffaber, University of Florida Case Study: A Ilm lthy Aquarium and the Nitrogen Cycle J effr ey J. Keaffabcr received hi s B.S. in bi ology and che mistry a t Man chester Co llege, Indi an a, a nd h is Ph .D . in ph ysical o rgan ic c he m istry a t th e Un iversity of Florid a. After fin ish ing hi s d o cto ral wo rk, Kea ffa bc r joined th e enviro n me n ta l research a n d d evel opmen t ar m o f \Valt Disn ey Im agineering . He h as wo rke d as a m ari ne environmental co nsu lta n t a n d h as tau ght che m istry and oceanog ra p hy in th e Califo rnia Co m m u n ity Co llege syste m. Hi s research is in th e fields of m arin e e nvi ron m ental che m istry a n d e ngin ee ring, and hi s co n tr ibution s h ave in cluded th e d esign of nitra te re d ucti o n a n d ozo n e di sinfection processcs fo r seve ra l la rge aquarium proj ect s.
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Eric Sccrri, University of California, Los Angeles Historical Perspeaiues: The ~)'tm)' of the Peri odic Table Er ic Sce rri is a co n tinu ing lec tu re r in th e De pa rtm ent of C he m istry a n d Bioc hem istry at Un iversi ty of Ca lifornia, Lo s Angelcs. Afte r o b ta in ing an und ergraduate degree fr om th e University of London , a nd a m ast e r 's d e gree fro m th e Un iversity of Sou tham p to n, he o btained hi s Ph. D. in th e hi story a n d philosophy of sc ience fr om Kin g 's Co llege , London , fo cusing on the que stion o f th e reduction of c h e m istry to q ua n tu m m echa n ics. Sce rri is the founder and editor of th e internat io nal journal Fou ndations of Chemistry an d rece n tly a u thore d The Periodic Table: Its Stot)' and Its Significance (Oxfo rd Un ivers ity Pr ess, 200 7), wh ich ha s bee n d escri bed as th e defin it ive ho ok o n th e p erio di c table . H e is a lso th e a u th o r o f more th an 100 a rticles on th e history an d philo soph y of che m istry, as we ll as che m ica l educatio n. At UCLA, Sce rri regu larly teach es ge ne ra l che m istry classes of 350 stude n ts a nd sma ller classes in the h istory a n d philo soph y o f sc ie nce. Felice Frankel, H arvard University Cover photog ra ph As a se n io r re search fellow, Fel ice Frankel he ad s th e En vision ing Science p rogram at H arva rd Un iversi ty's Initi ati ve in Innovati ve Co m pu ting (lie) . Frankel's imag es have appeared in m ore th an :100 articles a n d covers in journ als a n d general a udie nce publ icati on s. I IeI' awa rds include th e 2007 Lcnnart N ilsso n Award fo r Scie n tific Ph otograph y, a Gugge n he im Fellows h ip, and g ran ts from the National Science Fou nda tion, the Nation al En d owme n t for th e Arts, th e Alfre d P. Sloa n Fou nda tio n, the G ra h a m Fo u nd a tio n for the Advanc ed Stud ies in th e Fin e Arts, a nd th e Ca mi lle a n d Ilenry Dr eyfu s Foundation . Frankel 's books in clu de On the Surfare of l 1linf.fs, Images oj till' Extraordinary in Science; and Envisioning Science: 111e Desi6J71 and Craft of the Science Image, a n d she has a regu larly a p pearing co lu m n, "Sig h tin gs," in American Scientist magazin e .
About the Cover
T he lo tu s is the n ational flower of In d ia a n d Vietnam. T he flowe rs, seed s, yo ung leaves, and rhizomes of the p lan t a rc edibl e a n d h ave been used for centuries in Asia a nd In d ia. H in du s associate th e lotus blossom with the story of creatio n , a nd Buddh ists bel ieve it rep resents purity of body, speech. and m in d. T hese id eas h ave ro me in part from th e fac t th at the lotus flo we r grows in a muddy. watery e nviro n me nt, bu t, whe n th e flowe r
a n d leaves o pen , th e m ud an d water a re co mple te ly she d to leave a clea n surface . C hem ists rece ntl y discovered the un d erl ying reasons for th is p henomeno n . First, th e surface of th e leaves is no t smoot h; it is CO\'crcd with micro- and na no stru ctu rc d wax crysta ls, and the se tiny bumps allow only min im al contact between th e leaf su r face a n d the wate r droplet. T hus. on ly about 2% to 3 % of th e drop let 's surface is ac tu ally in contact with the leaf. Se cond. th e surface of th e leaf itself is hydrophobic, th ai is, th e forces of attraction between water molecules a n d th e surface o f the leaf a re relatively weak . Because of strong h ydrogen bonding, wate r molecules within a d roplet a re strongl y a tt rac ted 10 one a no th er instead of th e leaf's surface and so form spherical droplets. As th ese droplets roll off of the surface . a ny dirt on the su rface is swe p t away. On less hydrophobic su r face s, wat er molecules interact more strongly with the surface a n d drops g lide off rather th a n roll off. T hi s se lf-clea n ing property of lo tus leaves has been ca lle d th e "lo tus effect ," a n effect bea u tifull y illus trated by the photogra p h on t he coye r of th is book. Chemis ts are n ow tryin g to mimic this in new mat e rial s th a t can be in corp o rate d in to co nsu me r product s su c h as selfcleaning textil es, p ain t, a nd roofing tiles.
Preface
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CONCEPTS OF CHEMISTRY
1
Basic Concepts of Chemistry
Sports Drinks Sports drinks are popular among at hlete s and nonathletes alike. The original sports drink, Gatorade, contained sucrose, glucosefructose syrup, citric acid, sodium chloride, sodium citrate. potas-
sium dihydrogen phosphate, and flavoring and coloring agents. This is not unlike the usual soft drink, but the carbohydrates in the sports drink provide only about half of the calories in fruit juice or in an ordinary soda or soft dri nk. To understand the chemistry behind sports dri nks, we have to know the names and com position of the various organic and inorganic com pounds in the dri nk and understand such important areas of chemistry as therm odynamics and colligative properties. As you study chemistry you will learn about these compounds, their structures, and their functions, and you will come to understand more about the ingredients in sports drinks, among other things. For now, think how you would describe a sports drink. Questions: 1. What are its physical properties? Is it a homogeneous or heterogeneous mixture? Is the density of a sports dri nk more or Less than that of water? 2. What is the volume of a bott le of a typical sports drink in milliliters? In liters? In deciliters? Answers to these questions are in Appendix Q.
Chapter Goals
Chapter Outline
Set' Owllln Goals Revisited (pagf 20) for Stud)' flues/ions keyed to these bro(Jl~ and assignable in OWL
Unde rstand the differences between hypotheses, laws, and theories. Apply the kinetic-molecular theory to the properties of matter.
Classify matter. Recognize elements, atoms, compounds, and molecules.
• Identify physical and chemical properties and changes.
1.1
Chemistry and Its Methods
1.2
Classifyi ng Matt er
1.3
Elements and Atoms
1.4
Compou nds and MoLecules
1.5
Physical Properties
1.6
Physical and Chemical Changes
n Fchruarv 2006 a n ath lete fr om th e U.S. was se n t home from the win ter Ol ympics Ga mes in Ital y becau se he had us ed a co m mo n treatment for hi s baldn ess. The re ason was, as the O lympic co m m ittee sta ted, th at t he rem ed y ca n al so he used to mask o the r, illegal drugs. Sim ilarly, ath letes were han ned from the Summ er O lympics Games in Greece in 2004 a nd th e win ne r of the 2006 Tour d e France was strip pe d of hi s titl e fo r usin g banned stero ids . How ca n these drugs he de tected o r identified? On Jun e 13 , 200 3, a co lo rless liquid arrived a t th e O lymp ic Anal ytica l Laboratory (O AL) in Los Angel es , Ca lifornia. T his labo rato ry an n ually tests a bo u t 25, 000 samples for the presence of illega l drugs. Am ong its cl ients are th e U.S. Ol ympic Co m m ittee, the Na tio nal Co lleg ia te Ath le tic Associatio n , and th e Natio nal Football League. Abo u t the tim e o f the U.S. O utdoor Track and Field C ham pio nsh ips in th e su mmer of2 003, a co ach in Co lorado tipped off th e U.S. An ti-Dn ping Agen cy (US ADA) t hat seve ra l at hle tes were usin g a new ste ro id . The coach had found a syr inge co ntainin g an unknown su bsta nc e and had se n t it to th e USAOA. T he USAOA che mists di sso lved the co n te nts of the syringe in a few millilite rs of an alco ho l a nd se n t th e so luti on to th e OAL. T hat su bm issio n initiated week s of intense wo rk that led to th e identification o f a previo usly u n known stero id th at was presumab ly being used hy athle tes (Figu re 1.1) .
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ChemistryJl"Now~
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Throughout the text th is icon int roduces an opportunity for self-study or to explore interactive tutorials by signing in at www.th omsc nedu.co m/ Iogt n.
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o The stero id testosterone. All steroids, including cholesterol, have the same four-ring structure.
I~ '---_ _
A molecular model of testosterone.
A photo of crystals of the steroid cholesterol taken with a microscope using polarized light.
FIGURE 1.1 The steroid te sto sterone. The unknown compound discussed in the text is a steroid closely related to testosterone.
... .... r f_
Dr. Donald Catlin, the director of the Olympic Analytical l aborat ory in l os Angeles, California.
To iden tify the u n know n su bsta nce, chem ists at th e Ol ym pic Analytical Lab oratory used a G C- ~'1S, an instr u me n t wide ly employed in for e nsic scien ce wo rk (Figure 1.2) . Th ey first pa ssed th e sample throug h a gas chro matograp h (GC ) , an instr u men t th at ca n se pa ra te different chemical co m po u nds in a mi xture . A GC has a smalldiam eter, coiled tube (a typical in side di ameter is 0.02 5 mm ) in which th e in side su rfac e has been specially treat ed so th at che m icals are attrac ted to th e su rfac e. T hi s tube is placed in an oven an d heate d to temperatures of 200 °C o r h igh e r. The su bsta nces in a sam p le are swep t alo ng th e tube wit h a stre am o f heliu m gas. Becau se eac h com po ne n t in th e sam ple binds diffe rentl y to the mat erial on the insid e su rface of th e tube , each co m ponent moves through the co lu m n at a differe n t rate and ex its from th e en d o f th e co lu m n at a diffe rent tim e . T hus, sepa ra tio n of th e co mpon ents in th e mi xtu re is ac h ieved. Afte r ex iting th e GC , eac h co m pou nd is routed d irectly into a mass spe ctro me te r (!\..'1S) . In a ma ss spec trome ter, th e compounds are bombarded wit h high -en ergy e lectro ns and each co m pou nd is turn ed into ion s, a che mical spe cies with an electri c charge . Th ese io ns are th en passed th rough a stro ng magn etic field, causing th e ion s to be defl ected. T he path an ion take s in th e magn eti c field (the exte n t of d efl ection ) is re la te d to its mass. The masses o f th e io ns are a key p iec e of info rm ati on that helps to identify th e co mpo u nd. Su ch a stra igh tfo rwa rd proc ess: separa te th e compounds in a GC and identify th em in an ~'l S . Wh at can go wro ng? In fact, man y th ings ca n go wrong th at re q uire in genuity to o vercom e . In thi s case, th e unkn own ste ro id d id not su rvive th e high temperatures of th e GC . It broke apa rt into pie ces, and so it was only po ssibl e to stud y th e pi eces o f th e o rigin al molecu le . However, th is gave e nough evide nce to co nvinc e scien tists th at th e com pou nd was in deed a ste roi d . But what ste ro id? Dr . Catlin , th e dire cto r of OAL, said th at one hypothesis was th at "the ne w stero id was mad e by people who knew it was not goi ng to be detect ab le ," tha t th e molecul e had been d esigne d in a way th at wo uld guara n tee th at it wo uld not be d et ected by th e sta nd ard GC-MS procedure .
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FIGURE 1.2 AGC·MS (gas chromatograph-mass spectrometer). AGC· MS is one of the major tools used in forensic chemistry. The GCportion of t he inst rument separates the compo nents in a mixt ure of volatile compounds. and the MS portion then analyzes and identifies them. The GC· MS pictured here has an automated sample changer (carousel. center). An operator willload dozens of samples into the carousel, and the instrument will then process tile samples automatically, with tile data recorded and stored in a computer.
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Basic Concepts of Chemistry
So , Ca tlin and hi s co lleague s set out to id en tify the stero id . The first thing th ey did was to make th e molecul e stable during the analysis. This was d one by attachin g new atom s to th e mol ecul e to mak e what che m ists call a devivatioe. A number of ap proaches to m aking d erivatives wer e tested, and , within a few wee ks, they beli eved they knew th e identity of the unkn own ste ro id . The final ste p in so lving the myste ry was to try to mak e a sam ple of the co mpou nd in th e laboratory and th en to use th e ga s ch ro ma tog ra p h and m ass spectr omct e r o n this sam pic . If th e mat erial behaved the same way as th e unknown sam ple , th en th ey co uld be as ce rtain as possibl e th ey knew the identify of what th ey had received from th e tr ack coac h. These ex peri men ts worke d , co n fir ming the identity o f th e co m po u nd . It was an e n tirely ne w ste ro id , never seen before in natu re o r in th e laboratory. Its form u la is C2 1I L.! 80 ~b a nd its nam e is tct rah ydrogcstrino ne or TII G. It resem bled two well-known stero ids : ge str in o ne, used to tr eat gynecological proble ms. and tr enbolone , a stero id used by ra nc he rs to beef up ca ttle. Th ere are two se q uels to th e story. Fir st, a scien tific problem is no t so lved un til it has been ve rifie d in ano the r laborat ory. No t o n ly was thi s d on e , but a test was so o n d evised to find TIIG in u rin e sam ples. Sec o nd , with new a nalytical procedures, th e USADA as ke d O AL to re test 550 u rin e sa m ples- an d THG was fou nd in several. Throughout this mystery, che m istry played a critica l role , from u sing the ana lytical technique of m ass spectrometr y, to using che m ica l intuition to d et ermin e po ssible arra ngemc n ts of th e molecul ar fragme n ts, and fin ally to syn thesizing th e d eriv ative and th e proposed co m po u nd. A knowledge of che mistry is cr ucia l to so lving problem s n o t only like this on e but man y othe rs as wel l.
1.1
• Athletes and Steroids What is t he problem with athletes taking steroids? THG is one of a class of steroids called anabolic steroids. They elevate the body's natural testosterone levels and increase body mass, muscle strength, and muscle definition. They can also improve an athlete's capacity to train and compete at the highest levels. Aside from giving steroid users an illegal competitive advantage, the potential side effects of steroids are liver damage. heart disease, anxiety. and rage. A check of the internet shows that there are hundreds of sources of steroids for athletes. The known performanceenhancing drugs can be detected and their users banned from competitive sports. But what about as-yet unknown steroids? The director of the OAL, Dr. Donald Catli n. believes there are other steroids out in the market, made by secret labs without safety standards, a problem he calls horrifying.
Chemistry and Its Methods
Che mis tr y is abo ut change . It was on ce o nly abou t cha ng ing on e natu ral su bs ta nce into an ot her-wood and oil bu rn ; grape juice turns into wine; and cin nabar (Figu re 1.3) , a red mineral fro m th e eart h, ch an ge s ultimately into shiny quicksilver (mercu ry). Che m istr y is still abo ut change, but now che mis ts fo cu s o n th e change of on e pure su bstanc e, whether natural o r syn the tic, into ano the r (Figu re 1.4) .
FIGURE 1.3 Cinnabar and mercury. (a) The red crystals of cinnabar are the chemical compound mercury(II) sulfide. (b) It is heated in air to change it into orange mercury oxide, which, on further heating, is decomposed to t he elements oxygen and mercury metal. (The droplets you see on the inside of the test tube wall are mercury.)
(a)
(b)
1.1
I
Chemist ry and Its Meth ods
3
Solid sodium, Na
Sodium chLoride solid, NaCl
• Ch lorine gas, Cl2 FIGURE 1.4 Forming a chemical compound. Sodium chloride, table salt, can be made by combining sodium
metal (Na ) and yellow chlorine gas (el2 ) . The result is a crystalline solid, common salt. (The tiny spheres show how the atoms are arranged in the substances. In the case of the salt crystal. t he spheres represent electrica lly charged sodium and chlorine ions.)
Althoug h che mistry is end lessly fascin ating- at least to che m ists-w hy sho u ld you study che m istry> Eac h pcrso n p robabl y has a d iffer ent answer; but man y stud en ts take a che mistry co u rse because so meo ne else has d ecid ed it is an impo rt ant part o f p re parin g fo r a particular ca reer. Che m istry is especially useful becau se it is ce n tra l to ou r u nd erstand ing o f d isciplines as d iverse as biology, gcoI06'1', mat erials science, m edicin e , p h ysics, and m an y b ra nch es of e ngi nee ring . In ad dition, chem istry p lays a maj o r role in the eco nomy o f deve lo ped nations, an d che mistry and che mica ls affec t ou r daily lives in a wide varie ty of ways. Fu rt he rmore, a co urse in che mis try ca n hel p you see ho w a scie n tist thin ks about the worl d an d how to so lve problems . Th e know ledge and skills d evel oped in suc h a co u rse will benefit you in man y ca reer paths and will hel p you becom e a better-in forme d citizen in a world that is becom ing technologically mo re co mp lex-and mo re in teresti ng .
Hypotheses, Laws, and Theories As scientists, we stu dy question s o f our own choosing o r ones tha t so meo ne el se poscs in th e hope of find ing an answer or o f d iscove ri ng so m e useful informati o n . In the story of the ba n ned ste roid, TH G , the chemists at the O lym pic Ana lytica l Labo rato ry were hand ed a p roble m to so lve, an d they fo llowe d the usu al m ethods o f scie nce to ge t to th e answe r. After some prelimin ar y tests th ey recognized th at the mystery su bstance was pro ba bly a ste roi d . That is, they formed a hypothesis, a ten tative explanation or prediction based o n experimental o bse rva tio ns. After formulating o ne o r more hypotheses, scien tists perfo rm ex pe r ime n ts designed to give res ults tha t con firm or inva lida te these h ypot heses. In che mistry th is usually requi res that bo th quan titative and q uali tative inform ation be co llected .
4
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Basic Co ncepts of Chemistry
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-- 140mL
....... 120~"
_100
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Quantitative informatio n is nu me r ical d at a, suc h as th e temperature a t wh ich a c he m ica l su bsta nce m el ts or its m ass (Fig u re 1.5 ) . Qualitative in fo r m a tio n, in co ntrast, co nsis ts o f n o n num e ri cal o bse rva tio ns, suc h as th e co lor of a su bsta nce o r its ph ysical a p pearance. The ch emists a t th e GAL asse m b led a grea t deal of qu alitative a n d qu antitative information on various d rug s. Based o n th e ir ex perie nce. a n d o n publish ed re po rts of ex periments don e in th e p ast by other c hem ists who stud y ste ro ids , th e y becam e more certain th at th e)' knew th e id entity of th e su bsta nce. T he ir preliminary exp e rim en ts led th em to perform still more ex p e rime n ts. suc h as lo okin g for a way to sta bilize th e molecul e so it wo u ld no t d eco mpose , an d they look ed fo r a way to make th e molecu le in the laboratory. Fin al confirmati on carne wh en their wo rk was re p ro d uce d by scie n tists in othe r la bo rato ries. After scie n tists h ave do n e a number of ex perime n ts a n d th e re sults h ave been ch ecked to ensure th e y a re reproducible , a pattern of behavior or results m ay e merge . At thi s point it m ay be possibl e to su m marize the observations in the form of a ge ne ra l ru le or co n clusio n . After makin g a nu mber of ex perime n ta l observations, th e che m ists at O AL could co n clu d e , for ex am p le , that th e unknown su bsta nce was a stero id be cau se it h ad p rope rti es chara ct e ri stic of man y other stero ids th ey h ad o bse rve d . Fin ally, aft er numerous ex p e r ime n ts by man y scie n tists over a n extended period of time , th e orig in al h ypoth esis may become a law- a co ncise ve rb al o r math ematical sta te me n t of a be h avio r or a rel ation th at is co nsiste nt ly observed in n ature with o u t con tra d ictio n . An example mi ght be the law ~fm ass conservation in c he m ica l react ion s. \Ve base much of what we do in science on laws bec au se th ey h elp us p redict what may oc cur under a new set of circumstance s. For example, we know from ex perie nce that if th e c he m ica l element so d iu m co mes in co n tac t with wate r, a vio le nt reactio n occu rs a n d new substa n ces are for med (Figu re 1.6 ) , a n d we know that th e mass of th e su bs ta nc es produced in the reacti on is ex ac tly th e samc as th e m ass of so d iu m a n d water used in th e reaction . Th at is, r!lass is alwa)'.'; conserv ed in chemical reactions. Rut th e resu lt of a n expe rim ent mi ght be di fferent from what is ex pecte d ba sed o n a general rule . when th at happen s, ch e m ists ge t excited bec au se ex pe ri me nts th at do not follo w our ex p ec tatio ns a re often the most interesting. \ Ve know th at understandi n g the exc e p tio ns a lm ost in variabl y g ives n e w in sights.
FIGURE 1.5 Qualitative and quanti tative observations. Anew substance is form ed by mixing two known substances in solution. We can make several observations about the substances involved. Qualitative observations: The solutions before mixi ng are colorless and yellow; a yellow, fluffy solid is formed on mixing. Quantitative observations: Mixing measured volumes of the solutions produces a measureabte mass of solid.
FIGURE 1.6 The meta llic element sodium reacts with water.
1.1
I
Chemistry and Its Methods 5
O nce eno ug h re pro d uc ib le ex perime nts h ave been con d ucted, an d ex pe rime ntal results h ave been gene ra lize d as a law o r ge nera l rul e , it may be po ssibl e to co nceive a th eory to ex p lain th e obse rvatio ns. A theory is a well-tes te d, unifyin g p rin ciple th at ex p lains a body o f facts a n d th e laws b ased o n th em . lt is ca pab le o f suggesting n ew hypotheses th at ca n be tested ex p e rime n ta lly. So me times n o nscie n tists u se the word "theo ry" to im p ly th a t so meo ne h as m ade a guess an d that an id ea is not ye t su bstantiated. Bu t, to scie nt ists, a th eory is based on carefu lly d etermin ed and reproducible ev ide nce. T heo ries are th e corn e rsto ne of ou r u nde rsta ndin g of th e n atural world a t a ny give n tim e . Rem embe r, th ough , th at the o ries a re in ven tion s of th e hum an mi nd . Theories ca n a n d d o change as ne w fac ts a re un cove red.
Goals of Science Th e scie nces, in cluding che mistry, h ave seve ra l goals. T\\TO of th ese are p redicti on a n d con tr o l. \Ve do e xp eriments a n d see k ge ne ralities b ecause we want to b e a b le to predict what m ay oc cur und er a give n se t o f circ u ms ta nces. \Ve also wa n t to kn ow h ow we mi ght co n tr o l th e outcome of a c he mical reac tio n or process. A thi rd goal is ex p lanatio n a nd u nd erstandin g . \Ve know, for exam p le , th at certain el ements suc h as so d iu m will react vigo rously with wa te r. But why sh o u ld thi s be true ? To e xp lain an d unde rstand thi s, we turn to the o ries suc h as those d eveloped in C hap te rs 6 a n d 7,
Dilemmas and Integrity in Science
• The Stem Cell Scandal of 2005 In 2004 -2005 researchers at Seoul National
University in South Korea published several papers in which they claimed to have cloned DNA from human embryonic stem cells. Because it was such an important discovery, scientists around the world eX amined the data closely. Within months, however, it was discovered that many of the reported results were based on fabricated data, and the results were retraded. Misrepresentation of information does great harm because it misleads scientists into spending time, energy, and funds in trying to replicate the infurma tion and using that information in other experiments. The stem cell scandal illustrates, however, the ways in which the scientific community ensures that scientific research is correct and accurate. 4
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You m ay think re search in sc ience is stra ig h tfo rwa rd: Do an experiment; draw a co n clus io n. But, rese a rch is se ldo m tha t easy. Frustra tion s an d di sappoin tm e n ts are com m o n e n o ug h, a n d results ca n be in co n cl usive. Experim en ts so me times co ntai n so me level o f uncertainty, a n d sp u r io us o r co n tra d icto ry d at a can be co llect ed . Fo r exam p le , su p p ose you do a n ex pe ri men t ex p ecting to find a di rect re latio n b etween two ex per ime n ta l quantities. 'rou co llect six d at a sets. Wh en p lotted . fou r of th e se ts lie on a stra igh t lin e , but two others lie far away from th e lin e. Should yo u ign ore th e last two points? Or sh o u ld you do m ore ex perim e n ts whe n yo u know th e tim e they ta ke will m ea n so m eo n e el se cou ld publish th eir resu lts fir st an d th us get th e cr edit for a new scie n tific principle? Or sho u ld you co ns ide r that th e two points not on the lin e mi ght indicat e th a t you r o riginal hypothesis is wro ng, a n d th at you will h ave to a ban d o n a favo rite id ea yo u h ave worked 011 fo r a year? Scientists hav e a re sp onsibi lity to remain o bjec tive in these situa tio ns, but it is so metimes h ard to do. lt is important to reme mber th a t scien tis ts are hum an a n d th erefore subject to th e sa me moral pressures a n d dil emmas as any other person. To h e lp ensure integ rity in scien ce , so me sim p le principles h ave e merged ove r tim e th a t gu ide scie nti fic p racti ce:
• Ex pe rime n ta l resul ts sh o u ld be reproducible . Fu rt h e rm o re , th ese resu lts sh o u ld be reported in th e scien tific lite ra tu re in su fficie n t deta il that th ey can he u sed or re prod uce d by others. • Con cl us io ns sh o u ld be reaso n abl e a n d unbi ased . • Cr edit sh ou ld be g iven whe re it is due.
Basic Concepts of Chemistry
Chemical Perspectives
Moral Issues in Science
Moral and et hical issues frequently arise in
The problem did not end there, however. Small lizards also normally ate t he caterpillars and helped to keep t hat population in check, and the lizard population was controlled by cats. However, DDT was passed up t he food chain from caterpillars to lizards to cats, and cats began to die. When the cat population declined, this led to an infestatio n of rats. Un intended consequences, indeed. DOTuse has been banned in many parts of the world because of its veryreal, but unforeseen, environmental consequences. The DDT ban began in the United States in 1972 because evidenceaccumulated that the pesticide affected the reproduction of birds such as the bald eagle. As expected, the ban on DDTaffected the control of ma laria-carrying insects. Several million people, primarily children in subSaharan Africa, die everyyear from ma laria.
science . One of t hese concerns t he use of the
pesticide DDT. This is a classic case of the law of "unintended consequences." The pesticide was developed during World War II and promoted as effective in controlling pests but harmless to people. In fact, it was thought to be so effective t hat it was used in larger and larger quantities around the world. It was
especially effective in controlling mosquitoes carrying malaria, although it was soon evident t here were consequences. In Borneo, the World Healt h Organization (W HO) used large quantities of DDT to kill mosquitoes . The mosquito population did indeed decline and so did malaria. Soon, however, the thatc h roofs of people's houses felt down because parasitic wasps that ate thatch -eating caterpillars were wiped out by the DDT.
(a) The moLecu lar structure of DDT.
1.2
(b) Amo lecu lar model of DDT.
The chairman of the Ma laria Fou ndation International has said that "the malaria epidemicis like loading up seven Boeing 747 airliners each day and crashing them into Mt. Kilimanjaro." Co nsequently, there has been a movement to return DDT to the arsenal of weapons in fighting t he spread of ma laria, and in 2006 t he WHOapproved indoor spraying of DDTin some areas in Africa. There are many, many moral and et hical issues for chem ists. Chem istry has extended and improved lives for millions of people. Bu t just as clearly, chemicals can cause harm, particularly when misused. It is incumbent on all of us to understand enough science to ask pertinent questions and to evaluate sources of information sufficiently to reach reasonable conclusions regarding t he health and safety of ourselves and our commu nities.
(c) DDT can be used to control malaria carrying insects such as mosq uitos.
Classifying Matter
T his cha p ter begin s o ur di scussio n o f how che m ists think abou t scie nce in ge neral and abou t m atter in pa rticular. Afte r lo oking at a way to classify m atter, we will tu rn to so me basic id eas abo u t eleme n ts, atoms, co m po u nds, and mol ecul es and describe how che m ists cha rac terize th ese building bloc ks o f matter.
States of Matter and Kinetic-Molecular Theory An easily observed p ropert y o f matter is its state-that is, whether a su bsta nce is a so lid , liquid , o r g as (Figure 1.7) . 'Vi.)U re cogn ize a m at erial as a so lid because it has a rigid shape an d a fixed volu me tha t cha nges littl e as temperat u re and p ressure
1.2
I
Classifying Matter 7
Active Figure 1.7 States of matter-solid , liquid , and gas . Elemental bromine exists in all three states near room temperature. The tiny spheres represent bromine (Br) atoms. In elementa l bromine, two Br atoms join to form a Br2 molecule. (See Section 1.3 and Chapter 2.)
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• Water-Changes in Volume on Freezing Water is an exception to the general statement t hat a given mass of a substance has a smaller volume as a solid than as a liquid. Water is atmost unique in that, for a given mass, the volume increases on changing from a liquid to a solid. (That is, its density decreases. See page 15.)
• Gases, liquids, and Solids Gases and kinetic-molecular theory are discussed in detail in Chapter 11, liquids in Chapter 12, and solids in Chapter 13.
Bromine gas and liquid
change. Like so lids, liq uid s have a fixed vo lu me, bu t a liquid is fluid-it takes o n the shape of its co n ta ine r a nd has no d efi nite shape of its own , Gases are flu id as well, b u t the volum e of a gas is d e term ined by the size of its co n ta ine r. T he volum e o f a gas varies mo re th an th e volu me o f a liquid with temperatu re and pressu re . At low en oug h te m pera tu res, virtually all ma tter is found in th e so lid sta te . As the tem pe rat ure is rai sed , so lids usuall y me lt to fo rm liqui d s. Even tua lly, if the tem pe ra ture is high en o ug h, liqu id s evapo rate to fo rm gases . Volu me change s typically accom pan y cha nges in state . For a give n mas s o f mat eri al, th ere is usually a small in cr ease in volum e 0 11 me lting-wat e r be in g a sign ifica n t excep tio n-an d th en a large incre ase in volu me occu rs upo n evaporation. T he kinetic-molecular theory o f m atter he lps us interpret the p ro pe rt ies of so lid s, liq uid s, and gases. Accord ing to this theo ry, all m atter co nsists of ext reme ly tin y part icles (ato ms, mol ecul es, or io ns) , which are in co ns ta n t mo tion . • In so lids th ese particle s are packed clo sely together, usually in a regular array. The pa rticl es vibrate back and forth abo ut their average positions, but se ld o m does a par ticle in a so lid sq ue eze past its im m ed iat e neighbors to co me in to co n tact with a new se t of part icles. • T he a to ms or mo lecu les of liquid s are arrang ed ra nd o m ly ra ther than in the reg ular patterns fou nd in so lids. Liq uids and gases are fluid because the particle s are not co nfi ned to specific locatio ns and ca n move past o ne ano ther. • U nder no rmal co nd itio ns, the part icles in a gas are far apart. Ga s mo lecules move ex tre me ly rapid ly because they are not co nstrain ed by their nei ghbors. T he mol ecul es of a gas fly abo u t, co llid ing with o ne ano the r and wit h th e co ntainer walls. Th is rando m motio n allows gas molecul es to fill their co nta iner, so the vo lu me o f the gas samp le is t he vo lu me of the co n ta iner, An im portan t aspect of the kinet ic-mo lecular theory is t hat th e high er the te mpe ratu re , th e Jaste r th e pa rticle s move . T he energy o f motion of th e part icles (their kinetic energy) acts to over co me th e fo rces o f attraction between particl es. A solid mel ts to fo r m a liqu id whe n the tempe rat u re of the so lid is ra ised to the poin t at which th e particles vibrate fast e no ug h and far eno ug h to push one ano the r o u t of the way an d mo ve out of their reg u larly sp aced po sitio ns. As th e tem perat u re increases eve n m o re , the parti cles m ove even faster u n til fin ally they ca n escape th e clu tc he s of the ir co mrades and en te r th e gaseo us sta te. t ncwasine lemjwm lu re
8
Chapter 1
I
Basic Concepts of Chemistry
conrspimds to Jaster and Jaster motions oj atoms and molecules, a ge neral rule yo u will find useful in man y fu ture d iscussio ns.
Matter at the Macroscopic and Particulate Levels The characte ristic p roperties of gases, liquid s, and so lid s are observed b y the u naided h um an senses. T hey are determined using samp les of mat ter large e noug h to be see n , measured, and handl ed . Using suc h samp les, we ca n also dete rmin e , for exam ple , wha t the co lor of a su bsta nce is, whe ther it di ssolves in water, or whether it co nd ucts e lect ricity or reacts with oxygen . O bservatio ns suc h as these ge nerally take p lace in the macroscopic worl d of chemistry (Fig ure 1.8) . T his is th e world o f experiments and observations. Now le t us m ove to the level of atoms, m olecules, and io n s- a wo rld of che mistry we ca n not see . Ta ke a macro sco pic sam p le o f m at e rial a nd d ivid e it , agai n and ag ain, past the po in t where the am o u nt of sam ple can be seen by the nake d eye , past the po in t where it can be seen using an optical mi cr o scope. Eventually you reac h th e level of in di vidu al particles th at ma ke u p all m atte r, a level th at che mists re fer to as the submicroscopic or particul at e world of ato ms and m olecu les (Figu re s 1.7 and 1.8) . Chem ists are interested in th e struc tu re of matter at th e pa rticu late level. Ato ms, mo lecules, and ion s can not be "seen " in the same \\'ay that o ne views the m acr osco pic wor ld , b u t they are no less real. Chemists imagin e what ato ms m ust lo o k like and how th ey m ight fit togeth er to fo rm mol ecules. T he y create mod el s to re present atoms and mo lecules (Fignres 1.7 and 1.8) -where tin y sp heres are used to re p resent atoms-and the n use these mode ls to think about che mistry and to ex plain th e obse rvatio ns th ey have mad e abou t th e m acro sco pic wo rld.
Active Figure 1.8 Levels of matter. We observe chemical and physi cal processes at the macrosco pic level. To understand or illustrate these processes, scientists often try to imagine what has occurred at t he particulate atomic and molecular levels and write symbols to represent these observations. A beaker of boilingwater can be visualized at t he particulate level as rapidly moving H20 molecules. The process is symbolized by the chemical equation HzO(liquid) ~ HzO(gas).
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H20 (liquid) ------> H20 (gas)
Symbolic
1.2
I
Cla ssifying Matter
9
HETEROGENEOUS MAnER
Nonumform composition
Physically separable
Chemically separable
into...
into...
Combine chemically to/arm ... ELEMENTS
Physically separable
into...
Cannot be subdivided
by chemical or physical processes
SOLUTIONS
Homogeneous mixtures; uniform compostttons that may varywidely
Classifying Matter.
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II has been said that chem ists ca rry o ut cx pcrirnc rus at the m acr o sco pic level , but they think a bo ut c hemis try a t the p a rti culate level. T h ey th en wr ite down th ei r observatio ns as "sym bo ls," the lett ers (suc h as 1120 fo r wa ter or Bf2 for bro mi ne molecul es) an d d rawings th at sign ify th e el em en ts and co m pou nd s involved . This is a useful perspect ive th at will help you as yo u study che m istry. In d eed, o ne o f our goa ls is to help you make th e connectio ns in you r O\\Tn m in d am o ng th e symbo lic, particul at e , and macro scopic wo rld s o f chem istry.
Pure Substances A chemist lo oks at a glass o f drinkin g wate r a nd se es a liquid . T h is liquid could be the pu re chemical com po un d water. Mo re likel y, th ough , the liquid is a homogeneo us mi xtu re of water and di ssol ved su bs ta nces-t hat is, a so lution. It is a lso po ssibl e th e wate r sam p le is a heterogeneous mi xture , wit h so lid s suspend ed in th e liquid . The se d escr ip tions represent so me o f the ways we ca n clas sify matter (Fig u re 1.9) . Ever;' su bstance has a set of uniqu e properties by which it ca n be re co gni zed . Pu re wate r, for exa m ple , is co lo rle ss and odorl ess an d certai n ly does not co ntain suspen de d so lids . If yo u wanted to identify a su bstance con clusively CL"i wate r, you wou ld have to exam ine its properties carefu lly an d co m pare th em against th e known p roperti es o f pure water. ~l eltin g poi n t a nd boilin g po in t se rve th e p urpose well here. If yo u co uld show th at th e su bstance mel ts at 0 °C and boils at 100 °C at atmospheri c pressu re , yo u ca n be ce rtain it is water. No other kn own su bsta nce melts and boil s at precisely th ese te m peratu res. A second feature o f a pu re su bsta nce is that it ge ne ra lly can no t be separa ted into two o r more di fferent species by an y physical technique at o rd ina ry temperatu res. If it co u ld be se parated , o ur sam ple would be classified as a mix ture.
10
Ch apt er 1
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Basic Concepts of Ch emi stry
00
00
W
FIGURE 1.10 Mixtures. (a) Acup of noodle soup is a heterogeneous mixture. (b) Asample of blood may look homogeneous, but examination with an optical microscope shows it is, in fact, a heterogeneous mixture of liquids and suspended particles (blood cells). (c) Ahomogeneous mixture. here consisting of salt in water. The model shows t hat salt in water consists of sepa rate, electrically cha rged particles (ions), but t he particles cannot be seen with an optical microscope.
Mixtures: Homogeneous and Heterogeneous A cu p of no odle so u p is obviously a mi xtu re o f so lids and liquid s (Figu re 1.1Oa) . A mi xtu re in \I,..hi ch the un even te xtu re o f the material ca n be d etect ed is calle d a het erogeneous mi xture. l Ic tcrogcn cous mix tures suc h as bloo d may ap pear co mpletely unifo rm but o n clo ser exa mina tion are no t (Figu re 1.1OU) . Milk , fo r example , appears smo o th in texture to the un aided eye, but magnifica tion would revea l fat and p rot ei n globu les with in th e liqui d . In a het erogen eous mixture the p roperties in o ne region are differe nt fro m those in ano ther regio n. A homogeneous mixture co nsists of two o r more su bsta nces in th e same phase (Figu re 1.1Oc) . No amo u n t o f o p tica l magnific at io n will reveal a homogen eous mi xtu re to have d iffe ren t properties in di ffere n t regions. Homogeneous mi xtures are o fte n called solutions. Co m mon examples include air (mostly a mixture of nitrogen and oxyge n gases) , gasoline (a mi xtu re of carho n- and hyd rogen -eon tainin g co m po un ds called hJflrocarbons) , an d an u nopen ed soft d rink. When a mi xture is separa te d into its pure co m po ne n ts, th e co mpo nen ts are said to he purified. Effo rt"> at se paration are o fte n no t co mple te in a sing le step, however, and re pe tition almos t always g ives an incr easingly p u re su bstance , Fo r exam ple, so il particles ca n be se pa ra te d fro m water by filtrati on (Figu re 1.11 ). when th e mi xture is passed th rough a filte r, m any o f the part icles are rem oved . Repeat ed filt ratio ns will give water a high e r an d higher state o f puri ty. This purification p rocess u ses a property of th e mi xture, its clarity, to measure th e ex te n t o f purifica tio n. Wh en a perfe ctly clear sam ple o f water is o b ta ine d , all o f the so il particles are assu me d to have been re move d .
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1.2
I
Classifying Matte r
11
(0)
(b)
FIGURE 1.11 Purifying wat e r by filtration. (a) A laborat ory set up. A beaker full of muddy water is passed t hrough a paper filter, and the mud and dirt are removed. (b) Awater treatment plant uses filtration to remove suspend ed particles from t he wate r.
•
Module 1
1.3 Elements and Atoms Passing an el e c tri c cu rre n t through wa te r ca n de com p o se it to gaseous hyd ro gen
and o xyge n (Figu re 1.12a). Sub stan ce s like hydroge n and o xyge n that are co mposed o f o n ly o ne typ e of a to m a rc cl assified as e le men ts. C u rre n tly, 11 7 cl e m e n ts arc kn own . Of th e se , o n ly a bo ut 9 0-so me of wh ich a rc illustrate d in Fig ure 1.12are found in natu re . T he rem ain d er have be en created by scie n tists. T he nam e an d
symbo l fo r each elem en t are listed in th e tabl es at the fro n t an d back of th is boo k. Ca rbo n (C) , su lfu r (S) , iro n (Fe) , co p pe r (C u) , silve r (Ag) , tin (Sn), go ld (Au) , m ercu ry (IIg) , and lead (Pb) were known to the early Greeks and Ro man s an d to th e alchc m ists o f anc ien t Ch in a, th e Arab wo rl d, and medieval Eu ro pc . Ho wever, Oxygen-gas
Hydrogen- gas
Mercury- liquid
Powdered sulfur-solid
Copper wiresolid
Iron chipssolid
Aluminumsolid
Water- liquid
(0)
12
(b) FIGURE 1. 12 Elements. (a) Passing an elect ric current through water produces the elements hydrogen (test tube on the right) and oxygen (test tube on t he left). (b) Chemical elements can ofte n be distinguished by t heir color and t heir stat e at room temperature.
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m any o the r el em en ts- su ch as alu m in um (AI), silico n (Si) , iodin e (I) , and hel ium (He) -we re not d iscovered unti l th e l St h and 19th ce n tu r ies. Finall y, scien tists in th e 20th and 2 1st cen turies have m ad e ele me n ts th a t d o not exis t in natu re , such as tec hnetium (Tc), pl ut o n iu m (Pu), and americiu m (Am ). The tabl e in sid e th e front cover o f thi s bo ok, in which th e sym bo l and oth er inform ation fo r th e ele men ts are en clo sed in a box , is called the periodic table. \Ve will describe th is im portan t tool o f che mist ry in more detail begi nn ing in Chap ter 2. An atom is th e smalles t partic le o f a n e lem ent that re ta ins the charac teristic che mical p roperties o f that eleme n t. Modern che m istry is based o n an u nderstanding and ex plo ratio n o f nature at th e ato m ic level (.... Chap te rs 6 and 7).
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EXERCISE 1.1 Ele me nts
Using the periodic tabte inside th e front cover of thi s book: (a) Find th e names of the elements having the symbols Na. CL and Cr. (b) Find the symbols for the elements zinc, nickel, and pota ssium.
1.4 Compounds and Molecules
• Writing Ele ment Symbols Notice that only the first letter of an element's symbol is capitalized. For example, cobalt is Co, not CO. The notation CO represents the chemical com pound carbon monoxide. Also note that the element name is not capitalized, except at the beginning of a sentence. • Origin of Element Names and Symbols Many elements have names and symbols with la tin or Greek origins. Examples include helium (He), from the Greek word helios meaning "sun," and lead, whose symbol, Pb. comes from the la tin word for "heavy," plumbum . More recently discovered elements have been named for their place of discovery or for a person or place of significance. Examples include americium (Am), callfornium (Cf), and curium (Cm ; for Marie Cu rie). • Periodic Tables Online Sign in at www.t homsonedu.comj login and go to Chapter 1 Contents to see Screen 1.7 or the Toolbox. See also the extensive information on the periodic table and t he e{e~ meats at the Ame rican Chemical Society website: • acswebcontent.acs.org/games/pt. html
A pure su bst ance like sugar, salt, o r wate r, whic h is co m posed of two or m ore differen t e le me n ts held together by chemical bonds, is referred to as a chemica l compoun d . Even th ough only 117 el em ents are known, there ap pears to be no limit to th e numbe r o f co m po u nds that ca n be made fr om those ele me nts. More than 20 mill ion co m po u nd s are now kn own , wit h abo u t a half milli o n ad ded to the list eac h yea r. Wh e n e le ments become part of a co m pou nd, th eir original proper ties, such as th eir co lor, hard ness, a nd mel ting point, are re placed by th e characteristic properties o f th e co mpo un d. Co nside r common tabl e salt (sod ium ch lo ride) , whi ch is co m pos ed o f two ele me n ts (see Figure 1.4) : • Sodiu m is a shi n)' metal th at reacts vio len tly with water. Its solid sta te str uc ture has sodium ato ms tightly pack ed together. • Ch lo rine is a lig h t yellow ga s th at has a di stinctive, su ffocating o d o r an d is a powe rful irritan t to lungs and other tissues. Th e e le me n t is composed o f Cl2 u ni ts in whic h two ch lo rine ato ms are tightly bound tog et her. • Sodium chlo ride, o r commo n sa lt, is a co lorl ess, crysta lline so lid co m posed of sod iu m and ch lo rid e ions bound tigh tly to gether (NaCl) . Its pro pe rt ies are co mp lete ly un like those of the two e leme n ts from whic h it is mad e . It is importa nt to di stingui sh bet ween a mi xture of ele me n ts an d a che mi ca l co m pou nd of two o r more el ements. Pure met all ic iro n and ye llow, powdered sulfur (Figu re 1.13a) ca n be m ixe d in varying p ro portio ns . In the chem ica l co mpou nd iron pyrit e (Figure 1. 13b ), howeve r, there is no varia tio n in co m positio n . No t o n ly d o es iron pyrit e ex hibit properties peculi ar to itsel f and di ffere nt fro m those of e ither iro n o r su lfu r, o r a mi xtu re of these two eleme nts , b u t it also has a d efi nite perce nt age co m positio n by m ass (46.55% Fe and 53 .45 % S) . Thus, two 1.4
I
Compounds and Molecules 13
FIGURE 1.13 Mixtures and compounds. (a) The material in the dish is a mixture of iron chips and sutfur. The iron can be removed easily by using a magnet. (b) Iron pyrite is a chemical compound composed of iron and sulfur. It is often found in nature as perfect, golden cubes.
...
'
:
.
.~
(a)
(b)
maj o r differences exist bet ween mixtures an d pu re co mpou nds: Co m poun ds have disti nct ly d ifferent characteris tics fro m their paren t eleme nts , and they have a d efin ite percentage co mpositio n (by m ass) o f th e ir co mb in ing cleme n ts. Some compoun ds-such as ta ble salt, NaC I-are composed o f ions, whic h are e lectrically charged atoms or groups of ato ms [-. Chapter 2] . Other co m pou ndssuc h as water and sugar- co nsist of molecules, the sma lles t d iscre te uni ts that re tain the co m position and chemical charac teristics of the co mpoun d. T he com pos ition of an y compound is represen ted by its che mical formula. In the form u la for water, H 20 , for exam ple , th e symbol for hydroge n, H , is fo llowe d by a su bscri p t "2" indicating th a t two atoms of h yd rog en occur in a sing le water mol ecule . The symbo l for oxygen appears without a subscript, indicating that one oxygen a to m occu rs in th e mol ecule . As you shall see throug ho ut thi s boo k, molec u les ca n be represented with mo de ls th at de pict the ir co m position and str ucture. Figu re 1.14 illustrates the names, fo rmu las, and mod el s of th e structu res o f a few co m mo n mol ecul es.
1.5
PhysicaL Properties
'iou recognize yo ur frie nds by th ei r p hysica l ap peara nce: the ir heigh t and weig ht and th e co lor of th eir eyes and hair. T he same is tru e o f che m ica l su bstances. Y ou can te ll the di ffere nce be tween an ice cube an d a cube oflead of the same size no t o n ly becau se of the ir appearance (o ne is clear and co lorless, and the other is a lustro us m et al ) (Figure 1.15) , bu t also because o ne is m ore d en se (lead) th an th e other (ice ) . Pro per ties suc h as these, whic h ca n be observed an d measured withou t chang ing the co mposition ofa su bstance, are ca lled physical properties. T he che mical eleme n ts in Figure 1.12, for examp le, clearl y di ffer in terms of th ei r co lo r,
FIGURE 1.14 Names, formulas, and models of some common molecules. Models of molecules appear throughout this book. In such models, Catoms are gray, H atoms are white, Natoms are blue, and a atoms are red.
14 Ch apt er 1
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Basic Concepts of Ch emist ry
NAME fORM ULA
MOOEl
Water
H,o
Methane
Ammonia
Ca rbon dioxide
NH,
co,
TABLE 1.1
Some Physical Properties
Property
Using the Property to Distinguish Substances
Cotor
Is the substance colored or colorless? What is the colo r, and what is its intensity?
State of matter
Is it a solid, liquid, or gas? If it is a solid, what is the sha pe of th e parti cles?
MeLtin g point
At what temperature does a solid meLt?
Boiling point
At what temperature does a liquid boil?
Density
What is th e substance's density (mass per unit volume)?
Solubility
What mass of substance can dissolve in a given volume of water or other solvent?
Electric conductivity
Does the substance conduct elect ricity?
Malleability
How easily can a solid be deformed?
Ductility
How easily can a solid be drawn into a wire?
Viscosity
How easily will a liquid flow?
FIGURE 1.15 Physicalproperties. An ice cube and a piece of lead can be differentiated easily by t heir physical properties (such as density, color, and metting point).
ap pe a ra nce, a n d state (so lid , liqui d , or gas). Ph ysical p roper ties allow us to classify a n d id entify su bsta n ces. Ta ble 1.1 lists a few physical properties of matte r th at che m ists co m mo n ly use. • Units of Density As described on page 25, t he 51 unit of mass is the kilogram and the 51 unit of length is t he meter. Therefore, the 51 unit of density is kgjm l . In chemistry, t he more commonly used unit is q/ cm'. To convert from kgj ml to gjcml , divide by 1000.
EXERCISE 1.2 Physical Properties Identify as many physical properties in Table 1.1 as you can for th e following common substances: (a) iron, (b) water, (c) ta ble salt (chemical name is sodium chloride), and (d) oxygen.
Density, th e rati o o f th e mass o f a n o bjec t to its vo lu me, is a physical property useful fo r id e n tifying su bstances.
. mass Density ~ - - volume
(1.1)
For exa m p le, you can readil y tell th e d ifference be tween an ice cube a nd a cube of lead of identical size (Figu re 1.15) . Lead has a h igh de nsity, 11.35 g/cm 3 ( 11.35 grams p er cubic cen time ter) , whereas the d ensity of ice is sligh tly less tha n 0.9 17 g/cm 3 . An ice cu be with a vo lu me of 16.0 em " has a mass of 14.7 g , whe reas a cu be of lead with th e same vo lume has a mass of 182 g. T he temperature o f a sam p le of matte r oft e n affe cts th e num e rical values of its pro pe rties. De n sity is a pa rticul arly importa nt exa m p le . Alt ho ug h th e c hange in wate r den sity with tempe rat ure seems sma ll (Ta ble 1.2) , it affects o u r enviro n me n t profo u ndly. For exam p le , as the water in a lak e cools, th e d e n sity o f th e water increases, a n d th e denser water sin ks (Figu re 1.16a ) . T h is contin u es u ntil the water temperatu re rea ch es 3.98 "C, th e point a t wh ich wate r ha s its max imum d ensity (0.9999 73 g/c m 3 ) . If the water tempe rat ure drops fu rth e r, the de nsity d ecr eases slig h tly, a n d the co lder wate r flo at s on top o f water at 3.98 "C. If water is cooled bel ow a bou t 0 "C , so lid ice forms. Water is unique a mo ng su bstances in th e un iverse : Ice is less d ense tha n wate r, so the so lid ice flo ats o n liquid wa te r. Becau se the d ensity of liquids c hanges with tempera ture , th e volu me of a given mass of liquid also c hanges with te mperature. T h is is the reason laborat ory glassware used to m easure p recise volum es o f so lut io ns always specifi es th e tem perature at wh ic h it was calib ra te d (Fig u re 1.16h) .
• Calculations Involving Density and Mathematics Review See Let's Review beginning on page 24 for a review of some of the mathematics used in introductory chemistry. • Temperature Scales Scientists use the Celsius (0C) and Kelvin scales (K) for temperature. See page 26.
Temperature De pendence of Water Density
TABLE 1.2
Temperature (0C)
Density of Water (g/cm')
o (ice)
0.917
o (liq water)
0.99984
2
0.99994
4
0.99997
10
0.99970
25
0.99 707
100
0.9583 6
1.5
I
Physical Properties
15
Chemical Perspectives
Thermophilic Bacteria
When you think of chemistry, you probably think of colored liquids bubbling in flasks and maybe a fire or even an explosion. That is not what we usually see in a university laboratory, but pay a visit to Yellowstone National Parkin Wyoming (or to areas of the North Island of New Zealand) and you will see just that: bubbling. steaming hot water springs with colorful substances in a natural "laboratory." Yellowstone Park is unique in having one
with brilliant cotors. Different bacteria grow at different temperatures, usuallyin the range of 50 °Cto 70 °C, and the colors of the pools and streams can change with temperature. In the Gra nd Prismatic Spring shown in the photo, you can see that bacteria grow in the slightly cooler water around the edge of the spring, but the deep blue, very hot water in
the center is devoid of livi ng organisms. The bacteria growing around the hot springs are single-cell organisms ranging in size from 0.2 to 50 um in diameter. They are often highly colored, owing to pigments such as carotenoids and chlcrophylls, and different organisms with different colors grow at different temperatures. Some are anaerobic bacteria and use sulfur instead of oxygen for respiration (a chemical process). A study of the hot springs of Yellowstone National Park is a good example of the intersection of chemistry, geology, and biologyand of aU of their subdisciplines (biochemistry, geochemistry, bacteriology, and mineralogy, among others) . Our goal in this book is to introduce the chemistry background you will need to study more chemistry or to carve out a career in another field of science.
of the highest concentrations of geysers, hot
springs, steam vents, and mudpots on the planet. The reason rests in the geology of the area-the earth's crust is thinner here (about 64 km) compared with the crust covering the rest of the earth (144 km). Hot magma lies not far belowYellowstone's surface (6-16
km), and it heats the rocks above and the reservoirs of water closer to the surface (a physi cal process). The superheated water dissolves some of the minerals (another physica l process), and it is forced upwards through fissures in the rocks and sometimes explodes through the surface as geysers and hot springs. The hot water shooting to the surface carries with it dissolved minerats such as limestone (catcium carbonate), and they are deposited around the geysers and hot springs as limestone and travertine. Once the hot water reaches the surface, it can harbor thermophilic or "heat-loving" bacteria. Th ese can grow in enormous colonies
A travertineformation
in Yellowstone National Park. The formation consists largely of limestone (calcium carbonate) mixed with silica.
~
J!
•
i ~....,; Grand Prismatic Spring, Yellowstone National Park.
Extensive and Intensive Properties Extensive properties de pe nd on th e amo u n t of a su bsta nce p resent. T he mass and volu me o f the samp les of e lements in Figu re 1.12, o r the amou nt o f heat o b tai ne d from burn ing gasoline, are ex te nsive p roperties, for exam ple. In co ntrast, intensive properties d o not d epe nd o n th e amo u n t of su bsta nce . A sam ple of icc will melt at 0 "C, no m att e r whether you have an ice cu be or an ice berg. Density is also an in te nsive p rope rty. T he d ensity of go ld , fo r exam ple, is the same (19.3 g/c m 3 at 20 °C) whe the r you have a flake of pu rc g o ld o r a so lid go ld ring . Int en sive properties are often u seful in iden tifying a material. For examp le , the te m pera tu re a t whic h a material melts (its m elt ing point) is often so charac teristic tha t it can be used to ide n t ify th e so lid (Figure 1.17) .
16 Ch apt er 1
I
Basic Concepts of Chemistry
1000 ml TC 20'C \ ...,....----:,....-_ ..1
I
J
_
(a)
-;::;
:;
;
.
(b)
FIGURE 1.16 Temperature dependen ce of physical propertie s. (a) Change in density with temperature. Ice cubes were placed in the right side of the tank and blue dye in the left side. The water beneath the ice is cooler and denser than t he surrounding water, so it sinks. Theconvection current created by this move ment of wate r is traced by the dye movement as the denser, cooler water sinks. (b) Temperature and calibration. labo ratory glasswareis calibrated for specific te mperatures. The pipet will deliver and the volumetric flask will contain the specified volumeat the indicated temperature.
1.6
Physical and Chemical Changes
Change s in p h ysical properties are called physical changes. In a physical ch ange , the id en tity of a su bsta nce is p re se rved eve n though it may have change d its ph ysica l state o r the gross size and shape of its pieces. A ph ysical change d oes no t re su lt in a new chem ical su bst ance being p roduced. T he su bst an ce s (ato ms, mo lecules , o r ion s) presen t be fo re and after t he chan ge are the same. An exa m ple of a p h ysical chan ge is th e me lting of a so lid (Figu re 1.17) . In th e case o f ice mel ting, th e mo le cu les p resent bo th before and after the change a re 11 20 mo lecu les. T heir chem ica l iden tity has not change d ; they are now sim ply able to flo w pas t one ano the r in th e liqu id state instead of be ing lo cked in position in th e so lid .
FIGURE 1.17 A physical property used to distin guish compounds. Aspi rin and naphthalene are both white solids at 25 °C. You can tell them apart by, among other things, a difference in physical properties. Atthe temperature of boiling water, 100 0(, naphthalene is a liquid (left), whereas aspirin is a solid (right).
Naphthalene is a white solidat 25 °C but has a melting point of 80.2
-c,
Aspirin is a white solidat 25 °C. It has a melting point of 135 -c.
1.6
I
Physical and Chemical Ch anges
17
A ph ysical properly of hydrogen gas (l iz) is its lo w d en sity, so a balloon filled with H z floats in air. Suppo se, however, th at a lighted ca nd le is br ought up to th e balloon . \Vhen the hea t ca uses the skin of the balloon to rup ture, the h ydrogen com bines with the oxyge n (0 2 ) in th e air, and the heat o f th e ca nd le sets off a che mica l reaction , p rodu cin g water, H 20 (Figu re 1.18) . T his re action is an ex amp le of a ch emical change, in whic h o ne or mo re su bsta nces (the reactan ts) are tra nsfo rmed into one o r more differen t su bsta nces (t he products) . A che m ica l cha nge at th e part icu late level is illust rat ed by th e re action o f hydroge n and oxyge n molecul es to form water mol ecul es. 2 H,(gas) + O,(g05)
~
2 H,O(g05)
+ Reactants
Case Study
Ancient and Modern Hair Coloring
Humankind has always been interested in pigments for artistic uses and to decorate their bodies. One of the oldest pigments is galena, PbS (Figure 1), which was first brought from Asia to Egypt thousands of years ago. When powdered, it is black and has been widely used as a cosmetic, particularly for dyeing hair. Some evidence for this is that small piles of galena were found next to the skeleton of a young Egyptian woman whose remains were buried in 3080 Be ± 110 years. Ana lysis by chemicalarcheologists found that the galena in this tomb was not from Asia, though. Instead, using modern forensic techniques, they found it came from nearby cities on the Red Sea. According to recent research, the same effect on hair that galena has could be achieved by applying a mixture of the chemi-
cal compounds PbD and Ca(DHh. This was originallydescribed by Claudius Galen, a Roman physician who lived between about 130 and 200 AD. Similar formu lations have been used t hrough the centuries for dyeing wool, and the present-day hair-coloring product Grecia n Permute" still uses this technique. In recent research, chemists in France found that only a few hours after applying the PbD and Ca(D Hh mixture to hair (Figure za). the hair was blackened (Figure 2b), and that the blackening came from tiny particles of PbS (about 5 nanometers in diameter). But where did the sulfur in PbS come from? The French research found it came from the amino acids in hair. (The amino acids cysteine and methionine both have sulfur as part of their structure.)
(aJ
(b)
Photographs of strands of hair before (a) and after (b) treating with a mixture of PbO and Ca(OH h. From C& EN, Vol. 84, No. 37, p. 12, 2006, "Still Dyeing After 2,000 Years," Dr. B. Halford.Copyright © 2006 American Chemical Society. Used with permission. FIGURE 2
Questions: 1. What is the nameof the element Pb? OfCa? 2. What is the density of Pb? 3. What is the symbol for the element sulfur? 4. Can you find (on the World Wide Web, for example) the common nomefor the compoundCa(OH)z? 5. A particulate view of galena is illustrated in Figure lb. Briefly describe the shape of this tiny piece of PbS. 6. Compare the particulate view of galena with that of NaCl on page 4. Are theresimilarities? Anydifferences? Answers to these questions are in Appendix Q.
mL has a volume of 125 em". Altho ug h not widely used in th e Un ite d States, t he cu bic d ecim eter (d rn") is a com mo n u n it in th e rest o f th e wo rl d . A len gth of 10 ern is called a d ecimeter (d m) . Becau se a cu be 10 cm on a side defin es a vo lu me o f 1 liter, a liter is equiuatent to a cubic decim eter: I L = 1 dm ", Prod ucts in Europe , Africa, and o th er parts of the wo rld are oft en so ld by th e cu bic d ec ime te r. T he deriliter; dl: whi ch is exactl y eq u ivale n t to 0.100 L or 100 mL , is wid ely used in med icin e . Fo r exam ple , sta nd ards fo r co nc entra tio ns of e nviron me ntal co nta minan ts are o ften se t as a cer ta in mass per d ecilit er. T he state of Massachusetts re com me nd s th at ch ild re n with more th an 10 m icro grams (10 X 10- 6 g) o f lead per d ecil iter o f blood und ergo fu rth er test ing for lead poi so ning. EXERCISE 3 Volume (a) Astandard wine bottle has a volume of 750 ml. Wha t volume, in liters, does t his represent? How many deciliters? (b) One U.S. gallon is equivalent to 3.7865 l. What is the volume in liters of a z.n -quart carton of milk? (There are 4 quarts in a gallon.) How many cubic decimeters?
Fina lly, whe n chem ists pr epare che m ica ls for re act ions, th ey often tak e g iven quantities o r masses of mat e rials. The mass of a bod y is th e fu nd amen tal measu re of th e qu antity o f m atter, and th e SI unit o f m ass is th e kilogram (kg) . Smalle r masses are ex p resse d in gra ms (g ) o r milli grams (mg ) . 1 kg = 1000 9 and 1 9 = 1000 mg 1
I
Units of Measurement
29
EXERCISE 4 Mass
(a) A new U.S. quarter has a mass of 5.59 g. Express this mass in kilograms and milligrams. (b) An environmental study of a river found a pesticide present to the extent of 0.02 microgram per
liter of water. Express this amountin grams per liter.
Making Measurements: Precision, Accuracy, Experimental Error, and Standard Deviation 2
The pre cisio n of a measurement indicates how well seve ra l determin ati ons of th e sa me qua n tity agre e. T hi s is illustrated by th e re sults of throwi ng darts at a target.
In Figu re Sa, th e d a rt thrower
\\'35
ap pare n tly not skillfu l, and th e p re cision of th e
dart 's p lacement 0 11 the target is lo w. In Figures 5 b a n d 5c, th e darts a rc clustered to gether, in d ica tin g much better co ns iste ncy on the part of th e thrower-that is,
g rea ter precision .
• Accuracy and NISTThe National Institute for Standards and Technology (NIST) is an important resource for the standards used in science. Comparison with NIST data is a test of the accuracy of the measurement. See www.nist.gov.
Accu racy is the agreement o f a m easurement with the ac cepted va lue of th e qua ntity. Fig ure ?lc shows that our thro wer was acc u ra te as wel l as pre ci se-the aye rage of all sho ts is close to th e ta rgeted position , th e bull's eye . Figure fib sh ows it is possible to be pre cise withou t be in g accurate- the t hrowe r has co ns iste n tly mi ssed the hull 's eye , al th oug h all th e d arts are clustered pre cisel y a round o n e poin t on the targe t. Th is is a nalogo us to a n ex perime n t with so me flaw (e it he r in d e sign or in a me a su rin g devi ce ) that causes a ll re su lts to diffe r from th e co r rec t va lue by the sa me amou nt. The accuracy of a re su lt in th e labo ra to ry is ofte n e xpresse d in terms of percent e rro r, whereas th e pre cision is ex p ress ed as a sta n dard deviation .
ExperimentaL Error If yo u measure a q uan tity in th e laboratory, yo u may be re qu ire d to report the error in th e re sult, th e difference be twe en yo ur resu lt a nd the accep te d va lue ,
Error
(a) Poor precision and poor accuracy
= experimentaLLy dete rmined value -
(b) Good precision and poor accuracy
FIGURE 5 Pred sion and accuracy.
30
L e t ' s R evi ew
I
The Tools of Quantitative Chemistry
accepted value
(c) Good precision and good accuracy
o r th e percent error.
Percent error
error in measurement x 100% accepted value
Precision, Accuracy, and Error
• Percent Error Percent error can be positive or negative, indicating wh ether the experim ental value is too high or too low compared to t he accepted value. In Example 2, Student B's error is - 0.2%, indicating it is 0.2% lower t han the accepted value.
Problem A coin has an "accepted" diameter of 28. 054 mm . In an experiment. two students measure this diameter. Student A makes four measurements of t he diameter of t he coin using a precision tool called a micrometer. Student B measures the same coin using a simple plastic ruler. The two students report t he following results: Student A 28. 246 mm
Student 8 27 .9 mm
28.244 28.246 28.24 8
28. 0 27.8 28 .1
Wh at is the average diameter and percent error obtained in each case? Wh ich st udent's data are more accurate? Strategy For each set of values, we calculate t he average of t he results and then compare this average with 28. 054 mm. SoLution The average for each set of data is obtained by summing t he four values and dividing by 4. Average value for Student A = 28.246 mm Average value for Student B = 28 .0 mm Although Student Ahas four results very dos e to one another (and so of high precision), st udent A's result is less accurate t han that of Student B. The average diameter for Student A differs from t he "accepted" value by 0. 192 mm and has a percent error of 0.684%: Percent error
=
28. 246 mm - 28. 054 mm 28.0 54 mm
X
100%
=
0.684%
Student B's measurement has a percent error of only about - 0.2%. Comment We noted t hat Student A had precise results; the standard deviation calculated as described below is 2 X 10- 3 • In contrast, Student B had less precise results (standard deviation = 0. 14) . Possible reasons for the error in Students A's result are incorrect use of the micromete r or a flaw in t he inst rument .
Standard Deviation Lab orat ory m easu rem en ts can be in erro r for two basic reasons . First, the re may be "d ete rm inate " errors caused by faulty ins tru me n ts or human e rrors such as incorrect record keeping . So-ca lled "ind ete rm inate" errors arise fro m un certai n ties in a measurem en t whe re the ca use is no t known an d canno t be co ntrolled by the lab wor ke r. One way to j udge th e indet erminat e e rror in a result is to ca lculate th e standard d eviat ion . Th e standard deviation o f a se ries o f meas u re men ts is equ al to th e sq uare roo t o f th e su m o f th e squ ares of th e d eviati on s for eac h measu re men t fro m th e average d ivid ed b y one le ss tha n th e numbe r o f m easu r em en ts. It has a precise statistica l sig n ifica nce : assu m ing a larg e n u m ber o f m easurem en ts is used to calcu la te th e ave rage, fiR% of th e values co llected are ex pected to be with in on e sta ndard d evia tio n o f the value deter mi ne d , a nd 95 % are wit hi n two sta nd a rd d eviation s. Sup pose you ca re fu lly measured th e mass o f wate r deli vered by a IO-mL pipet. (1\ pipet co ntaining a gree n so lution is shown in Figure 4. ) Fo r five attempts a t the
2
I
Ma king Measurements: Precision, Accuracy, Experimental Error, and Standard Deviation 31
measureme nt (shown in col umn 2 of th e table below) , the standard d eviatio n is fo und as fo llows: First, the ave rage of the measu reme n ts is calcu lated (here , 9.9 84 ) . Nex t, th e deviation of eac h individual measu rem e n t fr o m this value is determined (co lu mn 3) . T hese values are sq uared, giving the values in co lu m n 4, and the sum of th ese values is d et ermin ed. T he sta n da rd deviatio n is then ca lcu lated by d ividi ng thi s sum by 4 (the n umbe r of d etermin atio ns minus I) and takin g the sq uare root o f the resu lt. Measured Mass (g)
Difference between Average and Measurement (g)
1
9.990
- 0.006
4 x 10- 5
2
9.993
- 0.009
8 x 10- 5
3
9.973
0.011
12 x 10- 5
4
9.980
0.004
2 x 10- 5
5
9.982
0.002
0.4 x 10- 5
Determination
Square of Difference
Average mass = 9.984 9 Sum of squares of differences Standard deviation
~
=
26 X 10- 5
J26X410 - 5
~
:to .008
Rased o n this ca lcula tio n , it wou ld be ap p ro p riate to represent the m easu red m ass as 9.9 84 ± 0.00 8 g. This wo uld tell a read er tha t if thi s ex perime n t we re repeat ed , a maj o ri ty o f the values wou ld fa ll in th e ra nge o f 9.976 g to 9.99 2 g.
EXERCISE 5
Accuracy, Error, and Standard Deviation
Two students measured the freezing point of an unknown liquid. Student A used an ordinary leboratory thermometer calibrated in 0.1 O( units. Student B used a thermometer certified by NIST (National Instit ut e for Standards and Technology) and calibrated in 0.01 O( units. Thei r results were as follows: Student A: - 0.3 °C; 0.2 Student B: - 0.02
0(,
0( :
0.0
0( :
and - 0.3 O(
+0 .02 °( , 0.00
0( ,
and +0.04
O(
Calculate the average value, and, knowing that t he liquid was water, calculate t he percent error and sta ndard deviation for each student. Which student has the more precise values? Which has the smaller error?
FIGURE 6 Lake Otsego. This lake, with a surface area of 2.33 x 107 m", is located in northern New York State. Cooperstown is a village at the base of the lake where the Susquehanna River originates. To learn more about the environ mentalbiology and chemistry of the lake, go to www.oneonta. eduJacademicsJbiofld
32
L e t ' s R evi ew
I
3
Mathematics of Chemistry
Exponential or Scientific Notation Lak e Otsego in northe rn New York is also ca lled Glimmergiass. a name sugges ted byJ am es Fenimore Cooper ( 1789- 185 1), th e great Am erican au th o r and a n early resid e nt o f the village no w known as Co o pe rstow n . Ex tens ive en viro n me n ta l st ud ies have be en do ne a lo ng this lak e (Fig u re 6 ) , and so me qu an titat ive inform ation us efu l to che m ists, biologists, and ge o logists is given in the foll o wi ng table:
The Tools of Quantitative Chemistry
Quantitative Information
Lake Otsego Characteristics Area
. ' -'!.-- - - . .
FIGURE 9 Data used to determine the density of a metal.
• Zeroes and Common laboratory Mistakes We often see students find the mass of a chemical on a balance and fail to write down trailing zeroes. For example, if you find the mass is 2.340 g, t he final zero is significant and must be reported as part of t he measured value. The number 2.34 9 has only three significant figures and implies the 4 is uncertain. when in fact the balan ce rea ding indi-
cated t he 4 is certain.
In t he case o f the widt h o f the p iece, you fo und it to be 2.5 0 em , where 2.5 is kn own with ce rta in ty, but th e fin al 0 is u n certain. T he re a re three sign ifica n t figu re s in 2.50. Wh e n you first read a n umbe r in a problem , or collec t d ata in t he lab orato ry, how do yo u d et ermine h ow man y sign ifica n t figu res it con tains? First, is th e numb e r a n exact n umbe r o r a measu red qua ntity? If it is an exac t n um ber, yo u don't h ave to wo rry a bo ut th e n u m be r o f sign ificant fig ures. For exa m p le , th ere a re exactly 100 em in 1 m . \Ve co u ld add as man y ze ros afte r th e d ecim al pl ace , a n d th e ex p ression wou ld st ill be tr ue . Usi ng th is numbe r in a ca lcu lat io n will n o t affect h ow m an y sig n ifica n t figure s yo u ca n re p ort in yo u r a nswe r. If, h oweve r, th e numbe r is a measured value, you must take into accou n t sign ificant figures. The number of sig n ifica n t fig u res in our data a bove is clear, wit h th e po ssibl e exce p tio n of 0.31 a n d 2.50 . Are th e zeroes sig n ifica n t? L Zeroes benoeen two other significant digits are significant. Fo r exam p le, th e ze ro in 103 is sign ifica nt. 2. Zeroes to the right of a non zero number and also to the right (if a decimal place are significant. Fo r exa m p le, in th e number 2.50 em, the zero is sign ifica n t. 3. 7rrocs that are placeholders are not signijicant. The re a re two types o f nu mbers th at fall und er th is rule . a) T he first are d ecim al numbe rs with ze roes th at oc cur before th e first nonzero di gi t. Fo r exam p le , in 0 .001 3, on ly the I a n d the 3 are sign ifica nt; the zeroes are no t. This number h as two sign ifica n t figures. b) T he seco n d a re numbers with trailiug zeroes th at must be th ere to indicat e th e magnitude of th e num be r. For exa m p le, the zeroes in th e numbe r 13,000 may o r may n ot be sig n ifica n t; it d epends O il whethe r the y were measured or no t. To avo id confus io n with regard to suc h numbers, we shall assu me in this book that trailing zeroes are significan t when there is a decimal/JOint to the ligh t of the last zero. Thus, we wo u ld say th at 13,00 0 h as o n ly two sign ifica n t figures but th at 13,000. h as five. We suggest th at th e best way to be u n amb iguous whe n wri ting n u m bers with trailing zeroes is to use scie nt ific notatio n. For exa m p le 1.300 X 104 clearly indi cat es fou r sign ifican t figures, whereas 1.3 X 10-1 indicat es two a n d 1.3000 X 10' in d icates five . Using Significant Figures in Calculations Wh en doi ng ca lcu latio n s usin g measured qua ntities, we follow so me basic rul es so that th e res u lts re flect the precision of all the measure me n ts that go into th e ca lcu lat ions. The rules used for signiJiwnt ,figures in this book are as follows: Rule 1. \Vhe n ad di ng or sub tracti ng num bers, th e n umber o f deci mal places in th e answe r is eq ua l to th e nu m be r o f d ecimal places in th e n u m ber wit h the fe west di g its afte r th e d ecim al. 2 decimal places
2 significa nt figures
+ 1.9
1 decimal place
2 significant figures
+ 10.925
3 decimal places
5 significant figures
12.945
3 decimal places
0.12
Th e su m shou ld be reported as 12.9 , a nu m b e r wit h o ne deci ma l pl ace , b ecause 1.9 h as o n ly one d ecim al p lace . 36
L e t ' s R evi ew
I
The Tools of Quantitative Chemistry
Rul e 2. In m ul tipli catio n or d ivisio n , tile n u m ber o f sign ificant figu res in the answe r is d et ermined by th e q uan tity with th e fewes t sign ifica n t figures. 0.01208 === = 0.512, or in scientific notation , 5.12 X 0.02 36
10- 1
Because 0.0236 has on ly three sig nifica n t d igits, while 0.0 1208 has fou r, the answer sho u ld have three sig n ifica n t di gits. Rul e 3. Wh en a nu mber is roun d ed off, the last d ig it to be re ta ine d is incr eased by o ne o n ly if th e fo llowing d igit is 5 or grca tcr.
Full Number
Number Rounded to Three Signifi cant Digits
12.696
12.7
16.349
16.3
18.35
18.4
18.351
18.4
Now let us app ly the se rul es to calc u late the d cnsity of th e piece o f metal Figu re 9.
III
lengt h X wi dth X t hickness = volume 6. 45 em X 2.50 em X 0.3 1 em
= 5.0 em"
) :c = 13.56 9 = 2.7 9/ cm3 _a_ss--,(",g'Density = _ -cm -volume (em") 5.0 em! T h e ca lcu late d d en sity has two sign ific an t figu res becau se a calculated result mtl be no more precise than the least precise data used, an d here the thickn ess has o n ly two sign ifican t figures. One last word o n sig n ificant figu res and calcu la tio ns: wh en wo rkin g problem s, yo u sho uld d o th e calculatio n with all the d ig its al lowed by yo u r calc ula tor an d rou nd off o n ly a t th e e nd o f th e calcu latio n . Rou nding off in the middle ~f (J calculalion (an in troduce errors.
Glassware and si gnificant figures. The 10-ml graduated cylinder is marked in O.l-m l increments. Graduated cylinders are not considered precision glassware, so, at best, you can expect no more than two significant figures when reading a volume. Conversely, a 50-ml buret is marked in 0.10 ~ml increments, so it can be read to the nearest 0.01 ml. Avolumetric flask is meant to be filled to the mark on the neck. When you have this volume, it is known to the nearest 0.01 ml, so a 250-ml volumetric flask contains 250.00 ml when full to the mark (or five significant figures). Finally, a pipet is like a volumetric flask in that the volume is known to the nearest 0.01 ml.
ChemistrY.{J.Now'" Sign in at www.thomsonedu .com /Ioqtn and go to Screen 1.17 for a self-study module on using numerical informati on.
Using Significant Figures Problem An example of a calculation you will do later in the book (Chapter 11) is
Vo Iume 0 f gas (l) =
(0.120 )(0.08 206)( 273.15
+ 23)
(230/760. 0)
Calculate the final answer to the correct number of sig nifica nt figures. Strategy Let us first decide on the number of significa nt figures represented by each number a nd then app ly Rules 1-3.
• To Multiply or to Add? Take the number 4.68. (a) Take the sum of 4.68 + 4.68 + 4.68. The answer is 14.04, a number with four significant figures. (b) Multiply 4.68 times 3. The answer can have only three significant figures (14.0). Yo u sho uld recognize that different outcomes are possible, depending on the type of mathematical operation. 3
I
Mathematics of Chemistry
37
SoLution
Number
Number of Significant Figures
Comments
0.120
3
The trailing 0 is significant.
0.08206
4
The first 0 to the immediate right of the decima l is not significant.
3
23 has no decima l places, so t he sum can have none.
2
230 has two significant figures because the last zero is not significant. In contrast, there is a decima l point in 760.0, so there are four significantdigits. The quotient may have only two significant digits.
273.15 + 23 = 296 230/760.0
~
0.30
Analysis shows that one of the pieces of information is known to only two significant figures. Therefore, the volume of gas is 9.6 L, a number with two significant figures.
EXERCISE 6 Using Significant Figures (a) How many significant figures are indicated by 2.33 X 107 , by 50.5, and by 200? (b) What are the sum and the product of 10.26 and O.063? (c) What is t he result of the following calculation?
x
(110.7 - 64) ~
(0.056)(0 .00 216)
Problem Solving by Dimensional Analysis Fig ure 9 illustrated th e d ata th at were co llec ted to d etermi ne th e de n sity of a piece of m etal . T he th ickn ess was m easu red in millimete rs, wh ereas th e le n gth a nd wid th we re m easu red in ce n ti m e te rs. To fin d the vo lu me of th e sam p le in cub ic ce n tim e te rs, we first h ad to h ave the le ngth , width , a nd th ick ness ill the sa me uni ts an d so co nverte d the thi ckn ess to cen time te rs. /' 1 em 3.1 mm X - - / 10 mm
~
0.31 em
He re , we m ulti pl ied the n um be r we wish ed to co n ve rt (3 . 1 mm ) by a amuersion fa ctor (I e m / IO mm ) to pro d uce th e result in th e desired u n it (0.3 1 em ) . No tice tha t u nits are treated lik e num bers. Becau se th e unit "m m " was in bo th the num era to r a nd the de no m in ator, div id ing on e by the ot her leaves a q uotie n t of 1. T he units arc sa id to "ca n cel o ut." Here , this leaves th e a ns we r in cc rui ruc rc rs , th e des ire d un it. T h is ap p ro ac h to p roblem so lving is ofte n called dim ensional analysis (o r sometim es th e factor-label method ). It is a ge n eral pro blem-so lvin g a pproac h th a t u ses th e d im ensions or un its of ea c h value to gu id e u s through calc u lation s. An d, it is ofte n the case th at conversio n fac tors a re used to c ha n ge measu red quanti ties to chem ica lly u seful in fo rm ation . A conversion factor ex p resses th e equiva len ce o f a m easurement in two diffe rent u ni ts (I em 10 mrn ; 1 g 1000 m g; 12 eggs 1 do ze n ; 12 inc h es 1 foo l) . Beca use th e num e rat o r a n d th e den omi nato r d escr ibe th e sa me q ua n ti ty, th e co nversio n facto r is equivale n t to the nu m be r 1. The refore , m u lt iplica tio n by this fac to r does no t c han ge th e m easu red q uantity, o n ly its uni ts. A co nversio n factor is a lways wri tte n so tha t it h as th e form "new un its divid ed by units of o rigin al n u m ber. "
=
38
L e t ' s R evi ew
I
The Tools of Quantitative Chemistry
=
=
=
• Using Conversion Factors and Doing Calculations As you work problems in t his book and read Example problems, notice that proceed ing from given information to an answer very often involves a series of multiplications. Thatis, we multiply the given data by a conversion factor, multiplythat answer of that step by another factor, and so on to the answer.
Number in originalunit [ new unit ] = new number in new unit originalunit
i
Quantity to express in new units
I
Conversion factor
Ouantity now expressed in new units
ChemistrY-9-NDW~ Sign in at www.thomsonedu .comjlogin and go to Screen 1.17 for a self-study module on dimensional analysis and using numerical information .
. Using Conversion Factors and Dimensional Analysis Problem Oceanographers often express the density of sea water in units of kilograms per cubic meter. If the density of sea water is 1.025 gjcm3 at 15 °C. what is its density in kilograms per cubic meter? Strategy To simplify this problem, break it into three steps. First, change the mass in grams to kilograms. Next. convert t he volume in cubic centimeters to cubic meters. Fi nally, calculate t he density by dividing the mass in kilograms by the volume in cubic meters. Solution Fi rst convert the mass in grams to a mass in kilograms. 1.025 9 X ~ 1000 9
~
1.025 X 10- ' kg
No conversion factor is available in one of our tables to directly change units of cubic centimeters to cubic meters. You can find one, however, by cubing (raising to the third power) the relation between the meter and the centimeter:
lem3 X (~)' 10 0 cm
= 1 cmJ
x
1 m' ) = 1 X 10- 6 ( i x ios cm '
m'
Therefore, t he density of sea water is Density
EXERCISE 7
=
"l.ccO,,25'--:c Xcclo0 -_'~k~g 1 X 10- 6 m'
1. 025 X 10 3 kgjm3
Using Dimensional Analysis
(a) The ann ual snowfall at lake Otsego is 198 em each year. What is t his depth in meters? In feet (where 1 foot = 30.48 cm)? (b) The area of l ake Otsego is 2.33
x
10 7 rn ", Wh at is this area in square kilometers?
(c) The density of gold is 19,320 kg/m3 • What is this density in g/cm 3? (d) See Figure 7. Show that 9.0
x
10 3 pc is 2.8 X 1017 km.
Graphing In a num ber of insta nce s in th is text , gra p hs arc used whe n ana lyzing cx pcri mc ntal d ata with a go al of o b taini ng a math e matical equa tio n that m ay help us predict new res u lts. T he proced ure used will o fte n result in a stra igh t lin e , wh ich has the eq ua tion y
~
mx + b
• Who Is Right-you or the Book? If your answerto a problem in this book does not quite agree with the answers in Appendix Nthrough a, the discrepancy may be the result of rounding the answer after each step and then using that rounded answerin the next step. This book follows these conventions: (a) Final answers to numerical problems in this book result from retaining four or more digits past the decimal place throughout the calculation and rounding only at the end. (b) In Example problems, the answer to each step is given to the correct number of significant figures for that step, but a numberof digits are carried to the next step. The numberof significantfigures in the final answer is dictated by the numberof significant fig ures in the originaldata. 3
I
Mathematics of Chemistry 39
FIGURE 10 Plotting data . Data for t he variable x are plotted on the horizontal axis (abscissa}. and data for ya re plotted on the vertical axis (ordinate). Th e slope of the line, m in the equation y = mx + b, is given by ayj a x. The intercept of the line with t he y-axis (when x = 0) is b in the equation. Using Microsoft Excel with t hese data , and doing a linear regression analysis, we find y = - 0.525x + 1.87.
3
Experimental data
2.5
x 3.35 2.59 1.08 - 1. 19
y
0.0565 0.520 1.38 2.45
Using the points ma rked with a square. the slope ofthe line is: 5lope - .iy _ 0.82 - 1.87
0.5
-
.lx - 2.00
=
- 0.525
0.00
o -2
• Determining the Slope with a Computer Program-least-Squares Analysis Generally, the easiest method of determining the slope and inte rcept of a st raight line (and t hus t he line's equation) is to use a program such as Microsoft Excel. These programs perform a "least squares" or "linear regression" analysis and give the best st raight line based on the data. (This line is referred to in Excel as a trendline.)
-1
o
4
In this eq uatio n , y is u su ally refe rred to as th e d e pendent variab le; its value is determin ed from (t hat is, is d ependen t 0 11) the values of x, m, a n d b. In th is eq ua tion, x is ca lle d the indepe n dent varia b le, a n d Tn is th e slo pe o f th e lin e . The para m eter b is the j -in te rce p t-s-th a t is, the value o f y wh e n x = O. Le t u s use a n exam p le to investigat e two thi ngs: (a) how to construc t a graph from a set o f d at a p oints, an d ( b) how to d erive a n equati on for th e lin e ge n e ra te d by th e d at a. A se t of d ata poin ts to be grap hed is p rese nt ed in Figure 10. \Ve first m a rk off eac h ax is in increments o f th e value s o f x an d y. H e re , o ur x-d a ta are with in th e ran ge from - 2 to 4 , so the » axi s is m arked o ff in in crements o f 1 unit. T he j-data fall s with in th e range fr om 0 to 2.5 , so we m a rk offthe j-axts in in creme n ts of 0 .5 . Eac h d at a poi n t is m arked as a circle o n th e grap h . Afte r pl otting th e points on th e gra p h (ro u nd circles) , we draw a stra igh t lin e th at co mes as close as possibl e 10 rep resentin g th e tre n d in th e d at a . (Do n ot co nn ect the do ts!) Beca u se the re is a lways so me in acc uracy in ex per ime n ta l d a ta, th is lin e will not pass exac tly through eve ry point. To id en tify th e specific equatio n correspo ndi ng 10 ou r d at a, we must d et ermi n e th e j-h ue rcc pt (b) a nd slo p e (m) fo r th e eq ua tio n y = mx + b. The j-inte rcc pt is th e point a t wh ich x = O. (In Figu re 10, J = 1.87 wh en x = 0 ). T he slope is d etermin ed by se lec ti ng two p oints on the line (m a rked wit h sCJuares in Figure 10) a n d ca lcu lating th e d iffere nce in values of y (a y = Y2 - Yl) and x (ax = X2 - x, ) . The slo pe o f th e line is then th e ratio of these differences, m = d y/ d x. Here, th e slo pe h as the value - 0.:' 25. \Vith the slo p e a n d inte rcept now known , we ca n write th e eq ua tion fo r th e lin e y = - 0.525x
+ 1.87
and we ca n u sc this eq ua tio n to ca lcu late j-valu cs fo r points th at a re not. part o f ou r o rigin al set o f x-y d at a. For exam p le , whe n x = 1.50 , we find y = 1.08.
40
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The Tools of Quantitati ve Chemist ry
EXERCISE 8 Graphing
To find the mass of 50 jelly beans, we weighed several samplesof beans. Number of Beans
Mass (9)
S
12.82
11
27.14
16
39.30
24
59.04
Plot t hese data wit h the number of beans on the horizontal or x-axis , and the mass of beans on the
verticalor y-axis. What is the slope of the line? Use your equation of a straight line to calculate the mass of exactly 50 jelly beans.
Case Study
Out of Gas!
On Ju ly 23, 1983, a new Boeing 767 jet air-
lated. and that mass converted to a volume of fuel to be added. The Fi rst Officer of the plane asked a mechanic for the conversion factor to do t he volume-to -mass conversion, and t he mechanic replied "1.77." Using that number, the First Officer and t he mechanics calculated that 4917 L of fueLshould be added. But late r calculations showed that this is only about one fourth of the required amount of fuel! Why? Because no one thought about the units of the number 1.77. They realized later that 1.77 has units of pounds per liter and not kilograms per liter. Outof fuel, the plane could not make it to Winnipeg, so controllers directed them to the town of Gimli and to a smallairport abandoned bythe Royal Canadian Air Force. Aftergliding for almost 30 mi nutes, the plane approached the
craft was flying at 26,000 ft from Montrea l to Edmonton as Air Canada Flight 143. Warning buzzers sounded in the cockpit. One of the world's largest planes was now a glider-the plane had run out of fuel! How did this modern airplane. having the latest technology, run out of fuel? A simple mistake had been made in calculating the amount of fuel required for the flight! Like all Boeing 767s, this plane had a sophisticated fuel gauge, but it was not working properly. The plane was still allowed to fly, however, because there is an alternative method of determining the quantity of fuel in the tanks. Mechanics can use a stick, much likethe oil dipstick in an automobile engine, to measurethe fuel levelin each of the three tanks. The mechanics in Montreal read the dipsticks, which were calibrated in centimeters, and translated those readings to a volume in liters. According to this, the plane had a total of 7682 L of fuel. Pilots always calculate fuel quantit ies in units of mass because they need to know the tota Lmass of the plane before take-off. Air Canada piLots had always calculated the quantity of fuel in pounds, but the new 767's fuel consumption was given in kiLograms. The pilots knew that 22,300 kg of fuel was required for the trip. If 7682 L of fuel remained in the tanks, how much had to be added? This involved using the fuel's density to convert 7682 Lto a mass in kilograms. The mass of fuel to be added could then be calcu-
Gimli runway. The runway, however, had been converted to a race course for cars, and a race was underway. Furthermore, a steel barrier had been erected across the runway. Nonetheless, the pilot managed to touch down very near the end of the runway. The plane sped down the concrete strip; the nose wheelcollapsed; several tires blew-and the plane skidded safely to a stop just beforethe barrier. The Gim li glider had made it! And somewhere an aircraft mechanicis paying more attention to units on numbers.
Question: 1. What is the fuel density in units of kgjL? 2. What mass and what volume of fuel should have been loaded? ( l ib ~ 453.6 g) (See Study Question 58, page 48. ) Answers to these questions are in Appendix O.
t
gj
~
•
~
11
H l~
~~
Ii
@~
The Gimli glider. After running out of fuel, Ai r Ca nada Flig ht 143 glided 29 minutes before landing on an abandoned airstrip at Gtmli, Manitoba, near Winnipeg .
3
I
Mathematics of Chemistry 41
ProbLem SoLving and ChemicaL Arithmetic
Problem-Solving Strategy So me of the ca lcu latio n s in che m istry can be co m p lex . Stude nts fre que nt ly fin d it is helpful to fo llow a de fini te p lan of attac k as illu str at ed in exam p les throughout thi s boo k. Step 1: Problem. Sta te the problem . Read it carefu lly- and then read it again .
Step 2: Strategy. Wh at key princ ip les are invol ved? Wh at info rm ati on is known or not known ? \Vhat in fo rma tio n might be th ere just to p lace th e questi on in th e co n tex t of c he m istry? O rga n ize the in form ation to sec what is requi re d a nd to disco ver th e rel at io nships amo ng the data g iven. Try writing th e inform at ion d own in ta ble form . If it is n u me rica l in fo rma tio n, be su re to in clude u nits. O ne of th e g reate st diffi culties for a stu de n t in introductory che mis t ry is p icturin g what is bein g asked for. Try ske tc h ing a picture of th e situ atio n in vo lved . For exa mp le , we ske tched a pi cture of th e piece o f met al wh ose d en sity we wan te d to calc u late , a n d put the dim ensions on th e drawin g ( page 36) . Devel op a p la n . I l ave you d on e a p roblem o f thi s typ e before? If n ot , pe rh ap s th e proble m is really j us t a com binatio n o f several sim p le r ones yo u have see n before . Break it d own into th o se sim p le r co m pone n ts. Try reasoning backward fro m th e units of the a nswe r. Wh at dale'! do yo u nee d to find an answe r in those u nit s? Step 3: Solution. Execu te th e p lan. Ca re fu lly write down eac h ste p of th e problem , being sure to keep track o f th e units o n n umbe rs. (Do th e u ni ts ca nce l to give yo u th e an swer in th e d esired u nit s?) Don 't skip ste ps. Don 't d o a nyth in g excep t th e sim plest ste ps in yo u r h ead . Stu d e n ts often say th ey go t a proble m wro ng becau se th e y "made a stu p id mi stak e ." Your in st ruct or- and bo ok a ut h o rs-ma ke th em , too , a n d it is usu ally becau se th ey don 't rake th e tim e to wr ite down th e ste ps o f th e problem clearly. Step 4: Comment and Check Answer. As a fin al c hec k, ask you rsel f wh e the r th e an swer is reason abl e .
Problem Solving ProbLem A mineral oil has a density of 0.875 q/cm'. Suppose you spread 0.75 g of this oil over the surface of water in a large dish with an inner diam eter of 21.6 cm. Howt hick is the oil layer? Express the t hickn ess in centimete rs. Strategy It is often useful to begin solvin g such problems by sketching a picture of t he sit uation.
f-= :===~ 21.6 em
---
This helps recognize that t he solution to t he problem is to find the volume of the oil on t he water. If we know the volume, then we can find the t hickness because Volume of oil layer
= (thickness of layer) X (area of oil layer)
So, we need two things: (a) the volume ofthe oil layer and (b) t he area of the layer.
42
L e t ' s R evi ew
I
TheTools of Quantitative Chemist ry
SoLution First, calculate the volume of oil. The mass of the oil layeris known, so combining the mass of oil with its density gives the volume of the oil used: , 1 em! 0.75 g X - - -. = 0.86 em" 0.875 9 Next, calculatethe area of the oil layer. The oil is spread over a circular surface, whose area is given by Area
= 11" X (radius)"
The radius of the oil layeris half its diameter ( = 21.6 cm) or 10.8 em. so Area of oil layer = (3.142)(10.8 ern)"
= 366 em"
With the volume and the area of the oil layer known, the thickness can be calculated. Thickness
3
= Volume = 0.86 cm = 0.0023 ern Area
2
366 cm
Comment In the volumecalculation, the calculatorshows 0.857143.... The quotient should have two significantfigu res because 0.75 has two significant figu res, so the result of t his step is 0.86 ern'. In the area calculation, the calculator shows 366.435.... The answerto this step should have three significant figures because 10.8 has three. When these interim results are combined in calculating thickness, however. the final result can have only two significantfigures. Premature rounding can lead to errors.
EXERCISE 9 Problem SoLving A particular paint has a density of 0.914 gJcm 3• You need to covera wallthat is 7.6 m tong and 2.74 m high with a paint layer 0.13 mm thick. Wha t volume of paint (in liters) is required? What is the mass (in grams) of the paint layer?
STUDY QUESTIONS
4. Make the followin g temperat ure conve rsio ns:
On line homework for t his chapte r may be assigned in OWL. JJ;. denotes challenging questions.
•
denotes questions assignable in OWL.
Blue-numbered questions have answers in Ap pendix a and fullyworked solutions in t he Student Solutions Manual.
°C (a) _ _ (1)) 63 (c) _ _
K
77 14.>0
Le ngth , Volume, Mass, an d Density (See Exa mple I and Exercises 2-4) 5. A mara thon d ist an ce race cove rs a d ista nce of 42 . 195 km . Wh a t is th is d ista nce in met ers? In mi les?
Practidng Skills
6. • T he ave rage lead pen cil , ne w a nd unused , is 19 em long. What is its length in milli m et e rs? In meters?
Temperature Scales (Exnrise 1) 1. Man y labo ra to ries u se 25 "C as a standard temperatu re . Wh at is thi s te mpe ratu re in kelvin s? 2. The te mpe ratu re on th e su rface of the SU ll is 5.Ei X 103 "C. Wh at is this tempe rature in kelvins? 3• • Make th e follo wing temperature co nve rsi o ns: °C (a) 16 (1)) _ _
K
:l70
7. A sta ndard U.S. posta ge stam p is 2.5 em long an d 2. 1 cm wide. Wh at is th e area of the sta m p in sq uare ce nti meters? In square me ters? 8• • A co m pac t disc h as a d iameter of 11.8 em. Wh at is th e su r face area of th e d isc in sq uare ce n t imeters? In sq uare m e te rs? [Area of a circle = (1T) ( ra diusj e.] 9. A typi ca l lab oratory be ake r has a vo lu me of 25 0. mL. Wh at is its vo lu me in cu hi c ce n time te rs? In lite rs? In cu bic meters? III cu bic de cimeters?
(c) 40
I
43
10. • Some soft drinks a re so ld in bo ttles 'with a vo lu m e of 1.5 1.. Wh at is th is vol ume in m illili te rs? In cubic cen um et ers? In cu bic d ec im eters? 11. A book h as a mass of 2.52 kg. Wh at is thi s ma ss in grams? 12. A new U.S. dim e ha s a m ass of 2.265 g. What is its mass in kilo g ram s? In m illig ram s? 13. • Et hylene glyco l, C 211 1,0 2' is an ingre dient of autom obile an tifree ze . Its densi ty is 1.11 g/c m 3 a t 20 "C. If yo u need 500. m l , of th is liq u id , what m ass of th e compou nd , in grams , is re q u ired? 14. • A pie ce of silver m et al has a ma ss of 2.365 g. If th e d en sity of silver is 10.5 g/c m 3 , wh at is the vo lum e of th e silver? 15. • You can id e ntify a metal by ca re full y det e rmi nin g its den sity «(l) . An unknown pi ec e of m etal , with a mass of 2361 g , is 23 5 em lo ng, 1.34 em wide , an d 1.05 mm th ick. Whi ch of th e foll o wing is th e elemen t? (a) Nickel , d = 8.9 1 g/c m 3 (b) Ti tan ium , d = 4.5 0 g/cm 3 (c) Zinc , d = 7.14 g/ cm 3 (d ) Tin , d = 7.23 ginn ] 16• • Whi ch occu p ies a larger vol ume , 600 g of wa ter (with a density of 0.995 gl ern") or 600 g of lea d (,v-ith a d e nsity of 11.35 g/cm 3 ) ?
Accuracy, Precision , Error, and Stand ard Deviation (Sri' Example 2 and Exercise 5) 17. \'bu a nd your lab part n er are aske d to d etermi ne th e d en sity of an alum in u m ba r. The m ass is kn o wn acc ura rely (to fo ur sign ifican t figu res) . You use a sim ple me tric ru ler to m easu re its dim en sions and fin d th e results in A. Your partner uses a p re cisio n mi cro meter, an d o bta ins th e res ults in B. Method A (g/cm ')
Met hod B (gjcm 3 )
2.2
2.703
2.3
2.70 1
2.7
2.705
2.4
5.811
Th e ac ce p te d den sity of alu mi num is 2. 702 g/cm 3 . (a) Calculate th e ave rage de n sity fo r e ac h m eth od. Sh ould all th e ex pe rim e n ta l res u lts be includ ed in yo ur calcu latio ns? If not, j us tify an y o m issio ns. (b) Calc u late th e percent e rror for each met hod 's average value. (c ) Ca lculat e th e stan dard d eviation fo r each set of dat a . (d ) Wh ich meth od 's ave rage valu e is more pre cise ? Which m eth od is m ore acc u ra te?
44 1
18. • Th e acc e pte d value of the me lting point of pu re asp irin is 135 °C. Tr ying to verity th at value, you obta in 134 °C, 136 -c, 133 °C, a nd 138 °C ill fou r se para te trials. Y,
3.00 2.00
Solving Equations 1.00
27 . Sol ve th e followin g equation to r th e unknown val ue, (; (0.5 02) ( 123)
0.50
x values
(7.>0.) C
28. So lve t he foll owing equation for th e unknown va lu e , n. (2 .34) (15.6)
25 . • Use the g raph bel ow to ans wer th e followin g que stion s. (a) Derive th e e qu atio n fo r the stra igh t lin e , )' = mx (h) What is th e value of )' when x = (i.o?
~
~
,,( 0.08 2 1) (273)
29 . So lve th e foll owing eq ua tio n for the unknown va lue, T.
+
25.00'---------------...,
b.
(4.J84)(2 44)( T - 292. 0)
+
(0 .44!I)(88.5)( T - 369.0)
~
0
30, Sol ve th e foI1O\\'in g eq ua tion for the unknown value, n. - 246.0
~ 1 31 2[~, -
,:']
General Questions 20.00
T hese questions
(JJt'
not desig nated as to t)1)f' or location in the
chapter: T he)' Inri)' combine sroeml concepts.
3 1. Mo lecul a r d istanc es a rc usu all y give n in n an ometers ( 1 11 m = I X 1O- y m ) o r in picomct ers ( I pm = I X IO- l t m) . However, the angs tro m (A) u nit is so met imes used , where I A = I X 10- 10 Il l. (T he a ngstrom un it is not an SI unit. ) If the di stance betwee n th e PI a to m an d the N ato m ill the cancer chemotherapy drug cispla tin is 1.97 A, wh at is thi s di stance in n anomc - ' tcrs ? In picomctcrs?
xvaLues
Cisplatin .
.... more challenging
• i n OWL BLue-numbered questions answered in Appendix0
I
45
32. • Th e separatio n bet ween ca rbo n ato ms in d iamo nd is 0.154 run . Wh a t is th ei r se paration in met ers? In p icom ctcrs (pm )? In Angs tro ms (t\.)? (See Study Q uestio n 3 1.)
A portion of the diamond structure.
38. • You have a wh ite crys tallin e sol id , kn own to be o ne of th e potassium co m po un ds listed below. To dete rmin e wh ich , yo u m ea su re it" de n sity. You me a su re o u t 18.82 g an d tran sfer it to a graduated cylinder co n tai ning kerosene (in wh ich sa lts will not d issolve ) . T he level of liqu id kerosene rises fro m 8.5 mL to 15.3 mL. Calc ul at e the density of th e so lid , and id en tify th e compo u nd fro m th e foll owing list. (a) KF, d ~ 2.4H g /Cl n ' (h) KC!, d = 1.98 g/Cln' (c) KRr, d = 2.75 g/c m J (d ) KI, d = 3. 13 g/ Cln' 39 . • .A Th e sma lles t re peat ing un it of a crystal of comm o n salt is a cu be (ca lle d a unit ce ll) with a n e dge leng th of 0.56 3 nm .
33 . • A re d blood ce ll h as a d ia m et e r of 7.5 um ( m icrom e ters) , Wh a t is th is dim e n sio n in (a) m et ers, (h) na n o m et e rs, an d (c ) picom et e rs? 34. • Th e pl at in u m-con ta in ing canc er d ru g cisplatin (Study Q u estion 31 ) co ntains 65. 0 ma ss-pe rcen t of the me tal. If you h ave 1.53 g of the co mpou nd , wh a t m ass of p la tin u m (in g ra ms) is con tai n ed in thi s sa m ple? 35 . • T h e a nesth e tic pro ca in e h yd ro chl o ride is often used to deaden pai n d u ring den tal surg ery. T h e co m pou nd is packaged as a 10.% sol ut io n (by mass; d = 1.0 g / m L) in water. If yo ur d entist injec ts 0.50 mL of th e so lu tio n , wha t m ass of pro caine h yd ro chlo rid e (i n mi llig ram s) is inj ec ted ?
10.563
om
Sodium chloride, NaCl.
36. • You need a cu be of alu min um 'with a m ass of 7.6 g . Wh at m u st he th e le ngth of th e cu be 's e dge (in em)? (T h e density of al um in um is 2. 698 g /cm 3 . )
(a) What is the vol ume of th is cu be in cu bic na nom ete rs? In cu bic ce n tim e te rs? (b) T he de ns ity of NaCI is 2. 17 g / cm 3 . Wh at is th e ma ss of th is smallest repea ting uni t ("'u ni t ce ll")? (c) Each re peatin g un it is co m posed of four Na C I "m ole cu les." wh a t is the mass of o n e NaCI m ole cu le?
37 . • 'rou have a 250.0-m L grad uate d cylin der co n taini ng so m e wa te r. Vi m d ro p 3 m a rble s with a tota l ma ss of 95.2 g into the wate r. Wh at is the ave rage de n sity of a ma rb le?
40 . • Diam o nd has a d e nsity of 3.5 13 g /cm 3 . T he mass of d iam o nd s is ofte n me a sured in "carats," whe re I ca ra t e q ua ls 0.200 g. Wh a t is the vo lu me (in cuh ic cen timeters) of a 1.5 O-Ca rat di amond ? 41. T he ele m en t ga lliu m ha s a m el ting po in t of 29.8 "C. If yo u hel d a sam ple of ga lliu m in yo u r h a nd , sho u ld it melt? Expla in hricfl v,
(a)
(b)
Determining density. (a) Agraduated cylinder with 61 mLof water. (b) Three marbles are added to t he cylinder. Gallium metal.
46
I
... more challenging • in OWL Blue-numbered questionsanswered in Appendix 0
42. • lit. The de nsity of pure wate r is give n at va rio u s te mperatu res. t
CC)
d (g/cm')
4
0. 99997
15
0.99913
25
0.99707
35
0.99406
Suppose yo u r laboratory partn er tell s yo u the d ensity of water a t 20 °C is 0.99910 g/on 3 . Is thi s a re asonable n u m ber? Wh y or why not?
43. \Vh e n
y O li heal pop corn , it pops be cause it lo se s wa ter explosivel y. Assu me a kernel of co rn , with a m ass of 0,1 25 g, ha s a m ass o f o nly 0. 106 g aft er popping. (a) What percentage of its m ass d id th e ke rn el lo se on
popping? (h) • Popco rn is so ld hy the pou nd ill th e U n ite d Sta tes . Using O.12fi g as the avnagc m ass of a popco rn kern el, h ow m a ny kernels arc th ere in a poun d of popco rn ? ( 1 II, = 453.6 g )
44. • ... Th e a lu m in u m in a pac kage co n tain ing 75 ft 2 of kitch en foil wei ghs ap prox ima tely 12 oun ce s. Alumi num h as a d en sity of 2.70 g / cm:l . What is th e approxim at e thickness o f th e a lu m in u m foil in milli m ete rs? ( 1 07. = 2H.4 g ) 45. • .... T h e fluorid ati on o f city wate r su p plies h as be en p racti ced in th e Uni te d Sta te s fo r seve ral d ecad e s. It is don e hy co n tin uo usly ad d in g sod ium fluoride to wa ter as it co mes fro m a rese rvo ir. Assum e yo u live in a m edium-sized city of 150 ,000 pe ople an d th at 660 L (170 ga l) of wa te r is con sumed pe r perso n per d ay. wh at mass of so d iu m flu o rid e (in kilogra m s) must be add ed to the wate r su p ply e ach year (3 65 d ays) to ha ve th e re q u ired fluoride conce ntra tion of I ppm (part pe r million )-that is, 1 kilo g ra m o f fluorid e pe r 1 million kilogram s o f wate r? (So d iu m fluo ride is 45 .0 % flu o rid e , and water has a d ensity o f 1.00 g /Cl1l3 . )
46. • ... Ab out two cen tu ries agu , Benjamin Fra nklin showed th at I teaspoon of oil would cove r abo u t 0.5 acre of still wa ter. If yo u kn ow th at 1.0 X l O' m'' = 2.4 7 acr es, a nd th at th e re is ap proximately 5 ern " in a teaspoon , wha t is th e thi ck ness of th e laye r of oi l? How mi ght thi s thi ckne ss be rel ated to the sizes o f molecules?
47. • ... Automobil e ba tte ries are till ed with an aq ue ous so lu tio n o f sulfu ric acid . Wh at is the m ass o f the a cid (in gram s) in 500. mL of th e batte ry ac id so lu tio n if til e density of th e so lu tion is 1.2H5 g/c m :l a nd if th e solu tion is 38.08% su lfu ric acid by mass? 48 . • A 26-me te r-ta ll sta tu e o f Buddh a in T ib et is cove red with 27 9 kg of go ld . If th e go ld was applied to a thi ckn ess of 0.001 5 m m, what sur fac e area is cove re d (in sq uare in cu-rs}? (G o ld d e n sity = 19.3 g/c m 3 )
.... morechallenging
• in OWL
BLue-numbered questions answered in Appendix 0
49 . At 25 DC, th e den sity of water is 0.997 g /nn \ where as th e den sity of ice a t - 10° C is 0 .917 g/ cm 3 • (a) If a soft-d rink can (vo lu me = 250. mL) is fill ed co m ple tely with pure wate r a t 25 °C a nd then fro zen a t - 10 DC, wh at vo lu me doc s th e so lid oc cupy? (b) Ca n th e icc be co n ta ined with in th e ca n ? 50 . • Suppo se yo u r bedroo m is I H ft lo n g , 15 ft wid e , ami th e di stan ce fro m floo r to ceiling is 8 ft, 6 in . 'rou need to kn ow th e vo lu me o f th e room in met ric u ni ts for so m e scie n tific ca lcu la tio ns . (a ) Wh at is th e ro om 's vo lum e in cu bic meu-rs? In lite rs? (b) What is th e m ass o f air in the roo m in kilo g ra ms? In poun d s? (Assu m e th e d en sity of air is 1.2 giL a nd that th e room is e m p ty o f fu rniturc. )
51. • A sp herica l ste el hall ha s a mass o f 3.47.e.. g an d a diam cu-r of 9.40 mm . Wh at is th e d e nsity of th e steel? [Th e vo lu me o f a sp here = (4/ 3) 7Tr3 where r = ra d ius.]
52. • .... T he su bstances listed below a re cl ear liquid s.
Y{Hl
are ask ed to id e ntity an u nknown liquid th at is known to he o ne of these liquids. You pipet a 3.50-m i. sam ple into a be aker. The e m p ty beaker h ad a ma ss of 12.20 g , an d th e be ak e r plus th e liquid wei g he d 16.0 8 g. Substance
Density at 25 °C (g{cm 3)
Eth yle ne glycol
1.1088 (the major component of antifreeze)
Water
0.9971
Etha nol
0.7893 (t he alcohol in alcoholic beverages)
Acetic acid
1.0 492 (t he active component of vinegar )
Glycerol
1.2613 (a solvent, used in home care products)
(a) Calcula te th e d en sity a nd id entify th e u n kn own . (h) If yo u were ahlc to m e asu re th e vo lume to o n ly two sig n ific an t figures (th a t is, 35 m l ., not 3.50 mI. ) , will the re su lts be su fficie n tly accurat e to id entify th e u nknown ? Ex plai n.
53. • ... You h ave an irregularly shaped piece of a n unk nown metal. To id e n tify it, yo u d ete rmin e its d e nsity an d the n co m pare th is value with kn own value s th at yo u look up in th e che mi stry lib rar y. T he mass of th e m etal is 74 .122 g. Becau se of th e ir reg ular sha pe , yo u m ea su re th e vo lu m e hy submergi ng th e metal in wa ter in a graduat ed cylind e r. Whe n you d o thi s, the water le vel in th e cylinder rises from 28.2 m l , to 36 .7 m l; (a) What is th e d en sity o f the m etal ? (Us e the correct number of sig n ifican t figu re s in yo u r a nswcr.) (b) Th e u n known is o ne of th e seven m etals listed helo w. Is it possi bl e to id entify the m et al base-d o n the d ensity y()U have calcu la ted? Explain . Metal zinc
Density (g{cm 3)
Metal
7.13
nickel
Density (gjcm 3) 8.9 0
iron
7.87
copper
8.96
cadmium
8.65
silver
10.50
cobalt
8.90
I
47
54 . • A Th ere ar e 5 hydroca rb on compounds (co m pound s of C an d II ) th a t h ave the fo rmul a C l;II 14 • (T h ese ar e iso me rs; they differ in th e way th a t C and II a to ms arc a ttac h ed. See C hap ters 8 and 10.) All are liquid s a t roo m te mpe rature but h ave slig h tly different d en siti es. Hydrocarbon
Density (g{mL)
hexane
0.66 00
2,3-dimethylbutane
0.6616
1-methylpentane
0.6532
2,2-dimeth ylbutane
0.6485
2-met hylpentane
0.6645
(a) 'rou have a purl' sam ple o f o ne of these hvdrocarbOIlS, an d to identify it you d ecide to measure its d e nsity. \ bu determ ine th at a Ei.O-m L sam ple (m ea sured in a g rad ua ted cylinder) has a mass o f 3.2745 g (measured O il an analytical bal ance. ) Assum e th at th e accu racy of th e values for mass and volu me is expressed by th e number of sig n ifican t figures, th at is, plus or m inus o ne (:::!: I ) in th e last sig n ifican t figure . \\'Iat is th e den sity of the liq uid ? (b ) Expre ss th e est imated un certai nty of yo u r value in two oth er ways: i) Th e value you have calcu la ted fo r th e de ns ity is uncertain to _ _ g / m L ii) Th e val ue calc ula te d fo r density is betwee n x g / m L a nd J g/ m L (c) Ca n you id entify th e unknown hydroca rbon ba se d on you r ex pe rim e n t? (d) Can yo u e limi na te an y of the five po ssibili tie s base d o n th e data? If so , wh ich on e (s) ? (e ) \ hu need a more acc u ra te vo lu m e m eas uremen t to so lve thi s problem , ami yo u redete rmin e th e vo lu m e to be 4.93 mL Rased o n these new d ata , what is the u n kn own co m po u n d? 55 . • A Su p pose you have a cylindrical glass tube with a th in capillary o peni ng , and yo u wish to d etermin e th e diam eter of the capillary. You ca n do thi s experim entally by we igh ing a piece o f th e tubi ng before a nd afte r fillin g a portion of th e ca pillary wit h m e rcu ry. Using the follo wing informat ion , calc u late th e di amet e r of th e capillary,
57 . ... CO PPER: Sec the illustratio n of the copper lattice on page 24. (a ) Suppose yo u h ave a cu be of coppe r metal th at is 0.236 em on a side with a m ass of 0.120 6 g. If yo u kn ow that eac h co p pe r atom (rad ius = 128 pm ) h as a m ass o f 1.0;')5 X 10 - 22 g (yo u will learn in Ch a p ter 2 how to find the mass of o ne a to m ) , how lTI an y a to m s are the re in thi s cube? Wh at fr action o f the cu be is filled with a to m s? (O r conve rsel y, hoc...' muc h of the lattice is em p ty spa ce ?) Wh y is the re "e m p ty" space in the lat tice? (b ) Now look a t th e smalles t , repe ating un it of th e crystal laui cc of co p pe r. Knowing that a n edge of thi s cu be is 36 1.4 7 pm an d the d en sity of co p pe r is 8 .960 g /cm 3, estim at e th e number o f copper a toms in thi s sm a lle st, repeating un it. 58 . . ... CASE ST UD\~ In July 19 83, a n Air Ca nada Bo ei n g 767 ran out of fuel o ver ce n tral Ca na da 011 a trip fro m Montreal to Edmonton. (T he pl ane g lid ed safe ly to a lan ding a t an aban d o ned ai rstri p .) The pilots knew th at 22 ,300 kg of fuel we re required fo r th e trip , and th ey kn ew that i 682 L of fuel were alrea dy in the tank . Th e g ro u n d cr ew added 4916 L o f fuel , wh ich was on ly abo u t o ne fifth o f what was requ ired . Th e crew mcmhers used a fact o r of l ,77 fo r the fu el d en sity-the p roblem is that 1.77 has u nits o f pounds pe r lite r and not kilogra ms pe r lite r ! Wh at is th e fuel de nsi ty in uni ts of kg / I.? What m ass an d what vo lu m e o f fuel shou ld h ave bee n lo aded ? (I III = 453.6 g )
In the Laboratory 59 . • A sa m ple o f un known m etal is placed in a grad ua ted cylin der con tain ing wa ter. Th e mass of th e sa m ple is 3 7.5 g , a nd th e wate r le vel s befo re and aft er ad ding the sam ple to th e cyli nd e r ar e as sho wn in the figure . Wh ich metal in th e following list is most likel y th e sampl e? ( d is the d ensit y of th e mc tal .) (a ) ~l g , d ~ 1.74 g/e m' (d) AI, d ~ 2.70 g/e m 3 3 (h ) Fe , d = 7.87 g /c m (e ) Cu , d = 8.96 g /cm 3 (e) Ag, d ~ 10 .5 g/ em' (I) Ph , d ~ 11.3 g/ em'
25
Mass o f tube before ad d in g m ercury = 3.263 g Mass o f tube afte r addin g m ercu ry
=
= 3.4 16 g
20 15 10
Volu m e of cylind rica l cap illary filled wit h m ercury (7T) (ra d ius )2(le ngth )
48
t:::
E .-
20 15
~
13.546 g / ern "
56 • • COPPER: Co p per h as a d e nsit y of 8.9 6 g/c m ~ . An ingot o f co p pe r with a m ass o f 5 7 kg (1 26 Ib ) is drawn int o wire with a di am eter of 9.50 mm . Wh at length of wire (in m eters) ca n he produced? [Vo lu me of wire = (1T ) (ra d ius ) 2 (le ng t h)]
25
~
Len gth o f ca p illary filled with m ercury = 16.75 mm Den sity of m ercury
~
5
~
~
10 5
Graduated cylinders with unknown metal (right).
... more challenging
• in OWL
Blue-numbered questionsanswered in Appendix 0
60 . • Ir on pyrite is o ften called "fo o l' s gold " beca use it looks like g o ld (sec page 14 ) . Supp ose yo u h ave a soli d th at lo o ks like go ld , bu t yo u beli eve it to be fo o l 's go ld . Th e sam ple has a m ass of 23.5 g. Whe n th e sa m p le is low ered into th e wa te r in a g rad u a ted cylin de r (se e Study Q ue stio n 37 ) , th e wate r le vel rise s from 47.5 m L to 52.2 mL. Is th e sa m ple fool 's go ld ( d = 5. 00 g/cm 3 ) or "re al" gold (d = 19.3 g /cm 3) ? 61. Yo u ca n analyze fo r a co pper co mpo un d in water using a n in stru m e nt ca lled a spect ro p ho to me te r, In th is tec hn ique , th e lig ht p assing th rough an aq ueous sol ut ion o f a co m pou nd ca n be ab sorbed , a n d th e am o un t of light abso rbed (a t a g ive n waveleng th of ligh t) d e pen d s direc tly o n th e am o u n t o f co m po un d pe r liter o f so lulio n . To ca librate the spec trop hoto me te r, yo u col le ct th e fo llowin g d a ta : Absorbance (A)
Concentration of Copper Compound (giL)
0.000
0.000
0.257
1.029 X 10 - 3
0.518
2.058 X 10 - 3
0.77 1
3.087 X 10 - 3
1.021
4.116 X 10 - 3
62. A gas c hromatogra p h ( pag e 2 ) is calib ra ted I'Of the analysis o f iso o ctaue (a m aj o r gasoline com ponen t) in g th e foll owin g d at a : Percent Isooctane
Instrument Response
(x-data)
(y-data)
0.352
1.09
0. 803
1.78
1. 08
2.6 0
1. 38
3.0 3
1. 75
4.01
u s~
If th e in stru men t re spon se is 2.75 , what pe rce ntage o f iso octa nc is p re se n t? (Da ta a rc tak e n from Analytical ChnniJlry, A n l ntrodu ainn, hy D .A . Sko og, D ,l\L West, F. .J. Hol ler; a nd S. R. Crouc h , T h omson-Brookx/ Cole , Be lmo n t, CA , 7th Edi tio n , 2000. )
Plo t the abso rbance (A) again st th e m ass o f co p pef com po u nd pe f lit er (g i L), and find th e slo pe (m) a nd in te rcep t (b) (ass u m ing th a t A is )' an d the amo u n t in sol u tio n is x in t he eq u ation fo r a stra igh t line , y = mx + b). What. is the a m o u n t of co p pe f com po u n d in th e so lu tion in g i L a nd m g /mL when th e absorban ce is O.635 ?
.... morechallenging • in OWL
BLue-numbered questions answered in Appendix 0
I
49
CONCEPTS OF CHEMISTRY
2
Atoms, Molecules, and Ions
Nineteen t h-century
chem-
/
\
TABELLE II.
The Periodic Table, the Central Icon of Chemistry
GRUPPE I.
GRUPPE II.
GRUPPE III.
GRUPPE IV.
GRUPPE V.
GRUPPE VI.
GRUPPE VII.
I
-
-
-
RH'
-
R'O
RH' RO'
RH
w
RH ' RO'
R20 7
RO'
z
w
~
R20 3
RO
1 2 3 4
H=l li = 7 Be=9,4 l Na=23 Mg=24 Ca =40 K 39
5 6
Zn=65 (Cu=63) Sr=87 Rb=B5
R20 5
B= 11 C=12 N=14 AI=27,3 Si=28 P=31 -= 44 Ti = 48 V=51
0=16 5=32
F=19 Cl=35,5
Cr=52
Mn=55
Fe=56, Co =59,
Se=78 Mo=96
Br=80 - =100
ists such as New lands, Chancourtois, Mayer, and others
Ni=59 1 Cu=63. 1 - = 6B
--i Ir
?Yt=88
- -72 90
As= 75
Nb= 94
devised ways to organize the chemistry of t he elements with varying degrees
of success. However, it was Dmitri Mendeleev in 1870 who first t ruly recognized
GRUPPE VIII .
Ru= 104, Rh= 104,
Pd=106, Ag=10S. (Ag = l OB)
7 S 9 10
-
11 12
-
(s=133
In = 113
Cd=112 Ba=137
-
(Au=199)
?Er= 178
Hg=200 -
11= 204 -
-
?La= 180
Te= 125
Sb=122 -
?Ce=140
?Di=138
-
( -)
Sn = 118
Ta = 182
Pb= 207 Th = 231 -
H
127
-
-
----
-
W=184
-
Bi=208
J
U=240
05=195 ,Ir = 197, Pt=198, Au=199.
-
----
the periodicity of t he chemist ry of t he elements, who proposed t he first periodic ta ble, and who
he left a place for them in the table (marking t he empty places with
used t his to predict the existence of yet- unknown elements.
a - ). For example, Mendeleev concluded that "Gruppe Nil was miss-
Mendeleev placed the elements in a ta ble in order of increasing
ing an elementbetween silicon (5i) and tin (5n) and marked its posi72." He called the missing element eka-silicon and
atomic weig ht. In doing so Li, Be, B, C, N, 0, and F became the
tion as "-
first row of t he ta ble. The next element then known, sodium (Na),
predicted the element wou ld have, for example, an atomic weight of
had properties quite similar to th ose of lithium (Li), so Na began
72 and a density of 5.5 g/cm'. Based on this and other predictions,
t he next row of the table. As additional elements were added in
chemists knewwhat to lookfor in mineral samples, and soon many of
order of increasing atomic weight, elements with similar proper-
the missing elements were discovered.
ties fell in columns or groups.
Questions: 1. What is eka-silicon, and how close were Mendeleev's predictions to t he actual values for t his element? 2. How many of t he missing elements can you identify?
If you compare the periodic table published by Mendeleev in 1871
(shown here) wit h the table in the front of this book, youwill see t hat many elements are missing in t he 1871 table. Mendeleev's genius was that he recognized there must be yet-undiscovered elements, and so 50
=
Answers to these questions are in Appendix Q.
Chapter Goals SPf' Cha/lln Goals Revisited (pagr 72) for Stud)' flueslions keyed to these bro(Jl~ and fl.uigna!JlP in OWL
Chapter Outline 2.1
Atomic Structure- Proto ns, Electrons, and Neutrons
2.2
Atomic Number and Atomic Mass
2 .3
Isotopes
t he isotopic masses and abundances. Know t he term inology of the periodic ta bLe.
2.4
Atomic Weig ht
2.5
The Periodic TabLe
Interpret, predict , and write formulas for ionic and molecula r compounds.
2.6
MoLecules, Compounds, and Formulas
2.7
Ionic Compou nds: Formulas, Na mes, and Properties . . . ~
Describe atomi c structure, and define ato mic number and mass number. Underst and t he nat ure of isot ope s, and calculate at omic weights from
Na me ionic and molecular compounds.
~~
Understand some properties of ionic compou nds.
2.8
MoLecu lar Com pounds: Formulas and Names
ExpLain the concept of the mole. and use molar massin calcu lations.
2.9
Atoms, Molecules, and t he Mo le
Calculate percent composition for a compound and derive formulas from experimental data.
2.10 Describing Compound Formulas 2.11
t"
~:s;
Hydrated Compou nds
T
he che mical elemen ts are fo rged in sta rs, and from the se elem en ts m olecul es suc h as water and am mo nia are mad e in o u ter spac e. T he se sim ple molecules an d much more co mplex o ne s suc h as DNA and hemoglohin are fo u nd o n earth. T
(19.91) 80.09) X I 0. 0129+ ( - X 100 100
l 1. 0 0 9 3 ~ 1 0.8 1
Eq uatio n 2.2 g ives an ayerage mass, weig ht ed in terms o f the a bu nda nce o f ea ch isotope fo r th e e le me nt. As illu strated by the d ata in Table 2.2, the atomic mass oJ an element is usually closerto the mass oJ the most abun dant isoto/Ie or isotopes. T he ato m ic weigh t of ea ch stable e leme n t has been de term ined ex perime n ta lly, and these n u m bers appear in th e pe riod ic table in side the fro n t coyer of thi s book. In the period ic tab le, eac h elem ent's bo x co n ta in s the ato m ic n u mber, the ele me n t sym bo l, and th e atomi c wcig h t. For u nstable (radioactive) clem en ts, th e atomic weigh t or mass number of the most stable isot ope is g iven in pare n theses. ChemislrY.{J.NDW~ Sign in at www.t homsonedu.comjlogin and go to Chapter 2 Contents to see Scree n 2.13 for an exercise and a t utorial on mass s pect rometers and on calculati ng at omic mass.
56
Chapter 2
I
Atoms, Molecules, and Ions
CalcuLating Atomic Weight from Isotope Abundance Problem Bromi ne has two naturally occurring isotopes. One has a mass of 78.918338 and an abundance of 50 .69% . The other isotope, of mass 80 .916291, has an abundance of 49.31%. Ca lculate the atomic
weight of bromine. Strategy The atom ic weight of any eleme nt is the weighted ave rage of the masses of the isot opes in a
representative sample. To calculate the atomic weight, multiply t he mass of each isotope by its percent abundance divided by 100 (Equation 2.2).
SoLution Atomic weiqht of bromine
~ ~
(50.69/100 )( 78. 918338) 79. 90
+ (49.3 1/ 100)(80.916291)
CalcuLating Isotopic Abundances Problem Antimony, Sb, has two stab le isotopes: m Sb, 120.904 ative abu ndances of these isot opes?
U,
a nd 123Sb, 122.90 4 u. What are th e rel-
Strategy The atomic mass of antimony is 121.760 u (see t he periodic table). Befo re we do t he calculation we can infer that the lighter isotope (l21Sb) must be the more abundant because the atomic weight is closer to 121 than to 123. Next, to calculate the abundances we recognize there are two unknown but related quantities, and we can write t he following expression (where the fractional abundance of an isotope is the percent abundance of the isotope divided by 100):
Elementalbromine. Bromine is a deep orange, volatile liquid at room temperature. It consists of BT2 molecules in which two bromine atoms are chemically bonded together. There are two, stable, naturallyoccurring isotopes of bromine atoms: 19Br (50.69% abundant) and Bl Br (49.31% abundant) .
Atomic weight = 121.760
=
(fractional abundance of l21Sb)(120.904)
+ (fractional abundance of m Sb)(122.904)
or 121.76 0 ~ x(120 .904)
+ y(122.904)
where x = fractional abundance of l21Sb and y = fractional abundance of 123Sb. Because we know that the sum of the fractional abundances of the isotopes must equal 1, x + y = 1, and we can solve the simultaneous equations for x and y. SoLution Becausey
= fractional abundance of 123Sb = 1 ~ x(120 .904) + (1 - x)(122.904)
x, we can make a substitutio n for y.
121.760
Expanding this equation, we have 121.760
= 120.904x + 122.904
- 122.904x
Finally, solving for x, we find 121.7 60 - 122 .904
x
~
(120 .904 - 122.904 )x
= 0.5720
The fractional abundance of l21Sb is 0.5720, and its percent abundance is 57.20%. This means that t he percent abundance of m Sb must be 42.80%. The result confirmsour initial inference that the tighter isotope is the more abundant of t he two.
EXERCISE 2 .3
Calculating Atomic Weight
Verify that the atomic weight of chlorine is 35.45, given the following information: 35 ( t mass = 34.96885; percent abundance 37 (l
= 75.77% mass = 36.96590; percent abundance = 24.23%
2.4
I
Atomic Weight 57
Catching Cheaters with Isotopes
Case Study The U.S. Anti-Doping Agency is responsible for testing for performance-enhancing drugs such as synt hetic testosterone that has been used by athletes (page 1). But, if an athlete were to take this synthetic steroid, how can chemists tell t he difference between t hat and the testoste rone normally occurring in a male at hlet e's body? Test ost erone (T) and epitestosterone (E) are closely related steroids, and bot h are produced naturally. (The latter is an isomer of testosterone, a compound with an identical formu la to testoste rone but with a difference in the way the molecule fills space.) In most
adult men the compounds are found in about equal amounts, although the natural TI E ratio can be as high as 4/1. To allow for elevated natural testosterone levels, the Anti-Doping Agency considers that a ratio of 6/1 is an indication an athlete must be using synthetic testosterone. This situation arose when the winner of the 2006 Tour de France bicycle race, Floyd l andis, was found to have a ratio of T/ E = 11/ 1 based on a urine test after one stage of the race. When the race was over, and the drugtesting results were announced, Landis strongly denied taking testosterone. He
" "3
e
~
argued that his body may have produced more testosterone than normal because of the rigors of the race. This prompted the doping-cont rol laboratory for the Tou r to seek additiona l evidence. They used a technique developed by the U.S. Anti-Doping Agency called "Isotope ratio mass spectrometry," which measures the ratio of carbon isotopes, 12( and 13C, in compounds. Carbon-t S is a naturally occurring isoto pe of carbon. When most plants grow using ( 0 2 from the atmosphere, about 1% of the ( atoms incorporated in the plant are 13e. The 13( is ingested either directly by humans when eating plants or indirectly when eating meat
from grazing animals, and it is t hen incorporated into the carbon-containing molecules, including testosterone, that our bodies build. Synthetic testosterone is made from wild yams and soy, plants that are so-called warm climate NC3 plants" that take up atmospheric CO 2 differentlythan temperate -zone "(4 plants." One important result of this difference is that C3 plants have a lower 13( j12( ratio than (4 plants. Because diets in most industrialized countries derive from a mixture of (3 and ( 4 plants, the naturaLtestosterone in male ath letes in most countries wi ll have a different 13( / 12( ratio tha n synthetic testosterone . A skilled scientist with a mass spectromete r (Figure 2.3), can relatively easily detect t he difference. In the Tour de France case, the 13( j12( ratio added further evidenceto t he case t hat illegal steroid use had occurred.
Questions: 1. How many neutrons are there in atoms of He? 2. Heis a radioactive isotope of corban that occurs in trace amounts in all living materials. How many neutrons are in a HC atom? 3. Use your library or the Warld Wide Web to find the source of 14C in living materials. Answers to these questions are in Appendix Q.
2.5 The Periodic Table
Module 1
T he pe riod ic tabl e of cle me nt s is on e o f th e mo st useful tool s in chem istry. No t o nly d oes it co n ta in a wea lth of inform ation , b u t it can also be used to o rgan ize many of the ideas o f che m istry. It is impo rtan t to beco m e fami liar with its main fea tures and termino logy.
Developing the Periodic Table • About the Periodic Table For more information on the periodic table, we recommend the following : The American Chemical Society has a description of every element on its website (http ://pubs.acs.org/cen / acth/etements. html). J. Emsley: Nature's Building Blocks-An A-Z Guide to the Elements, New York, Oxford University Press, 2001. O. Sacks: Uncle Tungsten-Memories of a Chemical Boyhood, New York, Alfred A. Knopf, 2001.
58
Although th e arra nge me nt of ele me nts in the pe rio d ic tab le is now understood o n the basis of a to m ic struc tu re (... Cha p te rs 6 and 7) , the tab le was o riginally d eveloped fro m many ex pe rimen tal o bservati o ns o f the chem ica l and p hysical p ro pe rti es of ele men ts and is the resu lt of the ideas of a nu mbe r o f che m ists in th e 18th and 19 th cen tu ries. In I H69, at th e Un ivcrsity o fS r. Pe tersburg in Ru ssia, Dmitri Ivan ovitch Mcndclccv ( 1834- 190 7) was pondering the properti es o f th e e leme n ts as he wro te a te xtbo ok o n che m istry. O n studying the chemica l an d p hysical pro pe rties of the elements, he realized th at , if th e cle me n ts were a rran ged in orde r o f in creasing ato mic weigh t, ele me n ts with similar pro perties appeared in a regular patte rn . T ha t is, he saw a p eriodicit y o r pe rio d ic repe titi o n of the p roperties of el em en ts. Me u d elee v o rgan ized th e kn own cleme nts in to a tabl e by lin ing th e m u p in a horizontal row in
Sign in at www.t homsonedu.comj login to download the Go Chemistry modu leforthis section or go to www.ichapters.comto purchase modules.
Historical Perspectives
The Story of the Periodic Table
by Eric R. Scerri, UCLA
In essence, the periodic ta ble groups togethe r sets of elements wit h similar properties into vertical columns. The underlying idea is that if the elements are arranged in order of increasing atomic weights, the re are approximate repetitions in their chemical properties after certain intervals. As a result of the existence of the periodic table, students and even professors of chemist ry were no longer obliged to learn the properties of all the elements in a disorganized fashion. Instead, they could concentrate on t he properties of representative members of the eight columns or groups in the early short-form periodic table, from which they could predict properties of other group members. Mendeleev is j ustly regarded as the leading discoverer of the periodic table since he conti nued to champion the finding and drew out it s consequences to a far greater extent tha n any of his contemporaries. First. he accommodated the 65 or so elements that were know n at the time into a coherent scheme based on ascending order of atomic weight while also reflecting chemical and physical similarities. Next. he noticed gaps in his system, which he reasoned would eventually be filled by elements t hat had not yet been discovered. In addition, by judicious interpola tion between the properties of known elements, Mendeleev predicted the
Dimitri Mendeleev was probably the greatest scientist produced by Russia. The youngest of 14 child ren. he wastaken by his mother on a long journey, on foot, in order to enroll him into a university. However, severalattempts initially proved futile because, as a Siberia n, Mendeleev was barred from attending certain institutions. His mother did succeed in enrolling him in a teacher training college, thus givi ng Mendeleev a lasting interest in science education. wh ich contributed to his eventual discoveryof the periodic system that essentially simplified the subject of inorganicchemistry. After com pleting a doctorate, Mendeleev headed to Germany for a postdoctoral fellowship and then returned to Russia, where he set about writing a book aimed at summarizing all of inorganic chemistry. It was while writi ng this book that he identified the organizing principle with which he is now invariably connected-the periodic system of the elements. More correctly, though, the periodic system was developed by Mendeleev, as well as five other scientists, over a period of about 10 years, after the Italian chemist Cannizzaro had published a consistent set of atomic weights in 1860. It appears that Mendeleev was unaware of t he work of several of his codiscoverers, however.
Statue of Dmitri Mendeleev and a periodic table mural. This statue and mu ral are at the Institute for Metrology in St. Petersburg, Russia. nature of a number of completely new elements. Within a period of about 20 years, three of these elements-subseq uently called gallium, scandium, and germanium-were isolated and found to have almost the exact properties that Mende leev had predicted. What is not well known is that about half of the elements that Me ndeleev predicted were never found. But given the dramatic success of his early predictions, these later lapses have largely been forgotten . Eric Seem. The Periodic Table: Its Story and Its Significance, Oxford University Press, New York, 2007.
order of in creasing a to m ic weig ht (page 50) , Every time he ca me to a n ele me n t with p roperties sim ilar to one already in th e ro w, he sta rted a new row. For exam ple, th e elements Li, He , H, C, N, 0 , and F were in a row. So d iu m was th e next e leme nt then kn own ; becau se it... p ro per ties clo sel y rese mbled those ofLi , Mende leev started a new ro w. As mo re an d more ele ments were adde d to th e tab le, new ro ws were ad ded, and eleme n ts with sim ilar pro perties (suc h as Li, Na , and K) were fo un d in the same vertical co lum n . An im po rtan t feature of Men d el eev's tab le-and a mark o f hi s gen ius-was that he left a n e m pty sp ace in a co lu mn whe n an cleme nt was not known h u t shou ld exist and have p roperties sim ila r to th e elemen t above it in h is ta ble. lIe d educed th at th ese spaces wo uld be filled by undiscovered ele me n ts. For example, he le ft a spa ce be twee n Si (silico n) and Sn (tin) in Gro u p 4A fo r a n cl em ent he ca lled ekasilicon. Based on the progressio n o f p rope rti es in thi s group, Mendel eev was ab le to predi ct th e prope rties of thi s mi ssin g e leme nt. With the discove ry of germaniu m (Ge ) in 1886, Mend cl ccv's predicti on was co nfirmed.
• Mendeleev and Atomic Numbers Me ndeleev developed the periodic table based on atomic weights because the concept of atomic numbers was not known untilafter the development of the structure of the atom in the early 20th century.
2 .5
I
The Periodic Ta ble 5 9
1
2 3 4
5 6 7
·· ·· ··
n I I
I
I I
·
I
•
4A
6A
I
Periods 1A
8A
In Mend eleev's tab le , th e eleme nts were ordere d by in cr easin g mass. A g lance a t a m ode r n tabl e , h o weve r, shows th at , if so me ele me n ts (suc h as Ni a nd Co, AI' and K, and Te and I) were ordere d by mass an d not che mical an d p h ysical properties, th ey would be reve rsed in the ir order of ap pearance. Me nd el eev recognized th ese di screpan cies and sim ply assu m ed th e ato m ic we igh ts kn own a t th at tim e were in accu rat e-n ot a bad assum pti o n based on th e ana lytica l methods then in use . In fact , hi s o rder is co rrect, and what was wro ng was hi s assu m pt io n that element p ro perties were a fun ction of th e ir m ass.
groups i s to number t hem 1 t hrough 18
In 19 13, II. G. J. Mosel ey ( 1887- 1915), a young Eng lish scie n tist wor king with Er nest Ru th e rford ( 187 1- 1937) , co rrecte d Mendeleev's assu mptio n . Moseley was doing ex perime n ts in which he bombard ed ma ny diffe re n t me tals with e lec tro ns in a catho de-ray tube (page 343 ) and exami ne d the x-rays emitted in the p rocess. In see king some orde r in his d a ta, he realized th at the wavele ng th of the x-rays e mitte d by a given eleme nt was related in a precise ma n ner to th e ato m ic number of the element. Indeed , o nce the co ncep t of an a tomic n u m ber was recognize d early in the 20t h ce ntu ry, chemists realized that organ izing th e elements in a tab le by inc reasing ato mic number co rrec ted th e inconsiste nc ies in th e Mc ndc lecv tab le . T h e law of chem ical periodicity is now sta te d as the jm1Jerties oJ the dements are jwri-
from left to right. This method is gener-
odic Junctions oJ a tomic number:
•
2A
•
3'
48
68
88
28
l'15.' 1! m '! I
l'
I I" . SA "
..
7
I: Groups or Fa mi lies Periods and group s in the periodic table. One way to designate periodic
ally used outside t he United State s. The
system predominant in the United States labels main group elements as Groups lA-SA and transition elements as Groups 18-8B. This book uses th e AlB system.
Features of the Periodic Table T he ma in orga nizati o na l features of th e peri od ic tabl e are th e follo wing: • Ele men ts are arra ng ed so th ose with sim ilar che m ica l and p hysica l pro perties lie in vertical co lum ns calle d groups or families. T he periodic tabl e common ly used in the Uni ted States has gro ups n u m be red 1 thro ugh 8, with each number fo llowe d by a le tte r: A or B. T he A grou ps are o fte n ca lled th e main group elements, and th e B gro ups are the transition elements. • The horizon tal rows of the ta ble are ca lled p eriods, an d th ey are numbered beg inning with 1 fo r th e peri od co n ta ining o nly H and H e . For exam ple, sodi um , Na, in Grou p l A, is the fi rst e leme nt in the thi rd period. Me rcu ry, I-Ig, in Gro u p 21\, is in th e sixt h peri od (or sixt h row ) .
o Mai n Group Metals o Transition Metals o Metalloids
D Non metals 60
Chapter 2
I
The peri od ic tab le can be divided in to several regions accord ing to the p ro perties o f th e eleme n ts. O n th e tabl e insid e th e fro n t cover o f thi s bo ok , ele ments th at behave as metals are indica ted in pu rp le , tho se that are nonmetals are in d icat ed in yellow, an d ele me n ts called metalloids appear in gree n . Elements gradua lly become less meta llic as o ne moves fro m left to rig h t across a pe rio d , a nd th e met alloids lie alo ng th e metal-n o nm e tal boun d ary. So me e leme n ts are shown in Figu re 2.4 . \'( HI are p robabl y fam iliar with m an y prope rties o f metals fro m everyday ex perie nce (Figu re 2.5a) . Metals are so lids (exce p t fo r me rcury), ca n con d uc t electrici ty, are usually du ctile (can be d rawn into wires) and malleable (can be ro lled into sheets), and can form alloys (so lu tions o f one o r more me tals in anot her me tal) . Iron (Fe) an d alu m in u m (AI) are used in au to mo bile parts becau se of th eir du ctility, malleability, and lo w cost relat ive to other me tals. Cop per (Cu ) is used in electric wiring becau se it co nd ucts electricity better than most o ther m etals. T he no nmetals lie to th e right of a diago na l lin e th at stre tc hes fro m B to Tc in th e pe riodic ta ble and have a wide var iety o f p ro perties. So me are so lids (carbon , su lfur, pho spho ru s, and iodin e ) . Five eleme n ts are ga'ies at ro om te mpe ratu re (hydroge n , oxyge n , nitrogen , flu o rin e, a nd chlorine) . O ne no nm et al , bro mine , is a liqu id at room tem peratu re (Figu re 2.5 b). \Vith the exceptio n o f carbon in the fo rm of grap hite , no n metals do not co nduct elec tr ici ty, which is one o f th e main fea tu res th at di stin gui sh es th em fro m m et als.
Atoms. Molecules. and Ions
Group 1A Lithium- Li (top) Potassium -K (bottom)
Active Figure 2.4
Some of the known 117 elements.
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in at www.thomsonedu.comjlogin
• Chapter Contents menu to explore an interacand go to the
Group 28 Zinc-Zn (top) Mercury-Hg (bottom)
tive version of this figure accompanied by an exercise.
Transition Metals Titanium-Ti, Vanadium -V, Chromium-Cr, Manganese-Mn, lron-Fe. Cobalt- Co, Nickel- Ni, Copper-Cu
2A
2 II
Mg 38 48 4
K
•
58 68
78 ,...-- 88 -----, 18 28
n
5i
P
6A
N
"
5
"1
N.
5n • • Pb
• • • • • • • • • Hg
Group 4A Ca rbon-C (top) Lead-Pb (left) Silicon-Si (right) Tin-Sn (bottom)
SA
• • 5. 8r
• !=:!.!.=:!~.=!;.=i!!=.;!!=.!!=:!. •
• •••
"(
(6A) (7A) (3A)
Group SA. Noble Gases Neon -Ne
Group 7A Bromine-Br
Group SA Nitrogen-N 2 (top) Phosphorus-P (bottom)
2 .5
I
The Periodic Table
61
Bromine, BTz
o
\
(b) Nonmeta ls
Metals
(e) Meta lloids
FIGURE 2.5 Metals, nonmetals , and metalloid s. (a) Molybdenum (Mo, wire). bismuth (BL cente r object), and copper (Cu) are metals. Metals can generally be drawn into wires, and they conduct electricity. (b) Only 15 or so elements can be classified as nonmetals. Here are orange liquid bromine and purple solid iodine. (c) Only six elements are generally classified as metalloids or semimetals. This photograph shows solid silicon in various forms, including a wafer that holds printed electronic circuits.
T he elem e n ts n ext to the di ago n al lin e fro m bo ro n (B) to tellurium (Te) h ave p rope rt ies th at mak e th em d ifficult to classify as metals o r nonmeta ls. C he m ists ca ll the m metalloids or, som etimes, semimetals (Figu re 2.5c) . Yo u shou ld kn ow, h oweve r, th at c he m ists often d isagree about wh ich el e me nts fit in to this ca tego ry. We will defin e a m e tallo id as an element th at has some o f th e physical cha rac te ristics o f a met al bu t so me of th e c h e m ica l c h ara cte ristics o f a n o nm et al ; we includ e o n ly B, Si, Ge , As, Sb , a nd Te in thi s ca tegory. This defi nition reflect s the am b igu ity in th e behavior of these ele me n ts. An timon y (Sb) , for exam p le , co n d uct s electr icity as we ll as m an y cl em ents th at a re true met als. Its ch e m istry, h owever, re sembles tha t of th e n o nm e tal p hosp horus.
A Brief Overview of the Periodic TabLe and the ChemicaL ELements • Alkali and Alkaline The word "alkali" comes from the Arabic language; ancient Arabian chemists discovered that ashes of certain plants, which they called al-qali. gave water solutions that felt slippery and burned the skin. These ashes contain compounds of Group lA elements that produce alkaline (basic) solutions.
62 Chapter 2
I
The metals in the leftm ost co lu mn , Group lA , are kn own as th e alkali metals. All a re met als an d are so lids at room temperature . T hey a rc all very rea ctive . Fo r exa mp le, th ey rea ct with water to p ro duce h ydroge n and alkalin e so lut io ns (Figu re 2.6). Because of their reactivity, these metal s a re only fou nd in n atu re com bine d in com po un ds (suc h as NaC I) (Figu re 1.4) , ne ve r as th e fre e clement. The sec o n d g rou p in the pe riodic tab le , Group 2A, is also composed e n tirely o f me tal s th at occur n atu rally o n ly in co m p o u n ds . Except fo r be ryllium (Be) , these c leme n ts react with water to produ ce alkalin e so lut io ns , a n d mo st of th eir ox id es (su ch as lime , CaO) form alkalin e so lu tio ns ; h e n ce , th ey a re known as th e alk aline ear th m etals . Magn esiu m (Mg) a n d ca lciu m (Ca) are the se ve n th a n d fift h mo st ab u nda n t cle me n ts in th e earth 's crus t, respect ively (Ta b le 2.:1) . Ca lciu m is o n e o f th e impo rtant el em e nts in teeth and bon es, a n d it o ccu rs n atu rall y in vast limeston e d e posits. Ca lciu m carb onate (CaCO :l ) is th e ch ief constitu ent of lim esto n e a n d o f corals, sea sh ells, marb le, a n d ch alk. Radiu m (Ra) , th e h eaviest alka lin e eart h clemen t, is rad io active a n d is u sed to tre at so me cancers by rad iation .
Atoms, Molecules, and Ions
The 10 Most Abundant Elements in the Earth's (rust
TABLE 2.3
Rank
(a) Cutt ing sodium.
(b) Potassium reacts with water.
FIGURE 2.6 Alkali metals. (a) Cutting a bar of sodium with a knife is about like cutting a stick of cold butter. (b) Whe n an alkali metal such as potassium is treated with water, a vigoro us reaction occurs, giving an alkaline solution and hydrogen gas, which burns in air.
Group 3A co n ta ins o ne clem ent of grea t importan ce , alu min um (see Figu re 2.4) . This e lement and three others (gallium , indi um , and thalli u m ) are me tals, whereas bo ron (B) is a me ta llo id. Aluminu m (AI) is the m ost abu ndant me tal in the earth 's crust at 8.2 % by mass. It is exceeded in abu nda nce only by th e nonm et al ox yge n an d m etall o id silico n. Th ese three elements are fo u nd co mbined in clays and other com mon min erals. Boron occurs in the mine ral bo rax, a co m pound used as a clean ing age n t, an tise p tic, and flu x fo r met al work. As a metall o id , boron has a different chemistr y than the other eleme n ts of the gro up , all of whic h are metals. Nonethe less, all fo r m compo un ds with analogous formu las such as BClj an d AlClj , and thi s similar ity m ark s th em as m embers of th e same periodic grou p. T h us far, all the el em en ts we have describe d , exce pt boron , have been me tal s. Begin ni n g with Group 4A, ho wever, th e grou ps co n tain mo re and rno re no n m eta ls. In Group 4A, th ere are a non me tal , ca rbon (C) , two metallo ids , silico n (Si) an d germani u m (Ge) , and twn m etals, tin (Sn) and lead (Pb) (Figure 2.4 ) . Because o f th e change fro m no n m etallic to m et alli c chara cte r, more vari ati on oc cu rs in the pro perties of th e eleme nts o f this group than in m o st othe rs. No nethe less, there are simi larities. Fo r example , these e leme nts form co mpo un ds with analog o us fo rmulas suc h as CO 2, Si0 2, Ge 0 2, an d Pb 0 2. Carbon is th e ba sis for the g reat variety o f chemical co mpo unds th at make u p living things. It is fo und in Ea rth's at mosp here as CO 2, on the surface o f the earth in ca rbonate s like lim estone and coral (calciu m ca rbonate , CaC0 3 ), a nd in coa l, petroleum , and natu ral gas- the fossil fue ls. One inte re sting aspect of the chem istry of the nonme tals is th at a par ticul ar e leme n t ca n often ex ist in seve ral differen t and d istinc t fo rms, ca lled allotropes , eac h having its own properties . Carbon has a n um ber of allotro pe s, the best kn own of whic h are grap hite and d iamond . Grap h ite consists of flat sheets in whic h each carbon ato m is co n nec ted to three o the rs (Figu re 2.7a) . Becau se th e shee ts o f carbon ato ms cli ng o n ly wea kly to o ne ano ther, o ne layer ca n slip easily over ano the r. This ex plains why gra p hite is soft, is a good lubrican t, an d is used in pen cil lead. (Pe ncil "lead" is not th e cle me n t lead (Ph ) hut a co m pos ite of clay a nd grap hite that leaves a trail of grap hi te o n the page as yo u wrire.)
Element
Abundance (ppm)"
1
Oxygen
474 ,000
2
Silicon
277 ,000
3
Aluminum
82,000
4
Iron
41,000
5
Calcium
41 ,000
6
Sodium
23,000
7
Mag nesium
23,000
8
Pota ssium
21,000
9
Titan ium
5, 600
10
Hydrogen
1,520
"p pm - g per 1000 kg.
liquid gallium. Bromine and mercuryare the only elements that are liquids under ambient conditions. Gallium (29.8 °C) and cesium (28.4 "C) melt slightly above room te mperature. Here gallium melts when held in the hand.
2. 5
I
The Periodic Ta ble
63
(b) Diamond
(a) Grap hite
(c) Buckyballs
FIGURE 2 .7 The allotropes of carbon. (a) Graphite consists of layers of carbon atoms. Each carbon atom
is linked to three others to form a sheet of six-me mber, hexaqonal rings. (b) In diamond, the carbon atoms are also arranged in six-mem ber rings, but the rings are not planar because each Catom is connected tetrahedrally byfour other Catoms. (c) Buckyballs. Amember of the family called buckminsterfullerenes. (60 is an allotrope of carbon. Sixty carbon atoms are arranged in a spherical cage t hat resembles a hollow soccer ball. Notice that each six-member ring shares an edge with three other six-member rings and three five-member rings. Chemists call this molecule a "buckyball." C60 is a black powder; it is shown here in t he tip of a pointed glass tube. • Sped al Group Names Some groups have common and widely used names. (See Figure 2.4) Gro up lA: Alkali metals Group 2A: Alkaline earth metals Group 7A: Ha logens Gro up 8A: Noble gases
FIGUR E 2.8 Compounds containing silicon. Ordinary clay, sand, and many gemstones are based on compounds of silicon and oxygen. Here, clear, colorless quartz and dark purple amethyst lie in a bed of sand. All are silico n dioxide, Si0 2• The different colorsare due to impurities. 64
Chapter
2
I
In di amond, each carbo n at o m is co n nec te d to fou r o th ers at th e co rners of a tetrahedron , and thi s ex te nds throughout th e so lid (see Figure 2.7b ) . This struc tu re causes di amonds to be extreme ly hard, d en ser th an grap hite (d = 3.5 1 g /cm 3 fo r diamond and d = 2.22 g/ cm 3 fo r grap hi te) , and che m ica lly less reactive. Because diamonds are not o nly hard but are ex ce llent co nd uc to rs o f heat, they arc used o n th e tips of metal- and ro ck-cullin g tools. In the late 1980s , an o the r fo rm o f ca rbo n was id en tified as a co m ponent of black soot, th e stu ff th at co llec ts whe n ca rbo n-co n tain ing materials are burned in a d eficien cy of ox yge n . This su bst a nce is made up o f molecu les with fiO carbo n a to ms arrange d as a sp he rical "cage" (Fig u re 2.7c) . 'ro u m ay recog ni ze th at th e sur face is made up o f five- and six-me m be r rin g s and re sembles a hollow so ccer ball. The shape also re mi nded its di scove re rs of an a rch itectu ral dom e co nceived several d ecad es ago by t he Am eri can philo sopher and engineer, R. Buckminster Fulle r. T his led to th e o fficial name o f the allo tro pe , buckminsterfu llerene , although che mists oft en simply ca ll these Ceo molecu les "b uc kyba lls." O xid es of silico n are th e basis of man y min erals such as clay, quart z, and beautifu l ge msto nes like ame thyst (Fig u re 2.8). T in and lead have been kn own for ce n turies becau se th ey are easily obtain ed from thei r o res. Ti n alloyed with co p pe r mak es bron ze , which was use d in anc ie nt tim es in ut ensils and weapo ns . Lead has been used in wate r pipes and paint, even th ough th e el em en t is toxic to hum ans. Ni troge n in Gro up 5A occurs nat u ra lly in the fo rm o f th e diato mic molecul e N 2 (Fig u re 2.9 ) an d makes u p abo u t three fou r th s of Earth 's atmosp here . It is also inc orporat ed in bio ch emically important su bstances suc h as chlo ro p hyll, p rotein s, and DNA. Th eref ore , scien tists have lon g soug h t ways to mak e co m po u nd s fro m at mos p heri c n itrogen , a process referred to as "n itrogen fixation ." Na ture acc o mplish es thi s easily in so me pro karyotic o rganis ms, but severe co nd itio ns (h igh temperatu res, fo r exam ple ) must be u sed in th e lab o rato ry an d in ind ustry to ca use N 2 to reac t with o ther ele me nts (suc h as 11 2 to mak e am m o nia, NI 13 , which is wide ly used as a fertilizer) .
At oms, Molecules. and Ions
N,
0,
°3
F,
FIGURE 2.9 Elements that exist as diatomic or triatomic molecules. Seven of t he e lement s in th e periodic table exist
Cl,
sr,
as diato mic, or two- ato m, molecules.
Oxygen has an additional allotrope, ozone, with three a atoms in each molecule. See also Chemist ryNow Screen 2.16, Elements that Exist as Molecules.
• Plad ng H in t he Periodic Table Where to place H? Tables often show it in Gro up lA even though it is clearly not an alkali metal. However, in its reactions it forms a 1 + ion just like the alkali metals. For this reason, H is often placed in Group l A.
I,
Ph ospho rus is a lso esse ntia l to life . It is an im po rtan t constitue nt ill bo n es, te eth , a n d DNA. T he ele m e n t g lows in th e d ar k if it is ex pose d to a ir (owin g to its re action with O 2 ) . a n d its n a m e , based o n Gree k word s m ea n in g "ligh t-b ear ing," re flec ts thi s. This e le me nt also h as seve ral allotro pes, the m o st importa n t be in g while (Fig u re 2 .4) a nd red phospho rus. Both form s o f pho sphorus a rc used com mercially. Whi te ph osph o ru s igni tes spo n taneou sly in air a n d so is no rm ally stored u nde r wa te r. \Vhe n it d oes react wit h a ir, it fo rm s P40 1O, whic h ca n react wit h wate r to fo rm ph osphoric ac id ( H 3 P0 4 ) , a co m p o u n d used in food products su c h as so ft d rinks. Red phosph o ru s is u sed in th e striking strips o n m at ch books. wh en a matc h is str uck , po tassium c h lo ra te in th e m a tc h h ead mix es with so me red phosphorus on the strikin g strip, a n d th e fri cti on is e n o ug h to ignite this mi xture . As wit h Gro u p 4A, we agai n see nonm etals (N a n d P ) , m etalloids (As a n d Sb ) , a n d a m e tal (Bi, Figu re 2.5a ) in G rou p 5A. In sp ite of th ese va riations, th e y also form a nalogo us co m pou n ds suc h as th e ox id es N 20 [1' 1'205 and As20 [,. O xygen , whic h co ns titutes ab ou t 20 % of Earth 's at mosp here a n d whic h co m b ines readi ly wit h most oth er elements, is a t the to p o f Group 6A. Mo st of th e ene rgy th at powers life on Ea rt h is d erived fro m reaction s in wh ich o xyge n co m bines with oth er su bs tances. Sulfur h as b een known in el em enta l fo rm sinc e ancie n t ti m es as brimstone or "bu rn ing sto n e" ( Figu re 2.10 ). Su lfur, sel e n iu m , a n d tellu rium a re o fte n referred to co llectively as chalcogens (fr o m th e G ree k wo rd , khalhos, fo r cop per) b ecause most co p pe r o res con tai n these el em e nts. Thei r co m p o u n ds ca n be foul sm e lling a n d poisonous; nevertheless, su lfu r a n d sele n iu m are essential co m p o n e n ts of th e hum an di e t. By far th e most important co m pou n d of su lfur is su lfu ric aci d (I 12S0 4 ) , wh ich is m anufactured in large r a m ou n ts th an a ny other co m p o u n d . As in Group 5A, th e se co n d- and third-period elements of G ro u p 6A have differen t stru c tu res . Like n it rogen , o xygen is also a di atomi c m o lecule (see f igure 2.9 ) . Un like nitroge n , how ever, o xygen h as a n a llo tro p e, th e tr ia tomi c m o lec u le ozone , 0 3' Su lfur, wh ich ca n be fou n d in n atu re as a ye llow so lid, h as m an y a llotropes. Th e most co m m o n allo tr o pe co ns ists of e igh t-m e m be r, cr ow n-shap ed rings of su lfu r a to m s (see Figure 2.10 ) . Polonium , a radioactive el e ment in Group 6A, was iso lated in 189 8 by Marie and Pierre Cu rie , who se pa ra ted it from ton s of a u ranium-containin g o re a n d n ame d it fo r Mad am e C urie 's n ati ve co u n try, Pol and. In Gro u p 6A, we once ag a in o bse rve a va ria tio n o f properties. O xygen , su lfu r, a n d se le n iu m are nonm eta ls; telluriu m is a m etall oid; a nd po lon ium is a m etal. No n e th eless, th ere is a fami ly rese m b lance in th eir c h e m istries. All form ox yge nco n ta in ing co m po u n d s suc h as S02, Se02, and Te02 a n d sod iu m-co n tain ing co mpounds (Na20 , Na2S, Na2Se, a n d Na2Te). At th e far righ t of th e periodic ta ble a re two gro u p s co m p osed e n tire ly of non meta ls. The Group 7A clcmcnts-c-Ilu oriuc, c h lo rine, bromine, iodine, and radi o ac-
FIGURE 2 .10 Sulfur. The most common allotrope of sulfur consists of S atoms arranged in eight-member, crown-shaped rings. 2. 5
I
The Periodic Ta ble
65
rive astati ne-are no nmetals, and all exi st as dia tom ic mo lec ules (see Figure 2.9) . At roo m te m perat ure , fluorine (1"2) and ch lo rine (CI2) are gases. Bromine (llr2) is a liqu id , and iodi ne (12) is a so lid, but bro mi ne and iodine vapo r are cle a rly visible over the liqui d or so lid (see Figure 2.5b) . T he Gro u p 7A clemen ts are amo ng the m o st reac tive of all cl em ents, an d all co mbi ne vio le nt ly with alka li metals to form sa lts suc h as tab le salt , Na CI (see Figure 104 ) . Th e na me for t his gro up , t he halogens , comes fro m th e Greek wo rds hals, me anin g "sa lt," an d gen es, fo r "fo r m in g." T he hal o gens a lso react wit h othe r m e tal s a nd with m o st nonmetals to fo r m co m po u nd s. The Group SA e lemerus-s- he liu m . neon , argon, krypto n , xe non , and ra d ioactive rad o n-are th e least re ac tive eleme n ts (Figure 2.11). All are gase s, and n one is ab undant o n Earth or in Eart h 's atmosphere. Because o f this , they were no t d iscovered u n til the end of the 19th cen tu ry. H eli um , the seco nd mo st ab undant elemen t in th e universe afte r hydrogen , was d et ect ed in th e sun in 1868 by analysis of th e so lar spec tru m . (T he na me of the element co mes fro m the Greek wor d for the sun , helios.) It was not foun d 011 Earth u ntil 1895, however. Until 1962, when a co mpo u nd o f xe no n was first pre pa re d, it was beli eved th at none of th ese elemen ts wou ld com bi ne chemi ca lly with an y o the r ele me n t. Th e nam e noble gas es fo r th is gro up, a term m ean t to d en o te the ir ge neral lack of reactivity, d erives fro m thi s fac t. For th e same reason , they are so m etimes ca lled th e inert gases o r, becau se of their lo w ab undance , the raregases. Stretching between Groups 2A and 3A is a seri es of eleme nts called the transition elem ents . T hese fill th e B-gro u ps (I ll through 8 B) in th e fou rth through th e seven th periods in the ce n te r of the periodic table. All are metals (see Figu re 2.4 ), and 13 of th em are in the top 30 e leme nts in terms o f ab u nd ance in the eart h 's crust. Some, like iron (Fe ), are abundan t in nature Cra ble 2.3). Most occu r natu rally in com binatio n with other el ement s, but a Iew-s-copper (Cn) , silve r (Ag) , go ld (Au ), and platinum (Pt)-are m uc h less reactive and so can be foun d in nature as p ure elements. Virtua lly all o f th e tran sition elem e nt s have co mmercia l uses. T he y are used as str uct u ra l mat erials (iro n , titanium , chromium , co pper); in pai nts (t itani um , chromi um ) ; in th e catalytic co n ve rters in au to mo bile ex haus t systems (platin u m an d rh odi um ) ; in co ins (cop per, ni ckel , zin c); a nd in batteries (manga nese, n ickel, cadmium, mercu ry) . A nu mb er of the transition e leme nt s pl ay impor tant biological roles. Fo r exam ple, iro n , a relatively abu ndan t el em en t (see Ta ble 2.3) , is th e ce n tra l element in th e che m istry of hem oglob in , the o xygen-carrying co mpo ne nt o f bloo d . FIGURE 2.11 The noble gases. This
kit is sold for detecting the presence of radioactive radon in t he home. Neon gas is used in advertising signs, and xenoncontaining headlights are increasingly popular on automobiles.
66
Chapter
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I
At oms. Molecules. and Ions
Two ro ws at th e bottom of th e tabl e accom mo date th e lanthanides [the series o f clements bet ween th e elements lanthanum (Z = 57) and hafnium (Z = 72) J and the actinides [ the series of e le me nts bet ween actin iu m (Z = 89) and rutherfo rdiu m (1 = 104) J. So me lanth anide co m po un ds are used in co lo r tel evision picture tubes; uranium (Z = 92) is th e fuel fo r ato m ic power plants, and ame riciu m (Z = 9Pj) is used in smo ke d et ectors. ChemiSlrY.{J"NOw'" Sign in at www.thomsonedu.comjlogin and go to Chapter 2 Contents to see: Screen 2.14 for an exercise on periodic table organization • Screen 2.15 for an exercise on chemical periodidty
EXERCISE
Lanthanides. If you use a Bunsen burner in the lab, you may light it with a "flint" lighter. The flints are composed of iron and "rniscbmetal.' a mixture of lanthanide elements, chiefly Ce, La, Pr, and Nd with traces of other lanthanides. (The word "mischmetal" comes from the German for "mixed metals.") It is produced by the electrolysis of a mixture of lanthanide oxides.
2.4 The Periodic TabLe
How many elements are in the third period of the periodic table? Give the name and symbol of each. Tell whether each element in the period is a metal, metalloid, or nonmetal.
2.6
Molecules, Compounds, and Formulas
A molecule is th e smalles t identifiabl e unit into whic h a pure su bsta nc e like suga r o r water ca n be divided and still ret ain the co m positio n and che mical prope rties of th e su bsta nce. Suc h su bsta nc es are co m pos ed of identical molecul es co ns isting of ato ms of two or more cl ements bound firml y to gether. Fo r exa m ple, atoms o f th e ele me n t alu m in u m, AI, co m bine with molecul es of the ele me n t b romine , Br2' to produce th e co m po u nd alu m in u m bromide, Al2Brli (Figu re 2.12). 2 Al(s) + 3 Br, (C) --> AL, Br,(s) aluminum + bromi ne -) aluminu m bromide
(e)
(b)
(c)
Active Figure 2.12 Reaction of the elements aluminum and bromine. (a) Solid aluminum and (in the beaker) liquid bromine. (b) When the aluminum is added to the bromine, a vigorous chemical reaction produces white, solid aluminum bromide, Al2B r6 (c). TU
Chemislry.o.Now
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I
Molecules, Compounds. and Formula s
67
To d escrib e this che mica l change (o r che m ical reaction ) on pap er, th e co m position of ea ch cl ement a nd compound is re p res e n ted by a sym bo l o r formula. Here , one molecul e of AlzBr(i is co mposed of two AI atoms and six Br a to ms . How do co m po u nd s diffe r from elements? \Vhen a co m po un d is produced fro m irs clem en ts, the charac teristics of th e co nstitue n t clem en ts are lost. So lid , me talli c alu min u m and red-o ran ge liqu id bro mine , for exam ple , reac t to fo rm AI2Br(j, a white so lid .
Formulas
• Writing formulas When writing molecular formulas of organic compounds (compounds with C, H, and other elements), the convention is to write Cfirst, t hen H, and finally other elements in alphabetical order. For example, acrylonit rile, a compound used to make consumer plastics, has the condensed formula CHlCH CN. Its molecular formula is C3 H3 N.
Fo r molecul es more co m plicated th an water, th ere is oft en more th an o ne way to write the fo rmul a. Fo r exa m ple , th e fo rmul a of e thanol (also called e th yl alco ho l) can be represen ted as CzlI fiO (Fig u re 2. 13) . Th is molecular formula d escribes th e co mposition of ethan o l mol ecules- two carbo n atoms, six hydrogen ato ms, and o ne ato m o f ox yge n o ccur pe r rno lecu le-e-but it gives us no stru ctural info rmat io n . Str uctural infor matio n- how the a toms are co n nec te d and how the mol ecul e fills space- is important, ho wever, becau se it helps us unde rstand ho w a mo lecule can inte ract with o th e r molecu les, wh ich is th e esse nc e o f che m istr y. To provide so m e str uctu ral info rm ati o n , it is useful to write a condensed formula , which indicates how certain ato ms are g ro u ped togethe r. For exa mple , th e co nd en sed formul a of ethano l, ClI ;sC1I20 H (see Figure 2. 13) , Inform s us th at the molecul e co nsists of th ree "groups ": a C1I3 grou p , a Cl I, grou p , an d all OIl grou p . \Vriting th e formula as CH 3CH zO H also show s th at th e co m po u nd is not dimethyl ether, f: H 3 0CII 3 , a co mpo u nd with th e same mol ecul ar fo rmul a but with a diffe rent structu re and d istinctly different p roperties. That e thano l and d ime t hyl ethe r are diffe rent molecul es is furth e r ap pa re n t from th eir structural for mulas (Figu re 2. 13) . Th is type o f formula g ives us an eve n high er level o f structu ral d etail, showing ho w all of th e ato m s are atta ched withi n a molecu le . T he lines be tween atoms represent th e che mica l bonds th at ho ld ato ms together in thi s mo lecul e (... C hap ter 8 ).
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• Ethanol and Dimethyl Ether Are Isomers Compounds having the same molecular form ula but different structures are catted isomers . (See Chapter 10, and sign in to ChemistryNow and see Screen 2.17, Representing Compounds.)
NAME
Ethanol
MOLECU LAR FORMU LA
COND ENSED FORMULA CH ,CH,OH
STRU CTURAL FORMULA
H
H
I
I
MOLECU LAR MODEL
H- ( - ( - O-H
I
H
Dimethyl ether
CH, OC H,
I
H
H H I I H- ( - O- ( -H I I H H
FIGURE 2.13 Four approaches t o showing molecular formulas. Here, the two molecules have the same molecular formula . Condensed or structural formu las. and a molecula r model, showthat these molecules are different.
68
Chapter
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I
At oms, Molecules. and Ions
EXERCISE 2.5
• Standard Colors for Atoms in
MolecuLar FormuLas
Molecular Models The colors listed here are used for molecular models in this book and are generally used by chemists. Note that Cl and 5 atoms have the same color, but it is usually apparent what atom is being designated.
Cysteine, whos e molecular model and str uctural formula a re illu strated here , is an important amino
acid and a constituent of many livin g things. What is its molecular formula?
+
0"
NH, H
I I ( -( -( O-Y' I I
carbon atoms
S -H hydrogen atoms
H H
oxygen atoms Molecular model
Structural formula nit rogen atoms
MoLecuLar ModeLs
chlorine atoms
Molecul ar structu res are o ften beautiful in the same sense th at art is beau tiful , an d there is some thing intrinsically beau tiful abo ut the pauern created by water m olcc u lcs asse m b led in icc (Figu re 2. 14). Mo re im portant, however, is the fact that the p h ysical an d chemical properties of a mol ecular co mpo u nd are o fte n closely rel at ed to its structure. Fo r exam ple. two wel l-k nown fea tures of icc arc rel ate d to its struc tu re . The first is th e sha pe o f
ice crys ta ls: Th e sixfo ld sym metry o f macro sco pic ice cr ysta ls also ap pears a t the particu late level in the form of six-sid ed rings of hydrogen and oxyge n ato ms. T he seco nd is wate r 's unique property o f be ing less dense whe n so lid th an it is whe n liquid . T he lowe r d ensity of ice , whic h has enormo us co nse quences for Earth 's climate, re sults from the fact that mol ecules of water are not packed toget her tightly. Becau se mo lec ules are three-di mensio na l, it is o fte n diff icul t to represent their shapes o n pape r. Certain co nve ntio ns have been d evelo ped , ho wever, that help re p resen t th ree-d imensional struc tures on two-di me nsio na l surfaces. Simple perspective d rawings are o ften used (Figure 2. 15).
FIGURE 2.14 Ice. Snowflakes are six-sided structures, reflecting the underlying structure of ice. Ice consists of sixsided rings formed by water mo lecules, in which each side of a ring consists of two o atoms and an H atom.
2.6
I
Molecules, Compounds, and Formulas
69
Simp le perspective drawing
Plastic model
Ball-and-stick model
Space-filling model
All visualizing techniques represent t he same molecule.
Ways of depicting a molecule, here the methane (CH 4 ) molecule.
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Several kinds of mo lecular mo de ls exist. In the ball-and-stick model , sp he re s, usually in d ifferent co lo rs, represe nt the ato ms , and sticks represen t the bond s holding th em toget he r. T hese mo d el s make it ea sy to see how atoms are attache d to o ne another. Mo lecu les ca n also be rep resented using space-fi lling models. T hese mo d els are mo re realistic be cause they offer a better represen tation of rel ative sizes o f atoms and th e ir p ro xim ity to eac h o ther whe n in a molecul e . A d isadva ntage o f p ictures of space-tilling mode ls is th at ato ms can often he h id d e n fro m vie w.
2.7
Module 2 Module 3
Ionic Compounds: Formulas, Names, and Properties
T he co m po u nd s yo u have en co u n tered so far in thi s cha p ter are molecular comp ounds-th at is, co m pou nds that co nsist o f dis crete mo lecu les at the particula te level. Ionic co mpo und s co ns titu te ano ther m aj o r class of com po unds. T hey co nsist of ions, ato ms or gro up s o f ato ms th at bear a po sitive or negative electr ic charg e . Man y fa m iliar co mpou nds are co m pose d of ion s (Figu re 2. 16) . Table salt, or so d iu m ch lo ride (NaCI) , and lim e (Ca O) are j ust two . To recogn ize io nic co m po u nd s, an d to be able to write formulas fo r the se co m po unds, it is im porta n t to know th e fo rmulas and charges o f co m mo n io ns. You also need to know the nam es of io ns and be able to nam e the co mpo u nds they fo rm .
FIGURE 2.16 Some common ionic compounds.
Common Name Ca lcite
Ca lcium carbonate
Ca CO l
Ions Involved Ca z+ , CO/ -
Fluorite
Calcium fluoride
CaFz
Ca z+ , F-
Gypsum
Calcium sulfate dihydrate
CaS04 . 2 H2O
(a U , S04Z-
Hema tite
Iron(lII) oxide
FeZOl
Fe H , 0 2 -
Orpiment
Arsenic sulfide
ASzSl
As3+ , Sz-
70
Name
Formula
! e
~
~
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Ions Atom s of m an y cleme n ts ca n ga in o r lose electro ns in th e co u rse of a ch e m ical re actio n. To be able to p redict the ou tcome of che mical reactions ( 110- Sections 3. 1-3.9), yo u n e ed to kn o w whe ther a n el ement will li kely gain or lo se el e ctro n s and , if so, h o w many.
Cations If an ato m lo se s a n e le ctro n (wh ic h is tran sferred to a n a to m of a no t h e r cleme nt
in the co u rse o f a reaction ), the ato m now h as on e fewer negative e lec tro ns tha n it h as po sitive p rotons in th e nucle us. T he resu lt is a po sitively cha rge d io n ca lled a catio n (Figu re 2.17). (T h e n am e is pro nou n ce d "ca r-io n.") Beca u se it h as an excess of on e po sitive charge, we wr-i te the ca tio n's sym bo l as, fo r exa mp le, U +: Li atom -7 (3 protons and 3 electrons)
+
e
Li+ cation (3 protons and 2 electrons)
Anions Co nve rse ly, if a n a to m gains on e or m o re e le ctro ns, th ere is n o w a great e r nu m be r of n egativel y c h a rg e d el ectro n s th an pro to n s. T he re su lt is a n an io n (Fig u re 2. 17 ). (T he n a m e is p ro n ounce d "a n n 'cio n.")
o atom (8 proto ns and 8 electrons)
~ 0 2 - anion (8 proto ns and 10 electrons)
+
H e re , the 0 ato m has gai ned two electro n s, so we write th e a n io n's sym bo l as 0 2 - . H o w do yo u know wh e the r an a to m is likel y to fo r m a catio n or a n a n io n ? It d epends o n whe the r th e e le me nt is a m eta l o r a n on m etal. • Metal s gen erally lo se e lec tr o ns in th e cou rse of their reactions to form catio ns . • N o n m e ta ls freq uen tly ga in o n e or m o re el e c tro n s to fo rm a n io ns in th e course of th e ir reac tion s.
e
)
Li
u+
Jp
Jp
3n
3n
0-------GJ
the resulting lithium cation as u ". A fluorine atom is also electrica lly neut ral, having nine protons and nine electrons. A fluorine atom can acquire an electron
lithium ion, U+
l ithium, li
e
~
Active Figure 2.17 Ions . A lithium-f atom is electrically neutral because the number of positive charges (three protons) and negative charges (three electrons) are the same. When it loses one electron, it has one more positive charge than negative charge, so it has a net charge of 1 +. We symbolize
F
F-
gp
gp
IOn
IOn
~ Fluorine, F
to produce an F- anion. This anion has one more electron than it has protons, so it has a net charge of 1-.
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Fluoride ion, F2.7
I
Ionic Compounds: Formulas, Names, and Properties
71
Monatomic Ions Monat omic io ns a rc sing le atoms tha t have lo st or ga ined electrons. As ind ica te d in Figure 2. 18, m et als typicall y lose e lec tro n s to form mon a tomic catio n s, an d
nonmetal s typically ga in el ectrons to form monat omic an io ns. How ca n yo u predict the number o f el ec tro ns g a ine d or lost ? Like lithium in
Fignre 2. 1i , metals oj Groups JA- 3A fo rm jwsitive ions luroing a charge equal to the grmJ.jJ number C!l the metal. Electron • Writing Ion FormulasWhen writing t he formula of an ion. t he charge on t he ion must be included.
Change
Metal Atom
Group
IA
Resulting Metal Cation
Na (11 proto ns, 11 elect rons)
- I
--7
Na " (11 protons. 10 electro ns)
2A
(a (20 protons, 20 electrons)
-2
--7
Ca 2+ (20 protons, 18 elect rons)
3A
Al (13 protons. 13 elect rons)
-3
--7
AlH (13 protons, 10 elect rons)
Tran sition m e tals (R-g rou p el em en ts) also form ca tio ns . Un like th e A-gr o u p metal s, h owever, n o easily predictabl e patte rn of behavior oc curs for tran sition metal ca tio ns. In ad d itio n , tr ansition metals ofte n form seve ra l different ions. An iron-co ntaini ng com p ou n d , fo r exam p le, may co ntain eit her Fe 2 + o r Fe3+ io n s. Indeed, 2 + a n d 3 + ion s a re typi cal of m any tr an sition m etals (see Figure 2. 18) . Group
Electron Change
Metal Atom
Resulting Metal Cation
7B
Mn (2 5 protons, 25 electrons)
-2
--7
Mn2 + (25 protons, 23 electrons)
8B
Fe (2 6 protons, 26 electrons)
-2
--7
Fe2+ (26 protons, 24 electrons)
8B
Fe (2 6 protons, 26 elect rons)
-3
--7
FeH (26 protons, 23 electrons)
Nonmetal.. often Jorm ions having a nl'gative chavgr equal to the brnJ1lf) number oj the element minus 8. For exa m p le , nitrogen is in G ro u p 5A, so it forms a n ion h avin g a
3- c harge because a nitrogen a tom ca n gain th ree elec tr o ns . Group
Electron Change
Nonmetal Atom N (7 protons, 7 elect rons)
5A
+3
Resulting Nonmetal Anion ~
N3- (7 protons, 10 electrons)
Charge = 5 - 8
6A
5 (1 6 protons, 16 electrons)
+2
5 2 - (16 protons, 18 electrons)
Charge = 6 - 8
7A
Br (35 protons, 35 elect rons)
+1
Br- (35 protons, 36 electrons) Charge = 7 - 8
IA H+
D Transition metals D Metalloids D Nonmetals
2A
u+ Na+ Mg2+ K+
-7A
D Metals
CaH
3B
4B
El
5B
4A
6A
N'
0' -
r-
p' -
5' -
Cl-
5e2 - Br 5nH
••••••
H-
5A
1-
6B 7B ~ 8 B~ 1B Cu + CrH Mn2 f. FeH Co2 Ni H H H 3 C 0 + Cr Fe Cu '
Rb + 5rH
L.......L......-
3A
8A
Pb H
Te 2 -
j-
Bi 3+
FIGURE 2.18 Chargeson some common monatomic cations and anions. Metals usually form cations, and nonm etals usually form anions. (The boxedareas show ions of identical charge.) 72 Chapter 2
I
Atoms, Molecules, and Ions
No tice th at hydrogen a p pears at two lo cation s in Figu re 2.18. The II ato m ca n eithe r lose o r ga in el ectron s, d epending on th e oth er atoms it en counters. H (1 proton, 1 electron ) ---> W (1 proton , 0 electrons) + e: H (1 prot on, 1 electron) + e: ---> W (1 proton, 2 electrons)
Electron lost :
ELectron gained:
Finall y, th e n oble gases ca l reaction s.
ver)'
rarely form mon at omic catio ns or a nio ns in che m i-
Ion Charges and the Periodic Table
Cation charges and the periodic table
The metals of Grou ps l A, 2A, and 3A fo rm ion s having 1+ , 2 + , a nd 3 + charges (Figure 2. 18); that is, th eir ato ms lose on e , two , o r three elec tro ns, respec tively. Fo r Grouj) lA and 2A metals and aluminum, the number lif electrons remaining on the cation is the same as the number C!! electrons in an atom ~! the noble gas that precedes it in the periodic table. For exa mp le, fVlg2 + has 10 electrons, the same numbe r as in an at o m of th e no ble gas neo n (a to m ic n um be r 10) . An atom of a nonmetal nea r the right side of th e periodic ta ble would have to lose a g reat many electrons to ac hieve th e same number as a noble gas ato m of lower atom ic nu mb er. (For in stan ce, CI, whose ato mic number is 17, wo uld have to lose seven elec tro ns to have the same numbe r o f electrons as Ne .) If a nonmetal ato m were to gai n j us t a few el ect rons, however, it wou ld have the same n u mb er as a noble gas atom of higher ato m ic n u m be r. For exam ple, an oxygen ato m has e ig h t electrons. By ga in ing two elec trons per ato m, it forms 0 2 - , wh ich has tcn electr o ns, th e same numbe r as neon . Anions hrtoing the same number oj electrons as the noble gas atom succeeding it in the jm10dic table art' commsmls observed in chemical compounds.
1A 2A
3A
rn
Group lA. 2A. 3Ametals form Mn+ cations where n = group number.
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EXERCISE
2.6 Predicting Ion Charges
Predict formulas for monatomic ions formed from (a) K, (b) Se, (c) Ba, and (d) Cs. In each case, indicate the number of electrons gained or lost by an atom of the element in forming the anion or cation, respectively. For each ion, indicate the noble gas atom having the same total number of electrons.
Polyatomic Ions Pol yat omic io ns are made up o f two o r m o re ato ms, and th e co llec tio n has an e lectr ic charge (Figu re 2.19 and Table 2.4). For example , ca r bo nate ion , C0 32- , a common pol yatomic a nio n, co nsists o f o ne C a to m and three 0 ato ms. The ion has two u nits o f negative charge because th e re are two m o re el ect rons (a to ta l o f 32) in th e ion th an th e re arc protons (a total of 30) in th e nuclei o f o ne C ato m and three 0 atoms. The am mon iu m ion , N I l.," , is a com mon pol yatom ic ca tio n. In th is case, four H a to ms su rro u nd an N a to m, and th e ion has a 1+ el ec tric charge . This ion has 10 ele ctro ns, but th ere a re 11 positively cha rge d protons in the nuclei o f th e N and II ato ms (seve n and o ne each, res pectively). Chemistrv.o..Now~
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2.7
I
Ionic Compounds : Formulas, Names, and Properties
73
J Calcite, CaCO) _
..
Apatite. Ca SF(P04) ]
Caldum carbonate
0 .•
Caldum fluorophosphate
pO, J -
Celestite, srS04 Strontium suLfat e
Common ionic compounds based on polyatomic ions.
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Formulas of Ionic Compounds Com po un ds a re electrically n eutral ; that is, they h ave n o net electric c harge . Thus, in a n ionic compound , the numbers o f positive an d nega tive io ns mu st be suc h th at th e positi ve a nd n egati ve cha rges bal an ce . In so di u m ch lo ride, the sod iu m ion h as a I + c harge (Na"} , an d th e c h loride io n h as a ) - charge (C I-). These io ns must be p resent in a 1 : 1 ratio, an d th e formul a is NaC I.
TABLE 2.4
Formulas and Names of Some Common Polyatomic Ions
FormuLa
Name
FormuLa
Name
CATION: Positive Ion
NH4 +
ammonium ion
ANIONS: NegativeIons Based on a Group4A eLement CW
HC0 3 -
ClO -
hypochlorite ion
acetate ion
Cia,-
chlorite ion
carbonate ion
Cl03 -
chlorate ion
hydrogen carbonate ion (or bicarbonate ion)
CIO, -
perchlorate ion
Based on a Group SAeLement
Based on a tran sition metaL
NO ~ -
nitrite ion
CrO/ -
chromate ion
N03 -
nitrate ion
Cr2072-
dichrom ate ion
P0 4 3 -
phosphate ion
Mn 04-
perma nganate ion
HPO/ -
hydrogen phosphate ion
H2P04 -
dihydrogen phosphate ion
Based on a Group 6A eLement
ow
hydroxide ion sulfite ion sulfate ion hydro gen sulfate ion (or bisulfate ion)
74
Chapter 2
I
Based on a Group 7A eLement
cyanide ion
Atoms, Molecules, and Ions
The ge ms tone ru by is la rgely the com pound form ed from alum in u m ions (AP +) and ox id e io ns (0 2 - ) (bu t the co lo r co mes from a trace o f Cr H io n s.) Here , the io ns have positive and n egative charges that are of d ifferent abso lute value. To have a co m po u nd with the same n umber of positi ve an d negative charges, two AP + ions [to ta l charge = 2 X (3 +) = 6 +] mu st co mbin e with th re e 0 2 - ions [to ta l charge = 3 X (2 -) = 6 - J to give a formu la o f AJ,O ,. Ca lciu m is a Gro u p 2A me tal , an d it fo rm s a ca tio n havin g a 2+ charge. It ca n co m bine wit h a varie ty o f an io ns to fo rm ion ic com pou nds suc h as th o se in the fo llo wing table :
• Balancing Ion Charges in Form ulas Aluminum, a metal in Group 3A, loses three electrons to form the Al3+ cation. Oxygen, a nonmetal in Group 6A, gains two electrons to form an 02- anion. Notice that in the compound formed from these ions, the charge on the cation is the subscript on the anion, and vice versa.
Compound
This often works well, but there are exceptions. For example, the formula of ti tanium (IV) oxide is Ti02, the simplest ratio, and not Ti204.
Ion Combination
Overall Charge on Compound (2 +)
Cal + + COl 3 Ca 2+
+ 2 P0 4 3-
+2x
(1 -)
(2 +) + (2-)
~
~
0
0
3 x (2+) + 2 x (3-)
~
0
2 A[3 + + 3 02 -
~
Al 20]
Ti H + 2 02 - -e Ti02
In writing form u las o f ion ic compo unds, t he co nven tio n is th at the symbol (!! the ration is given [irst, followed fl)' the anion s)ltubol. Also notice th e u se o f parentheses whe n more than o ne o f a g iven polva tonn c ion is p resen t.
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. Ionic Compound Formulas Problem For each ofthe following ionic compounds, write the symbo ls for the ions present, and give the number of each: (a) li 2C0 3 • and (b) Fe2(S04h. Strategy Divide the formula of the compound into the cation and the anion. To accomplish this, you witt have to recognize, and remember, the composition and charges of common ions. SoLution (a) li 2C0 3 is composed of two lit hium ions, Li" . and one carbonate ion, COl -. Li is a Group 1Aelement and always has a 1 + charge in its compounds. Becausethe two 1+ charges balance the negative charge of the carbonate ion, the latter must be 2 - . (b) Fe2(S04h contains two iron ions, Fe 3+, and three sulfate ions, SO/ - . The wayto recognize t his is to recallthat sulfate has a 2- charge. Because three sulfate ions are present (with a total charge of 6-). the two iron cations must have a total charge of 6+. This is possible only if each iron cation has a charge of 3 +. Comment Remember that the formula for an ion must include its composition and its charge. Formulas for ionic compounds are always writtenwith the cationfirst and then the anion, but ion charges are not included.
Ionic Compound Formulas Problem Write formulas for ionic compounds composed of aluminum cations and each of the following anions: (a) fluoride ion, (b) sulfide ion, and (c) nitrate ion. Strateg y First decide on the formula of the Al cation and the formu la of each anion. Combine the Al cation with each type of anion to form an electrically neutral compound. SoLution An aluminum cation is predicted t o have a charge of 3+ because Al is a metal in Group 3A. (a) Fluorine is a Group 7A element. The charge of the fluoride ion is predicted to be 1- (from 7 - 8 = - 1). Therefore, we need 3 F- ions to combine with one Al 3 +. The formula of the compound is AlF 3 • (b) Sulfuris a nonmetal in Group 6A, so it forms a 2- anion. Thus, we need to combine two Al3+ ions [total charge is 6+ = 2 X (3+ )] with three S2- ions [total charge is 6- = 3 X (2 - )]. The compound has the formu la Al2S 3 • 2.7
I
Ionic Compounds: Formulas, Names, and Properties 75
(c) The nitrate ion has the formula N0 3 - (see Table 2.4). The answer here is t herefore similar to the AlF3 case, and the compound has t he formula Al(N0 3h. Here, we place parentheses around NO, to show that three polyatomic N0 3- ions are involved. Comment The most common error students make is not knowi ng the correct charge on an ion.
EXERCISE 2.7
Formulas of Ionic Compounds
(a) Give t he number and identity of t he constituent ions in each of t he following ionic compounds: NaF, Cu (N 03h, and NaCH 3C 0 2• (b) Iron, a transition metal. formsions having at least two different charges. Write the formulas of the compounds formed between chloride ions and t he two different iron cations. (c) Write the formulas of all neut ral ionic compounds that can be formed by combining the cations Na " and Ba 2+ with the anions 52 - and P043 - .
Names of Ions Naming Positive Ions (Cations) With a few exce p tio n s (suc h as N H 4 +) , th e positive io ns d escribed metal ion s. Po sitive io ns are name d by th e foll owing rul es: • "eous" and "-ic" Endings An older naming system for metal ions uses the ending -ous for the ion of lower charge and -ic for the ion of higher charge. For example, there are cobaltous ((0 2+) and cobaltic ((03+) ions, and ferrous (Fe2+) and ferric (Fe H ) ions. We do not use this system in this book, but some chemical manufacturers continue to use it.
III
th is text a re
1. For a mo na to mic po sitive io n (that is, a metal catio n ) the nam e is tha t o f th e met al plus th e wo rd "ca tio n ." Fo r exa m p le , we have alre ady re ferre d to AFH as th e alu m in u m ca tion. 2. So m e cases o ccur, especially in the tr ansition series, in whic h a me tal ca n fo rm more th an o ne type of positive io n . In th ese cases, th e charge of the ion is indi cat ed by a Roman numeral in pa rentheses immedi at ely followin g th e io n name . For exa m ple, C0 2 + is th e co ha lt( II) cation , and Co H is the co balu 1lI) cal io n . Finall y, yo u will e nco u n ter the am mo niu m ca tio n, N H 4 " , man y tim es in this bo ok and in th e lab o rat ory. Do not co nfuse th e am mo ni um ca tio n with th e am mo nia mol ecul e , N H:\> which has no e lec tr ic charge and o ne less H ato m .
Naming Negative Ions (Anions) Th ere are two types o f negat ive ions: th o se havin g on ly o ne ato m ( monatomi c) an d th o se having severa l ato ms (j)o[y alomir).
1. A m on atomi c nega tive io n is n amed by ad d ing -ide to the ste m of t he na me of th e no nm et al cl em ent from whic h th e ion is d eri ved (Fig u re 2.20 ) . T he a nio ns of th e Grou p 7A e lem ents, the hal ogen s, are known as th e fluo rid e , ch lo ride, b romid e , an d io d id e ion s and as a gro u p are called halide ion s. 2. l' ol yato m ic negati ve io ns are co m m o n, especially th o se co n taining o xygen (ca lled oxoanions ) . Th e n ame s o f so me o f th e mo st co m mo n o xo anion s a re g iven in Ta ble 2.4 . Althoug h most o f th ese nam es m ust sim p ly be learn ed , so m e gu id eli nes can help. Fo r exam p le, co n side r th e foll owin g pai rs o f ion s: NO j - is th e ni tr at e ion , whe re as N 0 2 - is the n itr ile ion SO / - is th e sulfa te ion , whe re as 50 :\2- is th e sulfite ion The oxoan ion having the g re ater n um ber o f o xyge n ato ms is give n th e suffix -ate , and th e ox oa n io n having the smaller number o f ox ygc n ato m s has th e suffix -ite . For a se ries o f o xoanio ns having more th an two m embe rs, the ion with the largest num ber o f oxyg e n a to ms has the p refi x per- and the su ffix -ate. The ion 76
Chapter
2
I
Atoms, Molecu les. and Ions
FIGURE 2 .20 Names and charges of some common monatomic anions.
N'- 0' nitride ion
oxide ion
Ffluoride ion
p' -
5' -
Cl-
phosphide ion
sulfide ion
chloride ion
5e' - Br-
•
selenide bromide ion ion
Te' -
l-
telluride ion
iodide ion
having t he sm allest numbe r of oxygen ato m s has th e prefix h)jJO- and the suffix -ite; T he ch lo rine o xoani o ns are th e mo st co mmo n ly e nco u n tered examp le. ClO, C10 3 C10 2 ClO -
perch lorate ion ch lo rat e io n ch lorite io n hJj)()ch lo rite ion
• Naming Oxoanions
per . .. ate ~
c; ~
~
Oxoan io ns th at co n tai n hydroge n are named by ad d ing th e wo rd "hyd roge n " befo re the name o f the ox oanlon . If two h yd rogen s are in the anio n , we say "di hvd rogcu ." Man y h ydroge n-con tain ing oxoan ion s have co m mo n names th at are used as wel l. For example , th e hyd ro gen ca rbonate ion , H C0 3 - , is ca lled th c bicarbonate io n. [on
Systematic Name
HP0 42 -
hydrogen phosphate ion
H2 P0 4 -
dihydrogen phosphate ion
HC0 3 -
hydrogen carbonate ion
bicarbonate ion
HS04 -
hydrogen sulfate ion
bisulfate ion
HS03 -
hydrogen sulfite ion
bisulfite ion
~
c
c
u
.~
c
ro
~
u .~
. .. ate
0
~ ~
» x
. . . ite
0
hypo . . . ite
Common Name
Names of Ionic Compounds T he nam e of an io nic com pou nd is built fro m th e n a m e s of th c POSItI\'C and negative io ns in the co m pou nd . T he name of th e positive cat ion is g iven firs t, followed by th e name of th e n eg ative anio n . Exa m ples of ionic co m po un d na mes are give n below. Ionic Compound
Ions Involved
Name
CaB r2
Ca z+ and 2 Br-
calcium bromide
NaHS04
Na " and HS0 4 -
sodium hydrogen sulfate amm onium carbonate
(NH4 hC0 3
2 NH 4 + and CO/ -
Mg( OH),
Mgz+ and 2 OW
magnesium hydroxide
nn,
Ti2+ and 2 cr
titanium(II) chloride
C0 20 3
2 ( 03+ and 3 0 2 -
cobalt(III) oxide
• Names of Compounds Containing Transition Metal Cations Be sure to notice t hat the charge on a transition metal cation is indicated by a Roman numeral and is included in the name.
2.7
I
Ionic Compounds: Formulas, Nam es, and Propert ies
77
Problem Solving Tip 2.1
Formulas for Ions and Ionic Compounds
Writing formulas for ionic compounds takes practice, and it requires that you know the formulas and charges of the most common ions. The charges on monatomic ions are
If you cannot remember the formula of a polyatomic ion, or if you encounter an ion you have not seen before, you may be able to figure out its formula. For example, suppose you are told that NaC H0 2 is sodium formate. You know that the sodium ion is Na". so the formate ion must be the remaining portion of the compound; it must have a charge of 1to balance the 1 + charge on the sodium ion. Thus. the formate ion must be (H0 2- .
often evident from the position of the ele·
ment in the periodic table. but you simply have to remember the formula and charges of polyatomic ions-es pecially the most common ones such as nitrate, sulfate, carbonate, phosphate. and acetate.
Finally. when writing the formulas of ions. you must include the charge on the ion (except in the formula of an ionic compound). Writing Na when you mean sodium ion is incorrect. There is a vast difference in the properties of the element sodium (Na) and those of its ion (Na ").
ChB mislry-
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Introduction to Chemical Equations
When a strea m of ch lo rine gas, C L~ , is direct ed o n to so lid pho sph orus, P4 , th e m ixture bursts in to n am e , an d a che mical reac tio n prod uces liq uid p hosp horus tri chlorid e , PCl 3 (Figure 3. 1). \" e ca n d epict thi s react ion using a balanced ch emical equation .
P, (s)
+ 6 Cl,(g) ~ 4 PCl3 ( £)
'----------'" Rea ctan ts
P,( s)
,
Product
+ 6 ( l,(g)
REACTANTS
4 PCl 3(£) PRODUCT
FIGURE 3.1 Reaction of solid white phosphorus with chlorine gas. The product is liquid phosphorus trichloride.
3.1
I
Introduction to Chemical Eq uations 113
Historical Perspectives
Antoine Laurent Lavoisier, 1743-1794
On Monday, August 7. 1774, the Englishman Jose ph Priestl ey (1733-1804) isolated oxygen. (The Swedish chemist Carl Scheele [1742-1786] also discovered the element, perhaps in 1773 or earlier .) Priestley heated solid mercury(II) oxide. HgO. causing the oxide to decompose to mercury and oxygen . 2 HgO(s) -> 2 Hg(f) + O, (g) He did not immediately understand the significance of the discovery, but he mentioned it to the French chemist Antoine lavoisier in October, 1774. One of l avoisier's contributions to science was his recognition of the importance of exact scientific measure-
ments and of carefully planned experiments, and he applied t hese methods to the study of oxygen . From this work, Lavoisier proposed that oxygen was an element, that it was one of the constituents of the compound water, and that burning involved a reaction with oxygen . He also mistakenly came to believe Priestley's gas was present in aU acids, and so he named it "oxygen," from the Greek words meaning "to form an acid." In other experiments, Lavoisier observed that the heat produced by a guinea pig when exhaling a given amount of carbon dioxide is similar to t he quantity of heat produced by burning carbon to give the same amount of carbon dioxide. From these and other expertme nts he concluded t hat "Res piration is a combustion, slow it is t rue, but otherwise perfectly similar to that of charcoal." Although he did not understand the details of the process, this was an important step in the development of biochemistry. Lavoisier was a prodigious scientist, and the principles of naming chemica l substances that he introduced are still in use today. Fu rther, he wrote a textbook in which he applied the principles of the conservation of matter to chem istry, and he used the idea to write early versions of chemical equations. Because Lavoisier was an aristocrat, he came under suspicion during t he Reign of Terror of the French Revolution. He was an investor in the Ferme Generate, t he infamous
The decomposition of red mercury(II) oxide. The decomposition reaction gives mercury metal and oxygen gas. The mercury is seen as a film on the surface of the test tube.
• Information from Chemical Equations The same number of atoms must exist after a reaction as before it takes place. However, these atoms are arranged differently. In the phosphorus/chlorine reaction, for example, the P atoms were in the form of P4 molecules before reaction but appear in the PCl3 molecules after reaction.
114
Ch ap t er
3
I
Chemical Reactions
tax-collecting organization in 18th-century France. Tobacco was a monopoly product of t he Ferme Ge nerate. and it was common to cheat the purchaser by adding water to the tobacco, a practice that Lavoisier opposed. Nonetheless, because of his involvement with t he Ferme, his career was cut short by the guillotine on May 8,1794, on the charge of "adding water to the people's tobacco."
Lavoisier and his wife, as painted in 17BB by Jacques-Louis David. lavoisier wasthen 45, and his wife, Ma rie Anne Pierrette Paulze. was 30. (The Metropolitan Museum of Art, Purchase, Mr. and Mrs. Charles Wrightsman gift, in honor of Everett Fahy, 1997. Photograph © 1989 The Metropolitan Museum of Art.]
In a ch emical eq ua tio n, th e formulas for the reactants (t he su bsta nc es combined in th e reaction ) are wri tten to th e left of th e arrow, a nd th e formulas of th e products (the su bsta nces produced) are written to the right of th e arrow. The ph ysical states o f re act ants and product s can also be indicated . The sym bo l (s) indicat es a so lid , (g) a gas, and (f) a liquid. A su bstance d issolved in wat er, th at is, an aqueous so lu tio n o f a su bsta nce, is indicat ed by (aq). In th e 18th ce n tu ry, th e French scie n tist An to ine Lavoi sier (17 43-1794) introduced th e law of conservation of matter, whi ch sta tes that matter can neither be created nor destroyed. This means th at if t he total m ass of react ants is 10 g, and if th e reacti on co m ple te ly co nve rts reactants to product s, yo u must e nd up with 10 g o f product s. Thi s also mean s th at if 1000 ato ms of a parti cu lar el em ent are co ntained in th e reactan ts, th en those 1000 ato m s must ap pear in th e p roduct s in so me fashi on . When ap plied to the reaction of phospho rus and ch lo rine, the law of co nservatio n o f matte r tell s us th at 1 molecu le of phosphorus, P 4 (with 4 phosphorus ato ms) and fi diatomic molecul es o f CI2 (with 12 atoms of Cl) are required to produce
2 Fe(s)
+ 3 (l , (g)
--~)
2 FeCl, (s) '-r----' PRODUCT
REACTANTS
FIGURE 3.2 The reaction of iron and chlorine. Here, hot iron gauze is inserted into a flask containing chlorine gas. The heat from the reaction causes the iron gauze to glow, and brown iron(IlI) chlorideforms.
fo u r molecul es o f PCI3• Because ea ch PCh mol ecul e co n ta ins I P a to m and 3 CI atoms, the fou r PCl 3 m olecules are need ed to acco u nt for 4 P ato ms and ] 2 CI a to m s in the pro duct.
4 P atoms
6X2 =
4X3 =
12 Cl atoms
12 Cl atoms
4 P atom s
Nex t , con sider th e ba lanced e q ua tio n for th e re ac tio n o f iron a nd an d c h lorine (Figu re 3.2). In this case , there are two iro n ato m s and six chlorine ato ms on bo th sid es of the equation .
2 Fe(s) + 3 Cl,(g) - > 2 FeCl, (s) t
t t stoichiometriccoeffi cients
T he nu mb ers in front of formulas in balanced che mical equa tio ns are req uired by the law of co nservation of matter, T he y ca n be read as a number of atoms (2 ato ms of Fe) , molecul es (3 mole cul es o f CI2) , o r form ula units (2 fo rmula un its of th e io nic co mpou nd FeC I3 ) . They ca n refer eq ually well to amoun ts of reacta nts and products: 2 mo les of so lid iro n combine with 3 m oles of chlo rine gas to produce 2 mo les of so lid FeC I3. T he rela tio nship between the qu antities of chemical reactan ts and pro dnets is called stoichiom etry (p ro nou nc ed "stoy-key-Al l jd-uh-tree ") (... Ch ap ter 4) , and the coefficie nts in a balan ced eq ua tio n are the stoi chiometric coefficie nts. Chemistry.o.Now~
•
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3 .1
I
I nt roducti on to Chemica l Equa tions
115
• The Importance of Balanced Chemical Equations Balanced chemical equations are fundamentally important for understanding the quantitative basis of chemistry. You must always begin with a balanced equation before carrying out a quan titative study of a chemical reaction.
EXERCISE 3 .1
Chemica l Reactio ns
The reaction of aluminum with bromine is shown on page 67. The equation for the reaction is
+ 3 Br,((:) --> AI, Br,(s)
2 Al(s)
(a) Wh at are the stoichiomet ric coefficients in this equation ? (b) If you were to use 8000 atoms of Al, how many molecules of Br2 are required to consume the Al completely?
3.2 Ba Lancing ChemicaL Equations Balan cing a c h e m ica l e q u a tio n e n su res th at th e sa me number of ato m s o f eac h el ement a p pe a rs o n both sides of th e e q u a tio n . Man y ch e m ica l e q u a ti o n s can be ba la n ce d by trial a nd e rro r, a lt h o u g h so m e will in volve m o re tri al tha n othe rs. One ge n e ra l cl ass of c h e m ica l re a cti ons is the re acti on o f metals o r nonm etal s with oxygen to give oxid es of the ge n e ra l formula 1\l xO )'" Fo r e xample, iron react s wit h o xyge n to give iro n (III ) o xide (Fig u re 3.3a) .
4 Fe(s)
+ 3 O, (g) --->
2 Fe,O, (s)
The nonmetal s su lfu r a n d oxygen re a ct to form su lfu r di oxide (Figu re 3.3 b),
5(s)
+ O,(g ) ---> 50,(g)
a n d pho sphorus, P 4 , re acts vigorously wit h oxygen to give tet raph o sphorus d ecaoxide , P,O IO (Figure 3.3e).
The eq uatio n s written a bove a re bal an ced. The sa m e number o f iron , su lfu r, o r phosphorus atoms a nd oxyge n a to ms occurs o n each sid e of th e se e q u a tio ns . The combustion, o r burning , of a fu el in oxygen is acco m pa n ie d by th e evo lution of e n e q ,'y. \ bu a re familiar with combusti on re a cti ons suc h as th e burning o f octan e , CSH llh a co m po ne n t of gas o line , in an au to m o b ile e ng in e :
2 C8 H18 (f:)
+ 25 O, (g) --->
16 CO, (g)
+ 18 H,O(g)
FIGURE 3 .3 Reactions of a metal and two nonmetals with oxygen . (See ChemistryNow, Screen 3.4, Balancing Chemical Equations, for a video of t he phosphorus and oxygen reaction.)
(a) Reaction of iron and oxyge n to give iron(II1) oxide, Fe203.
116
Ch ap t er
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Chemica l Reactions
(b) Reaction of sulfur (in the spoon) with oxygen .
(c) Reaction of phosphorus and oxygen to give tetraphosphorus decaoxtde. P4010.
In all co m b us tio n reaction s, some or all the el em en ts in th e reacta n ts en d up as o xid e s, compounds co n ta ining oxyge n. When th c reactant is a h ydro carbon (a co mpou nd suc h as gasoline, n atural gas, o r propan e tha t co n tai ns on ly C an d H ) , the p roducts o f co m p le te com bus tio n are a lways just carbo n dio xid e an d wa ter. Wh en bal an ci ng che mical eq u atio ns, th e re are two im portan t th ings to re m ember. • Fo rmul as for reac ta n ts a nd product s m ust be co rrect, o r the equat io n is meani ngless. • Su bsc ri p ts in th e form ul as of reactan ts and p ro duct s ca n no t be cha nge d to bal an ce e quatio ns . Chang ing the su bsc ri pts cha nge s th e id en tity o f th e su bs ta nce. Fo r exam p le , yo u ca n n o t c ha nge CO 2 to CO to bal an ce a n e q ua tio n; carbo n mono xide , CO , a nd ca rbon di o xid e , CO 2 , are d iff er e nt co m pou n ds . As an exam p le o f eq uation balan cing , let us write the balan ced eq uati o n for the com ple te co m bustio n o f propan e, C 3 H g•
Acombustion reaction. Here, propane, C3H s, burns to give CO 2 and H20. These simple oxides are always the productsof the complete combustion of a hydrocarbon.
Step 1. ll'rite correct formulas f or the reactants and products. ( 3H8(g) + O,(g) unbalanced equation ) (O,(g) + H,O(t )
Here , propan e and ox ygcn are th c re act ants, a nd ca rbon di o xid e and wate r are th e products. Step 2. Balance the C atoms. In co mb ustio n rea ctions suc h as thi s, it is usually best to balan ce th e carbo n atoms first an d leave th e o xyg en ato ms un til the e nd (because th e oxyge n atoms are often fou nd in m ore than o ne product ) . In this case, three carbo n ato ms are in th c reactants, so three mu st oc cu r in th c p roduct s. Three CO 2 molecul es are therefore re qu ired on the rig h t side: ( 3H8(g) + O,(g) unbalanced equation) 3 ( O,(g) + H, O( t)
Step 3. Balance the If atoms. Propan e , th e reac ta n t, co n tains 8 H ato ms. Eac h m olcc ulc o f water has two hydro gen ato ms, so four molecul es o f water acc o u n t for th e re quired eigh t hydrogen ato ms o n th e right side: ( 3H8(g) + O,(g) unbalanced equation ) 3 (O,(g) + 4 H,O(t )
Step 4. Balance the () atoms. Te n ox yge n ato ms are o n the right side (3 X 2 = 6 in CO 2 plus 4 X 1 = 4 in H 20 ) . The re fore , five O 2 molecul es arc needed to su p ply th e re q uire d 10 oxygen atoms: ( 3H 8(g)
+ 5 O,(g) -. 3 ( O,(g) + 4 H,O(t)
Step 5. \lerify that the number of atoms of each element is balanced. Th e eq uatio n shows three carbo n ato ms, c igh t hydrogen ato ms, and tcn oxygen at oms on eac h side . Chemislry.cl"Now
m
•
Sign in at www.thomsonedu.com/login and go to Chapter 3 Contents to see Screen 3.4 for an exercise and a tutorial on balancing the chemical equations for a series of combustion reactions .
3.2
I
Bala ncing Chemical Eq uations
117
Balancing an Equation for a Combustion Reaction ProbLem Write the balanced equation for the combustion of ammon ia (NH ]
+ O2) to give NO and H20.
Strategy First, write the unbalanced equation. Next, balance the Natoms, t hen the Hatoms, and finally, balance the 0 atoms. SoLuti on Ste p 1. Write correct formulas for the reactants and products. The unbalanced equation for t he combustion is NH3(g)
+ 02(g) unbalanced equation) NO(g) + H20(£ )
Step 2. Balance the Natams. The re is one Natom on each side of the equation. The Natoms are in balance, at least for the moment. NH3 (g)
+ 02( g) unbalanced equation) NO(g) + H20(£ )
Step 3. Balance the Hatoms. There are three Hatoms on the left and two on the right. To have the same number on each side, let us use two molecules of NH 3 on t he left and t hree molecules of H20 on the right (which gives us six Hatoms on each side). 2 NH3(g) + 02(g) unbalanced equation) NO(g) + 3 H20(£ ) Notice that when we balance t he Hatoms, the Natoms are no longer balanced. To bring them into balance, let us use 2 NO molecules on the right. 2 NH3 (g) + 02( g) unbalanced equation) 2 NO(g) + 3 H20(t:) Step 4. Balance the 0 atoms. After Step 3, there are two 0 atoms on t he left side and five on t he right. That is, t here are an even number of 0 atoms on t he left and an odd number on the right. Because there cannot be an odd number of 0 atoms on the left (0 atoms are paired in O2 molecules), mu ltiply each coefficient on both sides of the equation by 2 so that an even number of oxygen atoms (10) can now occur on the right side:
°()
4 NH ]() g+2 g unbalanced equation ) 4 N0(g) + 6 H2O(£ ) Nowthe oxygen atoms can be balanced by having five O2 molecules on the left side of the equation: 4 NHJg)
+ 5 02(g) balanced equation) 4 NO(g) + 6 H20(£)
Step 5. Verify the result. Fou r Natoms, 12 Hatoms, and 10 0 atoms occur on each side of t he equation. Comment An alternative wayto write this equation is 2 NH;(g)
+ 51, O,(g) --> 2 NO(g) + 3 H,O(C)
where a fractional coefficient has been used. This equation is correctly balanced and witt be useful under some circumstances. In general, however, we balance equations with whole-number coefficients. EXERCISE
3.2 Balancing the Equation for a Combustion Reaction
(a) Butane gas, (4H lO, can burn completely in air [use OAg) as the ot her reactant] to give carbon dioxide gas and water vapor. Write a balanced equation for this combustion reaction. (b) Write a balanced chemical equation for the complete combustion of liquid tetraeth yllead, Pb((2H s)4 (which was used until the 1970s as a gasoline additive). The products of combustion are PbO(s), H,O«(), and (O,(g) .
3.3
Introduction to Chemical Equilibrium
To thi s poin t, we have treat ed che m ica l reac tio ns as p ro ceed ing in o ne direction only, with reactants bc ing co nve rt ed co m ple te ly to product s. Na tu re , howeve r, is mo re com plex th an thi s. Che m ica l reac tio ns are reve rsib le , and ma llY reac tio ns lead to in complet e conversio n of reac ta n ts to product s. A good exam ple of a re versib le re act io n th at d o cs not pro ce ed co m ple te ly to p rod uct s is th e reacti on o f nitrogen wit h hydrog en to form a m mon ia gas, a 118
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Chemical Reactions
N,(g)
,
jj ~
a
_~
~
;;;
"-. a
u
~
< ~
, a
E
-c
a •
+
3H,(g)
~ ~
FIGURE 3.4 The reaction of N2 and H2 to produce NH 3 • N2 and H2 in a 1:3 mixture react to produce some NH 3 • As the reaction proceeds, the rate or speed of NH 3 production slows, as does the rate of consumption of Ne and H2• Eventually, the amounts of Nz and H2, and NH 3 no longer change. Att his point, the reaction has reached equilibrium. Nonetheless, the forward reaction to produce NH 3 continues, as does t he reverse reaction (the decomposition of NH 3 ) .
2 NH ,(g)
-. -- -
r-s- Equilibrium achieved
H, NH ,
(
"
N,
Reactants proceeding toward equilibrium
•
com po u nd used ex te ns ive ly both as a fer til izer and in the production o f other fe rt ilize rs.
N2(g)
+ 3 H2(g) --> 2 NH,(g)
• Progression Toward Equilibrium Reactions always proceed spontaneously toward equilibrium. A reaction will never proceed on its own in a direction t hat takes a system further from equilibrium.
Nitroge n an d h ydro gen react to form am mo nia , but, u nde r the con dit ions of the reac tio n , th e product am mon ia also b reak s down into nitroge n and hydrogen in th e IT Y CrSe re acti on.
2 NH,(g) --> N2(g)
+ 3 H2(g)
Le t us consider wha t wou ld happen if we m ixed nitr ogen an d h ydrogen in a clo sed co ntainer u nd er the p roper cond itions fo r th e reactio n to o ccu r. At fir st, N 2 and H 2 reac t to produce so me a m mo n ia. As th e am monia is produced, ho wever, some N B 3 molecul es dec o m pose to re-for m n itro gen an d h ydrogen in the reverse reac tio n . At th e begin nin g o f the process, the fo rwa rd reaction to give NI 13 p redom in at es, but, as th e react ants are co nsu med, th e rate of th e forward reaction is progressively slower. At the same tim e , the reverse reac tion speeds u p as the amou n t of am mo n ia in creases. Eventually, th e rate o r speed o f th e for ward reacti on will eq ua l th e rat e of th e ITYCrSe reaction . Once thi s occ u rs, no fu r the r macroscop ic change is o bserved ; the amou nts of nitrogen , hydrogen , and ammonia in the co ntain er sto p chang ing (Fig u re 3.4) . \Ve say th e syste m has reach ed chemical equilibrium. T he reaction vessel will co n ta in all th ree su bsta nces: nitro gen . hydrogen . an d ammo nia. Because bo th the forward a nd reverse processes are still o ccurring (bu t at equa l rates) , we re fer to this sta te as a d ynam ic equilibrium . \Ve represen t a ) co n nec ting syste m at d ynami c eq uilibriu m by wr iting a doubl e arrow sym bo l ( the reactan ts and product s.
N,(g) + 3 H,(g)
2 NH,(g)
The form ation of stalactites and stalagm ites ill a limeston e cave is another example o f a syste m th at d epends o n th e reve rs ibility of a chem ical rea ction (Figure 3.5 ) . Sta lactites an d sta lag m ites are made chiefly of ca lciu m ca rbonate , a m in eral found ill und ergroun d d eposits in th e fo rm o f lim eston e , a leftover from anc ie n t oceans. If wate r see ping through th e lim eston e co n tains di ssol ved CO 2, a reaction occurs in whic h th e mine ral di ssolves, giving an aqueo us so lu tio n o f Ca( IIC0 3 h .
Ca CO,( s) + CO,(aq) + H,O(f) --> Ca( HCO,),(aq ) 3 .3
I
FIGURE 3 .5 Cave chemistry. Ca lcium carbonate stalactites cling to the roof of a cave,and stalagmites grow up from the cavefloor. The chemistry producing these formations is a good example of t he reversibility of chemical reactions.
Int roduction to Chemical Eq uilibrium 119
a Reactants : Solutions of CaCl 2 (left) and NaH C03 (right). Na + and Cl- are spectator ions (page 129; not shown)
b The solutions are mixed.
Forward Reaction
> Products: H20,
a precipitate of CaC0 3, and COl gas
HCO, -(aq)
Equilibrium Equation:
(0,(9)
(
)
d TheCa C0 3 dissolves when the solution has been saturated with CO2,
c The reaction can be reversed by bubbling CO 2 gas into the CaC0 3 suspensio n.
CaCO,( s) + CO,(g) + H,O(f )
Cave che m istry ca n be d on e in a lab oratory (Fig u re 3.6) using reactio ns that further d em o nstrat e th e rcvcrsi blity of th e reactions involved . A key qu esti on that arises is, "w he n a react io n reaches eq ui librium , will the reactants be co nverted largely to p roducts, or will mo st o f the react an ts still be prese n t?" T he answer will d epend o n th e nature of th e co m po un d s invol ved, the te m pe ra ture , a nd o th er facto rs, and that is the su bject of late r chapte rs (.... Chapters 16-1 8 ). Fo r the p resent, tho ug h , it is use ful to d e fine product-favored reactions as reactions in whi ch reactants are comptn els or largely converted to products at equ ilibriu m. Th e co m bustio n react ions we have been stu d ying are e xa mp les o f reactions that are p ro d uct-favored a t eq u ilib rium , in co n trast to the N d II 2 reactio n in Figu re 3.4 . In fact , m o st of th e reacti ons th at we sha ll stud y in th e rest of this chap te r are p ro du ct-favo red reac tions a t eq u ilib ri um . We usu all y write t he eq uations for reactio ns that are very prod uct-favored usin g o n ly the single arrows we have been usin g up to this poi nt. T he opposite o f a prod uct-favore d react io n is o ne th a t is reactant-favo red a t eq uilib riu m . Suc h reactio ns lead to th e co nversion of o nly a sma ll am o u n t of the reac ta nts to p roduct s. An exam p le of suc h a reacti on is th e io niza tio n o f ace tic acid in water, in whic h on ly a tin y fractio n of th e acid reac ts to p roduce ions. CH,CO , H(aq) + H,O(f ) ,
3.4
H,O+(aq)
+ CH, CO, -(aq)
• Quantitative Description of Chemical Equilibrium As you shall see in Chapters 16-18, the extent to which a reaction is product-favored can be described by a simple mat hematical expression, called the equilibrium constant expression . Each chemical reaction has a num erical value for the equilibrium constant, symbolized by K. Product-favored reactions have large values of K; small K values indicate reactant-favored reactions. For the ionization of acetic acid in water, K = 1.8 x
1O - ~ .
• Acetic Acid, a Weak Add Acetic acid is an example of a large number of acids called "weak acids" because only a few percent of the molecules ionize to form ionic products.
Chemical Reactions in Aqueous Solution
Man y of the reactio ns yo u will st udy in yo u r chemis try co urse and th e reac tio ns th at occ ur in living system s are carried o ut in aq ueo us so lu tio n . Because reac tio ns in aq ue o us so lu tio n are so important, th e rem ainde r o f thi s chap ter is an in trod uction to th e behavior o f co m pounds in so luti on and to som e o f the types of reactio ns yo u will o bserve. A solution is a ho mo ge neous mi xt ure o f two or more su bsta nces. O ne su bsta nc e is ge ne rally considered the solven t, the medi um in which a no the r su bstance- the solute-is di ssolved . In th e hum an body, th e so lvent for chem ical reac tio ns is usually wate r. Wat er assists in tr an sporting n utri ents an d waste p rod uc ts in a nd o u t of ce lls and is necessary fo r digestive, abso rp tio n, circu lato ry, an d e xcr eto ry fu nct ions. In fac t, th e hu man body is two-third s wate r; Wat er is an ex ce lle nt so lven t to use for bio ch emical reactio ns an d also for m an y o ther che m ica l reactio ns. Fo r th e next several secti ons of th is chapter, we shall study chemical reactions that occ ur in aqueous solutions where wa te r is th e so lven t. So th at yo u are fam iliar with types o f reactio ns as yo u wo rk th rough th e bo ok, we also wan t to introduce yo u to fou r major ca tego ries of reactio ns in aq ueous so lutio n : precipita tio n , aci d-ba se , gas-fo rm ing, and o xidatio n-re d uctio n reactio ns. As yo u learn abou t these reactions, it will be u seful to loo k for patterns th at a llow yo u to pred ict the re actio n p ro ducts. 'You will notice th at man y o f th e re actio ns are exchange reactions in wh ich the ions oj the muta nts change partners. ~
A+B- + (+D -
~
A+D-
+ ( +B-
'-------'" 3.4
I
Chemical Reactions in Aqueous Solution
121
It:JI
FIGURE 3.7 Predpit ation of silver chloride. (a) Mixing aqueous solutions of silver nitrate and potassium chloride produces white, insoluble silver chloride, Agel. In (b) through (d), you see a model of the processat the molecular and ionic level.
. ~.
(b) InitiaLLy, the Ag + ions (silvercolor) and Cl" ions (green) are widely separated.
-
I -
. 0' . ' ;0:
O · ,. -,
.'
. p ' l,' ,.
~
"
~
(c) Ag+ and cr ions approach and form ion pairs.
'l
r
(d) As moreand more Ag + and Cl" ions come together, a precipitate of solid AgCl forms.
[a]
For exam p le, aq ueo u s so lution s of silver nitrate and potassium c h lo ride react to prod uce so lid silver c h lo ride an d aq u eous potassium n itra te . (Fig u re 3.7a)
AgNO,(aq) + KCl(aq ) -> AgCl(s) + KNO, (aq) Recogn izin g th a t catio ns ex c hange anions in many ch e m ica l reacti o ns g ives u s a good way to predict th e prod u cts of precipitation , acid-base , a nd man y gas-fo rm in g reactions.
•
ModuLe 5
3.5
Ions and Molecules in Aqueous Solution
To u nderstand reactio n s o ccurring in aq u eou s so lu tio n, it is important first to u n derst a nd someth ing abo ut th e behavio r of compound s in water. T he wa ter yo u drin k cvery day, the oceans, a nd the aq ueou s so lut io n s in you r body co nta in many ion s, most of whic h resu lt from di sso lving so lid m at eria ls present in th e envir o nme n t (Table 3.1) .
TABLE
3.1 Concentrations of Some Cations and Anions in the Environment and in
Living CeUs ELement
Dissolved Species
Red Blood CeUs
BLood
Sea Water
Valonia t
Chlorine
Cl-
550
50
50
100
Sodium
Na +
460
80
11
160
Magnesium
Mg 2+
52
50
Calcium
(a
2+
Pota ssium
K·
Carbon
HCO] -,
Phosphorus
H2P0 4 -
10
cov,
HPO/ -
1.5
Plasma
2.5
2
10 - 4
2
10
400
92
10
30
< 10
< 10
30
Ba' +(aq) + 2 NO,- (aq) No tice th at th e polyato m ic ion stays to gether as o ne un it, N0 3 "", and th at th e two ni tr at e io ns se parate fr o m each othe r. Compo unds whose aq ueo us so lutio n s do not cond uct e le ctricity are ca lle d nonel ectrolytes . T he so lu te pa rt icle s p resent in th e se aq ueo us so lu tio ns are molecu le s, not io ns. M ost molecular compounds that dissolve in water are nonelearolstes. Fo r e xam p le , whe n th e mo le cu lar co m po u nd e than o l (C 2I I5 0 JI) di sso lves in wate r, each mol ecul e of etha no l stays to geth e r as a sing le uni t. vVe do not g e t io ns in the so lu tio n . C,HsOH(C) -> C, H,OH(aq)
124
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Chemical Reactions
Other exa mp les of nonel ectrol ytes are sucro se (C I2II 220 11) and an tifre eze (et hyle ne glycol, HOCH 2 CH 20H ) . Some molecul ar co m po u nd s (str o ng acids, weak acids, and weak bases) (.... Sectio n 3.7), ho weve r, react wit h water to form io ns and are t hus electrolytes . Hydrogen ch loride is a mol ecul ar com po u nd , but it react s wit h wate r to for m io n s, and t he so lu tion is refe rr ed to as hyrlmchloric acid.
HCl(g) + H,O(C) -> H,O+(aq)
+ Cl -(aq)
This reactio n is very p ro duct-favored . Eac h molecul e o f II CI produces ions in so lution so hydrochloric acid is a stro ng elect rol yte . A weak electrolyte is a molecul ar su bstance in whose aq ueous so lutio ns so me o f the mol ecules reac t with water to fo rm ion s b u t whe re so me o f the molecul es (usually most) re ma in as molecul es. T he ir aq ucous so lu tio ns arc poor co nd ucto rs o f electricity (see Figu re 3.9). As d escri bed o n page 135, the in teraction of ace tic aci d with water is very reactant-favo red . In vin egar, an aq ueo us so luti on o f ac et ic acid, fewer th an 100 molecul es in every 10,000 mo lecul es of ace tic acid are ionized to fo rm ace tate and h ydron iu m ion s. Thus, aqueo us ace tic acid is a weak electrolyte. • H+ Ions in Water As illustrated for acetic acid in Figure 3.9, the H+ ions from the acid are surrounded by water molecules. When writing an equation for acid ionization, we symbolize this with the H30+ or hydronium ion. For more on this, see page 134.
H,O( e)
water
acetic acid
acetate ion
hydronium ion
< 1% ionized
= weak electrolyte
Sign in at www.thomsonedu .rom/Ioqtn and go to Chapter 3 Co ntents to see: Screen 3.5 for exercises on the dissolvingof an ionic compound • Screen 3.6 for information on the types of electrolytes
EXERCISE 3 .3
Electrolytes
Ep som salt. MgS04 • 7 H20. is sold in drugstores and, as a solution in water, is used for various medical purposes. Metha nol. CH 30H, is dissolved in gasoline in the wi nter in colder climates to prevent t he formation of ice in automobile fuel lines. Which of t hese compounds is an electrolyte, and which is a nonelectrolyte?
Solubility of Ionic Compounds in Water Marry ion ic co m pou nd s dissolve co mp letely in water, but so me di ssol ve o n ly to a small ex te n t, a nd still others are esse n tially inso luble . Fortu na tely, we ca n make so me gene ral state me n ts abo u t whic h ionic co mpo un ds are wate r so lu ble . In th is cha p te r, we co ns id er so lu bility as an "eithe r- o r " qu estio n, referring to th o se mat eri als that arc so lu ble be yond a ce rtain ex te n t as "solu ble " an d to those th at do not dissolve to th at exte n t as "in so lu ble ." To get a better id ea o f t he amo u n ts th at will actua lly di ssolve in a g iven q ua n tity of water, we co uld d o a n ex pe rime n t or pe r fo rm a calcu latio n th at uses th e co ncept o f eq u ilib riu m (.... Chap te r 1R). 3.5
I
Ions and Mo lecules in Aqueous Solution 125
• Solubility Guidelines Observations such as t hose shown in Figure 3.10 were used to create the solubility guidelines. Note, however, that t hese are general guidelines and not rules followed under all circumstances. There are exceptions, but the guidelines are a good place to begin. See B. Blake, Journal of Chemical Education, Vol. 80, pp. 1348- 1350, 2003.
Figure 3. 10 lists broad g uid eli nes that help predi ct whe the r a particular io ni c compo un d is so lub le in wate r. For exam ple , sod iu m nitrat e , NaN 0 3 , co n ta ins both an alkali me tal catio n , Na ", an d the n itrate anio n, NO j - . Th e presence o f either of th ese io n s en sures that the co m po u nd is so luble in water. By contrast, calciu m hydro xid e is poorly so lub le in wate r. If a spoo nful o f so lid Ca( O H h is added to 100 rnl, of wate r, on ly 0. 17 g, o r 0.002 3 mo l, will d isso lve at 10 "C. Nearly all o f the Ca (O II ), remains as a so lid (Figu re 3. IOc) . Chemislry.o..Now~
•
Sign in at www.thomsonedu .com/Ioqtn and go to Chapte r 3 Contents to see Screen 3.7 for a tutorial and simulation on the solubility of ionic compounds in water.
SILVER COMPOUNOS SOLUBLE
COMPOUNOS
Almost aU salts of Na +, K+, NH"+ Salts of nitrate, N0 3 chlorate, Cl0 3perchlorate, CIa"acetate, CH 3C02Ag N0 3
AgCL
AgOH
(a) Nitrates aregenerally soluble, as are chlorides (except AgCL). Hydroxides are generally not soluble.
,-...._-
SULAOES
EXCEPTIONS Almost aUsalts of CI ". Br" , I" Fluorides of Mg 2+ , Ca 2+ , Sr2+ , Ba2 " . Pb2+
Salts containing FSalts of sulfate, SO/ -
INSOLUBLE
(NH 4)zS CdS
Sb zS3
PbS
(b) Sulfides aregenerally notsoluble
COMPOUNOS
Most salts of carbonate, CO/ phosphate, PO,,3oxalate, C20 "2chromate, CrO/ sulfide, S2-
Salts of NH" + and the alkali metal cations
Most metat hydroxides and oxides
Alkali metal hydroxides and Ba{O Hh
(exceptions include salts with NH4 + and Na +).
HYOROXIOES
EXCEPTIONS
Guidelines to predict the solubility of ionic compounds. If a compound contains one of the ions in the column on the left in the top chart, it is predicted to be at least moderately soluble in water. There are exceptions, which are noted at the right. Most ionic compounds form ed by the anions listed at the bottom of the chart are poorlysoluble (with the exception on compounds with NH 4 + and t he alkali metal cations). Active Figure 3.10
NaOH Ca(OH)2 Fe(OHh Ni(O H)z (c) Hydroxides aregenerally not soluble, except when the cation isa Group lA metal. 126
Chapter 3
I
Chemical Rea ctions
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• an interactive version of this figure accompanied by an exercise. menu to explore
Solubility Guidelines Problem Predict wheth er t he following ionic compounds are li kely to be water-sol uble. List th e ions pres-
ent in solution for soluble com pounds. (a) KCl
(c) F" O,
(b) MgCO,
(d) Cu (NO , ),
Strategy You must first recognize the cation and anion involved and then decide the probable water solubitity based on the guidelines outlined in Figure 3.10.
SoLution (a) KCl is composed of K+ and Cl- ions. The presence of either of these ions means that the compou nd is likely t o be solubLe in wat er. The solution contains K+ and t r ions dissolved in wate r. KCl(s) --> K'{aq) + Cl-(aq) (The solubility of KCl is about 35 9 in 100 mLof water at 20 °C. )
(b) Magnesium carbonate is com posed of Mg2+ and ( 0 32 - ions. Salts containing t he carbonate ion are usually insoluble, unless combined with an ion like Na + or NH4 + . Therefore, MgC0 3 is predicted to be insoluble in water. (Th e solubility of MgC03 is less t han 0.2 g/ 100 ml of water.) (c) Iron(III) oxide is composed of F e 3 ~ and 0 2 - ions. Oxides are soluble only when 0 2 - is combined with an alkali metat ion; Fe 3 ~ is a transition meta l ion, so Fe203is insoluble. (d) Copper(II) nitrate is composed of Cu2 +(aq) and N0 3-(aq) ions. Nitrat e salts are soluble, so t his compound dissolves in wate r, giving ions in solution as shown in the equation below Cu(NO, ),( s) --> Cu' +(aq)
+
2 NO," {aq)
EXERCISE 3.4 Solubility of Ionic Compounds
Predict whether each of the following ionic compounds is likely to be soluble in water. If it is soluble, write the formulas of t he ions present in aqueous solution.
(a) LiNO,
3.6
(b) CaCl,
(c) CuO
(d) NaCH,CO,
o
Precipitation Reactions
Module 6
A precipitation reaction produces a water-inso lub le so lid product , known as a precip itate. The reactants in suc h re actions are ge nerally water-so luble ion ic com pou nds . When th ese su bsta nces dissolve in wate r, they d issociate to g ive the ap pro p riate cations and an io ns. Ifthe catio n from o ne co mpou nd can fo rm an insoluble co mpo u nd with th e anio n from th e other co mpo un d in th e so lutio n, p recipi tatio n occu rs. As describe d earlie r, both silver nitrate and potassium chlo ride are water-so lu ble ionic co mpo u nds . \Vhen co mbined in wate r, they undergo an exchange reaction to produce inso luble silver chlo ride and so lu ble po tassium ni trate (Fig ure 3.7) . AgN O,(a q) + KCl(aq) -> AgC l(s) + KNO,(aq) Reactant s
+ NO, -( aq) + Cl -(a q)
Products
Ag '(a q)
Insoluble AgCl(s)
K+(aq)
K' (aq)
+ NO," (aq)
Predicting the Outcome of a Precipitation Reaction Man y co m binatio ns o f po sitive and negat ive ion s g ive in so luble su bstance s (see Figure 3. 10) . For exa m ple , th e so lu bility gu id el ines indicate th at mo st co m po un ds co n taining th e ch ro mate ion arc not so lu ble (alka li metal chro ma tcs and am mo n ium ch ro mate are excep tio ns) . Thus, we can pred ict that yellow, so lid lead (II ) Sign in at www.tho msonedu.comj login to downloadthe GoChemistry modulefor this section or go to www.ichapt ers.com to purchase modules.
12 7
FIGURE 3.11 Predpitation reactions. Many ionic compounds are insoluble in water. Guidelines for predicting the solubilities of ionic compounds are given in Figure 3.10.
IL
I
(a) Pb(N03h and K2Cr0 4 produce yellow, insoluble PbCr0 4 and soluble KN0 3•
$
~.
I
I I
l~
(b) Pb(NO,), and (NH,),S produce black, insoluble PbS and soluble NH 4N0 3.
(c) FeCl3 and NaOH produce red. insoluble Fe(O Hh and soluble NaCL
(d) Ag N0 3and K2Cr04 produce red, insoluble A9 2Cr04 and solubte KN0 3. See Exampte 3.3.
ch ro mate will precIp ita te whe n a wate r-soluble lead (Il ) co m po u n d is co m b in ed wit h a water-so lu b le ch ro ma te com pound (Figu re :\. 11a) .
Reactants Pb2+ (aq)
+ 2 NO ]- (aq)
2 K' {aq) + Cr0 4 l-(a q)
Products Insoluble PbCr04(s) 2 K' (aq) + 2 NO, "(a q)
Simi la rly, we know from th e so lu b ility guide lines tha t almo st all me tal su lfi d es are inso luble in wate r (Figure :\.1 1b ). If a so lution of a so lu b le me ta l co m poun d comes in co n tac t with a sou rce of su lfide ions, th e met al su lfide p reci p itates.
Reactants Pb2+(aq) + 2 NO ]"(eq) 2 NH 4 +(aq)
Black tongue. Pepto-Bismof has antidiarrheal, antibacterial, and antacid effects in the digestive tract. and has been used for over 100 years as an effective remedy. However, some people find t heir tongues blackened after taking t his over-the-counter medicine. The active ingredient in Pepto-Bismol is bismuth subsalicylate. (It also contains pepsin, zinc salts. oil of wintergreen. and salol. a compound related to aspirin.) The tongue blackening comes from the reaction of bismuth ions with traces of sulfide ions found in saliva to form black BilS]. The discoloration is harmless and Lasts only a few days. 128
Chapter 3
I
Chemica l Reactions
+ Sl- (aq)
Products Insoluble PbS(s) 2 NH 4 +(aq)
+ 2 NO ]- (aq)
In still a no th e r exam p le , th e so lu b ility gu ideli nes ind ica te that with th e exce ption of th e alkali me ta l cati o ns (an d BaH ), all metal ca tio ns fo rm inso luble h yd roxides. T h us , wate r-solu b le iron (IlI ) ch lo rid e an d sod iu m hydro xide react to g ive insoluble iro n (III) h ydrox id e (Figu res :t IOc a nd 3. 1Ic) . FeCl,(aq ) + 3 Na OH(aq) --> Fe(O H),(s) + 3 NaCl(aq) Reactants Fet ' {aq)
+ 3 Clfa q)
3 Na t feq}
+ 3 0 W(aq)
Products Insoluble Fe(OH )](s) 3 Ne '{aq )
+ 3 Clf eq)
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Writing the Equation for a Precipitation Reaction Problem Is an insoluble product formed when aqueous so lutions of pot assium chromate a nd silver nitrate a re mixed? If so, write t he bala nced equa tio n.
Strategy First decide what ions are formed in solution when the reactants dissoLve. Then use information
• Hair Coloring and Black Smokers The hair-darke ning reaction described on page 18 is a precipitation reaction and is much like that in "black smokers" and in the formation of Bi25 3 with Pepto-Bis mol.
in Figu re 3.10 to determi ne whether a catio n from on e reacta nt will combi ne with an anio n from t he
other reactant to form an insoluble com pound. SoLution Both reactants-AgNO] and K2Cr04 -are water-so luble. The io ns Ag+, NO] -, K ~, and (r O/- are
released i nto solution when th e compounds are dissolv ed.
AgNO,(s) --> Ag" (aq) K,CrO, (s) --> 2 K'{aq}
+ NO, -(aq) + CrO,' - (aq)
Here, Aq " could combine with crO/-, and K+ could combine with NO ]- . Based on the solubilityquidelines, we know t hat the former combination, Ag2Cr04, is an insoluble compound, whereas KN0 3 is soluble in water. Thus, the balanced equation for t he reaction of silver nitrate and potassium chromate is 2 AgNO ,(aq)
+ K,CrO, (aq) --> Ag ,CrO,(s) + 2 KNO ,(aq)
Comment This reaction is illustrated in Figure 3.11d.
EXERCISE 3.5 Precipitation Reactions In each of t he following cases, does a precipitation reaction occur when sotutions of the two water-soluble reactants are mixed? Give the formu la of any precipitate that forms, and write a balanced chemical equation for the precipitation reactions that occur. (a) Sodium carbonate and copper(II) chloride (b) Potassium carbonate and sodium nitrate (c) Nicket(II) chloride and potassium hydroxide
Net Ionic Equations \Ve have see n th at when aq ucous so lu tio ns o f silve r nitrat e a nd potassium ch lo rid e are mi xed , in sol ubl e silver chlorid e form s, leavin g potassium nitrat e in so lu tio n (sec Figu re 3.7). T he balan ced chem ical eq uatio n for thi s process is AgNO, (aq)
+
KCl(aq) -> AgCl(s)
+
KNO,(aq)
\Ve can represent this reactio n in ano the r way by writing an eq uatio n in which we show th at the so lu ble ionic com pou nds are p re sent in so lu tio n as di ssoci ated ions. An aqueous so lu tio n of silver nitrate contains Ag + and N0 3 - io ns, a nd an aq ue o us so lut ion o f po tassium ch lo ride co n ta ins K+ and CI- io ns. In th e p roducts , th e potassiu m nitrat e is present in so lu tio n as K+ and 1'\'0 3 - ion s. The silver ch loride, how ever, is in soluble and thus is not present in the so lu tio n as di sso ciated ion s. It is shown in th e eq uatio n by its entire fo rmul a, AgCI.
before reaction
after reaction
Thi s type o f eq uat io n is ca lled a complete ionic equation. The K+ and N O,I- ions are pr esent in so lu tio n before and after reaction and so appear o n both th e reactant and pr oduct side s of the co m ple te ionic eq ua tio n. Such ion s are oft e n called spectator ions becau se th ey do not part icip at e in the net reaction; th ey o nly "lo o k on " fro m th e sid el ine s. Little chem ical information is lost if th e eq ua tio n is writte n withou t th em , and so we can simplify th e equation to Ag +(aq) + Cl-(a q) -> AgCl(s) 3.6
I
Preci pitation Reactions 129
Problem Solving Tip 3.1
Writing Net Ionic Equations
Net ionic equations are common ly written for chemical reactions in aqueous solution because they describe the actual chemica l species involved in a reaction. To write net ionic equations, we must know which cornpounds exist as ions in solution.
prim arily as molecules. (See Section 3.7.) Insoluble salts such as ( aC0 3(s) or insoluble bases such as Mg(O Hh(s ) should not be written in ionic form, even though they a re ionic compounds. The best way to a pproach writing net ionic equations is to follow precisely a set of ste ps:
1. Strong acids, strong bases, and soluble
salts exist as ions in solution. Examples include the acids HCl and HN 03• a base such
as NaO H, and salts such as Na Cl and ( uCl z. 2. All other species should be represented by their complete formu las. Weak acids such as acetic acid (CH 3C02 H) exist in solutions
• Net Ionic Equations All chemical equations, including net ionic equations, must be balanced. The same number of atoms of each kind mustappear on both the product and reactant sides. In addition, the sum of positive and negative charges must be the same on both sides of the equation.
1. Write a complete, balanced equation.
Indicate the state of each substance (aq, s, C. g). 2. Next, rewrite the whole equation, writing a ll stro ng acids, strong bases, and soluble
salts as ions. (Consider only species labeled with an "(aq)" suffix in this step.) 3. Some ions may remain unchanged in t he reaction (the ions that appear in the equation bot h as reactants or products). These "spectator ions" are not part of the chemistry that is going on. You can canceLthem from each side ofthe equation. 4. Li ke molecular equations, net ionic equations must be balanced. The same number of atoms must a ppear on each side of the arrow, and the sum of the ion charges on the two sides must also be equal.
T he ba la n ce d equ atio n that results fro m leavin g out the spec tato r ions is th e n et ionic eq ua tion fo r th e reac tio n. Only the aqueous ions, insoluble com/murals, and weakor non electrolstes (wh ich can be so lu b le m o lecu lar com p o u n ds suc h as sugar, wea k acid s, wea k b ases, or ga ses) that participate in (J chemical reaction are included in the net ionic equation . Leaving o u t th e spectato r io ns do es n o t imply th at K+ a n d 1':0 3 - io n s a re u nimpo rta n t in the AgN 0 3 + KCI reaction . In d eed , Ag + io ns ca n no t ex ist alo ne in so lution ; a n egati ve ion , in th is case I\'0 3", must b e p resent to bala n ce th e po sitive ch a rge o f Ag+ . An y a n io n will do, however, as lo ng as it fo rms a wate r-so lu b le co mpound with Ag ": T h us, we co u ld h ave used Ag CI0 4 inste ad o f AgN 0 3. Sim ila rly, th ere must b e a positive io n prese nt to bal an ce th e negativ e charge of Cl' . In thi s case , the p ositive ion prcsen t is K+ in KCI, b u t we cou ld have used NaC I in stead o f KCI. T hc n et io nic equ atio n wo u ld h ave bee n the sa me. Fina lly, n o tice that th e re must always be a charge balance as well as a mass bala n ce in a bala n ce d che m ica l equ ation . T h us, in the Ag + + CI- ne t io n ic eq ua tio n, th e ca tio n a nd a nio n c h a rg es o n th c left ad d togeth c r to givc a n et c h a rge of ze ro , th e same as th e ze ro charge on Age l(s) o n th e righ t.
Chemistry.c\ .NOw
N
•
Sign in at www.thomsonedu.com/Ioqtn and go to Chapter 3 Contents to see Screen 3.10 for a tutorial on writi ng net ionic equations•
. Writing and Balancinq Net Ionic Equations ProbLem Write a balanced, net ionic equation for t he reaction of aqueous solutions of BaCl2 and Na2S04' Strategy Follow the strategy outlined in Problem Solving Tip 3.1.
Predpit ation reaction.The reaction of barium chloride and sodium sulfate produces insoluble barium sulfate and watersoluble sodiu mchloride.
130 Chapter 3
I
Chemica l Reactions
SoLution Step 1. First, notice that this is an exchange reaction. Thatis, the Ba2'" and Na " cations exchange anions (Cl" and 5042-) to give BaS04and NaCl. Now that the reactants and products are known , we can write an equation for the reaction . To balancethe equation, we place a 2 in front of t he NaCl. BaCl2 + Na 2S04 ~ BaS04 + 2 NaCl
Step 2. Decide on the solubility of each compound (Figure3.10). Compounds containing sodium ions are always water-soluble, and those containing chloride ions are almost alwayssoluble. Sulfate salts are also usually soluble, one important exception being BaS04' We can therefore write BaCI,(aq)
+ Na ,50,(aq ) --> Ba50,(s) + 2 NaCl(aq)
Step 3. Identify the ions in solution. AU soluble ionic compounds dissociate to form ions in aqueous sotution.
+ 2 CI- (aq) Na ,50,(s) --> 2 Na t faq) + 50/ -(aq) NaCI(s) --> Na t faq) + CI -(aq) BaClAs) --> Bav '{aq)
This results in the following complete ionic equation: BaH(aq)
+ 2 CI -(aq) + 2 Na t faq) + 50/ -(aq) --> Ba50,(s) + 2 Na '(aq) + 2 CI -(aq)
Step 4. Identify and eliminate the spectator ions (Na" and Cl -) to give the net ionic equation. Bav'{aq)
+ 50/-(aq)
--> Ba50,(s)
Comment Notice that t he sum of ion charges is the same on both sides of the equation. On the left, 2+ and 2- give zero; on the right, the charge on BaS0 4 is also zero.
EXERCISE 3 .6 Net
Io nic Equati ons
Write a balanced net ionic equation for each of the following reactions: (a) AlCl 3
+ Na lP04 --? AlP0 4 + NaCl (not balanced)
(b) Solutions ofiron(III) chloride and potassium hydroxide give iron(III) hydroxide and potassium chloride when combined. See Figure 3.10c. (c) Solutions of lead(II) nitrate and potassium chloride give tead(II) chloride and potassium nitrate when combined.
3.7 Adds and Bases Acids and bases are two important classes o f com poun ds. 'rou m ay already be famili ar with so me common properties o f acid s. T hey produce bubbles of CO 2 gas whe n ad ded to a metal carbo na te suc h as CaC 0 3 (Figure 3. 12a) , and th ey reac t with m an y me tals to produce hydro gen gas (II,) (Figure 3. 12b) . Althoug h tastin g su bsta nces is never don c in a che m istry lab o rat ory, yo u have pro ba bly ex pe rie nce d th e sour ta ste of aci ds suc h as ace tic aci d in vinegar and citric aci d (co m mo nly found in fr uits and ad d ed to can d ies an d soft d rinks) . Acid s and bases have some re late d properti es. Solutions o f ac ids o r bases, for example , can cha nge th e co lo rs of vege ta ble pigm e nts (Figu re 3. 12c) . You may have seen aci ds cha nge the co lo r of litmus, a dye d eri ved from ce rta in lich ens, from blue to re d . If an acid has mad e blu e litmus paper turn red , addin g a base reverses th e e ffec t, making th e litmus blue agai n. Thus, acids and bases see m to be opposites. A base ca n neutr ali ze th e effect of an ac id, and an ac id ca n neutrali ze the effec t of a base . See Tabl e 3.2 for a list o f co m mon acids a nd bases. Over the years, che m ists h ave exam ine d th e p rop ertie s, che mical struc tures, and reaction s of ac ids an d bases an d have p ropo sed differen t d e fin iti o ns o f th e term s "ac id" an d "base." In this sectio n , we sha ll exam ine th e two most co mmonly use d d efin iti ons, o ne p rop o sed by Svan te Arrhe ni us ( 1859 - 1927) an d ano t he r proposed by J ohann e s N . Bro n sted (1 87 9-1 9 47) an d Th om as M. Lowr y (18 74- 1936).
3.7
I
Adds and Bases 131
~
(a) Apieceof coral(mostly CaC0 3) dissolves in add to giveCO 2 gas.
More acidic
More basic
(b) Zinc reactswith (c) The juice of a red cabbageis normally blue-purple. On adding hydrochloric acidto acid,the juice becomes more red. Adding base produces a produce zincchloride and yellow color. hydrogen gas. FIGURE 3.12 Some properties of acids and bases. (a) Acids react readily with coral (CaC0 3) and other metal carbonates to produce gaseous CO z (and a salt). (b) Acids react with many metals to produce hydrogen gas (and a metal salt). (c) The colors of natural dyes, such as the juice from a red cabbage, are affected by acids and bases.
Acids and Bases: The Arrhenius Definition T he Swed ish che m ist Svan te Arrhen ius mad e a number o f important contri butions to che m istry, bu t he is pe rha ps best known fo r his stud ies of th e properties of so lutio ns o f salts, acids, and bases. In the late 1800 s, Arr hen ius p roposed th at these compo un ds di ssolve in wate r and ultimately fo rm io ns. This th eo ry o f e lectro lytes
TABLE 3.2
Common Adds and Bases
Oxalic acid H2C20 4
Ca rboxyl group
Acetic acid CH3C0 2H
• Weak Adds Common acids and bases are listed in Table 3.2. There are numerous other weak acids and bases, and many of these are natural substances. Oxalic and acetic acid are among them. Many of these natural acids contain COzH groups. (The H of this group is lost as W.) 132
Chapter
3
I
Chemical Reactions
Strong Acids (Strong Electrolytes)
Soluble Strong Bases
HCl (aq)
Hydrochloric acid
UOH
Lithium hydroxide
HBr (aq)
Hydrobromic acid
Na OH
Sodium hydroxide
HI (aq)
Hydroiodic acid
KOH
Pota ssium hydroxide
HN O,
Nitric acid
Ba(O H);
Barium hydroxide
uno,
Perchloric acid
H5 0 4
Sulfuric acid
Weak Acids (Weak Electrolytes)'
Weak Base (Weak Electrolyte)
H3 P0 4
Phosphoric acid
NH,
HZC O,
Carbonic acid
CH] C02H
Acetic acid
HZCl O4
Oxalic acid
HlC4H 4 0 6
Tartaric acid
H3C 6H 5 0 7
Citric acid
HC gH g 0 4
Aspirin
:1
W( aq) + Cl- (aq) • A base is a su bsta nce th at , whe n dis solved in wate r, in creases th e co nc en tration o f hydroxide ion s, OH - , in th e so lu tio n.
Na OH(s) --> Na+(aq)
+ OW(aq)
• The reac tio n o f an aci d and a base p roduces a salt and wate r. Because th e cha rac te ristic properties of an aci d are lost when a base is ad de d, and vice versa, acid-base reactions were logi cally described as resulting from th e co mbin ation of II + and OII - to form wate r.
HCl(aq) + NaOH(aq) --> NaCl(aq) + H,O(C) Arrhe n ius fu rth er proposed th at acid str en gth was rel ated to th e extent to which th e acid ionized. Some ac ids suc h as hydrochloric ac id (HC l) and nitric acid (I I NO ~d ion ize co mple te ly in wate r; th ey are str o ng e lec tro lytes, and we now ca ll them strong acids. Other acids suc h as ace tic ac id an d hydrofluoric acid are inc o mpletely ionized; th ey arc weak electro lytes and are weak acids. Weak acid s ex ist in so lu ti o n primarily as aci d mol ecul es, a nd onl y a fr action o f th ese mol ecul es ion ize to produce 11+(aq) ion s along with th e a p pro priate anio n in so lut ion . Wat er-soluble co m po u nds th at contain hydroxid e ions, suc h as so d iu m hydro xide (NaO Il) or po tassium hydro xid e (KO Il), are stro ng elect rol ytes and str o ng bases. Aq ueous ammonia, N H ~~( a q), is a wea k el ectrolyte. Even though it does not have an OIl - ion as part of its fo rmul a, it does produce ammonium ion s and h ydr o xid e ions from its re action with wate r and so is a base (Fig u re 3. 13) . The fact th at thi s is a weak electrolyte indicat es th at only a fra ction o f am mo nia molecules react with water to fo rm ion s; mo st of th e base remain s in so lut io n in mol ecul ar form . Altho ug h th e Arrhe n ius theory is still used to so me ex te n t and is interesting in an historical context , modern cOiKep ts of acid-base chem istry suc h as th e Brousted-. Lowry th eory have ga ine d p refe ren ce amo ng che m ists.
FIGURE 3.13 Ammonia,a weak electrolyte. Ammonia, NH], interacts with water to produce a very small number of NH 4 + and OH- ions per mote of ammonia molecules. (The nameon the bottle, ammonium hydroxide, is misleading . TIle solution consists almost entirely of NH ] molecules dissolved in water. It is better referred to as "aqueous ammonia.")
EXERCISE 3.7 Adds and Bases (a) What ions are produced when nitric acid dissolves in water? (b) Barium hydroxide is moderately soluble in water. What ions are produced when it dissolves in water?
Acids and Bases: The Bronsted-Lowry Definition In 1923, Bronstcd in Copen hagen (De n ma rk) and Lowry in Ca mb ridge (Eng land) independently sugges ted a ne w co nc ep t of acid and ba se behavior. They viewed aci ds and bases in terms of th e tr an sfer of a proton (I1+) from o ne species to an3.7
I
Acids and Bases
133
A Closer Look
The Hydronium lon-The H + Ion in Water
The H+ ion is a hydrogen atom that has lost its electron. Only the nucleus, a proton, remains. Because a proton is only about 1/ 100,000 as large as the average atom or ion, water molecules can approach closely, and the proton and water molecules are strongly attracted. In fact, the H+ ion in water is better represented as H30 +, called the hydronium ion . This ion is formed by
There will be insta nces when. for simplicity. we wi ll use H' {aq). However, we will usually use the H3 0+ symbol to represent the hydrogen ion in water in t his book. Thus, hydrochloric acid is better represented as a sc lution of H30 + and Cl-.
combin ing H+ and H20. Experiments also
show that other forms of the ion exist in water. one example being [H30(H 20 h] +.
+ hydroniumion H30 +(aq)
chloride ion W (aq)
When HCl ionizes in aqueous solution, it produces the hydronium ion, H30 +, and the chloride ion, u.
o th er, and th ey d escr ibed all ac id- base reactions in terms of equilibria. The Branstcd-Lowry theory ex pande d th e sco pe of the d efinition o f ac ids and bases and helped che m ists mak e prediction s of p roduct o r reac ta n t favo rability based o n aci d and base str engt h. \Ve will d escrib e thi s th eo ry here qualit atively; a more com p le te di scu ssion will be given in Cha p ter 17. The main conce pts of th e Brenstcd-Lowry th eo ry are th e foll owing: • An acid is a [mnon donor This is sim ila r to th e Arr he n ius d efin ition . • A base is a proton acalltor. This defini tion includes th e OH - ion , but it also b road en s th e n u mbe r and type o f bases. • A n acid-base reaction inuolues the transfer oj a [nnton Jrom an acid to ([ base to[orm a new acid and a new base. The reaction is written as an equilibrium reaction, and the equilibrium frWOH the ueaker acid and base. T his allows th e p rediction o f product- o r react ant-favored reaction s based o n aci d and base stre ng th .
Fro m th e poin t of view o f th e Brrsnsrcd-Lowry th eory, th e beh avio r o f ac ids suc h as H Cl o r CI-I :1COzII in wa te r is seen to in vol ve an ac id - base reaction . Bo th species (bo th Bron sted aci ds) d onat e a p roton to wate r (a Bron srcd base ) , forming II , O +(aq ) . H yd ro ch lo ri c aci d , II CI(aq ) , is a stro ng e le ct ro lyte be cause it io n ize s co m p le tel y in aq ueous so lu tion ; it is thus cl assified as a stro ng ac id. H),d rof!/tl chloride, a strong acid. ]00 % ionized. Eouil ilnium stronj.{l)' [auors produ cts.
HCI(aq)
+
hydrochloric acid st rong electrolyte = 100% ionized 134
Chapter 3
I
Chem ical Rea ctions
+
H, O(€)
water
hydronium ion
chloride ion
Sulfuric Acid
Chemical Perspectives
For many years, sulfuric acid has been the chemical produced in the largest quantity in the United States (and in many other industrialized countries). About 40-50 billion kilo - E grams (40-50 million metric tons) are made annually in the United States. The acid is so important to the economy of industrialized nations that some economists have said sulfuric acid production is a measure of a nation's industrial strength . Sulfuric acid is a colorless, syrupy liquid Sulfur. Much of the sulfur used in the u.s. is with a density of 1.84 q/rnt. and a boiling produced by the Frasch process. This works by point of 337 0 (, It has several desirable propinjecting superheated water into pockets of the erties that have led to its widespread use: it element deep in the earth. The sulfur is forced is generally less expensive to produce t han to the surface in the molten state by comother acids, is a strong acid, can be handled in stee l contai ners, react s readily with many organic compounds to produce useful prod-
pressed air. ucts, and reacts readily with lime (CaD), t he least expensive and most readi ly available base, to give calcium sulfate, a compound used to make wall board for t he constr uction indust ry. The first step in the indust rial preparation of sulfuric acid is combustion of sulfur in air to give sulfur dioxide.
...._-------_...
~ Asulfuric acid plant.
S(5) + O,(g) --> SO,(g)
This gas is then combined wit h more oxygen, in t he presence of a catalyst (a substance t hat speeds up a reaction), to give sulfur trioxide,
2 SO,(g) + O,(g) --> 2 SO,(g) which can give sulfuric acid when abso rbed in water.
SO,(g) + H,O(f ) --> H,SO, (aq) Currently, over two t hirds of th e production is used in the fertilizer indust ry. The remainder is used to make pigments, explosives, a lcohol, pulp and paper, and detergents, and is employed as a component in storage batteries.
Some products that require sulfuric acid for their manufacture or use.
In con tras t, CH 3C 0 2 II is a weak elect ro lyte , evide nce th at it is ionized to a small exten t in water and therefo re is a wea k acid . Acetic acid, a ueak acid, «
]00 % ionized. Equilibrium favors reactants.
CH,CO, H(aq) + H,O(f ) ,
H,O+(aq) + CH, CO,-(aq)
Weak Bransted acid
Bransted base
T he di ffe ren t e x ten t of io n izat ion for th ese two ac ids re la te s to th ei r acid st re ng ths . I Ivd ro chl o ric aci d is a stro ng acid, an d th e equ ilibriu m str o ng ly favors the p roduct s. In co n trast, ace tic acid is a weak aci d, and eq uilib riu m for its reactio n wit h water is react ant -favor ed . Su lfu ric acid , a diprouc acid (an acid ca pa ble of tra nsfe rrin g two 11+ ion s) , reacts with wat er in two ste ps . T he first ste p stro ng ly favo rs products, whe reas the secon d step is reacta nt-favor ed . ---+ H,O+(aq) Strong Acid : H,SO, (aq) + H,O(f ) NH,CI(aq)
ammonia
hydrochloric acid ammonium chloride
Notice that the reaction produces a salt, NH 4Cl, but not water. An H+ ion from the acid transfers to ammonia, a weak Brensted base, to give the ammonium ion. To write the net ionic equatio n, start with the fact that hydrochloric acid is a st rong acid and produces H,O+ and Cl" ions and that NH 4Cl is a soluble, ionic compound. On the ot her hand, ammonia is a weak base and so is predominantly present in t he solution as the molecular species, NH 3 •
NH ,(aq)
+ H,O"(aq ) + ([-(aq) --> NH , "(aq) + ([-(aq ) + H,O(f)
Eliminating the spectat or ion, Cl-, we have
NH ,(aq) + H,O" (aq) -> NH, " (aq) + H,O(f) Comment The net ionic equation shows that the important aspect of the reaction between the weak base ammonia and t he strong acid HCl is the t ransfer of an H+ ion from the acid to the NH 3 • Any st rong acid could be used here (HBr, HN0 3 , HCl0 4 , H2S 0 4 ) and the net ionic equation would be the same. EXERCISE 3.9
Acid-Base Reactions
Write the balanced, overall equation and the net ionic equat ion for the reaction of magnesium hydroxide wit h hydrochloric acid. (Hint: Think about the solubility guideline s.)
co, Oxides of NonmetaLs and MetaLs Eac h acid shown in Ta ble 3 .2 h as one or m ore H atoms in th e m olecular fo r mu la th at di sso ci at e in water to form H 30 + ions. Th ere are , h o weve r, le ss obvious co mpounds that fo rm a cid ic so lu tio ns . O xid e s o f n onmetal s, suc h as ca r bo n dioxid e and su lfu r tri oxide. h ave no H a to ms but re act with wate r to produce H 3 0 + ion s. Carbo n d iox ide, for exam p le , di ssol ve s in wa ter to a small ex tent, a n d so me of th e di ssol ved mole cule s re act with wate r to form th e weak acid, ca r bon ic a cid. This ac id th en ionize s to a sma ll e xtent to form the hydronium ion , H 30 +, a n d th e h ydrogen carbo nate (bicar bo na te) ion , JI CO j - .
• so,
CO, (g)
+
H,O(t )
H,CO, (aq)
+ H,O(t)
HCO, - (aq)
so,
H,cO, (aq) NO, Some common nonmetal oxides that form adds in water. 138
Chapter 3
I
Chem ical Reactions
+ H,O+(aq)
Th e II C0 3 - ion ca n also functio n as an acid, io n izing to prod uce II 3 0 + and the carbonate ion , C0 32- .
The se reactio ns are importan t in o u r enviro n me nt and in t he hum an bo d y. Ca r bon di o xid e is norm all y found in small a mou n ts in th e a tmosp he re, so rainwate r is a lways slig h tly ac idic. In th e human body, carbo n di o xide is di ssolved in body flui d s, whe re the H CO :I - and CO l - io n s perfo r m an importan t "b uffe ring" ac tion ( ~ Chap te r 18 ) . Oxides like CO 2 th at ca n react with wate r to produce H 30 + io ns are know n as acidic o xid es. O ther ac id ic ox ides in clud e t hose of su lfur and n itrogen . Fo r exam ple, su lfu r d iox ide, S0 2, fro m h um an and natural so u rces, can react with ox yge n to give su lfu r trioxide, S03, whic h th en re act s with water to form sulfuric acid. 2 50,(g) 50,(g)
Nitro gen dio xide , N0
2,
+
• Add Rain Oxides of sulfur and nitro-
gen are t he maj or source of t he acid in what is called add rain. These acidic oxides arise from the burning of fossil fuels such as coal and gasoline. I he gaseous oxides mix with water and ot her chemicals in the t roposphere, and the rain t hat falls is more acidic tha n if it contained only dissolved CO 2• When the rain falls on areas t hat cannot easily tolerate this greater-than-normal acidity, serious environmental problems can occur.
+ O,(g) --> 2 50,(g) H,O(f ) --> H,50, (aq)
react s with water to give ni tric and nitro us acid s.
2 NO,(g) + H,O(f ) --> HNO,(aq) + HN O,(aq) nitric acid
nitrous acid
Oxides of met al s are called basic oxides becau se th ey give ba sic so lu tio ns if th ey dissolve ap p reciab ly in wate r; Perh ap s the best exam p le is calc ium oxi de, Ca O , o ften ca lled lime, or quicklime; Alm os t 20 billi on kg of lim e is prod uced an nua lly in th e Un ite d Sta tes fo r use in th e met als and co ns truc tio n industri es, in sewage and po llutio n contro l, in water trea tme nt , and in agricu ltu re. T his me tal oxide react s with water to give calcium hyd ro xide , co m mo nly ca lled slaked lime. Altho ug h o nly sligh tly so lu ble in water (ahou t 0.2 g / IOO g H 2 0 at 10 ' c) , Ca(OH 12 is wid e ly used in industry as a base becau se it is in expen sive . CaO(s) + H,O(f ) --> Ca(O H) ,( s) lime
slaked lime
EXERCISE 3.10 Addie and Basie Oxides For each of th e following, indicat e whet her you expect an acidic or basic solutio n when the compound dissolves in wat er. Remem ber t hat compounds based on elements in the same group usually behave similarly.
(a) SeD,
3.8
(b) BaD
Gas-Forming Reactions
Seve ra l di ffe rent che m ical reac tio ns lead to gas fo rmatio n Crable 3.3) , but th e mo st com mon are those lead ing to CO 2 fo rmatio n . All me tal ca rbonates (and bicarbonates) reac t with acids to produce ca rbo nic aci d , II 2CO:1> whic h in turn d ecompo ses 3 .8
I
Gas-Forminq Reacti ons
139
TABLE 3.3
Gas-Forming Reactions
+ acid ---+ metal salt + (02(g) + Na ,CO,(aq) + 2 HCl(aq) --. 2 Na Cl(aq) + CO ,(g) + H,O(f) NaHCO ,(aq) + HCl(aq) --. NaCl(aq) + CO,(g ) + H,O(f)
Metal carbonate or bicarbonate
Metal sulfide
Na,5(aq)
+ acid
H20«(:)
---+ metal saLt + H25(g)
+ 2 HCl(aq) --. 2 NaCl(aq) + H,5(g )
+ acid --. metal salt + 50,(g) + H,O(f) Na,50 ,( aq) + 2 HCI(aq) --. 2 NaCI(aq) + 50,(g) + H,O(f)
Metal sulfite
Ammonium salt
NH,CI(aq)
+
strong base ---+ metal salt
+ NaOH (aq) --. NaCI(aq) + NH ,(g)
+ NH 3(g) + H20( i) + H,O(f)
FIGURE 3.15 Dissolving limestone
(calcium carbonate, (aC03 ) in vinegar. Notice the bubbles of CO 2 rising from the surface of the limestone.This reaction shows why vinegar can be used as a household cleaning agent. It can be used, for example, to clean the calcium carbonate deposited from hard water.
rapidly to ca rbo n di o xide a n d water. Fo r example, the reaction o f ca lciu m ca rbo na te a n d hydrochlori c acid is:
CaCO,( s) + 2 HCI(aq) -> CaCl, (aq) + H,CO,( aq) H,CO,(aq) -> H,O(f ) + CO,(g) Overall reaction:
CaCO,(s)
+ 2 HCl(aq) -> CaCl,(aq) +
H,O(f)
+ CO,( g)
If th e reaction is don e in a n o pe n beaker, mo st o f th e CO 2 gas bubbles o u t of th e so lu tio n. Ca lciu m carbon at e is a common res id ue from ha rd wate r in hom e heatin g systems a n d cooking ut ensils. Washing with vin egar is a go o d way to clean th e syste m or utensils because the insoluble calciu m ca rbo na te is turn ed into water- soluble calcium ace ta te in th e followin g gas-fo rm ing reaction (Figure 3. 15).
Wh at is the net ionic eq ua tion for thi s reaction ? Ace tic acid is a weak acid. Calciu m ca rbo na te is in soluble in wate r. Therefore, th e rea ctants are sim p ly CII 3C O,II(aq) a n d Ca C0 3(s). On the product s sid e , ca lciu m aceta te is wate rso lu b le a n d form s aqueous calcium and acetate ions. Water a n d ca rbo n dioxide arc molecular co m p ou n ds, so the n et ionic equation is
FIGURE 3.16 Muffins rise because of a gas-forming reaction. The acid and sodium bicarbonate in baking powder produce carbon dioxide gas. The acid used in many baking powders is Ca HP04, but tartaric acid and NaAl(S04h are also common. (Aqueous solutions containing the aluminum ion are acidic.) 140
Chapter
3
I
Chemical Reactions
Th ere are no specta to r ion s in this rea cti on . Ha ve you e ve r made biscuits or muffins? As you bake the dou gh, it rises in th e oven (Figu re 3. 16) . But what makes it rise? A gas-fo rm ing rea cti on occurs between a n acid a n d baki ng soda, so d iu m h ydro gen ca rbonate (b ica rbo nate of so da, N"a HC 0 3 ) . One acid used for thi s purpose is tartari c acid, a weak acid found in many foods. The n et ionic equation for a typical reacti on wo u ld be
H,C,H,O,( aq) tartaric acid
+
HCO, -(aq)
->
hydrogen carbonate ion
HC,H,O,"(aq) hydrogen tartrate ion
+ H,O(€) + CO, (g)
In dry bak ing powder, th e aci d an d Na IIC O :~ are kept apart by using sta rc h as a filler. When mi xe d in to th e mo ist batter, however, the aci d and sod ium hyd ro gen carbo na te diss o lve a nd come int o co ntact, Now they ca n react to p roduce CO 2 , ca using the doug h to rise . Chemislry.o..Now~
•
Sign in at www.thomsonedu .com/Ioqtn and go to Chapter 3 Contents to see Screen 3.14 for a tutorial identifying the type of reaction that will result from the mixing of solutions and to watch videos about four of the most important gases produced in reactions.
. Gas-Forming Reactions Problem Write a balanced equation for the reaction that occurs when nickel(lI) carbonate is treated with sulfuricacid. Strategy First, identifyt he reactants and write their formulas (here Ni C0 3 and H2S04 ) , Next, recognize this as a typical gas-forming reaction (Tab le 3.3) between a metal carbonate and an acid. The products are water, CO 2, and a salt. The anion of the salt is t he anion from the acid (50/-), and the cation is from the metal carbonate (NiH). SoLution The complete, balanced equation is NiCO,(s) + H,SO,(aq ) --> NiSO,(aq) + H,O(t) + CO,(g) EXERCISE
3.11 Gas-Forming Reactions
(a) Barium carbonate, BaC0 3 , is used in the brick, ceramic, glass, and chemica l manufacturing industries. Write a balanced equation that showswhat happens when barium carbonate is treated with nitric acid. Give the name of each of the reaction products. (b) Write a balanced equation for the reaction of ammonium sulfate with sodium hydroxide.
3 .9
Oxidation-Reduction Reactions
T he te rms "ox ida tio n" and "re d u ctio n" co rnc from re act ions that have been known fo r ce ntu ries. An cien t civilizatio ns learn ed ho w to c hange me ta l o xid es and su lfides into the metal , t hat is, how to "red uce" o re to th e me tal. A mo d er n exa m ple is the re d uction of iron (I II) ox id e with carbo n mo no xide to g ive iron m et a l (Figu re 3. 17a) . Fe20 3 loses oxygen and is reduced.
I
1
Fe,O, (s)
+ 3 CO(g)
-----+
2 Fe(s)
+ 3 (O, (g)
J
I co is the reducing agent. It gains oxygen and is oxidized.
In this react ion , ca rbo n mo noxide is th e agc n t th at brings abou t the redu ctio n o f iro n ore 10 iron metal, so carbo n mono xid e is called the red uci ng age nt. Wh en Fe 20 :~ is reduced by ca rbon mon o xid e , o xygen is rem oved fro m th e iro n ore and ad d ed to the ca rbo n mon o xid e . T he ca rbo n mon oxid e, therefo re , is "oxid ized" by the ad d itio n of o xygen to g ive ca rbo n d ioxide . An)' [nncess in which oxygen is added to another substance is an oxidation. In the reaction of o xyge n with mag nesium , 3 .9
I
Oxidation-Reduction Reactions 141
FIGURE 3.17 Oxidation -reduction. (a) Iron ore, which is largely FezO " is reduced to metallic iron with carbon or carbon monoxide in a blast furnace, a process done on a massive scale. (b) Bu rning magnesium metal in air produces magnesium oxide.
j 0
~
~
~
~
i
s• ~
]
3
•
'I
~
,
(b)
(a)
for ex am ple (see Fig ure 3.17b) , oxygen is the o xid izing age nt be cause it is th e age n t respo nsib le for th e ox id atio n o f m agnesium . Mg combines wit h oxygen a nd is oxidized.
I
2 Mg(s) + O, (g)
1 ~
2 MgO(s)
i
02 is t he oxidizing agent
Oxidation-Reduction Reactions and Electron Transfer No t all red ox react ion s in volve oxygen , bu t all oxidation an d reduction reactions can be accounted for II)' considn i ng them to occur by means of (J transfev of electrons betioeen su bstances. Wh en a su bsta nce acc ep ts el ectrons , it is said to be red uc ed be cause there is a red uctio n in th e n u merical value o f th e charge o n an ato m o f th e su bstanc e. In the net ionic eq ua tio n fo r th e reaction o f a silver salt with co p pc r metal , positively charged Ag " ions accep t e lec tro ns fro m cop pe r metal and are re d uce d to un ch arged silver ato ms (Figu re 3. 18). Ag+ ionsaccept electrons fromCu and are reduced to Ag. Ag + is the oxidizing agent. Ag+ (aq) + e" ~ Ag(s) I ~
2 Ag+(aq)
+ Cu(s) ------+ 2 Ag (s) + Cu2+ (aq)
I t Cu donates electrons to Ag+ andis oxidized to Cu2+. Cu is the reducing agent. Cu(s) ~ Cuz+(aq) + 2 e"
Because co p per metal su p plies the e lectron s an d ca uses Ag " io ns to be reduced , Cu is th e red uci ng age n t. \Vhe n a su bstance loses electrons, the n u meri cal valu e o f the charge o n an atom of t he su bstan ce in cr eases. T he su bstance is said to have been o xid ized . In ou r exa mple , co p pe r met al releases el ectron s on go ing to Cu 2 + , so th e metal is ox id ized . For thi s to hap pe n , so m eth ing must be availab le to acce p t ih e el ectron s from co p142
Chapter 3
I
Chemical Reactio ns
\
Pure copper wire
Copper wire in dilute AgN0 3 solution; after several hours
Blue color due to Cu 2 + ions formed in redox reaction
Silver crystals formed after several weeks
FIGURE 3.18 The oxidation of copper metal by silver ions. Aclean piece of copper wire is placedin a solution of silver nitrate, AgNO]. Over time, the copper reduces Ag + ions, forming silver crystals, and the copper metal is oxidized to copper ions, Cu 2+ . The blue color of the solution is due to the presenceof aqueous copper(II) ions. (Sign in to ChemistryNow and Screen 3.15, Redox Reactions and Electron Transfer, to watch a videoof the reaction.)
per. In th is case , Ag + is th e el ectron acce p tor, and its ch arge is reduced to zero in silver met al. Therefore , Ag + is th e "age n t" th at ca us es Cu metal to be o xidiz ed ; that is, Ag" is the oxidizing agent. In every oxidation-reduction reaction, one reactant is reduced (and is th erefore the oxidi zing agent) and one reactant is oxidi zed (and is th e re fore th e reducin g agen t). \Ve ca n show th is by d ivid ing th e general re d ox reaction X + Y ---7 XJl+ + Y'": into two parts o r Iw(frrYlctiol/..\: Half Reaction
ElectronTransfer
Result
Xtransfers electrons to Y
Xis oxidized to X"+. Xis the reducing agent
Yaccepts electron from X
Yis reduced to Y"-. Yis the oxidizing agent
• Balancing Equations for Redox Reactions The notion that a redox reaction can be divided into an oxidizing portion and a reducing portion will lead us to a met hod of balancing more complex equations for redox reactions described in Chapter 20.
In th e reaction of magnesium and oxygen , O 2 is reduced because it ga ins clcctroll s (fo u r el ect ro ns pe r molecul e) on go ing to two o xide io us. Thus, O 2 is th e oxidizin g age n t. Mg releases 2 e- per atom. Mg is oxidized to Mg2+ and is the reducing agent . ~
I
2 Mg(s) + O,(g) I
----->
2 MgO (s) t
O2 gains 4 e per molecule to form 2 0 2- . 02 is reduced and is the oxidizingagent.
In th e same reaction , mag nesium is th e reducing age n t becau se it releases two elec tro ns per at om o n being ox id ized to the ~I g2 + ion (and so two :vIg ato ms a rc re qu ired to su p p ly th e four el ectr on s re qu ire d by on e O 2 mol ecul e ). All redo x reac tio ns ca n be analyzed in a sim ilar man ner. 3.9
• Chemical Safety and Redox Reactions A strong oxidizing agent and a strong reducing agent may react violently. For this reason, it would not be a good idea to store a strong oxidizing agent next to a strong reducing agent. Certain chem icals are often stored separately from the bulk of the other chemicals. Examples of chemicals that might be stored in their own separate areas are strong oxidizing agents, acids, water-reactive chemicals, and highly flammable materials.
I
Oxidation-Reduction Reactions
143
T he o bservatio ns outlined so fa r lead to several im portant co nclusions: • If one substance is oxidized, another subs ta nce in th e sa me reaction must be reduced . For th is reason , such reactio ns are ca lled ox ida tion- re duc tio n reactions, o r redox reactions fo r sho rt. • The red ucing agen t is itse lf ox id ized, and th e oxidizing agen t is re d uced. • O xid ation is th e o p posi te of red uctio n , For examp le , th e rem oval of ox yge n is reductio n , and th e ad di tio n of oxyge n is oxid ati o n . T he ga in of electrons is red uct ion , and th e lo ss o f electro ns is oxidation .
Oxidation Numbers H ow call you tell an oxid ation-reduction react io n whe n you see o ne? Ho w can you tell whic h su bstance has ga ine d or lo st electro ns an d so d ecid e which su bstance is th e oxidizing (o r reducin g ) age n t? Sometimes it is obvious. Fo r example , if an u ncombined elemen t becomes part o f a co m pou nd (!\.fg becomes part o f (\1g0 , for exam p le), th e reaction is d efinitely a redox process. If it' s not o bvio us, then th e answe r is to look for a change in the oxidation number of an element in the coune of the reaction. T he oxidation number of an atom in a m olecul e or io n is defined as th e charge an atom has, orapl)('ars to have, as d et ermin ed by th e following gu id el ines fo r assign ing ox ida tion n u mbers. • Writing Charges on Ions Conventionally, charges on ions are written as (number, sign ), whereas oxidation numbers are written as (sig n, number). For exa mp le, the oxidation number of the Cu 2 + ion is + 2 and its charge is 2 +.
• Peroxides In peroxides, the oxidation num ber of oxygen is - 1. Fo r exam ple, in hydrogen peroxide (H20 2) , each hydrogen atom has an oxidation nu mbe r of + 1. To balance t his, each oxygen must have an oxidation nu mber of - 1. A 3% aqueous solution of H20 2 is sometimes used as an antiseptic.
1. Each atom in a pure elemen t has an oxida tion number o f zero. T he oxidation nu mber of Cu ill me tallic co pper is 0, and it is 0 fo r eac h atom in 12 and Sil. 2. For monatomic ions, the oxidation number is equal to th e charge on the ion. You kn ow that magn esium for ms ions with a 2 + charge (T\fg:H); the o xid ation nu m ber o f magn esium in thi s ion is therefore + 2. 3. 'When combined with another element, fluorine always has an oxidation number of -1. 4. The oxidation number of 0 is -2 in most com p ou n ds. The exce ptions to th is ru le occu r a) when oxyge n is co m bined with flu orine (whe re oxyge n takes o n a positive ox ida tio n n umber ) b) in compounds ca lled perox ides (suc h as Na 20 2) and su pc rox id cs (such as KO ~J in whic h o xygen has an o xid ation n u m ber of - I and - 1/ 2, resp ectively 5. CI, Br, and I have oxidation numbers of -1 in compounds, except when com b in e d with o xyge n and fluorin e . Th is m ean s that CI has an oxida tio n
A Closer Look
Are Oxidation Numbers " R e a l " ?
Do oxidation numbers reflect the actual electric charge on an atom in a molecule or ion? With the exception of monatomic ions such as n- or Na ", the answer is no. Oxidation numbers assume that t he ato ms in a molecule are positive or negative ions, which is not true . For example, in H2 0, the H atoms are not H+ ions, and the a atoms are not 0 2 - ions. Th is is not to say, however, that atoms in molecules do not bear an electric charge of any kind. Ca teulations on water
indicate the atom has a charge of about - 0.4 (or 40 % of the electron charge) and the H atoms are each about + 0.2. So why use oxidation numbers? Oxidation numbers provide a way of dividing up the electrons among the at oms in a molecule or polyatomic ion. Because the distribution of electrons changes in a redox reaction, we use this method as a way to decide whether a redox reaction has occurred and to distinguish the oxidizing and reducing agents.
144
Chapter 3
I
Chemical Rea ctions
°
Charge on 0 atom = - 0.4
Charge on each Hatom = +0.2
n umber of - I in NaCI (in whi ch Na is + I , as p re di ct e d by the fact t ha t it is an clement of Gro u p IA) . In th e ion CIO -, however, th e CI ato m h as a n o xidatio n n umb e r o f + I (and 0 has an oxidatio n nu mbe r of - 2; see gu idelin e 4) . 6. The oxidation number of H is + 1 in most compounds. T he key exce p tio n to this guideli ne occurs when I I fo rm s a bin ary compound with a metal. In suc h cases , the metal fo rm s a pos itive io n , and I I becom es a hydri d e io n , 1-1 - . T h us , in Ca lL, th e ox idatio n num ber of Ca is + 2 (eq ual to th e gro u p n um ber ) , and that o fH is - I. 7. The algeb raic sum of the oxidation numbers for the atoms in a neutral compound must be zero; in a polyatomic io n, the sum must be equal to th e ion charge . Fo r e xamp le, in II CIO -t the II at om is assign ed + I, and each 0 ato m is assigned - 2. Th is means the CI atom must be + 7.
• Why Use Oxidation Numbers? The reason for learning about oxidation numbers at this point is to be able to identify which reactions are oxidation-reduction processes and to identify the oxidizing agent and the reduci ng agent in a reaction. We return to a more detailed discussion of redox reactions in Chapter 20.
ChemislrY;J.Now~
•
Sign in at www.t homsonedu.comjlogin and go to Chapter 3 Contents to see Screen 3.16 for exercises and a tutorial on oxidation numbers.
Determining Oxidation Numbers Problem Determine t he oxidation number of the indicated element in each of the followi ng com pounds or ions: (a) aluminum in aluminum oxide, Al2 03 (b) phosphorus in phosphoricacid, H3P04 (c) sulfur in t he sulfate ion, 50/(d) each Er atom in t he dichromate ion, ( r20 / Strategy Follow the guidelines in t he text paying particular atte ntion to guidelines 4, 6, and 7. SoLution
°
(a) Al2 0 3 is a neutral com pound. Assuming that has its usual oxidation number of - 2, we can solve t he followi ng algebraic equation for the oxidation number of aluminum. Net charge on Al2 0 3
o~
2(x)
= sum of oxidation numbers for two Al atoms + three Datoms
+ 3( - 2)
O ~ 2x +( - 6)
+6
=
2x
x= +3 The oxidation number of Al must be
+ 3, in agreement with its position in the periodictable.
(b) H3 P04 has an overall charge of O. If each of the oxygen atoms has an oxidation number of - 2 and each of t he Hatoms is + 1, t hen we can determine the oxidation num ber of phosphorus as follows: Net charge on H3P0 4
o~
= sum of oxidation numbers for t hree H atoms + one P atom + four 0 atoms
3(+ 1) + (x) + 4( -2)
x = +5 The oxidation number of phosphorus in t his com pound is therefore + 5. (c) The sulfate ion, 504 2 - , has an overall charge of 2- . Oxygen is assigned its usual oxidation number of - 2. Net charge on 50/ 2-
x
~
(x)
+
= sum of oxidation numbers of one 5 atom + 40 atoms
4(-2 )
= +6
The sulfur in t his ion has an oxidation number of + 6. 3.9
I
Oxidation-Reduction Reactions 145
(d) The net charge on the Cr2 0/ - ion is 2-. Oxygen is assigned its usual oxidation number of - 2.
= sum of oxidatio n numbers for two Cr atoms + seven 0 atoms
Net charge on Cr20/ 2-
~
2-
~
2(x) 2x
+
7( - 2)
+ (-14)
= 2x = +6
12
x
The oxidation number of each chromium in this polyatomic ion is + 6. EXERCISE 3.12
Determining Oxidation Numbers
Assign an oxidation number t o the underlined atom in each ion or molecule.
(a)
~, o ,
(d) tlO, +
Recognizing Oxidation-Reduction Reactions You can always tell if a reactio n involves ox idatio n and red uction by assessin g the oxid ati on nu mbe r of each ele me n t and noting whe ther any of th ese numbers ch an ge in th e cou rse of the reaction . In many cases, how ever, th is will not be ne cessary. Fo r exa m ple, it will be o bvio us that a red ox react ion has oc cu rred if an u nco mb ined e le me n t is co nverted to a co m po u n d or if a we ll-known o xidi zin g o r red uc ing age n t is in vol ved (Ta ble 3.4) . Like ox ygen (0 2), th e halogen s (1"2, C12, Br2' and 12) are ox id izing agen ts in thei r reac tio ns with m et als and no nm e tal s. An exa m ple is the re ac tio n of chlorine with sod iu m metal (see Figure 1.4) . Na releases 1 e" per atom. Oxidation number increases. Na is oxidized to Na + and is the reducing agent. 1
2 Na(s)
+
1 Cl2( g)
~
2 NaC l(s)
J
I
(l2 gains 2 e" per molecule. Oxidation number decreases by 1 per Cl. Cl2 is reduced to Ct" and is the oxidizing agent.
TABLE 3.4
Copper metal oxidized to green Cu(N0 3h Active Figure 3.19 The reaction of copper wit h nitric acid. Copper (a reducing agent) reacts vigorously with concentrated nitric acid (an oxidizing agent) to give the brown gas N0 2 and a deep green solution of copper(lI) nitrate. N
ChemislrY! J.Now Sign in at
• www.t homsonedu.com/ login and go
to t he Chapter Contents menu to explore an interactive version of this figure accompanied by an exercise. 146
Chapter 3
I
Chem ical Reactions
Common Oxidizing and Reducing Agents
Oxidizing Agent
Reaction Product
Reducing Agent
Reaction Product
02' oxygen
02- ,
H2, hydrogen
H'' {aq). hydrogen ion or H combined in H20 or othe r molecule
M, metals such as Na, K, Fe, and AL
M'J +, metal ions such as Na " , K+, Fe2 + or Fe3 +, and Al3+
C. carbon (used to
CO and CO 2
oxide ion or
°combined in H 0 2
Ha logen, F2 , Cl2< Br2' or 12
HNO " nitric acid
Ha lide ion, F-, Br- ,orl-
cr .
Nitrogen oxides* such as NO and N02
dichromate ion
( r H , chromium(III) ion (in acid solution)
Mn0 4 - , permanganate ion
Mn 2 " , manganese(II) ion (in acid solution)
Cr20/ - ,
* ~()
is p rod uced with d ilute
H ~ () :h
whereas
reduce metal oxides)
~ () 2
is a product of concentrated acid .
A chlo rine molecul e ends up as two CI- ion s, having acq uire d two electro ns (fr o m two Na ato ms). T hus, th e oxid ation nu mber of eac h CI a to m has decrea sed fro m o to - I. Th is means Cl:! has been re d uced , and th erefo re it is th e o xidizi ng agen t. Figure 3. 19 illustrates th e che mistry of anot her exce llent ox id izing age n t, nitric acid , H N0 3 . Here, co p pe r met al is oxidi zed to g ive co p per (lI ) nitrat e , and th e nitrat e ion is reduced to th e bro wn gas N0 2 • The ne t ion ic equatio n for th e reaction is Oxidation number of Cu changes from 0 to + 2. Cu is oxidized to Cu H and is the reducing agent.
1
1
Cu ts) + 2 NO, - (aq)
+ 4 H,O+(aq)
------+
Cu2+ (aq)
+ 2 NO, (g) + 6 H,O(€)
1
1
N in N03- changes from + 5 to +4 in N02. N03is reduced to N02 and is the oxidizing agent.
Nitro gen has been reduced fro m + 5 (in th e N0 3 - ion ) to + 4 (in NO:!); th erefore , the nitrate io n in aci d so lutio n is the oxid izing ag ent. Copper metal is the re d uci ng agent; ea ch met al atom has g ive n up two el ectrons to produ ce th e Cu H ion . In th e reacti ons o f sodi u m with ch lo rine and co p pe r with nitric acid, th e metals are oxi di zed . This is typical o f metal s. In ye t ano t he r exam ple o f th is, alu m in u m metal, a goo d reducing age n t, is capable o f redu cin g iron (lIl ) o xide to iron met al in a re action ca lled th e thermite reaction (Figu re 3.20 ) .
FIGURE 3.20 Thermite reaction. Here. FezO] is reduced by aluminum metal
to produceiron meta l and aluminum oxide.
Fe,O,(s) + 2 Al(s) ---> 2 Fe(C) + AI,O, (s) oxidizing agent
reducing agent
Suc h a large quant ity o f energy is evolved as heat in th e reaction that the iro n is pro d uced in th e m olt en state . Ta bles 3.4 a nd :1.5 may help yo u o rga nize yo u r thinkin g as you loo k fo r o xid ation-reductio n reactions and use th ei r term ino logy. ChemislrY.ci.Now~
•
Sign in at www.thomsonedu .comjlogin and go to Chapter 3 Co ntents to see Screen 3.17 for an exercise on redox reaction.
, Oxidation-Reduction Reaction Problem For the reaction of iron(II) ion with permanganate ion in aqueous acid. 5 feH(aq)
+ MnO, - (aq) + 8 H,O"(a q) -7 5 fe H (aq) + MnH(aq) + 12 H,O(t)
decide which at oms are undergoing a change in oxidation number, and identify t he oxidizing and reducing agents.
TABLE
3.5
- --l--
Fe2-(aq) reducing agent
L---=::::...J
Recognizing Oxidation-Reduction Reactions Oxidation
Reduction
In terms of oxidation number
Increase in oxidation number of an at om
Decrease in oxidation number of an atom
In terms of electrons
l oss of electrons by an atom
Gain of electro ns by an at om
In terms of oxygen
Gain of one or more 0 at oms
loss of one or more 0 at oms
Thereaction of iron (II) ion and permanganate ion. The reaction of purple permanganate ion (Mn04 " , the oxidizing agent) with the iron(II) ion (FeH • the redu cing agent) in acidified aqueous solution gives the nearly colorless ma nganese(ll) ion (M n2 +) and the iron(III) ion (Fe H ) . 3.9
I
Oxidation- Reductio n Reactio ns 147
Strategy Determine the oxidation numbers of t he atoms in each ion or molecule involved in the reaction. Decide which atoms have increased in oxidation number (oxidation) and which have decreased in oxidation number (reduction). Solution The Mn oxidation number in Mn0 4- is + 7, and it decreases to + 2 in the product, the Mn2+ ion. Thus, the Mn04- ion has been reduced and is the oxidizing agent (see Table 3.4). 5 Fe' +(aq) +2
+ MnO,-(aq) + 8 H,O+(aq) --> 5 Fe' +(aq) + Mn' +(aq) + 12 H,O(f ) + 7, - 2
+ 1, - 2
+3
+2
+ 1, - 2
The oxidation number of iron has increased from + 2 to + 3, so each Fe2+ ion has lost one electron upon being oxidized to Fe3 + (see Table 3.5). This means the Fe2+ ion is the reducing agent. Comment If one of the reactants in a redox reaction is a simple substance (here Fe2+), it is usually obvious whether its oxidation number has increased or decreased. Once a species has been established as having been reduced (or oxidized), you know another species has been oxidized (or reduced). It is also helpfu l to recognize common oxidizing and reducing agents (Ta ble 3.4).
EXERCISE 3.13 Oxidation-Reduction Reactions
FIGURE 3.21 Th e redox reaction of ethanol and dichromate ion is the basis of the test used in a Breathalyzer. When ethanol, an alcohol, is poured into a solution of orange-red dich romate ion, it red uces the dichromate ion to green chromium(III) ion. The bottom photois a breath tester that can be purchased in groce ry or drug stores. See Exercise 3.13.
Case Study
The followi ng reaction occurs in a device for testing t he breath for the presence of ethanol. Identifythe oxidizing and reducing agents, the substance oxidized, and the substance reduced (Figure 3.21).
3 CH,CH,OH(aq) + 2 Cr,O,' - (aq) + 16 H,O+(aq) ethanol
3 CH,CO, H(aq) acetic acid
Ag (s) --> Ag +(aq) + e: Th is is a simple electrolysis procedure. (More about that in Chapter 20.) And 100 quadrillion silver ions? This is 100 X 1015 ions. Do silver ions act as a bacteriocide? There is plenty of medical evidence for this property. In fact, when you were born, the physician or nurse may have put drops of a very dilute silver nit rate solution in your eyes to treat neonatal conjunctivitis. And severely burned patients are treated with silver sulfadiazine (Cl0HgAgN402S) to prevent bacterial or fungal infections.
Chapter
3
+ 4 Cr3+ (aq) + 27 H,O(C) chromium(III) ion; green
Killing Bacteria with Silver
We recently read about a new washing machine that injects silver ions into the wash water. t he purpose being to kill bacteria in the wash wate r. The advertisement to ld us t hat 100 quadrillion silver ions are injected. How does this machine work? Is 100 quadrillion silver ions a lot? Does the silver kill bacteria? The washing machine works by using electrical energy to oxidize silver metal to give silver ions.
148
-
dich romate ion; orange-red
I
Chem ical Rea ctions
Silver ions as a bacteriodde in dental floss.
The use of silver to prevent infections has a long histo ry. Phoenicians kept wine, water, and vinegar in silver vessels. Early settlers in America put silver coins into water barrels. And you might have been born with a "silver spoon in your mouth." Babies in wealth ier
homes, who were fed from silver spoons and used a silver pacifier, were found to be healt hier. The historical uses of silver carry over to modern society. One can buy many different kinds of silver-containing water purifiers for the home, and dental floss coated with silver nitrate is available. Although silver does have health benefits, beware of fraudulent claims. For example, consuming large amounts of "colloidal silver" (nanosized particles of silver suspended in water) is claimed to have health-g iving properties . One person who tried this, Stan Jones, ran for the U.S. Congress in 2002 and 2006. From consuming silver, he acquired argyria, a medically irreversibLe condition in which the skin tu rns a gray-blue color.
Ouestions: 1. How many moles of silver are used in a
wash cycle? 2. What mass of silver is used? Answers to these questions are in Appendix Q.
EXERCISE 3.14
Recognizing Oxidation-Reduction Reactions
Decide which of the following reactions are oxidatio n-reduction reactions. In eac h case , explai n
your choice, and identify t he oxidizing and reducing agents.
(a) Na OH(aq) + HNO,(aq) --7 NaN O,(aq) + H,O(t) (b) Cu(s) + Cl,(g) --7 CuCl,(s) (c) Na,CO,(aq) + 2 HClO,(aq) --7
co,(g) +
(d) 2 S,O,'-(aq) + r,(aq) --7 S,O.'- (aq)
3.10
H,O(t ) + + 2 r-( aq)
2
NaCIO,(aq)
Classifying Reactions in Aqueous Solution
On e go a l of th is c h a p ter h as be en to ex p lo re th e mo st co m mo n typ e s of reac tio ns tha t ca n o ccur in aq ueo us so lu tio n. T hi s helps yo u d ecid e , fo r ex am ple. th at a ga s-
fo rmi ng reactio n occu rs when a n Alka-Se ltzer tabl et (co n tain ing citric aci d an d NaHC 0 3 ) is d ro p ped in to water (Figu re 3.22) . H3C ,H sO , (aq) citric acid
+
HC0 3 - (aq)
~
FIGURE 3.22 A gas-forming reaction. An Atka-Seltzer tablet contains an acid (cit ric acid) and sodium hydrogen carbonate (NaH C03) , the reactants in a gas-forming reaction.
hydrogen carbonate ion
+
H,C,HsO, - (aq )
H,O(€)
+ CO, (g)
dihydrogen citrate ion
\Ve have exam ined four types o f reactio ns in aq ueo us so lutio n : precipitatio n re actio ns, acid-base reactions, gas-fo rm ing reactions, and oxid ati on-redu ctio n reaction s. Three of these fou r (preci pitat ion , acid -base , and g as-fo r ming ) fall into th e ca te go ry of exc hange re act io ns. Precipitation Reactions (see Figu re 3. 11): Ions com bi ne in so lution to for m an insoluble reac tio n prod uct. Overall Equation Pb(NO, ), (aq)
+ 2 KI (aq) --->
PbI, (s) + 2 KNO, (aq)
Net Ionic Equation Pb' +(aq) + 2I-(aq) ---> PbI, (s)
Acid-Base Reactions (see Fig u res 3.12 and 3. 14): Water is a product of many acidbase reactions, and th e catio n o f th e base and th e anio n o f th e acid form a salt. Overall Equ ation Jor the Reaction oJ a 5itmng Acid and a Stm ng Base HN O, (aq)
+ KOH (aq) ---> HOH(€) + KNO, (aq)
Net Ionic Equation Jor the Reaction oJ a Strong Acid and a Strong Base H,O+(aq)
+ OW (aq) -->
2 H,O(e)
Overall Equation Jor the Reaction oJ a Heak A cid and a
~)'trong
Base
CH ,CO,H(aq) + NaOH (aq) --> Na CH,CO, (aq) + HOH(e) 3.10
I
Classifyi ng Reactions in Aqueous Solution 149
Net Ionic Equation f or the Reaction of a lleak Acid and a Strong Base CH ,CO, H(aq) + OW (aq) -> CH,CO, -(aq) + H,O(f )
Gas-Forming Reactions (see Figu re s 3.15 and 3.22) : T he mo st co m mo n examp les involve me ta l ca r bo na tes and aci d s, but o the rs ex ist (see Table 3.3) . One pro d uct with a met al ca rbo nate is always carbo n ic acid, 1I 2CO:h mo st o f which d ecom pos es to 1120 and CO 2, Ca rbo n di oxid e is th e gas in the b ubbl es yo u see durin g these reacti on s. Ouemll Equation: CuCO, (s) + 2 HNO, (aq) -> Cu(NO,) , (aq) + CO, (g) + H,O(f )
Net Ionic Equation CuCO, (s) + 2 H, O+(aq) -> Cu' +(aq) + CO,( g) + 3 H,O(f )
Oxidation-Reduction Reactions (see Fig u re 3. 18) : T hese reaction s are not io n exchange re actions. Rather, el ectro ns arc tr an sferred fro m o ne material to ano the r. Overall Equation Cu(s) + 2 AgNO )(aq) -> Cu (NO, ),( aq) + 2 Ag (s)
Net Ionic Equation Cu(s) + 2 Ag+(aq) -> Cu' +(aq) + 2 Ag(s)
T hese fou r types o f re action s are usu ally easy to recogn ize, bu t kee p in mi nd tha t a reac tio n may fa ll in to more than o n e ca teg ory. For exam p le, barium h yd roxide react s read ily wit h sulfu r ic acid to give ba rium su lfate and water, a react ion th at is bo th a p recip itatio n and a n acid - base re actio n. Ba(OH),(aq ) + H,SO, (aq) -> BaSO, (s) + 2 H,O(f )
ChemiSlrY- MgO (s) + Fe fs} (b ) AICI3 (s) + NaO H( a q) ---> AI(O H ),( s) + NaC l (aq ) (c) NaN 0 3 (s) + II ,SO, (f) ---> Na ,SO ,( s) + Il NCJ,( t ) (d) NiCO , (s) + Il NO, (aq ) ---> Ni (N O ,),(aq) + CO, (g) + Il ,O ( f:) 6. Balan ce th e fol lowing e quations , a nd name ea ch reacta nt and product: (a) SF, (g) + H ,O (t ) ---> SO , (g ) + HF (t ) (b) N H 3(a q) + O , (aq ) ---> NO(g) + H,O (t ) (c) RF, (g ) + Il ,O (t ) ---> II F (a 'l ) + 1I 3 RO , (aq ) Chemical Equilibrium (Ser Sections 3.3 and 3.5. ) 7. Eq ual am o un ts of two aci ds-IICI an d II CO .J I (fo rm ic acid )- are placed in so lu tion. \ Vhe n eq u ilibrium has be en ac h ieved , the HCl so lu tio n has a much g rea ter electr ical co nd uc tivity th an th e HCO zH so lu tio n. Which reactio n is m o re product favo red at equilib rium ? II CI(aq ) II CO,Il (aq )
+
Il ,O (t ) ;:=::Z Il ,o +(aq )
+ C1 -( aq )
+ Il , O (t ) ;:=::Z II ,O +(a'l ) + II C O , - (a'l )
8. Equal amo un ts of two co m po u nds, Ag Rr and I13P O ", are pla ced in so lu tion . When eq u ilib riu m h as be en ac h ieved , th e 113 pn , sol u tio n h as a greate r electrical con d uc tivity th a n th e Ag Rr so lu tion (wh ich is alm os t a non ele ct rol yte) . Which re act io n is m ore produ ct favored a t e quili bri um?
+ Rr - (a'l ) + II , O (f:) ;:=::Z II , O +(a q ) + II ,P0 4- (aq )
Ag Rr (s) ;:=::Z Ag +(a q ) Il ,P0 4(aq )
Ions and Molecule s in Aqueous Solution (Sf'f' Exrrdse 3.3, F,xnm!JIR 3.2, and ChnnisfY)Nml! Sorens 3.5-3. 7.) 9. • ' Vha t is an elec tr o lyte? l Iow can yo u d iffere n tiat e expcrimen tally between a weak ele ctrolyte a nd a str ong elec tro lyte? G ive a n exam ple of e ac h.
10. Name two acids th at are strong elect ro lyte s and o ne ac id that is a weak electro lyte , Na me two hascs that arc strong ele ct ro lytes an d on e base that is a weak ele ctrolyte . II. • Whic h co m po u nd or com po un ds in e ac h of th e following grou ps is (are ) e xpec ted to be solu ble in wa ter? (a) CuO , CuCI" FeC O , (b) Ag l , Ag, I'O " AgN O , (c) K,CO" KI, K~I IlO "
12. Whi ch co m po u n d or co m po un ds in eac h of th e foll owing (a) (b) (c)
g ro u ps is (a re) ex pec te d to be so lu ble in wa ter? RaSO" Ra (NO, )" RaCO, Na 2S ? (b) Ca (N O , ),(aq ) + KF(aq ) ---> ? (c) Ca (N O , ),(a'l) + Na ,C, 0 4(a q ) ---> ? Acids and Bases and Their Reactions (See Exercise 3.8, Example 3.5, Exercise 3./0, and Chemistryblmu Screens 3. / / - 3.13.)
21. • Write a bal an ced equa tio n fo r th e ionizatio n of n itric ac id in wa te r. 22. Writ e a bala nced eq uation fo r th e ioni zatio n of perchl oric ac id in wa ter. 23. O xali c acid , H 2C 20 1 , wh ich is fo u n d in certa in pl ants, ca n provid e two hyd ro n ium ions in wa te r. Wri te bala n ced e q uatio ns (like those fo r sulfuric aci d on page 135 ) to sh ow ho w o xal ic acid can su p ply o n e a nd then a sec on d H:l () + ion .
24. Phosphoric ac id ca n su p ply one , two, o r th re e H:j() + ion s in aq ueo us sol u tion . Write balan ced eq uations (li ke th o se fo r su lfu ric acid on page 135 ) to show th is successive loss of h yd ro gen ion s. 25. Write a hal a uccd equat io n fo r reaction of th e basi c o xid e , m ag nesiu m ox id e , with wa ter.
26, • Write a bal a nced equati on fo r the react io n of su lfur triox ide with water. 27, • Co mple te an d balan ce th e foll owin g acid-base equatio ns. Nam e the re a ctants an d pro duct s. Dec id e whic h is th e Bra nxted aci d and wh ich is the Bran stcd base . (a) C Il 3CO , Il (a q) + Mg (O Il ), (s) ---> (b) II CIO , (aq ) + N II , (a q) --->
I
15 3
28 . • Com ple te a nd balan ce the foll owing acid- base equ ation s. Name the reac ta nts an d p roducts. De cid e wh ic h is th e Bro n sted acid an d wh ich is th e Bro nsu-d base . (a ) H,I'O, (aq ) + KOH (aq ) ---> (h ) II , C,O, (aq ) + Ca (OII ),( s) ---> (11:1C:10 4 is o xalic aci d , an acid ca pa ble of don atin g two 11 + io ns.}
38. • T h e be autifu l red m ine ra l rhod oc h ro site is man ga nese (I I) carbona te . Write an overall, bal a nced e q ua tion fo r th e reac tion of th e mineral wit h hydroch lori c acid , and nam e th e p roducts.
29. Write a bal a nced equation for th e reac tio n of ba rium hvd ro x id c with nitric ac id .
30. \Vritt, a bal a nce d equatio n for th e re acti on of al u mi num hydroxid e with sulfu ric acid .
31. Na me two stro ng Breustcd ac id s an d o ne stro n g Bren stc d base . 32. Na me th ree we ak Bron stcd aci d s an d o n e we ak Bre nstcd base.
Writing Net Ionic Equations (See Examples 3.4 a nd 3.5 and C1wmist,yN ow Screen. 3. 10.)
Rhodoch rosit e. a mi nera l consisting largely of MnC03 •
39. • Write an ove rall, ba lanced eq uatio n fo r the react ion of (N H 4 h S with HB r, a nd nam e th e reacta nts a nd products. 40. Write a n ove ra ll, bal a nced equat ion fo r th e re ac tio n of Na:1S0 3with C H3C0 :1 Jl , and n am e th e re acta nts a nd p roducts.
33 . • Bal a nce th e fo llo win g eq uat ion s, an d th en write th e ne t ion ic e q ua tio n . (a) (N II,),C0 3 (aq ) + Cu (N O , ),(aq) ---> CllCO , (S) + N II 4N O ,( aq ) (h ) Ph (OH ), (s) + HCI(aq ) ---> Ph CI, (s) + H,o (f) (e) l\aC0 3 (s) + II CI (aq ) ---> BaCI, (aq ) + II ,O (C) + CO ,(g ) (d) C II , CO, II( aq) + Ni (O II ),(s ) ---> l':i( C H , C O ,), (aq) + H,O (f )
Oxidation Numbers (See f.x amjJlR 3. 7 and Chr mistrybknu S(H'f'1IJ 3. 15 an d 3. 16.)
34. • Bal a nce the followin g e qua tion s, and then wr ite the net io n ic equati on: (a ) Zll (S) + HCI (aq ) ---> H, (g ) + ZnCI, (a q ) (h ) Mg (OII ), (s) + II CI(aq ) ---> Mg CI, (aq ) + II ,O (C) (e) H NO, (aq ) + CaC 0 3 (s) ---> Ca (N O ,), (a q ) + H,O (f ) + co, (g ) (d) (N H, ),S (aq ) + FeCI 3(aq ) ---> N H.,CI(aq ) + Fe ,S 3(s)
42. • Dete rmin e th e o x idatio n n u m ber of each cle men t in
35. Ba lance th e fo llo win g equatio ns, an d th en write the net io nic eq u ation . Sh ow states fo r all re ac tan ts and produ cts (.'I , t , g, aq ). (a ) th e react io n of silver nitrat e a nd pota ssium io d ide to give silver iod id e an d potassiu m n itrate (h ) th e re actio n of hariu m h vd ro xid c an d n it ric aci d to give bari um ni trate an d wate-r (c) the reacti on o f sodi um p ho sph ate and ni ckd (lI ) nitra tc to g ive ni ckd (lI ) p hosp ha te and so d iu m ni tra te 36 . Balance each o f th e followin g e q ua tion s, and th en write the net ion ic equat ion , Show sta te s for all rea ctants an d p roduct s (s, f , g , aq ). (a) th e reacti o n of so d iu m h ydro xide a nd iron ( II ) ch lo rid e to g ive iron ( lI ) h yd roxid e and so d iu m ch loride (b) the reac tio n o f ba rium chloride with so d iu m carbonat e to give barium carbo na te an d so d iu m chl o rid e (c ) the reacti o n of a m mo n ia with p hosph o ri c aci d
Gas-Fo rming Reactions (S('(' Example 3. (j and ChnniJtr)N ow Scnen. 3. 14.)
37. • Side rite is a mineral co n sisting la rg ely of iron ( lI ) ca rbo na te . \ Vrite a n over a ll, balan ced equati on for its rc actio n with nitric acid , an d name th e prod u cts.
154
I
41. • Det ermin e th e oxid ation n um ber of eac h cl e ment in the fo llowin g io n s or co m pou nd s. (a ) I1rO ,,(d) Ca H, (b ) C,O ,'(e ) II ,S iO , (e) F(I) HSO, th e fo llowin g io ns o r com pou nds . (a) PF,;(d ) N, O , (e) PO CI" (h ) H, AsO., (e) u o« (I) Xe O .,' -
Oxidation-Reduction Reactions (See EXrHfljJ/P 3.8 flnd ChnniJ(1)Now Sm'f'1I 3. / i .)
43, Wh ich two o f the followin g reac tion s arc ox idationredu ctio n reactio ns? Ex plai n yo u r answe r in e ach case . Clas sify the rem a in in g react ion . (a ) Zn (s) + 2 NO J- (aq ) + 4 H 30 ' (aq ) ---> Zn" (aq ) + 2 N O, (g ) + 6 1I, 0 (C) (h ) Zn (O H ), (s) + H,SO , (aq ) ---> Zn SO , (a'l ) + 2 II , O (C) (e) Ca t s) + 2 H,o (f ) ---> Ca (O H ), (s) + H, (g ) 44. • Whi ch two of the foll owin g re actio ns arc ox id atio nre d u ct ion react io ns? Explai n yo ur answe r b riefl y. Clas sify the re m ai ni ng reactio n. (a ) CdC I, (a q ) + Na,S (aq ) ---> CdS (s) + 2 NaC I(a q) (b ) 2 Ca( s) + O ,( g ) ---> 2 Ca O (s) (e) 4 Fe (O Il ),( s) + 2 11, 0 (1') + O , (g ) ---> 4 Fe (OH ),( aq ) 45. • III th e fo llowin g re a ct ion s, d e cid e whic h reactant is o x idiz ed an d wh ich is re d uc ed. Desig na te th e ox id izing ag en t an d th e reducing ag ent. (a) C, II, (g) + 3 O , (g ) ---> 2 CO , (g ) + 2 1I, 0 (f) (h) Si t s) + 2 Cl,( g ) ---> SiCL,(C) ... more challenging • in OWL Blue-numbered questionsanswered in Appendix 0
46 . • In the foll owing re actions, decid e wh ich reactant is o xidized an d wh ich is reduced. Design ate the o xid izin g ag ent a nd th e re d ucing age n t. (a) C,,07' - (a q) + 3 Sn2+(aq) + 14 1l 30 ' (aq ) ---> 2 CrH (aq ) + 3 SnH (aq) + 2 1 H,O (t ) (h) FeS (s) + 3 N O , -(aq) + 4 Il ,O ' (aq ) ---> 3 N O(~) + SO,' - (aq ) + FeH (aq ) + I) H ,O ( t ) Types of Reactions in Aqueous SoLution (Sf'(' Examl)[p J. 9 and Chf1llislryNo1l! Soeen J.8.)
47 . • Balan ce th e followin g eq ua tion s, a nd then classi fy each as a precipitation , a n acid-base . or a gas -fo rm ing re action. (a) Ra (OIl ), (aq ) + Il CI (aq ) ---> Ra CI, (aq ) + 1l,o (C) (h ) Il NO, (a q ) + CnC 0 3(s) ---> Cn(NO ,),(aq) + 1l,0 (C) + C O,(~) (e) Na31'0 ,(aq) + Cu(N0 3),(a q) ---> C u,( I'O,),(s) + NaNO ,(aq) 48 . • Balance th e followin g equat io n s, a nd then classify ea ch as a precipitation , a n acid-base , or a gas -fo rm ing reacti on. (a) K,CO, (aq ) + Cu(NO ,),(aq) ---> C uC O ,(s) + KNO, (aq ) (h) I'h (NO, ), (aq) + HCI (aq ) ---> I'hCI,(s) + Il NO, (aq ) (e) M ~CO , ( s ) + HCI (aq ) ---> M ~CI , ( aq ) + H,O (C) + C O ,(~) 49 . Balan ce the follo w... ing e q ua tio ns , a nd then clas sify each as a pr ecipitation , an acid-base, or a gas~fo r m ing reaction. Show sta tes fo r th e p ro duct s (s, f, g, aq }, a nd th e n balanc e the co m plete d e q uatio n. , ·\trit e th e n et ioni c e q ua tio n. (a ) Mn CI, (aq ) + Na-S'(aq ) ---> ~ln S + NaCI (h ) K,CO, (aq ) + ZnC I,( aq) ---> ZnCO , + KCI
50. Balance the foll owing equ ati on s, a nd th en classify each as a precipitation , an acid-base , or a gas-fo rm ing reaction. Write th e net ion ic equa tion . (a ) Fl'(OIl ),(s) + Il NO , (aq ) ---> Fl'(N 0 3), + 1l,0 (b) FeC O ,(s) + H NO , (aq ) ---> Fe (NO, ), + CO , + H,O 51. • Balance each of the followin g equat ions, and classify th em as pre cipitation , acid-base, gas-form ing . or ox id ation-rcducti on re action s. (a) C uC I, + Il ,S ---> C uS + II CI (b ) H,I'O., + KOH ---> H,o + K31'O, (c) Ca + lIBr ~ Ho! + Ca Br;,! (d) MgCI, + H,O ---> Mg (OH ), + HCI 52 . . .... Complete a nd bala nce th e equat io n s bel ow, a nd classify the m as pre cipitat io n , acid- base, gas-fo rm ing, o r oxidation-reduction re acti ons. (a) N iC 0 3 + I 12S0 4 ~ ? (b) Co( O H), + HRr ---> ? (e) A~ C H , C O , + NaC I ---> ? (d) Ni O + CO ---> ?
.... more challenging • in OWL Blue-numbered questionsanswered in Appendix 0
General Questions T hese questions are not desig nated as to f)j) e vr location in the ella/Her. T he)' ma)' combine concepts [rom several dw/Jlf'rs .
53. Balance the foll owin g e q uatio ns : (a) th e syn thesis of urea , a comm on fertili zer CO,(g)
+
N H 3(g) ---> N H, CON H,(s)
+
H,O (C)
(b) reactions used to m ak e uranium(VI) flu oride for th e e n rich me n t of natural uranium UO,(s) + HF (aq ) ---> UF,(s) + II ,O ( C) UF,(s) + F, (g ) ---> UFe(s) (c) th e re a ction to m ake tit a nium (IV) chlo ride , ,»hi ch is then converted to titanium metal TiO, (s) + C I ,(~ ) + C(s) ---> Ti CI , (C) + CO ( ~ ) Ti CI, (C) + Mg (s) ---> T i(s) + MgCI, (s) 54 . • Balance th e foll owin g equa tio ns : (a) re act ion to p roduc e "su perp hosphate" ferti lizer Ca 3(\' O,),(S)
+
H,SO, (aq ) ---> Ca( H , I'O , ),(aq )
+ CaSO , (s)
(b) reaction to produce di bora nc , B2H j ; Na RH ,(s)
+ H,SO, (aq ) ---> R,H,;(g )
+
H, (g )
+
Na,S O .,(aq)
(c) reaction to produce tungsten metal fro m tungsten (VI) o xide
(d) decomposition of a m mon iu m di chromate (N H.,),Cr, O ,(s) ---> N,(g)
+
H,O (C)
+
Cr, 0 3(s)
55. • Give a formul a for each of th e following : (a) a so lu ble co m pou n d co nta in ing the bromide io n (b) a n in soluble h ydroxide (c) a n in soluble carbo na te (d ) a soluble ni tra te-con tai ning co m poun d (e ) a ' v'c ak Bre nstcd ac id 56 . ( ;ive th e fo rmul a fo r th e followin g co m p oun ds: (a) a so lu ble co m pou n d containing the ace ta te ion (h) a n in soluble sulfide (c) a sol u ble hyd ro xid e (d ) a n insoluble chl oride (e ) a strong Brren stcd base 57 . • Whi ch of th e followin g co ppcr( II) salts a rc so lu ble in water a nd wh ich are in soluble: Cu(l\' 0 3h, CuCO j , CU3(J' 0 1h, CuCI;,!? 58 . Name two an ions th at co m bi ne with 1\1:1+ ion to p roduce wa ter-s o luble com po un ds. 59. Id entify the specta to r ion or ion s in the reacti on of uiu-ic aci d a n d magnesium h ydroxide , an d write th e net ioni c e q uatio n. Wh at type of reaction is th is? 2 II "O ' (aq )
+ 2 NO , -(aq) + Mg (OIl ), (s) ---> 4 H,O (t) + M ~2+ (aq ) + 2 NO ,-(aq )
60. • Id entify and name th e water-insoluble product in eac h re action and wri te th e n et ionic equatio n : (a) Cu CI,(aq) + f1,S (a 'l ) ---> C uS + 2 HCI (h) CaCI,(aq) + K, C0 3(aq ) ---> 2 KCI + CaC0 3 (e) AgN O ,,(aq) + Na l( aq) ---> Agi + NaN O ,
I
155
61 . • Bro mi n e is obtai ned fro m se a wate r by th e foll owing re act io n: Cl, (g )
+ 2 Na hrfaq) ---> 2 Na fl lfaq) + Ilr, (C)
(c) Is th e reac tion ob se rve d produc t-favo red o r re actant-favo red ? (d) Na me th e re act ants an d p ro d ucts.
(a) Wh at h as bccu o xid ized ? Wh at ha s bee n red uced ? (b) Identify th e o xid izin g a nd reducin g age n ts. 62 . Ident ify each of th e following su bsta nce s as a likel y o xid izing o r re ducin g agen t: H N0 3 , Na, Ch , 0 t, K ~l n0 1' 63. • Th e m ine ral dolomite co n tai ns m ag ne sium ca rbona te . T h is react s wit h hyd ro chl oric acid . MgCO,(s)
I
+ 2 Il CI (a'l ) ---> CO, (g )
+
~l gC I ,( a 'l )
,"
+ Il ,O (C)
~.,
(a) Wrill' th e net io n ic equation fo r th e re actio n of m ag ne siu m carbo na te a nd hydro ch lo ric acid, a nd nam e th e spec tato r ion s. (b) Wh at type of rea ctio n is thi s? 64 . Ammonium sulfide , (N H 4 h S, reac ts with Hg (N0 3 h to pro du ce FIgS a nd N l l ,N0 3 . (a ) Write th e over a ll bala nced e q ua tio n fo r the rc action . Indi cat e th e state (s, aq ) for eac h co m po un d . (b) Na me ea c h co m pou nd. (c) Wh a t type of rea ct ion is thi s? 65 . • \Vha t spe cies (a to ms , m o le cule s, o r ion s) a re pre se nt in a n aq ueo us so lution of each of th e following com po u nd s? De cid e wh ich a rc Bro n stcd acids o r bases and wh e th e r th ey are stron g or 'weak. (a ) N H, (b ) C H, CO , H (e) N aO H (d) H ll r
66 . (a) Name two wa ter-sol uble co m po un ds co n ta ining th e Cu H io n . Na me two wa ter-insol uble com po u nd s based o n th e Cu H ion. (b) Na me two water-so luble com pou nd s con tain ing the Ba' " ion , Na m e two wate r-inso lu ble co m po u n ds ba sed o n th e Ra 2 + ion. 67 . • Balance e q ua tio ns for th ese reac tion s th at oc cur in aq ueo us so lutio n, an d then classif y eac h o ne as a pn·· cip ita tion , acid-base, o r gas-fo rmi ng re action . Sh ow sta tes for the p ro ducts (s, C, g, aq }, give their na m es, an d wri te th e net ionic e q ua tio n. (a) K,CO, + HClO. ---> KClO , + CO, + H, O (h) FeC I, + (N Il .),S ---> FeS + N Il 4 CI (e ) Fe (NO , ), + Na, CO, (a'l ) ---> Fe CO , + NaN O , (d) Na O H + Fe CI, ---> NaC I + Fe (OIl ),
Preparation of iodine. Amixture of NaI and MnOzwas placed in a flask in a hood (left) . On adding concentrated H2S 0 4 (right), brown gaseous12 was evolved.
70. • A If YO U have "silverware " in you r home, YO U kn ow it tarn ish ~s easily. Tarn ish is fro m th e o xida tio;l of silver in th e presen ce ;)1' sulfur-con taining co m po unds (in th e atmos ph ere or in yo ur foo d ) 10 give bla ck Agt S. To remove th e ta rni sh , yo u ca n war m th e tarnished o b jec t with so me alum inum foil (in wa ter with a small a mo un t of bak ing soda). Th e silver sulfide reacts with alum in um to pro d uce silve r as well as alum in um ox ide and hyd ro gen sulfide. 3 Ag, S(s)
+
2 AI(s)
+
3 H, O (C) ---> 6 Ag (s) + Al,O, (s)
+ :l H,S (a'l )
Hyd ro ge n su lfide is foul sme lling , bu t it is rem ove d by react io n with th e haki ng so da. Na Il CO, (aq )
+ II ,S(aq ) ---> Na IlS(a'l)
+ II , O (C) + CO, (g)
Cla ssifv th e two reactio n s, a nd idcn t ifv a ny aci ds, bases, OXid izin g agen ts, or re d uc ing age n ts. ' ,
68 . • Fo r each react ion , write an ove rall , bala nced eq uatio n a nd th e ne t io n ic eq u atio n . (a ) th e re ac tio n o f aque ous le ad (l l) n itrat e a nd aq u eo us potassium hyd ro xide (b) th e re act io n o f aque ous co p pe r( lI) ni trate an d aque ous so d iu m carbon ate
In the Laboratory 69 . • Th e fol lowing reaction ca n be use d to p repa re iod in e in the laborat ory.
+ 2 Il ,SO . (a'l ) + MnO, (s) ---> Na,SO.( a'l ) + MnSO . (a'l ) + I, (g ) + 2 Il , O (C) (a) Dete rm in e th e oxidati o n n um be r of eac h a to m in the e q ua tion . (h ) What is the o xid izing age n t, a nd 'what has be en ox id izcd ? What is th e re d uc ing age n t, and wha t has be e n reduced ?
2 Na l(s)
156
I
(a) (b) Removing sHvertarnish. A badly tarnished piece of silver (a) is placed in a dish with aluminum foil and aqueous sodium hydrogen carbonate. The portion of the silverin contact with the solution is now free of tarnish (b). ... more challenging • in OWL
Blue-numbered questionsanswered in Appendix0
71. • lit. Suppose yo u wish to pre pa re a sample of magnesium ch loride. O ne way to do this is to use an acidbase reaction, th e reaction of magnes ium hyd ro xid e wit h hyd ro chlo ric acid . Mg (O H ),( s)
+ 2 Il CI (aq ) ---> MgC I,(aq ) + 2 Il , O (f )
When th e reaction is comple te , evaporatin g the wa ter will give so lid magn esium ch lo rid e. Can you suggest a t least one o ther way to prepare MgC12?
72. lit. Suggest a labo ra to ry method for p repa ri ng b arium p hosphate . (Se e Question 71 fo r a way to approac h th is question.) 73 . • O ne way to te st for th e prese nce of sugars (say in a urine sam ple) is to trea t th e sa mple with silver ion s in aq ue o us a m mo n ia. (T h is is calle d the Tol len 's tes t.) Using g lucose , CliH1tOr; to illu st rate th is te st , the ox id a tio n-red uction reaction occurring is C"H I2O " (a q)
+ 2 Ag +(aq ) + 2 O H - (aq ) ---> C"H I2O , (aq ) + 2 Ag (s) + H,o (f )
What has been oxidized , and wha t has been reduced ? What is the oxidizing agent , and wha t is the reducin g agent?
Summary and Conceptual Questions Th e folluwhll{ questions ma)' USf' concepts from this a nd [nruious chapters.
74. T here a rc m an y ionic compounds that diss olve in water to a ve ry sm all extent. One exa mple is lcad fll ) ch lori de . Sup pose yo u stir so me so lid PbC I\! into water. Expla in how yo u wo u ld p rove th at th e compound dissolves hut to a small cx u-n t? Is th e d issolving process product-favored (II' reactant-favored? PbC1 2 (s)
Pb 2+ (aq)
+
2 CI - (aq)
75. lit. Most natura lly occurri ng acids arc wea k acids . Lactic acid is o ne ex am ple. C II ,CH (O Il)CO, II (s) + II , O (f) H, O +(a q ) + C H 3C H(O H)CO, - (aq) If yo u pla ce some lac tic acid in wat er; it will ionize to a sma ll ex ten t, a nd an equ ilib rium will be esta blis hed even tually. Suggest so m e ex perimen ts to p rove t hat thi s is a weak acid and that th e establishme nt of eq ui librium is a re versible process.
H
OH 0
I I II
H- [ - [ - [ - OH
I I
H
H
Lactic acid
76 . • lit. You want to pre pare ba rium c hloride , BaCh, u sing a n exc hange reaction of some type . To do so, you h ave the fol loweing reagents from "... h ich to select the rcacrants : RaSO ~ , RaRrt, RaCO j , Ra (Olf h , lICl , H gSO ~ , AgNO j , and II N0 3 . Wri te a com plete, ba lanced equation for the reac tion cho se n . (Note: Th ere are seuem l jmssibilities.} (b)
(a}
Totten 's te st. The reaction of silver ions with a sugar such as
glucose produces metallicsilver. (a) The set up for the reaction. (b) The silvered testtube
77 . • Des cribe h o w to prepare RaSO~, barium su lfate , by (a) a precipitation reac tion a nd (b ) a g-as-forming- reaction. To do so, you have the foll owin g re ag-e n ts fr om which to sel ect th e reacta nts: RaC I2, RaC0 3 , Ba (O H h , H 2Sn" a nd Na 2S0... Writ e com ple te , balanced equations for the reac tions cho sen. (Se e pa ge 130 for an illu str a tio n of the prepar ation of a compound.) 78 . • De scribe h ow to pre pa re zinc ch lo ride by (a ) an acid- ba se reaction, (h) a gas -formin g reactio n , and (c) a n oxidation-reduc tio n reaction . T he availab le sta rting material s arc ZnC0 3, H C1, ci, HNO j , Zn (O H h , N'a C l, Zn (NO:1h , and Zn. Wr ite co m plet e, bala nced equatio ns fo r the reacti ons cho sen .
.... more challenging
• in OWL
BLue-numbered questions answered in Appendix 0
I
157
CONCEPTS OF CHEMISTRY
4
5toichi 0 metry: Qua ntitative Information About Chemical Reactions
The Chemistry of a Sparkler This "sparkler," like many forms of firewor ks, depends on some very straightforward chemistry. It consists of a mixture of finely powdered metals, such as Al and Fe, other substances such as KCl0 3, KN 0 3, and a binder that holds the mixture onto a wi re handle. When ignited. the powdered metaLreacts with oxygen in t he air (or extracted from KClO, ). and t he sparks fly!
Questions: 1. What ki nd of chemical reaction is occurring here? 2. What are the likely products from t he reaction of aluminum metal or iron with oxygen in the air or wit h KCl0 3? 3. If t he sparkler contains 1.0 g of Al, what is t he mass of the product of the reaction of alumin um with oxygen? Answers to these questions are in Appendix Q.
158
Chapter Goals
Chapter Outline
SPf' Cha/lln Goals Revisited (pagr 194) fo r St ud), Qu estions keJnl to tlIPS£, bro(Jl~ a lld assig nable in O\VL
4.1
Mass Relationships in Chemical Reactions: Stoichiometry
4.2
Reactions in Which One Reactant Is Present in Limited Supply
Calculate the theoretical and percent yields of a chemical reaction.
4 .3
Percent Yield
Use stoichiometry to analyze a mixture of compounds or to deter mi ne t he form ula of a compound.
4.4
Chemical Equations and Chemical Analysis
4 .5
Measuring Concentrations of Compounds in Solution
Perform stoichiometry calculations using balanced chemical equations. Understand the meaning of a limiting reactant in a chemical reaction.
Define and use concentration in solution stoichiomet ry.
4.6
pH, a Co ncentration Scate for Acids and Bases
4 .7
Stoichiometry of Reactions in Aqu eous Solution
4.8
Spectrophotometry, Another Method of Ana lysis
T
he o bj ec tive of this chap te r is to introduce th e quantitati ve study o f ch emica l reac tio ns . Quan titative stud ies a re needed to d etermine , for ex am ple , how much oxygen is required for th e co mp le te co mb ustion of a given quantity of ga so line and wha t masses o f ca rbo n dioxid e and wate r ca n be obtained . T his part of chemistry is fundam ental to much of what che m ists , che mical e ng inee rs, bioch emists, molecul ar biologists, ge oc he m ists, and m an y o the rs do.
4 .1
n
Mass Relationships in Chemical Reactions: Stoichiometry
ChemislrY-9-Now~ Throughout t he text this icon introduces an opportunity for self-study or to explore inte ractive tutorials by signing in at www.th omsc nedu.com/ Iogtn.
Mod ule 7
A balan ced che mic al equatio n shows th e qu antitati ve re lat io nsh ip bet ween reacrants and p roducts in a che m ica l reactio n . Let us ap ply this co nce p t to the reactio n o f phosphorus and chlo rine (... Figure 3.1) . P, (5)
+ 6 Cl, (g) ---; 4 PCl, (C)
Suppose yo u use 1.00 mo l of p ho sphorus (P4' 124 g / mo l) in th is reaction . The bala nced eq uatio n shows th at 6.00 mol (= 425 g) o fC I2 must be used for co m plete reaction with 1.00 mo l of P 4 and that 4.00 mo l ( = 549 g ) of PCI3 ca n be produced . Th e mo le an d mass relation ships of reactan ts and product s in a re ac tio n can be su m ma rize d in an amounts lab/e. 'rou will find suc h tabl es he lpful in identifying th e amou n ts o f react ants a nd p roducts a nd th e changes th at oc cur up on reaction . +
Equati on
P, (s)
Initial amount (mol)
1. 00 mol
6.00 mol
6 CI, (g)
---.
a mol
4 PCI, (t )
(124 g)
(425 g)
(0 g)
Change in amount upon reaction (mol)
- 1. 00 mol
- 6.0 0 mol
+ 4.00 mol
Amount after complete reaction (mol)
o mol (0 g)
o mol
4.00 mol
(0 g)
[549 9
~
124 9 + 425 g)
The ba lan ced eq ua tion fo r a reaction tells us the co rrec t mole ratios of reac ta n ts and p roducts . Here , 6 mol of Cf sho uld be used pe r mo le of Pi . Now, what if o nly n.o 100 mol o f P4 ( 1.24 g ) is available ? Six times as man y mo les of C12 are still required (0.0600 mo l of 0 , ; 4.25 g ) , and 0.0400 mo l of PCI, (5.49 g) will be form ed . Th e on ly requirem en t is th at th e re sho u ld be a ratio of 6 to 1 for the am o u n t o f CI2 relative to the amou n t of P4.
• Amounts Tables Amounts tables are useful not only here but will also be used extensively when you studychemical equilibria more thoroughly in Cha pters 16-18.
Sign in at www.thomsonedu.comj login to download theGo Chemistry module forthis section or goto www.ichapters.com to purchase modules.
159
• Stoichiometry and Reaction Completene ss Stoichiometry calculations
assume reactions are complete. That is, all of at least one of the reactants has been converted to the product or products. Reca ll from the discussion of chemical equilibrium in Chapter 3 that some reactions do not reach completion, however.
Fo llowing thi s line of reason ing, let us d et e rmi ne (a) wha t mass of Clz is req uir ed to react co m ple tely with 1.45 g of p ho sp ho rus and (b) wha t mass of PCI3 ca n be p ro d uced .
Part 2 CO,(g)
one mol of reaction occurs when 2 mol of CO and 1 mol of O2 produce 2 mol of CO 2 , If the reaction mixture consists of only 1 mol of CO and 0.5 mo l of 02, then only 1 mol of CO2 is produced, and 0.5 mol of reaction has occurred according to t his balanced equation. To pursue this examplefurther, suppose 9.5 9 of COand excess O2 are combined. What amount of CO2 (moles) can be produced?
9.5 g CO X 1 mo t CO X 1 mo l-om 28.0 9 CO 2 mol CO = 0.34 mol-rxn 0.34 mol-rxn X
2 mol CO 2 1 mol-rxn
= 0.68 mol CO2
1. Calculate the moles of reaction predicted
for each reactant, and decide on the limiting reactant. In the case of the NH 3/ 0 2 reaction, 4 NH ,(g)
+ 5 O,(g) ---> 4 NO(g) + 6 H,O(t )
1 "mole of reaction" uses 4 mol of NH 3 and 5 mol of O2 and produces 4 mol of NO and 6 mol of H20. In the example on page 163, we started with 44.0 mol of NH3 , so 11.0 moL of reaction can result.
of moles of reaction predicted by the lim iting reactant corresponds to the number of moles of reaction t hat can actually occur. Each reactant and product will undergo this number of moles of reaction, 4.68 rno l-rxn in this case. To calculate the change in amount for a given reactant or product, muttip ly this number of moles of reaction by the stoichiometric coefficient of the reactant or product. To illustrate t his step, for NH3 this corresponds to the followin g calculation: 4.68 moL-rxn X ( 4 mote NH3 ) = 1 moL-rx n 18.8 mol NH3 • The amount of each reactant and product after reaction is calculated as usual. Equation 4 NH](g) + 5 02(g) ~ 4 NO (g) Initial amount (mol) 44.0 23.4
_
44.0 mo l N H~ X
1 mol-rxn
_ = 11.0 mol-rxn
4 1110t-NH"3
4.68
23.4 mo l O2 X
1 mol-rxn _ = 4.68 mol-rxn 5 l11 0l O2
Fewer motes of reaction can occur with t he amount of O2 available, so O2 is the limiting reactant.
2. Calculate the change in amountand the amount upon completion of the reaction, for each reactant and product. The number
4.2
I
o
4.68
4.68
4.68
Change in amount (mol) - 4.68(4) - 4.68(5) + 4.68(4) +4 .68(6) = - 23.4
= + 18 .8
= +28.1
Amountremaining after complete reaction (mol) 25.2
Based on t he amount of O2 available, 4.68 mol of reaction can occur.
o
Moles of reaction based on limiting reactant (mol)
= - 18 .8
You can see in this examp le that the numberof moles of reaction that occurred is calcu lated by multiplyi ng the amount (moles) of the reactant CO by the factor. 1 mol-rxn/z mol CO (which amounts to dividing the amount of CO by its stoichiometric coefficient). All reactants and products involved in a chemica l reaction undergo the same number of moles of reaction because the reaction can only occur a certain numb er of times before t he reactants are consumed and the reaction reaches completion. If one of the reactants is in short supply, t he actual number of times a reaction can be
+ 6 H20(g)
0
18.8
28 .1
Finally, from t he amounts present after completion, we can calculate the masses of t he products and of any reactant remaining. You may find t his approach easier to use particularly when there are more than two reactants, each present initially in some designated quantity. Afinal note: t he concept of "moles of reaction" will be appliedin this text in the discussion of thermochemistry in Chapters 5 and 19.
Reactions in Which One Reactant Is Presentin limited Supply 167
4.3
•
Percent Yield
The maximum ma ss of product that ca n be o btain ed from a chemica l reaction is th e theoretical yield. Frequently, however, the actual yie ld of the product-the mass of mat eri al th at is ac tu ally obtained in th e laboratory or a che m ical plant-is less th an the theoretica l yield . Lo ss of product often occurs during the isol ation and purification ste ps . In addition , so m e reacti ons do not go completel y to product s, a n d reactions are so me times co m p licate d by giving more than on e se t of products . For all these re asons, the actual yiel d is a lmost a lways less than the th eoretical yiel d (Figu re 4. 3). To provide information to o the r c he m ists wh o might wa n t to ca rr y 011t a reaction, it is cus to ma ry to report a percent yiel d . Percent yie ld, which specifies how mu ch of the theoretical yiel d was o b taine d , is defined as
•
(a)
t vi e ld = Percen)'l
actua l yield X 100"./1" t heoretical yield
(4.1)
Suppo se yo u made asp ir in in the la bo rato ry by th e followin g reaction: C7H 60 3(S)
+
( , H, D,( I)
-y-?'
(b)
~
FIGURE 4.3 Percent yield. Atthough
salicylic acid
acetic anhydride
C9H S04 (s)
~
+
(H,CO,H ( C)
~ aspirin
acetic acid
not a chemical reaction, popping corn is a good ana logy to th e difference between
a theoretical yield and an actual yield. Here, we began with 20 popcorn kernels and found t hat only 16 of t he m poppe d. The percent yield from our "reaction" was
and that yo u began with 14.4 g of salicylic acid and a n ex ce ss of ac eti c a n hyd rid e . Th at is, salicylic acid is th e limiting reactant. If YOll o htain 0.26 g of asp irin , wh at is th e p e rcent yie ld of this p roduct? The first ste p is to find th e am o u n t of the lim itin g reactant, salicylic acid (C"H ,( O H) CO , H ) .
(16/20) x 100%, or 80%.
.... 1 mo l C,H, (OH)CO,H 14.4 g.c,H, (O H)CO,H x . 138.1g C,H,(OH)CO,H
~
0.104 mol C,H, (O H)CO,H
Nex t, u se the sto ich io me tric factor fr om th e bal an ced eq u atio n to find the am ou n t of asp irin ex pec te d based o n th e lim itin g reac ta n t, C oH 4 (O H ) C0 211.
__ 1 mol aspirin 0.104 mol C,H,(OH )CO,H x . = 0.104 mol aspirin . 1 mol C,H,(OH)CO,H Th e ma ximum a mou n t of aspirin th at ca n be produced-the th eoret ical yield- is 0.104 mol. Because the quantity you measure in th e lab oratory is th e mass of th e product , it is cus to ma ry to express the th eoretical yiel d as a ma ss in g rams .
. •-
0.104 mo1aspmn X
180.2 g aspirin .. . ~ __ = 18.7 9 aspmn 1 molasp irin
Fin ally, with th e ac tual yield known to b e only 6.26 g , th e percent yie ld of asp irin call be ca lcu la ted .
6.26 g aspirin obtained (actual yield) 18.7 g aspirin expected (t heoretical yield) 168
Chapter 4
I
Stoichiometry: Quantitative Information About Chemical Reactions
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EXERCISE 4.3 Percent Yield Alumin um carbide, A4( 3. reacts with water to produce methane. Al C,(s) + 12 H,O(f) ---> 4 AI(OH ),(s ) + 3 CH.( g) If 125 9 of aluminum carbide is decomposed. what is the theoretical yield of methane? If only 13.6 9 of meth ane is obtained, what is t he percent yield of this gas?
4.4
Chemical Equations and Chemical Analysis
Analytical che m ists use a vari et y of ap p roac hes to id entify su bsta nces as well as to mea sure the quantitie s o f co m pon e n ts of mi xtures . An alytical chemistry is often
don e no w u sing in strumental methods (Figu re 4.4) , but classical che mical reac tio ns and sto ichio met ry still playa central rol e .
Quantitative Analysis of a Mixture Quantitative ch emical analysis ge ne rally depends on one of th e follo wing basi c ideas: • A su bsta nce, present in unknown amo u n t, ca n be allowe d to re act with a known quantity of ano the r su bstance . If th e sto ichio metric ra tio for their reaction is kn own , th e unkn own am ount ca n be determined. • A mat e rial of unknown co m positio n ca n be co nverted to on e or more su bsta nc es of known co mpositio n . Those su bsta nces ca n be identified, the ir am ou n ts d etermined , and th ese a mou n ts rel at ed to the am ount o f th e o riginal, unknown su bsta nce.
FIGURE 4 .4 A modern analytical inst rument. This nuclear magnetic resonance (NM R) spectrometer is closely related to a magnetic resonance imaging (M RI) instrument found in a hospital. NMRis used to analyze compo unds and to decipher their structure. (The instrument is controlled by a computer and console not seen in this photo.)
An e xample of th e fir st type of analysis is th e analysis o f a sam ple of vin egar co n tain ing an unk no wn amou n t of ace tic acid, th e in gredi ent th at mak es vinegar acidic. The ac id re acts readily and co mple tely with sod iu m hydroxid e .
CH]CO,H(aq) + Na OH(aq) ---> CH]CO, Na(aq)
+ H,O(f )
acetic acid
If th e ex ac t amo u n t of so d iu m hydro xid e use d in th e reac tio n ca n be measured , th e amount of aceti c ac id pr esent can be ca lculate d . This type of anal ysis is th e su bject of a lat er sec tio n in thi s chap ter (... Secti on 4.7) . The secon d type of analysis is exe m plified by th e analysis o f a samp le of a mineral, th cnarditc , which is lar gel y sod iu m sulfate, Na 2S0 . j. (Figu re 4.5) . So d iu m su lfate is so lu ble in water. The refore , to find the quanti ty of Na 2S0 . j. in an impure min eral sam ple, we wo uld crush th e rock and th en wash th e powdered sam ple th oro ughly with wate r to di ssolve th e sod iu m su lfate . Nex t, we would treat thi s so lutio n of so d iu m sulfate with barium chlo rid e to precipitat e th e water-inso lu ble co mpo u nd barium su lfate . The barium sulfa te is co llec te d on a filte r an d weighe d (Figu re 4.6 ).
Na,50,(a q) + BaCl,(aq) ---> Ba50, (s) + 2 NaCl(aq) 4.4
I
FIGURE 4 .5 Thenardit e. The min eral thenardite is sodium sulfate, Na 2S0 4 • It is named after the French chemist l ouis Thenard (1777 - 1857 ), a co-discoverer (with J. l. Gay- lussac and Hum phry Davy) of boron. Sodium sulfate is used in making detergents, glass, and paper.
Chemical Eq uations and Ch emical Analysis
169
(c)
Na 2S0 4(aq ). clearsolution
BaCI2(aq). clear solution
BaS0 4, white solid
NaCl(aq) . clear solution
r
(d)
NaCl(aq). BaS04 • clea r solution whitesolid caughtin filter
Mass of dryBaS0 4 determined
Active Figure 4.6 Analysis for the sulfate content of a sample. The sutfate ions in a solution of Na2S04 react with barium ions (BaH) to form BaS04' The white. solid precipitate, barium sulfate (BaS0 4), is collected on a filter and weig hed. The amount of BaS04 obtained can be related to the amount of Na2S04in the sample.
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• Analysisand 100% Yield Quantitative analysis requires reactions in which the yield is 100%.
\Ve can the n find the amo unt of sod ium sulfa te in th e min eral sample because it is dire ctly rel at ed to the amount o f BaS04' 1 mo l Na, 50, (aq ) -; 1 mol Ba 50 , (5)
Exa m ple 4.3 illustr ate s ano the r in stan ce o f th e analysis o f a min eral in thi s way. ChemislrY.{J"Now~ Sign in at www.thomsonedu.comjlogin and go to Ch apter 4 Contents to see Screen 4.7 for a tutorial on chemical analysis.
Mineral Analysis ProbLem Nickel(II) sulfide. NiS. occurs naturally as t he relatively rare mineral millerite. One of its occurrences is in meteo rites. To analyze a mineral sample for the quantity of NiS, the sample is dissolved in nitric acid to form a solution of Ni(N03h.
Ni5(5) + 4 HNO,( aq) ---> Ni(NO;);(aq) + 5(5) + 2 NO,(g) + 2 H,O(t) The aqueous solution of Ni(N0 3h is then treated with the organic compound dimethylglyoxime (C4HsN 20 2 • DMG) to give the red solid Ni(C4H7N 20 2h.
Ni(NO ,);(aq) + 2 C,H,N,O,(aq) ---> Ni(C,H,N,O,),(s) + 2 HNO ,(aq) Suppose a 0.468-g sample containin g millerite produces 0.206 9 of red. solid Ni(C4H,N2 0 2h. What is the mass percent of NiS in the sample? A precipitate of nickel with dimethylglyoxime. Red, insoluble Ni(C 4H,N20 2h precipitateswhen dimethylq lyoxtme (C 4HsN 20 2) is added to an aqueous solution of nickel(II) ions. (See Example 4.3.) 170 Chapter 4
I
Strategy The balanced equations for the reactions show the following "road map": 1 mol NiS ----7 1 mol Ni(N0 3h ----7 1 mol Ni(C4 H,N20 2h If we know the mass of Ni(C 4H7N l0 2h, we can calculate it s amount and thus the amount of NiS. The amount of NiS allows us to calculate t he mass and mass percent of NiS in t he sampLe.
Stoichiometry: Quantitative Information About Chemical Reactions
Solutio n The molar mass of Ni (C4 H7N zOzh is 288 .9 q/mol. The amount of this red solid is H NO) 1 mol Ni(C,H,N,O, ), _ :::: 7.13 x 10-, rna l N"(C 0.20 6 9 N1"(C-4HNO")" 7 Z Z Z X 1 4 7 Z Z Z 288 .9 g_Ni(CI,HrN20 z)2 Because 1 mol of Ni(C4H7N 202h is ultimately produced from 1 mol of NiS , the amount of Ni S in the sample must have been 7.13 X 10- 4 mol. With t he amount of NiS know n, we calculate the mass of NiS. 7.13 X 10- 4 molNiS X 90 .76 9 NiS :::: 0.0 647 g NiS 1 mol NiS
Fina lly, t he mass percent of NiS in t he 0 .4 68-g sample is "
Mass percent NIS =
EXERCISE 4 .4
0 "0647 9 NiS 0 .4 68 9 sample
X 100% =
13.8% NiS
Analysis of a Mixture
One method for determining the purity of a sample of titan ium(IV) oxide, Ti0 2, an important industrial chemical. is to react the sample wit h bromine t rifluoride. 3 TiO,(s)
+ 4 BrF, (e) --> 3 TiF, (s) + 2 Br,(f) + 30,(g)
This reaction is known to occur completely and quantitatively. That is, all of the oxyge n in Ti0 2 is evolved as Oz. Suppose 2.367 9 of a TiOr containing sample evolves 0.143 9 of O2 • Whatis t he mass percent of TiOz in the sample?
Determining the FormuLa of a Compound by Combustion T he em pirical formula of a co m pou nd can be d e te rmined if th e pc rce nt. co mpositio n of th e co m po u nd is known (.... Sect ion 2.10) . But whe re d o the percent co mpositio n d at a co m e fro m? O ne chem ica l m ethod th at wo rk s wel l for co m poun d s that burn in o xygc n is ana lysis by co mbustio n. In thi s techn ique , each c le me n t in the co m po u nd co m bines wit h o xyge n to produce the ap p ro priate o xid e. Co ns id er an ana lysis of th e hyd ro carbon me tha ne , CH". A bala nce d eq ua tio n for thc combustio n o f methan e shows th at eye ry ato m of C in th e o rig ina l co mpo un d ap pears as CO 2 and every ato m of J I ap pears in the fo rm of wate r. In ot her words, fo r eve ry m ole o f CO 2 observed, th ere must have been one mol e of car bon in th e un kn own co m po u nd. Sim ilarl y, for cycry mole o f H 20 o bse rved from co mbustion , th ere m ust have been two mol es o f J I ato ms in the u n kn own ca rbo nhydrogen co mpoun d . CH,(g) + 20,(g)
+
---4
CO,(g) + 2 H,O(f)
+
In the com bustio n experime nt, gaseous carbo n d io xide a nd wate r arc se pa ra ted (as illu stra ted in Figu re 4.7) and th eir masses d ete rm ined . From th ese m asses, it is possible to calculate the am ou n ts o f C and II in CO 2 and H 20 , resp ecti vely, an d 4 .4
I
• Finding an Empirical Formula by Chemical Analysis Finding the empi rical formula of a compound by chemical analysis always uses the foHowi ng procedure: 1. Th e unknown but pure compound is converted in a chemical reaction into known prod ucts. 2. TIl e reaction products are isolated, and the amount of each is determined. 3. The amount of each product is related to the amount of each elementin the original compound. 4. Til e empirical form ula is determined from the relative amounts of elements in the original compound.
Chemical Equations and Chemical Analysis 171
Furnace
0,
~ ~I
H,Oa bsorber
CO,a bsorber
Active Figure 4.7 Combustion analysis of a hydrocarbon. If a compound containing Cand His burned in oxygen, CO 2 and H20 are form ed, and the mass of each can be determined. The H20 is absorbed by magnesium perchlorate, and the CO 2 is absorbed byfinely divided NaOH supported on asbestos. The mass of each absorbent before and after combustion gives the masses of CO 2 and H20 . On ly a few milligrams of a combustible compou nd are needed for analysis.
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• explore an interactive version of this figure accompanied by an exercise.
the rat io of a mo u n ts of C a n d H in a sam p le o f th e original compou n d ca n th e n be fo u nd . T h is ratio g ives th e e m p irical fo rm ula. [X 1
•
IIII
mol H 20 I 18.02 9
,
I
..
III
I
x
·
2
1
mol H 1
m ~l H,O •
'
I
..
I
•
: -.'
1 mol CO 2 J 1 mol C J x ~x ~
. . Using Combustion Analysis to Determine the FormuLa of a Hydrocarbon ProbLem When 1.125 9 of a liquid hydrocarbon, CxH y , was burned in an apparatus like t hat shown in Figure 4.7, 3.447 g of CO 2 and 1.647 9 of H20 were produced. The molar mass of t he com pound was found to be 86.2 q/mo l in a separate experiment . Determine the empirical and molecular formulas for t he unknown hydrocarbon, CxH.... Strategy As outlined in t he preceding diagram, we first calculate the amounts of CO 2 and H20. These are t hen converted to amounts of Cand H. The ratio (mol Hlm ol C) is used to determine t he empirical formula of t he compound. The molar mass of the compound and the molar mass of t he empirical formula are t hen used to deter mine t he molecular formula. Soluti on Th e amounts of CO 2 and H20 isolated from the combustion are 3.447 9 CO; x
1 mol CO 2 44 .010 9 CO,
=
0.078 32 mol CO 2
1.64 7 9 H20 X
1 mol H20 18.01 5 9 H,O
=
0.09142 mol H20
For every mole of ( 0 2 isolated, 1 mol of ( must have been present in t he unknown compound. 0.0 7832 mol CO; X
172
Chapter
4
I
1 mol ( in unknown
Stoichiometry: Quantitative Information Ab out Chem ical Rea ctions
1 mol CO2
=
0.07832 mol C
For every mole of H20 isolated, 2 mol of H mu st have been present in t he unknow n. 0.09142 motI H20
X
2 mol H in unkn own
1 mol H20
= 0.1828 mol H
The original 1.125 g sample of compou nd t herefore contained 0.07832 molof Cand 0.1828 mol of H. To determine the empirical form ula of the unknow n, we find the ratio of motes of H to moles of C ( 7) at 25 -c.
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179
A Closer Look
Serial Dilutions
We often find in the laboratory that a solution
or 1/10 of the concentration of t he original solution (because we diluted t he sample by a factor of 10). Now we ta ke 5.0 mL of t he diluted solution and dilute t hat once again to 100. 0 mL. The final concentration is
is too concentrated for th e a na lyti cal tech-
nique we want to use. You might want to analyze a seawater sample for its chloride ion content, for instance. To obtai n a solutio n
with a chloride concent ration of t he proper mag nitude for analysis by the Mo hr method (Case Study, page 186), for example, you
might want to dilute the sample, not once but severa l times.
0.00500 L X 5.50 X 10 - 2 mcl/L = 2.75 X 10- 4 mol NaCl CNaC[
Suppose you have 100.0 mLof a seawater sample t hat has a NaCl concentration of 0.550 moljL. You transfer 10.0 mL of t hat sample to a 100.0· ml volumetric flask and fill to the mark with distilled water. You then transfer 5.00 mLof that diluted sample to another 100.0 mL flask and fill to t he mark with distilled water. Whatis t he NaCl concentrat ion in t he final 100.0-mL sample? The original solution conta ins 0.550 mol/L of NaCL If you remove 10.00 ml., you have removed
= 2.75 ~
10- 4 mol/ 0.1000 L 2.75 X 10- ' M X
This is 1/200 of t he concentration of t he origina l so lution.
A fair question at t his point is why we did not j ust ta ke 1 mL of the original solutio n and dilute to 200 mL. The answer is t hat there is less error in using larger pipets such as 5.00or 10.00-mL pipets rat her than a 1.00-m L pipet. And then t here is a limitation in available glassware. A 200.00-mLvolumetric flask is not ofte n available.
Question: You have a 100.0-mL sample of blue dye having a concentration of 0.36 M. You dilute Q 1O.0-mL sample of this to 100.0 mL and then a 2.00-mL sampleof that solution to 100.0 mL. What is the final dye concentration? (Answer: 7.2 X 10- 4 M) Q
~--Tr-an-Sf·e~.-o-m-L-""" ' - -T-ra-nS-fe! o-om- L - -'"
o
NaCl concentration 0.550 mol/L
0.0 1000 L x 0.550 mcl/ L = 5.50 X 10- 3 mol Na Cl
o
Fill to mark with
Fill to mark with distilled water
distilled water
and the concentration in 100.0 mL of the diluted solution is CNa ct
= 5.50
X
1/200 original concentration
10- 3 mol/0.100 L ~ 5.50 X 10- 2 M
Original Solution 100.0 mL sea water sample
10.0 mL sample diluted to 100.0 mL
5.00 mLsample diluted to 100.0 mL
pH . Bloo d , whi ch you r co m mo n se nse te lls you is likely to be n either acidi c n or basic, has a p H n ear 7. Indee d , fo r aq u eo us so lu tio ns a t 25 "C, we ca n say th a t acid s will h ave p H value s less th an 7, bases will h ave value s grea te r t ha n 7, a n d a p H o f 7 re presents a neu tral so lu tio n (Figu re 4.11 ) .
o
14
pH = 3.8 Orange pH = 2.8
pH = 2.9
Vinegar
Soda
pH = 7.4 Blood
pH = 11.0 Ammonia
pH = 11.7 Oven cleaner
Active Figure 4.11 pH values of some common substances. Here, the "bar" is colored red at one end and blueat the other.These are the colors of litmus paper, commonly used in the laboratory to decide whether a solution is acidic (litmus is red) or basic (litmus is blue). tu
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180 Chapter 4
I
Stoichiometry: Quantitative Information About Chemical Reactions
FIGURE 4 .12 Determining pH. (a) Some household products. Each solution contains a few drops of a universal indicator, a mixture of several acid-base indicators. A color of yellow or red indicates a pH less than 7. Agreen to purple color indicates a pHgreater than 7. (b) The pH of a soda is measured with a modern pH meter. Soft drinks are often quite acidic, owing to the dissolved CO 2 and other ingredients.
(b)
A-(aq) + H,O(f)
Strategy The key to t his problem is to recognize that the molar mass of a substa nce is the ratio of the mass of a sample (g) t o the amount of substa nce (mol) in t he sample. Here, molar mass of HA = 1.056 9 HAJx mol HA. Because 1 mol of HA reacts with 1 mol of NaOH in this case, the amount of acid (x mol) is equal t o the amount of NaOH used in t he tit ration, which is determined by its concent ration and volume. SoLution Let us first calculate the amount of NaO H used in the titration. Am ount of Na OH
= CNaOHVNa OH
=
0.256 mol L
X 0.03378 l
= 8.65 X 10-3 mol NaOH 4 .7
I
Stoichiometry of Rea ctions in Aqueous Solution 187
Next, recognize that the amount of NaOH used in the titration is the same as the amount of acid titrated . That is. 8.65 X 10- 3 mol-NaO H X
1 moL HA 1 mol NaOFi
= 8.65
X 10- 3 mol HA
Finally. caLcuLate the moLar mass of HA . Molar mass of acid
=
1.056 g HA 8.65 x 10-3 mol HA
122 q/rnol
EXERCISE 4.14 Determining the Molar Mass of an Acid by Titration An add reacts with NaOH according to the net ionic equation HA(aq) + OH -(aq) --> A-(aq) + H,O(/)
Ca lcuLate the molar mass of HA if 0.856 g of the acid requires 30.08 mL of 0.323 MNaDH.
Titrations Using Oxidation-Reduction Reactions Analysis by titra tion is not limi ted to ac id-base chemistry. Many o xid at ion -re ductio n rea cti o ns go rapid ly to co m ple tio n in aq uco u s so lu tio n, a nd methods exist to d eterm in e th ei r eq uiva le nce poin t.
Using an Oxidation-Reduction Reaction in a Titration ProbLem The iron in a sampLe of an iron ore can be converted quantitatively to t he iron(II) ion, Fe 2- . in aqueous solution, and this solution can t hen be titrated with aqueous potassium permang anate, KMnD 4• The balanced, net ionic equation for the reaction occurring in the course of this titration is MnO,-(aq)
purple
+ 5 fe "(aq) + 8 H,O'(aq) --> Mn "(aq) + 5 fe H(aq) + 12 H,O(f) colorless
colorless
pale yellow
Case Study
Forensic Chemistry: Titrations and Food Tampering
The U.S. Food and Drug Admin istration (FDA) has recently discovered cases of product tampering involving the addition of bleach to products such as soup, infant formula. and soft drinks. Household bleach is a dilute solution of sodium hypochlorite (NaClO) . a compound that is an oxidizing agent and is dangerous if swallowed. One method of detecting bleach uses starch-iodide paper. The bleach oxidizes the iodide ion to iodine in an add solution,
1- in a ratio of 1 mol HClD to 2 mol 1-. The iodine formed in t he reaction is then titrated with sodium thiosulfate, Na 25203in another oxidation-reduction reaction (as in Exercise 4.15).
2 qaq)
+
HCIO(aq) I,(aq)
+ H, O'(aq)--> + 2 H,O(f) + (l-(aq)
and the I, is then detected by a deep blue color in the presence of starch .
This reaction is also used in the quantitative analysis of solutions containing bleach. Excess iodide ion (in the form of KI) is added to the sample. The bleach in the sample (which forms HClD in acid solution) oxidizes
188
Chapter
4
I
I,(aq)
+ 2 S,O,'-(aq)--> 2 I - (aq)
+ S,O. ' -(aq)
The amount of Na 2S20 3 used in the titration can then be used to determine the amount of NaClO in the sample.
Question: Excess KI is added to 0100.0 mL sample of a scft drink that hadbeen contaminated with bleach, NaClD. The iodine (12) generated in the solution was then titrated with 0.0425 MNO~203 and required 25.3 mL to reach the equivalence point. What mass of NaCiO was contained in the 100.0-mL sample of adulterated soft drink? Answer to this question is in Appendix
Stoichiometry: Quantitative Information About Chemica l Reactions
Q.
A distinctive blue color is generated when iodine reacts wit h water-soluble starch.
A 1.026· g sample of iron-containing ore requires 24.35 ml of 0.0195 MKM n04 to reach t he equivalence point. What is the mass percent of iron in the ore? Strategy Because the volume and concentration of the KM n04 solution are known, the amount of KMn04 used in t he tit ration can be calculated. Using t he stoichiometric factor, t he amount of KMn0 4 is related to the amount of iron(II) ion. The amount of iron(II) is converted to its mass, and the mass percent of iron in t he sample is determined. Soluti on First calculate the amount of KMn 04 • Amount of KM n04
= CKMnO,
X
VKMnO,
=
0.0195 mol KMnO. L
x 0.02435 l = 0.000475 mol
Use the stoichiometric factor to calculate the amount of iron(II) ion. _ 5 mol Fe'" 0.000475 mol KM nO, x 1 mcl Ketnn,
= 0.00237 mol Fe' "
The mass of iron can now be calculated, _ 55.85 g Fez+ 0.00237 molFe' " x _ 1 mol Fe"
= 0.133 g Fe"'
Fina lly, the mass percent can be determined. 0.133 g Fez", 1.026 g sample
x 100%
= 12.9% iron
Comment This is a useful analytical reaction because it is easy to detect when all the iron(II) ion has reacted. The Mn0 4 - ion is a deep purple color, but when it reacts with Fe2+ , t he color disappears because the reaction product Nn! " . is colorless. Therefore, KMn 04 solution is added from a buret until the initially colorless, FeH -containing solution just turns a faint purple color (due to unreacted KMn 04 ) , t he signal t hat the equivalence point has been reached.
Using an oxidation-reduction reaction for analysis by titration. Purple, aqueous KMn 04 is added to a solution containing Fe2+. As KMnO. drops into the solution, colorless Mn 2 + and paleyellow FeH form. Here, an area of the solution containing unreacted KMnO. is seen. As the solution is mixed, this disappears untilthe equivalence point is reached.
EXERCISE 4.15 Using an Oxidation-Reduction Reaction in a Titration Vitamin C, ascorbic acid (C6H ij06), is a reducing agent. One way to determine the ascorbic acid conte nt of a sample is to mix t he acid with an excess of iodine, C. H,O.(aq)
+ I, (aq ) + 2 H,O(f) ---> C.H. O.(aq) + 2 H,O" (aq) + 2 I - (aq)
and then titrate the iodine that did not react with the ascorbic acid with sodium t hiosulfate. The balanced, net ionic equation for the reaction occurring in t his titration is I, (aq)
+
2 S,O;'- (aq) ---> 2 I-( aq)
+ S,O. '-(a q)
Suppose 50.00 mL of 0.0520 M12 was added to the sample containing ascorbic acid. After the ascorbic acid/I, reaction was complete, the 12 not used in this reaction required 20.30 ml of 0.196 MNa ZS203for titratio n to the equivalence point. Calcu late the mass of ascorbic acid in the unknown sample.
4 .8
Spectrophotometry, Another Method of Analysis
Solutions of m an y co m po u nds are co lo red, a co nseq ne nce o f th e abso rp tio n o f lig h t (Fig ure 4. 15) . It is poss ible to me asure , quan titati vely, th e exte n t of ligh t abso rp tio n and to relate th is to the co nc e n tra tio n of the d issolved so lu te. T his kind of e xperime n t, ca lled spectrophotometry, is an im po rtant analytica l me thod. Every su bsta nce abso rbs o r tr ansm its ce rtain wavele ngt hs of rad ian t e nergy but no t o thers (Figu res 4. 15 and 4.16) . For examp le, n ickel (II) io ns (and chlo ro p hyll) absorb red and blue z'viol et ligh t, whil e tra nsm itti ng o r re flecting g ree n lig h t. Your eyes "see" the tran sm itted o r reflec te d wavele ng ths , th ose no t abso rbed, as th e co lor 4.8
I
Spectrophotometry, Anothe r Method of Analysis 189
FIGURE 4.15 Lig ht absorptio n and color, Abeam of white light shines on a solution of nickel(II) ions in water, and the light that emergesis green. The color of a solution is due to the color of the light not absorbed by the solution. Here, red and blue/violet light was absorbed, and green light is transmitted.
FIGURE 4.16 An absorptio n spectrophot ometer. A beam of white light passes through a prism or diffraction grating, which splits the light into its componentwavelengths. After passing through the sample, the light reaches a detector. The spectrophotometer"scans" all wavelengths of light and determines the amount of tight absorbed at each wavelength. The output is a spectrum, a plot of the amount of light absorbed as a function of the wavelength or frequency of the incoming or incident light. Here, the sample absorbs light in t he greenblue part of the spectrum and transmits light in the remaining wavelengths. The sample would appear red to orange to your eye.
Selected Transmitted
7""" I
l;gh1r -
Glowing filament
Prism or diffraction grating
Asolution of a colored compound
..
700
600
500
400
Wavelength of incident light (nm)
gree n . Fu rt he rmore , the specific wavelengths ab sorbed and transmitted are cha racteristic for a su bstance, and so a sp ectrum se rves as a "fingerprin t" of the su bsta nce that ca n hel p iden tify an unknown . Now su p pose you lo o k at two so lu tio ns of the same substa nce, o ne a d ee per co lo r than th e oth er. Your co m mo n sense te lls yo u that the inten sely colored one is the more co ncen trated (Figu re 4. 17a) . This is true, and th e int ensity of the co lor is a mea sure of the co nce n tratio n of the m at erial in the so lu tio n . In recent yea rs, spec tr o p ho to me try has become on e of the most freq uen tly used me thods of qu an titative an a lysis. It is app lica ble to man y ind ustri al and clinical p roble m s in volving the q uantita tive d et ermination o f compou nd s that are co lo red o r that reac t to form a co lored product.
Transmittance, Absorbance, and the Beer-lambert law To u nde rstand the exact relat ion sh ip o f ligh t abso rption and so lu tio n con centration , we need to defin e severa l ter ms. Tr an smittance ( T ) is the ratio of the am ou n t of ligh t tr ansmitted by o r passing throug h th e sam p le rela tive to th e am oun t o f light th at in itia lly fell o n the sam ple (the in cide nt ligh t) . P
P,
Incident light
Transmitt ed light Sample
Transmitt ance (T) 19 0
Ch a p t er 4
I
Stoichiometry: Quantitative Information About Chemical Reactions
P P,
inte nsity of transmitt ed light intensity of incident light
FIGURE 4 .1 7 lig ht absorption, concentratio n, and path length. (a) The test tube on the left has a solution of copper(II) sulfate with a concentration of 0.05 M. On t he right, the concentration is 1.0 Min copper(II) sulfate. More light is absorbed by the more concentrated sample, and it appears more blue. (b) The amount of light absorbed by a solution depends on the path lengt h. Here, both solutions have the same concentration, but the distance the light travelsis longer in one than the other.
(b)
(al
Ab sorbance is d efined. as th e ne ga tive loga rith m of th e tr an smittan ce, a n d yo u will not e th at absorbanc e and tr an smittan ce bea r an in verse relation ship. Th at is, as the abso rbance o f a so lu tio n increases , the tra nsmittan ce decreases
Absorbance
= - log T = - log PIPo
Goin g ba ck to o u r ex a m p le of a n a q u eo us so lu tio n of co p pc r( ll ) io n s in Figure 4. 17, if )' 0 11 h ave two co lored so lu tio ns, y O l1 may d educe that the blue r so lu tio n ap pears more blue becau se it absorbs mo re of the ligh t fallin g 011 it. T hat is, th e absorbance, A ,
~r a
sample in creases as the concentratio n increases.
Nex t, su p pos e th at the re are two lest tubes, both co n taining the same so lut io n at the same co nce n tra tio n . The only differen ce is th at o ne of the test tubes has a smalle r di am et er than th e o th er (Fig u re 4. 17b) . \Ve sh ine lig h t of th e sa me intensity ( Po) on both lest lubes. In th e first case, th e ligh t has 10 travel o nly a short distance through th e sam ple, whereas in the seco nd ca se it has to pass th ro ugh more o f the samp le. In the seco nd ca se more of the lig ht will be absorbe d be cau se th e path length is lo nger. In other wo rds , absorhrmce in creases as IMth length increases. T he two o bse rva tio ns d escri bed above co ns titu te th e Beer- Lambert law.
Absorbance (A) o: path length (t) X concentration (c)
• Beer- Lambert law The Beer- lambert taw applies strictly to relatively dilute solutions. At higher solute concentrations, the dependence of absorbance on concentration may not be linear.
(4.5)
A= e xe x c whe re
• A , th e abso rba nc e of the sa mple, is a dim en sionless number. • E , proportion alit y co ns ta n t, is ca lled t he molar absmtJtivit)'. It is a co nsta n t fo r a given su bsta nce , provid ed th e temperature and wave leng th are con stant. It has units of L/m ol ·cm . • t and c ha ve the units o f leng th (e m ) and co nce n tration (mo l/ L), respectively.
The Beer-Lambert law shows th at there is a linear relationship betuem a sample's absovban re and its concentration for a j..,riveJl !Jath u!ub,rth.
4.8
I
Spectro photometry, Anot her Met hod of Analysis
191
FIGURE 4.18 Spectrophotometers. The inst ruments illust rated here are often found in introductory chemistry laborat ories. (left) Spectronic 20 from Spectrontc Inst ruments. (right) Ocean Optics spectrometer (where t he digital data are collected by a computer).
Spectrophotometric Analysis Th ere are usu all y four ste ps in carrying o ut a sp ec trop ho to me tric ana lysis. • Record the absorption spectrum of the substance to be analyzed. In introd uctory c h e m istr y laborat ories, thi s is oft en do ne using a n in strumen t su ch as th e o nes sh own in Figu re 4. 18. The re sult is a spectru m suc h as th at for aq u eous perm an ga nai e ion s (]\(n 0 4- ) in Figure 4. 19. The spectru m is a plot of th e a bsorba nce of th e sam p le versus th e waveleng th of in cid e nt light. H ere , th e maximum in abso rba nce is at abo ut 525 nm . • Choose the wavelength for the m easurement. Accordin g to the Bee r-Lambert Law, th e absorba nce at eac h wavele ng th is proportional to co nce n tra tion . T he refore, in th eory we co u ld choose a ny wavelength for qua ntitative est im ations o f co nce n trati on . H o weve r, th e magn itude of the abs orba nce is im po rtan t, espe cia lly whe n yo u are tr ying to d etect very small amo un ts of mat erial. In th e spectra of perm an gan at e ion s in Figure 4.19, n ote th at th e difference in a bso rbance be twee n curves I a n d 2 is at a maximum at a bou t S25 nm , an d a t thi s wavelength th e change in a bsorba nce is grea tes t fo r a g iven change in concentration . That is, th e measurement of co ncen tra tio n as a fun cti on o f concen tratio n is most se nsi tive at th is wave le ng th . For th is reaso n , we gene-fa ll)!
1.0
CUlVe 1
0.8
-•s
~ 0.6
" ~
-e:
0.4
select the wavrlenglh cf maximum absorbance !or ou r measurements. 0.2
~--::-:-=---;::c-;:::::::~ 450 500 550 600 650 700
0.0 400
x.nm FIGURE 4.19 The absorption spectrum of solutions of potassium permanganate (KMn04) at different concentrations. The solution for curve 1 has a higher concentration t han tha t for curve 2.
192
Chapter 4
I
• Prepare a calibration plot. Once we ha ve c hose n th e wavele ng th, th e next step is to construc t a calibration curve or calibration plot. This co nsists of a plo t of a bso rba nce versus co ncentratio n for a se ries of sta n dard solution s wh ose con centrations are accurately known. Becau se of th e lin ea r rel ation bet ween conce n tra tio n a n d a bsorba nce (at a g iven wave lengt h a n d pathlength ) , this plot is a straig h t lin e with a positive slo pe . O nce th e plot has be en made , a n d the eq ua tio n for th e lin e is known, you ca n find th e co nce ntra tio n of a n un kn own sam p le fro m its a bsorba nce. Exam p le 4. 13 illustrates the prepa ration o f a ca libratio n cu rve a n d its use in d et e rmin in g th e conce n tra tio n of a speci es in so lu tio n .
Stoichiomet ry: Quantitative In formati on About Chemica l Reactions
Us;ng Spectrophotometry in Chemical Analys;s Problem Asolution of KMn0 4 has an absorbance of 0.539 when measured at 540 nm in a Lu-cm cell. Whatis the concentration of the KMn0 4?
Prior to determini ng the absorbance For the unknown solution, the following calibration data were collected For the spectrophot omet er. Concentrati on of KMn04 (M)
Absorbance
0.0300
0.162
0.0600
0. 330
0.0900
0.499
0.120
0.670
0.150
0.840
0.9 00 0.800
Strategy The first step is to prepare a calibration plot From the data above. You can then use the plot to estimate the unknow n concentration From the measured absorbance or, bette r, find the equation For t he straight line in t he calibration ptot (see pages 39 and 40) and calculate the unknown concentration. We shall do the latter . Solution Using Excel or a calculator, prepare a calibration plot from the experimental data. The equation for the straight line (as determined using Excel) is
y
~
Abso rbance
0.700
ec
•
~
~
~
0. 300 0.200
0.009
0.100
EXERCISE
0.00 0 0.0000
5.633 (Cone) - 0.009
Unknown concentration
0.400
«
If we put in the absorbance for the unknown solution, 0.539
0.500
~
5.633x - 0.009
= 5.633 (Co ne) -
0.600
0.0500
0. 1000
0.1500
0.2000
Concentration (M)
= 0.0973
4.16 Analysis Using Spectrophotometry
Using the following data, calculate the concentration of copper(II) ions in the unknow n solution. (The cell pathlength is 1.00 em in all cases, and the wavelengt h used in the det ermination was 645 nm.) Calibration data Concentration of Cu2+ (M )
Absorbance
0.0562
0.710
0 .0337
0.434
0 .0281
0.332
0 .0169
0.219
Absorbance of unknown solution containing Cu 2 + ions
= 0.418
4.8
I
Spectrophotometry, Another Method of Analysis
193
Chapter Goals Revisited
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Perform stoichiometry calculations using balanced chemicaL equations a. Understa nd th e prin ciple o f th e co nse rva tio n o f matte r, wh ich form s th e b a sis o f c h e m ical stoich io m e tr y. b. Ca lcu la te th e m a ss o f on e re ac ta n t o r product fro m th e m ass o f a no the r re act ant or produ ct by u sin g th e balan ced che m ica l e q uatio n (Se c tio n 4 .1) . Study Question(s) assignabLe in OWL: 2, 5, 8, 77, 81, 93, 95, 97, 99, 100; GoChemistry Module 7. c. Use a m o u n ts tab les to o rg a n ize sto ich io me tric info rm atio n . Study Question(s) assignabLe in OWL: 8. Understand the meaning of a Limiting reactant in a chemical reaction a. Dete rmin e whic h of two re a ct ants is th e lim iting re ac ta nt (Se c tio n 4. 2) . Study Question(s) assignabLe in OWL: 12, 14,96,132; GoChemistry Module 8. b. De te rmin e th e yiel d of a product based on th e limiting re ac tant. Study Question(s) assignabLe in OWL: 12, 14 , 16, 18. Calcul ate the theoreticaL and percent yields of a chemical reaction Ex p la in the d ifferences a m o ng a ctual yiel d , the oretica l yiel d, and pe rcent yiel d , a n d ca lcu la te percent yiel d (Sec tio n 4.3 ) . Study Question(s) assignabLe in OWL: 19 . Use stoichiometry to ana lyze a mixture of compound s or to determine the formul a of a compound a. Use sto ic h io me try princi p le s to a na lyze a mi xture (Se c tio n 4.4 ) . Study Question(s) assignable in OWL: 23, 123, 125, 127 . b. Find th e em pi ri ca l formula of an u nkn o wn co m p o u n d u sin g c he mi cal sto ic h io m e try (Se c tio n 4.4) . Study Question(s) assignabLe in OWL: 29, 34. Define and use concentrations i n solution stoichiometry a. Calc u la te th e co nce n tr a tio n o f a so lu te in a so lu tio n in u nits o f mole s pe r lite r (mo la rity) , a n d use co n cen trations in ca lcu la ti o ns (Se c tio n 4.5 ) . Study Question(s) assignabLe in OWL: 37, 39. 41. b. De scribe h o w to prepa re a so lu tio n o fa g ive n co n ce n tra tio n from th e so lu te a n d a so lven t or by di lut i o n fr o m a m o re conce ntra te d so lu tio n (Sec tio n 4.5). Study auestion(s) assignabLe in OWL: 46. 47 ,51. c. Calc u la te th e pH o f a so lu tio n fro m the co nce n tra tion o f hydronium ion in th e so lu tio n. Ca lcu late the hydronium io n co nc e n tra tio n o r a so luti on fr o m th e pH (Sect io n 4 .6) . Study auestion(s) assignabLe in OWL: 54. 55; GoChemistry ModuLe 9. d. So lve sto ic h io me try problems u sin g so lu tio n co nce n t ra tio n s (Se ctio n 4.7) . Study Question(s) assignabLe in OWL: 59, 62, 106, 107. e. Ex p la in how a titrat ion is ca r ried. o ut, exp lain t he procedure of sta n d a rd ization , a n d ca lcu la te co nce n tra tio ns or am ounts o f re a ct ants from titra tion d a ta (Se c ti o n 4.7) . Study Question(s) assignabLe in OWL: 67, 71. f. U n dersta n d an d us e the p rin ciple s o f spec tro p hotome try to dete rm in e th e co nc ert natio n of a spe cies in so lu tio n. (See to n 4.8 ) . Study auestion(s) assignable in OWL: 75.
194
Chapter 4
I
Stoichiometry: Quantitative Information About ChemicaLReactions
KEY EQUATIONS Equation 4.1 (page 168)
Percent yie ld
=
Percent yield
Equation 4.2 (page 175)
actu al yield x 100% t heoreti cal yield
Definition of mol arity, a measu re of th e co ncen tra tio n
o f a so lu te in a so lu tio n .
Molarity of x (c,) =
amount of solute x (mol) volume of solution (L)
A usefu l form o f th is eq uatio n is Amount of solute x (mol) = c, (moljL) X volume of solution (L)
Dilution Equation (page 179) T h is is a sho rtcu t to find, for ex ample , th e co nce ntrati on of a so lutio n (Cd) a fte r dilutin g so me vo lu me (Ve ) of a more co nce ntra te d so lut io n (cc) 10 a new vo lume ( Vd ) .
Equation 4.3 (page 179)
pH . T he pH o f a so lu tio n is th e negative logarithm o f
th e h ydronium ion co nce ntratio n .
pH ~ - log[H,O+ ] Equation 4.4 (page 181) Calc u la ting [113 0 +] fro m I' ll. The e q uatio n fo r calcu la tin g th e hydron ium io n co nce n tra tio n o f a so lu tio n fro m th e p H of th e so lu tio n . [H,O+]
~
10- pH
Equation 4.5 (page 192) Be e r-Lambert La w. T he a bsorban ce of ligbt (/I) by a su bs ta nce in so lu tion is eq u al to th e mol a r a bso rp tivity of th e su bs tance (e) , th e pa th le ngth o f th e ce ll (C), a n d th e co nce n tra tio n o f th e so lu te ( c). Absorbance (A) rye pat h lengt h (I') X concentration (c)
A =e X C Xc
STUDY QUESTIONS Onli ne homework for this chapter may be assigned in OW L. £ denotes challenging questions.
•
denotes questions assignable in OWL.
Blue-numbered questions have answers in Appendix a and fully-worked solutions in the Student Solutions Manual.
Practidng Skills Mass Re latio nsh ips in ChemicaL Reacti ons: Basic St oich io metry (SI'I' Examl,[f 4. 1 a nd Chl'l llisfryNow Screens 4.2 and 4.3.)
1. Alu min um re acts with oxyge n to give alumi n u m oxide. 4 AI(s ) + 3 O ,(g ) --> 2 AI, 0 3(S)
Wha t amount of ()1, in moles, is neede d fo r com plete react io n with 6.0 mo l o f AI? Wha t mass of AI1()3, in gr ams, can he p rod uce d? 2. • What mass of lI e l, in gr a ms , is re q uired 10 reac t with 0.750 g of AI(O lIb? W hat mass of water, in gram s, is produced? AI(O H), (s) + 3 H CI(aq ) --> AICI,(aq ) + :> H, O (f)
I
19 5
3. Like man y me tal s. a lu m in um re ac ts with a h al o gen to g ive a m et al h a lid e (see Figure ~ . 1 2) . 2 Al (s)
+
:1 I\r , (I ) -> Al,l\r,;{s)
Wh at m ass o f fi r2, in g ra ms, is required for co m p le te re act ion wit h 2.56 g of AI? wh at m ass of wh ite , so lid A1 2Br(j is ex pe cted ?
4. Th e bal a n ced eq u a tio n for a re ac tion in th e p ro cess o f th e redu ction of iron o re to the metal is Fe,03(s)
+
3 C O(g ) -> 2 Fe (s)
+
3 CO, (g )
(a) Wh at is th e max im u m m ass o f iron, in g ram s. that ca ll be obtained fro m 454 g ( 1.0 0 Ih ) of iron ( lll) o xide ? (b) Wh at mass of CO is required to re act with 4 54 g o f Fe 2() 3?
5. • Methan e , C H I, h u rns in ox yg e n . (a) Wh at arc th e p roduct s of the reaction ?
(h) Write th e balan ced eq ua tio n fo r the re action . (c) Wh at mass of 0 2, in g ra m s. is re qui red fo r co mple u- com bustion of 25.5 g of meth ane? (d) Wh at is th e total mass of p roducts ex pec te d from the co m b us tio n o f 25 .5 g of meth an e ?
6. T he form at io n of wa te r-ins o lu ble silver ch lo rid e is uscful in th e analysis of c h lo ride-co n ta in ing su bsta n ces. Co ns id er th e followin g unbalanced. eq ua tion : l\aC l, (aq )
+
Ag;'\ ( l., (a q ) -> Ag CI (s)
+
l\ a (NO"j,(aq )
(a) Write the b ala n ced eq ua tio n . (h) Wh at m ass of AgN 0 3• in gram s. is re q u ired fo r
co m plete re ac tio n \'I.·. ith 0 ,156 g of Rae l2? Wh at mass of Agel is produced ?
Amounts Tables a nd Chemical Stoichiometry For fYu:h question. hl'low, set I1j) an (111101111 1.\ taUI' that lists the initial amount or all/aun ts f!f reacta nts, the d Ul1IKf'J in amou nts of reaaa nts and products, a nd IIt I' amounts of reactants and [nndncts after reacti on. SI'f' jHl{;" / 5 9 and Example 4. /.
7. A m aj or so u rce o f air poll ution ye ars ag o was th e m etals in dustry. On e com m on pro n .'ss involv ed "roastin g" m e ta l su lfi d es in th e air: 2 PbS (s)
+
3 O , (g ) -> 2 I' bO (s)
+
2 SO ,(g)
If yo u h eat 2.50 mol of Pb S in th e air, wha t am o u n t o f O ~ is re q uired fo r co m p le te reac tion ? Wh at am o un ts of Pb O a n d S0 2 are e x pe cted?
8. • Iro n ore is co n verted to ir on metal in a re actio n wit h car b o n . 2 Fe,O,,(s)
+3
C (s) -> 4 Fe( s)
+3
CO, (g )
If 6.2 mol of Fe20 J(S) is u sed , what am o u n t o f C (s) is n e ed ed . a n d wha t a mo u n ts o f Fe a n d CO 2 arc p roduccd?
196
I
9 . Ch rom iu m m etal re act s with o xygen to g ive chromium t l l l ) o xid e , Cr ~0 3 ' (a) Wri te a b alan ced eq ua tio n for th e re actio n . (h) If a pi e-ce o f ch ro m iu m h as a m ass o f 0 .175 g, wha t mass (i n g ra ms) o f C r20 J is produce d if the metal is co n ve rte d co m p le te ly to th e o xide ? (c) Wh at mass o f O 2 (in g ra ms) is req u ire d fo r th e reac tion ?
10. Etha n e, C 211 6 , burns in o xyge n . (a) Wh at arc th e p roducts of th e re ac tion ? (h) "Vrite th e balan ced eq ua tio n fo r the re actio n . (c) Wh at mass o f 0 2, in g ra ms , is req u ire d fo r complete co m b us tio n of 13.6 o f eth ane ? (d) Wh at is th e total m ass o f p roducts expect ed from th e co m b u st io n of 13.6 g of eth an e ? Limiting Reactants (.\'('1' J.,'X(lIIljJiI' 4.2 and Exercise 4.2. SI'(, also Chl'1ui.\lry.No1ll Sorens 4.4 and 4.5.) II. Sod iu m su lfid e , Na2S, is u sed in the leath e r industry to re m ove h a ir from h id es. Th e Na 2S is m ade by the re action Na, SO , (s)
+
4 C (s) -> Na,S (s)
+
4 CO rgi
Su p pose yo u mi x 15 g of Na2SClI and 7.5 g of C. Which is th e lim iting react an t? What m ass of Na2S is p ro duced ?
12. • Ammon ia g as ca n he p repare d hy th e react io n of a me tal oxide su ch as ca lciu m o xid e with am mo n iu m chlo rid e . CaO (s )
+
2 N II , CI(s ) -> 2 N H , (g)
+
H,O (g )
+
CaCI, (s)
If 112 g of Ca O an d 224 g of N H ,CI are mi xed , wha t is the lim itin g rea ct ant. an d what m ass of NII J ca n b e p ro du ced? 13 . The co m p o un d SF6 is made by burning su lfu r in a n a tmosp here of fluorin e . T he balanced eq u ation is S, (s )
+
24 F, (g ) ->
~
If yo u begin with 1.6 mol o f su lfu r, F2, wh ic h is th e lim itin g re agent?
SF, (g )
s;
and 35 mol of
14. • Disulfur d ichl oride , S ~C I ~ , is used to vu lca n ize rubber. It ca n be m ad e by tre ati n g molten su lfu r with g aseous chlorine:
Sta rt in g with a m ix ture of 3 2.0 g of sulfu r a nd 7 1.0 g o fCl 2, (a) Which is th e limitin g re ac ta n t? (b) Wh at is the th e oret ica l yiel d of S2C12? (c) wh at m ass o f th e excess re acta n t re m ai n s when the react ion is co m ple te d ?
... more challengi ng • in OWL
Blue-numbered questionsanswered in Appendix0
15. T h e reactio n of meth an e and wa ter is o ne way to p re pa re h yd rog e n tor use as a fuel: C H., (g)
+
H, O (g ) -> CO (g )
+ :J
H, (g )
If you begi n with 99 :) g or en , a nd 25 10 g of wa te r, (a) Whi ch reactant is th e limitin g rea ct an t? (b) Wh a t is the m axi m u m m ass of H 2 th a t can be prepa red? (c) Wha t mass o f the excess re actant rem ai ns whe n the re ac tio n is co m ple ted?
16• • Alu m in u m c h lo rid e, AICI:h is m ade by treatin g scrap alu m in u m with ch lorin e .
2 Al (s)
+ 3 Cl,( g ) -> 2 AICI, (s)
If yo u be gin with 2. 70 g o f AI and 4.0.5 g o f C12, (a) Whi ch react ant is lim itin g ? (b) Wha t m ass of AlCl j can be p roduced? (c) Wh at ma ss of th e ex cess re ac ta nt re m a ins when th e react ion is co m p le ted? (d ) Se t up an a mo u nts table fo r th is proble m . 17. I [ ex am- (C{;l II ~ ) b urns in air (0 2 ) to g ive CO 2 a nd H 20 . (a) Write a balan ced eq ua tio n fo r the reacti on . (b) If 2 15 g of C\iH 1,/ is m ixe d wit h 2 15 g of O 2, wha t m asses of CO 2 ami 11 20 an' p roduced in th e react ion ? (c) Wha t ma ss o f the e xce ss reacta n t rem a ins afte r th e he xan e h as been hurtl e d? (d ) Set up a n amou nts tabl e for thi s problem . 18. • Asp irin , Cr;Il/(O COC l l j )C 0 211, is produce d by th e re actio n o f salicylic ac id , Cr;H 4 (O H)C0 2H , ami ace tic a n hyd ride , (C H, C O),O (page \(;8 ) . C,;I I, (O Il)C O, II (s ) + (C II ,CO ),O (I)-> C"H ., (OCOCH , )CO,H (s) + C H , CO , H (I) If yo u mi x 100 . g of e ach o f th e re actan ts, wha t is the m a ximu m m ass of aspirin th a t can he obt ain ed?
Percent Yield (See Exercise 4.3 a mi ChrmiJtt)'Now Screen 4.3.)
19. • In Examp le 4. 2, yo u fo u n d th at a parti cular m ixtu re of CO and H 2 co uld p roduce 4 07 g C l1jO II . C O(g )
+
20 . Am mo nia gas can be p repa red by the following reacti on :
+ 2 N H ,CI(s)
-> 2 NII , (g)
+
II , O (g )
+
CaC I, (s)
If 112 g of CaO and 224 g of N I J4CI are m ixe d , the theoretical yie ld of N H 3 is 68. 0 g (Study Que stio n 12 ) . If on ly 16.3 g of NH ] is ac tually obt a ined , wh at is its percent yiel d? ... more challenging • in OWL
CuSO , (aq )
+
4 N H,,(aq ) -> C u (N H"lrSO ., (aq )
(a) If yo u use 10.0 g o f c u s n j an d excess N H 3 , wha t is the th e o re tical yiel d of Cu(N lh ).,SO ..? (b ) If yo u iso la te 12.6 g ofCu (NH 3) 4S0 4, what is the pe rcent yiel d of C u( N l l j) 4S04?
22. Bla ck sm o ke rs are found in th e depth s o f the ocean s (pa ge 112 ) . T hinking th at th e condition s in th e se sm o ke rs mi gh t he co n d ucive to th e form at ion o f o rga n ic co m po u nd s, two che m ists in Germa ny fou nd th e fo llow in g reaction co uld o ccu r in sim ila r co ndi tio ns . 2 C II "SII
+
CO -> C II "CO SC II "
+
II , S
If yo u hegin with 10.0 g o f C H]S H an d ex ce ss C O , (a) Wha t is th e th eo retical yie ld of C H 3C O SC H]? (h ) I f 8 .65 g of C l l:\CO SC H 3 is iso la te d , what is its p e rce n t yie ld?
Analysis of Mixtures (Sf'!' Exa m/llr 4.3 a nd CllPmi.\ l1y,vow Screen 4. 7.)
23 . • A m ix tu re of CUS0 4 an d CUS0 4 . 5 1120 h as a mass of J .245 g . Aft er hea tin g to drive o ff all th e wa ter, th e m ass is o n ly 0.832 g. Wha t is the mass pe rcent of c' USOl . fJ H ~O in the mi xture ? (Set' page 97 .) 24. A 2.634-g sa m p le con ta ini ng impure CuCI2 ' 2 H 20 was hea le d . T h e sam ple m ass after hea tin g to d rive off th e wa te r was 2.12fJ g . Wha t was the m ass perc ent o f CuC l2 . 2 H 20 in th e original sam ple? 25 . A sa m ple of lim es to ne an d o ther so il ma terials \v'as h ea te d , and the limesto ne d e composed to g ive ca lciu m ox id e an d ca rb on d io xid e .
+
CaCO ,,(s) -> CaO (s)
CO ,(g )
A 1.50 6-g sam p le of lim esto ne-containing material gave 0.558 g o f CO 2, in ad di tio n to Ca O , afte r be in g hea te d a t a h ig h te mpe rature . \ Vh a t is the ma ss pe rc e-nt of c' aC,0 J in th e o rig in al sam ple? 26 . At hi g her te m p e ra tu re s, Na I IC0 tively to Na2C0 3'
3
is co n vert ed quantita-
2 Na HCO ,,(s) -> Na, CO ,,(s) + CO ,(g ) + H 20 (g )
2 1I, (g ) -> C H, O Il (I)
If on ly :·t12 g o f C H 30 H is ac tua lly produced , what is th e perc ent yiel d of th e compou nd ?
Ca O (s)
21. Th e d eep blu e co m po u nd Cu (N lh) 4S0 4 is m ad e by th e re action of co pperf l l ) sul fa te an d am m o n ia.
Blue-numbered questionsanswered in Appendix 0
He a tin g a 1.718 4-g sam ple o f impure Na HC0 3 g ives 0.19 6 g of CO 2, Wh at was th e m ass pe-rcent o f Na IleO] in th e origi nal 1.7 184-g sa m ple? 27 . A pe stic ide co n tai ns th all ium (I ) su lfa te , T 12So.1• Disso lvin g a 1O.20-g sam ple o f im pu re pesticid e iii wate r a nd ad d ing so d iu m iod id e precipitate s 0.19 64 g of thalli um ( I) io d id e , T ll. Tl , SO, (aq )
+
2 Na l (aq) -> 2 T il (s)
+ Na ,SO, (a q )
Wh at is th e mass pc rc e nt of T I2SO j in the original 1O.20-g sa m p le?
I
197
28 . ... T he aluminum in a 0.764-g sam p le of an unknown m aterial was precipitat ed as alu min um h ydroxide . AI(O lI ):" whi ch was th e n co nverted to AI ~/ ) 3 by he ating stro ng ly, If 0.127 g of A1 20;1 is ob tain ed from th e 0.764g sa m ple , wha t is th e m ass pe rcent of alum inum in th e sam p le?
Using Stoichiometry to Determine Empirical and Molecular Formulas (."1,,(' Example 4.4, Exercise 4.6, and Chpmi.\fJ )lNow Sorcn 4.8.) 29 . • Styre ne. the buildin g block of polystyre ne , co ns ist." of only C a nd H . If 0.43H g o f styre n e is b u r n ed in oxygen and produces 1.48 1 g of CO 2 a nd 0.303 g o f 1120. wh at is th e empirica l fo rm ul a of styre lle? 30. Mc sitylcn c is a liq ui d hyd roca rbo n . Bu rn in g 0.11 5 g of th e compound in o xyge n g ive s 0.379 g o f CO ;! an d 0.10 35 g of H 20 . Wh at is th e e m p ir ica l for m u la o f m esitvlc ne? 3 1. Cycl o pe manc is a simple hydrocarbo n . If 0.0956 g o f th e compound is bu rn ed in ox ygen , O.:WO g o f CO 2 and 0.12 3 g of H 20 a re isol at ed . (a) Wlrat is till' e m pirica l formul a of cyclo p cn ta ne? (b ) If a se para te ex pe rim e n t ga ve 70 . 1 g /mol as the mola r m ass of th e compound , wh a t is its molecul ar fo rmula? 3 2. Azulcuc is a beau tiful blu e hyd ro ca r bo n . 11'0. 10 6 g o f the compound is burned in ox ygen . 0.364 g o f CO 2 and 0.0 59 6 g of H 20 are isolated. (a) Wh at is th e e m pir ica l fo rmula of azu lc ne? (b) If a s e p a ra tl~ experimen t gave 12 H.2 g/ m o l as th e m o la r mass of the compound, what is its molecul ar fo r mula? 33 . An un known co m po u nd has the formu la C ~H /) ~. You hurn 0.0956 g o f th e co m p o u nd a nd iso lat e 0 .1 ~~ 56 g of CO 2 and 0.08 33 g o f H 2 0 . \ Vhat is the e mp iri ca l form ula of till' compound ? If th e molar m ass is 62.1 g / m ol , wha t is the m olecular fo rmu la? (See Exe rci se 4 .6.) 34. • An un known co m po u nd has th e formula C..H .)O ~. 'hu hum 0.15 23 g of the co m po u n d and iso la te 0 371 8 g of CO t and 0.152 2 g o f H 20 . What is the empirical formul a of the compound? If th e molar m ass is 72.1 g / mol, wha t is the molecular formula ? (Se e Exercise 4. 6.) 35 . Nick el forms a compound wit h carbo n monoxide , NU CO )y. To d e term ine its formu la, yo u ca re fu lly h ea t a O.097 3-g sam ple in a ir to convert the ni ckel to 0.0426 g o f NiO and the CO 10 0.100 g of CO 2, Wha t is th e empiri cal form ul a of N i..{CO)i 36. To find the fo rm u la o f a compoun d composed o f iron and carbon mon oxide , Fe ..(CO) ,., the compound is burn ed in pure oxyg en to g- ive Fe 203 and CO 2, If yo u burn 1,959 g of Fe ..{CO ) .~ and o b ta in 0.799 g of Fe20:1 and 2.200 g o f CO 2, wha t is the empi rica l formul a of
Fe . (CO ) ,? 198
I
Solution Concent ration (See Ex ample 4.5 and ChemiJll)'Now Screen 4.9.) 3 7. • If 6.7:1 g of Na2CO :l is di ssolved in enoug h water to m ake 2S0. m L of so lu tio n , what is th e m o la r con cen tratio n of th e so d iu m carbonate ? What are th e molar co ncen tra tio ns of the Na " and C0 32- io ns? 38. Som e po ta ssiu m di chro ma te ( K2Cr20i), 2. 33 5 g. is d isso lved in enough wa ter 10 m ake e xac tly SOO. rn l , of xoIuti o n . \Vhat is the molar co nce u tra tio u of th e potassiu m di chromat e? Wh at a re the mol ar con cent ration s of the K+ a nd (:r20i2 - ion s? 39 . • What is the mass of so lu te , in gra ms , in 250 . m L of a 0.01 25 ~1 so lu tio n o f KMno.j? 40 . Wh at is t he m ass of so lu te, in g rams, in 125 mL o f a 1.0 2 :~ X 10 - 3 M so lu tio n of Na:lP0 1? Wh a t is the mo la r con centration o f the Na+ and PO l - ion ? 41 . • Wh al vo lume of 0.12:1 M NaO l l , in millilin-rs. contains 25. 0 g of Na O ! I? 4 2. \Vh at vo lu me of 2.06 g o f so lu te?
~1
KMno.j, in lite rs, co n ta ins 32 2
43. For ea ch so lu tio n , idcntitv th e io ns that exist in aqueous so lu tio n. a nd spec ify the co nc e n tra tion o f e ac h ion . (a) 0.25 M (NH,),SO , (h) 0.123 M Na, CO" (c) 0.056 M HNO , 4 4. Fo r each so lut ion . id en tify th e io ns th at exi st in aq ue· ous so lution , and s p eci f~' tl u- con ccu u-atiou o f eac h ion . (a) 0.12 M RaCI, (h) 0.0125 M CnSO, (c) 0.500 M K2Cr20 i
Preparing Solutions (See Exercises 4.7-4.9, Exam/J[e 4.6, an d Chemisl1)'NotIJ S C1"PP lI
4. JO.)
45. An ex pe rime n t in yo ur laboratory require s 500. m l. of a 0.0200 .\1 so lu tio n o f 1'\a2CO :l' ' htl a rc given so lid Na2CO:h d istill ed water, an d a 500.-mL vo lu m etric fl ask. De scribe how to prepare th e requi red so lu tio n . 46 . • Wh at mass o f o xali c aci d , H 2C 20 4• is required to prepare 250. m l , o f a so lution th a t has a con cen tration of 0.1.0 M II, C,O ,? 47 . • If yo u d ilute 25.0 m L of 1.50 M h yd rochlo ric acid to 500. ml .. wh at is the molar co nce n tra tio n of the d ilu te acid? 48 . If 4.00 m L of 0.02 50 ~'1 coso, is diluted to 10.0 m l, with pure wa ter, wha t is th e molar concen tra tio n of co p per( II) sulfate in the dilu ted so lu tion? 49 . Whi ch of the following me thods wo u ld YO ll use to prepare i.oo L of 0.12 5 M II 2SO ..? (a) Dilute 20.8 m l , of (l.OO M H 2S0 4 to a vo lu m e of 1.00 I.. (il) Ad d 950. m l, ofwater to 50.n m l , o f 3.00 M 11250.,. ... more challenging
• in OWL
Blue-numbered questions answered in Appendix 0
50. Whi ch of th e fo llowi ng m e th o ds w ould yo u use to prepa re :~ O O. III L of 0.500 ~1 K;?Cr;?07? (a) Add 30.0 m L of 1.50 M K;?Cr ;?07 to 270 . m l . of water.
(b) Dilut e 250 . mL or 0.600 M K;?Cr ;?07 to a vo lu me of 300 . mL.
Stoichiometry of Reactions in SoLution (Set' f ,x fl m!JI(' 4.8 fi nd ChemiJ{,)'N01L) S Cn'(' 1I 4. /2 .) 59. • \-Vhat vo lum e of 0.109 .\1 II NO ,I' ill m illili ters, is rcq uircd to re act co m plet ely wit h 2.50 g of Ba (O H h ?
2 11:\0 3(a'l )
Serial Dilutions (See A Closer Look: Serial Dilu tions, !Hlgp I RO. )
5 1. • You have 250 . mL 0£0.136.\1 Hf .I. Using a vo lu metri c pipe t, yo u take 25. 00 m l . of th at so lution an d di lu te it to 100.00 m l. in a vo lu metri c flask. No w yo u tak e 10.00 m l. of th at sol u tio n, using a volumetric pip et , a nd di lute it to 100 .00 m L in a vo lu m e tric flask. Wha t is the co nce n tra tio n of h yd ro ch lori c ac id in the fin al solu tion ?
52. .A. Suppose yo u haw 100.00
m l.. a so lu tio n o f a d ye and
tr an sfer 2.00 mL of the so lu tio n to a 100.00-m!. vo lumetri c flask. After add in g wa ter to th e 100.OO mL ma rk, yo u take 5.00 m l , of th at so lu tio n a nd aga in dihu e to 100.00 ml, If yo u find the d ye co nce n tra tio n in the final d iluted sa m p le is 0.000 158 M , what was th e d ye co uc cu tra tio u in the orig in al solution ?
{See Exmnpl(, 4. i and ChemislryNow Screen 4. 11.) 53 . A ta ble wine has a pH of 3.4 0. Wh at is th e hyd ro n iu m ion con centratio n of th e winc? Is it acid ic or basic?
l\ a (O H ), (s) ---> 2 11, 0 (1) + Ba(N0 3), (a'l )
60. \ Vha t ma ss of Na;?CO :I , in g ra m s, is required for co mplete reaction with 50.0 mL o f O.l25 M H N0 3 ?
Na,CO j( a'l)
+ 2 II NO j (aq ) ---> 2 NaNO ,,(aq )
+
CO , (g)
+
H, O (f )
6 1. When a n e lectric curren t is passe d th ro ugh a n aqueous solution of i\'aCI. the valua ble indust ria l ch e m ica ls H;? (g ), CI;? (g ) , and NaO H are produced.
2 NaCI(aq)
+ 2 H, O (I) ---> 11, (g)
+ CI, (g) + 2 :\aO I1(a 'l )
\ Vha t ma ss of NaO H can he formed from 15.0 L of 0.35 ~1 Na CI? What m ass of ch lorine is obtained?
62. • l lvd razinc , N;? I1.j, a base-lik e ammoni a , ca ll react with sulfu ric a cid .
2 :\, H, (aq )
Calculating and Using pH
+
+ H,SO , (aq ) ---> 2 :\, H, +(aq ) + SO/- (a'l )
Wh a t ma ss of hyd razin c re acts with 250. m L of 0. 146 M II ,SO , ? 63 . In th e p ho to graphi c developi ng proce ss, silver bromide is di sso lved by add ing sod iu m th iosu lfat e .
54. • A saturated sol u tio n of m ilk of m agn esia , Mg (O H h , has a pH of 10.5. Wh at is th e hyd ro ni u m ion cou ccntrat iou of th e solu tion? Is th e so lutio n aci d ic o r bas ic?
Agl\ r (s)
55 . • Wh at is th e hvdronium ion con ce n tra tion of a 0.0013 i\.I sol ution of HNO:l ? Wha t is its p H ?
If yo u wa n t to d issolve 0.225 g of AgB r, what vo lu me of 0.0 138 M Na;?S;?03. in m illiliters, sh o u ld be used?
+ 2 Na,S,0 3(aq) ---> ~a:IAg (S ~ O :l h (aq )
+ Na Rr (aq )
56. Wh a t is the h ydronium ion co n ce n tra tio n of a 1.2 X 10- 4 M solu tion of H el O t? What is its pll ? 57. Make the fo llowin g co nve rsio ns . In e ach case , te ll whether th e so lu tio n is acidi c o r ba sic.
pH (a) 1.00
(h) 1050 (c) _ _ (d ) _ _
1.3 X 10-" M 2.3 X 10- ' M
58. Make the fo llowing couvcrsi ou s. In e ach ca se , tel l whethe r th e solu tio n is acidi c o r ba sic.
pH (a) _ _ (h ) _ _
6.7 X 10- '0 M 2.2 X 10- " M
(c) 5 .25 (d) _ _
... morechallenging • in OWL
Blue-numbered questionsanswered in Appendix 0
_----_....
~ ...
(a>
(b)
Silver chemistry. (a) A precipitate of AgB r formed by adding AgN0 3 (aq) to KB r(aq). (b) On adding NaZS203(aq), sodium thiosulfate, the solid AgBr dissolves.
I
199
64 . You ca n di ssolve an alum inum so ft-d rin k can in a n aq ueo us b ase suc h as potassium h ydroxide . 2 AI(s )
+ 2 KOH (aq ) + 6 H,o (1) -> 2 KAI (O ll) , (aq )
+ 3 1I, (g )
If you place 2. 05 g of aluminum in a be aker with 18 5 mL of 1.35 M KOH , will a ny aluminum remain ? 'Vh at m ass o f KAI «)II )~ is produced? 65 . \Vllal vo lu me of 0.750 :VI Ph ( i\03h , in m illiliters, is requi red to re ac t comple tel y with 1.00 L of 2.25 M NaCI sol ution ? T he halanced e q ua tion is I' h (1\0 3),( aq )
+ 2 NaO H (aq ) -> Na,C, O ,(aq )
+ 2 11,0 (1)
Titrations (SPf' Exa mples 4.9-4. J2 and ChnniJll )Nm.l! Sur-en 4.1 J. )
67 . • What vo lu me 01'0 .812'\1 HC l, in milli liters, is re quired to titra te 1.45 g of Na O H to th e eq u ivalenc e poi nt? NaO H( aq)
+
HCI (aq ) -> H,O (l )
+
NaCI(aq)
68. Wh at vo lu me of 0.955 M H Cl, in m illilite rs, is requi red to tit rat e 2.1.")2 g of Na2C O j til the eq uivale nce p oint?
+ 2 II CI (a'l ) -> H, O (l)
+ CO, (g ) + 2 NaC I(a q )
69 . If 38.55 mL of HCI is requi red to titrat e 2. 150 g o f N a ~C O :\ a ccord ing to the fo llo wing eq ua tio n , wha t is the co nce ntration (m ol/ L) of th e Hel so lu t io n? Na ,C O , (aq )
+ 2 IICI (aq ) -> 2 NaCI(aq )
+ C O,(g) + 11,0 (1)
70. Po tassiu m hyd rog e n p ht halate , KHC ~J-I.l() .lo is u sed to sta nd a rd ize so lu tio ns of base s. Th e aci dic an ion re acts with stro ng bases ac cording to th e follo wing net ion ic e qu a tio n : II C,II,O ,- ("q )
+ Oll - (aq ) -> CsH..D,' - (a q)
+ H ,O (l )
If a 0.90 2-.g sa m ple o f potassium h ydrogen phth a late is d issolve d in water an d titrated to th e equ ivalen ce po int with 26.45 mL of NaO H, what is th e molar concentration of th e NaO I I?
71. • Y(m have 0,9 54 g o f an unknown aci d , 11,;,!f\ , wh ic h re acts with Na O H a cco rd ing to th e balanced eq uatio n H,A (aq )
+ 2 Na O H (a q ) -> i'ia ,A (a q ) + 2 H,o (1)
If 36. 04 m L of 0.509 M NaO n is req u ired to tit rate th e aci d to till' sec o nd equivalen ce po int, what is th e mol a r m ass o f the ac id?
200
I
+ 3 i'ia O Il (aq ) ->
113 C"I1'()7(aq )
'I H,O (l )
+ Na"C"H , 0 7(aq )
Tartm i c arid:
+ 2 NaO Il (aq) ->
II , G ,II.o,,(aq )
2 11,0 (1) + Na,C, II , O,,(aq )
+ 2 NaI'O ,(aq )
66. Wh at vo lu me of 0.125 M o xalic acid , H2 C 20~ is re qui red to re act with 35.2 mL o f 0.54 6 M NaO H?
Na, CO, (aq )
Citric acid:
+ 2 NaCI(a q) -> l'bCI, (s)
H,C,O., (aq )
72 . An unknown so lid acid is e ith er citric ac id o r tarta ric aci d. To determi ne which aci d yo u h ave, you tit rat e a sam ple o f th e so lid with aq ueo us Na O l 1 an d from th is det erm ine the molar m a ss o f the unknown aci d . Th e appropriate equation s a re as follows:
A O.956-g sa m ple requires 29. 1 mL o f 0.5 13 M l\'a O H to consum e the acid co m ple te ly. Wh at is the unknown acid? 73. To ana lyze an iron -con tain ing com po u nd . yo n co nvert all th e iron to Fc/ " in aqueous so lu tio n an d th en tit rate the so lu tion with sta nd ardized K'\fnCl,. The bal an ced , n e t io n ic e q ua tio n is MnO , - (aq )
+ 5 Fe2+(aq ) + 8 H 30 +(a q )-> Mn ' +(aq )
+5
Fe3+(aq )
+
1211,0 (1)
A 0 .598-g sa m ple of the iron-co n ta in in g co m po u nd requires 22.25 m L ofO .01 2 :~ M K..\fno.. for titrat ion to th e equivale n ce point. Wh at is the mass pe rce-nt of iro n in the sam p le ? 74. Vitamin C h as th e form ula C1j H I'.0 ij. Resid es being an aci d , it is a re d ucing agen t. On e m et h od fo r determinin g the a mount of vita mi n C in a sam ple is the refo re to titr a te it with a so lut io n of bromin e , Rr2' all oxi di zing agen t. Cr,H, O ,( aq )
+
IIc, (aq ) -> 2 Hllr (aq )
+
C"H ,,(),,(aq )
A I .OO-g "c hewab le" vita m in C tablet requi res 27.8:> mL of 0.102 1\1 Br2 fo r tit rat ion to the eq uivalence po int. Wh at is the m ass of vitami n C in the tablet? Spectrophoto metty (See Sedion 4.8. T IIP /n'Oh!n m belmo are mIa/Jln l from Funda mentals sa, pd., 0' D , A . Skoog, D , st. Hlfst, F.J. I iollr r, and S. R. Crou ch, Tlunnson /B rooks-Cde, Belmont, CA 2()()4.)
(!{ A nalJfim l ClwmisllJ,
75. • A so lutio n of a d ye was a na lyzed by spectrop ho to m etr y. and the followin g ca lib ra tio n d at a w ere co lle cte d. Dye Concentration
0.50 X 10- " M
Abso r ba nce at 475 nm
0.24 0.36 0.44 0.59 0.70
1.5 X 10- (,,\1
2.5 X 10 - 6 ,\.1 3.5 X l o- liM 4.5 X IO- fj M
(a) Co ns truct a calibrati on plot, an d dete rmin e t he slo pe a nd in te rce pt. (h ) What is the d ye co nc e n tra tio n in a so lu tio n with A = 0.52?
... more chaLLenging
•
in OWL
BLue-numbered questions answered in Appendi x 0
76. Th e nitrite ion is in vol ved in th e bio ch emical nitrogen cycle . \ iHI ca n an alyze fo r th e nitrite io n co n te n t of a sam ple usin g spec tro ph o tome try by fir st usin g seve ra l o rganic co m po un ds to crea te a co lored co m po u nd fro m th e ion. The fo llowi ng dat a we re co llec te d . NO~-
Ion Concentration
Absorbance at 550 nm of Nitrite-Ion Containing Solution
2.00 X 10- '; M 6.00 X 1O- 1i M 10.00 X Io- ti M 14.00 X 100tiM
0.065 0.205 0.:l3H
0.474 0.598 0.402
I H.OO X 10- " M
Unkn own so lu tio n
(a) Co nstruc t a ca libra tio n pl ot , and d etermin e th e slo pe ami intercept. (b ) Wh at is th e nitrite io n concentratio n in th e un kn own so lution ?
General Questions on Stoichiometry 771£'v '111£'JIi01B a n' not d£'Jigu tttnf as to
lYlu'
or
taouion
i n lit £'
dmpter: Tlu)" may combine seueml COl/UPls f rom the dutinn: 77 . • Su p pose 16.04 g of be nzen e , C1;H 6 , is burned in o xyge n . (a ) What arc th e prod ucts o f the reac tion? (b ) What is th e balan ced e q ua tio n fo r the re action? (c) What m ass of O 2, in grams, is re q uir ed fo r co mplete combu stion of ben zene ? (d ) What is th e tot al m a ss of product s expected from 16.04 g of be nze ne?
79.
body d ea ls with e xce ss nit ro gen by excre ting it in th e fo rm of urea, N I 1 2 CO ~;'I1 2 ' Th e reaction produ ci ng it is th e co m bin atio n of arginin e (Clif Tl'IN.10 2) with wato r to give urea an d o rn ithin e (C"HI2N20:t).
" (lUI'
l,l i H I .I N 40 2
Argin ine
+
H 20
~
NH 2CON H 2 + C .~HI 2 N :t0 2 Urea Ornithine
If you excre te 95 mg of ur e a , what m ass of argi n in e mu st ha ve be en u sed ? \Vhat m ass of ornithin e mu st h ave been produce d ? 80. Th e re action of iro n m etal an d ch lo rine ga s to g ive iro n (III ) ch lo ride is illu strated in Figure 3 .~ . (a) \Vrit e th e bal a nced che m ical eq ua tion fo r th e reaction . (h) Reg inn in g with 10.0 g of iron, wh at m ass of e 12, in g ra ms . is required fo r complet e rea cti on ? Wh at mass of FeCI] ca n be p ro duced? (c) If on ly l R !i g of Fr-Cl, is obta in ed from 10.0 g of iron and excess C1 2 , wha t is the percent yiel d ? (d ) If 10.0 g eac h of iro n a m i c hlo rine are co m bi ned , what is the th e oretical yiel d of iron ( II I) ch lorid e ? 81. • Som e m etal hal ides reac t with wa te r to produ ce th e m et al o xid e an d th e a pp ro pria te h ydro ge n h alide (se e ph ot o ) . Fo r e xa m ple,
T iCl, (f)
+ 2 Il ,O (I) -> T iO,(s) + 4 Il CI(g)
78. Th e metabolic dis orde r di ab et es ca us es a b u ild u p of acetone , C H:lC OC H 3 , in th e bl ood. Acetone , a vo la tile com po u nd , is ex haled , g iving the brea th of untreated d iabet ics a d istin ctive o d o r, T he aceton e is produced by a break down o f fat s in a se ries of reactions . Th e e q u atio n for til e las t su- p. the b reakdown of acetoac et ic ac id to give acetone a nd CO 2 , is C H,C OC H,CO, H -> CH ,COCH,
+ CO ,
"......;;;iiiii ~ The reaction of TiCk. with the water in moist air.
(a) Name th e fou r co m po un ds in volved in thi s
re acti o n. (b) If yo u be gin with 14.0 m l , of T iCI4 ( d = 1.7:{ g /mL ), what m ass of wate r, in grams, is requ ired fo r co mple te reactio n? (c) What ma ss of eac h product is e xpe cte d?
82. T he re action of 750 . g eac h of N H 3 a nd O 2 was found to produce 5 6 ~ g of NO (sec page s 163-1(5) , \·Vhal mass of ace to ne ca n he produced fro m 125 m g of acetoacetic a cid?
.... morechallenging • in OWL
Blue-numbered questionsanswered in Appendix 0
4 NIl , (g ) + 5 O , (g ) -> 4 NO (g) + Ii Il , O (l') (a) What m ass of wate r is produced by thi s re actio n ? (h) Wh at m ass of O 2 is req uired to co ns u me 750 . g o fN H 3?
I
201
83 . So d iu m azid e , the ex plosive ch em ica l used in automobile airbags . is ma de by t he fo llowing reac tion : NaNO j
+ 3 NaN H z ~ NaN j +
:~
NaO H
+ N H:1
If you co mbine 15.0 g of NaNO:1 (85.0 g /mol) with 15.0 g of NaN H z, what m ass of NaN j is p rodu ced? 84 . Io d in e is ma de by the fo llo win g reaction 2 Na IO , (aq) + 5 Na IlSO, (a q ) ---> 3 Na HS O ,(aq ) + 2 Na, SO.,( aq) + H,O (l) + (,(aq ) (a ) Na me th e two reacta n ts. (h ) lfvo u wish to prepa re 1.00 kg of It , wha t mass of I':a10 3 is required ? Wh at m ass of Na HS0 3? (c) Wha t is the th e oreti ca l yiel d of It if yo u m ixed 15.0 g ofNa IO .~ wit h 125 mL o f 0.853 M Na HS O :\? 85 . Saccharin , an artifi cial swee te ner. has th e form ul a Ci J I"N 0 3S. Su ppose yo u have a sam ple of a sacc hari ncon ta ini ng sweete ner wit h a m ass of 0.2 140 g. After deco m posi tio n to free th e su lfu r a nd convert it to th e SO / - ion, th e su lfa te ion is trapped as wa ter-i nsol ub le BaSo.j (se e Figure 4.6 ). T he quan tity of BaSel. obta ine d is 0.2070 g . Wh at is th e m ass pe rcen t of sacc harin in th e sa m ple of swee ten er? 86 . . ... Bo ro n fo rms a n ex te ns ive se ries of com po u nds with hyd ro gen, a ll with the general form u la BJ I)' II, Hy(s) + excess O , (g) ---> :'. 1I,o, (s) +
2
2
2
H ,o (g )
If 0.148 g of BxHJ give s 0.422 g o f HtO j when burn ed in excess 0 t, what is th e e mpi rical form u la of BxH.I ? 87 . ... Silicon an d hyd ro gen form a seri es of com pounds with th e general formula SiJ I)" To find th e fo rm ula of o ne of th em, a 6.22-g sam ple of th e co mpo u nd is b urned in oxyge n . All of th e Si is converted to 11.64 g o f SiO t , and all of the H is converted to 6.980 g of HtO . What is th e empirical for m ula of the silico n co m po u nd ? 88 . ... Men tho l, fro m oi l of min t, h as a charac teristic o do r. T he co m pound co ntains o n ly C, H , a nd 0 . If 95.6 mg of men tho l burns comple tely in 0 t , an d give s 269 mg of CO t and 1 J 0 m g of li tO , wha t is the em pirical form ula of men tho l?
89. ... Q ui no ne , a chemical used in th e d ye indust ry a nd in ph ot ogra phy, is an organic co m poun d co ntaining only C, H, a nd O . What is th e em pirical for m ul a of th e co rnpo un d if 0.105 g of th e com pound gin'S 0.257 g of COt and 0.0350 g of lI tO when burned com pletely ill o xygcn ? 90 . ... Iro n (lI ) chloride an d so dium su lfi de react to form iro uf l l j sulfi dc a nd so d iu m chloride (Ch cmistryN ow Screen 4.g .) (a ) Wri te the balan ce d equa tion for th e re acti o n . (b ) If yo u combi ne 40 g each of Na tS an d FeCit , what is th e limit ing reac tan t? (c) Wh a t ma ss of FeS is prod uce d? (d) \Vha t m ass of Na2S or FeC It re mains after th e reac tion ? (e) Wh a t m ass of FeCIt is req uired to rea ct co mpletely ,.. . ith 40 g of Na tS? 202
I
91. Sulfuric ac id ca n be p rep ar ed st arting with th e su lfide ore. cu pri te (C UtS). If ea ch S a tom in CUtS leads to on e molecule of 11 2SO j • wha t is th e th eo reti cal yield of 112S0 4 from 3.00 kg of CUtS?
92 . ... In an experime n t, 1.056 g ofa me tal carbonate, contai n ing an u n kn own m et al M. is he a ted to give t he me tal oxi de a nd 0.376 g CO t . MC O ,,(s) + he al ---> MO (s) + CO, (g ) Wh at is th e iden tity o f th e met al M? (a) M ~ Ni (c) !VI ~ ZIl (h) M ~ Cu (d ) M ~ Ba
93 . . ... An un k nown met al react s with oxygen to give the me tal oxi de . MO t. Id e n tify th e m et al based O il th e following informa tion: Mass o f meta l = 0.:356 g Mass of sam pl e after converting metal co m pletel y to o xide = 0.452 g
94 . ... T itani um(lV) o xide , T i0 2 , is he at ed in hyd ro gen ga s to g ive wate r a nd a ne w titanium ox ide , Ti xO )' If 1.598 g of TiO t pro duce s 1.438 g of TixO ), what is th e empirica l fo rmul a of the new oxi de ? 95 . . ... Po tassium pe rchlorate is pre pared hy the following seq uen ce of reac tions: CI,(g) + 2 KO H (aq ) ---> KCI(aq) + KCIO (aq ) + H,O (l) 3 KCIO (aq ) ---> 2 KCI (aq ) + KCIO , (a q ) 4 KCIO , (aq ) ---> 3 KCIO, (aq ) + KCI(a q ) Wh at m ass of C it (g ) is require d to p ro duc e 234 kg of KCIO ,?
96 . . ... Com mercia l sod iu m "hyd rosu lfiu-" is 90. 1% NatStO,. T he sequencc o f reactions used to p re pare th e com pou n d is ZIl(S) + 2 SO, (g) ---> ZIlS,O,(S ) ZIlS,O ,(S) + Na,C0
3(a q )
---> ZnCO,(s) + Na ,S,O,(aq )
(a ) Wh a t mass of pur e Na tS tO, can he pre p ared from 125 kg of Zn , 500 g of SO t, and a n excess of NatC03? (b ) Wh a t m ass of th e com merc ial product wo uld conta in th e NatS2o.1 pro duc ed using th e amou n ts of reac tan ts in p art (a) ? 97 . • Wh at mass of lim e , Ca O , ca n be ob tained by he at in g 125 kg of lim eston e th at is 95 .0% by mass CaC0 3 ? CaCO ,,(s) ---> Ca O (s) +
C O , (~)
98 . ... T he elements silver, molybde num , an d sulfur combin e to form Ag2 ~loS" . Wha t is th e maximu m m ass of AgzMoS" th at ca n he obtain ed if 8.63 g of silver, 3.36 g of molybden um , an d 4.81 g of su lfur are com bined?
... more chaLLenging
•
in OWL
BLue-numbered questions answered in Appendi x 0
99 . . ... A m ixt ure of bute ne , C 4 1Ifl , a nd buta ne , C 4 H ]I). is bu rned in ai r to g ive CO z a n d wa ter. Su p po se you burn 2.R6 g o f th e m ixture an d obt ai n fUW g o f CO z a nd 4.14 g o f H 20 . Wh al a re th e m ass perce n tage s of b un-n o an d b u tan e in th e mixtu re? 100. • •
Clo th ca n be wa te r p ro o fe d b y coa ting it wit h a silicone layer. T h is is d o n e by e x po sing th e cl o t h to (C H 3 h S iCl 2 va por. T he silico n co m po u nd re ac ts with O H g ro u ps o n th e d ot h 10 form a waterproo fi ng fi lm (d e ns ity = 1.0 g/c m 3 ) o f ( (C II 3 h SiO ],,, whe re n is a la rge intege r num be r. n (C H gh SiCI2 + 2 n OH -
~
2'1 CI- + '1 11,0 + [ (C II , 1,S iO J"
Th e coa ling is ad d ed layer hy laye r, ea ch laye r of [( C H 3 b S iO ] " be ing 0.60 nm th ick. Su p pose yo u wan t to wate r p ro o f a p iec e o f d oth that is 3.00 m sq uare, and yo u wa n t 250 laye rs o f wate rproo fin g co m po u nd OIl th e clo th . Wh at m ass o f (CII 3h SiC I2 do yo u ne ed ?
101. . .. Co p pe r me tal ca n be pre pa red b y roastin g co p per o re , which ca n con ta in cu p rite (CU2S) a m i co p pe r( lI) su lfide . Cu,S (s)
+ O , (g ) ---> 2 Cu (s) + SO, (g )
+
CuS (s)
O , (g ) ---> Cu (s)
+
SO, (g )
Su ppo se a n ore sa m ple co n ta ins 11.0 % im p urity in add ition to a m ixtu re o f Cux a n d Cu:,!S. He ating 100.0 g o f th e m ixture p rod uces 75.4 g o f co pper m et al with a p urity o f 8 ~Uj % . Wha t is the wei g h t percen t o f CuS in th e o re ? T he we igh t pe rce n t of CUzS? 102. Whi ch has lh e larger conc en tra tion o f h yd ron ium io ns, 0.01 5 ,\ 1 HC I or aqueou s HC I with a pH o f 1.2? 103 . T he mine ra l d o lomi te co n tai ns mag ne sium ca r bo nate . MgC O , (s)
+
2 II CI (a'l ) ---> CO , (g )
+
MgCI, (a 'l )
+
II , O ( C)
(a) Write th e n e t ionic equa tio n fo r the re ac tion o f Mg CO , an d II CI(a'l ) ' (1)) What type of re ac tio n is th is? (c ) Wha t ma ss o f MgC0 3 will re ac t with 12:) mL or HCI (aq ) wit h a pH o f 1.56 ? 104 . An Alka-Scl tzc r tabl et co n tai ns e xactly 100 . mg of ci tric acid , 1I3C j1150 j , plus som e so di u m bicarbona te. If th e foll owin g reacti on occu rs, wh at mass o f so d iu m bica rb on ate m ust the ta blet a lso co ntain if citric acid is co m ple tely co ns u m ed by th e fo llowing reaction? f1, C,II, O , (a'l ) + :I Na IlCO, (a'l ) ---> :I H ,O ([) + :I CO , (g)
+
]'\a,C"II , O , (a'l )
10 5. • Sod ium b icar bo nat e an d ac e tic aci d re act according to the eq uatio n Na IlC(}, (a 'l)
+
C II 3 CO , II (a'l) ---> NaC H 3CO ,( a'l) + CO, (g )
+
H,O ( I)
Wh at mass of so d iu m ace ta te ca n be ob ta in e d from mi xing 15 .0 g of Na HC0 3 with 125 ml. o f 0. 15 M acetic ac id ? .... morechallenging • in OWL
Blue-numbered questionsanswered in Appendix0
106. • A no n carbon at ed so ft drink co n tains an un kn own a m o u n t of ci tric ac id . H 3C 6I1 "O j . If IOO. m l , o f th e so rt drin k req u ire s 33.5 1 mL of 0 .0 102 M NaO H to ne u tra lize the ci tri c ac id co m ple tely, wha t m ass o f citric aci d does th e soft d ri nk co n tain p e l' 100 . m L? Th e reac tion o f citric ac id and Na O H is H ,C"H,O, (a'l )
+ :l Na O H (aq) ---> Na , C, H , O ,( a'l)
+
:I H, O (f)
107 . Sodi um th io sulf a te , N"a2S20 3, is u sed as a "fi xe r " in hlack-aurl -whit c ph o to g raphy. Su p pose yo u have a bot tle o f so d ium th io su lfa te an d wan t to d e te r m in e its p urity. T he thi o sul fat e io n can be o xidi ze d with 12 accordi n g to th e bala nced , ne t ioni c e q uatio n
If you use 40 .2 1 mL of 0.246 M 12 in a tit ra tio n , wh at is the we igh t perce nl o f N aZS203 in a 3.23 2-g sam pl e o f im pure mate rial ? 108 . \ i.m h ave a m ixt u re o f oxalic aci d . 112C z0 4 , an d a u o the r so lid th a t docs n ot re act with so d iu m h yd roxide . If 29.58 m L o f 0.550 M NaO Il is re qui red to tit rate th e o xalic ac id in th e 4.554-g sa m ple to th e sec o nd eq u ivalence point. what is th e m ass percent o f o xalic aci d in th e m ix ture ? O xa lic acid an d Na O H re act accordi ng to the e q ua tio n Il , C, O , (a 'l )
+
2 NaO H (a 'l ) ---> Na ,C, O ,( a'l)
+ 2 11, 0 (£)
109 . (a) Wh a t is the p l I o f a O.lO;') M IIel so lu ti o n? (b) What is th e h yd ro n iu m io n conc e n tra tio n in a so lu tio n with a pH o f 2.56? Is th e so lu tio n acid ic o r ba sic? (c) A so lu tio n has a p H o f 9 .6 7. Wh at is th e h yd ro nium io n concen t ratio n in th e so lution? Is th e solu tion ac idic or basi c? (d ) A IO.O-m L sa m ple of 2.56 M HCl is d ilu ted with wa te r to 250. mL. What is th e pH of the d ilute so lu tio n? 110. A so lu tio n of hyd rochl o ric ac id ha s a vo lu me o f 125 m L and a p H o f 2.56 . Wha t m ass o f Na I-lC0 3 must he ad ded to co m ple tely co n su m e the I ICI? Ill . .. O ne halfli n-r (5 00. m l .) o f 2.5 0 M II CI is mi xe d with 250 . m L o f 3.75 M H C!. Assumi n g th e total so lutio n vo lu m e aft er mi xin g is 750. mL, wha t is the co nce n tra tio n o f h yd ro chl oric a cid in th e res u lting so lutio n? Wh a t is its p l l? 112. A so lut io n of h yd ro ch lo ric acid has a vo lume of 250. ml , and a pH of 1.92. Exactly 2:">0. m l, of 0.010 51\1 NaO H is ad de d. Wh at is th e p H of th e re sulti n g so lu tio n? 113. ... Yi:lll pla ce 2.5 6 g o f CaC0 3 in a bea ke r co ntai n in g 250 . ml . ofO.12.ii M lI Cl . Wh e n the reacti o n has ceased , d o e s a ny ca lciu m carbo nat e re m a in ? Wha t m ass o f CaCl z ca n be p ro duced ? CaC0 3 (s)
+
2 HCI (a'l ) ---> CaCI,( a'l)
+ CO, (g ) + Il , O ( f)
I
20 3
114. Th e canc e r ch e m o th e ra py drug cisp la ti n , Pt (NI -13)1C11. ca n h e m ade b y rea ctin g (N I I..h P tCl., with am m o n ia in aque ous so lu tion . Be sid e s cis p la tin, th e oth er product is N H ,C I. (a) Wri te a balan ced eq ua tio n fo r thi s rea ction . (h) To o b ta in 12 ..~0 g o f cisp latin , wh a t mass of (:'\ H 1hPtC L, is req u ired? ' Vhal vo lu me o f 0 .12 5 t\.1 N H :J is re q u ire d ? (c ) A Cisp la tin ca n reac t with th e o rg an ic co m p o u n d pyridin e , C!}H "N , LO fo rm a n ew co m p o u n d. 1', (N H , j,CI, (aq )
+
x
C ,H,N (a q) ---> P, (N H , ) , C I, (C"H , N ) J s)
Suppose yo u t re a t 0 .150 g of cisplatin wit h wh at yo u bel ieve is an excess o f liqu id p yridine ( l5 0 ml .: d = 0 .979 g / m L). When th e rea ct ion is com plete , yo u ca n find o u t h ow m u ch pyridi n e was no t u sed hy tit ratin g th e sol u tio n with sta n dard ize d lI Cl. If 3 7.0 mL of 0 .47 5 M (-ICI is req u ire d to tit rate the exce ss pyridi n e ,
wha t is th e formula of th e u nkn own co m po u n d p, (N I I, j,CI,( C, I I, N) ,? 115. A \bu ne ed to kn ow th e vo lu me of wa ter in a sm all swim m ing poo l, h u t, owin g to th e po ol's irregular sh ap e , it is not a sim p le m a tter to d e te rm ine its d imensio ns a n d calcula te the vol um e. To so lve th e p roblem . yo u stir in a so lu tio n of a d ye ( l. O g o f methylen e b lu e , C 16 H II;CIN 3S, in 50 .0 m L o f wa te r ) . After the d ye h as mi xed with th e wa ter in th e p ool , yo u tak e a sam p le o f th e wate r; Using a spectro p h o to m eter, yo u de termine th at th e co n ce n tration of th e d ye in the po ol is 4. 1 X 10- 8 M. Wh a t is the vo lu me o f wa ter in th e p o ol ? 116 . A Calc iu m an d m ag nesium ca rb o n a tes oc cu r to ge t h e r in th e m ine ra l dolo mite . Suppose yo u h ea t a sa m p le o f th e mi n eral to o b tain th e oxides, CaO a n d MgO , an d th e n tr ea t the oxide sa m p le wit h hydro ch lo ric acid . If 7.69 5 g of th e o xid e sa m p le req u ires 125 mL of 2.55 ~1 lI Cl, CaO (s )
+
MgO (s)
+ 2 II CI (aq ) ---> MgCI, (aq ) + 1I, CaCI,( aq )
+
1I, 2 Cu l (s)
+
I, - (a'l )
T h e libera te-d 13 - is tit rated with so d iu m thiosul fate ac cord ing to th e eq u a tio n
(a) De sign ate t h e ox id izing an d re du cin g age n ts in th e two re ac tio n s ab ove . (b ) If 26.:12 m L of 0.10 1 ~1 Na1S10 Jis req u ired fo r titr ation to th e equivalen ce p oint, wh at is tlu- weig h t percelll o f Cu in th e alloy? 122. A A co m po u n d h as be en iso la ted th a t ca n h ave e ithe r o f two po ssible fo r m u las: (a) K[ Fe (C10" h (1I10 )Zl or (b ) K3 [Fe (C 1o. l h l . To find wh ich is correct, yo u di sso lve a weigh ed sa m p le o f th e co m p o u n d in ac id an d th en titr at e th e o xala te io n (C 1n I1- , whic h in ac id be... more chaLLenging
•
in OWL
BLue-numbered questions answered in Appendi x 0
comes (l 1C10.,) wit h p ot assiu m pc rm an gan at c , Ki\1 nG. , (the source of tlu- .\l nO-\ - ion ) . T he bala nced . net io ni c equation fo r the titra tio n is 5 Il , C, O , (aq )
+ 2 MnO, - (aq) + 6 1l 3 0 ' (aq) -> 2 Mn 2 +(aq ) + 10 CO , (g ) + 14 H 20 (f )
Ti tra tion o f 1.356 g" of th e co mpou nd re qu ires :14.50 rn l , of 0. 108 1\.1 KMn (},. Whi ch is the correct formul a o f th e iro n-con tai n in g co m p o un d : (a) or (b ) ? 123 . ... Chro m ium f l ll ) io n form s ma n y co m poun ds with am monia. To fi n d th e formula of o ne of these com pou nd s, yo u tit ra te th e N H 3 in th e com pou n d with sta ndardi zed aci d. Cr (N II ,) ,CI, (a q) + x Il CI (aq ) -> x 1\ fl,'(aq) + Cr3+(a q)
+
(x
+
3) CI- (a q )
Assume that 24 .26 m l . o f 1.50 0 M H e l is used to titrau- 1.580 g of Cr(N I 13 ) ~C h \Vhat is th e valu e o f x? 124 . ... T hioridazine, C~21 1 1 2l,N2S2 . is a ph armace uti cal ag en t used to regu late dopami ne. (Do p am in e, a n eu ro tran smi tte r, affects brain p rocesses th at con trol movem e-n t, e motio na l resp o n se , a n d abi lity to ex perie nce pleasure a nd p ain .) A chem ist can ana lyze a sam p le o f th e ph armaceu tical fo r the thi orid azin c con tent by dccomposing it to convert th e su lfur in th e co m p o u n d to sul fa te ion , T hi s is the n "tra p ped" as watc r-insuluhlc bariu m sul fate (see Figu re 4.6).
sol-(aq . from
th io rid azin c)
+ RaCI 2 (aq ) ----7 Ila50., (s )
+ 2 CI - (a q )
Su ppose a 1 2~tablet sa m p le of th e d rug yielded 0 30 1 g" of RaS(},. \Vhat is th e thiorid azin e con le n t, in milligrams, of eac h table t? 125 , . ... A h e rbicide co ntains 2,4-1) ( 2,4 -d i c h l o ro p h e n ()xy~ acetic aci d ), C!'l H hC120 3• t\ 1.236-g sa m ple of the h e rb icide was d eco mposed to liberat e th e chlorine as CIio n . This was precipi ta ted as AgCl , with a mass of O.I R40 g . Wh at is th e m ass p er ce nt of 2,4-D in th e sam p le?
OCH ,CO,H
I
H"
/ C"" / Cl
C
II
H/
I
C
" c"" C" I
126 . ... Sulfuric ac id is listed in a catalo g with a concc n tration of 95-98 %. A bott le of the ac id in th e sto ckro o m sta tes th at 1.00 L h as a mass of I .H4 kg . To determimth e co n ce ntration of su lfu ric ac id in th e stockroom bo tt le , a stu den t d ilutes 5.00 m L lo !lOa. mL. Sh e th en takes fou r 10.OO~ m L samples and tit ra tcs eac h with standardized so diu m h yd ro xid e ( c = O.17GO mol /L ). Blue-numbered questionsanswered in Appendix 0
2
21. 30
3 20.40
4 20 .35
127. .. An h ydrous calcium ch loride is a good d ryin g agent as it will rapidly pick u p wa ter. Su ppose yo u h ave sto re d some ca refu lly dri ed CaC12 in a d e ssicato r; Un fortunately, SOITH'OTH' did not close th e to p of the d essicator tightly, and the Cae}2 bec a me par tiall y hvd rat ed. A 150-g sample of thi s p art iall y h yd rated material was d isso lved in 80 g of h o t wa ter. Wh en the sol u tio n was cooled to 20 °C , 74.9 g o f CaCI 2 . 6 1J10 preci p itate d . Knowi ng th e so lu bility o f calc iu m c h loride in wa ter a t 20 °C is 74.5 g CaC12 /100 g ' v'at er, d e te r mi n e the wa ter con ten t of th e 15 0~g samp le of par~ tial ly h ydrat ed calcium ch lori de (in m o le s o f wa ter per mole o f Cae12 ) . 128. ... A sa m p le con sistin g o f a m ixture of iro n an d iro n ( III ) oxide was di sso lve d com ple tel y in acid (wh ich co nverted th e iro n to iro n ( lII ) ions.) Ann ad ding a re ducing agen t to ensure th at all of the iro n was in the fo r m of iron (I I) ions, the so lu tio n was titrate d with th e sta n d a rd ized KMn o.l (0.0424 0 M) ; 37 .50 ml . o f IIH' K~111 0 .1 solution was required . Calcu la te th e mass p e rce nt of Fe a nd FC2()3 in th e 0 .55 10-g sa m ple. (See Exam ple 4.12 for th e reaction o f iro n ( II) a nd K;\ln O ,.) 129. ... Ph o sph at e in urine ca n be de te rm ined by s p e c t ro~ p ho to me try. Aft er re m ovin g protein fro m th e sa m p le , it is trea ted with a molybd e n u m com pou nd to give , ultim atel y, a dee p blue polvm ol vhd a u-. Th e absorbance of th e blue polvmolyhdatc can he m e asured a t 65 0 run and is di re ctl y relat ed to the u rin e ph o sph a te co n cen tration , A 24 ~ho llr urin e sa m p le was co llected fr om a pa tient; th e vo lu m e o f u rin e was 1122 m L. Th e p h os p hate in a I .OO m l . po rt io n of the mille sa m p le was converted to th e b lue polym ol ybdate (P) and dilu ted to 50.00 mL. A ca libratio n curve was prepar ed us ing ph o sphate-con taining so lu tion s. Solution (m ass P I L )
H
1 20. 15
(a ) What is t he average conce ntra tion of t he dil u te d su lfuric ac id sa mple? (b) What is t he mass p e rce nt o f H 2SO -\ in the original bo ttle of th e aci d ?
C
Cl
... morechallenging • in OWL
Sam ple Volume i\aO Il (m f.)
1.00 X 10- " g 2.00 X lO - ij ~ 3.00 X IO- l ' g 4.00 X lO - ij s U rine sa m p le
Abso r b an ce at 650 om in a 1.0 em ce ll
0.230 0.436 0.638 0.848 0.5 18
(a) What are th e slo pe an d interce p t of th e ca libratio n curve? (1)) What is the mass of ph o sph o rus p e r lit e r of u rin e ? (c) What mass o f p hos p ha te: d id th e p at ie nt excrete per d ay?
I
205
130 . ... A 4.00 0-g sam p le con taining KCl a n d KClCl , was dissolved in sufficie n t water to give 250.00 ml , of solu tion . A 50 .0o-mL portion o f the so lu tion re quire d 4 1.00 mL of 0.0 750 M AgNO.'l ill a Moh r titrati o n (p ag(' 19G) . Next, a 25 .0o-m L porti on of the original solutio n was treated wit h V2 ( SO ~ h to reduce the p er ch lo rat e ion to ch loride, 8 VJ+(aq )
+ C10.,- (aq ) + 12 H , O ( I) -> Cl Iaq) + 8 VO " (aq ) + 8 II ,O ' (aq ) r
an d the res ulting so lu tion was titrated with AgNO:t • T his titra tio n req u ired 3R 12 mL o f 0 .0750 M AgN 0 3 . 'Vh at is th e m ass p erce nt of KCl an d KCI O ~ in th e mixt ure?
Summary and Canceptual Questians The jiJllowi ng qu estions ma)' UJf'
COf/(f'jJ[J
from this a nd preceding
rhrljJ fn.\.
131. Two beake rs sit on a balance; th e total mass is 167.170 g. On e be ak er con tai n s a so lu tio n of KI; th e o ther co n tai ns a solu tion o f Pb (NO:1h . Wh e n th e so lu tio n in on e b eaker is poured co m p le te ly into th e o ther, th e followin g reac tion occ u rs: 2 KI (aq )
+ Ph (NO, ), (aq ) ->
2 KNO , (aq )
+ Phl , (s)
(a) Wh at mass of Br2 is used whe n the reac tio n cons umes 2.0 g o f Fe? (h ) Wh at is th e mole ratio o f Rr2 to Fe ill th e rea ction ? (c ) Wh at is the e mpiric al fo rm u la of the p roduct ? (d ) Wr ite th e b al a n ced che m ica l eq u at ion fo r th e reac tio n of iron a nd bromin e . (e) What is the n am e of th e reacti o n p roduct ? ( f) Whi ch s ta temen t or state men ts h est descri be th e exper im e n ts su m marized by the gra p h? (i) When 1.00 g o f Fe is ad d ed to th e fir2, Fe is the lim it in g reage n t. (i i) \Vh en 3.50 g o f Fe is added to th e fir2, the re is a n ex cess o f Rr2' (iii) Wh en 2.50 g of Fe is ad d ed to th e Rr2, bo th re actants are us ed lip com p le te ly. (iv) Wh e n 2.00 g- of Fe is add ed to th e Br 2, 10 .0 g of p ro duct is formed . T h e perc ent yiel d must t he refore he 20 .0 % . 133. Let us e x p lo re a reaction with a limiting rea ct a nt. (See Cln-misn-ylcow Scr een s 4 .4 an d 4.5 .) He re , zin c metal is add ed to a flask con ta in ing aque ou s H el , a n d 112 gas is a p roduct. Zn (s)
+
2 II CI (a'! ) -> ZnCI, (a q)
+
II , ( ~)
T h e three fla sks eac h contai n 0 ,100 m ol o f Hf"I. Zinc is ad d ed to ea ch flask in th e followin g qu antit ies. Fla sk I: 7.00 ~ Zn Flask 2: 3 .27 g Zll Flask 3: 1.31 g Zn
16 -, 110
II.. -, I 10
Solutions of KI and Pb(N03)z before reaction.
.,
Solutions after reaction.
Wh at is th e total mass of th e be a kers an d so lu tio n s after reactio n? Explai n co m p lete ly. 132 . ... A we ighed sa m ple of iron (Fe) is ad ded to liq ui d bromin e (Br2) an d a llowe d to react com p letely. The reactio n p rodu ces a single p roduct . wh ich ca n he iso la te d an d we ig he d. Th e ex perime n t was re peate d a nu mber of tim es with d iffere n t m asses o f iron but with th e sa m e ma ss o f b ro m ine . (Se c th e g rap h bclow.)
When the reac ta n ts arc co mbined , the H 2 in fla tes th e
ba lloo n attac he d 1.0 th e Flask. T he rc sulus arc as follows: Flas k 1: Balloon ma ins whe n Flask 2: Ball oon Fla sk 3: Ball oon m ain s. 2 Mass of Fe (g)
206
I
J
inflates com p le tely, hut so me Zn rc infl atio n ceases. in flates com p le tel y. No Zn remains. d o e s no t in fla te co m p le te ly. No Zn re -
4
Exp la in th e se res u lts. Pe rform calcu la tio ns th at su p p o n yo u r expl an at io n . ... more chaLLenging • in OWL
Blue-numbered questionsanswered in Appendix 0
134. T h e reaction of alu mi n um a nd bro m ine is p ictured in Figu re 2.12 and below. Th e whiu- so lid o n till' lip of th e b eaker a t th e end of th e re act ion is AI2Rrl;' In th e reac tion p ictured bel ow, whi ch was th e lim itin g reactan t, Al or Rr2? (Se c C he m istryN ow Scree n 4.2.)
Before reaction .
After reaction .
... morechallenging • in OWL
Blue-numbered questionsanswered in Appendix0
135 . Ii. Two stude n ts titra te di ffe rent sa m ple s of th e sa me so lutio n of H e} usin g (J.lOO 1\.,1 Na O Il so lu tio n an d phenolph thalein in d icat o r (se e Figu re 4. 14) . Th e first stu den t pipets 20.0 m l , of the l IC l soh Ilion in to a fl ask, add s 20 m l, of di stilled wa ter a nd a few drops of phenolphthalein so lu tio n , a n d titra tcs unti l a lastin g pin k color ap pea rs. Th e secon d stu d en t pipe ts 20.0 ml, of th e HCI so lut io n in to a fla sk. ad ds 60 m l, of d istilled wate r an d a few d ro ps of phe nolp hth alein solu tion , a nd titran-s to the first lastin g pink co lor. Each studen t correc tly calcula tes the m o la rity of a ll I l e i solutio n . Wh at will th e sec on d stu den t's res u lt he? (a) fo u r times less th a n lh e first stu de n t's n ..su lt (b) fou r times grea ter th an th e first stud e n t's re sult (c ) two tim e s less than the first stud e n t's result (d ) tw o Lime s g re ate r than the first stu de n t's re su lt (c ) th e sa m e as th e fir st stud en t's result
136. A vid eo on Scree n 4.12 o tClu-misuvlvow sh ows the re ac tio n of Fe 2 + with ~l nO 'I - in a q ue o us so lutio n . (a) Wh at is th e balan ced equa tion fo r th e reac tio n th at oc curred ? (h) wh at is th e o xid izing agen t, a n d wh a t is the re duci ng agen t? (c) Equal vo lu mes of FeH -co n tain ing so lu tio n a nd MrrO," -co n ta in ing so lu t ion were m ixed. The a moun t of Fe H was just su fficie n t to co nsu m e a ll of the M nO ~ - . Which ion (Fe2+ o r M n O ~ " ) was in itially presen t in larger con centration? 13 7. In so m e sta tes , a pe rson will rece ive a "d riving wh ile in tox ica ted " (DWI) ticke- t if the blood alc o h ol level (HAL) is 100 m g per deciliter (d L) of blo od or hi gh er. Suppose a pe rso n is found to h ave a RAt of 0.033 m ol of e tha nol (C 2H :;O H ) per lite r of blood. Will th e person rec eive a D\,VI ticket ?
I
207
CONCEPTS OF CHEMISTRY
5
PrincipLes of ChemicaL Reactivity: Energy and ChemicaL Reactions
A Hot Air Balloon Question: These colorful balloons usually consist of a gas bag or envelope of You have a balloon with a volume of 1100 m' and wa nt to heat the nylon wit h a basket suspended below for passengers. A propane air inside of the envelope from 22 O( to 110 °C. What mass of propane must you burn to accom plish this? (The sped fic heat capacity burner sits on top of t he basket and below the gas envelope. When of air is 1.01 Jig ' K, and the density of dry air [at sea level] is the air inside the envelo pe is heated by burning propane, t he bal- about 1.2 kq/rn'. Otherinformation you need is in t his chapter or loon can ascen d (because th e density of heated air in the bag is less Appendix L. ) than that of the cooler surrounding air). Unde r norma l conditions, about 3 m! of envelope volume is required to lift 1 kg of mass. Thus, to carry one person and the needed equipment, most balloons have a volume of about 1000 m3• 208
Answer to this question is in Appendix Q.
Chapter Goals Set' Cha/lln Goals Revisited (/H1gf' 24 1) lor Stu d), Questions kp)'f'd to these bJ'(}al\ ruul a.nigllahlf' i n OH'L
Chapter Outline 5 .1
Energy: Some Basic Princip les
5 .2
Specific Heat Capacity: Heating and Coo ling
Assess the tra nsfer of energy as heat associated with changes in temperature and changes of state.
5.3
Energy and Changes of State
Understa nd and apply t he first Law of thermodynamics.
5 .4
The First Law of Thermodynamics
Defi ne and understa nd state functions (enthalpy, internal energy).
5.5
Entha lpy Changes for Chemical Reactions
Learn how energy changes are measured.
5.6
Calori metry
Ca lculate the energy evolved or required for physical changes and chemical reactions using tables of the rmodynamic data.
5 .7
Enthalpy Calculations
5 .8
Prod uct- or Rea ctant-Favored Reactio ns and Thermodynamics
T
he im portan ce of e nergy is evide n t in ou r d aily lives-in heating and cooling our ho mes, in powering o ur ap pliances, and in pro pe lling ou r vehicles, amo ng o ther things . Mo st o f th e energy we use for th ese p urposes is o btained by ca rrying ou t chem ica l reac tio ns , mostly by burnin g fossil fuel s. We use natural ga s for heating, coal and natu ral gas to ge nerate m ost of ou r electric po wer, and fue ls derive d fro m petrol eum fo r automobiles and fo r heat. In addition , e neq,ry is re q uired for all life processes. Che mical reactio ns in o ur bod ies provide th e energy for all bo dy fu nctio ns, fo r move me nt , and 10 main tain body tempe ra tu re . It is not su rprising that t he topic o f energy is a pro minen t pa rt o f our di scussion of che mi stry. To scien tists, however, ene rgy has sig n ifica nce th at goes well be yond th ese man y p ractical uses. In this chapter, we will begin the disc ussion of therm o dynamics , the science of heat and wo rk . T h is su bject will p rovid e im po r tant in sights o n the followin g questions:
..~
N
Chemislry.o.Now
•
Throughout the text this icon introduces an opportunity for self-study or to explore interactive tutorials by signing in at www.t homsonedu.comj login.
• World Energy Con sumption Burning fossil fuels provides about 85% of the totat energy used by people on our planet. Nuclear and hydroelectric power each contribute about 6%. Th e remaining 3% is provided from biomass, solar, wi nd, and geothermat sources.
• How do we measure and calc u late th e enerh'T changes th at are associated with p h ysica l changes and che m ica l reactions? • \Vhat is the relati o nshi p between e nergy changes, heat, and work? • H ow ca n we de termi ne whe the r a che m ica l reacti on is p roduc t-favored o r reactant-favored a t equi libri um? • H ow can we dete rm ine whether a che mical reaction or ph ysical process will occ u r spo n ta neo usly, th at is, withou t o u tside inte rvention ? \Ve will concent ra te atten tio n o n th e firs t two qu estio ns in th is chap ter and add ress the last two questions in Chap ter 19.
5.1
Energy: Some Basic Principles
Ene rgy is d efin ed as the capacity to d o work. You d o wo rk again st the force of gravity whe n carry ing yourself an d h iking eq ui p ment up a mountain . T he energy to do this is provided by th e foo d you have eate n. Food is a source of che m ical ene rgy-e ne rgy sto red in che mi ca l co m pou nds and rel eased whe n th e co m po u nd s un dergo th e chem ical reactio ns of me tab o lism in yo u r bod y.
5 .1
I
Energy: Some Basic Principles
209
(a) Gravitational energy
(b) Chemical potential energy
(c) Electrostatic energy
En ergy and its conversion. (a) Water at the top of a waterfall represents stored, or potential. energy. As water falls, its potential energyis converted to mechanical energy. (b) Chemical potential energyof t he fuel and oxygen is converted to thermaland mechanical energy. (c) Lig htning converts electrostatic energyinto radiant and t hermal energy. Active Figure 5.1
ChemislrY.c\Now'" Sign in at www.tho msonedu.comj login and go to th e Chapte r Co nte nts menu to
• expLore an interactive version of t his figure accompanied by an exercise.
En ergy can be classified as kinetic o r pot ential. Kineti c energy is energy assoc iated with moti o n, suc h as: • The moti o n of ato m s, mo lecu les , o r ion s at th e su b m icrosc o pic (pa rt icula te) level (thermal energ)') . All mail er has thermal e ne rgy. • T he m otio n o f macro scopic o bj ec ts like a movin g te nn is ball o r au to mo bile
i meclum ical
energy).
• The movement of e lectro ns th rough a co nd uc to r telectricnl ener[!J') . • T he co mpressio n and expan sion of th e spaces between mo lecu les in the tran sm ission o f so u nd (aco us tic enerJ!J). Potential energy resu lts from an o bject's position and includes: • Energy po ssessed by a ba ll hel d above th e floor and by wate r at th e top o f a wat erfall igraoitn tiona! energy) (Figu re 5. 1a) . • Energy sto red in fuel s (chemical nter[!J') (Fig u re 5. 1b) . All chem ica l reac tio ns invo lve a cha nge in chem ica l e nef!.,Ty. • The e nerg y associated with th e separa tio n oftwo elec tr ica l cha rges ( electrostatic PIlelgy) (Figu re 5. 1c) . Pot ential e nergy and kin e tic energy can be interco n verted. Fo r exam ple , as wate r falls over a wate rfall, its po te n tial e ne rgy is co nverted into kinetic ene rgy. Sim ilarly, kinetic e nergy can be co nverted in to pot e n tial ene rgy: Th e kinet ic en ergy of fa llin g wate r ca n tu rn a turb in e to pr oduce elec tr icity, wh ich ca n then be used to co nver t wate r into 1-1 2 and () 2 b y el ectrolysis. Hydro gen ga s co n ta ins sto red chem ica l po ten tial e nergy becau se it can be bu rned to p ro d uce heat an rl lig ht o r e lec tricity.
210
Chapter 5
I
Princi ples of Chemical Reactivity: Energy and Chemical Reactions
Potential energy (energy of position)
ChemislryJ J,.Now'"
•
Sign in at www.thomsonedu .com/Ioqtn and go to Ch apter 5 Contents to see Screen 5.3 for an exercise that examines examples of energy transfer.
Kinetic energy (energy of motion)
Conservation of Energy Stand ing on a d iving board , y O ll have co nsid erable po ten tial energy becau se of yo u r po sitio n above th e wa ter. Once yo u jump off the board, so me of tha t po te n tia l ene rgy is converted int o kine tic energy (Figure 5.2) . Du rin g th e dive, the force o f gravity acce lerates you r bo d y so th a t it HlO\"CS faster an d faster. ' 'bu r kinetic ene rgy in cre ase s, a n d your po te ntia l ene rgy decrease s. At th e mo m e nt yo u hit th e wa te r, yo u r velocity is ab ruptly re d uce d , and m uc h of yo u r kine tic e nergy is tr ansferr ed to the water as yo u r bo d y moves it asid e . Even tually. yo u float to the surface, an d th e wate r becom es still aga in . If you cou ld see th em , however, you wo u ld find th at the wate r mo lec ules are moving a little faster in the vicinit y of yo u r e ntry int o the wate r; th at is, the kinet ic energy of the water mo lecules is slight ly high e r. T h is ser ies of e ne rgy co nve rs io ns illustrat es th e law of conservation of energy, whic h states that energy can neither be created nor destroyed. Or, to state this law di ffere n tly, the total enn g)' (if the un iverse is constan t. T he law of co nservatio n of ene rbl)' su m marizes th c res u lts o f a g reat m an y ex perime n ts in whic h th e amou n ts of e ne rgy transferre d have been measure d and in wh ich th e to tal energy con tent has been fou nd to be the same before and after an eve nt.
Heat and work (thermaland mechanicalenergy) FIGURE 5 .2 The law of energy conservation. The diver's potential energy is converted to kinetic energy. and this is t hen transferred to the water, illustrating the law of conservation of energy. See ChemistryNow Screen 5.2 Energy, to view an animation of this figure.
Temperature and Heat T he te mperature of an o bject is a measure o f its ab ility to tr an sfer cne rb'1' as heat. Whe n two objec ts a t di ffe rent temperatures are b rough t int o co ntac t, e nergy will be tran sferred as heat from the o ne at th e highe r te m perature to the one at the lower te mperature . O ne way to m easu re tem pera tu re is with a th e rm om et er C O Iltaining mercu ry or so me o the r liqu id (Fig ure 5.3 ) . when the the rmometer is pl aced
FIGURE 5 .3 Measuring temperatur e. The volume of liquid mercury in a thermometer increases slightly when immersed in warm water. The volum e increase causes the mercury to rise in the thermometer, which is calibrated to give t he temperature.
r"-JI r -- 'II"""I! "
~.
_ _ _ 20"
Immerse thermo meter in warm water
I 0
~
~
"
~
tI I
~
..1
•• •. ,.•-.f ~
•
-,. '"
5 .1
I
En ergy: Some Basic Principles
211
in hot wate r, th erm al en ergy is transferred from th e wate r to th e th ermomet er (heating th e th ermomet er an d cooling th e wate r) . This ca uses th e ato ms o f liquid mercury to move more rapi d ly (increasi ng th e ir kin et ic energy) and th e space bet ween t he m to increase sligh tly. Th e res ulting increase in volu me ca uses th e colum n o f liq uid to rise highe r in th e th ermo meter tube . Several im portan t aspects o f th erm al energy and te m pe rature sho u ld be recognized : • Tempe ratu re de term ines th e direction of th erma l ene rgy tran sfer. • The h igh er th e temperature of a given o bj ect, th e g rea ter th e th erma l en ergy (e nergy associate d with mo lecul ar mot ion ) o f its ato ms , io ns, o r mo lecul es. • I leatin g an d co o ling are processes by wh ich en ergy is tr an sfe rr ed as heat from an o bject at a higher temperature to one at a lowe r temperature . Heat is not a su bstance. (See A Closer r ook: H,7Ull Is lleat ?)
Systems and Surroundings
FIGURE 5 .4 Systems and their surroundings. Earth can be considered a thermodynamic system, with the rest of the universe as its surroundings. Achemical reaction occurring in a laboratory is also a system, with the laboratory its surroundings.
In th erm odynam ics. th e terms "syste m " and "su rro u nd ings" have preci se and important scie n tific meanings. A system is d efined as an o bject, o r co llectio n of obj ect s, being stu d ied (Figure 5.4 ) . The surroundings include everyth ing o u tside the syste m th at ca n exch an ge energy and /o r mailer with th e system. In th e d iscu ssion th at fo llows, we will need to d efin e systems p reci sely. If we are studying the en ergy evo lved in a che m ical reaction ca rried out in so lutio n , for exam ple , th e system might be define d as th e reacta n ts, products, and so lven t. The su rro u nd ings wo u ld be th e reac tio n vessel and th e air in th e room and anyth ing el se in contact with th e vessel th at mi gh t excha nge energy or matt e r. At the ato mic level , th e syste m could be a sing le ato m or mo lecu le , an d th e su rro u ndings wo uld be th e at o ms o r mo lecu les in its vicinity. How we choose to defin e th e system and its surro u nd ings fo r eac h situa tio n d epends o n th e info rma tio n we are tryin g to o b ta in or co nve y. This conce p t o f a syste m and its su rro u nd ings ap plies to non ch emical situ ation s as wel l. If we wan t to study the e neq,T)' balance on thi s plan et , we might cho ose to d efin e Earth as th e syste m a nd o uter space as the surrou ndi ng s. O n a cos m ic level, th e so lar syste m mi ght be de fined as th e systc m bein g stud ied , and th e rest of th e ga laxy wo uld be th e su rro u nd in gs.
Directionality and Extent of Transfer of Heat: Thermal Equilibrium Energy is transfe rred as heat if two obj ects at different temperatures are b rought into co n tac t. In Figure 5 .:'1, for exam ple , th e beak er o f wate r and th e pi ece of mc ta l being heat ed in a Bun sen bu rner flam e have diffe rent temperatures. When the hot
FIGURE 5.5 Energy tra nsfer. Energy transfer as heat occurs from the hotter metal bar to the cooter water. Eventually, the water and metal reach the same temperature and are said to be in thermal equilibrium . See Chemi st ryNow Screen 5.4, Energy Transfer Between Substances, for a simulation and tutorial. c ~
g....._ - - - - - - _..... 2 12
Ch a p t er
5
I
Principles of Chemical Rea ctivity: Energy and Chemical Reactions
A Closer Look
What Is Heat?
Two hundred years ago, scientists characterized
From our discussion so far, we know one thing that "heat" is not- but what is it? Heat is said to be a "process quantity" as opposed to a "state quantity." Th at is, heating is a process that changes the interna l energy of a system. It is the process by which energy is transferred across the boundary of a system owing to a difference in temperature between the two sides of the boundary. In this process, the energy of one object increases, and the energy of another object decreases. Heating is not the only way to transfer energy. Work is another process by which energy can be transferred between objects. The idea of energy transfer by t he processes of heat and work is embodied in the definition of t hermodynamics: the science of heat and work.
heat as a real substance called a caloric fluid.
The caloric hypothesis supposed that when a fuel burned and a pot of water was heated, for example, caloric fluid was transferred from the fuel to the water. Burni ng the fuel released caloricfluid, and the temperature of the water increased as t he caloricfluid was absorbed. Over the next 50 years, however, t he caloric hypothesis lost favo r, and we now know it is incorrect . Experiments by James Joule (1818 -1889) and Benj ami n Thompson
(1753-1814) tha t showed t he interrelationship between heat and other forms of energy such as mechanica l energy provided the key to dispelLing this idea. Even so, some of our everyday language retains the influence of t his early theory. For exam ple, we often speak of heat flowing as if it were a fluid.
Work and heat . Aclassicexperimentthat showed the relationship between work and heat was performed by Benjamin Thompson (also known as Count Rumford) (1753- 1814) using an apparatus similar to that shown here. Thompson measured the rise in temperature of water (in the vessel mostly hidden at the back of the apparatus) that resulted from t he energy expended to turn the crank.
metal is plunged into th e cold wate r, e nergy is tran sferred as h eat from th e me tal to th e wate r. Th e th ermal e lle rgy (m o lec u lar m o tio n) o f th e wate r molecul es in creases; the thermal energy o f th e metal ato ms d ecreases. Eventually, th e two obj ects reach th e sa me temperature . At th at point, th e syste m h as reach ed thermal equilibrium . TIl e di stinguishi ng feature of th ermal equilibrium is th at , Oil th e macrosc o p ic scale, no further temperatu re cha nge occu rs; both th e m et al and water are a t th e same temperature. Putting a hot met al b ar into a b eak er of wate r a n d followin g th e tempera ture cha nge may see m like a rath er sim p le exp eriment with a n o bvious outcome . Illustrated in th e ex perime nt, howeve r, a re th ree principles that a re important in our furth er d iscussion :
• Thermal Equilibrium Although no change is evident at the macroscopic level when thermal equilibrium is reached, on the molecular level transfer of energy between individual molecules will continue to occur. A general feature of systems at equilibrium is that there is no change on a macroscopic level but that processes still occur at the particulate level. (See Section 3.3, page 118.)
• En ergy tr an sfe r as heat will o ccur sp o nta n eo usly from all object at a high er temperature to an o bject a t a lower temperature . • Tran sfe r of ene rgy as heat con tin ues un til both o bjects a re at th e sa me temperature a n d th erm al eq u ilib riu m is ac h ieved . • After thermal equilibrium is a ttain ed, th e object wh os e temperature increased has ga in ed th e r ma l en e rgy, a n d th e o bj ec t wh ose temperature d ecreased has lo st th ermal e nergy. For th e sp ecific case whe re e ne rgy is tr an sferred as heat with in all iso lat ed system (tha t is, a syste m that ca n n o t transfer e ith e r e nergy or matter with its su rro u n d ings) , we can also say that the quanti ty o f thermal e ne rgy los t by a hotte r o bject a n d the quan tit y of the rm al e ne rbf)' ga ine d by a coo le r o bject a re num erically equal. (T h is is required by th e law of co nse rvatio n of e n e rgy.) Wh en en ergy is transferred as 5 .1
I
Energy: Some Basic Principles
21 3
Exothermic q,y; < 0
Endothermic SYSTEM
q>ys
SURROUNDINGS
SYSTEM
>0 SURROUNDINGS
Exothermic: energy transferred from system to surroundings
Endothermic: energytransferred from surroundings to system
Active Fi gure 5.6 Exothermic and endothermic processes. The symbol q represents the energy transfered as heat, and the subscript sys refers to the system.
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• Exothermic and Endothermic The terms "endothermic" and "exothermic" apply specificalty to energy transfer as heat . The more general terms "endoerqic" and "exoerqic" are sometimes used, encompassing any type of energy transfer between system and surroundings.
heat be tween a systc m and its su rro u nd ing s, wc d escribe th e directionality o f thi s tr an sfer as exothe rmic or endo the rm ic (Figu re 5.fi) . • In an exothermic process, e nergy is tran sfe rr ed as heat fro m a syste m to its su rro u nd ings. T he e ne rgy of the system d ecreases, and th e e ne rgy of th e su rroundin gs increases. • An end o ther mic process is th e o ppo site of an ex o the rm ic p ro cess. Ene rg y is tr a nsferred as heat from the su rrou nd ing s to th e system , increasing th e en e rgy of the syste m , de creasin g the energy o f th e su rro u nd in gs.
Energy Units • James Joule The joule is named for James P. Joule (1818-1889), the son of a wealthy brewerin Manchester, England. The family wealth and a workshop in the brewery gave Joule the opportunity to pursue scientificstudies. Among the topics that Joule studied was the issue of whether heat was a massless fluid. Scientists at that time referred to this idea as the caloric hypothesis. Joule's careful experiments showed that heat and mechanical work are related, providing evidence that heat is not a fluid. (See A (loser Look: What Is Heat?)
Wh en ex p res sing e ne rgy qu ant ities, most chem ists (a nd m uch o f th e wo rld outsid e th e United States) use th e joule U), the SI unit. The j oule is rel at ed d irectl y to th e "i units used for mech anical en crgy: 1.J eq ua ls 1 kg . m 2 / s2 • Because th e joule is 8 in con venie ntl y sma ll fo r most uses in che mi stry, th e kilojoule ( ~ J), eq uivale n t to 15 -E" 1000 joules. is ofte n th e unit of cho ice . f To give you so me fc eling for j oulcs, su p pose yo u drop a six-pack of so ft-d rin k ~ ca ns, ea ch full o f liquid, o n yo n I' foo t. Altho ug h yo u probabl y will no t ta ke time to -~ ca lculate th e kineti c e ncq,J} at the mom ent o f impact, it is between 4 .J and 10.J. ~ T he ca lo rie (cal) is an o ld er energy unit. It is d efined as th c en ergy transferred § as heat that is re q uire d to raise the tem perature of 1.00 g of pure liquid wate r from :§. 145 °C to 15.5 °C. A kiloc alorie (kcal) is equivalent to 1000 ca lo ries. T hc couversio n fact or rel atin g jou les and calo ries is
1 calorie (caL)
= 4.184 joules (J)
The dietary Ca lo rie (with a ca pita l C) is o fte n used in the U nitcd Stat es to re pre sent th e c ncrgy co n te n t o f foods. Thc di et ary Ca lo r ic (Cal) is eq uiva le n t to th e kilocalorie o r 1000 calo ries. Thus, a break fast cereal th at gives yOll 100.0 Calories o f nutrition al cne rb'Y per se rving pro vides 100.0 kcal or 4 18.4 k]. ChemislryJ :>"Now'"
•
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214
Chapte r 5
I
Principles of Chemical Reactivity: Energy and Chemical Reactio ns
Chemical Perspectives
Food and Calories
Th e U.S. Food and Drug Administration (FDA) mandates t hat nutritional data, including energy content, be included on almost aU packaged food. The Nutrition Labeling and
Because carbohydrates may contai n some indigestible fiber, the mass of fiber is subt racted from the mass of carbohydrate when calculating the energy from carbohydrates. As an example, one serving of cashew nuts (about 18 g) has
Education Act of 1990 requires that t he total
energy from protein, carbohydrates, fat, and alcohol be specified. How is this determined? Initially, the method used was calorimetry. In this method, which is described in Section 5.6. a food product is burned, and the energy transferred as heat in the combustion is measured. Now, however, energy contents are estimated using the Atwatersystem. This specifies the follow ing average values for energy sources in foods: 1 1 1 1
9 9 9 9
prot ein ~ 4 kcal (17 kJ) carbohydrate ~ 4 kcal (17 kJ) fat ~ 9 kcal (38 kJ) alcohol ~ 7 kcal (19 kJ)
EXERCISE 5.1
Nutrition Facts
St'rviny Silt' 1 cup (30g) Children Under 4 • ¥4 cup (20g) ServIngs Per Container About 19 Children Under 4 • About 28
'-"'Calori e.
14 9 fat ~ 116 kcal 6 g protein = 24 kcal 7 g carbohydrates - 1 g fiber = 24 kcal Tot al ~ 174 kcal (718 kJ)
CaJorieS fromFal
110 15
- ....... -_.
~.~ UlIIl .....
150 20
70 10
'" DallJYaIM"
Total Fat 2g' ~lurate H,( g)
+ If, O,(g)
!J.,W
= + 241.8
• Notation for Thermodynamic Parameters NIST and IUPAC (International Union of Pure and Applied Chemistry) specify that parameters such as !i H should have a subscript, between !i and the thermodynamic parameter, that specifies the type of process. Among t he subscripts you
wi ll see are: a lower case r for "reaction." f for "formation," c for "combustion," fus for "fusion:' and vap for "vaporization."
kJj mol-rxn
T h e positive sig n o f !!J.J t " ill this case ind icates tha t th e d ecom p ositio n is a n e n d oth e rm ic p rocess. The re are seve ral impo rt a nt thi n gs to know ab ou t aJIo. • The d esign ati o n o f aJr as a "sta n da rd e nt h alp y cha nge" mea ns tha t th e pure, unmi xed re actants in th e ir sta n d a rd stat es have fo rmed pu re , unmi xed products in th eir sta n d a rd states (whe re th e su pe rscr ip t ° indicates sta n dard co n d iti o ns) . The standar d state o f a n ele me n t o r a co m po u n d is d e fin ed as th e mo st sta b le form of th e substan ce in th e ph ysical sta te that ex ists at a pressure o f I bar a n d at a sp ecified tem perature . [i\l os t so urc es repo rt sta nd ard reactio n e n th alp ies at 25 °C (29R Kj .] • T h e "pe r mol-rxn" d esign a tion in th e u n its fo r a Jlom ea ns thi s is th e en th alp y c ha nge fo r a "mo le of react ion " (where rxn is a n a b bre viatio n fo r reac tio n ). For ex a m p le , fi n' th e reacti on I J, O (g ) ---> I1, (g ) + 1/ 2 O , (g ) , a mol e of reaction h as occu rred whe n I m ol o f wate r vapor h as bee n con verte d co m p letel y to 1 mol of Hz a nd 1/ 2 mol o f O z. Now con side r th e op po site rea ctio n , th e com bina tio n o f hydrogen a n d oxyge n to fo rm I mol of wate r. T h e mag nitude o f th e e n th alpy change fo r thi s reacti o n is th e same as th at for the d eco mpo sition reactio n , bu t th e sign of ! J.J r is reversed . T h e exo t he rm ic formati o n of 1 mo l of wate r ya p or fro m 1 m ol o f Hz a n d 1/ 2 mol of O 2 tra nsfe rs 24 1.8 ~J to th e su rr o u n d ings (Figu re 5. 13).
H,(g) + 'I, O,(g) ---> H,O(g)
!J.,W
= - 241.8
kJj mol-rxn
T h e value of aJ 1° d epends o n th e c he mica l eq uatio n used. For exam p le , aJfO for 1 mol e o f th e re ac tio n
2 H,( g) will b e twice th at o f
+ O,( g) ---> 2 H,O(g)
!J.,W
~
• Moles of Reaction. Mol-om One "mole of reaction" is said to have occurred when t he reacti on has occurred according
to the number of moles given by the coefficients in the balanced equation. (This concept was described as a way to solve limiting reactant problems on page 167.)
• Fractional Stoichiometric Coefficients When writing balanced equations to define thermodynamic quantities, chemists often use fractional stoichiometric coefficients. For example, to define .l ,H for the decomposition or forma tion of 1 mol of H20, the coefficient for O2 must be 1/ 2.
- 483.6 kJjm ol-rxn
aJr for th e reaction
H,(g) + 'I, O,(g) ---> H,O(g)
!J.,W
= - 241.8
kJj mol-rxn
Th is h appe ns beca use 1 mol e o f reac tio n for the first eq ua tio n uses twice the amo u n t o f reactants a nd p roduces twice the am ount o f product as th e sec o n d equatio n . 5 .5
I
Ent halpy Changes for Chemical Reactions 227
(b) When t he balloon breaks, the candle flame ignites the hydrogen .
(a) Alighted candle is brought up to a balloon filled with hydrogen gas.
il rHO = - 241.8 kJj mol-rxl
'12 O,( g)
+
H,O(g)
HAg)
Active Figure 5.13 The exothermic combustion of hydrogen in air. The reaction transfers energy to the surroundings in the form of heat, work, and light.
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• Standard Conditions The superscript indicates standard conditions. It is applied to any type of thermodynamic data, such as enthalpy of fusion and vaporization (il fusW and il vapW ) and enthalpy of a reaction (il ,W). Standard conditions refers to reactants and products in their standard states at a pressure of 1 bar. One bar is approximately one atmosphere (1 atm = 1.013 bar; see Appendix B). 0
• Enthalpy of Fusion and Enthalpy of Vaporization Previously, we called ilfusW and iloapW the heat of fusion and heat of vaporization, respectively. You can see now t hat, based on the way that the process is carried out (at constant pressure) and by the use of H in their symbols, these are more properly referred to as the enthalpy of fusion and t he enthalpy of vaporization. From this point on, we will refer to them by these designations. 228
Ch apter
5
I
It is importan t to id en tify the states o f reactants an d prod uct"; in a reaction beca use th e magnitud e of !:J. I'H o d e pe nd s o n whet he r th ey are so lids, liquids, or gases. For the formation of I mo l of liquid wafer fr o m th e e le ments, th e en tha lpy change is - 285.8 ~J .
H,(g)
+ 'I, O,(g) --->
H,O(C)
!J. ,W
~
- 285.8 kJjm ol-rxn
No tice tha t thi s value is no t th e same as !:J. J / 0 fo r the fo rmatio n of umlrr va/JOI" fro m hydrogen and ox ygen. T he d iffere nce betwee n th e two value s is equal to th e en tha lpy change for the con de n satio n of I mo l of water vapor to I mol of liqu id wate r; T hese exa mp les illustra te seve ra l ge neral fea tures o f the entha lpy changes fo r chem ica l reaction s. • En tha lpy changes are spe cific to th e reactio n bei ng ca rried o ut . T he id en tities of reactants and products and th eir states (s, C, g ) are important , as a re the amou nts of reactan ts an d prod uct s. • T hc e nt ha lpy change depen ds o n th e n u m ber of mol es o f reaction; th at is, th e nu mber of tim es the reaction as unitten is carried out. • !:J.J JO has a negative value for an exothermic reaction . It has a pos itive value for an endo therm ic react ion . • Values o f !:J. J l " are n um e ricall y the same , but opposite in sig n, for chem ical reactio ns that are the reverse of each ot her. Standard reaction en thalpies ca n be used to calculate the quan tity o f eneq,l)' transfc rrcd as heat un d er co nd itio ns of co nsta n t pressu rc by any given mass of a reactant or p ro d uct. Su p pose yOll want to know the e ne rgy transferred to th e surrou ndin gs as heat if 454 g of pro pane, C :~ H R , is burn ed (a t co nstant pr essu re ), given the equation for the ex ot hermi c co mbustio n and th c entha lpy cha nge for th e reaction . C,H, (g)
+ 5 O,(g) --->
3 CO, (g)
Principles of Chemical Reactivity: Energy and Chemical Reactions
+4
H, O(C)
tJ.,W
= - 2220 kJjm ol-rxn
Two ste ps are need ed. First, find the am o un t of propan e present in the samp le: , ( 1 mol C,H, ) 454 9 C,H, ' = 10 .3 mol C,H, 44.10 9 C,H,
Seco nd , m ult iply 6. ,.Ho by th e amo un t of pro pan e: . ' t. ,H0 _ - 10. 3 mol C,H,
(
1 .mot"rx"' )(- 2220 _
1 mol C3Hs
kJ) -_- 22,900 kJ
1 mot-om
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• CalcuLating the Enthalpy Change for a Reaction Problem Sucrose (sugar, C12 HnO n) can be oxidized to CO 2 and H20 and the enth alpy change for the reaction can be measured (under conditions of constant pressure).
C"H" O,,(s) + 12 O,(g) --> 12 (O,(g) + 11 H,O(C)
.l ,W = - 5645 kf/mo l-rxn
Whatis t he energy t ransferred as heat by burning 5.00 g of sugar? Strategy We willfirst determine the amount (mol) of sucrose in 5.00 g, t hen use t his with the value given for the enthalpy change for the oxidation of 1 mol of sucrose. Solution _ 1 mol sucrose ~ 5.00 q s ucrcse X
342 .3 q.sucrose
d ,W
• Chemical Potential Energy Gummi Bears are mostly sugar. See Example 5.5 for a calculation of the energy in a spoonful of sugar, and see ChemistryNow Screen 5.2 for a video of a Gummi bear consumed by an oxidizing agent.
= 1.46 X 10- 2 mol sucrose
- 5645 kJ) = 1.46 X 10- ' mol sucros·e, ( 1 mol-rxn_. ) ( --=c _=-:.:..:::. 1 mol sucrose
1 mol-rxn
.l ,W = - 82.5 kJ Comment A person on a diet might note t hat a (level) teaspoonful of sugar (about 3.5 g) supplies about 15 Calories (dietary Calories; the conversion is 4.184 kJ = 1 Cal). As diets go, a single spoonful of sugar doesn't have a large caloric content. But willyou use just one level teaspoonful? EXERCISE 5. 5
Enthalpy CalcuLation
The combustion of ethane, C,He, has an enthalpy change of - 2857.3 kJ for t he reaction as written below. Calculatethe value of energy transferred as heat when 15.0 9 of C,He is burned.
2 (,H. (g) + 7 O,(g) --> 4 (O,(g) + 6 H,0(9)
5 .6
.l ,W = - 2857.3 kJ/mol-rxn
Calorimetry
T he energy evo lved or required as heat in a che mic al or ph ysical process can be measured by calo rime try. The ap para tus used in th is kin d of ex pe rimen t is a calori meter:
Constant Pressure CaLorimetry, Measuring AH A cons ta n t pressure calo rime te r can he used 10 measu re th e ene rgy cha nge fo r a chemic a l rea ctio n as e ne rgy is tran sfer red as heat und e r co ns ta n t press ure co nd itio ns , th at is, it measures th e en tha lpy change. 5 .6
I
Calorimelry
229
Thermometer Cardboard or ~_-'ll tn:"", / Styrofoam lid
~:;: ,
cam cups
-~'--
Reaction occurs in solution.
FIGURE 5.14 Acoffee-cupcalorimeter. Achemical reaction produces a change in temperature of the solution in t he calcrirneter. The Styrofoam container is fairly effective in preventing the t ransfer of energy as heat between the solution and its surroundings. Because the cup is open to the atmosphere, this is a constant pressure measu rem ent.
In gene ra l che m istry labo rato ries, a "co ffee-cu p ca lorim eter " is often used to estimate e n t ha lpy chan ges for che mica l reactions. T his in expensive d evice co ns ists of 1\"'0 nested Styrofoam co ffee cu ps with a loosc-fl nl ng lid and a temperatu remeasu rin g device suc h as a thermom et e r (Fig u re 5.14) or th ermo cou pl e . Styrofoam , a fairly good in sul ator, minimi zes en er6'1' tr ansfer as heat bet ween th e syste m an d th e surroundings. The re action is carried out in so luti o n in th e cu p. If th e reactio n is ex o the rm ic, it rel eases ene rgy as heat to th e so lu tio n , and th e temperatu re o f th e so lu tio n rises. If th e reactio n is e ndo the rm ic, e ne rgy is a bso rbe d as heat fro m th e so lutio n , a nd a d ecr ease in th e te mperatu re of th e solution will be see n . Th e chang e in temperature o f th e so lu tio n is measu re d . Kno win g th e mass and specific he at ca pacity of t he so lu tio n a nd th e temperatu re cha nge, the e n tha lpy cha ng e for th e reactio n can be calcu late d . In th is ca lo ri m e tr y ex pe rime n t, it will be co nve nien t to define th e che m ica ls and th e so lu tio n as th e system . T he su rr o u nd in gs are th e cu p and everyt hin g beyond th e cu p. As noted above, we assu me tha t the re is no e ne rgy tr a ns fer to th e cu p o r beyond and th at e ne r6'1' is transferred o nly as heat with in th e syste m . Two e ne rgy cha ng es oc cu r with in th e system . One is t he cha nge that tak es p lace as th e che m ical re action o ccurs, ei the r rel easin g th e potenti al en e rgy sto red in th e reacta n ts or abso rbing ene rgy and co nve rting it to potentia l e ne rg y stored ill th e p roducts. \Ve label thi s e ner6'1' as 1,. The other en er6'1' cha nge is th e e nergy ga ine d o r lost as hea t by th e so lut io n ( q so luti o ll ) ' Based on the law of con se rvation of ene rgy,
The value o f 'l"".Jhllioll can be ca lcu lated from th e spec ific he at capac ity, mas s, and change in temperature o f the so lut io n . The qu antity o f ene rgy evo lved o r absorbe d as heat for th e reacti o n (qr) is the unknown in the eq ua tio n . The acc u racy of a ca lo rim e try exper imen t d epends o n the accuracy of th e measured quantities (tem peratu re, mass, specific heat capa city). In addition , it d epends 011 how clo sely th e assu mp tio n is followed t hat th ere is no ene rgy tr an sfer beyond th e so lu tio n . A coffee-cup calo rime ter is a n un sophisticat ed ap pa ra tus, and th e result s obtain ed with it are not h ighl y accura te, la rgel y because th is assu m p tio n is poorl y met. In research laboratories, calorime ters are used th at mo re effectively lim it the en ergy transfer between system and su rro u nd ings. In add ition , it is also possibl e to estima te and co rrec t fo r th e mi nimal e ne rg y transfe r th at o ccurs between the system an d th e su rro u nd in gs.
. Using a Coffee-Cup Calorimeter Problem Suppose you place 0.050 0 g of magnesium chips in a coffee-cup calorimeter and t hen add 100.0 ml of 1.00 MHCl. The reaction that occurs is
Mg(s) + 2 HCI(aq) ---> H, (g) + MgCl,(aq ) The temperat ure of the solution increases from 22.21 °C (295.36 K) to 24.46 °C (297.6 1 K) . What is t he ent halpy change for the reaction per mole of Mg? Assume t hat t he specific heat capacity of t he solution is 4.20 J/g ' Kand t he density of the Hctsolution is 1.00 q/ml.. Stra tegy The energy evolved in the reaction is absorbed by the solution. Solving the problem has t hree steps. Fi rst. calculate q,olution from t he values of t he mass, specific heat capacity, and j. Tusing Equation 5.1. Second, calculate o. assuming no energy t ransfer as heat occurs beyondt he solution, t hat is, q, + q , olution = O. Third, use the value of qr and the amount of Mg to calculate t he enthalpy change per mole of Mg.
230
C h apte r
5
I
Principles of Chemical Reactivity: Energy and Chemical Reactions
Solution Step 1. Calculate q solution' The mass of t he solution is the mass of t he 100.0 mLof HCl plus the mass of magnesium.
q;o'""," ~ (1 00 .0 9 HCI solution + 0. 0500 9 Mg)( 4.20 Ji g K)(297 .61 K - 295.36 K)
=
9.45 X 10 2 J
Step 2. Calculate q,. q,
+ q lotution = 0 =0
q, + 9.45 X 102 J q,
=
- 9.45 X 10 2 J
Step 3. Calculate t he value of !iH per mole of Mg. Note t hat q. found in Step 2 resulted from the reaction of 0.0500 g of Mg. The enthalpy change per mole of Mg is t herefore .l,H ~ (- 9.4 5 X 10' J/O.0500 9 Mg)(24 .31 9 Mg/l mol Mg) ~
- 4.60 X 10' J/m ol Mg ( ~ - 460. kJ/m ol-rxn Mg)
Comment The calculation gives the correct sign of q, and !i ,H. The negative sign indicate s t hat this is an exothermic reaction.
EXERC iSE 5 .6
Using a Coffee-Cup Calorimeter
Assume 200. ml of 0.400 MHCl are mixed wit h 200. mL of 0.400 MNaOH in a coffee-cup calorimeter. The tempe ratu re of the solutions before mixi ng was 25.10 DC; after mixing and allowing t he reaction to occur, t he te mperature is 27.78 0 [, What is the ent halpy change when one mole of acid is neutralized? (Assume t hat t he densities of all solutions are 1.00 q/m l, and their specific heat capacities are 4.20 Ji g ' K.)
Constant Volume Calorimetry: Measuring J1U Co ns ta nt volum e ca lorime try is oft en used to evalua te hea ts of co mb ustio n o f fu el s and th e ca lo ric value of foods. A weighed sam ple o f a co m bustible so lid or liquid is placed inside a "bo mb," o fte n a cylind e r ab ou t the size ofa large frui t juice can with th ick stee l walls an d e nds (Figu re 5. 15) . Th e bomb is placed in a water-filled co n ta ine r with well-insu late d walls. After fillin g th e bomb with pure oxyge n , th e sam ple is ignited , u suall y by an el ec tric spa rk. T he heat ge ne ra te d by th e co mbustion reacti on warms th e bomb an d the wate r aro u nd it. T he bom b, its con tents, and th e wate r are d efined as th e system . Assessme n t of en eq.,Y ' tran sfers as heat within th e syste m shows th at
• Calorimetry, .lU, and j, H The two types of calorimetry (constant volume and constant pressure) highlight t he differences between enthalpy and internal energy. The energy transferred as heat at constant pressure, q~, is, by definition, j, H, whereas the energy transferred as heat at constant volume, q.. is .l U.
where 'Iris th e e ne rgy p roduced by the reaction , q ho m h is th e energy invo lved in heating th e ca lo rime te r bo mb, and q.,.. ~ ( n is the e nerh'Y involved in heating th e wate r in th e ca lori me te r. Because th e volu me d o cs not change in a co ns ta n t volu me ca lori meter, ene rgy transfer as wor k ca n not occ ur. Th erefo re , the e nergy tran sfe rre d as heat at co nsta n t volume (qJ is th e cha nge in in te rna l e nergy, dUo
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5 .6
I
Ca lorimet ry 231
Active Figure 5.15 Con stant volume calorimet er. Acom bustible sample is burned in pure oxygen in a sealed metal containeror "bomb." Energy releasedas heat warms the bom b and the water surrounding it. By measuring the increase in temperature, the energy evolved as heat in the reaction can be determined.
Water
Stirrer
Thermometer
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Steel Sample Steel Insulated outside container dish bomb container
Thesample burns in pure oxygen. warming t he bomb.
The heat generated warms the water, and .lTis mea sured bythe thermometer.
Constant Volume Calorimetry ProbLem Octane, ( sH 1s, a primary constituent of gasoline, burns in air: e,H,,(t) + 25/2 O,(g) ---> 8 eO,(g) + 9 H,O(t)
A 1.00-g sample of octane is burned in a constant volume calorimeter similar to that shown in Figure 5.15. The calorimeter is in an insulated contai ner with 1. 20 kg of water. The temperature of the water and the bomb rises from 25.00 O( (298.15 K) to 33.20 O( (306.35 K) . The heat ca pacity of the bomb, Cbomb, is 837 J/K. (a) What is the heat of combustion per gram of octa ne? (b) What is t he heat of combustion per mole of octane? Strategy (a) The sum of aU the energies tra nsferred as heat in the system will be zero; that is, q, + q bomb + qw~tef = o. The first term, q, is the unknown. The second and third terms in the equation can be calculated from the data given: q homh is calculated from the bomb's heat capacity and .iT, and qw~te , is determined from the specific heat capacity, mass, and .i Tfor water. (b) The value of q, calculated in part (a) is the energy evolved in the combustion of 1.00 g of octane. Usethis and the molar mass of octane (114.2 g/mol) to calculate the energy evolved as heat per mole of octane. Solution
(a)
qw~te '
= Cwaw ~
X m w~tef X .i T
(4.184 J/g ' K)(1.20 X 10' g)(306.35 K - 298.15 K)
~
+41.2 X 10' J
q' om' ~ (e' om') ( ~7) ~ (837 J/K)(306.35 K - 298.15 K) ~ 6.86 X lO' J q, +
q wate,
+
qh omh
q, + 41.2 X 103J + 6.86 q, Heat of combustion per gram
=
~
=0 x 10l J = a
- 48.1 X 10' J (or - 48.1 kJ )
- 48.1 kJ
(b) Heat of combustion per mol of octane
= (-48.1
kJ/g)(114.2 g/mol)
=
- 5.49 X 10l kJ/mol
Comment Because the volume does not change, no energy transfer in the form of work occurs. The change of internal energy• .iV, for the combustion of ( eH ls(t') is - 5.49 X 10 3 kl/mc l. Also note that Cbomb has no mass units. It is the heat required to warm the whole object by 1 kelvin. 232
Ch apter 5
I
Principles of Chemical Reactivity: Energyand Chemical Reactions
EXER CISE 5 .7
Constant Volume Calorimetry
A 1.00-9 sample of ordinarytable sugar (sucrose, C12H n OIl ) is burned in a bomb calorimeter. The temperature of 1.50 X 10 3 9 of water in the calorimeter rises from 25.00 O( to 27.32 "C. The heat capacity of the bomb is 837 J/K, and the specific heat capacity of t he water is 4.20 Jig ' K. Ca lculate (a) t he heat evolved per gram of sucrose and (b) the heat evolved per mole of sucrose.
5.7
Enthalpy Calculations
Module 10
Entha lp y changes for a n e norm o us number of che m ica l a n d ph ysica l p rocesses a re ava ila ble 0 11 th e Wor -ld \ Vi d e ' Ne b a n d in referen ce books. T h ese d at a h ave been co llec ted by scie n tists ove r a number of yea rs fro m m an y ex pe rime n ts a n d arc u sed 10 ca lcu la te e n tha lpy c hanges for c he mical p ro cesses. Now we wan t 10 di scuss how to use suc h d at a.
Hess's Law Th e e n th a lpy c ha nge ca n be m ea sured by ca lo rime try for m an y, but n ot all , chemica l processes. Co ns ider, { H,O(f)
.l ,Hl
~
- 285.8 kf/rncl-rxn
Equation 3:
CH ,(g) + 2 O,(g) --> CO,(g) + 2 H,O(f)
.l ,H,
~
- 890.3 kJ/mol-rxn
Use this information to calculate J.,Ho for the formati on of methane from its elements.
Strategy The three reactions (1, 2, a nd 3), as they are written , cannot be added together to obtai n the equation for the forma tion of CH 4 from its elem ents. Met hane, CH4, is a productin the reaction for which we wish to calculate J.,W , but it is a reactant in Equation 3. Water appears in two of these equations although it is not a component ofthe reaction forming CH 4 from carbon and hyd rogen. To use Hess's law to solvet his problem , we will first have to manipu late the equations and adjust t he ~ ,W values accordingly before adding equations together. Recall, from Section 5.5, that writing an equation in the reverse direction changes the sign of .1. ,W and t hat doubling t he amount of reactants and products doubles the value of j" W . Adj ust ments to Equations 2 and 3 will produce new equations that along with Equation 1, can be combi ned to give the desired net reaction.
SoLution To have CH 4 appear as a product in the overall reaction, we reverse Equation 3, which changes the sign of .1.,W . Equation 3':
CO,(g) + 2 H,O(f) --> CH,(g) + 20,(g) j" H]'
=
- j" H'l = + 890.3 kJjmo l-rxn
Next, we see that 2 mol of H2( g) is on t he reactant side in our desired equation. Equation 2 is written for only 1 molof H2(g) as a reactant. Therefore we multiply the stoichiometric coefficients in Equation 2 by 2 and multiply the value of j" H'" by 2. Equation 2':
2 H,(g) + O, (g) --> 2 H, O(f) ~ ,H r
=
2 j, rH't = 2 ( - 285 .8 kJjmo l-rxn)
=
- 571.6 kJjmo l-rxn
We now have three equations that when added toget her, will give the targeted equation for the formation of methane from carbon and hydrogen. In this summation process, 02(g). H20( !'). and COAg) aU cancel. Equation 1:
C(s) + O,(g ) --> CO,(g)
j" Hi = - 393.5 kf/rncl-rxn
Equation 2':
2 H,(g) + O, (g) --> 2 H,O(f)
Equation 3':
CO,(g) + 2 H,O(t) --> CH,(g) + 2 0, (g)
= .1,H3' = j" H'r
.l ,H:"
Net Equation: C(s) + 2 H, (g) --> CH, (g)
.l ,H:"
~
=
~
= - 571.6 kl/mcl-rxn - ~,Hl = + 890.3 kJjmol-rxn
2 .1.,H'2
.l ,H; + 2 .l ,Hl + (- .l ,H;)
( -393 .5 kf/mc l-rxn) + ( -571.6 kJ/mol-rxn) + ( +890.3 kf/mc l-rxn) - 74.8 kJj mol-rxn
Thus, for the formatio n of 1 mol of CH 4(g) from the elements, we find j" H'"
=
- 74.8 kJjmol- rxn.
EXERCISE 5.8 Using Hess's Law Use Hess's lawto calculate the enthalpy change fur the form ation of CS2(f ) from C(s) and 5(s) [c(s ) + 2 S(s) --> CS,( f) ] from the following entha lpy values. C(s) + O,(g) --> CO ,(g)
.lo W
~
- 393.5 kJ/mol -rxn
S(s) + O,(g) --> SO,(g)
.lo W
~
- 295.8 kJ/ mol-rxn
CS,(f) + 3 O, (g) --> CO,(g) + 2S 0,(g)
.lo W
~
- 1103.9 kf/mol-rxn
5 .7
I
Enthalpy Calculations
235
Problem Solving Tip 5.2
Using Hess's Law
How did we know how the three equations should be adjusted in Example 5.B? Here is a general strategy for solvi ng this type of problem.
are reactants in Eq uations 1 and 2, and the product, CH4(g). is a reactant in Equa tion 3. Equation 3 was reversed to get CH 4 on t he product side.
Step 1. Inspect the equation whose J ,Ho you
Step 2. Get t he correct amount of t he substances on each side. In Example 5.8, only one adjustment was needed. There was 1 mol of H2 on the left (reactant side) in Equation 2. We needed 2 mol of H2 in the overall equa-
wish to calculate, identifying the reactants and products, and locate those substances in the equat ions available to be added . In
Example 5.8. the reactants. C(s) and H, (g).
tion; this required doubling the quantities in Equation 2. Step 3. Make sure other substances in the equations cancel when the equations are added. In Exam ple 5.8, equal amounts of O2 and H20 appeared on the left and right sides in the three equations, and so they canceled when the equations were added together.
Standard Enthalpies of Formation Values Consu lt the National Institute for Standards and Technology website (webbook. nist.gov/ chemistry) for an extensive compilation of enthalpies of formation. •
,),/W
Calo rime try and the appli cation of Hess's law h ave made ava ilab le a g reat many for che m ica l reaction s. Often , th ese values a re asse m b led in to tabl es. Th e tab le in Appendi x L, fo r exam p le , lists standard molar enthalpies of formation, ajH o. Th e standard molar p.uthal!)y of"[ormation is the ruthalj)y chau;.,1'£' for the formation of J mol of a compound diJutl), f rom its component elements in their standard states. Se veral exa m ples o f sta n da rd mo lar c n t ha lp ics of form ati on will be he lp ful to illu st rat e thi s definition . tJ.JH o for NaCl(s): At 25 °C a n d a p ressu re o f I bar, Na is a so lid, an d CI2 is a gas. The sta n d a rd e n tha lpy o f form atio n o f NaC l(s) is d efined as th e en th alpy change that occu rs wh e n 1 mol o f NaC I(s) is formed from 1 mol o f Na(s) a n d 'I, mo l o f CI2(g).
aJr values
Na(s)
+ 'J, Cl, (g) --> Na Cl(s)
tJ.fW = - 411.12 kJ/mo l
air for NaCI(aq): The enthalpy o f fo rm atio n fix a n aq ueo us so lu tio n o f a com p o u n d refers to th e e n tha lpy c hange for th e form at ion of a I mo l/ I. so lu tio n o f th e compound sta rting with th e elemen ts m akin g up th e compound. It is thus th e enthalpy of formation o f the com pou n d p lus th e e n th alpy c h a nge that occu rs whe n th e su bs ta nce di ssol ves in water. Na(s) + 'I, CI,(g) --> Na Cl(aq) • Units for Enth alpy of Formation The units for values of .1JH" are usually given simply as kJjmol where the denominator is really mol-rxn. However, because an enthalpy of formation is defined as the change in enthalpy for the formation of 1 mot of compound, it is understood that "per mol" means"per mol of compound, " which, in this case, is the same thing as
tJ.fW = - 407.27 kJ/ mol
afio for C 2H50H(f) : At 25 °C and 1 bar, the sta nd a rd sta tes o f th e cle me n ts a re C(s, gra p h ite), Il, (g ) , an d 0 2(g ) . The sta n dard e n tha lpyof formatio n o fC 2Il , O Il (C) is defin ed as th e e n th alpy c ha nge th at oc cu rs wh e n 1 mol o f C 2 H..,O H (t ) is formed from 2 mol o f C(s) , 3 mol o f H , (g ) , and 1/ 2 mol of 0 2(g ) . 2 C(s)
+ 3 H, (g) + 'I, O, (g) --> C,HsOH(C)
No tice th at th e reacti on d efining th e e n tha lpy o f form ation for liquid. et ha no l is n ot a re act ion that a ch emist ca n ca rry o u t in th e laboratory. This illu strat es an important point: the enthaljJ) of fonuation of a compinmd does not necessarily correspond to a reaction that ran b" carried ou t. Ap p e n dix L lists valu es of a rHo for so me co m mo n su bsta nces, and a review o f th ese values leads to som e im p o rta n t obse rvat ions.
"per mot-rxn."
• 771e standard rnthalj)_v offonnation f or an element in its standard state is zero. • Most arH o values are n egative , indicatin g th at fo rmatio n o f mo st compounds from the clements is ex othermic. A ve ry few vallie s a re po sitive , a n d th ese 236
C h apte r 5
I
Principles of Chemical Reactivity: Energy and Chemical Reactions
represent compounds th at a re u ns table wit h re spect to d ecomposition to th e elements. (O n e exa m p le is NO (g ) with D./r = + 90 .29 k]/mo l.) • Valu es o f tJ.tl t " ca n o fte n b e used to co m p a re th e stab ilities o f rel at ed co mpounds. Con side r th e valu es of tJ.l T" fo r the hydroge n h alid es. I lydrogen flu oride is th e mo st sta ble of t hese co m pou n ds with res pe ct to d ecomposition to th e el em ents, wh e re as III is th e least sta ble (as indicat ed by tJ.lr o f HF bein g the most nega tive value a n d that o f I II bein g th e most po sitive ) . EXERCISE 5 .9
• AfW Values of Hydrogen Halides
Compound HF(g) HCI(g) HBr(g) HI(g)
(kJj mol) - 273.3 - 92.31 - 35.29 + 25.36
j, JW
Standard EnthaLpies of Formation
Write equations for the reactions t hat define t he standard enthalpy of formatio n of FeCl](s) and of solid sucrose (sugar, C12Hn 0 l1 ) .
Enthalpy Change for a Reaction Usin g sta n da rd mo lar e n th aipies o f form ati on a n d Eq ua tio n 5.6, it is po ssib le to ca lcu late th e e n thalpy ch a nge for a rea ct ion u n d er sta n da rd co nd itio ns . (5.6)
In thi s equation . th e sym bol ~ (t he G reek ca pi ta l lette r sig ma) m eans "ta ke th e su m ." To find !!J.J-jO , add up th e mola r e n th aIpies o f formation of th e produ ct s, eac h multiplied by its sto ich io met ric co efficie n t, a n d su btrac t from this th e su m o f t he molar c n th alp ies of formati o n o f th e rea ctants, each mult ip lied by its sto ich iometric coeffic ien t. T h is equatio n is a logic al co nsc q uc nce ofthc d efi nitio n of !!J.1J-f' a n d Hess's law (see A Closer Look: Hess's Iru o and Equation 5.6) . Suppose yo u wan t to know h ow much heat is requ ired to d ecompo se 1 m ol o f ca lciu m ca rbo n ate (limestone ) to calc iu m ox id e (lime) a n d ca rbo n dioxide under sta n dard co n d itio ns :
j, = Final - Initial Equation 5.6 is another example of t he convention that a change P ,) is always calculated by subtracting the value for the initial state (tile reactants) from the value for t he fi4 nal state (the products).
•
CaCO,(5) ---> CaO(5) + CO,(g) 'rou wou ld use th e fo llowin g ent h alpies of fo rm ation (fro m Ap pend ix L): Compound
~lI o
Ca CO,(s)
- 1207.6
CaO(s)
- 635.1
(O,(g )
- 393.5
(kJ/ mol)
an d the n use Equation 5. 6 to fin d th e sta n d ard e n th alp y c ha n ge fo r t he reaction , tJ. ,.J-f' .
a
r
w _ [(1 -
=
mOl CaO) ( -6 35.1 kJ)
1 mol-rxn
mol
Caa
+
( 1 rT]ol CO; ) ( - 393.5 kJ )] 1 mol -rxn
1 mol CO;
_ [( 1 rT]o l caco; ) ( - 1207.6 kJ_)] 1 mol CaCO, 1 rno l-rxn + 179.0 kJ/ mol- rxn
T he d ecompo sition of lim esto n e to lime a n d CO 2 is e n d o the r m ic. T h at is, e nc rbT)' (179.0 ];)) mu st be su p p lied to decom po se 1 mol o f CaCO ,,(s) to Ca O (s) and CO 2 (g ) .
Chemistry-9. NowN Sign in at www.t homsenedu.com/ Ioqtn and go to Chapter 5 Contents to see Screen 5.16 for a tutorial on calculati ng the sta ndard enthalpy change for a reaction.
5 .7
I
Ent halpy Calculations
237
A Closer Look
Hess's Law and Equation 5.6
Equation 5.6 is an application of Hess's law. To illustrate this, let us look further at the decomposition of calcium carbonate.
Energy level diag ram for the decomposition of CaC0 3(s)
Ca ts) + C( s) + + o,(g)
CaCO,(s) ---> Ca O(s) + CO,(g) Because enthalpy is a state function, the change in enthalpy for this reaction is independent of the route from reactants to products. We can imagine an alternate route from reactant to products that involves first converti ng t he react ant to elements in their standard states, then recombining these elements to give the reaction products. Notice that the enthalpy changes for these processes are the enthalpies of formation of the reactants and products in the equation above: Ca CO,(s) ---> Cats) + C(s) + 3/2 O, (g)
- ~fW [ Ca CO , ( s ) ] ~ ~ ,H ;
C(s) + O,(g) ---> CO, (g)
~fW [ CO , ( g ) ] ~ ~ , H ;
Cats) + 'I, O,(g) ---> CaO(s)
~fW [ Ca O ( s )1 ~ ~ ,H ;
.1rH ~
=
;;l,Hi
CaO( s) + CO,(g)
+
~fW [ CO , ( g )l -
+ .1 r H ~ + .1rHi + 179.0 kJ
.l rH ~et = .l rH ~ =
+ :l,H'2 + !l ,H'3
~ , W ~ ~fW [ Ca O ( s ) 1
=
.lr H ~ = - ~fH ' [ C a C 0 3 ( s ) 1 = +1207.6 kJ
Ca CO ,(s) ---> Ca O(s) + CO,(g) J. rH ~et
+ .1rHi
(-635.1 kJ) + (-393.5 kJ)
~f W [Ca CO , ( s ) 1
CaCO, (s)
That is, the change in enthalpy for the reaction is equal to the enthetpies of formation of products (C02 and CaD) minus the enth a lpy of formation of the reactant (CaC0 3) , which is, of course, what one does when using Equation 5.6 for t his calculatio n. The relationsh ip among t hese ent halpy quantities is illust rated in the energy-level diagram.
. . Using Enthalpies of Formation ProbLem Nitroglycerin is a powerful explosive that forms four different gases when detonated :
2 C,H,(NO,),(f) ---> 3 N,(g) + 'I, O,(g) + 6 CO,(g) + 5 H,O(g) Calculate the enthalpy change that occurs when 10.0 g of nitroglycerin is detonated. The standard enthalpy offormation of nitroglycerin , llJW , is - 364 kJjmot. Use Appendix L to find other J.fW values that are needed. Strategy Use values of j.f W for the reactants and products in Equation 5.6 to caLculate the enthalpy change produced by one mote of reaction P,W). From Appendix L, J.f W [CO Ag)] = - 393.5 kJjmol, .lf W [ H20 (g)] = - 241.8 kJjmot, and .lfW = 0 for N2(g) and 02(g). Determine the amount (moL) represented by 10.0 9 of
nit rogLycerin; then use this value with.l ,W and the balanced chemical equation to obtain the answer. SoLution Using Equation 5.6, we find the enthalpy change for the explosion of 2 mol of nitrogLycerin is
s.n: ~
( 6 mOICO' ) d W [CO,(g)] + (5 mol H'O) d W [H,O(g)J 1 moL-rxn f 1 moL- rxn f _ (1
d ,W
1 moLCO2
Ch a p t er 5
I
mo -rxn
= ( 6 mol CO, ) ( -393.5 kJ) + ( 5 mol H,0 ) ( -241.8 kJ ) 1 mol-rxn
238
m~1 C,~,(NO,), ) d fW [C ,H,(NO,),(f)]
1 mol-rxn
1 mol H20
kJ
1 mol C,H,( NO,), ) ( - 364 ( 1 met-on 1 mol ( 3HS(N03h
Principles of Chemical Reactivity: Energy and ChemicaL Reactions
) = - 2842 kJ/ moI-rxn
The problem asks for t he enthalpy change using 10.0 g of nitroglycerin. We next need to determine t he amount of nit roglycerin in 10.0 g. . . . ~.~ ( 1 mot nitrOglYCerin ) 10.0 g ni troqlycerin . _. _ 227 .1 9 mtroglycenn
.. = 0.0440 mol mtrogtycenn
The enthalpy change for t he detonation of 0.0440 mol of nitroglycerin is dW
. qtycerin _._ ( = 0.0440 mol.nitro ~
.
1 riiot-rxn )(- 2842 kJ ) . . .~ _
2 mol .nitrc qtycerin
1 mol-rxn
- 62.6 kJ
Comment The large value of ;l W is in accord with t he fact that t his reaction is highly energetic.
EXERCISE 5. 10 Using Enthalpies of Formation Calculate t he standard enthalpy of combustion for benzene, C6 H6 •
+ 15/2 O, (g) -> 6 CO, (g) + 3 H,O(f) ~I W [ C. H . ( f ) ] ~ +49.0 kJ/mat.
C.H. (f )
Other values needed can be found in App endix L.
Product- or Reactant-Favored Reactions and Thermodynamics 5 .8
At th e b cg innin g o f th is c h a p te r, we no ted th at the rmodyn am ics wou ld provid e an swers to fou r qu estions. H ow m u c h e nergy is evo lved o r re q ui re d in ph ysical ch a nges a n d in c he m ica l react ion s, a n d th e re lati onship of h ea t a nd work h ave bee n th e prima ry topics of thi s c h a p te r. T he first two que st io n s we re ad d ressed in this c ha p te r. bu t th ere a re 1\ \'0 oth e r im po rt an t q u esti ons: II ow ca n we d e termine whe the r a reac tio n is p rod uc t-favo red o r re act a nt -favored a t eq u ilib riu m ? An d wha t d e termi n e s whe th e r a c he m ica l reac tion will occ u r spo n ta neous ly; th at is, with o ut ou tsid e in te rven tio n ? In C ha p te r 3, we lea rn ed th at c hemica l reac tio n s pro cee d to ward eq u ilibriu m, a nd sp o n tan c ous c h a nge s o cc u r in a way th a t a llows a systc m to a p p ro ac h eq u ilib riu m . Reac tio n s in whic h re acta nts a re la rg e ly co nv e n e d to p roduc ts whe n eq u ili b riu m is reac h ed a re sa id to b e jJrodurl{'lVornl. Re act io n s in wh ic h o n ly a sma ll a mou n t of pro d u c ts a re p rese n t at e q u ilib r iu m are ca lle d reaaant-faoored ( ~ page 12 1) . Let us loo k back a t th e m a ny c he m ical react ion s th at we have see n . Fo r e xa m p le, a ll co m bu stio n reactio n s a re ex o th e rm ic, a n d th e oxi datio n o f iron (Fig u rc 5.17) is clearly e xot h e rm ic.
4 Fe(s) + 3 O,(g) -> 2 Fe,O,(s) 'HO -U r
[Fe20 3 (s)] -- ( 2 mo l Fe,O, )( - 825.5 kJ ) -- - 1 651. 0 kJ/ mal-rxn 2 u'HO f 1 rn ol-rxn
1 mo l Fe 20 3
T he reactio n has a n e gat ive value for t1 J r , a n d it is a lso spo n ta n eo us an d p rod n et-favo red . Co n versely, th e de co m position of ca lciu m ca rbo na te is e ndo thermic .
CaCO,( s) -> CaO( s)
+ CO,(g)
f). ,W = + 179.0 kJ/ mol-rxn 5.8
I
FIGURE 5 .17 The product-favored oxidation of iron. Iron powder, sprayed into a bunsen burner flame, is rapidly oxidized. The reaction is exothermic and is product-favored.
Product- or Reactant-Favored Reactions and Thermodynamics
239
The Fuel Controversy: Alcohol and Gasoline
Case Study
For more information, see the references in Wald's Scientific American article.
It is clear that supplies of fossil fuels are decli ning, and their price is increasing, just as the nations of the earth haveever greater
Ouestions: For the purposes of this analysis. let us use octane ((aH18) as a substit ute for t he complex mixture of hydrocarbons in gasoline. Data you will need for this question (in addition to Appendix L) are:
energy needs. We wilt have more
to say about this in the Intercha pter (Energy) that follows. Here. however, let's analyze ~ the debate about replacing qaso- ..§: line with ethanol (C2HsOH). ~ As MatthewWald says in the £ article "Is Ethanol in for the lo ng .3~ Haul?" (Scientific American ,
j,f W
Ja nuary 2007), "The U.S. has
gone on an ethanol binge." In 9 2005, the U.S. Congress passed an Ethanol available at a service station. E85 fuel is a blend of 85% energy bill stating that ethano l pro- ethanol and 15% gasoline. Be aware that you can only use E85 in duction shouldbe 7.5 billion galvehicles designed for the fuel. In an ordinary vehicle, the ethanol lons a year by 2012, up from about leads to deterioration ofseals in the engine and fuelsystem. 5 billion gallons a year presently. The goalis to Beyond this, there are other problems at least partially replace gasoline with ethanol. associated with ethanol. One is that it cannot Is the goalof replacing gasoline with ethanol be distributed through a pipeline system as reasonable? This is a lofty goal, given that presgasoline can. Any water in t he pipeline is ent gasoline consum ption in the U.S. is about misci ble with ethanol. which causes t he fuel 140 billion gallons annually. Again, accordi ng to value to decline. Instead. ethanol must be Matthew Waldo"Even if 100 percentof the U.S. trucked to service stations. corn supply was distilled into ethanol, it would Fi nally, E85 fuel-a blend of 85% ethanol supply onlya small fraction of the fuel consumed and 15% gasoline-cannot be used in most by the nation's vehicles." Wald's thesis in his current vehicles because relatively few vehiarticle, which is supported by numerous scientific des as yet have engines designed for fuels studies, is that if ethanol is to be pursued as an with a high ethanol content (so-called "flexialternative to gasoline, more emphasis should be ble fuel" engines). The number of these vehiplaced on derivi ng ethanol from sources other des would need to be increased in order for than corn, suchas cellu lose from cornstalksand E85 to have a significant effect on our qasovarious grasses. line usage.
• Reactant-Favored or Product-Favored? In most-but not all-ceases exothermic reactions are product-favored at equitibrium and endothermic reactions are reactant-favored at equilib rium.
[C.H,,«() ]
~
- 250. 1 kJ/mo l
Density of ethano l = 0.785 gjmL Density of octane = 0.699 gjm L
1. Calculate ~ rH o for the combustion of ethanol and octane, and compare the values permol and pergram . Which provides more energy pergram? 2. Compare the energy produced perliterof the two fuels. Wh ich produces more energy for a given volume (something useful to know when jillingyourgas tank)? 3. What mass of COlt a greenhouse gas, is produced perliter of fuel (assuming com plete combustion)? 4. Now compare the fuels on an energyequivalent basis. What volume of ethanol would have to be burned to get thesame energy as 1.00 Loj octane? When you bum enough ethanol to have the same energy as a liter of octane, which fuel produces moreCOl? 5. On the basis of this analysis and assuming the same price per liter; which fuel will propel yourcarfurther? Which will produce lessgreenhouse gas? Answers to these questions are in Appendix Q.
Th e deco m position o f Ca C0 3 p ro ceeds spo n ta ne o usly to a n equ ilih riu m th at favors th e reac tan ts; th at is, it is reactant-favo red . Are all ex othe rmi c reacti ons p rodu ct-favored and all endo the rm ic reacti ons reactan t-favo red ? From these exam ples , we mi gh t form u late that idea as a hypothesis th at ca n be tested by experimen t and by exami natio n o f othe r exa m ples. \Ve wou ld find th at in most cases, product-faoored reactions have negative values ofa rH o, an d veactant-faoored reactions have [msitiue values ofa rllo. Bu t th is is not fllwfl)'s true ; there are ex ce p tio ns . Clea rly, a fu rt her discussion o f th e rmo dynam ics must be tie d to the co nc e p t o f eq uilib riu m. This rel atio nship , and the co mp le te di scussio n of the thi rd an d fourt h question s, will be p resen ted in Cha p ter 19. ChemislrY.ci.Now~
•
Sign in at www.thomsonedu.com/Ioqtn and go to Chapter 5 Contents to see Screen 5.17 Product-Favored Systems, for an exercise on the reaction when a Gummi Bear is placed in molten potassium chlorate.
240
Ch apter
5
I
Principles of Chemical Reactivity: Energy and Chemical Reactions
C hapter Goals Revisited N ow that you ha ve studi ed this chapter; you should ask uihether you have met the chnp ter
Chemislry.o.Now"" Sign in at www.
goals. In porticulai; ) 'OU shou ld be able to:
t homsonedu:com/log in to: Assess your understanding with Study Questions in OWL keyed to each goal in t he Goals and Homework menu for this chapter For quick review, download Go Chemistry mini-lecture flashcard mod ules (or purchase them at www.ichapters.com) Checkyour readiness for an exam by taking t he Pre-Test and exploring the modules recom mended in your Personalized Study plan.
Assess the t ransfer of energy as heat as sodated wit h changes in tem perature and changes of stat e a. Describe various fo rm s of ene rgy a n d the n atu re of e nergy tr an sfe rs as h eal b. c.
(Se c tio n n.n. Usc th e m o st co m m o n e ne rgy unit , th e j oule, a n d co nve r t betwe en other ene rgy units a n d j oules (Se c tio n 5. 1) . Study Question(s) assignable in OWL: 5. Re co gnize and use th e la ng ua ge o f thermodynam ics: th e syste m and its su r rou ndi ngs; exothe r mic and e n d o ther m ic reactions (Sec tio n ~) . l ) . Study Question(s) assignable in OWL: 61 , 92.
d.
Use specific heat ca paci ty in ca lcu latio ns of e n e rgy tran sfer as h eat an d of tem p erature c han ges (S e c tio n :> .2 ) . Study Ouestion(s) assignabte in OWL: 8, 10, 12, 13, 16,18,83 ,
e. f.
Un dersta n d the sig n co nve n tio ns in th ermodynami cs. Use e ntha lpy (h eat ) o f fu sion a n d enthalpy (beat) o f vap o rizatio n to find th e quan tity of e nergy tr an sfe rred as h eat th at is invo lved in chan ges of state (Sectio n fi.3 ). Study Ouestion(s) assignable in OWL: 20, 22, 23, 24, 28, 68, 70, 71, 88, 93, 97.
'6 Access How Do I Solve It? t utorials on how to approach problem solving using concepts in this chapter.
Understand and apply t he first law of t hermodynamics a. Un dersta n d the basis o f th e first law of th ermodynamics (Sec tio n 5.4) . b. Recognize ho w e nergy tr a n sferred as h eat a n d work don e o n or by a syste m co n trib ut e to changes in th e in te rnal e ne rgy of a syste m (Sectio n 5.4 ). Define and understa nd state fu nctions (enthalpy, internal energy) a. Recognizc sta te fu n cti ons whose valu es are d etermin ed only by th e state o f th e syste m a n d no t by th e path way b y whic h th at sta te was achi eved (Sec tio n 5.4) . Learn how energy cha nges are measured a. Recogn ize th at wh en a process is ca rried out und er consta n t p ressu re co n di tion s, the energy tr an sferred as h eat is th e e n thalpy c ha nge , aH (Secti o n 5.5 ) . Study Ouestion(s) assignable in OWL: 28, 29, 30, 52, 54. b. Describ e h ow to measure th e qu antity of e n e rgy tran sfe rred as heat in a reaction by ca lorime try (Sec tio n 5.6) . Study Ouestion(s) assignable in OWL: 32, 33, 34, 36, 38 ,40,42.
Calculate the energy evolved or required for physical changes and chemical reactions using tables of thermodynamic dat a. a. Apply H ess's law to find th e e n th alpy change for a re action (Sectio n 5.7) . Study Ouestion(s) assignable in OWl: 44, 73, 74, 79; Go Chemistry Module 10. b. KIH)\\r h ow to draw and in terpret e n e rgy le vel di agr ams (Sectio n 5.7) . c. Usc sta nd ard molar e n th alp ies offormation, ai r , to calcu late the enthalpy cha nge to r a re ac tio n aJr (Sectio n f>.7 ) . Study Ouestion(s) assignable in OWl: 49,53 ,58.
KEY EQUATIONS Equation 5.1 (page 215) The e n e rgy tra n sferred as h eat whe n the te m pera ture o f a substance changes. Ca lcu lated from th e sp ecific heat ca pacity (C) , mass (rn) , an d chan ge in temperature (6. "1' ).
q(J)
= [(JIg
' K) X m(g) X M (K) Key Equati ons
241
Equation 5.2 (page 215) Temperatu re changes are always ca lcula te d as fin al ternpcrature mi n us ini tial tcmpe ra tu re . ~T
= Tf ina l
-
Tiniti al
Equation 5.3 (page 217) If no c ncrgy is tra nsferred between a systc m and its surro u ndi ngs and if e ne rgy is transfe rred withi n th e syste m on ly as hea t, the su m of the therm al energy chang es with in the syste m eq ua ls zero .
q, + q, + q, + . . .
~
0
Equation 5.4 (page 223) The first law of thermodyn amics: The ch an ge in in ter nal c ne rbT)' (tiU) in a syste m is th e su m o f the e ncrgy t ransfe rred as heat (1) and the e ne rgy trans ferred as wo rk ( w) .
IW =q +w Equation 5.5 (page 224) Wo rk (w) at co nsta n t pressllr e is the p roduct o f p ressllre ( P) and cha nge in volu me ( ~ V)
w~
- P( ~ V)
Equation 5.6 (page 237) T his eq ua tio n is used to ca lcula te the sta ndard en thalpy ch ange of a reac tio n (ti rE-f") whe n the cn thalpi cs o f fo rmation (ti/ --f') of all of th e reac ta n ts and pro d ucts are known .
STUDY QUESTIONS On line homewo rk for this chapter may be assigned in OWL. .&. denotes challenging questions.
•
denotes questions assignable in OWL.
2. A solar panel is p ictu red in th e pho to . Whe n ligh t sh ines 011 th e pan el, it generates an el ec tri c curren t th a t is used Ill' a sm a ll el ectri c motor to propel th e car. Wh at types of ('>Tll'rgy an' invol ved in th is setu p?
Blue-numbered questions have answers in Appendix 0 and fully-worked solutions in t he Student Solutions Manual.
Practidng Skills Energy (,\'rr Section. 5. J and Chnni.\,tJ}Now Screen. 5.2.) l. Th e llash lig h t in the ph ot o do cs no t u sc ha ncrics.
Instea d , yo u ITI m 'e a lever, wh ich turn s a geared m ec hanis m an d re sul t." fin all y in lig ht fro m th e bu lb . Wh a t type of en e q.'.", y is use d to m ove the lever? Wh at t)'pe o r typ e s of en n gy are prod uced ?
A solar panel operates a t oy car.
Energy Units (See Exercise 5. J and
Chemi.~t'}'Now
Scnen 5.5.)
3 . You are o n a die t t h at ca lls for eating n o more th a n 1200 Cal i da )'. Wh at is th is e nefj..,'"y in joules ?
4. A 2-in . piece of chocola te cake with fr osting p rovid es 1670 ~ I of e nergy. What is th is in d iet ary Calories (Ca l)?
A hand- operat ed flashlight. 242
I
5. • On e fo od product has a n e ne rgy content of 170 kcal pe r serving. and an ot her h as 280 ~I pe r se rvin g. Wh ich food provides the grea ter e n ergy pe r serving?
6. Whi ch provid e s th e grea ter e neq.,,:' pe r se rvin g. a raw apple or a raw aprico t? Co to th e USDA Nutri ent Da tabase on th e Wo rl d Wide Web fo r th e info rm ation : h u p :I I \I,'\I/w,ars.usda ,govI m a ill/ sitt'_m a in , ht m ?m ode cod e = 12354500 . Repo rt th e ene rgy co n te n t of th e fr ui t in kca l a nd k].
Specific Heat Capacity (See ExmnlJ!fJ 5. 1 and 5.2 an d ClmnisftyNow Screens 5.6-5. 10 .) 7. T he mo lar heat capacity of me rcury is 28. 1 J / m ol . K. Wha t is the specific heat cap acit y of th is m eta l in J i g · K? 8• • T he specific hea t cap ac ity of benzene (Cl,l ll;) is 1.74J / g · K. \Vhat is its mo lar heat ca pac ity (in J l m o l . K) ? 9. T he sp eci fic h e at ca pa city of ('o p pn is 0 . 3 ~ 5 Ji g · K. Ho w muc h energy is requi red to hea t 168 g of coppe r from - 12.2 °C to + 25 .6 OCt 10. • How m u ch ene rgy is requi red to ra ise th e tem pera tu re of 50.0 0 m L of wa ter fro m 25.52 °C to 28. 75 °C? (T h e de nsity of water a t th is tem peratu re is D.!!!'7 g/ m L.) 11. T he initi al te mp e ratu re of a 344-g sa m ple of iro n is 18.2 °C. If th e sam ple abso rbs 2.2:'"> kJ of en n gy as he at . what is its final tem pera ture ? 12. • After abso rbing 1.850 ~ J of e ne rgy as he at , th e te mperatu re of a 0.500 -kg block o f co ppe r is 37 °C. Wha t was its initi a l te mpe ra ture ? 13. • A 45.5-g sam ple of co p per a t 99.8 °C is dro pped int o a lx-ake r co n taini ng 152 g of water a t 1 ~ .5 °C. W hat is the fin al tem pera tu re wh en th e rm al equil ibrium is reach ed? 14. A 182-g sam ple of go ld a t so me te mpe rat ure is a dded to 22.1 g of wate r. T he ini tial wa ter te mperatu re is 25. 0 °C. and the final te mperat ure is 27.5 -c , If th e spe cific hea t ca paci ty of go ld is O.128 J / g · K, what was th e in itial te mperatu re of th e go ld? 15. O ne be a ker conta ins 156 g of wa ter at 22 °C, a n d a seco u d beake r co n ta in s 85.2 g of wa te r a t 95 °c' T he wate r in the two be akers is m ixed . Wha t is the fin al wa ter u-rn pe rat u re ? 16. • Whe n 108 g of wa te r at a tem pe rature 0 1" 22.5 0(: is m ixe d with 65.1 g of wa te r a t an un kn own te mperatu re , the final tempe ratu re o f th e re sult ing mix ture is 47.9 "C. What was th e initial te mpe rat u re of the se con d sa m ple of water? 17. A 13.H·g p ie ce of zinc was heat e d to 98.8 °C in ho i1ing water a nd th en droppe d in to a be ake r co n ta ining 45. 0 g of wa te r a t 25 .0 0(:. Wh en th e wa u-r a nd m et al co me to th e rm al eq uilib riu m , th e tempe ra ture is '1.7.1 "C. wh a t is the speci fic he a t ca pac ity of zin c?
18. • A '1.37-g pie ce of m o lybde num , in itially a t 100.0 DC, is d ro pp ed into 244 g of wa te r at 10.0 0(:. Wh e n th e system co mes to th er ma l eq uilibriu m . th e tem p era tu re is 15.3 °C , Wha t is th e speci fic he at capaci ty of molybde nu m ?
Changes of State (See E.\·(WIIJ!fJ 5.3 and 5.4 and (;hnni.\II)'Now Scr een 5.8.) 19. I lo w m uc h e nerh')' is evo lved wh e n 1.0 I. of wa te r a t o °C solid ifi es to ice? (T he he at of fu sio n of wa te r is 333 j!g.) 20• • T he e nc q,'J' req uired 10 m elt 1.00 g of ice a t 0 °C is 333 J. If on e icc cube has a mass of 62.0 g a nd a tray con tains 16 ice cubes. wha t q ua nti ty of e m'll D" is requ ire d to m e lt a tray of ice cu be s to fo rm liq uid wa ter a t 0 DC? 2 1. • How much energy is req uired to vaporize 125 g of be nze ne , C(,H l ; , at it." boi ling po int , 80. 1 °C? (T he heat o f vapo riza tio n of beu zeru- is 30.H ~J / m ol. ) 22• • Ch lo ro me th an e , C H 3C I. arises fro m mi crobial fe rme n tati o n an d is fou nd t h ro ug ho u t t he e nviro nment. It is also p ro d uced indust riall y an d is use d in th e m an u factu re of vario us che m ical s a nd h as be e n u sed as a to p ical anest het ic. Ho w m uch en ergy is re q u ired to co nvert 92 .5 g of liquid to a va por at its bo iling poin t, - 24.09 OC t (T he heat o f va po riza tio n of C H jC I is 2 1,40 kl/ mol. ) 23 . T he fre e zing po int of mercury is - :l8.8 -c . Wh at q u antity of e llergy, in j oules. is rel ease d to th e su rroun d ings if 1.00 m l. of m ercury is cook-d fro m 23. 0 °C to - 31"L8 °C and th en fro zen to a so lid? (T h e d en sity of liqu id m e rcu ry is 13.6 g/c m 3 . Its sp eci fic h e at ca pa city is 0.140 J i g ' K an d its hea t of fusio n is 11.4 .JI g.) 24. • Wha t q uan tity of en e rgy. in j o u les , is re q ui re d to ra ise the te m pe ratu re of 454 g of tin fro m roo m te mperatu re , 25 .0 °C, to its melting po in t, 23 1.9 °C, a nd th en m elt the tin a t that tem pe rature? (T h e speci fic h eat ca pac ity of tin is 0.227 J i g ' K, a nd th e h eat of fusio n of th is metal is 59 .2 .JI g.) 25 . Ethan ol, C;?lI :;O Il . boils a t 78.29 °C. 1I0w m uch en ergy, in j o ule s, is re q uire d to ra ise th e te mp eratu re of 1.00 kg of et hano l fro m 20.0 °C to th e boi ling poi n t a nd th en to cha nge th e liqu id to vapor at th at te m pe rat u re ? (T he sp ecific heat ca paci ty of liqu id e tha no l is '1. ,44J /g · K, an d its en tha lpy of va po riza tio n is 855 JIg .) 26. • A 2Pd J-m L sam ple ofhenzen c a t 19.9 °C was cooled to its me lting point , 5.5 °C, and th e n fro ze n . How m uc h energy as hea t was give n off in th is p ro cess? (T h e de n sity o f be n zene is O.HO g / m L; its spec ific h eat ca pac ity is 1.74 Ji g ' K, a nd its he at of fu sion is 1'1.7J / g.)
I
243
Enthalpy Changes (S('(' f :xm nj Jlp 5.5 find
Ch ('misl1 }Nm.l! Screens
5. /2
fwd
5./ J .)
27. Nitrogen mo noxide , a gas re cen tly fou nd to he involve d in a wide range o f b io logical p ro cesses, reac ts with ox yge n 10 give bro wn NO ';! gas.
2 NO (g ) + O , (g ) -> 2 NO ,(g)
s.u-»
- 114.1
~ J /mo l-rx n
Is this reactio n e ndoth erm ic o r exot hermic? What is th e enth alpy c h ange if 1.25 g o f NO is co nve r ted cornplct cl v to NO ';!? 28 . • Calc iu m carbide , Cae:!> is m anufa ctured by th e reaction of CaO with ca rbon a t a h igh temperature . (Cal ciu m ca rbi d e is th en u sed to m ak e acc tvlcne. ) Ca O (s )
+
3 C (s ) -> CaC, (s)
+
CO (g)
I1,H o = + 464.H ~ J /mo l-rx n Is th is reactio n e ndoth ermi c o r ex o thermic? What is th e e n thal py change if 10.0 g of CaD is a llowe d 10 rcact with an excess of ca rbo n ? 29 . • lsooctan c (2,2,4-trim e thylp c n tan e ), on e of th e man y h yd ro ca rbo n s th at make up gaso line. b u rn s ill air to give wa ter an d ca rb on d io xid e. 2 C, H 18 (f )
+
+
25 O ,(g ) -> 16 CO ,(g )
s.u-» -
18 H,O (f )
10,922
~J / mol-rx n
Wh at is th e e n tha lpy change if yo u h u m l. OO L o f iso o ct an e (d e ns ity = 0.69 g/ m L)?
30 . • Ace tic aci d , C1I3C 0 2 H , is ma de indu stri all y hy th e reaction o f m e th an ol and ca rbo n mon oxide. C H , O H( f )
32 . • ' hu m ix 125 mL of 0.250 M CsO IJ with 50. 0 mL of 0.625 ;\,1 IIF in a co ffee-cu p calorime te r, an d the rem p e ra ture o f h o th so lu tions rises from 2 1.50 °C before mixi ng to 24.40 °C after the rea ction . CsO H (a q)
+
H F(aq ) -> C,F (aq )
+
H , O (C)
Wh at is t he e n tha lpy o f reac tio n p e r mo ll' o f CsO H? Assu me the densi ties o f tl n- solutions ar e a ll 1.00 g / m I. an d th e sp ecifi c heats o r th e so lu tio n s arc 4.2.J /g . K. 3 3. • A pie ce o f titanium metal with a m ass o f 20 .8 g is h e al ed in h oil in g wa ter to 99.5 °C a n d th en d ropped into a coffee-cu p ca lo rime ter con ta ining 75.0 g o f wa ter a t 2 1.7 "C. When th ermal eq uilibri um is rea ched, th e fin al tempe ratu re is 24 .3 "C. Calc u la te th e sp ecific h eat ca pacity of titan ium .
34. • A p iec e of chro m iu m metal with a mass o f 24.26 g is h e at ed in hoiling wa ter to 9H.3 °C a nd th en d ropped int o a coffee-c u p ca lo rime ter con ta in ing 823 g o f wa ter a t 23 .3 °C . Whe n th e r mal equili briu m is re a ch ed, th e fin al tempe rat ure is 25.6 "C. Calcu la te th e sp ecific h e at capaci ty o f chro mium . 35 . Ad ding 5.44 g of i\' H 1i\'O :\(s) to 150 .0 g of water in a co ffee-c u p cal o rimete r (with stirring to d isso lve th e salt) re sult ed in a decrease ill lempe rature from 1H.6 °C 10 16.2 " C. Calcu la te the e n t ha lpy ch ange for d isso lving N lI -tN0 3(s ) ill wa ter. ill k.l/m ol. Assume that th e so lu tio n (wh os e m ass is 155.4 g ) h as a spec ific h eat capa city of 4.2.J/ g . K. (C old p acks take advan ta ge o f th e fa ct th at d isso lvin g a m mo n iu m n it rate ill wa te r is a ll cn dothermi c p ro cess .)
+ C O(g) -> C H"CO , H (C)
s.u-»
- 355 .9
~J /mo l-rx n
If yo u prod u ce 1.00 I. of ac etic acid (d = 1.044 g / m L ) by th is reaction , h ow much energy as h e at is evol ve d ?
CaLorimetry (SPf' f ;xm lljJ!I'J
5.6
fwd
5. 7 fwd
Chem i,\'f')No1J) ,"":l1'(,IIJ
5.8, 5.9,
ruul 5. 14.)
31. Assume yo u m ix 100.0 ml , o f 0 .200.\1 CsO H wit h 50.0 mL o f 0.400 M H CI in a co ffe e-c u p calorimete r. T he follo win g reac tio n o ccu rs:
C, O H (a q)
+
H C1(aq ) -> C,Cl(aq )
+
H,O (C)
Th e tempe ratu re of hoth so lu tion s before m ixing was 22 .50 °C , a nd it ri ses to 24 .28 °C a fter th e acid-base rcaction . Wh at is th e en tha lpy change for th e reac tion pe r mole o f C."Oll? Assu me th e densities of th e solutio n s a rc all 1.00 g / m L a nd th e spe c ific h eat ca paci t ies o f th e so lution s a re 4.2 Ji g ' K.
244
I
Acold pack uses the endothermic enthalpy of solution of ammonium nitrate.
36. • ' h u sh o u ld USI,' care whe n dissolving I1 2S0 1 in wa ter be cau se th e p ro ce ss is h igh ly exot h erm ic. To m easure th e ent hal py ch an g e, 5.2 g I12SO -t ((') v..-as ad d ed (with stirring ) to 1:t~ g of water in a co ffee-c u p calorimeter. Th is resulted in an incre ase in te mperatu re fr om 20.2 °C to 28.8 "C. Calcu la te th e en tha lpy chan g e fo r the p ro cess 112S0 4 ( (') -7 112So.l(aq ), in kJ / mol.
... more chaLLenging
•
in OWL
BLue-numbered questions answered in Appendi x 0
37 . Su lfur (2 .56 g) is burned in a co ns tan t vo lu m e ca lo rimeter with ex cess 0 2(g ). T h e te mpe rature increases from 2 1.25 DC to 26 .72 DC. T he hom b h as a h ea t ca pacity of 923 .I/ K, an d tile ca lori me te r con tains H15 g of water. Calcu la te d U per mole o f S02 for med , for th e rea cti on
S, (s) + 8 O ,(g ) -7 8 SO ,(g)
40. • A 0.692-g sa m p le of g luc ose , C tj[J 120 6, is bu rn e d in a consta nt volum e calorime te r. T h e tem pe rature rises from 21. 70 °C to 25.22 DC. Th e ca lorim eter co ntai ns 575 g of wa ter, a nd the ho m h has a he at ca paci ty of 650 .l /K. \Vh at is ~ U per mole of g luc ose? 41. An "ice calori me te r " can be use d to de term ine the specific he at ca pa city of a meta l. A piece o f h o t m e tal is d ro pped onto a weig hed q u a ntity of ice . T h e ene rgy tran sfe rred from th e m e tal to th e icc call he de te rm in ed fro m th e am o u nt of ice mel ted . Suppose yOIl h ea t a 50. 0-g piece of silver to 99.8 "C a nd th en drop it onto ice . \ Vhe n th e m e tal' s tempe ra ture has dro p ped to 0.0 °C, it is fo u nd th a t 3.54 g of icc h as m e lted. What is th e speci fic he at ca paci ty of silver?
42. • A 9.36-g piece of plati n um is heat ed to 98 .6 °C in a bo iling wate r bath an d th en d ro p ped o n to ice . (Se e St udy Question 4 1,) Wh en th e m etal's te mperatu re has d rop ped to 0 .0 DC, it is foun d th at 0.37 g of ice h as m el ted . Wh a t is the spe cific h ea t ca paci ty o f pl at in um ?
Hess's law Sulfur burns i n oxygen wit h a bright blue flame to give S02(g).
38• • Suppose yo u burn O. :~OO g of C (g ra phi te ) in a n e xcess of 0 2(g ) in a consta n t vo lume ca lo rime ter to give
CO , (g) . C (g rap h ite )
+ O , (g ) -7 CO , (g)
T he te mperatu re of th e ca lorim e te r, whi ch co ntains 775 g of wa te r, incr eases from 25.00 °C to 2738 "C. T he h ea t ca pacity of t he bomb is 8 9:~ .11 K. Ca lcu late d Upe r mol e of ca rbo n. 39 .
SUPPO Sl~ yo u bu rn 1,500 g o f be n zoi c aci d, C ji H 5C 0 2 H , in a co nsta n t vo lume cal orime te r a nd fi nd th at the n-mpcra tun- increases fro m 225 0 °C to :~ 1.69 °C. T ile ca lorim e te r co n tai ns 775 g of ,...' ate r, a nd the bo m b h as a h ea t ca pac ity o f 893 .1 1 K. Calc ulate d U pe r mol e of be nzoi c ac id.
(See Exa m/)lf 5.8 and CllPlll iJl1y,vow Screen 5.15.)
43 . T he ent hal py ch a nges for th e fo llowin g reactio ns can be me asured : CH,(g ) + 2 O Ag ) -7 CO ,(g) + 2 H, O (g) ~ rH o = - 802.4 kJ /m ol-rxn CII 3 0 Il(g) + 0/, O ,(g ) -7 CO , (g ) + 2 11,0 (g) ~ rH o = - 676 ~J/mo l- rxn (a) Use these values an d He ss's law to det e rmin e the e n th al py change fo r th e reaction
CH ,(g) +
Y, O, (g ) -7 CH , O Il(g)
(h) Dra w an e ne rgy-leve l d iag ram that shows t he re latio n ship bet ween th e ene rgy q uantitie s involved in th is problem . 44 . • T he e n thalpy cha nge s of th e lullowi ng rea cti on s can be m easured:
C, II,(g) + 3 O , (g) -7 2 CO , (g) + 2 11, O(t) ~ ,H D = - 14 11.1 kJ / mol-rxn C, H , O H(t)
+ 3 O , (g ) -7 2 CO , (g) + 3 H, O (O dJr = - 136 7.5 kJ /mol-rx n
(a ) • Use th e se values and Hess's law en tha lpy ch an ge fo r the react io n
C, H ,(g) Benzoic add. C6H sC02H , occurs naturally in many berrie s. Its heat of combustion is well known. so it is used as a standard to calibrate calorimet ers.
... morechallenging • in OWL
Blue-numbered questionsanswered in Appendix0
10
determine the
+ 11, 0 (0 -7 C, II , O Il (t)
(h) Draw an e n e rgy-level di ag ra m th at shows th e rc latio n ship be twee n th e e ne rgy quantitie s invo lved in th is prohk-m .
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45 . En th a lpy cha n ges for th e foll owing re act ions call be d ete rrni ned ex pe ri men tally: N,(g)
+ 'I H , (g ) -> 2 NH , (g ) d ,lJO == - 9 1.8
~J / m ol-r xn
4 N Il,(g) + 5 O , (g ) -> 4 NO (g ) + (; 1l , 0 (g ) d ,H o == - 906.2 ~J /m nl -r x n H, (g )
+ Y, O , (g ) -> H , O (g ) !1J r
- 24 1.8
==
~J /m ()l ·r xl1
US(' these values to det erm ine the e n th alpy chang e fo r th e fo rm a tio n o f K O( g) fr om the e le m e n ts (a n e nth alpy cha ng e th at ca n n o t he measured direc tly heca u se th e reaction is re act ant-favo red ) .
Y, N A g ) + Y, O , (g ) -> NO(g) 46. \'Im \.. . ish
10 kn ow the e u th al pv change fo r the fo rm alio n of liq u id PCh from th e e le m ent s.
P, (s) + 6 CI, (g) -> 4 PCI, (£) T h e enthalpy c ha nge fo r th e form a tio n o f PCI:; from th e ele m en ts ca n he d vn-rmit u-d e xpe rim e ntally, as can th e e ntha lp y c hang e for the re action of PCh (f) with more ch lo rine to give PCl::,(s) :
P, (s)
+ 10 CI,(g)
-> 4 l'Cl-t s) !J. ,H"
~
- 1774.0 kJ/m ol-r xll
PC1, (£) + Cl,(g) -> PC1, (s)
6 ,1JO == - 123.8
~J /m o l·rx l1
Usc th e se da ta to calculate th e e ntha lp y c h a ng e for th e fo r ma tion of 1.00 m ol o f PCh (O from p hosp horus a nd ch lo rin e .
Standard Enthalpies of Formation (S('(' f :xmnj Jlp 5. 9 find Ch('miJl1}Nm.l! Screen 5./6.) 47 . Write a bal a n ced chem ica l equation for th e form atio n o f C H 30 H (f) fro m th e e le me n ts in th eir sta n dard sta tes. Find the value for ~IH o fo r C H 3 0 H (f) in Appendi x I.. 48. Write a b ala nced chem ica l eq uat io n fo r th e form ation o f CaC 0 3(s) from th e cleme nts ill their sta n d ard sta tes. Find th e va lu e fo r 1:1; 1/ 0 fo r CaCOj (s) in Ap p e n d ix L. 49 . (a) Writ e a bala n ced ch e m ica l eq u at io n for th e for m atio n o f I m ol o f Cr:t03(S) from Cr an d O a in th e ir stan dard sta tes. Find th e va lue fo r I:1JH o for Cr:tOJ(s) ill Append ix L. (h) . W h a t is the sta ndard e n th alpy ch an ge if 2.4 g of ch ro m iu m is ox id ize d to Cr20 3(S)?
50. (a ) Wr-ite a bala nced che m ica l equation for th e for matio n of 1 m ol of l\.'fgO (s) from th e clem en ts in th e ir stan da rd sta tes. Fin d th e va lu e fo r l:1/ r for MgO (s ) in Ap p en di x I.. (h) Wh a t is th e sta n dard e n t h alpy chang-e for th e reaction o f 2.5 mol o f ~r g wit h o xygen ? 5 1. Use sta n d ard c n th alp ies of form atio n in Ap pe ndix L to calc u late enth alpy ch an ge s fo r the following : (a) 1.0 g of whi le p h o sp horus hum s. fo rm in g P'IO W(S) (h) 0.20 mo l of NO(g) de com po se s to N 2(g ) ami 0 2(g ) (c) 2.40 g of Na CI(s) is fo rm e-d from Na (s) an d ex cess
CI,(g) (d) 250 g o f iron is o x id ize d with o xygen
10 Fe :t 0 3 (.~ )
52. • Us e sta n d ar d cr ultal p ics o f fo r m ation in Ap p endix L to calcu la te e n tha lpy ch ang-e s fo r th e following : (a) 0.054 g of sulfur burn s, formin g SO :t(g) (h) 0,20 mol of H g O (s) d eco mposes to Hg (f) and O , (g ) (c ) 2.40 g of N I I3(g ) is form e d from N 2(g ) an d exce ss H :t(g ) (d) 1.05 X 10 - 2 mol of carb o n is o xidized to CO 2(g) 53. T he fir st ste p in the production of n it ric acid fro m a mm o n ia in vo lves th e o xid a tio n of N H 3. 4 N JI ,(g)
+ 5 O , (g ) -> 4 NO( g) + 6 JI,O (g )
(a) Use stan dard cn tha lpies of fo rmat io n to ca lc u la te th e sta n d ard e n th alpy ch an g e fo r thi s re ac tio n . (h) . H o w mu ch en e rgy as h ea t is evo lved or abso rbed in th e o x ida tion o f 10.0 g o f N H ) 54. • Th e Rom a n s u sed calciu m o x ide , CaO , to p ro d u ce a stro ng mo rtar to build sto n e str uc tu res . Th e Ca O was mi xed with water to give Ca( O H )z, whic h re acte d slo wly with CO 2 in th e air to g ive CaC0 3. Ca (O Jl),(s)
+ CO, (g)
-> CaC0
3(s)
+ 1I,o (g )
(a) Calc u la te th e sta n dard en thalpy cha nge fo r thi s IT · act io n . (h) I low much en ergy as heal is evo lved o r abs o rbed if 1.00 kg o f Ca( O H ) 2 reac ts with a sto ich io m e tr ic amo u n t o f C0:t? 55 . T he stan d ard e n thalpy of fo rmation o f so lid bariu m ox ide , BaO , is - 553.5 ~ I / m() l , and th e sta n d ard en th alpy of form ation of b arium p ero xid e . Ra02' is - 634.3 ~J/1ll ()1. (a) Calcula te th e stan dard e n t h alpy ch a nge fo r th e lo llowin g reactio n . Is th e re ac tio n ex o th erm ic or end o th e rm ic? 2 RaO , (s) -> 2 RaO (s)
+
O , (g )
(h) Draw a n en erg-y-leve l d iag ra m th at sh ows th e re latio n sh ip be tween th e cu tha lpv cha ng e o f th e deco mpo sitio n of Ba():t to BaO a nd O 2 a n d t he ('11thaipi e s o f form ation of RaO (s) a n d BaOA s) .
246
I
... more chaLLenging •
in OWL
Blue-numbered questions answered in Appendix 0
56. An im po r tant ste p in th e p ro d uction o f sulfuric aci d is til l' o xida tio n o f SO ~ to S0 3' SO ,(g )
+ Y,
O , (g ) -> S0 3(g )
Form a tio n o f 50 3 fro m th e a ir po llutan t SO ~ is a lso a key ste p in th e fo rm at ion o f a cid rain . (a) Use sta nd a rd en thalpics of form a tio n to calculate the en th alpy chan ge fo r th e reac tio n. Is the reactio n exoth erm ic or en do th e rm ic? (b) Dr aw an e n ergy-level d iag ra m tha t shows the re latio ns h ip be twee n the ent hal py change fo r th e oxida tio n o f 5 0 ~ to 50 3 and th e cn thalpi es of fo rmation of SO ~ ( g ) an d S0 3(g) . 57 . T he e n thalpy cha nge for th e ox id a tio n o f na p h thale ne , C1oH fj, is measu re d by ca lo rime try. C ",II , (s)
+
+
12 O , (g ) ---> 10 CO,(g )
4 11,o (( )
D..,H o = - 5 156.1 kJ / m o l-rx ll Usc th is value , a long with the stan dard en th ai pie s o f fo rm a tion o f C O~(g) a n d lI ';.! O ( f) , to calcu late th e enth alpy o f for ma tio n o f n a p h thalen e , in ~J / m o l.
58. • T he e n thal py cha nge for th e o xida tio n of styre ne, CgIl g , is m e asu re d by ca lo rime try. CsH, (()
+ 10 O , (g ) --->
H
co, (g) + Ix.H"
=
4 H,o (( ) - 4395.0 kJ / m o l-rx n
Use thi s value , a lo ng wit h the sta nda rd e n tha ipie s o f form a tion of C O ~(g) an d 11 20 ( f), to ca lcu late til e cuth alpy o f fo rma tio n of styren e , in kJ / m o l.
General Questions on Thermochemistry These questions are nol designated as 10 1)'1)(> or location in the rlla/del: Tlu')' mil)' combine seteml concepts.
59. Th e foll owing te rm s are used ex te nsively in th erm odvna mi cs. Define eac h an d give an exam ple. (a) ex o th erm ic and eu do tlu-rmi c (b) syste m and su rrou nd ings (c) spec ific h eat ca pac ity (d ) sta te fu nc tio n (e) sta nd a rd sta te (I) e n thalpy ch ange , !1H (g) sta n d a rd e n tha lpy of fo rm ati on 60. Fo r ea ch of the fo llo wing , tell whet he r th e pro ce ss is exothe rmic o r endoth erm ic. ( No calcu latio ns arc rc q uirc d .) (a) 1I,o ( C) ---> 1I, 0 (s) (b) 2 H, (g ) + O , (g ) ---> 2 H, O (g ) (c ) II , O ( C, 25 "C ) ---> 11, 0 (( , 15 "C) (d ) H,o (C) ---> H, O (g )
... morechallenging • in OWL
Blue-numbered questionsanswered in Appendix 0
61. • Fo r eac h of the fo llow ing , defin e a syste m a nd its surro u n d ings, an d give the di rec tio n o f e n ergy tra n sfe r b et we e n syste m a nd su r ro u n dings. (a) Me th an e is bu rn in g in a gas furn ace in yo u r h o m e . (b) Wat e r drops, sitt ing on yo u r skin afte r a d ip in a swim m ing po o l, eva po rat e . (c) Wat er, a t 25 °C , is placed in th e fr eezi n g co m pa rttru-u t o f a refrigera tor, \.. . here it co ols a nd even tua lly so lid ifies . (d ) Alum inu m an d Fe:!O J (s) a rc m ixed in a flask sitting o n a la bora to r y be nch . A reac tion O CCllI'S, an d a large q ua ntity o f e n e rgy is evolved as h ea t.
62. Wh a t does the term "sta nd ard sta te" mea n? Wh at arc the sta n dard sta tes of the foll owing su bs tances a t 298 K: H';.! O , NaC I, Hg , CH.t ?
63. Use Ap pen d ix L ro find th e stan d ard c n thal pics o f fo rmatio n o f o xyg ell ato ms , oxygen m o le cu les ( O~), a nd oz one (O J) , Wh a t is th e stan d ard sta te o f o xyge n? Is th e fo rm a tio n o f o xygt.'n a to ms fro m O 2 exo therm ic? Wh at is th e e n thalp y cha nge for th e for m a tion o f 1 mol of 0 3 fr o m O ~ ? 64 . See the Chc mi strylvow website , Screen 5 .9, He a t Transfer Be twee n Su bs ta nc es . Use th e Simul a tio n se ctio u of th is scr een to do the fo llowing ex peri men t: Ad d 10.0 g of AI a t 80 °C to 10.0 g of wa te r a t 20 °C . Wh at is the fi nal te m pe ra tu re when equili bri u m is ach ieve d? Usc th is valu e to es timate th e spe cific heat ca pacity o f alu m in u m . 65 . See th e CIll' mistr yN ow websi te , Screen 5. 15, H ess's Law. Use the Simu la tion se ctio n of thi s scr een to li nd th e value o f !1,1 r fo r Snl\r, (s)
+ TiCl, (( ) -> Sn CI, «( ) + TiI\r, (s)
66. ' Vhich g ives li p m ore e ne rgy on co ol ing fro m 50 °C to 10 °C , 50.0 g o f water o r 100. g of et h an ol (spe cific h ea t ca paci ty of ethan o l = 2.46 Ji g . K)?
67 , You d e termine th at 187.J of hea t is req u ired to raise th e te m pe rature o f 93.45 g o f silver from 18.5 °C to 27.0 °C . What is th e specifi c he at ca pac ity of silver? 68 . • Ca lcul a te th e q u an tity of e l1eq.,~' req u ired to co nve r t 60.1 g of 1I,o (s) a t O.O "C to Il , O (g ) a t 100.0 "C. T he heat of fu sio n o f icc a t 0 0(: is :-t 13 Ji g; t he he at o f vapo riza tio n of liqu id water a t 100 DC is 2260 J i g.
69 . • 'rou ad d 100.0 g o f wate r a t (iO.O °C to 100.0 g of ice a t 0.00 "c. So me of the ice mel ts and cools the wa te r to 0.00 °C . Wl n-n th e ice an d water m ixtu re h as co m e to a u n ifo rm te m pe rat u re of () °C , how m uc h ice has m e lted ?
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247
70 . ... . T h re e 45-g- icc cubes a t 0 °C arc d rop ped into S.OO X lOt m l , of te a to m ake iced te a. T h e te a was in itially at 20.0 DC ; wh e n th erm al equi lib riu m was read ied , th e fi n al temperature was 0 DC. H o w muc h of th e icc melted , a n d h ow muc h re m ai n e d flo a tin g in the beverag e? Assu m e th e spec ific h ea t cap acity o f te a is t he sam e as t hat of pu re '»atc r, 71. ... . Su p p ose th at o n ly two 45-g- ice cubes h a d be en ad d ed to yo u r glass co n taini ng 5. 00 X lOt m l , o f lea (se e St udy Q u es tio n 70 ) . Wh en th erm al eq ui lib ri u m is reac hed, all o f th e icc will h aw mel ted, an d th e remp e ra tu re o f th e m ix tu re ,»ill b e so m ew he re be tween 20. 0 °C a n d 0 DC. Ca lc u la te th e fili a l tempe ratu re o f th e be ve rage . (No te: T h e 90 g of wa te r form ed when th e icc melts mu st be ' »a r me d fro m 0 °C to th e fin al tc mpc rature .) 72 . Yon tak e a d ie t co la fro m th e refrig erato r an d pour 24 0 mL o f it in to a g lass . Th e tem p e ra ture o f the beverage is 10.5 DC, Vi.m th e n ad d on e icc cube (45 g) . Wh ich o f th e foll owin g de scribe s th e syste m wh en th erm al c quililu-iu m is reac hed? (a ) T h e te mpe ra tu re is 0 °C, an d so me ice re m ain s. (b ) T he te mpe ra tu re is 0 DC , a n d no icc re mai ns . (c) T h e te mperatu re is h igh er th an 0 DC , a n d no ice re mains . Dete r m in e tl u - final temperat u re an d th e amount of ice re mai n ing, if an y.
(a) • Ca lc ula te th e en th alpy change for th e reaction o f C I '-, (g ) an d CI a to ms to g ive C l l j C1(g ) an d II Cl (g ) . Is th e reacti on p ro duct -favo red or re ac ta ntfavored? (h ) Draw a n energy-leve l d iag ra m th at sh ows h ow th e vari o us cnt hal pi es in thi s p ro b lem are re-lat ed . 75. Wh e n h eat e d to a h igh tempe ratu re , co ke (m ain ly car... b o n , obtaine d by h eati n g coal in th e abs en ce of air) a nd stea m p ro d u ce a mi x ture called wa ter gas, wh ich can he used as a fu el o r as a ch emi ca l fe edstock fo r ot her reactio n s. T h e eq ua tion fo r th e p rod uctio n of wau-r gas is C (s )
+
Il , O (g ) ---> CO(g)
+
Il , (g )
(a ) Usc stan dard en thalp ies of form at ion to d et erm in e th e e n th alpy ch ange fo r th is react io n . (b ) Is the reactio n product -favo red o r re ac ta n t...favored ? (c) W h at is th e e n thal py ch a n ge if 1.0 m etric to n ( 1000.0 kg ) o f carb on is co nverte d to water gas?
76. Cam p ing stov es arc fu ele d by p ro p an e (C:1IIl'! ) . b u ta n e [G1IIw(g), d/f' = - 127. 1 ~J/m o l ], g aso lin e , or e thanol (C t 11,,01 I ) . Ca lcula te the e n thalp y of com b us tio n p e r g ra m o f eac h of th e se fuels. [Assum e th at gaso line is rep re sented hy iso o ctan c , C~ II I l'!(f), with d/ JO = - 259.2 ~ J / m o l. ] Do yo u n otice any grea t di ffer en ces am o ng th ese fu r-ls? Are th e se d ifferences rel at e d to th e ir co m p os itio n ?
73 . ... . T h e sta n dard m o la r e n thalpy o f fo r m at io n o f di bo ra ne, Rt Il 6 (g ), can n o t h e de term ined d ire ctl y he ca u se th e co m p o u nd ca n not be prepar e d by th e re actio n o f b o ro n and h yd rog e n . It ca n h e ca lc u la ted from o th er e n tha lpy c hanges, h oweve r. T he followin g enth alp y changes ca n h e m easu red. 4 B (s) H,( g )
+
3 O , (g ) ---> 2 B,O, (s) d rH D
+ Y,
B, H H , O (g ) !!.,lr
=-
~ J / Ill o l -r xn
24 1.8 kJ/ lllo l-rxll
3 O , (g ) ---> B, O ,(s) + 3 H , O (g ) d,H D = - 20;t !.9 q /m ol-r xn
(a) Sh ow h ow th e se e q ua tion s ca n b e ad d ed toge th er to give th e equa tio n for th e form atio n of Rtl ll;(g ) fro m B (s) a n d H 2 (g ) in th ei r stan dard sta tes. Assig n e n tha lpy cha nges to eac h re action . (h ) Calc ulat e d / JOfo r Rt Il 1;(g ) . (c) D raw a n e IH.TbJT-Il'Ye l d iag ra m th at sh ows h ow th e va rio us enth alpies in th is p robl e m are rela te d. (d ) Is th e form at io n of Bt H ij(g ) fro m its e le m e nts p ro d u ct-favo re d or re ac tan t-favored? 74. C h lorome thane , C HJ-::I, a co m po u n d fo un d uhiq uito u sly in the e nviro nmen t , is fo r me d in the re a ct ion o f ch lo rin e a to ms with me tha n e. C I I,(g)
248
I
+
2 Cl (g) ---> C Il 3C l (g)
+
A camping stove that uses butane as a fuel.
77. Met h anol, C H 3 0 H , a co m p o und th at can be m ad e re latively in expe nsive ly fro m co al, is a p rom ising su bst itu te fo r gasoline. T h e alco ho l h as a sma ller ene rgy con te n t th an gaso line, b u t, with its h igh er oc ta n e ra ting, it h um s mo re effi cie ntly th an gas o line in co m b us tion engines. (It h as th e ad ded advan tag e of contri buting 10 a lesser degree to so me a ir pollu ta nts.) Com pa re th e en tha lpy o f co mbus tio n pe r g ra m of C1130 [ 1 a n d Ct\I IIH (iso oc ta nc ) , th e latter be in g represe n tative of th e com pou nds in gaso line . (D.; H = - 259.2 q / m o l for isoo ct a nc. ) D
I!Cl( g )
... more chaLLenging
•
in OWL
Blue-numbered questions answered in Appendix 0
78. Hydrazin c and Ll -dim ct hvl h yd ra zin e ho th 1"I..' act sp onta neously with O ~ a n d ca n he used as rocke t fu e ls. N, H, (f)
+ O , (g ) -> N, (g) +
~
N, II , (C II , ),(f)
+
~
O , (g ) -> ~ CO ,(g)
fo r a ncn diet ln-vcrage whose label ind icat es a ca lori c con tent of 2-to Ca lo ries.
II ,O (g )
hydrazine
I.I -dimet hylhydrazine
( c) Ca r r y o u t a com p ariso n sim ila r to th a t in p art (b )
8 1. ... Ch lorofo r m . C HCI:\. is Io rn u-d fro m meth ane a nd chlorine in the fo llowi ng reactio n .
+
~
H,O (g )
+ N, (g )
T he molar ent hal py of forma tion of :"\~H . (O is +:)0.6 ~ l /mol , aud thar of X~ 1I 2 (C Il 3 b ( O is + -tH.9 q /mol. Usc these values, with o th e r ,1/ val u es . to decide whet her the reacti on of hvdrazinc or Ll-dimethvlhydra zin e with ox ygt'l1 provides m ore enl'r~· per gr.lIl1.
r
C H , (g )
+ 3 CI,(g) -> 3 H CI(g ) + C HCI , (g )
Calculate J.rH o, the.' enthalpy cha n ge fo r th is re action , using th e cnthalpics of form at ion of CO~ ( g ). H 2 0 ({) , and C HCI , (g ) (!1JHo = - 1O:1.I kll mol ) , and t he enth alpy chan ges fo r the followin g reactions:
C H , (g ) ~
+
~
O , (g ) ->
~
11 ,0 (1) + CO, (g ) I1 rlr = - 890.4 k.J/mol-rxn
HCI (g ) -> H , (g ) + C I, (g} j. ,1fO
= + 184 .6 kJ/mnl-rxn
82 . Walc r gas, a m ixture of ca rbon mon o x ide a n d hyd rogen, is produced hy trea tin g ca rbon (in the form o f coke or co a l) with steam at h igh te m pe ra tu re s. [See Question 75 .)
A control rocket in the Space Shuttle uses hydrazine as the fuel .
79 . • (a) Calc u la te th e e n thal py ch ange, I1J/o. for th e fo rm a tion of 1.00 m ol o f stron tium ca rbona te ( tlnman-rial tha t g ives th e re d co lor in fi rewo rk s) from its clemen ts. Sr( s)
+ C(gra phi te) + Y, O ,(g ) -> Sr CO,,(s)
T h e cx pc rinn-ntal iuforma tio u avai lable is
Sr (s)
+ Y, O , (g ) -> S,O (s)
S,{) (s)
!1/1 " = -59~ ~l /l1l11l -rXll
+ CO, (g) -> SrC O , (s) !1 ,1f" = - ~ :14 kj/mol-rxu
C (gmphile)
+ O , (g ) -> CO ,(g ) !1JH " ~
-:19~ ~ J /l1l11l-rxll
(h) Draw an energy-level d iagram re la tin g the t'lwrgy qu antities in thi s problem .
80 . 'io u drink 350 m L of diet so d a th at is a t a tc.'mpt'f
~1g (O H ), ( s )
+
H , (g )
~ Ui n th e deco mposit io n o f 7.647 g of am mon iu m nitrat e ca n be me asured in a bo mb ca lori mete r. Th e re ac tio n th at occ urs is
87. Th e value of
NJI.,N O , (s) -> N 2 0 (g )
+
2 11, O (g)
T he tempe rat ure of th e calori meter, whi ch co n ta in s 4 15 g o f water; incr ea ses from HUlO "e to 20 .72 "C. T he he a t ca pacity of th e hom h is 15.1i .11K. What is the val ue of ~ Ufor th is reaction , in ~J / m ol ?
The " heat er meal" uses the reaction of magnesium with water as a source of energy as heat.
Calcula te th e e n thalpy change under sta nda rd co nditio ns, in j unks, for th is reaction . Wha t qua ntity of ma gnesiu m is ne ed ed to su p ply th e e nergy requi red to wa rm 25 mL of wa ter (d = 1.00 g / m L) fro m 25 °C to 85 °C ? (See W, J ensen : journa l of Chemical Education; Vol. 77 , PI' . 7 13-7 17,2000. )
90, On a co ld day, yo u ca n warm yo u r ha nd s with a "lu-a t p act " a dev ice th a t uses th e oxidation of iron to P!'> ducc e nergy as h ea l.
4 Fe (s)
+
3 O , (g) -> 2 Fe, O ,,(s)
The decomposition of ammonium nitrate is clearly exothermic.
88 . • A h om h calo rim e tr ic ex perimen t was run to de termi ne th e heat of com bustion of e tha no l (a com m o n fu el ad di tive ) . Th e rea c tion is C, II ,O JJ (C)
+
+ 3 JJ20 (() ca paci ty of 550 .11K, a nd the cal-
3 O , (g ) -> 2 CO,(g )
Th e bo mb h ad a he at ori me ter co n tai ned 650 g of wa ter. Bu rn in g 4.20 g of et hanol , C 2' h O B (f) resulte d in a rise in tem pe ra ture fro m 18.5 "C to 22.3 "C. Calc ula te the e n thal py of cornbustio n of e thanol, in ~J /mo l.
250
I
~
~ ...
_-----,;;;- -_...
A hand warmer uses the oxidation of iron as a source of thermal energy.
What mass of iron is n eeded to su pply th e ('He rb':' requ ir ed to wa rm 15 mL of water ( d = 1.00 g /m L) fro m 23 -c to 37 °C?
... more chaLLenging
•
in OWL
BLue-numbered questions answered in Appendi x 0
Summary and Conceptual Questions T Ill' [ollounng questions
Wi)'
use concepts from this fwd tnrnious
clutpters. 91. With out do in g ca lcu la tions, d e cide ,...h cth c r e ac h of the following is pro duct-favo red or reac tan t-favored. (a ) th e co m bus tio n of natu ral gas (h) til e d ecomp ositi on o f g luc ose , C.;l 11 20 (;' to carb on a nd water 92 . • Which of th e fo llo wing a rc sta te fu n ctio n s? (a) th e volume o f a ba llo o n (h ) th e Lim e it la ke s to drive fro m yo ur hom e to yo u r co llege o r university (c) th e tem perature o f th e wa u-r in a co ffee cu p (d ) th e p ot enti al e nergy o f a b all h el d in yo ur h an d 93 . ... . You wa n t to d ete r min e th e value for th e e n thalpy o f form ation o f CaSn1(s ) . Ca (s)
+
Sis)
+
2 O , (g ) ---> CaS O ,(s)
Th is reac tion can n o t h e do n e d irectly. You kn ow, however, that hoth calc iu m a nd su lfur reac t with oxygen to p ro duce ox ides in rea cti o n s th at ca n he stu d ie d cal o rime tricall y. \b u also kn ow tha t th e basic oxi de Ca O re ac ts with the aci d ic ox ide SO :l(g ) to p rodu ce Ca Sn l(s ) with 6,1 r = - 402.7 kJ. Ou tlin e a met hod fo r d etermini n g 6 JH o fo r CaS 0 4(s) , an d ide nti fy th e inform atio n that m us t h e co llec te d b y ex peri me nt. Usin g inform ation in Ap pe nd ix L, co n fi rm th at 6/ -fO for CaSO ,(s) ~ - 1433.5 kJ/mol.
96 . .... Suppo se yo u are a lle n di n g su mm e r sc h o o l an d are living in a ve ry o ld d ormitory. The day is o p p ressively h ot. Th ere is lIO ai r-co n d itio ner, a n d yo u can't o pe n th e windows o f your ro om b ecause th ey arc stuc k shu t fro m layers o f p aint. There is a refri ge rat or in the room , howeve r. In a stroke o f g e nius , '·Oll open the 0 0 0 1' of th e refrigerator, an d cool a ir cascades ou t. T he relief' d o e s 1I0t las t lon g , th ough . S O O ll th e refrige rator moto r a nd condenser he gin to rim , and not lon g the reafte r the ro o m is h otter th an it was befo re . Why d id the roo m wa r m up?
97 , • You wa n t to h eat th e ai r in yo u r hou se with n atural gas (C I-I.1) . Assu me yo u r h ouse h as 275 m 2 (abo u t 28 00 ft 2 ) of 110 0 r area an d that th e cei lings arc 2.5 0 m fro m th e 110 0rs. Th e a ir in th e house h as a m o lal' h e at ca pacity of 29 ,1 J lmol . K. (T he number o f m o le s of air in th e h ouse ca n he found by ass u mi ng that th e ave rage mol a r m ass of air is 28.9 g / m o l a n d th at th e d en sity of air at th ese temperatures is 1.22 giL. ) \Vhat mass of m e th an e d o yo u have to burn to heat th e air from 15 .0 °C to 22. 0 OCt 98. Wat e r ca n he d ec omp o sed to its e le m e n ts, H 2 a n d ()2' u sin g e lec trical energy o r in a se rie s o f c he mical re a ct ion s. T he fo llow ing se q ue nce o f reaction s is o n e p o ssib il ity:
+
Ca Rr,(s) II g (l:)
+
H gllr, (s)
94. P repa re a g ra p h o f specific h eat ca pac ities fo r me tal s versus th ei r a tomic weigh ts. Com b in e th e d ata in Figure 5.7 a n d th e va lu es in the foll owin g tabl e . wh at is th e rel atio nship bet ween speci fic h e al ca p acity and a to m ic we ig h t? Use this rel ation shi p to pred ict th e specific heat ca pacity of pl atin um . Th e specific h e at ca pacity fo r p lat inum is g iven in th e litera ture as 0 .133 Ji g ' K. 110\,,.' good is th e ag ree men t bet ween tilt' p re d ict ed a nd ac tu al va lues? Metal Ch romi um Le ad Silver T in T ita n iu m
+
2 II Rr (g ) ---> II gR r, (s)
+
CaO (s) ---> H gO (s)
H gO (s) ---> Hg (l:) +
+
2 I1Rr (g)
+
1I, (g )
Ca ll r, (s)
Y, O , (g )
(a) Show th at the n et result of th is se ries of reac tion s is th e d ecomp ositi on o f wate r to its elemen ts. (b) l f yo u use 1000. kg of wa ter, wh at mass of H 2 ca n be p rodu ced ? (c) Calcu la te the val ue o f 6 rH o for ea ch ste p in th e series. Are th e reactions pred icted to b e p rodu ctfavored or re act ant-favo red ?
Specific Heat Capacity U/g . K)
dJH' (Callr, (s) ]
~
- 683 .2 kJ/mol
0 .450 0 .127 0 .236 0 .227 0 .522
df H' [ Hg llr, (s) ]
~
- 169 .5 kJ/ mol
95 , O bserve th e m olar h eat ca pacity va lues fo r th e metal s in Figure 5.7. Wh at ohserva uon can yo u ma ke about these va lu es-speci fica lly, ar c th ey wide ly dif fe re nt or very similar? Usi ng th is informa tio n, e-s timate the sp ecific h e at ca pacity fo r silve r. Co m pare this es ti m a te with th e co rrec t valu e for silver, 0 .2:-36 J /g . K.
... morechallenging • in OWL
IIP (g ) ---> CaO (s)
Blue-numbered questionsanswered in Appendix 0
(e ) Com m en t 011 th e commercial fe asibilit y of usin g this series of reactio ns to p ro duce H 2 (g ) fr om wa te r.
99 . Suppose th at an inc h of ra in fall s over a squa re mi le of grou n d . (De n sity o f wa te r is 1.0 g/ cm 3 . ) Th e e n tha lpy o f va poriza tio n ofwater a t 25 °C is 44.0 q /m ol. 1I0 w much energy as h eat is tran sferred to th e su rro u n d in g s fr om th e co n d e nsatio n of wa te r va po r in fo rmi n g th is q uan tity of liqu id wa ter? (The huge numbe r tell s you h ow much e n erb'")" is "store d " in water va por a n d wh y we th ink o f sto r ms as suc h grea t fOIH ' S o f e nergy in na ture , It is in te res ting to co m pare this res u lt with th e cll e r!,..,.· give n off, 4.2 X ]()6 k ], whe n a to u of dy namite cx p lo dcs. )
I
251
100 . ... Pe an ut s a nd pe a n u t oi l a re o rg a n ic mate rials a n d bu rn in a ir. H ow m a llY hurni ng pean uts d oe s it take to p ro vide th e e n e rgy to ho il a cu p of wat er (25 0 m L of wa ter) ? To solve thi s probl em, we ass u me each pe an u t, wi th an average m a ss o f 0 .73 g , is 49% p eanu t o il an d 2 1 % sta rc h ; th e rem a inde r is no nco m bus tible . We fur th er assum e pea n ut o il is pa lm it ic aci d , CUiIl 3202. wit h a n e n thalpy of fo rm atio n o f - 84H.4 kJ/m o l. Sta rc h is a lo ng c ha in of CiiH w r (als o 10.0 g, C = 0.38 5 Ji g ' K) is chi lle d in an ice bath to 0 "C. Roth pieces of met a l are p lace d in a be ak er con tai n ing ISO. g H 20 a t 20 "C. \-Vill th e te m pera ture of th e wate r be g reate r th an or less th an 20 "C wh e n therm al equilibrium is reache d ? Ca lc ulate th e Fi nal temp erature .
... more chaLLenging • in OWL
BLue-numbered questions answered in Appendix 0
105 . Me th an e , CH " ca n be converte d to metha nol whic h , like et han ol . ca ll he used as a fue l. T he e nerb'J' level d iagram shown here pr ese nt s rel ation sh ip s bet ween ene rgies of th e fuels an d th eir o xida tion products. Use th e info rmat io n ill th e di agram to answer th e fo llow ing q ue stion s. (T he e llerg;' terms a re pe r m o l-r xn .)
(a) Whi ch fu el. me th an ol or m eth ane , yie lds th e m ost energy pe r m ol l' when bu rn ed? (h ) Wh ich fu el yields the mos t energ y per gram whe n
burn ed? (c ) Wh at is the en t hal py cha nge fo r th e co nversio n of m ethan e to meth ano l? (d) Eac h a rrow on th e di ag ra m re presents a chem ical reaction, \ Vrite tlu- e q uatio n fo r th e re ac tion th at co nverts met hane to m etha nol. 106. Calculate DoJrfo r th e rea ction
2 C (s) + 3 II, (g) + V, O, (g ) --. C, Il , O Il (f) given th e in form ation below C (s)
+
2 H, (g)
O , (g ) ---> CO , (g)
+
!J.,J-r
- :, 93.5 kjl mol-r xll
O , (g) ---> 2 H,O (f) Dol'Jr = - 571.6 kJ/ l1Iol-r xn
C, Il , O Il (1')
+ 3 O,(g) ---> 2 CO, (g) + 3 Il ,O (1')
tx.H' = - 1367.5
kJ/ mol-rxn
107 . You h ave the six pieces of meta l listed be low, plu s a be ake r of,v-a rcr con taining 3.00 X 102 g ofwa n-r. T Ile wate r te mperatu re is 2 1.00 -c. Metals
Spedfic Heat (Jig K)
Mass (g)
1. Al
0.9002
100.0
2. Al
0.900 2
50.0
3. Au
0.1289
100.0
4. Au
0,1289
50.0
5. Zn
0.3860
100.0
6. Zn
0.3860
50.0
... more challenging • in OWL
Blue-numbered questionsanswered in Appendix 0
(a ) In yo u r fi rst ex perime n t yo u selec t on e p iece of metal a nd he at it to 100 °C. a nd th e n selec t a se co nd pie ce of m et al and coo l it to - 10 °C. Roth piece s of m etal are th e n pl ace d in th e beaker of wa ter and th e te mpe rature s equi libra ted . '\hu wan t to select two p ieces of m e tal to use . suc h th at the fin al te mperature of th e wa ter is as h igh as poss ihie . Wh at p ie-ce of m et al will yOIl h eat? What piece of metal will you cool? What is th e fi nal te mpe rature of the water? (h) T he se co nd ex peri me n t is done in th e same way as th e fir st. H oweve r, yo ur goal now is to ca use th e te mp e rat ur e to ch ange th e lea st. th at is. the fin al te mp eratu re sho u ld be be as nea r to 21.00 DC as possib le . Wh at piece of m e tal will yo u h eat ? Wh at piece of m e ta l will yo u coo l? Wh at is th e fina l te mpe rat u re o f th e water? 108. In lab . you p lan to ca rry ou t a ca lo rim e try ex peri me nt to deter mi ne th e Do,-H for the exo thermic reac tion o f Ca (O Il h (s) and II Cl( aq ) . Pre d ict h ow each of th e fo llowing will a ffec t th e calc u late d value of tx.H, (T he value calculat e d hIT' Do,!! for thi s reaction is a negative value so choose yo ur an swer from th e following: Do,-H will be too low [tha t is. a la rger n ega tive val ue ], ts. H will be una ffecte d . Do JI will be to o high {t ha t is. a sm alle r negative valuc. j ) (a ) ' CO,(g)
Ii ,H' = - 393.5 kJj mol-rxn or - 32.8 kJj g C
Bu rni n g h yd rogen to fo rm water is much m ore exo th erm ic o n a per-g ram ba sis, with ab o u t 120 kJ evolved p er g ra m of h yd ro gen co ns u med.
H,(g) + '/, O,(g) -> H,O(g) Ii ,H'
=-
241. 8 kJj mol-rxn or - 119.9 kJj g H,
Co a l is mo stly carbo n , so its h ea t o u tpu t is sim ila r to that o f p ure ca rb o n. In co n tra st, m e th an e is 25 % hyd r oge n (b y m ass), and th e h igh e r-m o lecul ar-wei ght h yd rocarbon s ill pet rol eum a n d p r odu cts refin ed fro m p e tr ole11l11 ave rag e I fi- 17% h yd r og e n co n te n t. Th e re fore , th e ir h ea t o u tp u t o n a p er-gra m ba sis is g re a te r th an that o f p u re ca r b o n , bu t less tha n that of h yd rog en itsel f.
TABLE 2
Energy Released by Combustion of Fossil Fuels
Substa nce
Energy Released (kJjg)
Coal
29-37
Crude petroleum
43
Gasoline (refined petroleum)
47
Natural gas (methane )
50
258
I
The Chemistry of Fuels and Energy Resources
\Vhil e th e b asic che mical prin ciples for ex trac ting e nergy fro m fo ssil fuel s ar e sim p le , co m p lica tio ns arise ill practi ce . Let us look at eac h of these fu els in tu rn .
Coal Th e solid ro ck-like su bs tan ce th at we ca ll coal began to fo rm a lmost 290 m illion yea rs ago. Decomposit ion of pl ant m ail er occ urre d to a suffic ie n t ex te n t that th e p rim ary co m po nen t of coal is ca rho n . Descrihin g coal simply as ca rbo n is a sim plifica tion, h owever, Sa m p les o f coal \'a ry co nsi dera b ly in the ir co mpos itio n an d charac teristics. Ca rbo n con te n t m ay ra n ge from 60% to 95 %, with var iab le a mo u n ts of hyd rogen , oxygen , su lfu r, an d n itroge n p resent in th e coal in various form s. Sulfur is a co nu uon co ns tituen t in so me coals. T he el ement was in corporat ed in to the mi xture partly fro m d ecayin g pl ants an d p artl y fro m hyd ro gen su lfide , 1-1 2S, whi ch is th e waste product from ce rtain b acteri a. In ad d ition , co al is likel y to co n tai n tr ace s o f m an y o ther e le ments , includin g so me that are h azardous (such as arsen ic, lIIercury, cadmi um , an d lead ) an d so m e th at are not (su ch as iron ) . \Vh e n coa l is burn ed, so me o f the imp urities are d isp erse d in to the air, and so me e n d up in the ash that is funn e d . In th e U n ited Sta tes , coal-fire d po wer pl ant s arc re spo ns ible for 60% o f thc e m issio ns of S02 an d 33 % o f m ercury e m issio ns in to th e e nvi ro n III en t. ( LJ .S. pl an ts e m it abo u t :')0 to ns o f m ercllry p e r yea r in the U. S.; wo r -l dw ide, abo ut 55 00 tons are e mi tted.) Su lfu r di o xid e re acts with wa ter and oxygen in th e a tmos p he re to form su lfuric aci d , wh ich co ntr ib u tes (alo n g with nit ri c ac id) to th e ph cn omeno n known as aci d rai n. 2 50,(g) + O, (g) --> 2 50 ,(g)
50 ,(g) + H,O(C) --> H, 50 4 (aq) Becau se these acids are h armful to th e envi ro n me n t, legislatio n limi ts th e ex ten t o f su lfur oxide em issions from coal-fire d p lan ts. Chem ical scrubbers h aw be en de veloped th at ca n be attac hed to the sm okestacks of power p lan ts to reduc e su lfurbased em issions. Sim ply, th e co mb ustio n ga"ics arc pa ssed th rough a water spray with chem ica ls suc h as lim estone (calciu m ca rbonate) to form solid.s th at ca n be removed: 2 50,(g) + 2 (a CO,( s) + O, (g) --> 2 (a 50 4 (s) + 2 (O, (g) H owever, th ese d evices are ex pensive an d ca ll in crease th e cos t of th e e ne rb')' produc ed fro m these facili ties. Coal is class ified in to three ca teg o ries (Tab le 3) . An th ra ci te , or h ard coal, is the hi gh est-quality coal. Amo ng th e fo rms of coal, anthracite coal r el eases th e large st am o u n t o f h ea t per gram an d h as a low su lfur co u rc n r. Un fo rtu n a tely, anthracite coa l is fairly un commo n , with o n ly 2% of th e U .S. coal re se rve s occu rr ing in th is form . Bitu mi nous coal, also referred to as soft coal, acco u nts fo r abo u t 45 % of thc U .S. coa l reser ves a n d is th e coal m o st wid ely u sed in elec-
TABLE 3
Types of Coal
Type
Consiste ncy
SuLfur Conte nt
Heat Content (kJ{g)
li gnite
Very soft
Very low
28-30
Bit uminous coal
Soft
High
29- 37
Anthracite
Hard
Low
36-37
Figure 3 Bitu minous coal being extracted from a strip mine in Montana.
tri c power ge n eratio n (Fig u r e 3) . Soft coal typi call y h as th e hi gh est su lfur co n te n t. Lignue. also ca lle d b row n coal b ecause o f its p aler co lor, is geologically th e "you ngest" for m o f coal. It releases a smalle r a mou n t of h eat per gra m th a n the oth er forms of coal, oft en co ntains a sig n ifican t amo u nt o f wa ter, and is the least po p u lar as a fu el. Coal can b e co nve rt e d to co ke by h eatin g in the abse n ce o f a ir. Cokc is almost pure carb o n an d a n ex celle nt fu e l. In the p ro cess of coke for ma tio n , a vari ety of o rga n ic co mpo und s a re driven o rr. Th ese crnnpouuds are u sed as raw m at e rials in the ch e m ical industry for th c producti on o f polym ers, ph arm aceuti cals, syn the tic fa b rics , waxe s, tar, and IIll1I1 erOll S o th e r p r oduct s. Technology to cou ver t coa l into gaseo us fu els (co al gay ifica tio n] (Figu re 4) or liq u id fu els (liq uefaction) is unde r developm ent , hut h amper ed by cost. T hese p rocesses provi d e fuels th at burn mo re cleanl y th an coal, exce p t that 30-40% o f th e availab le eIltTb,)' is lost in th e process. As petroleum an d n atu ral gas re serves dwindle an d th e cos ts of th ese fuels in cre ase , liquid an d ga seous fuels d erived fro m coal are likel y to become mo re im portant.
Nat ural Gas Na tu ra l gas is fou n d d eep under th e earth's su rface , wh e re it was formed by bacteri a d ecom posing orga ni c m atter in an ana erob ic environ m en t (in wh ich n o O a is p re sent) . T he major co m po n e nt o f natu ra l gas (70-9.:1%) is m eth an e (C H-t ) . Lesser quantiti es of o th er ga ses suc h as ethan e (CtJ In) , propan e (C3Hl:\)' an d bu tan e (C -t1Iln) are also p resent, alo ng with ot her gases includin g N 2 , li e , CO 2 , an d Il tS. Th e impurities an d high er-m ol ecular -weight co m p one llt'i of
Fossil Fuels
il 'a
I
8
@ -
Figure 4 Coal gasification plant. Adva nced coal-fired power plants, such as this 2 544 ~to n -pe r~ da y coal gasification demonstration pilot plant. will have energy conversion efficiencies 20% to 35% higher than those of conventional pulverized-coal steam power pLa nts.
natu ral gas are se para ted o ut d uring th e refining process, so th at thc gas pi ped into ou r ho m es is pr imarily m eth ane. Nat u ra l gas is all in cr easingly po pular cho ice as a fue l. It b u rns mo re cleanly th an the o thcr fossil fuels , e mi ts fewer pollutants, and prod uces relative ly mo re eneq,')' th an th e ot her fossil fuel s. Na tu ral gas ca n be transp orted hy pipelin es over land and pi ped into b uildings suc h as yo u r hom e, whe re it is used to heat a mbient a ir, and to heat water for wash ing, hat h ing , o r for cooki ng. It is also a popular cho ice for II CW elec tr ical power plants, wh ich have hi gh efficiencies due to new ga s turbin e s and n.' Co\'C I} of waste heat.
Pet roleum Pet rol e um is o fte n fo u n d in porous r o ck fo rmati on s tha t are bounded hy imp erm eabl e rock. Pe trole u m is a co mpli cat ed m ixtu re of h ydro carbon s, wh ose m ol ar m asse s ra nge fro m low to \'ery hi gh (.... page 495). Th e h ydrocarbo ns m a), have a n ywhere from o ne to twen ty o r m o re ca rbon atoms in th e ir structures , a nd co m po u nds co nta inin g su lfu r, nit ro gen , a nd o xygen llIay a lso he present III small amo u n ts. Pe trol eum goes throug h e xte nsive pro ce ssin g a t refineries to se para te the var io us cOlllpo ne n ts an d co nve rt less valuabl e co m pou nds in to m ore val ua ble ones . Nca rly 85% o f th e cr ude pe tro le u m p u m ped from th e groun d e nds u p being used as a fu el , e ithe r for tra nsporta tio n (gaso line and di esel fuel ) o r for hca ting (fu el o ils).
Other Fossil Fuel Sources \\1 1Cn natural gas pi peli nes were laid across the Un ited Stat es and Ca na da , pipel ine o pe ra tor s soon fo un d that, u nless water was carefully ke p t o u t of th e lin e , ch u n ks of m eth ane h ydrat e wo uld fo rm an d clog the pipes. Methan e h y-
I
259
drat e was a com ple tely u ne x pecte d substa nce because it is m ade up o f meth ane and water, two chem ica ls tha t would ap pe ar to ha ve littl e affini ry for eac h o ther. In methan e h ydrate . m ethan e becomes trapped in cavities in the crystal str uctu re o f ice (Fig u re .1)) . Meth ane hydrate is stable o n ly at tem pcratur es below the fr eez ing poin t ofwater, If a sample o f me th an e h ydrate is warm ed above 00 C, it melts, a nd m ethan e is rel ea sed. T he volume of gas released (at norm al pr essllr e and tem perature ) is abo u t 165 tim es large r than th e volum e of th e h ydrate. If m ethan e h ydrat e forms in a pipelin e , is it found in nature as we ll? In May 19 70 , oceanogra p hc rs drillin g int o the sea bed off th e coast of Sou th Ca ro lina pulled up sam ples of a wh itis h so lid th at fizzed an d o ozed whe n it was re moved from th e d rill casing. Th ey qui ckly real ized it was methane h ydrat e . Since th is o r iginal d isco very, m ethane hydra te has been fo und in m a ny parts o f the ocea ns as wel l as under permafr o st in th e Arc tic. It is estima te d th at 1.5 X 10 13 to ns of m ethan e h yd rate arc buried un d er th e sea flo o r aro u nd th e wo rld . In fact , th e energy ava ilable from thi s so urce may su rpass th at of all th e ot he r known fossil fuel reserves by as m uc h as a fa ctor o f 2! Clea rly, th is is a po te n tial sou rce of an importa nt fuel in th e futu re . To day, h owever, th e tec h no logy to ex tract m ethan e from these hyd rate d e posits is very expensive , e specially i ll CO IlIpari son 10 th e well-develo ped technol o gies used to extrac t cru de o il, coal, a m i gas c o us m ethan e . There a re othe r so u rces of m ethan e in o ur environm ent, For exa mp le , methan e is ge nerate d in swa m ps , where it is cal led swamI) gas or marsh ga.5. He re , m eth an e is form ed by bacteri a wo rk ing o n o rga n ic m atter in an anaero bic environ men t-name ly, in se d im e n tary laye rs o f coasta l wa te rs a nd in m arsh es. Th e p n Kess of fo rm at ion is sim ilar to the processes occurring eo ns ago th at ge nerate d th e natural gas d epo sits th at we cu rre n tly use fo r fue l. In a m arsh , th e gas can esca pe if th e se d imen t layer is th in . You see it as bubbles risin g to th e su rface. Unfortu na tely, be cau se o f th e rel at ively sm all amou n ts genera te d , it is imp racti cal to co llec t and use thi s gas as a fue l. In a str iking analogy to wha t occ u rs ill nature, m ethan e also forms in hum an -made landfill site s. A g rea t d eal o f organ ic ma tte r is b uried in land fills. It re mains o u t o f co ntact with ox ygen in th e air, and m ethane is formed as th e o rgani c m atter is d eg rad ed by bac teria. In th e past, landfill gases have bee n d eemed a nuisan ce. To day, it is po ssible to co llect this m ethan e a nd u se it as a fuel. You m ig h t have see n pl astic pi pes in th e g ro u nd in a landfill th at ven t th e m e th an e 10 a ho lding tan k. An ot her source of fo ssil fuels, a nd o ne th at is be ing used right now, is o il fWIlI tar sands. Ta r san ds (also called oi l sand s) co ntain a \'er y \ i SCO HS organic liquid called "bitumen ." This is che mically sim ilar to th e hi ghest- mol ecul ar-we ight fraction o b taine d hy d istillation of crude o il. wh at mak es thi s source so e n ticing is th e huge qu antity of oil th at co uld he obtained fWIIl such sites. Th e lar gest re sou rce o f tar sands
260
I
The Chemistry of Fuels and Energy Resources
~
§
~ ~
-s,
§ ~
~
~. ~
~
~ ...,,-........ii....... (a) Methane hydrate burns as methane gas escapes from the solid hydrate.
(b) Met hane hydrate consists of a latti ce of water molecules with metha ne moLecu les tra pped in the cavity.
(c) A colony of worms on an outcropping of methane hydrate in the Gulfof Mexico.
Figure 5 Methane hydrate. (a) This interesting substance is found in huge deposits hundreds of feet down on the floor of the ocean. When a sample is brought to the surface, the methane oozes out of the solid, and the gas readily burns. (b) The structure of the solid hydrate consists of methane molecules trapped within a lattice of water molecules. Each point of the lattice shown here is an oxygen atom of a water molecule. The edges are O- H-O bonds. Such structures are often called "clathrates. ~ (c) An outcropping of methane hydrate on the floor of the Gulf of Mexico. See E. Suess, G. Bohrmann, J. Greinert, and E. Lausch: Scientific American, pp. 76- 83, November 1999.
in th e world is found in Alberta , Ca nada (the Atlutbasca .' ·a n rl\). This is followed clo sely hy those in Ven ezuela. Resources a pp roach in g 3.S trillion barrels of o il are estimated in these two locatio ns-c-twice th e world's knO\\'T1 rese rves of petroleum . Th e U.S. imports more oi l from Canada th an an y other country (0.8 million barrels per da y) , and most of thi s is from th e Athahasca Tar Sands! Ex tracting the oil from tar sa n ds is quite costly. Essentiall y, the sa n d s must be min ed and th en mixed with h ot wate r o r steam to e xtract the b itum en . In o rde r to avoid an environm enta l catastrophe, the mined lan d mu st he re stored (reclaimed) . This adds to the cost of the process. Also , most of th e Canad ian tar sa n ds arc lo cated in dry areas, so o b ta in ing all adeq ua te wa te r su p p ly for ex tr action mi ght pose a const rai nt on increased producti o n .
Environmental Impact s of Fossil Fuel Use As mentioned earl ie r, abo u t 85 % o f th e energy used in th e wo rld to d ay co mes from fo ssil fuels. We are a carbon-based soci ety. Whi le this percentage is relat ively stable, th e am ount o f gaseous em issio ns o f ca rb o n co m po u n ds into o u r cn viroument co nt in ues to rise. These include mainly CO 2 b u t also CH .\, CO, a n d ch loro fluorocarb o ns (CFCs) . T h c co rrel ation is quite distinct- rising e nergy use co rre la tes wel l with risin g carb o n e m issio ns (Figu re I ) .
Th e "g ree n h o u se e ffec t" is a nam e g iven to th e trappi ng of energy in th e ear th's atmosp here by a process ver y similar to th at used ill gree n h o uses (Figu rc 6 ) . Th e atm osp here , like win d ow glass , is tran sparent to in coming so la r rad iatio n. T h is is absorbed by th e eart h and re -ernitted as in frared r ad ia tio n. Gases ill th e a tm os phere, like win d ow glass, tr ap so me of th ese lo nger in frared rays, ke ep ing th e eart h wan ne r th an it wo u ld otherw ise be . III th e last ce ntury, th ere h as bee n an in crease in co nce n tr a tio n s in the atmos p he re o f ca rbo n di o xide a n d other so-cal led g ree n house gases (m et h ane , n itrogen o xides) du e to increase s in fo ssil fuel hurnin g . T here h as also been a co rr es p o n d ing increase in glo ha l a\'e rage temperatu res th at most sc ie ntists att rib u te to in creases in these gre e n house gas co nce ntration s (Figure 7) . This co rrelat es H:-ry we ll with increased co nce n tra tio ns o f CO 2 in th e a tmosp he re. Fo r th e n e xt two decades, a wa r m ing of about 0 .2 °C per d ecade is proj ected by so me model s. Su ch tcmpcratu re changes will affec t th e earth 's climat e in lIlan y ways, suc h as m ore inten se sto r ms , precipitation ch an ges, and sea level rise . H ealth issue s will also b e a facto r. Glo ba l warmin g-th e in cre ase in ave rage g lo b al temp e ratures, wh ic h is p robably owin g to hum an a cti viti es in creasi n g th e g ree n ho use e ffect-has b ecome o ne of the hi gg e st issu e s facing u s wo rl d wide . In d e ed , lIlan y of th e ste ps mad e in th e last de cad e to pu t in cr e ase d
Fossil Fuels
I
261
Figure 6 The greenhouse effect.
e m p has is on re newable e ne rg ies is due to thc co nce rn for th e earth's clima te. (Fo r m o r e o n th c g rec n ho usc e ffec t sec "T h e Che m ist ry o f thc Envi ron mc n r," pag c 949.) Anot hc r p robl c m d ue to incr eased bu rn ing of fo ssil fue ls is lo cal an d intcrn ati on al a ir po llut io n. T he h igh tc m pcratu r e an d p rc ssu rc u sed in th e coru bux tio n p r o ce ss ill auto m o b ile e ngi ncs have th e un fortun a te conseq uc nce o f also ca usi ng a reac tio n bet wee n a tmosp he ric nitrogen an d oxyge n th at re sults in so me NO forma tion . T he N O can the n r eact further wit h oxyge n to produce ni trog cn di o xid e . This poi son ous, b r own gas is fu rthe r ox idize d to fo rm nit ri c ac id, Il N0 3 , in th e pre se n ce of wat cr.
0.6 0.4
E -'" ro E
g
ro
GlobalTemperatures Annual average Five year average
0.2 0
~
N,(g)
+ O, (g) -> 2 NO(g)
B ~ - 0.2
1 ,W = 180. 58 kJ/ mol-rxn
2 NO(g) + O, (g) -> 2 NO,(g)
1 ,W = - 114 .4 kJ/ mol-rxn
3 NO,(g) + H,O( C) -> 2 HNO, (aq) + NO(g)
1 ,W = - 71.4 kf/mo l-rxn
ro
~
E
~
- 0.4 - 0.6
To so me ex te n t, th e am o u nts o f po llu tan ts re le ase d can be limited b y u sc of a u to mo b ile ca ta lytic co nve rters. Ca ta lytic conve r ters a re h igh - surface-area m etal g r ids th at arc coa te d with pl at inum o r pa llad iu m. T hese very ex pe-ns ive m e tal s ca n ca ta lyze a co m ple te co m b ns tio n r cac tio n, help in g to co mbi ne oxygc n in the ai r with 1lI1 b u r ned h yd ro carbo n s or ot hcr by-product s in th e ve hic le ex ha us t. As a resu lt, th e co m b us tio n p roducts ca n be co nve r tcd to wa te r a nd carbo n di o xide (o r o the r oxides), p rovided th cy land o n th e gr id o f the ca ta lytic co nve rte r before exiting th e ve hicle's tailpipe . H oweve r, so me n itri c ac id a n d N O :! in evitabl y remai n in a uto mo bi le e xh aust, an d th cy are m ajor co n tr ih u to rs to e nviro n me n ta l pollution in th e fo r m o f aci d rain an d smog. T he brown , ac id ic atmos p heres in hi ghl y co nges te d cities suc h as Beijing , Lo s Angeles, Me xico City, an d H ouston resu lt lar gel y from the e mis sio ns fro m au to mo biles (Fig u rc 8). Such pollution pro blem s have led to str icter e m issio n sta nda rd s for a u to m ob iles , an d a h igh p r io rity ill th e automobil e industry {m o tivate d b y impe ndin g emissio n sta nd ard s o f suc h sta tes as Ca lifo rn ia) is th e d evelopment o f lo w-emissio n or e m issio n-free ve hi cles.
--------------1860 1880 1900 1920 1940 1960 1980 2000
Figure 7 Variation in global mean surface s temperatures for 1850 to 2006 . These are relative to the period 1961-1990.
.t. @
Figure 8 Smog. The brown cloud that hangs over Santiago, Chile contains
nitrogen oxidesemitted by millions of automobilesin that city. Other substance s are also present. such as ozone (03), nitric oxide (N02) , carbon mon-
oxide(CO), and water.
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The Chemistry of Fuels and Energy Resources
Energy in the Future: Choices and Alternatives Fuel Cells In th e ge nera tio n of electri city, th e e nergy d e rived as h eat from co m bustio n of fossil fu els is used to prod uce high-p re ssu re steam , wh ic h sp ins a tu rb in e in a gene rato r. Un fo rt u n ately, not all o f th e e ner b')' fro m co m b ustio n ca ll b e co nverted to usable work . SOllie o f th e eneq,'"y sto re d in th e chem ica l bo n d s of a fu el is tran sfe rred as h eat to th e su rrou n d ing s. Illa kiIlg thi s an inh erentl y in efficient p n >eess. T h e e fficie ncy is ab o ut 35-40% for a coal-fired ste am turbine (an d 50-55 % fo r th e newer n at u ral gas turbines) . A much m o re efficien t p ro cess wou ld be possible if m obile elect ron s, th e ca rr iers of elec tr ici ty, co u ld be ge n e ra te d d irec tly from th e ch e m ical bon d s th em sel ves, ra ther than going th ro ug h an enerb,)' co nversion proce ss Front h eat to mech ani cal wo rk to elec tr icity Fuel ee l! te chll ol o~n' makes d irect co n versio n o f ch em ical po ten tial e n erb,)' to electr icity possibl e. Fuel ce lls are sim ilar to batteri es, ex cept tha t fuel is su p plied fro m a ll ex te rn al SOUlTe (Figu re 9 an d Sectio n 20.3). Th ey ar c more effici en t th an com bustion-ba sed en e r~D' production , with u p to 60% en eq,'y co nve rsio n. Fuel cells ar e not a n ew di scovery, In fac t, th e first fuel cell was demo n strat ed in 1R39 , an d fuel cells are used in the Spa ce Shuttle , Fuel cells arc curren tly under d evelopmen t as well as in usc for ho me s. b usin esses, and auto mob iles. Th e ba sic d esig n of fuel ce lls is quite sim p le. O xid ati on a n d red u ct ion (... page 141 ) tak e p lac e in two se para te co m part me n ts. T h ese co m part me n ts arc co n n ec ted in a way that allows elec tro n s to flow fro m th e o xida tion comp arlm en t to th e red uctio n co m partme nt th ro ugh a co nd u ctor suc h as a wire . In on e co m pa rt m c n t, a fuel is oxiElectrical energy output
; -1 J e
w -
+- fuel
CH30 H(C) + 'I, O, (g) --> CO,(g) + 2 H,O(C)
il ,W = - 727 kJ/ mol-rxn or - 23 kJ/ g CH30H Using e m h alp les o f for mation d ata (Se ctio n 5.8 ) , we can ca lc u late th a t the e n erg y ge n e r a ted is 727 kJ/lIl ol (o r 23 kJ/ g) of liq uid m eth anol. Th a t is eq u ivale n t to 200 \....a rt-h ou rs (\V-h ) of e n e rgy per mol o f meth a n ol ( I \\' = 1 J/ s), or 5 .0 k\V-h p e r lit e r of m e th an o l. Pr ot o types o f ph on es a nd lap top co m p u te rs powered by fue l ce lls have bee n d evel oped re cently. Th e small meth a nol ca rtrid ge s use d to fu el the m are no bi gger th an a sta n dard AA ba tte ry, ye t th ey a re longer las tin g . Man y au to m obi le m an u factu rers arc actively exp lo r ing the use of fuel ce lls th at use h yd ro gen or m e th an ol . Hond a 's FCX (Figu re 10) uses h ydro gen (s to re d in hi ghp ressu re ta nks ) a n d h as a range o f 350 mil es. Th e h ydro-
+-
H'
Unused ANODE
Oxygen from air
d ized, pro d u cin g elec tro ns . Th e electro n s 1lI 00'e through th e conductor 10 th e other co m partment whe re th ev react with an ox id izing age nt, typi call y 0 2' T he ~pontaIleo;ls flo ,.. . of el ectro ns in th e el ectrical circ u it co ns titutes th e el ectric cu rre nt. \\'h ile ele ct ron s flow th ro ugh the ex ternal circu it, io n s m ove b e twe en th e two co mpart m e nts so th a t th e charges in each co m pa rt m e n t re m ai n in bala n ce . T he n e t reactio n is th e oxidatio n o f th e fuel an d th e co nsu m p tion of th e ox id izin g ag en t. Be cause the fuel an d the ox id a n t n ever co m e d ir ectl y in contact with eac h o the r, th ere is n o com bustio n an d minim al loss of enerb,)' as h eal. T h e e n e rgy evolved in th e reactio n is co nve rt ed di rectly in to el ect ricity. l Ivdrogen is th e fuel em p loyed in th e fu el ce lls o n ho ard th e Space Shuttle . T h e overall react ion in th ese fue l ce lls involves the com bi na tion of h yd ro g en a nd o xyge n to fo rm \....ater (Figu re 9 ) . l Ivdro carbo n-based fu els such as me than e (C I I-\ ) a nd methanol (C1130 I I) are also ca n d idates fo r use as the fuel in fu el cells; fo r these COllipo u nds, th e react ion p rod ucts a re CO t an d li tO . Wh en me thano l is used iII fue l cel ls, for exam p le , th e n et reactio n in th e cel l is
PROTON EXCHANGE MEMBRANE
Water
-+
CATHODE
( ~
Figure 9 Hydrogen -oxygen fuel cell. Th e cell uses hydrogen gas, which is converted to hydrogen ions and electrons. The electrons flow through the external circuit and are consumed by the oxygen, which, along with H+ ions, produces water. (H, is oxidized to Wand is the reducing agent. O2 is reduced and is the oxidizing agent.)
..;;;;::::;;;._ _..1
Figure 10 A hydrogen fuel cell passenger car from Honda . The car is powered by a fuel cell using hydrogen and oxygen. The hydrogen is stored in a 171-l, high-pressure (350 atm) tank. It is scheduled for limited sale in Japan and the U. S. in 2008.
105 . Meth an e , C H 4 , ca n be converte d to me tha nol whic h, like et han ol , ca n be u sed as a fu el. Th e e nerb~' level d iagram show n here pr ese nt s relatio ns h ips between energies of th e fuels an d th ei r o xidation p rod ucts. Use th e info rmati on in th e di agram to answe r th e fo llowing qu est io ns. (T he e nergy te rm s arc pe r m ol-rx n. )
(a) Wh ich fu el , m eth anol or m ethan e , yiel ds th e mos t energy pe r mole when burned? (h) Wh ich fu el yields th e m os t en ergy per g ra m when burn ed ? (c) Wh at is th e en th al py cha nge fo r th e co nversio n of m ethan e to m eth an ol ? (d) Eac h a rrow on th e di ag ram represents a che m ica l reac tion. Write the e q uation fo r th e reac tion th at co nverts m etha ne to me than o l. 106. Calcula te Do,Jrfor th e re acti o n
2 C(s) + 3 Il, (g) + Y, O, (g) --. C, Il, O Il(f) giwn th e in formation helmv'. C(s)
+
2 H, (g)
O , (g) ---> CO ,(g)
+
IJ.,Fr
~
- 393..>kJ/ mol-rxn
O , (g) ---> 2 H,O (f ) Dorf f' = - 57 1.6 kJ/ m o l-rxn
C, Il , O Il(f)
+ 3 O, (g) ---> 2 CO,(g) + 3 1l,O(f) tx.lT
=-
1367.5 kJ / m o l-rx n
107 . You h ave th e six piece s of me ta l listed below, plus a be aker of wate r con tai ni ng 3.00 X lO2 g of wa te r. T he wa te r temperatu re is 2 1.00 °C. Spedfic Heat (Jig K)
Mass (g)
1. Al
0.9002
100.0
2. Al
0.9002
50.0
Metals
3. Au
0.1289
100.0
4. Au
0.1289
50.0
5. Zn
0.3860
100.0
6. Zn
0.3860
50.0
.4 more challenging
• i n OWL BLue-numbered questions answeredin Appendix 0
(a) In yo u r first ex perime n t yo u selec t on e p ie ce of metal an d heat it to 100 °C, a nd th e n selec t a seco nd pi ec e of m etal and coo l it to - 10 "C. Both p iece s of m etal are th e n pl aced in th e beaker of water and th e tempe ratu re s eq u ilib ra ted. Y ou wan t to select two pi e ce s of metal to u sc , suc h that th e final temperature o f th e water is as h igh as possihl e. Wh at pie ce of m et al will you h eat? Wh a t pie ce of metal will yo u coo l? What is th e fin al te mpe ratu re of th e water? (h) T he sec on d ex peri me n t is d o ne in th e same way as th e first. H oweve r, yo ur goa l no w is to ca use th e tempe rature to change th e lea st, th at is, the fin al temp eratu re sho ul d be be as n ea r to 2 1.00 °C as po ssible . Wh at piece of metal will yo u h eat ? What pi ece of me ta l will yo u coo l? wh at is th e fin al te mpe ratu re of th e water? 108. In lab , yo u p lan to carry o u t a ca lorimetry ex peri me n t to d etermi ne th e DorH for th e exo the rmic reac tion o f Ca (O Hh (s) an d HCI (aq ) . Predict how each of th e fol lowing will a ffec t th e calc u late d valu e o f tx.H, (T he value calculate d fo r Do,J/ for th is react io n is a negative val ue so choose your answer from th e following: DorH will he too low [tha t is, a large r n egati ve val ue ]' Do.,H will he unaffected, Do rH will be to o h igh [tha t is, a smaller negati ve valuc. j ) (a) \ {111 spi ll a litt le bit of th e Ca( O Hh o n the benc htop befo re adding it to th e ca lorime ter. (1)) Be ca use of a m iscalculatio n , yo u add an excess of Hel to th e measu red am ount of Ca( O Hh in the calori m e ter. (h) Ca( O H h readily a bsorbs wate r fro m th e air. T he Ca( O Hh sa m ple you weighed h ad bee n ex posed to th e ai r prio r to wcighi ng an d had a bso rbed so me wa ter. (c) Afte r weighi ng o u t Ca( O Hh, th e sam ple sat in an open be ake r an d ab so rb ed wa ter. (11 ) You del ay too lo ng in rec ording th e fin al te mpe ra tu re. (e) Th e ins ula tion in yo ur co ffee cu p ca lori me ter was poor and so so me energy as heat was lo st to th e surrou ndi ngs du ring th e ex perimen t. (e) \ {lU have igno red th e fact th at energy as heat a lso raised th e te mpe ratu re of th e stirrer an d the th erm omete r in your system.
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The Chemistry of Fuels and Energy Resources
4kg
Mg zNiH 4
laNi,Hc
liquefied hydrogen (beLow - 250 °C)
MetaL hydrides
Pressurized hydrogen gas (at 200 bar)
Figure 12 Comparison of the volumes required to store 4 kg of hydrogen relative to the size of a typical car. (l. Schlapbach and A. Ztittel: Na ture, Vol. 414, pp. 353-358, 2001.)
sto res h ydroge n fo r its fuel cell at h igh pressu re (350 au u) in a 171-L tank (Figu re 10 ). T his is larger th an the gaso line tan ks fou nd in mo st au to mo bi les, so ot he r sto rag e m eth od s th at have smaller volu mes and ye t ar e safe are so ug h t. One po ssibi lity fo r h yd ro ge n sto rag e rel ie s o n the fac t that ce rta in m et als will absorb h yd rog e n reve rsibly (Figu re l ~{ ) . \\'h en a meta l absorbs hyd rog en , II a to m s fill the h ol es, ca lle d interstices, betw een m e tal atom s in a m e tall ic crys tal la ttice. Pallad iu m , fo r exam ple , will abso rb up to 9~ti) tim es its volu m e of hyd ro ge n . T h is hyd roge n can he r eleased u po n heati n g , an d th e p ro cess of abso rp tio n and rele ase call be re pe at e d . No m at ter how hyd rogen is used , it has to he deliver ed to vehicles and homes in a sate and p ractical mann er. Work h as also bee n d one in this are a , but m an y problem s re main to he so lved. European research e rs have fou nd tha t a tan ke r tr uck that ca n d e liver 2'100 kg of com p ressed na tu ral gas (mostly m ethan e ) ca n del ive r o n ly 288 kg o f H ~ at the same Hz gas
Metal hydride
Electrolyte
I
Metaladsorbed ~ hydrogen .~ Solid solution « -phase . .
4\
Hydride phase {3-phase . .
Figure 13 Hydrogen absorbed by a metal or metal alloy. Many metalsand metal alloys reversiblyabsorb large quantitiesof hydrogen. On the left side of the diagram, Hz molecules are adsorbed onto the surface of a metal.The H2 molecules can dissociateinto Hatoms, which form a solid soLution with the metal (a-phase). Under higherhydrogen pressures, a true hydride forms in which Hatoms become H- ions (,B-phase). On the rightside, Hatoms can also be adsorbed from solution if the metalis used as an electrode in an electrochemical device.
~
~
e
Figure 14 Iceland, a "carbon-free:' hydrogen-based economy. A geothermal field in IceLand. The country plans to use such renewable resources to produce hydrogen from water and then to use the hydrogen to produce etectricityin fuel cells.
p ressu re . Although h yd ro gen ox idatio n d eli ve rs abo u t 2.4 tim es more ene rgy pe r gr am (1 19.95 kJ / g ) th an m ethan e ,
CH,(g)
+ 2 O,(g ) ---> CO,( g) + 2 H, O(g)
fi ,W = - 802.30 kJ/mol rxn or - 50.14 kJ/ g CH, th e tanke r can cany abou t eigh t times m o re m eth an e tha n II ~ . That is, it will take m o re tan ker tr ucks to deli ver the hyd roge n needed to p m.. .er the same n um ber of cars or ho m es ru nning Oil hydrogen th an those run ning o n m eth an e . Th ere is au int e resting exam ple in \.. . hi ch the h ydroge n eco nolllY has ga ine d a rea l toe ho ld. Icel and has an no u nced th at th e co u ntry will become a "carbo n-free eco no my..' Icelande rs plan to rely o n h vdroge u-powc rc d ele ct ric fuel cel ls to r un ve h icles and fish ing bo ats. Icel and is fortunate in that 1:\\ '0 thirds o f its e neq.,J]· and all o f its e lec tr icity already co m e fro m re newable so u rces-hydroelec tric a nd geoth e rma l enerb'T (Figu re 14). T he co u n try has d ecid ed to use th e elect rici ty p ro duced hy geo the rm al hea t o r h yd ro elect ric power to se pa ra te water into hydroge n and oxyge n. T h e hydrogen will th e n he used in fuel ce lls o r co m bine d with CO 2 to m ak e m ethanol , CI I 30 H , a liq uid fuel that ca n e ither be b u rn ed o r he u sed in diffe ren t types o f fuel cells.
Biosources of Energy Biofu el s now su p ply abo u t 1% of th e fuel used wo rldwide fo r tr ansportatio n, bu t so me p roj ect th at it lIlay co n trib u te 30% to U.S. tra ns po rtatio n needs by 2030 . Gaso line so ld to d ay o ften co n tains et hanol, C~ l l:-, O l I. In ad d itio n to be illg a fuel, ethanol imp ro ves th e burn in g chara cte ristics o f gaso line . Eve ry state in the U.S. now has ava ilable a ble nd o f at least 10% et han o l and 90% gaso lin e. (See "Case Study: T he Fuel Co urrovcrsy: Ak o ho l an d Gasol ine ," page 240, and the q uestions o n e tha no l o n page 86 0.) Ethanol is readil y mad e by ferm en tatio n o f g lucose fro m renewa ble resources suc h as corn, suga r cane, o r ag ricultural residues. Whil e it Illay not eme rge as th e so le fuel of th e
Energy in the Future : Choices and Alternatives
future, th is m ateri al is likel y to co n trib u te to th e p has ing-o ut process of fossil fuels and Illay be o ne of the fuel sources in th e fu ture. Whil e th e U.S. is ste pp ing up its program to produce m ore ethanol fro m corn o r o ther plant ma tter, Bra zil has m ade the prod uctio n of et ha no l fro m sugar cane a top prio rity. About ,10 % of its m oto r fuel is ethan ol. l\lo st new cars sold in Brazil are "flex-fuel " cars th at run o n eithe r gaso line o r ethano l. T he mo st co nu no n fuel used in suc h cars co ns ists o f85 % e th ano l and 15 % pe trol eum-based fuels and is labe led E85. Th e U.S. a nd Brazil p roduce 70% o f the world's et ha nol, with the U.S. havin g m oved into the to p po sition in 2000 . Whil e et han o l is cur re ntly th e p redo minan t hi o fuel , hi o die sel m ak e s up alm os t 80% of Euro pe's total bi ofuel pro duction . This co m es fr om su n flower see ds , rapeseed , soybea ns , o r used cooking o il. Biodiesel is a mi xtu re of meth yl es ters o f organ ic aci ds, fo rmed fro m var ious plantd e rived o ils (... page 479 ). There a re seve ra l poi nts to he m ade about th e use o f e tha no l as a fue l. Gree n plants use th e su n 's energy to m ak e biomass fro m COt and I-I ~O by ph oto syn th e sis. T he su n is a renewable resource , as, in p ri ncipl e , is the e th ano l d e rived from h io m ass. III ad d itio n , th e process recycles C O ~ . Plan ts u se CO :z to create hi o mass, wh ich is in turn used to m ak e et ha no l. In th e fina l ste p in thi s cycle , o xidatio n of ethanol retu rn s C O ~ to th e a tm osp here. O ne serious issue co nc e rn ing th e use o f co rn -de r ived e th ano l is th e net e ne rgy bala nce . G II C has to co nsi de r the e ncrbTY ex pe nded in th e fuel to run the tr act ors an d tr ucks, harvest th e co r n, m ake th e fertili zer, amo ng o the r th ings , ve rsus th e ene rgy ava ilable in th e ethanol produced as the end product. Re ce nt analyses an d im p ro vem e nts in co rn- to e tha no l pr eparat ion se e m to ind icat e more e ner!:.,'} is availab le than is used in p ro d ucti on , bu t no t by much . Whil e p r odu cti on o f et hano l fro m co r n has been increasing at 20-2.1") % per ye ar, e ne rgy a nalysts believe that no n-fo od plants that ca n g row on marg ina l lands with a minim al in p ut of fe rtil ize rs a re th e be st hope for bi ofuels. To re-e ngineer suc h cellu lose pl an ts as gr asses or trees will require a lo t of ch em ical an d hi ol o gical researc h . Recent researc h 0 11 ethan ol has take n this to pic in a new direction . Na mely, e thanol can he used as a source of hydroge n. It has been possible to create hydr ogen gas fro m ethanol by using a steam refonnl ug process like th e m e th an erel ated p ro cess. Th e recently d evelo ped m ethod. involves th e par tial ox idation of ethanol mixe d with water in a small fuel injector, like those use d in cars to d eliver gasoline, along with rhodium and ce r iu m ca talysts to cr eate hyd roge n gas exotherm ically (Figu re 15). Th e net reaction is
C2HsOH(g) + 2 H20(g) + 'I, 02(g) --> 2 C0 2(g) + 5 H2(g) Th e sta nd ard en thalpy o f th is re acti o n is approx ima tely - 70 kJ / mo l-rxn (or a bo u t 1.5 kJ / g o f et hanol) . For fu rther in sight in to thi s p rocess, let us analyze th e ove ra ll ene rgy cycle , starting with the ph o tosynth etic co m hi-
I
265
Figure 15 Hydrogen from ethanol. Ethanol can be obtained byfermentation of corn. In a prototype reactor (right), ethanol, water, and oxygen are converted by a catalyst (glowi ng white solid) to hydrogen (and CO 2) ,
+ 20 kJjmo L
t
(,HI20, + 4 H20 (+6 02)
- 140 kJjmo L
- 24 20 kJjmo L
Figure 16 An energy-level diagram for the reactions leading from the production of biomass (glucose) to CO 2 and " 2' (Based on a Figure in G. A. DeLuga , J . R. Salqe. L. D. Schmidt, and X. E. Verykios: Science, Vol. 303, pp. 942 and 993, 2004.)
natio n of CO:! and water to gene rdte g lll7 mp h ) . Ho w fast must th e ba ll travel to h ave a waveleng th of 5. 6 X 10- 3 nm ?
26. • A rifle b ulle t (m ass = 1.50 g) h as a vel oci ty of 7.00 X lO2 mp h (m iles per h our ) . Wh at is the wavelengt h associ a ted with thi s bulle t?
27 . (a) Vv11 en TI = 4, what are the po ssibl e valu es of (? (b) W hen (' is 2, wha t a rc th e possible values of JIl(? (c) Fo r a 45 orbi tal . wha t arc the possible values of II , f, and 111(? (d ) Fo r a 4/ orbita l. wha t a re th e po ssib le va lu es of n. l', and m(? A more chaLLenging • in OWL
BLue-numbered questionsanswered in Appendix0
28• • (a ) Wh en 11 = 4, f = 2, a n d 11/( = - I , to wha t o rb ita l typ e d o es th is re fer? (Ciw th e orb ita l lab el . suc h as lx) (h) I low ma n y orb itals o ccur ill th e 11 = S el e ctro n sh e ll? H ow m a ny subshell s? What a re the le tter labe ls o f th e subshclls? (c) If a su bshc ll is la bel ed f, how many orbita ls oc cu r in the su bsh ell? Wh at an' th e value s of III(? 29. A possib le ex cited sta le of th e II a tom h as th e e le ctron in a 4 !J o r bi tal. List a ll possib le se ts of qua m u m n u mhe rs n, t, a n d 1n( for th is el ectron . 30. • A possible excited s ta le fo r th e II a to m h as an elect ro n in a Sri orb ital. List all po ssib le se ts of q u antum nu mbers n, f , a n d 1n( for th is el ect ro n . 3 1. H o w ma ny suhshc lls occ ur in th e elect ron sh el l with th e p rin ci p al q u a n tu m n um ber 11 = 4? 32. • H ow m a n y su bsh cl ls o ccu r in the el ectro n shell with th e pri n cip al q u a n tu m n um ber 1/ = 5?
33 . Ex p la in b riefly why eac h o f th e following is n ot a p ossible se t o f qu an tu m n u mbe rs for a n e le ct ro n in a n a to m . (a) 1/ 2,( 2, m ( 0 ~,f 0 , UIf (h) 1/ -2 6, f 0, mf (c) 1/ 1 ~
~
~
~
~
~
~
~
34• • Whi ch o f th e fo llowing re presen t va lid se ts of qua nn un n um bers? For a se t th at is invalid , expl ai n bri efl y why it is n o t co r rec t. - I (a) 1/ = 3, f = 3, 1fI / 0 (c) 11 = 6, f = 5, '!It -4 (h) 1/ ~ 2, f ~ I, "" ~ 0 (d ) 11 = 4 , f = 3, 1fI ( 35, • Wha t is the m ax im u m nu mbe r o f o r bi ta ls th a t can b e id e ntifie d by eac h o f th e fo llo win g se ts of qua nt u m n umbers? Wh en "n o n e" is th e correct an swe r, ex p la in yo u r reaso ning . (a) 11 = 3 , ( = O, 11I( = + I (c) 1/ ~ 7, f ~ 5 (h) 1/ ~ 5, f ~ 1 (d ) 11 = 4, f = 2, tnf = - 2
36, • Wh a t is the m aximum num ber o f orb ital s th at can b e id entifie d by eac h o f th e fol lowin g se ts o f qua nt u m n u mhe rs? Wh e n "n o n e " is th e co r re ct answer, ex plain yo u r reaso n in g . (a) 1/ ~ 4, f ~ ~ (c) 11 = 2, f = 2 (h) 1/ ~ 5 ( d )I/ ~~, f ~ l , m ( = - 1
37, • Explai n h ric fl v why eac h of th e fo llowing is n ot a possibl e se t o f q uan tu m n u m be rs fo r a n el ectron ill a n a to m . In e ac h case , ch a n ge th e in co rrec t val u e (o r va lu es) to m a ke th e se t valid . (a) 11 = 4, f = 2, lII( = 0 , 1fI, = 0 (b)II= 3, (= 1, 1IIf (c)
11
= 3 , ( = 3,
m (
... more challenging • in OWL
- 3, m, =- Y2 - 1, 11/" =
+'12
Blue-numbered questionsanswered in Appendix 0
38. Ex p la in bri efl y wh y each o f th e foll o wing is n ot a p o ssibl c se t of q ua n tu m n u m b ers fill' a n electron in a n a to m . In e ach case , cha nge th e in co rrect va lu e (o r valu es) to m ake the se t va lid . (a) 1/ = 2, ( = 2,
lII,
= 0, ~
- I,
~
+ 2, m,
(h)
11
~
2,
c
~
I,
tn,
(c)
Il
~
~,
f
~
I,
m(
11/,
= 111,
+ '12 ~
~
0
+ \;2
39. St ate which o f the fo llowin g o rbita ls ca n not ex ist accord ing to th e q u a nt u m theo r y: 2J, 'J.d, 3/J, 3J, 4) ; and SJ. Br iefly ex p la in yo u r a nS' HT S. 40 . • Sta te wh ich o f th e foll owin g a re in correct desig n atio ns fo r o rb itals acco rd ing to th e q ua n tum th eory: 3p, 4s, 2f , and I p. Br iefl y ex p la in yo u r a nswe rs. 41 . Write a co m p le te se t of quan tum n um b e rs (n, t, a n d 1n( ) th at q u antu m th eory a llows for eac h of th e fo llowin g o r bi tals : (a) 2/), (b ) 3d, and (c) 1[,
42. • Writ e a com ple te set of q uan tu m n u m b ers (1/, f, an d !1If ) for eac h o f th e fo llow ing orbi tal s: (a) Sf, (h) 4d, a n d (c) 2J. 43. • A p a rti cula r o rb ital h as 11 = 4 a n d ( = 2. Wh at must th is orhi tal he : (a) 3/), (h ) 4/), (c) 5d, o r (d ) 4d? 44. • A g ive n o rb ital h as a magn eti c qua n tu m num be r o f !1If = - 1. Th is coul d not he a (an ) (c) !J o r bi ta l (a) f orb ital (h) d o rb ital (d ) J o rb ital
45. I lo w m all Y pl an ar n o d es a rt' associ a te d with each of th e fo llowing orb ita ls? (a) 2" (h) 5d (c) 5f 46. • H ow many pla na r nodes a rc asso ciate d with eac h of th e following a to m ic orbi tals? (a) 4f (h) 2/) (c ) 6"
General Questions on Atomic Structure T ht'JP questions
(ll"('
not dt'Jig1/atnl as to tJjJf or
dutpter: T he)' ma), combine seueml
{Of al ion
in the
COII U ! Jt.\.
47 . • W hic h of th e foll owing are ap p lica b le wh e n ex pl ain ing the ph otoelec tri c effe ct ? COTTec t any statem e nt s th at arc wro n g. (a) Light is el ec tromag netic radiat io n . (h) T h e inten sity of a light beam is re lat ed 10 its frcque n cy. (c ) Light ca n h e tho ug h t o f as co n sistin g o f rna.. .sk-ss part icle s wh os e energy is g ive n by Pla n ck's eq ua tio n , E = hv 48 . In wh a t region o f th e e lect ro mag ne tic sp e ctru m fo r h ydrogen is th e Lyma n se ries o f lin es fo u n d ? T he Balmer se ries>
I
299
49 . Give th e number o f n o d al surfaces th rou gh th e n u cle us (p la n ar no de s) fo r each orbi ta l type : s, I), d, and ;: 50. What is th e maxi mu m n umber of s orbitals fou nd in a give n e le ctro n sh e ll? Th e m aximu m n um be r o f I) orbita ls? Of rl orbitals? Of f orb ita ls? 5 1. • Match th e values of f sh own in th e tahlc wit h orbital typ e (J, I), d, or f).
c Value
Orbital Typ e
3
o 1 2 5 2. Sketch a p ictu re o f the 90 % boundary surface of an s orbital a nd the I):.; orbital. Be su re th e latt e r drawing sh ows why th e I) orbital is lab el ed pxa n d n o t Pol' for exam p le. 53 . • Comple te the foll o win g tab le .
O rbital Type
Number of Orbitals in a Given Subsh ell
Number o f Nodal Surfa ces through the Nucleus
I' d
} 54. Exci ted H atom s h ave m an y emission lines. One serie s o f lines. called th e Pfu n d series. o ccurs ill the in frare d region . IL re su lts when a n electron changes fro m h igh e r energy levels to a le vel with n = 5. Calcula te th e wavele n gt h ami fre q uency of th e lowe st energy lin e of th is se rie s. 55 . An ad ve rt isin g sign give s off red light a nd g ree n lig ht. (a) Wh ich light h as th e high e r-en e rgy p hoton s? (h ) One o f th e colors h as a wavele ngth o f 680 n m, and til e ot her has a wavele-ngth o f 500 nm . Wh ich co lo r lias wh ich wave lengt h ? (c) \Vhich light h as th e h igh e r trcqueucy? 56 . • Rad iat io n in the u ltraviole t re gion of the electromagne tic sp ectru m is qu ite energetic. It is th is ra diation th at ca us es d yes to fa de ami yo ur skin to d evel op a sunbu rn . If yo u a re bo mbarded wit h I .OO m o l o f ph o to n s with a wavelen g th of :n 5 11 m , wh at am o u n t of encrgy, in kilojou les pe r mol e of p hotons , a re yo u bein g su bjected to ? 57 . • A ce ll ph o n e sends sig n a ls a t abo u t 850 MHz (l MHz = I X IOh Hz or cycles per sec o n d). (a ) What is th e wave le ngth o f th is rad iatio n ? (b ) What is th e energy of 1.0 mol o f ph ot on s with a freque n cy o f 850 Ml l z? (c) Co mpare th e e n ergy in part (h) wit h th e energy of a m ol e o f p ho tons o f hluc ligh t (420 11 1Tl) . (d) Comrru-ru on th e difference ill en ergy between 850 MH z radiation an d bl u e light.
300
I
58 . • Assume yo ur eyes rece ive a sig n al con sisting of b lu e lig h t, A = 470 mn . T ill' e n e rgy o f th e sign al is 2.50 X 1O-H J. I-low man y ph o to n s reach yo u r eyes? 59 . If su ffi cie n t en eq..,,:· is absorbed by a n atom , an elec tron can be lo st by th e ato m a nd a positive io n fo rm e d . T he am o un t of e n e rgy required is ca lled th e ioniza tion ene rgy. In t he H a tom , th e ioniza tion energy is th at req ui red to cha n ge th e e lec tron from n = I to n = infinity. (Sec Ex erc ise 6.5 , p age 28 0.) Calcula te the ioni zatio n e n ergy fo r th e Hc " ion , Is the io n iza tion energy of th e He " more or les s th a n th at of II ? (Ro h r 's theory a p p lie s to He " becau se it, like th e H atom , h as a single electro n. Th e elec tro n eneq..,'} , h o weve r, is now given hy E = - Z2Rlu:! n'2, where II. is th e a tomi c n um ber of h elium .) 60 . • Su p pose h yd ro ge n a tom s absorb ellngy so that electron s are e xcited to th e 11 = 7 e n e rgy leve l, Electron s th e n u n dergo th ese tran sit ion s, among oth e rs: (a) 1/ = 7 ---71/ = 1; (h) 1/ = 7 ---7 11 = 6; and (c) 11 = 2 ---7 II = I . Whic h transition p ro duces a p hoto n with (i) th e sm allest e n ergy, (ii) tile hi gh est freq uenc y, a n d (iii ) the sh o rt es t waveleng th ? 61. Ran k the fol lo win g o rbita ls in the II a to m in order o f increasin g e n e rgy: :h, 2J, 2jJ, 4.~ , :\ jJ, I J, an d :id. 6 2. • H ow m a ny o rbi ta ls ing designatio ns? (a) 31' (d) (b ) 41' (e ) (e) 41~ (I)
correspon d to eac h of the fo llow-
lid 5d
(g ) n ~ 5 (b) 7"
5}
63 . • Co hah-titl is a radioactive iso to pe used in m ed icin e fo r the tr ea tm e n t of certain cancers. It prod uces {3 particl es and y ra ys, the latte r h avin g en erg ies of 1.17 3 a n d 1.:332 McV. ( I t"'fc\' = I (yi elec tron-vo lts and 1 ('\' = 9.6485 X lO ' J / mol. ) w hat art' the wavele n gt h and freque nc y of a y-r ay p hoton with an energy of 1.173 MeV? 64. ... . Exp os u re to h igh doses o f mi crowaves can cause tissu e damage . Est imate h ow many p ho ton s, with A = 12 ern , m ust he abso rbed to ra ise th e te m p e rature of yo u r eye by 3 .0 °C . Assu me th e mass o f an eye is 11 g a n d its sp ec ific h e a t ca p acity is 4.0J /g . K. 65 . When th e Sojourner sp ac ecraft lan de d 011 Mars in 19!)7, th e pla net was approxi ma tel y 7.8 X lOi km from Earth. H ow lo n g di d it ta ke for th e televi sion pic ture sig n a l to reac h Ea rth fro m Ma rs? 66 . • Th e most p ro min e nt lin e in the em issio n spectru m o f chrom iu m is found at 425 .4 n m . Othe r lin es in th e chro mi um sp ect ru m are fo u n d at :3:)7 .9 nm , 359.3 nm , 3GO.5 nm , 427.5 nm, 429.0 nm. a n d 520.8 urn . (a) Wh ich of these li n e s represen ts th e most energe tic ligh t? (h) W hat color is light of waveleng th 425.4 nm ?
... more chaLLenging
•
in OWL
BLue-numbered questions answered in Appendi x 0
67 . Answe r the fo llowi ng q uestio ns as a su m mary qu iz on til l' chap ter. (a ) Th e q ua n tum numbe r II des cr ibes th e of an a to mic orbi tal. (b) Th e sha pe of an atom ic o rbital is g- iven by th e quantu m num be r ' (less or more ) (c) A ph ot on of g-reen light has en ergy th an a phot on of o ra nge light. (d ) T he ma xi mu m num be r of orbita ls th at m ay be associa ted with the set of q ua n tu m nu mbers 11 = 4 and f = :) is (e) Th e maxim um number of orbi ta ls th at m ay be associa ted with th e quant u m num b er set n = S, f = 2, a nd m( = - 2 is ' (f) • Lab el eac h of th e fo llowing orbita l p ictu re s with the a ppro pria te lctu-r:
(h) Whi ch of th e foll o win g is n o t a valid se t of q uantum numbers? II me m, 3 2 1 _1-'2 2 I 2 +'1, 4 3 0 () 0 ) What is th e m aximu m number of orbi tals t hat ca n he associated with each of th e fo llowing se ts o f quantu m numbe rs? (O ne po ssible answer is "no ne ,") (i) ,, ~2 an d ~ (ii) 11 = 3 (iii ) 1/ = 3 a nd f = 3 (iv) 11 = 2, t = 1, a nd 1n( = 0 (j) A CU'H ion h as o ne unp air ed ele ctro n, Is a samp le of C uBr2 pa ram ag n eti c o r di am agn e tic?
e
e
69 . Th e di ag ra m s below re prese n t a sma ll se ctio n of a solid. Each circle re prese nts all a to m , an d all arrow re p resent'i all el ectr on .
(g) Wh en II = 5, th e possible values of ( are ' (h ) Th e n um ber of orbi ta ls in the n = 4 she-ll is _ _' 0) A Co 'H io n has th re e u npaired elec trons. A sa m ple of CoC I2 is ( param ag ne tic) (d iamag-netic), 68 . Answe r th e fo llowi ng questions as a sum mary quiz 0 11 th is cha p ter. of an (a) The quan tum n umbe r II de scr ibe s the a to m ic orbital , an d th e q uantum n u m ber f describes ito; (b ) Wh en II = 3, th e possib le val ues of f arc ' (c) Wh at typ e of orbital corres po nds to ( = 3? _ _ (d ) Fo r a 4 d orbit al. th e value of n is ' the val ue ' o f ( is , a nd a po ssible valu e of m~ is (c ) Eac h of the f(IIl O\.... ing drawings rep re se nt s a type of ato m ic orbi ta l. Give th e le tter designa tio n for th e o rbita l; give its value of t, an d speci fy th e n u mber of nodal surfaces .
Letter = f value = Plan ar nod es = (f ) An a to mic orb ital with th rel' nodal surfaces th ro u gh the nucleus is _ _' (g) Whi ch of th e followin g orb ital s ca n no t ex is t ac3f, co rd ing to m od ern qu antu m t he ory: 2s, 3/1, 51'. 61)?
(b)
(a>
(c)
(a) Wh ich repre sents a di am ag ne tic soli d , wh ich a paramagnet ic solid , an d wh ich a h..rrom agn et ic so lid? (b) Wh ich is m os t strong ly attracted to a m ag ne tic fiel d ? Wh ich is leas t strong ly au ra cu-d?
In the Laboratory 70. A so lu tion of K~1 nOl abso rbs ligh t a t 54 0 nm (page 192) , Wh at is the frequen cy of till' light abso rb edwh a t is the ene rgy of o ne m ole of ph ot on s wit h A = 540 n m? 71. A la rge pi ck le is a ttac he d to two ele ctrodes, wh ich a rc th en attached to a 110-V pow er supply (sec th e pr oble m on Scr e e n 6.7 of C hcmistryblow) . As the vo ltage is increased across th e pickle, it beg ins to glow with a ye llow co lor, Kn owin g th at p ickles are m ad e by soaking th e vege tab le in a co nce n tra te d sa lt so lu tio n , descr ibe why the pickle m ight e mi t ligh t when eh-ctrical eu ergv is added .
u:
.... morechallenging • in OWL
Blue-numbered questionsanswered in Appendix 0
The "electric pickle."
I
301
72 . • The spectrum shown here is fo r aspiri n . T he vert ical axis is the amo un t of light absorbed, and the ho r-izontal axis is th e wave le ng th of in cid ent light (in nm ) .
3.5 w v
c
•
D
~
3.0 2.S
D
-e: 2.0
(a ) O ne p o int o n th e ho rizon tal ax is is ma rked as 20 00 cm - ]. Wh at is th e wave le ngt h of lig h t a t thi s po int ? (h) Wh ich is th e low ene rgy c u d of th is spectru m (left or righ t) , a nd whi ch is the h igh e nergy en d ? (c) The broad absorp tion a t about 3300- ~HOO cm " ! indi cat es th at infrare d rad iatio n is in te rac ting- with th e 0 11 grou p o f th e m eth anol m ole cu le . T he narrower ab sorp tions a ro und 280 0-30 00 ern -] are for inte racti on s with C- J-I bonds. Wh ich inte raction req uires more elle q:,'J', with 0 -11 or with C- H?
1.5
Summary and Conceptual Questions 220
230
240
250
260
270
280
290
300
310
Wavelength (nm)
74 . Bo hr pi ctured th e elec tro ns of the atom as he ing loca te d in definite orbi ts abou t the nu cleu s, j ust a s th e plan ets orbit th e sun . Criticize th is m od el.
Wh at is th e fre q uency of light with a wave le ngt h of 278 nm ? W hat is the ('n ergy of on e m ole of photon s with A = 278 nm ? Wh at regi on o f th e e lectromagnetic spe ctru m is cove re d by the spectru m above? Kn o wing th a t aspirin o n ly absorbs light in th e region de pic ted by thi s spec tru m , wha t is th e co lor of asp ir in?
73. • Th e infrar ed spectru m for me than o l, C II :l)[-I , is illustrated below. It shovv's the a mo unt of light in th e in frared regi on th at me thanol tra n sm its as a fu nct ion of wavelen gth , T he ver tical axis is the amount of ligh t transmitted . At poin ts near th e top of th e g ra p h, mo st of th e inci den t light is be ing tra nsmi tted by the sam ple (or, conve rsely, little ligh t is absorbcd. ) Th e ref o re , th e "peaks" o r "ha nds" th at descen d fro m th e top in d ica te ligh t absorbe d ; th e lo nge r th e band , th e mo re light is bei ng absorbed (or, conversely, the less is bein g tra nsmi tted ) . T he horizontal scale is in u n its of "v...'avcn um h crs," abbreviated em - I . Th e e ne rgy of light is given by Planck's law as E = he/A ; that is, E is proportion al to l / A, Therefore , th e horizontal scale is in units of I/ A an d reflects the e ne rgy of the ligh t incident o n th e sam ple .
0.8
.~ ~
c
S
0.6 0.4 0.2 3000.
2000.
Wavenumber (cm- r)
302
I
75. Lig h t is given off by a sodiu m- or merc ury-con taini ng stre e tlig h t whe n th e a to m s are e xci te d . The lig-h t yo u sec arises for whic h of th e foll ow.ing reasons? (a ) Elec tron s are mov ing from a g iven e nerg y level to o ne of h igh er e ne rgy. (b ) Electro ns are be in g rem ove d fro m th e ato m , the reby creating a metal ca tion . (c) Elec tro ns are moving fro m a g iven en ergy level to one of lowe r ene rgy. 76 . H ow d o we inte rp ret th e ph ysical m e aning of th e sq u a re o f th e wave fun ctio n? Wh at a re th e u nits of 4 1Tr 'tIj/ ;/. ? 77 . Wh al d ocs "wave- p art icle dualit y" me a n ? Wh at are its impl icati on s in o ur modern view of a tom ic struct ure? 78 . • Whi ch of th e se ar e o bserv abl e ? (a ) position of a n elec tron in a n I I a tom (h) fre q uency of radia tio n e mi tted hy H at om s (c) path o f a n e lec tr on in a n I I a to m (d) wave mo tion of el e ct ro ns (e) d iffract ion pa ttern s produced by e lectrons (I) diffract io n pa tte rn s p ro duce d by light (g ) en eq.,y ' re quire d to re m ove electro ns from II ato m s (h) a n at om (i) a m ol ec ul e a wa ter wave
m
w v
c
Th e f ollowing qu estions use lXnu ryJt.\" from this a nd tnroious dlfljJII'rS.
1000.
79. In p rin ciple , wh ich o f th e fo llow ing can he de term ined ? (a ) the l' lH'l'gy of an elect ro n in th e I I a tom with high pre cision and accura cy (h) th e positio n of a hi gh -spee d e lec tron with hi gh precisio n a nd ac curacy (c) a t the same lim e , bo th the p ositio n an d th e en ergy of a hi gh -sp eed elec tron wit h hi gh pre cision an d accuraq
... more chaLLenging • in OWL
Blue-numbered questions answered in Appendix 0
80. ... . Su ppose you live in a d ifferen t universe whe re a differe nt se t of q u an tum num bers is re qu ir ed to d escribe th e atoms o f that u n ive rse . Th e se q ua n tum uumh e rs h ave the foll owin g rules:
»; p rin cipal L, orbi ta l AI, m ag netic
1, 2, 3, . . . , = N - 1, 0, + 1
YO
How man y orbi tals ar e th e re a ltoge ther in th e fi rst three ele ct ron sh el ls? 8 1. A p h o to n with a waveleng th of 93.8 nm strikes a hydrog'ell a to m , a n d liglu is emi tted by th e a to m . H ow man y emission lines wo uld be observed? At what wavek n g th s? Ex plai n bri efly (se c Figure Ii.I tl}.
82. Ex pl a in why yo u co uld or co uld n ot measu re th e wavele ngth o f a gol f ba ll in flig ht.
83. • T he rad ioac tive elemen t te chn etiu m is n o t foun d na turally o n earth ; it m us t he syn th es ize d in th e labo ralo ry. It is a valuab le element, h o wever, because it h a s medical uses . Fo r exam ple, the cl emen t in th e fo r m o f sodium p e rt cchn ctate (NaTco.j ) is used in imaging stu d ies of th e b rain , th yro id , a nd sa liva ry g lands a nd in re n al b lood flow stu d ie s, among o t her th in gs. (a) In what group and period o f th e periodic table is the el em ent foun d ? (b) T he valenc e ele ctrons of te chn etiu m are foun d in th e 5J a n d 4d su bs h c lls. Wh at is a set o f qu antu m num bers ( 11, f, an d 11l() fo r o ne of th e electrons o f the 5J su bshcl l? (c) • Te chnetiu m em its a y-ra y with a n energy o f 0.14 1 MeV. (l l\1cV = lOl; el ect ron -volts, where 1 eV = 9.6485 X 101.l/ m o!.) Wh at are the wavelength an d frequency o f a y-ra y ph o to n with an e nergy o f 0.14 1 tvfeV?
... morechallenging • in OWL
Blue-numbered questionsanswered in Appendix 0
(d ) To m ake Na Tc04, th e m etal is d isso lve d in n it ric ac id.
7 H NO,,(aq )
+
Tc (s) -> IITc(},(aq )
+ 7 N O , (~ ) + 3
II , O (f)
a nd th e p roduct , IITco.h is treated with Na O Il to ma ke NaTcO t. (i ) Write a balan ced e qua tion for the reactio n of irr -o, with NaO H . (ii) Ifvo u b egin wit h 4.5 Ill ~ o f Tc m etal , h ow much NaTcO j ca n b e made? Wh at m ass o f Na O II , in grams , is re q uired to co nvert all of th e IlT( 0 4 in to NaTc O~? (e) If yo u syn thesize 1.5 m icromol cs o f N"aTco." wha t mass of co m poun d d o you h ave? If th e co m p o u n d is d isso lved in 10.0 m L o f sol u tio n , w h a t is the co n ce n tra tio n?
84 . See Che m istr yNow Screen 6, 1, Ch em ical Pu zzle r. T h is screen sh o ws th a t lig h t of diffe rent colo rs can come fro m a "ne on " sign o r from ce rtain salts wheu th ey a re pla ced in a burning o rgan ic liquid . ("N eo n " sign s are glass tube s fille d with ne on , a rgo n , a n d o ther ga ses , an d th e ga ses arc exci te d by an e lectri c currcm.) What do th e se two sources of ligh t h ave in com mon ? [l ow is th e light g en e rated in ea ch ca se? 85 . Sec Chem ist ryNow Scre en G,7, Boh r 's Mo d el of the Hyd ro gen Atom , Simu latio n. A ph o to n wit h a wavele ngth of 973 nm is fired at a h yd ro ge n a to m and leads to the e missio n o f light. H ow man y e m issio n lines a re em itted? Explain wh y m o re th an 0 1\ (' line is emi tted.
I
303
ATOMS ANO MOLECULES
7
The Structure of Atoms and Periodic Trends
The ChromiumBearing Mineral Crocoite, PbCr04 Minerals containing t he chromate ion are extremely rare. The best exam pte is crocoite, lead(II) chromate, which is found almost exclusively in Tasmania. Alth ough it is nearly insoluble in water,t races wi ll nonetheless dissolve and contaminate groundwater with Pb2 + and ( rO/ - ions.
Ch romium compounds in groundwater can be a problem. In t he United States, 56% of the popula tion relies on groundwater for drinking
chrom ium is one of the essential elements, and it is implicated in
water, and the shallowaquifers from which the water is often obtained insulin regulation. In fact, t he recom mendeddaily dose of chromium are susceptible to contamination. One such contaminant is chromi um in its various ionic forms
(Cr3+ , (r042- ,
and
(r20 72- ).
is 5-200 JLg . A com pound called chromium(III) picolinate is mar-
The source of
keted as a "nutritional" supplement and is widely used as a weight-
such ions can be mineral deposits, but more significant sources are
loss aid. The majority of research has found, however, t hat it is
industries involved in leather tanning [which uses Cr(OH)S041 or elec-
neither helpful nor beneficialin weig ht-loss programs.
troplating of chromium for corrosion protection. In a case depicted in
Questions: 1. What is t he electron configuration for t he Cr atom, for t he CrH ion, and for the chromi um in the CrO/ - ion? 2. Is chromium in any of t he ionic form s paramagnetic? 3. What is t he electron configuration for t he lead ions in PbCr04?
the movie Erin Brockovich, a gas and electric utility wasfound to have contaminated groundwater in southern California with chromates that had been used in water-cooling towers to prevent rust. Chromium compounds can also be biochemically active, although they are not implicated in as many important processes as another element in Group 68, molybdenum . There is some evidence t hat 304
Answers to these questions are in Appendix Q.
Chapter Goals
Chapter Outline
See Chapl pr Goals Revisited (! Hlgf' 331) l or Stud)' Quesfion,\" Juyn[ to these goals find fl.uigllflhl1" i n OH'L.
7.1
The Pauli Exclusion Principle
7.2
Atomic Subshell Energies and Electron Assignments
Recognize the relationship of t he four quant um numbers (n, C, me, and ms) to atomic structure . Write the electron configuration for eleme nts and monatomic ions.
7.3
Electron Configurations of Atoms
7.4
Electro n Configurations ofIons
Rationalize tre nds in atom and ion sizes, ionization energy, and electron affinity.
7.5
Atom ic Properties and Periodic Trends
7.6
Periodic Trends and Chemical Properties
ChemistrY-9-NOw'
T
he wave mech an ical mod e l o f the at o m has spinn ing elec tro ns assig ne d to orbitals th at arc best d escribed as m atter waves. The o rbi ta ls are arran ged in su bshe lls that are in tu rn part of e lectro n she lls. O ne obj ective o f this chap ter is to app ly this model to the e lect ro nic st ructu re o f all of th e ele me nt" . A second objec tive is to ex plo re so me o f th e ph ysical properties of cleme n ts, am o ng th em the ease with whic h ato ms lose or ga in elec tro ns to fo rm io ns and the sizes of ato ms and ion s. T hese properties are d irectly relat ed to th e arrange me n t o f el ect ron s in ato ms and thus to th c chem istry o f th e elemen ts and th eir co mpounds.
7.1
Throughout t he text th is icon int roduces an opport unity for self-st udy or to explore int eractive t uto rials by signi ng in at www.thomso ned u.comj logi n.
The Pauli Exclusion Pri ndple
To make the CJllantum th eo ry co ns iste n t with ex pe rime n t, th e Aus tria n physicist \Volfg an g Paul i ( 1900- 1958) sta te d in 1925 his exclusion principle: no two electro ns in a n ato m ca n have the sa me se t of four quantu m numbers ( n, f , 111/0 and Tfl.;) . The co nse CJ ue nce of thi s is th a t no atomic orbital ran be assign ed more than two electrons, and th e two electro ns assigned to an o rbital must have diffe rent values o f fils. An e lec tron assigned to th e I s o rb ital o f th e JI ato m may have the se t o f quan tum n u m be rs n = 1, = 0 , and me = 0 , and Tfl.; = +1;2. Le t us re p rese n t a n orbital by a bo x and th e elec tron spi n by an arrow (t o r J,). A re p resen ta tio n of the hyd rogen a to m is then :
e
Electrons in l s orbital:
• Orbitals Are Not Boxes Orbital s are not boxes in which elect rons are placed. Thus, it is not concept ually correct to talk abou t elect rons being in orbitals or occupying orbita ls, altho ugh this is commonly done for tile sake of simplicity.
[] Quantum numbe r set l s n = 1, r = 0, m f = 0,
Th e choice o f m~ (ei ther +'l2 o r -'l2) an d the d irectio n o f th e elec tron spin arr ow are arbitrary; th at is, we co uld choose e ither value , an d th e arrow may poin t in eithe r d ire ction . Diag ram s such as th ese are ca lled orbital bo x diagrams. A helium ato m has two elec trons, both assigned to th e I s o rbital. The Pauli ex clusion pr inciple re qu ires that eac h electro n must have a d ifferent set of q ua n tu m numbers, so th e orbita l box di agram no w is:
L ls
Two electrons in l s orbital:
This electron has n = 1, This electron has n = 1,
e=
0, m, = 0, m, = _ 1/,
e = 0, me = 0,
ms
= + Vz
By having opposite spins , th e two e lec tro ns in the 1s orbita l of an I Ie atom have different se ts of the four Cjllantll m n umbers. 7 .1
I
The Pauli Exclusio n Principle
305
Number of Electrons Accommodated in Electron Shells and Subshells with n 1 to 6
TABLE 7.1
=
Electron
OrbitaLs AvailabLe (2t + 1)
Number of Electron s PossibLe in sub shell [2(2t + 1)1
Maximum ELectrons PossibLe for nth Shell (2n')
(n )
Subshells AvailabLe
1
5
2
2
2
5
1
8
Shell
P 3
5
6
6
3
6
5
10
2
5
P d 4
3
1
5
P d
3
6
5
10
f
7
14 1
5
p
3
6
d
5
10
f
7
14
g"
9
18
P d
3
6
5
10
r
7
14
1
5
g"
9
18
h"
11
21
18
31
50
71
*T hese o r bi ta ls an.' n o t o cc up ied in [h e g rou n d sta te of a n y kn own ele me nt.
• Spin Quantum Number and Arrows In this book, we arbitrarily use an arrow pointing up (t) to represent m, = +% and an arrow pointing down (J.) to represent m, = - %. We will usually designate t he first electron assigned to an orbital as having m, = + %.though it could just as readily have m, = - %.
Our unders tanding of orbitals and the knowled ge that a n orbita l can accommodate no more th an two electro ns te ll us the maximum n u m ber of electrons that ca n occupy each ele ct ron she ll or subshe ll. Fo r ex ample , because each of the three orbitals in a jJ su bshell can hol d two electrons, /) sub she lls can hold a max imum of six electro ns . By the same reason ing, th e five orbita ls of a d su bshc ll can accommodate a total of I() e lect ro ns , and the seven J orbitals ca n acco m mo d ate 14 electro ns. Reca ll th at there are n su bshel ls in the nth shell, and tha t there are n2 orbitals in that she ll (... Table 6. 1, page 286) . T hus, the max imu m n u mber oj electrons in any shell is 2 n2 • T he rel atio nship among the qu an tu m numbe rs and th e numbers of e lectrons is sho wn in Table 7. 1.
7.2 Atomic SubsheLl Energies and Electron Assignments Our goal in thi s sec tio n is to un derstand and predict the orbital distribution of e lectrons in a to ms with m an y electrons. T he procedure by whi ch elect rons are assigned to o rbita ls is know n as the (WjbflU principle (aufbau means "bu ild ing up" ) . 306
Chapter 7
I
The Str ucture of Atoms and Periodic Trends
Electro ns in an atom are assig ned to she lls (defi ned b y th e qu an tu m n um ber n) and su bshc lls (defi ne d by th e q uantu m nu m ber C) in order of increasing ly high e r energy. In this way, the total energy o f the a tom is as low as possible.
Order of Subshell Energies and Assignments
• Electrons are assigne d to su bshe lls in o rde r of increasing "n + C" value. • For two su bshells with the same value of " n + e," electrons are assigned first to the su bshe ll of lo wer n: T he foll owing are e xam ples of these rul es: the 2s subshe ll (n + C = 2 + 0 = 2) before the 2/) 3) . 2/10rhita ls ( n + f = 2 + I = 3) before the 3 s su bbe cause n for the 2/) electrons is less tha n for the 4 s orbitals ( n
because n
+(
+ t' = 4 + 0 = 4 )
3
3p-
-
3s-
Samen + f, different n
Zp -
Quantum theory and t he Boh r model state that the e nergy o f the II atom , with a sin gle electro n , d epe nd s o n ly on th e value of n (E,! = - Rhc/ n'2 ). For ato ms with mo re than one electron , ho weve r, the situ atio n is more co mp lex. T he o rder of su bshe ll e nergies for n = 1, 2, and 3 ill Fig ure 7.1 shows that su bshe ll e nergies ill mu ltic lec rron atoms depe nd on both n and C. Based o n theo retical a nd experimen tal stu d ies of o rbital e lectro n distribu tio ns in atoms, che m ists have fo und th at there are two ge neral rules that hel p predict the se arra nge men ts:
• Electrons are assigned 10 su bshc ll ( n + ( = 2 + 1 = • Electrons are assig ned to she ll (n + C = 3 + 0 = 3) 35 electrons. • Electrons are assigned to she ll ( n + C = 3 + 2 = 5 )
3d
-
[
Z
n+f
-
-
Samen, different f 2>-
1< -
Active Figure 7 .1 Order of subshell energies. Energies of electron shells increase with increasing n, and, within a shell, subshell energies increase with increasing t, (The energy axis is not to scale.) The energygaps between subshells of a given shell become smalleras n increases.
ChemislrY..,0..,Now'" Sign in at www. thom sonedu:comjtogin and go to the Chapter Contents menu to explore an interactive version of this figure accompanied by an exercise.
befo re the 3d suhis less for 45 than for 3d.
e
Figu re 7.2 summarizes th e assignme nt of e lectro ns accord ing 10 incr easing n + values, and th e discussio n th at follows explores the u nd erl ying cau ses an d th e ir consequen ces and connects at omic electron configurations to th e periodic table.
n value 8
I} > d > I T h us , th e effe ctive nuclear charge ex pe rie nc ed by el ec tron s in a mu hie lectron ato m is in the ord er ns > UI} > nd > n]: T he values of Z* for J a nd II e lectron s fo r t he se co nd -pe rio d el em e nts (Ta ble 7.2) illustrat e thi s. In each ca se , Z* is g re ate r for s elect rons th an fo r p el ect ron s. In a given she ll, s electrons always have a lowe r e ne rgy tha n I} electro ns; I} electro ns h ave a lowe r e ne rgy th an d e lec t ro ns, and d e le ctro ns have a lo wer e ne rgy tha n f el e ctro n s. A conseq uence o f thi s is th at su bshc lls withi n a n elec tro n shell are filled in th e o rde r ns before nl) be fore uri befo re nJ. Table 7.2 also show s tha t fo r the second-period ele ments the value o f Z* for the higher e nergy elec trons increases across th e perio d. As you will see in Sectio n 7.5 , thi s effe ct is im portan t in u nd e rstand in g tren ds in p roperties of e le me n ts acr oss a pe rio d. Wh at emerg es fro m thi s ana lysis is th e o rder of shell and su bshell en erg ies fo r an y given atom and the fillin g orde r in Figure 7.2. With th is as bac kgro un d, we turn to the perio d ic table and use it as a guide to e lectron arra nge ments in ato ms .
Atom
Z"'(ns)
Z'(np}
Vatue of n
C Si
3.22
2 3
Ge
8.04
3.14 4.29 6.78
4.90
4
Chemislry.o.NOw~
•
Sign in at www.t homsonedu.com/ login and go to Chapter 7 Contents to see Screen 7.2 and Screen 7.3, as well as Screen 7.4, which has a simulation and exercise exploring effective nuclear charge and shielding value.
7.3
Electron Configurations of Atoms
Arrangeme nt s of e lectro ns in the elemen ts u p to I09- their electron co n flgura tio n s-care given in Table 7.3. Specifically, th ese are tile gro und state elec tron co nfigu rat ion s, where elec tron s are fou nd in th e shel ls, su bsh clls, and o rb ita ls th at result in th e lowest energy fo r the ato m. In general, elec trons are assigne d to o rbit als in order of inc re asing n + T he e mp has is here , however; will be to co n nect th e con figu ra tion s o f th e eleme n ts with th eir po sitio ns in th e pe riodic table (Figu re 7.4) .
e.
ELectron Configurations of the Main Group ELements II yd rogen , th e first e lement in th e periodic tabl e , has o ne electron in a I s orbital. One way to de pict its electro n configu rat ion is with tile o rb ital bo x di agr am used ea rl ier, bu t an alterna tive a nd mo re fre q ue n tly used method is th e spdf notation . Using thi s me th od , the e lectron co nfignratio n o f II is 1.5 1, read "one s o ne ." T his ind ica tes that th ere is o ne electron (ind icat ed by th e su pe rscrip t) ill tile 1s orbital.
Hydrogen electron configuration:
~ 15
Orbital Box Notation
r or
number of electrons assigned to designated orbital
Z'(2p)
Atom
Z'(25)
Li
1.28
B
2.58
2.42
C
3.22
3.14
N
3.85
3.83
orbital type (f)
0
4.49
4.45
elect ron shell (n)
F
5.13
5.10
l: 15'
TABLE 7 .2 Effective Nuclear Charges, Z*, for n = 2 Elements
spdf Notation 7.3
I
Electron Configurati ons of At oms
309
TABLE 7 .3
Ground State Electron Configurations Configuration
Z
Configuration
Z
Element
Configuration
H
15'
37
Rb
[Kr] 5s'
74
W
[Xe ]4f" 5d' 6s'
2
He
Is'
38
ST
[Kr] 5s'
75
Re
[Xe ]4f" 5d' 6s'
3
Li
4
Be
[He]2s'
39
Y
[Kr]4d' 5s'
76
as
[XeJ4f" 5d'6s'
[HeJ2s'
40
ZT
[Kr]4d'5s'
77
IT
[Xe]4f" 5d'6s'
5
B
[He]2s'2p'
41
Nb
[Kr]4d' 5s'
78
Pt
[XeJ4f" 5 Fe ' +: [Ar]3d' + 2 e: Fe1+ : [Ar]3d' -> Fel+: [Ar] 3dS
+ e:
No te th at in th e ionization of transi tio n met a ls th e us elec tro ns are lo st befo re (n - l ) rI electro ns. All th e co m mo n tran sition met als lose th e ir us e lec tr o ns first, and th e catio ns form ed have el ectron co nfig u ra tio ns o f the ge neral typc [noble ga s co re] (n - I ),fo:. T h is poi n t is impo rtant to re membe r because the magn etic pro pe rties o f tran siti on metal ca tio ns are d et ermined by th e num ber o f unpair ed e lec tro ns in d o rbitals. For exam ple. th e Fe H ion is paramagneti c to th e ex te n t o f five unpaired e lec tr o ns (Fig ures 6. 18 ,7 .5 , an d 7.6 an d A Closer Look: Param agnetism, pa ge 292). If thre e 3d elect rons had been removed in stead o f two s e lec tr o ns an d o ne d el ectr on , th e Fe H io n wo uld still be paramagn et ic but o n ly to th e cx te nt o f three u npaired e lec tro ns. ChemislryJJ.Now~
•
FIGURE 7.5 Formation of iron{III) chloride. Iron reacts with chlorine (( 12 ) to produce FeCl3• The paramagnetic Fe3 + ion has the configuration [Ar]3ds.
Sign in at www.thomsonedu.comjlogin and go to Chapter 7 Conten ts to see Screen 7.6 for a simulation of the changes to an element's electron configuration when it ionizes, a tutoria l on determining an ion's box notation , and a t utorial on determining whether an element is diamagnetic or paramagnetic.
7 .4
I
Electron Configurations of Ions
317
FIGURE 7.6 Paramagnetism of Transition Metals and Their Compounds. (a) Asample of iron(III) oxide is packed into a plastictube and suspended from a thin nylon filament. (b) When a powerful magnet is broughtnear, the paramagnetic iron(III) ions in FezO) cause the sample to be attracted to the magnet. (The magnet is made of neodymium, iron, and boron (NdzFe14B]. These powerful magnets are used in acoustic speakers.)
(b)
(a>
Configurations of Transition Metal Ions ProbLem Give the electron configurations for Cu. Cu " and Cu2 many unpaired electrons does each have?
+ .
Are either of the ions paramagnetic? How
Strategy Observe the configuration of copper in Tab le 7.4. Recall that to form a transition metal ion.
5
and then d electrons are removed
SoLution Copper has only one electron in the 45 orbital and ten electrons in 3d orbitals:
Cu: [A r]3d10 4s1 3d
45
When copper is oxidized to tu ". the 45 electron is lost.
IHltII tJltl IHI D 3d
4s
The copper(II) ion is formed from copper(I) by removalof one of the 3d electrons.
Itml ltlltll t i D 3d
4s
Acopper(II) ion (Cu 2+) has one unpaired electron, so it is paramagnetic. In contrast Cu " is diamagnetic.
EXERCISE 7 .4
Metal Ion Configurations
Depictt he electron configurations for V2+, Vl +, and Co!" . Use orbital box diagrams and the noble gas notation. Areany of the ions paramagnetic? If so, give the number of unpaired electrons.
318
Chapt er
7
I
The Structure of Atoms and PeriodicTrends
7.5
Atomic Properties and Periodic Trends
Module 11
O nce electron co n figuratio ns we re u nd ersto od, che mists reali zed th at similarities in !nnjwrties of the elements are the result oj simila r -oatrnce shell elearon configurations. An o bj e cti ve o f this sec tio n is to descri be ho w ato m ic elec tro n co n fig ura tio n s a rc relat ed to so me of th e ph ysical and chem ical p ro perties of the e leme nts and wh y those p ro perties chang e in a reasonably predicta ble man lier when mov ing d own gro u ps and across pe riod s. T his backg rou nd sho uld mak e th e periodic tabl e an eve n mo re useful to ol in you r study o f che mistry. \Vith a n u nd erstandin g o f e lectro n co nfigurat io ns and their relatio n to p ro perti es, yo u sho uld be able to o rga n ize and predi ct many che m ica l and physical p rope r ties of th e elements and th eir co mpo u nds.
Atomic Size An o rbita l has no sharp bo u nda ry (.... Fig u re 6. 13a ), so how ca n we d efi ne the size o f an ato m? The re are ac tua lly several ways, and th ey can give sligh tly di ffere nt results. On e of th e sim p lest a nd m ost useful ways to d efi ne ato m ic size is to rel a te it to th e d istan ce bet ween atoms in a sa m ple o f th e cl em ent. Let us co ns ider a diato mi c mol ecul e suc h as CI:! (Fig ure 7.7a) . T he ra d iu s of a CI ato m is assu me d to be half th e ex per im en ta lly d ete rmin ed di stan ce bet we en th e ce n ters o f t he two ato ms (19 8 pm) , so th e radius o f one CI ato m is 99 pm . Similarly, th e C...-C d istan ce in d iam o nd is 154 p m , so a ra d ius o f 77 p ill ca n be assig ne d to carbo n . To test th ese estimates , we ca n ad d t he m to gether to es tim ate t he C-CI di stan ce in CC I-t. T he predi ct ed d istan ce o f 176 p m ag rees with th e ex perime ntally measu red C- CI distance o f 176 p m . (Radi i d e te r m in ed th is way are ofte n ca lled "cova le n t radii .") This ap pro ac h to det e rmi nin g ato m ic ra d ii ap plies o nly if mo lecu lar co m pou nd s of the e le me nt exi st (and so it is lim ited to no nm etals and met alloi d s) . For met al s, ato m ic ra d ii are so me tim es estima ted fro m m easurem ents o f th e ato m-to-ato m d istance in a crysta l o f th e eleme n t (Fig u re 7.7b) . Some int e restin g pe riodic tre nds are seen im med iately o n look in g at a ta ble o f ra di i (Figu re 7.8) . For the main brrlmjJ elements, atomic rad ii general(y increase goin g down
([
Cl (
(
154 pm
00
198 pm
176 pm
Adistance equivalent to 4 times the radius of an aluminum atom
00
FIGURE 7 .7 Determining atomic radii. (a) The sum of the atomic radii of [and Cl provides a good estimate of the [-Cl distance in a molecule having such a bond. (b) Pictured here is a tiny piece of an aluminum crystal. Each sphere represents an aluminum atom. Measuring the distance shownallows a scientist to estimate the radius of an aluminum atom.
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319
lA
Active Fig-ure 7.8 Atomic radii in picometers for main group elements. 1 pm = 1 X 1O- 1l m = 1 X 10 - 3 nm. (Data takenfrom J. Emsley: The Elements, Clarendon Press, Oxford, 1998, 3rd ed.)
H, 37
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lA
2A
3A
4A
SA
6A
7A
u. 152
Be, 113
B,8 3
C. 77
N, 71
0, 66
F, 71
Na. 186
Mg, 160
Al,143
Si, 117
P, 115
S,104
(I, 99
K, 227
Ca. 197
Ga, 122
Ge, 123
As, 125
Se, 117
Br,114
Rb, 248
Sr, 215
In, 163
Sn, 141
Sb,141
Te, 143
I, 133
CS. 265
Ba,21 7
H,170
Pb. 154
Bi,1 55
Po, 167
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~ o o
[§J o o
MAIN GROUP METALS
TRANSmON METALS
(J
METALLOIDS NON METALS
grout) in the periodk: table and decrease gning ((ctrJss a twnod. T hese tren d s reflec t two
important effe cts: • T he size of an a to m is d et e rm in ed by the outermost electrons. III go ing from th e top to th e botto m of a g ro u p in the periodi c tab le, the o u termost el ectro ns ar e assig ned to orbit als with increasin gly h igh e r values of the principal t'] uan tu m n um be r, u. Beca use the u nde rlying elect ro ns requ ire some space, these highe r e nergy e lec tro ns are, o n average, fur ther fro m the nucleu s. • For main grou p elem ents o f a given period , the principal q uantu m numbe r, n, of the va le nc e el ectron o rbitals is the same. In go ing fro m o ne el em ent to th e next across a peri o d, Z*, th e effective n uclear charge increases Cra ble 7.2). This resu lts in an increased att ractio n be twee n the n ucl eus and the valen ce electro ns , and th e ato m ic rad ius decreases .
• Atomic Radii-Caution Numerous tab ulations of atomic and covalent radii exist. and t he values quoted in t hem often differ somewhat. The variation comes abo ut because several method s are used to dete rmine t he radii of atoms. 320
Chapter 7
I
T he peri odi c tre nd in the ato m ic ra d ii of tr ansition met al ato m s (Fig u re 7.9) across a period is so mewhat differen t from th at for main g rou p cleme n ts. Go ing fro m left to rig ht in a g ive n period, the radii in itia lly d ecrease . H o weve r; the sizes of the elemen ts in the mi d dl e of a transition series change ver y littl e , an d a sm all increase in size oc cu rs at th e e nd o f th e ser ies. T he size of tran sition me tal ato ms is de termined largely by elec tr o ns in the o utermost shell- that is, by the e lec tro ns of the us su bshell- b u t e lec tro ns are bei ng ad ded to the (n - I ) d orbitals ac ro ss th e se ries. T he increa sed nu clear charge o n the atoms as o ne moves from left: to
The St ructu re of Atoms and Periodic Trends
Cs
250
E .eo ,• '6 •
FIGURE 7.9 Trends in atomic radii for the t ransiti on elements. Atomic radii of the Group lA and 2A metals and the transiti on metals of the fourth , fifth, and sixth periods.
250
6th Period c::::J 5th Period _ 4th Period
200
200
'"
150
100
Period
lA
2A
3B
4B
58
6B
7B ~ 1B
2B
8B
Transition met als
rig ht should ca use the radius to decrease . This e ffect , ho wever, is mostly cance led ou t by increa se d el e ctro n- el ec tron rep u lsio n . O n rea ch ing t he Gro u p I Ban d 2B cl e ment", a t th e e nd of the se r ies, th e size in cre ase s sligh tly beca u se th e d su bs hc ll is filled an d el ectron- e lectron re pu lsio ns do min at e.
• Trends in Atomic Radii General trends in atom ic radii of 5- and p-block elements with position in the periodic table. Increase
m
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•
J1
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EXERCISE 7 .5
Atomic radii
L...-_
PeriodicTrends in Atomic Radii
Place the t hree elemen ts At. C. and Si in order of increasing atomic radius.
Ionization Energy Ionizati on elle rbry (IE) is t he energy re q u ire d to rem ove an electro n fro m an atom in the gas phase. Atom in ground state (g) ---> Ato m+(g) il U == io nizati o n
e ne rgy,
+ e"
IE
To sepa ra te an electr on fro m an atom , energy m ust be su p plied to overcom e the attrac tio n of the nuclea r ch arge. Becau se e ne rg y must be su p plied , ioniza tion e nergies always have positive values . 7 .5
I
Atomic Properties a nd Periodic Tre nds
321
Atoms othe r than hydro gen have a se rie s of ioni za tio n e nergies as e lectrons are re moved seq ue n tia lly, For exa m ple , th e fir st th re e ionizatio n e nergies o f mag n esium are First ionization energy, lEI
• VaLence and Core ELectrons Removal of core electrons requires much more energy than removal of a valence electron. Core electrons are not lost in chemical reactions.
= 738 kJ j mol Mg(g)
• Measuring Ionization Energy Ionization energy values can be measured accurately as compared with the estimations that must be made when measuring atomic radii.
--->
Second ion izati on energy, IE,
= 14 51
Third ionization energy, IE,
7732 kJj mol
~
Mg'(g)
+
e-
kJj mol
MgH(g)
Mg"(g)
+ e-
Removing each su bse q uen t elec tro n re q uire s more ene rbT)' becau se th e el ect ro n is be ing rem oved from an increasi ngl y po sitive ion (Ta ble 7.5) , but there is a pa rticularly large increase in io n iza tion energy for re mov ing th e thi rd e lectron to g ive ~ lgH. T he first 1\ \'0 ion ization ste ps arc for th e rem oval of elec tro ns from th e o u termo st or vale nce she ll of electrons. The thi rd electro n , however, must come from the 2jJ su bs he ll, wh ich has a much lo wer e nergy th an th e 3s su bshell. T his large increase is experi men tal evidence for the electron shell stru cture '?! atoms. For mai n grou p (s - an d jJ-bloc k) elements, first ioniz ati on energies generall)' in crease a (TOSS a per iod and decrease down a J.,m m jJ (Figu re 7. 10, Tabl e 7.5, and Ap pe nd ix F) .
T he trend across a peri od co rrespo nds to th e in cre ase in effective nuclear cha rge, Z*, with increasing atomic nu m ber. As Z*i ncreases, the e nergy re q uired to re move all e lectro n in cr eases. T he ge neral dec rease in io ni zation e nergy d own a gro u p
TABLE 7.5
First, Second, and Third Ionization Energies for the Main Group Elements in Periods 2-4 (kJjmol) Li
C
N
0
F
Ne
1st
513
899
80 1
1086
1402
1314
1681
2080
2nd
7298
1757
2427
2352
2856
3388
3374
3952
3rd
11815
14848
3660
4620
4578
5300
6050
6122
3rd Period
Na
Mg
Al
5i
P
5
CI
AT
1012
1000
1251
1520
2nd Period
Be
1st
496
738
2nd
4562
3rd
6912 K
B
577
787
1451
1817
1577
1903
2251
2297
2665
7732
2745
3231
2912
3361
3826
3928
Ca
Ga
Ge
As
5e
BT
KT
1st
419
590
579
762
947
941
1140
1351
2nd
3051
1145
1979
1537
1798
2044
2104
2350
3rd
4411
4910
2963
3302
2735
2974
3500
3565
4th Period
322
Chapter
7
I
The Structu re of Atoms and Periodic Trends
Act ive Fi gure 7 .10 First ionization energies of the main group elements these fi rst four periods. (For data on these elements see Table 7.5 and Appendix F.)
2500
~ 0
2000
E
a ~
e'
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1500
•c
•c 0
"'•" "E
1000
.s
;;
'"
500
1A
2A
3A
4A 5A Group
6A
7A
SA
occu rs becau se the electro n removed is increasingly farth e r from th e n ucle us an d thu s hcld less stro ng ly. No tice th at at omic radius and io n izatio n energy are both link ed to Z*. T hey arc inversely relate d: as the atomic radius decreases, the ionization energy increases. A clo ser loo k at io n iza tion e nerg ies reveals that th ere arc exce p tio ns to th e ge ne ra l tre nd in a pe riod . O ne exce ptio » o ccurs o n go ing from v-block to I)·bloc k eleme n tx-c-fro m beryllium to bo ron , fo r exa mp le. The 21) elec tr o ns are sligh tly high er in c ncrgy th an th e 2s e lect ro ns so the io nization c nc rgy for bo ro n is slig h tly less th an tha t for bc ryllium. Ano ther di p to 100\'cr io ni zat io n cne rgy occu rs o n go ing fro m nitrogen to oxygen . No cha nge o ccurs in e ithe r n o r b u t e lec tro nelec tro n repu lsio ns in cr ease fo r th e following reason . In Groups 3A-5A, el ectro ns are ass ignc d to sc parate I) orbitals (j)x, f)y, a nd f)~). Beginn in g in Gro u p fiA, however, two electro ns are assigned to th e same I) o rbita l. The fo u rth I) elec tr o n sha res an orbita l wit h another electron and thus ex perienc es gre ate r repulsion than it wo u ld if it had bee n ass igned to an o rbi tal o f its own.
• Trends in Ionization Energy General trends in first ionization energies of s- and p-block elements with position in the periodic table . Increase
First ionization energy
e,
o (oxygen atom) [Ne][TI) 2s
+ 1314 kJ/mal
1111tit I 2p
0 + (oxygen cation) + e-
[Ne][TI) 2s
It l tl t l 2p
T he greater repulsio n ex perie nc ed by th e fo u rth 21) e lectro n makes it easier to re move . T he usua l tr end resu m es o n go ing fro m oxygen to fluo ri ne to neo n , however, re flec ting th e increase in Z*. 7 .5
I
Atomic Properties and Periodic Trends
323
Electron Affinity • Trends in fA General trends in electron affinities of A-group elements. Exceptions occur at Groups 2A and 5A and in moving from period 2 to period 3 in the n-block. Increase in affin ity for electron (EA becomes more negative)
Electron affinity
• fA and Nonmetals Nonmetals generally have much more negative values of EA than metals. This observation agrees with chemical experience; nonmetals form anions, and metals generally do not.
T he electron affin ity, EA, o f a n ato m is defin ed as th e e n eq,JT c h a nge for a process in wh ich a n el ectro n is ac q uire d by th e at om in th e gas phase (Figu re 7.11 and Appendi x F). tJ.U
T he g rea te r th e affin ity a n ato m h as for a n electro n , th e lower th e e ne rgy o f th e ion will be co m pa red to that o f th e ato m a n d th e fre e electron , and th e more nega tive th e value o f J;-:A. For exam p le, th e el ectr on affin ity of flu orin e is - 328 ~J /mol , indicatin g a n exo the rm ic reaction to for m th e a n io n , F- , from a fluorin e atom a n d a n electron . Bo ron has a mu ch lower affinity fo r a n electron , as indicat ed by a much less negative EA value of - 26.7 kJ/m ol. Becau se electron affin ity a n d ionization e n e rgy represent th e e neq,JT in volved in th e gain or lo ss of a n electron by a n ato m, it is n ot su rp rising th at periodic tr end s in th ese p ro p e r ties are also rel ated . The in cr ease in effec tive nuclea r charge o f ato ms ac ross a pe riod (Ta b le 7.2) makes it more d ifficult to ionize th e a tom an d also inc reases th e attracti on of the atom for a n additional electron . Thus, a n element with a high io ni za tio n e ne rgy ge nera lly h as a mo re n eg ative value fo r its e lectro n affin iry. As see n in Figu re 7.11 , the values o f J:,'A ge ne ra lly be come more n eg ative on moving ac ross a period, but th e tr end is n ot smoo th . The elem ents in Gro u p 2A a n d 5A a p pear as va riatio ns to th e ge ne ra l tr en d , corres po n d ing to cases where th e added electro n wou ld sta rt a j) su bs h ell o r wo u ld he paired with anoth e r el ectron in th e /) su bshe ll, res pectively.
Active Figure 7.11 Electron affinity. The largerthe affinity (EA) of an atom for an electron, the more negative the value. For numerical values, see Appendix F. (Data were taken from H. Hotop and W. C. Lineberger: "Binding energies of atomic negative ions, Journalof Physical Chemistry, Reference Data, Vol. 14, p. 731, 1985.)
.f
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H
T ..
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= electron affini ty, fA
- 350 - 300
'5
a• if
~
- 250 - 200 - 150
e
1::
O'!
- 100 - 50 0
1A
2A
3A
4A
Group
324
Chapt er
7
I
The Structure of Atoms and Periodic Trends
5A
6A
7A
T he value of e lect ron affin ity usually becom es less neg ative o n d escen din g a grou p o f the periodi c tab le. Elec tro ns arc ad de d increasingly fa rth e r fro m th e n uc leus, so the attractive for ce be twee n th e nucleus and elec tro ns d ecreases. Th is ge ne ra l tre nd d oes not ap ply to seco nd period ele me nts , ho wever. For exa m ple , the va lue of the electro n affin ity o f fluo rine is higher (less negative ) than the EA value for ch lo rine . T h e same p hen om eno n is o bse rved in Gro u ps 3A th ro ug h GA . O ne ex planatio n is th at sign ifican t electron-electro n repulsions occ u r in small anio ns such as F- . T hat is, ad d ing an electro n to the seve n e lec tro ns alread y p resen t in the n = 2 she ll o f th e small F ato m lead s to co nside rable re pu lsio n bet ween e lectro ns. C hlo rine has a la rge r ato m ic volu me th an fluo r ine , so ad d ing an elect ro n d ocs not result in suc h sign ifican t e lec tro n-e lec tron rep ulsio ns. A few ele men ts, such as n itr oge n a nd the C ro u p 2A e leme n ts, have no affinity fo r el ect ro ns and are listed as having an f:d\ value of ze ro. T he noble gases are ge ne rally no t listed in electron affin ity tables. They have no affini ty fo r el ectro ns, becau se a ny ad ditio na l e lectron must be ad d ed to the next higher e lec tro n she ll. No ato m has a negat ive el ectron affin ity for a seco nd e lect ron . So wha t acco u n ts for the existe nce of ions suc h as Q 2- th at occu r in m an y co m po u nd s? T he answer is th at d oubly charged an io ns ca n be stab ilized in crysta lline e nviro n me n ts by electros ta tic att ractio n to ne ighbo ring po sitive ion s (see Chap ters 8 an d 13 ).
• Electron Affinity and Sign Conventions Changes in sign conventions for electron affin ities over the years have caused confusion. For a useful discussion of electron affi nity, see J . C. Wheeler: "Electron affinities of the alkaline earth metals and the sign convention for electron affi nity, ~ Journal oj Chemical Education , Vol. 74, pp. 123- 127, 1997.
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• Periodic Trends Problem Compare the three elements C. 0, and Si. (a) Place them in order of increasing atomic radius. (b) Which has t he largest ionization energy? (c) Which has t he more negative electron affinity, 0 or C? Strategy Review the tre nds in atomic properties in Fi gures 7.8-7 .11, Ta ble 7.5, and Appendix F, SoLution (a) Atomic size: Atomic radius declines on mcvinq across a period, so oxygen must have a smatter radius than carbon. However, t he radius increases on movinq down a periodic group. Because Cand Si are in t he same group (Grou p 4A) , Si must be larger than C. Th e trend is a < ( < Si. (b) Ionization energy: Ionization energies generally increase across a period and decrease down a group. Thus, the t rend in ionization energies is Si (787 kJ/mo l) < C(1086 kJjmo l) < a (1314 kJjmol). (c) Electron affini ty: Electron affinity values generally become less negative down a group (except for the second period elements) and more negative across a period. Therefore, the fA for o ( ~ - 141.0 kJjmo l) has a more negative fA than C ( ~ - 121.9 kJjmo l). EXERCISE 7 .6 Periodic Trends
Compare t he t hree elements B, At. and C. (a) Pla ce t he t hree eLements in order of increasing atomic radius. (b) Ra nk t he elements in order of increasing ionization energy. (Try to do this wi thout looking at TabLe 7.5; then compare your prediction with th e table.) (c) Which element, Bor C, is expected to have t he more negative electron affinity value?
7 .5
I
At omic Properties and Periodic Trends 325
[§J 0 o
lA
2A
. - -li ' , 18
~ B e 2+ , 3 4 Be, 113
Li, 152
3A
5A
6A
1A 0 2 - ,140 0,66
N3 - , 146 N, 71
F- , 133
F.71
Main Group Metals
o Transition Metals o Metalloids
o Nonmetals
. -Na+, 98
. -Mg" , 79
Na, 186
Mg, 160
K+. 133 K. 227
(a, 197
Ca 2+, 106
Rb+, 149
5rH
Rb, 248
s-, 215
cs' . 165 «, 265
BaH. 143 6a.217
•
127
s>. 184 5.104
Cl-, 181 Cl,99
Ga. 122
Se2-,1 91 5e, 117
Br, 114
1n3+, 92
Te 2 - , 211
1- . 220
In, 163
Te. 143
I, 133
. - - AI" , 57 Al,143
. . -Ga
e---
H
, 62
Br- , 196
a..-nH , 105 n,170
•
Active Figure 7.12 Relati ve sizes of some common fons. Ra dii are given in picometers (1 pm = 1 X 10- 12 m). (Data taken from J. Em sley, The Elements, Cla rendon Press, Oxford, 1998, 3rd edition.)
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Trends in Ion Sizes Th e trend in th e sizes of ion s down a periodic g ro u p are the same as th o se fo r neutral atoms: Pos itive and negative io ns increase in size whe n descending the g ro u p (Figu re 7. 12) . Pause for a moment, however, and co m pare the ionic radii with th e ato m ic radii , as illustrat ed in Figure 7. 12. \Vhen an el ectron is removed fro m an a to m to fo rm a ca tio n , t he size sh rin ks co nsid era bly. The rad ius of a ca tio n is always smalle r th an that o f the ato m from wh ich it. is derived. For example, the rad ius of Li is 152 pm , whe reas th e rad ius of Li + is only 78 pm . \Vhen an ele ct ro n is re moved fro m all Li ato m , the att rac tive force of three pro tons is now exerted on o nly two el ectrons, so the rem aining el ectron s are drawn closer to the nu cleu s. The de cr ease in ion size is especially great whe n the last e lec tro n of a parti cu lar shel l is removed. as is the case fo r Li . T he loss of the 2s electron from Li leaves Li" with no electrons in the n = 2 she ll. U ato m (radius
= 152 pm)
U+ catio n (radius
- 1 elect ron) 15 326
Chapt er 7
I
The St ructu re of Atoms and Periodic Trends
25
= 78 pm)
ImD 15
25
Metals in Biochemistry and Medicine
Case Study Many main group and transition metals play an important role in biochemist ry and in medicine. Your body has low levels of the following metals in the form of various com pounds: Ca. 1. 5%; Na, 0.1%; Mg, 0.05%, and t he metals iron, cobalt zinc, and copper, all less tha n about 0.05%. (Levels are percentages by mass.) Much of the 3- 4 9 of iron in your body is found in hemoglobin, the substance responsible for carrying oxygen to cells. Iron deficiency is marked by fatigue, infections, and mouth inflammations. Iron in your diet can come from eggs, and brewer's yeast has a very high iron conten t. In additio n, foods such as ma ny breakfast ce rea ls are "fo rtifi ed" wit h metallic iron
[made by the decomposition of Fe(CO )SJ . (In an interesting experiment you can do at home, you can remove the iron by stirring t he cereal with a strong magnet.) Vi tamin pills often contain iron(II) compounds wit h anions such as sulfate and succinate
(C,H,O!-j. The average person has about 75 mg of copper, about one third of which is found in t he muscles. Copper is involved in many biological functions, and a deficiency shows up in many ways: anemia, degene ration of the nervous system, and impaired immunity. Wilson 's disease, a genetic disorder, leads to the overaccumulation of copper in the body and results in hepatic and neurological da mage. Like silver ions (page 148), copper ions can also act as a bacte riocide. Scientists
Questions: 1. Give the electron configurations for iron and the iron (II) and iron(III) ions. 2. In hemoglobin, iron can be in the iron(Il) or iron(IlI) state. Are eitherof these iron iansparomagnetic? 3. Give the electron configurations for copper and the capper(I) and copper(II) ions. Is copper in any of theseforms paramagnetic? 4. Why are copper atoms (radius = 128 pm) slightly larger than iron atoms (radius =
..0
~ ~
124 pm)?
I 9
Filling a brass wat er jug for drinking wate r in India. Copper ions released in tiny amounts from the brass killbacteria in contaminated water.
from Britain and India recently investigate d a long-held belief among people in India that storing water in brass pitchers can ward off illness. (Brass is a n a lloy of copper and zinc.) Th ey filled brass pitche rs with ste rile water inoculated with E. coli bacteria and filled ot her brass pitchers with contaminate d river water from India. In bot h cases, t hey found t hat fecal bacte ria counts dropped from as high as 1,000,000 bacteria per milliliter to zero in two days. In contrast, bacteria levels stayed high in plastic or earthenware pots. Appa rently, just enough copper ions are released by t he brass to kill the bacteria but not enough to affect humans.
5. In hemoglobin, the iron isenclosed by the porphyrin group, a flat grouping of carbon, hydrogen, andnitrogen atoms. (This is in turn encased in a protein.) When iron is in the form of the Fe3 +ion, itjust fits into the space within thefourNatoms, andthe arrangement is flat. Speculate on what occurs to the structure when iron is reduced to the Fel+ ion. Answers to these questions are in Appendix O.
_
(( CHZ
H,C
~~ "
'" \
HC
\
CH,
"' ( Hz
N
N-
"'/ Fe N
H,C ~ \
/
'. C""'
CH N
/; CH ,
H
HOzC
A large d ecrease in size is also ex pec ted if two o r more el ectro ns are rem oved . Fo r exa mple , an aluminum ion , AFH , has a radius o f 57 p m ; in co n trast, th e ato m ic ra d ius o f an alu minu m ato m is 143 pm . Al atom (radius = 143 pm)
[NelITil 11-,-1..J.......J w. 3s
3p
AlH cation (radius = 57 pm)
- 3 electrons
[NelD 3s
1 1 3p
YO l1 can also see by co m par ing Figu re s 7.8 a nd 7. 12 th at anions are always lar ger tha n the ato ms from whic h they are d e rived . H ere , the arg u men t is th e o pposite of that used to ex plain positive io n radii . Th e F a to m, fo r exa m ple, lias nin e protons and nine electro ns. On fo rm ing th e an io n , th e n uclear cha rge is still + 9, but th e
7 .5
I
Atomic Properties and Periodic Trends
327
anio n has ten e lectro ns. T he F- ion is larger than th e F atom because of in cr eased elect ron-el ectron repu lsio ns . F at om ( radius
= 71 pm)
F- ani on (ra dius
71 pm
= 133
pm)
133 pm
[ill ItllHl t I + 1 electron) [ill Itlltll tll 2s
2p
2p
2s
Fina lly, it is useful to co m pa re th e sizes of isocl cct ronic ions across th e peri o d ic table . Isoelectronic io ns ha ve the same number of electro ns (b ut a dif ferent numbe r o f pro tons). O ne suc h se ries o f io ns is N3- , 0 2 - , F- , Na +, and J\.lg2 +: Mg2 +
f-
[on
Number of elect rons Nu mber of nuclear proto ns Ionic radius (pm)
10
10
10
10
7
8
9
11
12
146
140
98
79
133
10
All these io ns have 10 electrons b ut they d iffer in the n um ber o f p ro tons. As the n u mber of proton s in cr eases in a se ries of isoelectron ic ions, the balan ce bet wee n e lec tro n- p ro to n a ttrac tio n and elec tro n-electro n repul sio n shift s in favo r o f attractio n , and the radius decreases. ChemislrY.ci.Now~
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EXERCISE 7 .7 Ion
Sizes
What is the tre nd in sizes oft he ions K+, S2- , an d Cl-? Briefly explain why t his tren d exist s.
7 .6 Periodic Trends and Chemical Properties Ato m ic and ionic rad ii, ionizatio n ene rgies, and el ectron affini ties are p roperties associat ed with ato ms and th e ir io ns. It is re asonabl e to ex pect th at kn owledge o f th ese properties will be useful as we ex plore th e chem istry invo lving formation o f ion ic co m po u nd s. T he periodic tabl e was created by grou p ing to gether ele me n ts havin g sim ila r chemical prope rties (Fig u re 7. 13). Alka li metals, for example , characteristically fo rm co m po u nds contain ing a 1+ io n , such as Li" . Na ", o r K+. Thus, th e reactio n
328
Chapter 7
I
The St ructu re of Atoms and Periodic Trends
IA
-
D MAIN GROU P METALS D TRAN sm ON METALS D METALLOIOS D NONMETALS
/' -
-
---.
7A
Elements of Group lA, the alkali metals, all undergo similar reactions with water. Elements of Group 7A, the halogens. aUundergo
similar reactions with metals or other nonmetals.
IA
7A
3
17
Li
Cl
lithium
Chlo rine
2li(s)
+ 2 H20 (f )
_
2liOH(aq)
+ H2 (g)
II
6 CI 2 (g)
+
P, (s)
-
4 PCI,,!')
35
Na
Br
Sodium
Bromine
2 Na(s)
+ 2 H20 (!')
-
2 NaOH(aq)
+ H2(g)
19
4 PB r,( !')
53
K
I
Potassium
Iodine
]2(S) + I n(s)
o
2 K(s)
+ 2 H20(!')
-
2 KOH(aq)
-
l nI2(s)
+ H2(g)
Active Figu re 7.13 Examples of the periodidty of Group lA and Group 7A elements. Dimitri Mendeleev developed the first periodictable by listing elements in order of increasing atomic weight. Every so often. an element had properties similar to those of a lighter element, and these were placed in vertical columns or groups. We now recognize that the elements should be listed in order of increasing atomic number and that the periodic occurrence of similar properties is related to the electron configurations of the elements.
Chemislry.o.Now'" Sign in at www.thomsonedu.comjlogin and go to the Chapter Contents menu to explore an i~teractive version of this figure accompanied by an exercise.
7 .6
I
Periodic Tre nds and Chemica l Prope rties
329
between sod iu m a nd chlorine g ives the ionic co m po und, NaC I (co m pos ed o f Na " and CI- ion s) [Figu re 1.4, pa ge 4], and po tassiu m and wate r react to fo rm an aq ueous so lutio n o f KO II , a so lut io n co n ta in ing th e hydra ted ions K+(aq) and O ll - (aq ) . 2 Na(s) + Cl,(g) --> 2 Na Cl(s) 2 K(s) + 2 H,D(C) --> 2 K+(aq) + 2 DW (aq) + H, (g)
T he faci le form ation of Na " and K+ io ns ill che mica l reactio ns ag rees with th e fact t hat a lkali meta ls have low io ni zat ion ene rg ies . Io n izatio n e ne rg ies also acc o u n t fo r th e fact that these reac tio ns o f so di u m and potassium do no t produce com pou nd s suc h as NaCI2 o r K(O II h . T he format ion of an Na 2+ or K2+ ion would be a ve ry u n favo rab le p ro ce ss. Removing a seco n d elec tr o n from these metals re qu ires a g reat deal o f e nergy because a core el ectron wou ld have to be removed. T he ene rge tic barrier to th is proc ess is the u nd e rlying reason that main J.,mmp metals generally/arm cations with an electron configuration equivalent to tha t 0/ the preceding noble gas. \\l hy isn't Na2CI ano t her possib le p rod uct from the sodi um and ch lo rine reactio n? T hi s fo rm u la would imply th at t he co m pou nd co n ta ins Na " and C12- io ns. Ch lorine ato ms have a re latively negative value fo r elec tro n affin ity, bu t o n ly for the ad di tio n o f o ne el ectron . Ad di ng two e lect ro ns pe r ato m means that th e secon d elect ro n mu st en ter the next h igher shell at mu ch h igher e nergy. An io ns suc h as C12- are not kn own . T h is ex am ple lead s us to a ge ne ra l state me n t: nonmetals gt'twrall)' acquire enough electrons to form an anion with the electron configura tion 0/ thr next noble gas.
We ca n use sim ilar log ic to ration alize o ther obse rvati on s. Ion iza tio n e ne rgies in crease o n go ing from le ft to rig ht across a pe riod . \\!e have see n that ele me n ts fro m Gro u ps lA and 2A form ionic co m po u nds , an observat ion d ire ctly rela te d to the low io nizatio n e nergies for these ele men ts. Io nizatio n e nergies for ele me n ts toward the m idd le an d r ight side of a perio d , however, are sufficie n tly large that catio n for mat ion is u n favo rab le . T h us, we generally do no t expect to e nc ou n te r io nic co m po u nds co n taini ng ca rbo n; in stead , we find ca rbo n sharing elect ro ns with other el em e nts in com po u nds su ch as CO 2 and CC l4. O n th e rig ht side of the seco nd period , oxygen an d flu o ri ne much p refer taking o n electro ns to g iving th em u p; these ele me n ts h ave h igh ionizati o n e nerg ies and relat ively large , negative e lectro n affin ities. Th us, oxygen and fluo rin e fo rm an io ns and not catio ns whe n the y react.
Chemislry.o.Now'"
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EXERCISE
7.8 Energies and Compound Formation
Give a plausible explanation for the observation t hat magnesium a nd chlorine react to form MgCl2 and not Mgct3 •
330
Chapter 7
I
The Structure of Atoms and Periodic Trends
C hapter Goals Revisited N ow that you ha ve studi ed this chapter; you should ask uihether you have met the chnp ter
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goals. In porticulai; ) 'OU shou ld be able to:
t homsonedu:com/log in to: Assess your understanding with Study Questions in OWL keyed to each goal in t he Goals and Homework menu for this chapter For quick review, download Go Chemistry mini-lecture flashcard mod ules (or purchase them at www.ichapters.com) Checkyour readiness for an exam by taking t he Pre-Test and exploring the modules recom mended in your Personalized Study plan.
Recognize t he reLationship of the four quant um numbers (n, €, m" and m.l to atomic structure a.
Reco g nize th at eac h elec tr o n in an aro rn h as a diffe rent set of th e four qua ntum numbe rs, n, C, Tn/, a nd m; (Sectio ns 6 .:')-6.7, 7.1, and 7.3) . Study Questions assignabte in OWL: 11, 13, 35, 37, 52.
b.
Unde rsta n d that th e Pauli e xclusion p rin ciple lead s to th e concl usio n th at no a to m ic orbital ca n be assigned more th a n two electrons and that th e two electrons in an orbital mu st have oppo sit e sp in s (d iffe re n t va lues o f TIl.; ) (Sectio n 7. 1) .
Write the electron configuration for atoms and monatomic ions a.
b.
c.
Reco gni ze that el ectrons are assigned to th e su bs he lls o f a n a to m in order of in creasin g su bshc ll e ne rgy (Au fb a u pri n ciple, Section 7.2) . In the H ato m, th e su bs hell e n ergies in crease with in c reasing n, bu t, in a man y-elect ron ato m , the e n e rg ies d epend on both n a n d C (see Figu re 7.2) . Un d erstan d effe ctive nuclea r charg e , 1 *, a n d its ability to ex p lai n wh y d ifferen t su bsh ells in th e same sh ell o f multielect ron a to ms h ave diffe rent e n e rg ies . Also , u nd erstand th e rol e o f 1* in d et e r mining th e prope rti es o f ato ms (Sectio n 7.2) . Usi ng the perio d ic tabl e as a gui de , d epict electr on co nfigu rat io ns of n eutral ato ms (Sectio n 7.3 ) and mon at omic io ns (Se c tio n 7.4 ) usin g th e orbital box or st)(l/ n o ta tio n. In bo th cases, co n figu ra tio n s ca n be a bb reviate d with th e nob le gas notati on . Study Question(s) assignable in OWL: 2, 3, 6, 10.13, 15, 18. 20, 21, 33 ,
'6 Access How Do I Solve It? tutorials on how to approach problem solving using concepts in this chapter.
34.35. 36,39.44. 52,59. 7l.
d. e.
Wh en assig n in g el ectrons to ato mic orb ita ls, a p p ly th e Pauli ex clus io n p rin cip le a n d lIund 's rul e (Se ctio n s 7.3 a n d 7.4) . Unde rst a nd th e rol e m agneti sm p lays in re vealing ato mic stru ctu re (Sec tio n 7.4 ) . Study Question(s) assignable in OWL: 20, 21, 33, 34, 39, 52.
Rationalize trends in atom and ion sizes, ionization energy, and eLectron affinity a. Predict how p roperties of a toms- size , io ni zat ion e ne rgy (IR), a n d e lec tro n a ffin ity (EA)-c ha nge on movin g down a grou p or ac ross a period o f the periodic table (Sectio n 75 ) . The ge n e ral p eriodic trends for th ese properties are as follows: Study Question(s) assignable in OWL: 24, 26, 28, 30, 32 , 40-43, 45-50,52,56-58.64.
b.
(i) Atomic size d ecreases across a p eriod a n d in cr eases d own a grou p. (ii) IR in c reases across a period a n d d ecreases d own a gro up. (iii) Th e value of EA becomes more n egative acr oss a period an d bec omes less ne gati ve down a gro u p. Recog nize th e ro le th at io niz a tio n e nergy a nd e lec tro n affin ity play in fo rmi ng ionic co m p o u nd s (Sectio n 7 .6 ) . Study Question(s) assignable in OWL: 72.
Chapt er Goals Revisit ed
331
STUDY QUESTIONS Online homewo rk for this chapte r may be assigned i n OWL. .... denotes challenging questions.
•
denotes questions assignable in OWL.
Blue-numbered questions have answers in Appendix a and fully-worked solutions in the Studen t Solu tions Manual.
Practidng Skills Writing Electron Configurati ons of Atoms (S()(> Examples 7. 1-7.3; Tnbles 7.1, 7.3, rlud 7.4; nsul Sa ren 7.5 and till' Toolbox in ClwmiJtJ)'Now.) l. \Vrite th e el ec tro n con fig u ra tio ns fo r P a nd CI using
both sjJ({!no ta tio n a nd o rb ital bo x d iag ra ms. De scr ibe th e rel a tion shi p betwe e n eac h atom 's elec tro n co nfig ura tio n a nd its position in th e pe riod ic table .
2. • \Vrill' th e ele ct ro n con fig ura tions for Mg an d Ar using ho th sjJdj no ta tio ll and orbita l box diagrams. De scr ibe th e rel a tio nsh ip o f th e ato m's e le ct ro n co nfig urat ion to its po sition in th e periodic ta ble. 3. • Using s/J({f n o ta tio n , write th e ele ctron co nfig ura tio ns fo r a to ms o f ch rom ium and ir o n , two o f th e majo r co m pon e n ts o f stainless steel. 4. Using .\/)(lf n o ta tio ll , give the c k-c rrou co nfig u ra tion of vanad iu m, V, an el ement foun d in so me b rown and red al ga e an d so m e to ad stoo ls. 5. Dep ic t the e le ctro n co nfig u ra tio n for each of th e fo llowing a to ms using slId/an d noble g as no tatio ns. (a) Arse n ic, As. A deficie ncy o f As ca n impa ir g row th in an imals eve n th ough large r a mo u nts arc poiso nous. (h) Krvp tou , Kr. It ra n ks seve n th in ab un d ance o f til e g ases in Ea rt h 's at mos p he re. 6. • Using .5/Jd/ an d n o ble gas nota tion s, write el ec tro n co n fig ura tio ns fo r a to ms of th e fo llowin g cle me n ts, and th e n check yo ur answe rs with Tabl e 7.3. (a) St ron tiu~n , Sr. T his d e m e n t is nam ed fo r a to wn in Scotl an d . (b) Zirco n iu m. Zr. T he meta l is ex cept io nally resistan t to co r rosion a n d so has im por ta n t indust ri al a p plica tio ns. 1\.1oo11 rocks show a surp risingly h igh zircon ium co n ten t co m pa re d with ro cks o n Earth . (c) Rho di um , Rh . This m e ta l is u sed in jewelry a nd in cata lvsts in ind ust ry. (d) Ti n , 'S I1 . Th e m et al was used ill th e ancie n t wo rld . Alloys of tin (so lder, b ron ze , a nd p ewte r) arc im · p(lrt a n L 7. Usc no hlc g as a nd .\l J(if n o la tio ns to d e pic t e lec tron co nfi gu ra tio ns fo r the fol lowin g meta ls o f the th ird tra nsitio n series . (a ) Tan talu m , Ta. T he metal and its alloys resis t CO IT Osio u a nd arc oft en used in su rg ical and dent al tools. (b) Platinum , PI. T hi s m e tal was used by P'''> Co lum bia n Ind ia ns in j ewe lry. It is u sed now in jewel ry a nd for an tica n ce r d rugs and ca ta lysts (such as th o se in au to mo bile ex h au st syste ms). 332
8 . Th e lanthanides, o n ce ca lled the rare earth c le m e n ts, a rc re ally o n ly "m ed ium rare ." Usin g no ble gas a nd .5/)({! n o ta tions, d e p ict reaso nable elec tro n co nfig urations fo r th e fo llowin g- elem ents. (a) Sam a riu m , Sm . Th is lan th an ide is u se d in m ag ne tic m a te rials. (h) Ytterbi um , \h . Th is el cuu- n t was nam ed fo r th e villag e of Yn crb y in Swed e n , whe re a mine ra l so u rce of th e el e men t was fou nd . 9. T he ac ti n id e am eric iu m , A m, is a ra d ioac tive el em e n t th a t ha s foun d u se in ho me sm o ke d e te c tors. De pi c t i ts e le ctron co nfig u ra tion usin g noble gas a nd .\'/)(1/ no ta tion s. 10 • • Pre d ict re aso n able el ectron config uratio ns fo r the followin g cl e m e n t" of tilt' actin ide series of e lemen ts. Use no hlc g-as and s/Hifno ta tio ns. (a) Pluto n iu m , Pu. T h e c le m e n t is bes t known as a byp ro d uct o f n u clea r po wer plan t ope rati on s. (h) Curi u m , Cm. T h is ac tinide was named fOT" Ma rie Cu rie (pag-e 342). Quantu m Numbers and Electron Configuratio ns (See Exa m/)!f' i .2 and Chnni.\'tJ)'Noll) ,\' m 'PII.\ 6 !2 nml i.2- i.5.)
II. • Wlt at is th e ma ximu m n um ber o f electron s that can he id en tifie d with ea ch of th e fo llo win g sets o f q llall· tum numbe rs? In o ne case, th e an swer is "non e ." Explain why th is is true . (a) II = 4, ( = S, 11/( I i;2 (b) II = 6. ( = I , 11/( = - I , m, = - i;2 (c) II = 3, f = 3, m ( = -:~ 12. Wh at is th e m ax im u m n um ber of e le ct ro ns th at can be id cn ti ficd with eac h of th e foll owi n g- se ts of qu a ntu m n u m be rs? In so m e cas es, the a nswe r m ay be "n o n e ." In suc h cases, e x plai n why "n o n e" is th e correc t answer. (a) /I ~ :J (h) /I ~ :J a nd f ~ 2 (c) 11=4 , (= 1, 1II( - l, a nd m,, = +\;2 (d) n = 5, ( = 0 ,
IJ/ (
= - 1,
fII ,
=
+~
13. • Depict th e ele ctr o n configura tio n fo r m agnesiu m using an orh ita l bo x diagram and no ble gas no tat io n . Give a co m p le te se t o f fo ur qua n tu m n umbe rs fo r ea ch of th e ele ctro ns beyond th ose o f th e preceding noble gas. 14. Dep ict th e e lectron co n figu ra tio n fo r ph o sph o ru s u sing a n o rbit al bo x d iag ram a nd noble g as n o ta tion . Give o ne p o ssi ble se t o f fou r qua n tum n u m bers for e ac h o f th e e lec t ro ns beyo n d th o se of th e preced ing no b le gas. 15. • Using an o rb ital ho x d iag ram a nd no bl e gas not atio n , sh ow th e ele ctro n co nfigu ra tio n o f gall iu m, C a . ( ;'iv(' a se t o f q uan tu m numbe rs to r th e h igh est-('n e rh')' el ectron . 16. Using a n orbi tal box d iag ram a nd no bl e gas nota tion, show th e e le ctro n co n figu ra tio n of tita niu m . G ive on e possible se t of fou r q uantum n u mbe rs fo r eac h of th e el ec trons beyo nd th o se o f th e pre-cedin g no ble gas.
ELectron Configurations of Atoms and Ions and Magnetic Behavior
28 . • Arra nge th e foll owing ato m s in orde r of in cre asing io n iza tio n e neq..'-Y: , Li, K, C , a nd N .
(,";,,1' Fxam/JI" 7.4, Section 6.7, and Ch"mL~b)'N()w Scnrns 6. 16,
7.5, and 7.6.)
17. Us ing orb ita l b o x di agram s, d e pict a n el e ct ro n co nfi gu ra t ion for e ac h o f th e followi n g io n s: (a) Mg:.! +, (h) K +, (c ) CI - , an d (d ) 0 ' - .
18. • Usin g orbita l bo x d iag ra m s, d ep ict a n elect ro n co n fig urat io n for e ach of th e foll owin g io ns : (a) Na" , (h) AI3+, (c) (; e H , a n d (d) r -. 19 . Usin g o rbita l box diagra m s a n d noble gas n o ta tio n, d epict the e lec tro n con fig u ra tio n s of (a) V, (b) V 2+ , an d (c ) VH . Are a ny o f the ion s pa ra m agn eti c?
20• • Usin g orbi ta l bo x di agram s a nd noble g as no ta tion , d e pict th e elec tro n con fig u ra tio ns o f (a) Ti , (b ) a n d (c ) T iH . Arc a ny o f th e ions p a ram agn et ic?
Ti H ,
2 1. • Man g an ese is fou n d as ~l n0 2 in d e ep oc ea n d e po sits. (a ) De pi ct th e elec tro n co n lig u rat io n of thi s c lem ent u sin g th e n o b le gas n ota tion an d an orbi tal bo x d iag ra m. (h ) Usi ng an orb ital box d iagram , sh ow th e ele ctro n s bey o n d th o se of th e p rece d in g n oble gas for th e 4 + io n , (c) Is th e 4 + ion pa ram agn etic? (d ) H ow ma n y un p a ire d el ectro ns d o c s th e Mn H io n h ave ?
22. O ne com p oun d fou nd in alka line ba tteri es is NiO O Il, a co m po u n d con ta in ing Ni3+ io n s. Wh e n th e b atte ry is d isch arg ed , th e Ni 3 + is re d u ced to ~ i H ion s [a s in Ni (O H h l. Us ing orbita l b o x di ag ra m s a nd the n ob le gas n ot ati on , show elect ro n co n fig ura tio ns o f th ese ions. An ' ei ther of tlu-se io n s pa ram ag n et ic?
29. Co m pare th e elements Na , Mg , 0 , an d P. (a ) Whi ch h as the la rgest a to mi c radius? (b) Whi ch h as th e most n ega tive e le ctron affini ty? (c) Place th e el e m e nts in ord e r o f in creasin g io n izati on e n erh'")'· 30 . • Co m pa re th e el em ents R, AI, C, an d Si. (a) Whi ch h as th e most m et alli c ch a ra ct er? (h) Whi ch h as the la rgest a to mic radiu s? (c) Whi ch h as the m o st n ega tive e le ctro n a ffin ity? (d ) Place th e three e le me n ts R, AI, an d C in orde r o f increas ing fi rst io ni za tio n e n e rgy.
3 1. Ex p la in each an swer b riefl y. (a) Plan ' th e fo llowing elem en ts ill o rd e r of in cre asin g ion iza tio n e nergy: F, 0 , a n d S. (b) Which h as th e la rgest ion iza tion en e rgy: 0 , S , or Sc ? (c) Whi ch h as th e m o st n ega tive e le ct ron affi ni ty: s-, CI, o r fi r? (rl ) Wh ich h as th e la rge st ra d ius: ()2- , F- , or F?
32. • Exp lai n e ach a nswer b riefly. (a ) Rank th e fo llowin g in o rd er o f in creasin g a to mic radius: 0 , S, a nd F. (b) Whi ch h as the la rge s t ioni za tion e ne l"l..,")-': , P, Si. S, o r Se ? (c) Plan ' th e fo llowing in o rd er o f increasing ra di us : 0 2- , i'P- , a n d F- . (d ) Place th e followi n g in o rd e r o f in creasing io n ization e l\er!..'")': c-, s-, a n d Ra .
General Questions T hese questions are not designated as fo
'JPe or location
in the
chapter: T hf)' nUl)' combine several concepts.
Pe riodic Properti es (See Section 7.5, Exa m/lie 7.5, an d Chnnistr)'Now Screens 7.7-7. 13.)
33. • Us ing an orb ital b o x d iag ra m a n d n o b le gas nota tio n .
23 . Ar ra nge th e fo llow ing c lem ents in o rder o f in creasing
34. • T h e rare ea rt h ele m en ts, or lanth ani d e s, co m mo n ly ex ist as 3 + ion s. Using an orbi tal bo x di agram a nd n o-
size: AI, fi, C , K, and -Na . (Try d o ing it wit h ou t lo o ki n g a t Figure 7.8, an d th e n chec k yo ursel f b y lo o ki n g up th e n ec e ssary a tom ic radii. ) 24 • • Arra n ge th e fo llowi ng c le men ts in o rde r o f in creasing s ize : Ca , Rb , P, Cc , an d Sr. (Try doin g it withou t lookin g at Figu re 7 .8 ; th e n c h ec k yo ursel f by lo o kin g up th e n ecessa ry a tom ic ra dii.)
25 . Se le ct th e a to m or ion in ea ch p air th at h as th e lar ge r rad iu s. (a) Cl or
et -
(h) Al o r
°
(c ) In or I
26. • Se le ct th e a to m or io n in e ac h p air th at ha s th e la rge r radius. (a) Cs or Rb (h) 0 2 - o r 0 (c) fir or As
show the electro n con figuratio n s o f uran ium an d of th e u ran ium (IV) ion . Is ei th er o f thes e p ara magn etic?
ble th e (a) (h)
gas n otat ion , sh ow th e e le ctro n co n fig u ra tio ns o f fo llm.. . ing e le m e n ts an d ion s. Ce a n d Ce:l+ (c eriu m) 11 0 an d 110 :1+ (h o lm iu m )
35 . • A neu tra l atom h as two el ectro n s with 1/ = I , e ig h t electron s wit h II = 2, eigh t elect ron s wit h II = :), a n d two electro n s with II = 4. Assu m ing th is c leme n t is in its g ro u n d sta te , su p p ly th e fo llo wing in fo r ma tio n : (a ) a to m ic n um ber (b ) to tal nu m b er o f J el ectron s (c) tota l nu mbe r of p elec tron s (d) tota l nu mbe r o f d e le ctron s (I..') Is th e el e me n t a m etal, m et a llo id , o r n on m etal ?
27 , Which of th e fo llo wing g ro u ps o f e lem en ts is arran ge d correc tly in order o f in creasing ioni za tio n e n eq..,Ty? (a) C < Si < Li < N e (c) Li < Si < C < Ne (h) Nl' < Si < C < l.i (d ) N t' < C < Si < l.i
•
more challenging
• in OWL
Blue-numbered questions answered in Appendix 0
I
333
36 . • Ele me n t 109, now named meit ncri um (in ho nor o f th e Aus tria n-Swed ish physicist, Lise Mci mcr [ 187819fi8] ) , was p ro d uce d ill Augus t 19H2 hy a team a t Germany's Insti tute for Heavy Io n Rese arch. Dep ict its electron co nfig ura tio n using sj)({{ an d nobl e gas not atio ns. Nam e a nother d e m ent found in th e sa me group as meitnet-ium .
(a ) W r-ite th e electron co n fig u ra tio n o f eac h of these clemen ts u sin g a n o rbital bo x d iag ram and noble g as notat io n . (b) Are th ese cleme n ts par am agn etic or d ia magne tic ? (c) ' Vri te th e electro n co nfig uratio ns o f Nd 3 + and Fe 3 + using orbi tal bo x diagrams and no ble gas nota tion . Are th ese io ns paramagne tic o r di am ag neti c? 40 . • Name the e lemen t corresponding to eac h characteristic below. (a ) th e c leme nt with the e lectron co nfi gu ra tio n I s22s 22p(i3s2 3 j):1 (b ) th e a lkali ne ea rth elem en t wit h th e sm allest atom ic rad iu s (c ) th e ele m e n t wit h th e la rgest io n izat io n en ergy in Croup 5A (d) the e le men t whose 2+ io n has th e co nfiguration [ Kr ]4 d; (e) th e eleme n t wit h the m o st n egati ve e le ctro n affin ity in Grou p 7A (f) th e eleme n t whose electro n co nfigu rat ion is [Ar]3 d W4 s2 41. • Arrange th e foll owing a toms in th e order of increasing ioniza tio n energy: Si, K, P, and Ca.
~ec
@, -'I_ lise Meitner (1878-1968) and Otto Hahn (1879-1968). Element 109 (Mt) was named afte r Neitner. She earned her Ph.D. in physics under Ludwig Boltzmann at t he University of Vienna, the first woman to do so at t hat university.
42 • • Ra n k th e foll o win g in order of incr e asing ionization energy: C l, Ca 2 + , and 0 - . Brief l y e x plain yo u r answer. 43 . • Answer th e q ue stio n s below about. till' ele men ts A a nd B, whic h have th e e lec tro n co n fig u ra tio ns shown . A
37 . Wh ich of th e foll owin g is not an allowable se t of quan tum n u m bers? Expl a in yo u r answer b riefl y. For th o se sets that are valid , id e ntify a n cle men t in wh ich a n o uterm ost valence elect ro n could have that se t of q uantu m n u m bers. 11
(a) (h ) (c) (d)
1IIi
2
0
I
2
I I
0 0 - 1
4
2
+2
B
=
[Ar ]3d I04 s24ji1
affini ty?
111,
-~
[Kr ]5 s 1
(d) Whi ch elem e nt has th e larger a to m ic radius? (e) \Vha l is th e fo r mula fo r a co m po un d fo rm ed betwee n A an d B?
_ 1
0 +12 - 12
=
(a) Is elemen t A a metal. no n metal , or metalloid? (b) Which el em ent has th e g reater ioni za tion e n ergy? (c) Wh ich elem e n t has th e less neg ati ve electron
44 . • Answer th e fo llowin g que stio ns aho u t th e cle ments with th e elec tro n co nfig uratio ns shown h e re :
A
38. A po ssible exc ited sta te for the [J atom h as all electro n in a 4/) orbital. List all poss ible sets of q ua ntum n um he rs ( n, t', mi, m,) fo r this e le c tron .
~
[Ar]4s'
1\ ~ [Ar)3d"'4s'4p"
(a) Is c lemen t A a metal, m etallo id , or non me tal? (b) Is el e ment B a me ta l, meta lloid , o r nonm e tal ? (c) Wh ich el em ent is e xpected to have the la rger io nizat io n c nergy? (0.) Wh ich e lem ent has th e sm alle r a to mi c ra dius?
39 . • Th e magne t ill th e ph oto is made from neodymium , iro n , a nd boron .
45 . • Whi ch of th e followin g ion s arc unl ikel y to be fo u nd in a c hem ica l co m poun d : Cs ' , In H , Fe'!" , Tc 2 - , Sn S+ , an d I- ? Ex p la in b riefl y. 46 . • Place th e fo llowi ng ions in o rder o f d ecr easing size: K+, CI - , 5 2 - , and Ca 2+.
A magnet made of an alloy containing the elements Nd, Fe, and B.
334
I
47 . • Answer eac h of the followin g q uestio ns: (a ) O f the e lemen ts S, Se , and CI, whic h h as til e largest a to mi c radius? (b ) Which has th e lar g e r ra d iu s, HI' o r Br" ? (c) Wh ich should have th e largest difference be twee n th e first and secon d io nizat ion e ne fhJy: Si, Na , P, Of Mg? (d ) W hi ch has the largest io n izat io n e lH')'h'J': N, P, o r As? (e ) Which o f th e fo llowing has th e la rges t ra di us: 0 2 - , N 3-, or F-?
•
more chaLLenging •
in OWt Blue-numbered questions answered in Appendi x 0
4 8• • T h e fo llo win g are isocl cctronic sp ec ies: CI -, K+, a n d Ca ' " : Rank them in o rder of increasing (a) size, (h) ioni zat ion e n e rgy, a n d (c) e lec tron affi ni ty. 49. • Co mpare th e clements Na , R, AI, an d C wi th regard to th e foll owing pro perties: (a) W h ich h as the large st ato mic rad iu s? (b) Wh ich h as the most nega tive e lec tro n affi n ity? (c) Plan' th e elements in orde r of increas ing io n iza tio n e n e rgy.
54 . .... Sp inel s are so lids with th e gen eral fo r m ul a M 2 + (M ' :H hO ~ (wh e re ,\12+ an d M'3+ are m etal ca tion s of the sam e or diffe rent m et al s) . The best-kn own exampie is comm on mag n e ti te , Fe 3o.1 [which yo u ca n fo rm ul ate as (Fe H ) (Fe3+hn 11.
50. . .... Two dem ents in th e second tran sition se ries (Y th ro u g h Cd ) h ave four unpaired e lec trons in t heir 3 + ion s. Wh a t cl e ments fit thi s description ?
5 1. T he configuratio n for an ele men t is g ive n h ere .
[Ar]
A crysta l of a spinel .
lillIf] 11111 1 [ill 3d
4,
(a) Wh a t is th e id en ti ty of th e elem ent with thi s co n figu ra tio n ? (b) Is a sam p le of th e cle m e n t p ara ma gn eti c or diamagne tic? (c) How ma n y unpaired electro n s doc s a 3 + io n of thi s e le m e n t h ave ?
(a) G iven its n ame , it is evid e n t th at m agnetite is ferromagne tic. [l ow man y u n p a ire d elec tro ns are there in iron ( I1 ) an d in iro n (II I) ions? (h) Two o th er sp inels a re C oA I ~ O ~ a n d S n c'o ~o.l ' Wh at m e tal io n s a re inv olved in each ? Wh a t a re th e ir e lec tro n config ura tion s? Are th e metal ion s a lso pa ram agn etic, and if so h ow man y un p a ired e le ctron s are involve d ?
52• • T h e configu ra tion of an dem ent is give n h ere .
[Ar] lliIIC-'I.L - 1--'----'
Summary and Conceptual Questions TIll'fiJll-Owing questions use (:01/(:1 )1"'/11)111 (1I i.\ a nd tnroious (haJlliT-i.
3d
(a) What is th e idcn ti tv of the elem ent ? (h) In what g ro u p and' period is th e ele men t fo un d ?
(c) Is the e lem e n t a nonmeta l, a main g ro u p dem en t, a transition metal , a lant h ani de , o r an act in ide ? (d) Is th e d ement di am agne tic o r p aramagn e tic? If paramagnetic , h ov.. m an y unpa ired electrons an ' th ere? (c) Write a co m p le te se t of qu antu m n u m b e rs (n, f, nI( , m~) fo r e ac h of th e va le n ce elec tr on s. (f) What is th e configurat ion of th e 2+ ion form ed from thi s eleme nt ? Is th e io n d iamagn eti c or paramag netic?
53 . Ni ckc l f l l ) fo rm at e [Ni(l -IC0 2h J is wid clv used as ca talyst precursor a nd to m ake metallic n i ck ~·1. It can he p repared in th e general ch e mi stry labora tory b y treatin g readily ava ila b le ni ckd (l l) ac etat e with formic ac id ( H Ca,H).
+
2 Il CO ,Il (aq ) --> Ni ( IlCO, l,( aq ) + 2 CI L,CO, 1l (aq )
( ~rl'l'n crystallin e 1': i( H C O ~ ) 2 is precipitated after a d din g eth ano l to the so lu tio n. (a) Wha t is the th eo retica l yield of nickd (lI ) fo r ma te from 0.500 g o f n ickel (I1) ace ta te a n d excess fo rmi c aci d ? (h) Is ni ckd (Ir ) form ate paramagnet ic or d iam agnet ic? If it is param ag netic, h o w Illany unpaired ele ctro n s wo u ld yo u expect? (c) If nickel ffl ) form a te is lu-at ed to 300 °C in th e abse n ce o f air for 30 minutes, th e salt d ecomposes to for m p ure n ickel po wder. \Vhat mass of n ickel pow der sh o u ld be produced by h eat ing' 253 mg' o f n ickel (I I) fo r mat e? Are nickel a toms p aram agn etic?
•
morechallenging
• in OWL
56. • Whic h io n s in th e foll owin g list are no t likel y to h e foun d in chemica l com pou nds: K2+, Cs ". AIH , F2- , ami Se 2-? Ex p lain briefl y.
57 . . ... 1\.,.0 c le men ts h ave th e foll owing li rst th rou g h fourth io n iza tio n e n e rg ies. Deduce th e g ro u p in the periodic table to which th ey prob ably belong. Exp lai n hric tly. Io n izat io n En erj..,'} (kJ / m o l)
In the Laboratory
Ni( Cll ,CO , l, (aq )
55 . W hy is th e rad iu s of Li " so m u c h sm a lle r tha n th e radiu~ of Li? '\Th y is th e ra d ius o f F- so much large r than th e radius of F?
Blue-numbered questions answered in Appendix 0
Elem e-nt 1086 ,2 2352 4li20 6222
I " IE
2"" IE 3 cd IE 4(11 IE
Elerru-nt Z 577.4 1816.6 2 74 4.6 11 575
5 8. . ... T he ionizat ion of th e hydrog e n a to m can be ca lcu la ted fro m Bo h r 's eq u a tio n for th e e le ctro n ene rg:'.
E
= - (N, lIhc)(7:'I n')
wh er e N 4. RlIc = 1312 ~ J /mo l and Zis th e atomic number. Le t us use th is ap p roach to ca lcu late a possible ionization ene q.,. Y ' fo r h eli um . First, assu me th e electron s of tile l lc ex perien ce th e fu ll 2 + nucle a r cha rge. This gives us th e up per limi t fo r th e ion izati on e n e rgy. Next, assume one electro n of He co m ple tely screens th e nu clear ch arg e from the o th e r ele ctrons , so Z = 1. T h is g ives us a lower lim it to th e io n iza tion energy. Compar e th ese ca lcula ted valu es for th e upper and lower limits to th e ex perimen tal valu e o f 2372.3 U /mol. What d ocs th is tell us about the ab ility of on e electron to scree n the n ucl ear cha rge ?
I
335
59 . • Co m pa re th e co n fig u ra tio ns bel ow with two elect ro n s lo cated in /J orbi tal s. Whi ch wo u ld he th e m o st sta ble (have th e lowest e lle q.,"y)? Wh ich wo u ld h e th e le ast stah ie ? Ex p la in yo u r an swers.
(.J (b)
(c)
(d)
[ff[IJ [illIJ [llITJ [[[ill
(a) J;,. '\Th y do th e orbit al en ergies gen erally h e co me m o re n ega tive on p ro cee d in g across th e secon d p e ri o d ? (b) H ow a re th ese valu es rel a te d to th e ioniza tion energy and ele ct ron affini ty o f th e elem ents? (c ) Use th ese en erg y values to e xp lain th e obse rvat ion th at th e ioniza tion energies o f th e first fo ur seco ndpe r iod c le ments a rc in th e ord er Li < Be > B < C. No te th a t th ese e n eq,,; ' va lues are th e basis for th e d iscussi o n in th e Sim u lation on Ch e m istryNow Screen 7 .8. (D ata fro m .J. R. Ma nn , T. L. Me ek, a n d L. C. Allen: [ournnl of tllP Amnimn Chrmical Sodfl)', Vol. 122 , p. 2780,
2000 .) 68. J;,. Th e iouizatiou ene rgies fo r th e re moval o f the fi rst electron in Si, P, S , a nd C I a rc as listed in th e tab le b elow. Bri e fly rati o n a lize th is tr e nd .
60 . T h e b o n d le n g th s in C12, Br2' a n d 12 are 200 , 228, an d 266 pm , resp ecti vely. Knowin g th at th e tin ra dius is 141 p m , es ti ma te the b ond dis ta nc es in Sn - CI, Sn - Rr, an d Sn - I. Co m pare th e es timate d values wit h th e exp eri m e nta l values of 23:1, 250, a n d 270 pm , respe cti ve ly. 6 1. Write e le ctr o n co n fig u ra tions 10 sh ow th e first two ionization p ro cesse s for p ot assiu m , Ex pla in why th e seco n d io n iza tio n e ne rgy is m uc h g reate r th an the firs t.
62. Ex p la in how the io nizatio n en ergy o f a to ms ch an g es and wh y th e cha nge oc cu rs whe n proc e eding d own a g ro u p o f th e p e riod ic table.
Fi rs t Ionizatio n Energy (kJ / m o l)
Elem ent
780
Si
r s
1005
CI
125 5
I 06!l
69 . Us ing yo ur kn owledge of th e trends in c le men t sizes o n goi ng across th e p er io d ic tabl e , expl ain briefly why th e d ensity of th e cle m en ts in creases fro m K th rough V. 8
63 , (a) Explain why th e sizes of ato ms change 'whe n p roceed ing across a p eriod of the periodi c tab le . (h) Expla in why the sizes of tr a n sit ion m e tal ato ms cha n ge ve ry littl e acro ss a pe rio d .
V 6 E
0;)
.~
15 (k Il m o l)
- I'l l :) - 2373
u
2s (kIlmol)
Il C N
o
336
-
1356 1875 2466 3124
oln n
- 38 76
- 17!19
- 4677
- 2083
I
19
20
21
22
23
Atomic num ber 70 . T he den siti e s (i n g/cm 3 ) of e le m e n ts in Grou ps 6B, 8 B, and I B a rc give n in th e table below. Pe rio d 4 Pe riod 5 Pe riod 6
Cr, 7.19 Mo , 10.22 ' \T, 19 .30
Co , 8 .90 Rh , 12.41 II', 22.56
Cu ,8.96 Ag, 10 .50 Au , 19.3 2
Tra n sit ion metals in th e six th p e rio d all h ave m uc h grea ter d ensi tie s th an th e ele ments in th e sa m e grou ps in th e fou r th an d fift h p eriod s. Refer to Fig ure 7.9, a n d ex plai n this o bse rvat io n .
- 800.8 - 1029 - 1272 - 1526
F
c.
2
2b (kIlm o l)
Nl"
sc r--
~
- 5 20 .0 - 899 .3
Be
4
c
66 . Ex p la in why tlu- first ioniza tion e n ergy o f Ca is grea ter th a n th at of K, whe reas th e sec o n d ionizat io n e nergy of Ca is 10\\"(.'1' th a n th e sec o n d ion iza tion e nergy o f K.
II Il l"
-
-c,
65 , J;,. Wh at a rg u me nts would yo u u sc to co nvi nce a not her stu de n t in g e neral ch e mi stry th at MgO co ns ists o f th e ions Mg 2 + a nd 0 2- an d n o t th e io ns Mg " an d o - ? Wh at ex peri m en ts co u ld b e don e to p rovid e so me ev id e n ce th at th e correc t fo r m u la tion of mag n e siu m o xid e is M g 2+0 2- ?
Element
n
e:;-
64, • W h ich o f th e followin g el e m en ts h as th e greatest d ifferen ce be twee n its first a n d secon d io n iza tio n e nergies: C, u. N, Be ? Ex plain yo ur answer.
67 . T h e energies of th e orb ita ls in llla n y e le men ts h ave be en determined. For th e fi rst two periods they h ave th e foll owing va lu es:
-
•
more chaLLenging • in OWL
Blue-numberedquestions answered in Appendix a
71. • T he di sco very o f two n ew el e m ents (a to m ic num bers 113 and 115 ) was a n n o unc ed in Feb ru a ry 200 4.
~
~
.
8
j
)
J I
~
74. So di u m m etal reacts re adily with ch lo rin e g'a s to g ive so di u m ch lo ride. (See ClwmiJlr)'iVoUi Screen 7. J 7 Chemical fluv .ln .)
Na(s)
+ V,
CI,(g ) -> NaCI(s )
(a ) Wh a t is th e reduc ing age n t in th is rea ction ? What p roperty of the clem e nt con tri b u tes to its a bi lity to ac t as a re d u cing' ag e n t? (b) Wh at is the ox id izing age nt in thi s reactio n ? wh at property of th e cl ement con tributes to its ab ili ty to ac t as an o xid izin g age n t? (c) Wh y does th e react io n p roduce NaCI and no t a co m po u nd suc h as Na2Cl o r NaCl 2?
's
i
8
Some members of the team that discovered elements 113 and 115 at th e Lawrence Livermore National Laborat ory (left to right) : Jerry Landrum, Dawn Shaughne ssy, Joshua Patin , Philip Wilk, and Kent on Moody.
(a ) Use J/)(lfan d nobl e g as nota tions to give the el ectron con fig uration s o f th ese two elem en ts. (b ) Na m e an el e ment in the sa me pe rio d ic g ro u p as th e two el erru-nts. (c) Ele m en t 113 was made by firin g a ligh t a to m a t a he avy a merici u m a to m . T he two co m bine ( 0 g ive a nucle us with 113 proto ns. Wha t ligh t a to m was u sed as a project ile ? 72• • Expla in why th e react io n of ca lciu m am i fluori ne d o e s no t fo rm CaF J .
73. ... T hi onyl ch lo rid e, SOC1:!. > is an important chlorinating and ox id izing age nt in o rgani c chem istry. II is prepared industrially by o xyg('n a to m tran sfe r from S03 10 SCI2. SO , (g )
+
SCI, (g ) -> SO, (g )
+
SOCI, (g )
(a) ( ;ive the ele ctro n co nfig uratio n fo r an ato m of sulfur usin g an orbi ta l box di agram . Do not u se th e noble gas no tat io n . (b) Using th e con fig u ra tion given in part (a ) , wri te a se t of qu antu m nu mb ers fo r the h igh est-e nergy electron in a su lfu r a to m. (c) Wha t elem e n t invol ved in th is reaction (0 , 5, C I) should h ave the smalles t io n iza tio n en er~' ? T h e smalles t radi u s? (d ) Whi ch should be smalle r: the sulfi de ion , 5 2 - , or a sulfu r a to m, S? (c ) lfyou wan t to mak e 67 5 g o f SOCI2 , wha t mass o f SCl 2 is requ ired? (I) If yo u usc 10.0 g o f S03 an d 10.0 g of SCl 2, wh at is th e t heoretical vic ld of SOc:I 2 ? (g) ts.H" for the re ;lctio n of 50 3 a nd SCI2 is - 96.0 kJ/mol SO Cl 2 produced. Using d a ta in Ap pe ndix L, ca lcu la te the stan dard m o la r e n thalpy o f form ation o f SCI2 .
•
more challenging
• i n OWL Blue-numbered questions answered in Appendix 0
75. ... Sla te r's ru les are a sim ple way to es tima te the effe ct ive nuclear charge ex perienced by an e lec tron. III th is approach , the "sh ie ld ing co ns tan t," a , is ca lcu la ted. The effe ctive nucle ar cha rge is then th e d iffe re n ce be tween c and the at omic number, 7.. (No te that th e res u lts in Tabl e 7.2 we rt.' calc ulated in a sligh tly diffe rent way.) Z* =7. - fT
T he shie ld ing con stant, a , is calcu la ted usin g the followin g rules: I , T he ele ct ron s of a ll ato m are grou pe d as follo ws: ( 1.\) (2s,2jJ) (3s,3jJ) (3d) (4,, 4/» (4d), and so on . 2. Electro ns in h igh e r g rou ps ( to th e rig ht ) d o not s hield th o se in the lowe r grou ps. 3. For 1/5 an d np vale nce elec tro ns a ) Ele ct ron s in th e sa me n5, nl) g rou p con tr ib ut e 0.35 (for 1.\' 0 .30 works hcncr}. b ) Elec tron s in the 11 - 1 g ro u p co n trib u te 0.85. c) Ele c tron s in the /I - 2 g'HHIp (a nd lowe r) co ntribut e 1.00. 4 . For nd an d 'if ele c tron s, elec trons in th e sa me nd o r n] grou p co n tr ib u te 0 ,35 , and those in g rou ps to the left co n tribu te 1.00 . As an exam ple, let us calcu la te Z* for th e out erm o st el e ctron o f o xygen : "Now'·
•
Th roughout t he text this icon int roduces an opportunity for self-study or to explore interactive tuto riaLs by signing in at www.thomscnedu .com/ Iogtn .
ChemicaL Bond Formation
Wh en a che m ical react io n occurs be tween two ato ms , th e ir valence electrons are reorganized so that a net attra ctive fo rce-a chemical bo nd-o ccu rs bet ween ato ms . There are t\VO gene ra l types of bonds, ionic and covale n t, and th ei r formation ca n he d e picted using Lewis sym bo ls. An ionic bo nd forms whe n one or more valence electrons is transferred from OJ/P atom to anoth er, crea ting po sitive and negative ions. Wh en so d iu m and chlo r ine react (Figu re S. la) , an elec tro n is tran sferre d from a so d ium ato m to a ch lo rine atom to form Na " and CI- .
Na · + MetaL atom
. Cl : Nonmetal atom
~
[ Na 'r>. . El .. :]
~
Electron transfer from reducing agent to oxidizing agent
[ Na+
:ci :-]
Ionic compound. Ions have noble gas electron configurations.
The "bo nd " is th e attrac tive force between th e po sitive and nc gative ion s. Covalent bonding, in co n trast, involves shmi ng of valence electrons between atoms. Two chlo ri ne ato ms, for exam p le, share a pair of el ectrons , one e lectro n from each at om , to form a covalen t bond .
: Cl· +· Cl :
~
• Valence Electron Configu rations and Ionic Compound Formati on For the formation of Na Cl: Na changes from ls 22s 22p63s 1 to Na " with ls 22s22p6, equiva lent to the Ne configuration. Cl changes from [Ne]3s 23 pS to Cl" with [ Ne ] 3sl3 p6, equivalent to the Ar configuration.
:Cl :Cl : 8 .1
I
Chemical Bond Formation
349
FIGURE 8 .1 Formationofioniccompounds. Both reactions shown here are quite exothermic, as reflected by the very negative molar enthalpies of formation for the reaction products. (See ChemistryNow Screen 8. 3 for a video of the formation of sodium chloride from the elements.)
(a) The reaction of elemental sodium and chlorine to givesodium chloride.
(b) The reaction of elemental calcium and oxygen to give calcium oxide.
~fHo
~fHo
[NaCl(s)] ~ - 411.12 kf/mol
[CaO( s)] ~ - 635.09 kJ/ mol
As bo ndin g is d escrib ed in greater detai l, yo u will d iscover th at th e two types o f bo ndin g- complet e electro n transfer and th e eq ua l sha r ing o f elec tr o ns-are extr em e cases. In mo st che m ica l compounds, electro ns are sha re d un equally, with th e extent o f sharing varyi ng widely from ve ry little shari ng (largely io nic) to co nside ra ble sharing (largel y cova len t) . Io n ic bonding will be d escribed in m ore d et ail in Chap te r 13, whi le th e present chap te r focu ses o n bo nd ing in covale nt compoun ds .
8 .2
Module 12
Th ere are m an y exam ples of co m pou nds having co vale n t bonds, including th e gases in o ur a tm os p here (° 2 , 1'\ 2' 1I20 , and CO 2 ) , co mmo n fu el s (C lI _I ) , and mo st of th e c o rn pounds in yo u r body. Covalen t bond ing is also respons ible fo r th e ato mto-atom co n nectio ns in co mmon ions suc h as CO j 2- , CN - , NI I"" . N0 3 - , a nd PO ,,:~ - . \Ve will d evelop th e basic pri nciples o f structu re an d bon di ng usin g th ese and o ther sma ll molecu les and ion s, but th e same prin cipl es ap ply to larger mol ecu les from aspir in to p rot eins and DN A with th ousands of atoms. The mol ecul es and io ns j ust mentioned are co m pose d en tire ly of nonmetal ato ms . A point th at needs special emphasis is th at , in molecul es o r ions mad e up onl)' of no n me tal atoms, the atoms are attac he d by covalen t bonds. Conve rsel y, the prese nce of a metal in a formula is o fte n a signal th at th e co mpou nd is like ly to be io nic.
:0:
I .. I ··
II I
II
H
C- O- H :0: H
H 'c ~c ,c "A~ -C- T-H
I
Valence Electrons and Lewis Symbols for Atoms
W,C ""C /C ' H
I
H Aspirin
.I
One goal of this chapter is to understand whya molecule such as aspirin has the shape t hat it exhibits . 350
Covalent Bonding and Lewis Structu res
Th e electrons in a n atom are of two types: valence electrons and core electrons. Che mical reac tio ns result in th e loss, gain , o r rearrange ment of vale nce electron s. Th e co re electro ns a re not involved in bondin g o r in che m ical re actions. Fo r main gro u p el e ments (e le me n ts of the A grou ps in th e periodi c table ) , th e valence el ect ro ns are th e s and I) e lectro ns in th e o u termost shel l (Ta ble 8 .1). All e lec tro ns in inne r sh ells are core el ectrons. A useful gu idel ine for main grou.t) elements is th at the number oj valence electrons is equal to the f.,rmul) number. Th e fa ct th at all elemen ts in a pe riodic g ro up have the same n u m ber o f valence elec tro ns acco u n ts for the sim ilari ty of che m ical properties amo ng m em bers of th e g rou p.
Sign in at www.thomsoned u.comjlogin to download the Go Chemistry module for this section or goto www.ichapters.com to purchase modules.
TABLE Element
8 .1 (are and Valence Electrons for Several (ammon Elements Periodic Group
Core Electrons
Valence Electron s
Total Configuration
Main Group Elements
Na
lA
l s22s22p6 = [Ne ]
3s'
[Ne]3s'
Si
4A
l s22s22p6 = [Ne ]
3s23p2
[Ne]3 s'3p'
As
5A
l S22s22p63s23p63dlO = [A r]3d1O
4s24p 3
[Ar]3d 104 s24p l
Transition Elements
Ii
4B
l s22s22p63s23p 6 = [Ar]
3d24s2
[A r]3 d'4s'
Co
8B
[A r]
3d74s2
[Ar]3o"4s'
Mo
6B
[Kr)
4d5 5s1
[Kr)4d'5s'
Valen ce electrons for transition elements in clud e the e lec tr ons in t he ns an d (n - 1) d orbita ls (see Table 8. 1) . T he rem ainin g electro ns are core electro ns. As with main group e le ments, th e valence e lec tro ns for tr an sitio n metals d et e rm in e the che m ica l p roperties of the se eleme n ts. T he Ame rica n che mist Gilbe rt New to n Lewis (1875- 1946 ) introduced a useful way to represent elec tro ns in the vale nce she ll of an ato m . The e leme nt 's sym bo l represents th e ato m ic n ucleus to geth er with the core electrons. Up to four vale nce elec tr o ns , re prese n te d by d ots, are placed one at a time arou nd t he sym bol; th e n , if any valence electrons re mai n , they are pai red with o nes alread y there. Chemis ts now refer to these pi ctures as Lewis electro n dot symbo ls. Lewis d o t symbols for main group el em ents o f th e second an d third peri ods arc shown in Ta ble 8.2 . Arra nging the vale nce electrons of a main grou p ele ment aro u nd an a to m in four gro u ps sugges ts tha t the vale nce she ll ca n acco mmodate fou r pairs of el ectr o ns. Because thi s re prese n ts e ight elec tro ns in all, thi s is referred to as an o ctet of electro ns . An oc te t o f e lectro ns surro u nd ing an ato m is regard ed as a stable co n figuration . Th e no ble gases, with th e exce pt io n of hel iu m , have eight valence electro ns and demo n str ate a not abl e lack o f reactivity. (H el iu m, neon , and argo n do not u nde rgo any chemical reactions, and the ot he r noble ga ses have very limi ted che mical react ivity) Because che m ica l reac tio ns in vol ve cha nges in the valence electro n shc ll, th e limited react ivity of the noble gascs is tak en as evid ence of th e stability of th e ir noble gas (w,2njf) e lectro n con figuratio n . H ydrog en , whic h in its co mpo un d s has two electro ns in ito; vale nce she ll, obeys th e spir it of thi s rul e by match in g th e electro n con figura tio n of He . ChemistrY-9-NDW~ Sign in at www.thomsonedu.com/Iogtn and go to Chapter 8 Contents to see Screen 8.2 for more on the correlation of the periodic table and valence electrons.
TABLE 8 .2
Lewis Dot Symbols for Main Group Atoms
lA
2A
ns2np 2
ns 2np 3
ns 2np 4
8A
ns'
ns 2np1
7A
ns'
3A
4A
SA
6A
ns2np s
ns 2 np 6
Li ·
· Be ·
. B·
. C'
. N·
: 0·
: F·
: Ne :
Na'
' Mg'
· AI·
. Si·
.p.
: S'
: Cl ·
: t« :
8.2
I
Covalent Bon ding and Lewis Structures
351
• Gilbert Newton lewis (1875-1946) Lewi s introduced the
theory of '0 .0= the shared .§,~ electron-pair ~ ~ -5~ 0,~ chemical bonds in a
paper published in the Journa l of
Lewis Electron Dot Structures and the Octet Rule In a sim ple d escri p tion o f covalent bo ndin g, a bond resu lts when o ne o r mo re e lec tro n pairs are sha re d between two ato ms . T hc e lectro n pair bond bet ween th c two ato ms of an 1-1 2 m olecu le is re p rese n ted by a pair of dots o r, a lternative ly, a lin e.
'!_ "'j"
Electron pair bond
I
~€
H: H
j!
the American
Gilbert Newton Lewis
Chemical Society in
1916. Lewis also made major contributions in add-base chemistry, thermodynamics, and on the interaction of light with substances. lewis was born in Massachusetts but raised in Nebraska. After earning his B. A. and Ph.D. at Harva rd, he began
I
H-H
T he representati on of a mo lecul e in th is fashio n is ca lled a Lewis electron dot structure o r just a Lewis structure in hon o r o f G. N. Lewis. Simple Lewis struc tu res, suc h as that for F2 , ca n be drawn starting with Lewis dot sym bo ls fo r ato ms and arra ng ing th e valence el ectrons to form bonds. Fluorine, an ele me n t in Grou p 7A, has seve n valence e lec tro ns. The Lewis sym bo l shows th at an F atom has a sing le unp aired elec tro n alo ng with th ree electron pairs. In F2 , th e sing le el ect rons, o ne o n eac h F ato m, pair lip in th e covale n t bond .
his career in 1912 at the University of
California at Berkeley. He was not only a productive researcher, but was also an influential teacher. Among his ideas was the use of problem sets in teaching, an idea still in use today.
l one pair of electrons "-
:F · + ·F:
~
:F:F :
or
:F -F:
.. I··
Shared or bonding electron pair
In th e Lewis str uctu re for F2 th e pair of elec tr o ns in th e F-F bond is the bonding pair, o r bond pair. T he o ther six pairs resid e on sing le atom s an d are called lone pairs. Beca use th ey are not involved in bonding, they arc also ca lled nonbonding elec tro ns. Ca rbon di o xid e , CO 2, and din itro gen , N z, are exam ples of mol ecul es in which two ato ms a rc mu ltiply bonded; tha t is, th ey share more th an o ne el ectron pair. • H Atoms and Electron Octet s Hydrogen atoms cannot be surrounded by an octet of electrons. An atom of H, which has only a ts valence electron orbital. can accommodate only a pair of electrons.
Octet of electrons around each 0 atom (four in double bond and four in lone pairs)
-,
0=(=0
:N=N:
I
Octet of electrons around each Nato m (six in triple bond and two in lone pair)
• Importance of Lone Pairs Lone pairs can be important in a structure. Since they are in t he same valence electron shell as the bonding electrons. they can influence molecular shape. See Section 8.6.
352
Chapt er
8
I
I
Octet of electrons around the Catom (four in each of two double bonds)
In car bo n dio xid e , the ca rbo n a to m shares 1\\' 0 pairs of e lec tro ns with eac h o xyge n and so is linked to eac h ato m by a d ouble bond . T he vale nc e shell o f ea ch o xygc n ato m in CO 2 has two bonding pa irs and two lon e pairs. In din itro gcn, the two nitrog e n ato ms sha re th ree pa ir s o f e lec tro ns, so they are link ed by a tr ip le bond . In ad d itio n, each N ato m has a sing le lon e pair. An impo rta n t o bse rva tio n can be mad e abo u t th e molecul es you have seen so far: each a tom (excep t I I) has a sha re in fou r pairs o f e lectrons, so eac h has ac h ieve d a noble gas configuration. Each atom is surrounded bJ an ode! of fight electrons. (I Jydrogen typically form s a bond to only one o the r ato m, re sulting in two electr ons in its vale nce she ll.) The tendency of molecules an d pols aunnic ions 10 have structu res in which eight electrons surrourul each atom is kn own as the octet rule. As an examp le , a tr ipl e
Bo nding and Mo lecular Structure
°
bo nd is n ece ssary in d in itrogen in o rd e r to h ave a ll o ctet a ro un d each n it rogen a to m . T h e ca r bo n ato m a n d both ox ygen ato ms in CO 2 ac h ieve th e o ctet co n figuratio n hy formin g d ouble bo nds. Th e octet ru le is ex tre me ly useful, but keep in m ind that it is mo re a guideline than a rul e. Particu la rly for th e sec o nd pe rio d elemen ts C, N , 0 , an d F, a Lewis struc ture in whic h eac h ato m ac h ieves a n oc te t is like ly to be correc t. Alth ou gh there are a few ex ce pt io ns, if an atom such as C, N, 0, o r F in a Lewis str uctu re does n o t follow th e octet rul e , you sh o u ld question th e str uc tu re 's valid ity. If a stru c tu re o be yin g th e octet rul e ca n no t be wri tte n , the n it is po ssib le an incorrect formul a h as been assigned to th e com po un d or th e ato ms have be en assem bled ill a n in co rrect way.
• Exception s to th e Octet Rule Although the octet rule is widely applicable. there are exceptions. Fortunately. many will be obvious. such as when there are more than four bonds to an element or when an odd number of electrons occur. See Section 8.5.
Drawing Lewis Electron Dot Structures Th e re is a syste m atic a p p roac h to cons truc tin g Lewis struc tu res of mo lecu les and ion s. Le t us ta ke formaldeh yde , CI-I 20, as a n e xample . 1. Determine (he arraneement oJatoms within a molecule. Th e ce n tral ato m iS 1.lS1Ulll)' th e on e with th e lowest el ect ron affin ity. In C112 0 , the ce n tr al at o m is C. You will co me to recognize tha t ce rta in el ements o fte n a p pea r as th e ce n tra l atom , amo ng the m C, N, P, a nd S. Iialo ge ns a re ofte n te r m in al ato ms fo rmin g a sin gle bo nd to o ne ot h er ato m , bu t the y ca n be th e ce n tral ato m whe n co m bine d with 0 in oxoacids (su c h as H CI0 4 ) . O xygen is th e ce n tra l ato m in water, but in conj u nc tion with nitrogen , p hosp horus , a n d th e halogen s it is us u all y a te rmi n al ato m . I lyd ro gen is a te rm inal a to m becau se it typ ica lly bonds to o n ly o ne o the r a to m. 2. Determine the total n um ber oJ -aalenre electrons in tile molecule or ion. In a n eut ral molecu le , thi s nu m be r will be the sum o f th e valen ce electron s for eac h a to m . For a n an ion, add th e n umber o f el ect ron s eq u al to th e n egati ve c h a rge ; fo r a ration, su btract the n u m be r of e lec trons eq u al to th e positive c harge . T h e numbe r o f vale nce electron pairs will be h al f th e to tal n umber of vale n ce electro ns. For C H 2 0 ,
Va lence electrons
• Choosing th e Central Atom 1. The electronegativities of atoms can also be used to choose the central atom. Hectroneqativity is discussed in Section 8.7. 2. For simple compo unds. the first atom in a formula is often the central atom (e.g.. SOb NH4 + . N0 3 - ) . This is not always a reliable predictor. however. Notable exceptions include water (H20) and most common acids (HN0 3• H2S04) . in which the acidic hydrogen is usually written first but where another atom (such as Nor S) is the central atom.
= 12 electrons (or 6 electron pairs) = 4 for ( + (2 X 1 for two H ato ms) + 6 for 0
3. Place oue pair oj electrons beuieen each prur oj bonded atoms to Jorm a single bond.
H Single bond - \
( -0
I
H
He re , three e lectro n pairs are used to m ake th re e sing le bond s (whic h a re represen ted by sing le lin es) . T hree p airs o f el ectro ns rem ai n to be used . 4 . U5e any remaining pairs as lone pairs aroun d each term inal atom (excep t H) so tha t eacli terminal atom is surroun ded 'I)' eight electnms. If, afte r th is is d one , th e re a re electro n s left over, assign th e m to the ce n tral ato m . (If the ce n tra l ato m is a n cleme nt in the third o r h igh er period, it ca n have mo re than eig h t elec tr o ns. See p age 364. ) H Single bond - \
/
lone pair
( -0 :
I
H
8 .2
I
Covalent Bonding and l ewis Structures
353
I le re , all six pairs have been assig ne d, but no tice that the C ato m has a share in only thre e pairs. 5 . If the central atom has [noer tha n eight electrons at this IJOin t, change one 01' more of the lone pa irs on the termin al atoms in to a bonding pair betuxni the central and terminal atom to fonn a multiple bond.
Move lone pair to create double bond and satisfy octet for C. • 0 as a Terminal Atom Oxygen atoms are usually terminal at oms when combined with many other atoms such as B. C. Si, S, N. P, and the halogens.
As a general r ule , double or tr iple bo nds are mo st often encountere d whe n both at oms arc from th e fo llowing list: C, N, o r O. T ha t is, bon d s such as C=C, C=~ , a nd C=O will be enco untered Frequently.
Chemistry.{:>.Now" Sign in at www.t homsonedu.com/ Ioqtn and go to Ch apter 8 Contents to see: Screen 8.4 for an animation of the factors influ encing bond formati on Screen 8.5 for a tutorial on Lewis structures Screen 8.6 for a tutorial and an exercise on drawing Lewis st ructures
Drawing Lewis Structures ProbLe m Draw lewis structures for the chlorate ion (Cl03 -) and t he nitronium ion (N0 2+). Strategy Follow the five steps outlined for CH 20 in t he preceding text. SoLution fo r chlorate ion
1. Clis t he central atom, and the 0 atoms are t erminal at oms. 2. Valence electro ns = 7 (for CL)
+
= 26 (13
pair s)
18 (six for each 0 )
+
1 (for the negative charge)
3. Th ree electron pairs form single bonds from Clto the 0 term inal atoms.
o I
0 - ([ -0 4. Distribute three lone pairs on each of the te rminal 0 atoms to complet e the octet of electro ns aroun d each of these atom s.
[: ~ -il -~ ' .. 0.
]-
5. One pair of electrons remains, and it is placed on the central Cl atom to complete its octet.
[
..
: .. 0:
I ..
]-
: ~ - C l - ~:
Each atom now has a sha re in four pairs of electr ons. and the l ewis structu re is complete. Solution for nitronium ion
1. Nitrogen is the central atom , because its electron affinity is lower than that of oxyge n.
354
Chapter 8
I
Bonding and Molecular Structure
2. Valen ce electrons = 16 (8 valence pairs)
= 5 (for N) + 12 (sixfor each 0) - 1 (for the positive charge) 3. Two electron pairs form single bondsfrom the nitrogen to each oxygen: O-N-O
4. Distribute the remai ning six pairs of electrons on the terminal 0 atoms:
[ ~ -N-~r 5. The central nitrogen atom is two electro n pairs short of an octet. Thus, a lone pair of electrons on each oxygen atom is converted to a bonding electron pair to give two N=O double bonds. Each atom in the ion now has four electron pairs. Nitrogen has four bonding pairs, and each oxygen atom has two lone pairsand shares two bond pairs. Move lone pairs to create double bonds and satisfy the octet for N.
Comment Why don't we take two lone pairs from one side and none from the other? We shall discuss that after describing charge distribution in mo lecules and ions (page 377). EXERCISE 8 .1 Drawi ng Lewis Stru ct ures
Draw Lewis structures for NH 4 " , CO, NO +, and SO/ - .
Predicting Lewis Structures Lewis struc tu re s are useful in gai ni ng a perspective on th e str uc tu re an d che mist ry o f a molecu le or ion . T he gui del ines for drawing Lewis struc tu res are helpful , but che m ists a lso rel y on patterns o f bonding in re lated molecules. Hydrogen Compounds
Som e co m mo n co mpou nds an d ion s form ed from second -period non met al elemen ts and hydro gen are shown in Table 8.3. Their Lewis struc tures illustrate the fac t th at the Lewis symbol for an clement is a u seful gu id e in d etcrmining the number o f bo nds fo rm ed by th e ele ment. Fo r exam ple , if th ere is no charge , n itrogen has five valence el ect ron s. Two elec tro ns occur as a lon e pair ; th e other three occu r as unpaired electron s. To reach a n o ctet, it is nece ssary to pai r eac h of th e unp aired e lec trons with an e lec tro n from ano ther ato m . Thus, N is p redict ed
Problem Solving Tip 8.1 The octet rule is a useful guideline when drawi ng Lewis structures. Ca rbon forms four bonds (four single bonds; two single bonds and one double bond; two double bonds; or one single bond and one triple bond). In uncharged species, nitrogen forms t hree bonds and oxygen forms two bonds. Hydrogen typically forms only one bond to another atom.
Useful Ideas to Consider When Drawing Lewis Electron Dot Structures When mu ltiple bonds are formed, bot h of the atoms involved are usually one of the following : C, N, and O. Oxygen has the ability to form multiple bonds with a variety of elements. Carbon forms many compou nds having multiple bonds to another carbon or to Nor O. No nmetals may form single, double, and triple bonds but never quadruple bonds.
8.2
Always account for single bonds and lone pairs before determining whether multiple bonds are present. Be alert for t he possibility t he molecule or ion you are working on is isoelectronic (page 358) with a species you have seen before.
I
Covalent Bondi ng and lewis Structures 355
TABLE 8 .3
Lewis Structures of Common Hydrogen-Containing Molecules and Ions of Second-Period Elements
Group 4A
[H,
Group SA
H
met hane
NH,
I
ammonia
H- [ -H
I
Group 6A
H- N-H I H
water
H- N- N-H I I H H
hydrogen peroxide
Group 7A
H-O -H
H2O
HF
H-F:
hydrogen fluoride
H ( 2 H6 eth ane
H H
I
I
I
I
N2 H4
H-[ - [ - H
hydrazine
H-O -O - H
H20 2
[..r
H H H- [ = [ - H
C2 H4 eth ylene
( 2 H2 acetyle ne
I
I
H H
H- [ = [ - H
NH 4 +
ammo nium ion
[--1-.1'
NH 2 -
[H -~ - Hl
amide ion
H3O +
H - ~ -H
hydro nium ion
[~- Hl
OW hydroxide ion
to fo rm th re e bonds in u n ch arged m o lec u les , a n d thi s is indeed t he cas e . Sim ilarly, ca rb o n is e xpe cted to fo r m four bonds, oxyge n two , a n d flu o rin e o n e . Group4A
I
- [-
I
Group5A
Group 6A
- N-
- 0-
Group 7A
: F-
I
Predicting Lewis Structures Problem Draw Lewis electron dot structures for CCl4 and NF 3 • Strategy One way to answer th is is to recognize that eCl.. and NFl are similar to CH 4 and NH ] (in Tab le 8.3), respectively, except th at H atoms have been replaced by halogen atoms.
SoLution Recallt hat carbon is expected to form four bonds and nitrogen t hree bonds to give an octet of electrons. In addition, halogen atoms have seven valence electrons, so both Cland F can attain an octet by forming one covalent bond, just as hydrogen does. : [I:
..
I
..
: F- N-F :
: Cl - [ - Cl:
I
I
:[I:
: F:
carbon tetrachloride
nitrogen trifluoride
As a check, count the number of valence electrons for each molecule, and verify that all are present.
eeL: Valence
electrons
= 4 for C + 4
X 7 (for el)
= 32 electrons (16 pairs)
The str ucture shows eight electrons in single bonds and 24 electrons as lone pair electrons, for a tota l of 32 electrons. The structure is correct. NF3 : Va lence electrons
= 5 for N + 3 X 7 (for F) = 26 electrons
(13 pairs)
The st ructure shows six etectrons in single bonds and 20 etectro ns as lone pair electrons, for a totat of 26 electrons. The structure is correct. 356
Chapter 8
I
Bonding and Molecular Structure
EXERCISE 8 .2 Predicting Lewis Structures Predict Lewis structures for methanol. H3COH, and hydroxyla mine. H2N OH. (Hint: The formulas of these compou nds are writte n to guide you in choosi ng the correct arrangement of atoms.)
Oxoacids and Their Anio ns Lewis str uc tu res o f common acids and their a n io ns a re illu stra ted in Ta b le 8.4. In the absence of water, th ese acids are cova len tly bo nd ed molecu lar compounds, a
• Lewis Structures for Anions of Oxoadds Stuctures for oxoanions such as PO,3-, 50, 2-, and elO, - are sometimes drawn with multiple bonds between the central atom and oxygen. Theory suggests that this does not accurately represent the bonding in these species, and that structures obeying the octet rule are more appropriate. See l. Suidan, J. K. Badenhoop, E. D. Glendening, and F. Weinhold, Journal of Chemical Education, Vol. 72, pages 583-585, 1995.
co nclusio n that we sho u ld draw because all elements in th e formu la are no nm e tals. (N itr ic acid, for exa m ple, has properti es tha t we assoc iate with a co valen t co mpound: it is a co lo r less liquid with a boi lin g point o f 83 °e .) In aq ue o us so lu tio n, however, lI NO J , I-J 2 SO h and II CIO -t are io ni zed to give a hyd ron ium ion an d the app ropriate a nio n. A Lewis stru cture for the ni trate io n , for ex amp le , can be create d using th e gui d e line s o n page 353 , a nd th e result is a struc tu re with two N - O sin gle bonds and one 1"= 0 double bond . To form nitric acid from the n itrate io n , ato ms that has a sing le bo nd to the a hyd rogen io n is att ac hed to o ne of t he ce n tral N.
°
:O O:] ..- N I = .. [ :0 :
H-O -N =O :
nitrate ion
nitric acid
I
:0 :
A charac teristic p ropert y of aci ds in aCll leo us so lu tio n is th e ir ability to donat e a hyd rog en ion (IJ +, wh ich co mbines with water to g ive the h ydronium ion ) . The r\ 0 :1- anion is formed when th e ac id, H NO ;3, loses a hyd rogen ion . T he H + ion
TABLE 8 .4 HN0 3
Lewis Structures of Common Oxoadds and Their Anions H-O - N=O :
I
nit ric acid
:0 :
HlO, phosphoric acid
: O- H
H2S04 sulfuric acid
I : O- p- O: .. I I
: O- H
I .. I ..
: O-S -O: : O- H
H- O: H NO,nitrate ion
HCIO, perchloric acid
:O - N= O:] ·. I [ : 0:
: O- H
·.
: O-
I .. CI - O: I ..
P043phosphate ion
HSO, hydrogen sulfate ion
: O- p- O:
[.. HOCI hypochlorous acid
.I .. ]3-
' 0'
..
I ..
: 0:
SO/sulfate ion
H-O -CI :
: 0:
ClO,perchlorate ion
]:O - JI -O : [ ' .. 0'
·.
I
OClhypochlorite ion
. [
~-H ] -
:os-o : ·. I .. :0 : '0'
]'-
·. ..I .. :o-s-o : ·. I ..
[
: 0:
H-ciT
..
: 0:
8 .2
I
Covalent Bonding and lewis Structures
357
separa tes fro m the acid by bre aking the II- O bo nd, the el ectrons of the bond staying with th e 0 ato m . As a re su lt, H N O J and N O :I - have th e same n u mber o f electron s, 24, an d the ir struc tures are closely rela ted.
EXERCISE 8 .3 Lewis Structures of Adds and Their Anions Draw a Lewis structure for the a nion H2 P0 4 ", derived from phosphoric acid.
Isoelectronic Species • Isoeleetronic and Isostruetural The term isostruetural is often used in conjunction with isoelectronic species. Species that are isostructural have the same structure. For example, the P04 3- , 50 4 2 - , and Cl04 - ions in Table 8.5 all have four oxygens bonded to the central atom. In addition, they are isoelectronic in that all have 32 valence electrons.
T he spe cies NO " . N 2 , CO , and CN - are similar in t hat th ey each have two ato ms an d the same total nu mbe r of valen ce electron s, 10, which lead s to the same Lewis struc tu re fo r each mo lecu le or ion. T he two ato ms in eac h are lin ked with a trip le bond . \Vith th ree bon d ing pairs and one lo n e pair, ea ch ato m th us has an o cte t o f electron s. [: N= O :]+
: N==N :
:(= 0:
Molecul es an d ions having th e same number oj valence eleclnms find the same l .exois struct ures ere: sa id to be isoelectronic Cra ble 8.5). You will find it hel pful to recogn ize
isoel ectronic mol ecul es and io ns becau se thi s is ano the r way to see relatio nsh ips in bonding am ong common che mical su bs tances. T he re are similar ities and im po rtan t di ffe ren ces in che m ical properties o f isoelectr on ic spec ies. For exam ple , bo th carbon mo no xide, CO, and cyanid e io n, CN - , are ver y to xic, which resu lts from the fact tha t they can hind to the iro n of hem og lo bi n in blo od and block th e up take of o xygen . T hey are di ffe ren t, t houg h, in th e ir acid-base chemistry. In aq ueo us so lu tio n, cyan id e ion readil y adds H + to fo rm h ydrogen cyan id e , whereas CO does not p rot o nat e in wat er,
EXERCISE 8 .4 Identifying Isoelectronic Species (a) Is the acetylide ion, C/ -, isoelectronic with N2? (b) Identify a common molecular (uncharged) species that is isoelectro nic with nitrite ion , N02 Identify a common ion that is isoelectronic with HF.
.
TABLE 8.5 Some Common Isoelectro nic Molecules and Ions Formulas
Repre sentative lewis Structure
[--1-.J
Formulas
Representative lewis Structure
:O O:] ..- N I = .. [ :0 :
.. ]3-
'0 '
H- N-H
I
H
co,
OCN - , 5C N-, N,O NO,.. OC5, C5,
358
Chapter 8
I
Bonding and Molecular Structure
0 =( =0
:O- ~-O I ..
[..
:0:
8 .3
Atom Formal Charges in Covalent Molecules and Ions
'You have seen that Lewis struc tu res sh ow h ow e lectro n pairs a re p lac ed in a covalen tly bonded species, whe ther it is a ne ut ra l molec u le o r a pol ya tomi c io n . Now we turn to one o f th e co n se q u e nc e s o f th e pl acem en t of el e c tr on p airs in thi s way: individ ual atoms can be n ega tively o r po sitively c ha rge d or h ave n o e lec tric charge . T he loca tion of a po siti ve o r n egat ive c harge in a mo lecu le or io n will in flue nce, amo ng o t her th ings , th e a to m at wh ic h a re action occ urs. Fo r exa m p le, docs a positive II + io n a ttac h itself to th e C1 Of the a o f CIO -? Is the p roduct H elO or H O CI? It is reasona ble to expect H + to at tac h to th e mo re n egatively c ha rged ato m . \\" c ca n pred ict thi s by eva luati ng ato m formal c h a rges in mol ecul es an d ions. T he formal ch arge is the charge o n a n atom in a mo lecu le or pol vatomlc ion , a n d th e su m of th e forma l c ha rges fo r t he atoms in a species equa ls th e ove rall charge o n th e ion or is zero (for a n un charged mol ecule ) . T he forma l charge for a n atom in a mo lec u le or io n is calc u la ted b ased o n th e Le wis str u ctu re of th e mo lecu le or io n , usin g Equatio n 8.1.
Formal charge of an atom in a molecule or ion group number of the atom - [LPE + 'I,(BE)]
(8.1)
In this equa tio n : • The gro u p n u mber gives the n um ber of vale nce elec tro ns bro ug h t by a pa rtic u lar atom to th e mol ecul e or ion . • LPE = numbe r of lon e pai r electrons o n a n ato m . • BE = n u m ber o f bonding elec trons aro und a n a to m . Th e term in sfJuare brackets is the n u m be r of e lectro ns assigned by the Le wis stru ctu re to a n atom in a m olecul e o r ion . T he diffe ren ce be twee n th is term an d th e g rou p number is th e formal c h a rge. An ato m in a mo lec u le o r io n will be p ositive if it "co ntribu tes " mo re electr o ns to b ondin g th an it "gets ha ck." T he ato m 's fo rmal c h a rge will be n egative if th e reve rse is true . T here a re two important assu m ptions in Equa tion 8. 1. First, lone pairs are assu med to be long to the ato m o n whic h they resi de in the Le wis stru ctu re . Seco nd , bond pa irs are assu med to be divid ed equa lly between th e bonded ato ms. (T he facto r of Y2 divid es the bon d in g elec tro ns equa lly between th e ato ms lin ked by th e bo nd.) T he su m o f th e forma l ch a rges on th e ato m s in a mo lec u le or io n always eq ua ls t he net c harge o n th e mol ecul e o r ion . Co nside r th e hypo chl o rit e ion . Ox yge n is in Gro u p 6A a nd so h as six vale nce elect rons. H oweve r; oxygen ca n la y cla im to seve n e lectro n s (six lon e pair e lectro ns an d one b o nd in g electron ), an d so th e a to m has a formal c ha rge of - 1. T he 0 atom h as "fo rm ally" gai n ed an elect ro n as part of th e io n . Formal charge = - 1 = 6 - [6 +
-8J-
[:ci .. I " I
Formal charge = 0
=
~ (2 ))
) Sum offi"mol chorqes
~-1
Assume a covalent bond,
7 - [6 + i (2)] so bonding electrons are
divided equally between Cl and O.
Th e formal cha rge o n th e C I ato m in CIO - is ze ro . So we h ave - 1 for ox ygen an d o for chlori ne, a n d th e sum of these equa ls the net charg e o f - I for th e ion . An impo rtant co n clusio n we can draw fro m the fo rma l c ha rges in CIO- is tha t, if all 8 .3
I
Atom Formal Charges in Covalent Molecules and Ions
359
II + io n ap p roaches the io n , it shou ld attac h itself to the nega tively charg ed 0 ato m to give hypoch lorous ac id, HOCI.
ChemistrY.{J.Now'" Sign in at www.thomsonedu.comjlogin and go to Chapter 8 Co ntents to see Screen 8.11 for practice determining formalcharge.
CalcuLating FormalCharges Problem Calculate formal chargesfor the atoms of the ClO]- ion.
• HClO..Ad ds and Formal Charge Both ClO- and ClO l - ions attract a proton to give the corresponding acid, HClD and HClO]. In all of the HClD. acids, the W ion is attached to an 0 atom, owing to the negative formal charge on that atom . See Table 8.4.
Strategy The first step is always to write the Lewis structure for the molecule or ion. (The Lewis structure for the Cl0 3 - ion is in Example 8.1.) Then Equation 8.1 can be used to calculate the format charges. SoLution Formal charge = - 1 = 6 - [6 + ~ ( 2 ) ]
.1.
..
:0:
]_
I ..
[ : ~ - C l - ~:
I
Formal charge = +2 = 7 - [2 + t (6)] The formal charge on each a atom is - 1, whereas for the Cl atom it is + 2. The sum of the atom's formal chargesis the charge on the ion, which is - 1 for ClO ]- .
EXERCISE 8.5
CalcuLating FormalCharges
Calculate formal charges on each atom in (a) CW and (b) SO /-.
A Closer Look
Comparing Formal Charge and Oxidation Number
In Chapter 3, you learned to calculate the oxidation number of an atom as a way to tell if a reaction involves oxidation and reduction. Are an atom's oxidation number and its formal charge related? To answer this question, let us look at the hydroxide ion. The formal charges are - 1 on the 0 atom and Don the H atom. Recall that these format charges are calculated assuming the O- H bond electrons are sha red equally in an O- H covalent bond.
In contrast, in Chapter3 (page 144), you teamed that 0 has an oxidation number of -2 and H has a number of + 1. Oxidatio n numbers are determined by assuming that the bond between a pair of atoms is ionic, not covalent. For OH- this meansthe pair of electrons between a and His located fullyon O. Thus, the oatom now has eight valence electrons instead of six and a charge of - 2. The Hatom now has no valence electrons and a charge of + 1. Oxidation number
Forma l charge = - 1 = 6 - [6 + t (2)]
/
HJ[:0.. 1
Sumatformal charges =
Chapter
8
I
[OJ iT
-1
Formal charge = 0 = 1 - [0 + t (2)]
360
/
Assume an ionic bond
Bonding and Molecular Structure
=
-2
Sumof oxidation numbers = - 1
Oxidation number = + 1
Formatcharges and oxidation numbers are calculated using different assumptions. Both are useful, but for different purposes. Oxi dation numbers allow us to follow changes in redox reactions. Formal charges provide insight into atom charges in molecules and polyatomic ions.
8.4 Resonance Ozone, O :h an un stabl e, blu e, di am agnetic gas with a cha rac teristic p ungent odor, pro tcctt;; th e eart h a nd it" inh ab itan t" fro m intense ul tra viol e t rad iati on from th e SU Il. An im portan t fea tu re of its struc tu re is th at the two oxyge n-oxygen bo nds are the same length , whic h suggests th at th e two bo n ds are equivalent. T ha t is, eq ua l 0 - 0 bond lengths im p ly a n eq ual numbe r of bon d pairs in ea ch 0 - 0 bo nd. Using th e gu idelines for d rawin g Lewis struc tures, h owever, yOll might co me to a different co nclusio n. There are two possib le ways of writing the Lewis struc ture for the mo lecule:
Alternative Ways of Drawing the OzoneStructure Do uble bond on the left:
0 =0 - 0 :
Ozo ne, 0 3, is a bent molecule wit h oxygen-oxygen bonds of the same length .
Double bond on the right: : 0 -0 =0 These stru ctu res are eq uiva lent in th at eac h h as a doub le bond on one side of th e cen tra l ox ygen ato m and a single bond 011 the ot her side . If eit her were th e actual str uc ture o f ozo ne , o ne bon d (0=0) shou ld be sho rter th an th e o ther (0- 0) . The actual structu re of ozone shows th is is no t the cas e. The ine sca pa ble co nclusio n is that th ese Lewis structures do n o t correctly re presen t th e bondin g in ozone. Linus Paul in g ( 190 1-1994) propo sed the theory o f resonance to so lve th e pr oblem . Resonance structures are used to represent bonding in a molecule or ion when a single l rutis structu refails to describe accurately the actual electronic structure. T ile alterna tive structu res show n for ozo ne a re ca lled resonance structures. T hey have ide n tica l patte rn s of bonding a nd equal energy. The actual structure of this mol ecule is a composi:«, or resonance hyb rid, o f the eq u ivalent resonance structures. In thi s hybri d , the bo nds betwee n th e oxyge ns are between a sing le bond a nd a double bond in leng th , in this case correspo nding to one a n d a half bo nds. Th is is a reaso na b le conclusion becau se we see that the 0 - 0 bonds both have a leng th o f 127.8 p m, intermed iate be tween th e a\'erage length of a n 0=0 dou b le bond (12 1 p m) a nd an 0 - 0 sing le bond ( 132 pm ) . Because we can not accura te ly draw fraction s of a bo nd , che m ists draw the re so nanc e stru ctu res a n d con nect th em with dou ble-headed a rrows (f----7) to indi cate th at the true str uc tu re is so mew he re in bet ween th ese ex tre mes. Lone pair becomes a bond pair. Bond pair becomes a lone pair.
I
I
:o'lob6:
~ : o ~ o - o:
• Depicting Resonance St ructures The use of an arrow (ot---+) as a symbol to link resonance structures and the name "resonance" are somewhat unfortunate. An arrow might seem to imply that a change is occurring, and the term resonance has the connotation of vibrating or alternating back and forth between different forms. Neithe r view is correct. Electro n pairs are not actually moving from one place to another.
Be n zene is th e classic exam ple of th e use o f res o na nce to re presen t a struc tu re . T h e b e n ze ne mo lec u le is a six-mem ber ring of carbon atoms with six equivale n t carbon-carbon b onds (and a hyd rogen a to m attac he d to eac h carbo n atom ) . The carbon-carbon b o nds a re 139 p m lo ng , interm ediat e be twee n th e average length ofa C=C d ouble bon d ( 134 p m ) a n d a C-C single bo n d ( 154 pm ) .
Resonance structures of benzene, C6H 6
Abbreviated representation of resonance structures 8 .4
I
Resonance
361
Problem Solving Tip 8 .2
Resonance Structures
Resonance is a means of representing the bonding when a single l ewis st ructure fails to give an accurate pictu re. The atoms must have the same arrangement in each resonance structure. Attaching the atoms in a different fashion creates a different compound.
Resonance structures differ only in the assignment of electron- pair positions, never atom positions. Resonance structures differ in the number of bond pairs between a given pair of atoms. Even though the forma Lprocess of converting one resonance structure to anot her seems to move electrons
about. resonance is not meant to indicate the motion of electrons. The actual structure of a molecule is a composite or hybrid of the resonance structures . There will always be at least one multiple bond (double or trip le) in each resonance structure.
Two resona nce struc tu res th at differ o nly in doub le bond placement ca n be writte n for th e molecul e . A hybrid of th ese two struc tu res , ho wever, will lead to a mo lecule with six eq uivale n t ca r bo n- ca rbo n bonds. Let us a p ply th e co nce p ts of re son ance to d escribe bonding in th e ca r bo na te ion, CO l - , an ani on with 24 valence electrons (12 pairs) .
" ]2-
:0 =( -0 :
"" I ""
[
:0:
"" [
]2-
:0 - ( =0 : :0-(-0 :]2"" II "" " " I" " [ :0 : :0 :
Three eq u ivalen t struc tu res can be drawn for thi s ion , differing o nly in th e lo ca tio n of th e c= o d ouble bond. This fits the classica l situ ation for reso nan ce , so it is ap pro pria te to co ncl ude tha t no single struc ture correc tly d escribes this io n . Inste ad , th e ac tua l struc tu re is the composite o f the three struc tu res, in good agreement with ex pe rime n ta l res u lts. In th e CO}!- ion , all th ree carbo n-oxygen bond di stan ces are 129 pm , int ermediat e bet ween C-O sing le bond ( 143 pm ) and C= O doubl e bond (1 22 pm ) di stan ces. Fo rm al charges ca n he calcula ted fo r ea ch a to m in the reso nance structure for a mol ecul e o r ion . For exa m ple . usin g one of the res o nanc e struc tures for th e nitra te ion , we find th at th e ce n tra l N ato m has a formal cha rge of + 1, and th e sing ly bonded 0 atoms are both - I. The doubly bonded 0 ato m has no charge. The net charge for the ion is thus - I. Formal charge = 0 = 6 - [4 + t (4)]
I
"0 "
"" II
""
]-
[ : O- N- O:
"" I ""
I
Format charge = + 1 = 5 - [0 + t (8)] Formal charge = - 1 = 6 - [6 + t (2))
Is th is a reasonabl e charge distribution for th e nitrat e ion? The a nswer is no . Th e ac tual str uc ture of th e nitrate ion is a res o na nce h yb rid of three eq uivale n t res o nance stru ctures. Becau se th e th re e oxygen ato ms in N0 3 - are eq uivalent, th e charg e on o ne o xyge n ato m shou ld not be different Irom the o ther two. Th is ca n be res o lved, however, if th e form al charges are ave raged to give a fo rmal charge o f - % o n th e o xygen at o ms. Summing th e cha rges on the three o xygen ato ms an d th e + I ch arge on th e nitrogen ato m th en g ives - I , th e ch arge on th e ion . 3 62
Chapter
8
I
Bonding and Molecular Structure
In th e resonan ce structu res for O:h C0 32 - , and N 0 3 - , fo r exa m ple , all th e possible reso nan ce structures are equa lly likely; they are "eq uivale n t" struc tu res . T he mol ecul e o r ion the re for e has a sym metrica l di stributi on o f e lec tro ns over all th e ato m s involved-c-rhat is, its e lec troni c structu re co ns ists o f an equa l "m ixtu re," o r "hyb r id ," o f th e re sonance struc tures .
Sign in at www.thomsonedu.com/login and go to Chapter 8 Contents to see Screen 8.7 for a tutorial on drawing resonance structures.
, . Drawing Resonance Structures Problem Draw resonance structures for the nitrite ion, N0 2- . Are the N-O bonds single, double, or intermediate in value? What are the formal charges on the Na nd 0 atoms? Strategy Draw the Lewis structure in the usual ma nner. If multiple bonds are required, resonancestruct ures mayexist. This wiLL be the case if the octet of an atom can be completed by using an eLectron pair from more than one terminal atom to form a multiple bond. Bonds to the central atom cannot t hen be "pure" single or double bonds but rather are somewhere between the two. SoLution Nitrogen is the central atom in the nitrite ion. which has a tota l of 18 vaLence electrons (nine pairs). Valence electrons
= 5 (for the Natom) + 12 (6 for each 0 atom) + 1 (for negative charge)
After forming N- O single bonds, and distributing lone pairs on the terminal 0 atoms, a pair remains, which is placed on the central Natom.
O:J[:O ..- N- .. To complete the octet of electrons about the Natom, form an N=O double bond.
Because there are two ways to do this, two equivalent structures can be drawn , and the actual structure must be a resonance hybrid of these two structures. The nitrogen-oxygen bonds are neither single nor double bonds but have an intermediate value. Taking one of t he resonancestructures, we find t he formalcharge for the Natom is o. The charge on one a atom is 0 and - 1 for the other 0 atom. Because the two resonance structures are of equal importance, however, t he net forma l charge on each 0 atom is - %. Formal charge =
Formal charge =
0 =6 - [4+ W)]
I
- 1 = 6 - [6 +t(2)]
I
[~=N -~ J I Formalcharge = 0 = 5 - [2 + t (6)] EXERCISE 8 .6 Drawing Resonance Structures Draw resonance structures for t he bicarbonate ion, HCOl "" . (a) Does HCO l - have the same number of resonance structures as the CO / - ion? (b) Wh at are the formal charges on the 0 and Catoms in HCa l -? What is the average formal charge on the a atoms? Compare this with the 0 atoms in CO l 2 - . (c) What do formal charges predict about the point of att achment of the H atom in HeO l -?
8.4
I
Resonance
363
8 .5
Exceptions to the Octet Rule
Although t he vast maj o rity o f molec u lar co m po u nd s a nd ion s o bey th e o ctet rul e, th ere arc exce ptio ns . T h e se incl u de molecule s and io n s th at have fewer th an fo u r pa irs o f elec tro ns on a ce n tr al a torn, th ose tha t have more th an four pairs, an d tho se th at h ave a n o d d n u m ber o f el e c trons.
Compounds in Which an Atom Has Fewer Than Eight Valence Electrons Boron , a metalloid in Gro u p 3A, has three vale nce electro ns and so is ex pec te d to fo rm th ree cova le n t bon d s with othe r non met a llic e le men ts. Th is resu lts in a valen ce shel l fo r boron in its co mpo u nds with on ly six elec tro ns , two sho rt o f an octe t. Man y bo ro n co m pou nds o f th is type are known, in cluding such co m mo n co mpo u nds as boric ac id (B(O Il) ,,), bo ra x (Na 2B,O,,( 0 1l) , . 8 11, 0 ) (Figure 8.2), and the boron trihalides (BF3 , Be l" BBr,,, and BI3 ) . H
I
: F- B- F:
'O -B - O:
I
.- / 8 atom surrounded /
_ bY4 electron pairs
e-4lIt-'1II!
I
I
: F:
H ' O-H
boront rifluo ride
boric add
Bor o n co mpo u nd s suc h as BF:1 th at are two electrons sho rt o f an o ctet are q uite reactive. T he boron at om can acco mmodate a fourth e lec tro n pai r whe n th at pair is p rovid ed by a no ther ato m , an d molecules o r ion s with lo ne pairs ca n fu lfill thi s role . Am mon ia , fo r ex am plc . react s wit h BF3 to fo rm H 3 N ~B F:) .
"?" H
I H-N : + I
B atom surrounded by 3 etectron pairs
H
FIGURE 8 .2 The anion in Borax. Borax is a common minera l, which is used in soaps and contains an interesting anion, B405 (OH)42 - . The anion has two B atoms surrounded by four electron pairs, and two Batoms surrounded by only three pairs.
covalent bond
:F:
I .. B- F: I
:F: ..
~
..
H :F: ~
I I .. I I
H- N- B-F : H :F: ..
If a bon di ng pair of e lec tro ns o riginat es o n o ne of the bonded ato ms, the bo nd is called a coordinate covalent bond. In Lewis struc tu res, a co ord inate co vale n t bo nd is o fte n d esign ated by an arr ow that po int s away from the a to m d o n atin g the e lectro n pair.
Compounds in Which an Atom Has More Than Eight Valence Electrons Ele me n ts in th e th ird or h ighc r periods oft e n form co m po u nds and io ns in which the ce n tr al ele me n t is su rr o u nded by more than fo ur vale nce elec tro n pai rs (Ta ble 8.6). \Vith mo st co mpo un ds an d ion s in this ca tego ry, th e cen tra l ato m is bonded to flu ori ne , chlorine, o r oxyg en. It is ofte n o bvio us from the fo rm u la of a co m pou nd t ha t a n octe t aro u nd a n ato m has been ex ceed ed. As an exa m p le, co n side r su lfur he xafl uoride , SFr" a gas 364 C h a p t e r 8
I
Bonding and Molecular Structure
TABLE 8 .6
lewis Structures in Which the Central Atom Exceeds an Octet
Group 4A
Group SA
Group 6A
SiFs-
PFs
SF,
[ r
: F:
"F'
" ' I '#'F: : F-Si :: " 1' F:
' F-
'"
"
1'" F: P' " 1"':F:
Group 7A
: F: " 1" F: ' -S "
'
ClF,
Xe F2
: F:
:r:
1 -,, :
: - ( 1 .. 1 ' " F: : F:
I ",'F:
: F: "
: F:
SiF6 2 -
PF. -
SF,
" : F: " :..F, • PI " ..F:
: F: " " ".' F. ....., 51,,# •F' • •
BrFs .. : F: .. :,F.... 1 "",F:
:'F "' I "''F:
::F"' I "''F :
:,F"' I "' F:
['r ",' F. "'-.. S'1_ot'••F'. .. 1 .. : F'" I ' F: .. : F: ..
[..
: F: ..
: F:
J
..
Group 8
Sr
1
", :
: -Xe ' 1 "
:r:
..
: F:
fo rm ed by th e react io n of su lfu r a n d excess fluo rin e . Su lfur is th e ce ntral a to m in th is co m p o u n d . a n d flu o ri n e typic ally bo nds to on ly on e o the r a to m with a sing le el e ct ro n pair bo nd (as in H F and C F-j). Six S- F bonds a rc required in SF 6 , meaning th e re will b e six e lec tron p airs in th e valence she ll o f th e su lfur a to m. If th ere a rc m ore th an [our g ro u ps bonded to a ce n tral a to m , thi s is a reli able sig na l th a t the re are m o re th a n e ig h t e lectrons aro u n d a ce n tra l a to m . Bu t be ca re fu l- t he ce n tra l a to m octet ca n a lso be e xceeded with fou r o r fewe r a to ms b ond ed to th e ce n tra l a to m. Co ns ide r th ree exa m p les fr om Tab le 8 .6: th e centra l at om in SF-t, C. IF3 , a n d XeF 2 has five e lec tro n pairs in its vale nce she ll. A u seful observati on is th a t oul)' elements (if the third and higher periods in the periodic table nUl)' [orm compounds and ions in which all octet is exceeded. Second -period e le me n ts (B, C , N , 0 , F) a re restricted to a max im um o f eig h t e lec tro ns in thei r co m p o u n ds. For exa m p le , nit roge n for m s co m po un ds a n d ion s suc h as N H :~, N H.t + , a n d NF:I> bu t N F5 is u n kn own . Ph o sp h o ru s, th e th ird-p eriod cl e m en t j us t b elo w nit ro gen in th e period ic tab le , fo r ms m an y co m p o u nds simi lar to ni tr oge n (P H:l , PIJ -t + , PF:~ ), bu t it a lso readi ly acco m modates five or six vale nce e lec tro n p air s in co m p o u n ds suc h as PF 5 o r in io n s suc h as PF 6 - . Arse n ic, a n tim o ny, a n d b ism uth , the eleme nts b elow p hosph orus in Gro u p 51\, rese m b le ph osp h orus in th eir b eh avio r. The u su al ex p la na tion for t he con tras ting behavior o f secon d- a n d th ird -period ele me n ts cen ters on th e n u m be r o f orb ita ls in th e valen ce she ll o f a n a to m. Secon dp e r io d el em en ts h ave fou r valence orbita ls (o ne 2s an d th ree 21) orbi tals) . Two elec tron s pe r orbital result in a to tal o f e igh t elec tro ns being acco m modate d a ro u n d a n a to m. For e leme nts in the thi rd and h igh e r p e riods, the d orbi ta ls in th e o uter shel l a re tr ad it ionall y in cl uded a mo ng valence o rb ita ls for th e e le me n ts. Th us, for ph osp ho ru s, th e 3d o r bita ls a re incl uded wit h the Ss a nd 31) o rb ital s as vale nce o rb ita ls. T h e ex tra orb ita ls provide th e e leme nt with a n o p portu n ity to accommod at e u p to 12 e lec tro ns.
• Xenon Compounds Compounds of xenon are among the more interesting entries in Table 8.6 because noble gas compounds were not discovered until t he early 1960s. One of the more intriguing compounds is XeFz, in part because of the simplicity of its synthesis. Xenon difluoride can be made by placing a flask containing xenon gas and fluorine gas in the sunlight. After several weeks. crystals of colorless XeF 2 are found in the flask (see page 404).
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8 .5
I
Exce pti ons to t he Oct et Rule 365
• Lewis Structures in Which the CentraL Atom Has More Than Eight Electrons ProbLem Sketc h the lewis structur e ofthe [ClFd - ion. Strategy Use the guide lines on page 353.
SoLution 1.
The (1 atom is the central at om.
2.
This ion has 36 va lence electrons [= 7 (for Cl)
3.
Draw th e ion with
fO UT single
+4
X 7 (for F)
+ 1 (for ion charge)}
or 18 pairs.
covalent Cl- F bonds .
H-J 4. Place lone pairs on t he terminal atoms. Because two electron pairs remain after placing lone pairs on the four Fatoms, and because we know that Cl can accommodate mo re than four pairs, these two pairs are placed on the central Cl atom. The last two electron pairs are added to t he
central Cl atom.
EXERCISE
8 .7 Lewis Structures i n Which th e Centra LAt om Has More Than Eight
Electrons Sketch the Lewis structures for [ClF2 the Cl atom in each ion?
r
and [CtF2J-. How many lone pairs and bond pairs surround
Molecules with an Odd Number of Electrons Two nit rogen oxides- NO , with 11 valence electrons, a n d N0 2 , with 17 va len ce electrons- arc amo n g a yc ry small g ro u p of stab le molecules with a n o d d number of electron s. Becau se th ey have an o d d nu mbe r o f electrons , it is impos sib le to d raw a struc tu re o be ying th e o ctet rul e ; at least one e lec tro n must be u npaired . Even th ough N0 2 d oes not o be y th e oct et rule, a n electro n d ot struc ture ca n be writte n tha t a p proximate s the bonding in the m o lecul e. This Le wis str u cture p lace s the u npaired electron on nitro gen . Two re son an ce struc tu res sh ow th at th e nitrogen-oxygen bo n ds a rc ex pecte d to b e equiva le n t. N ·· O~ ' 0··
.
.. .
Expe rime ntal evide nce for NO indicat es th at th e bo nd in g bet ween Nan d 0 is in te r med iate be tween a d ouble a nd a tripl e bond . It is not po ssib le to write a Lewis str uct ure fo r NO th at is in acco rd with th e p ro perties of th is su bstan ce, so a different th eory is n eed ed to und ersta nd bondin g in thi s mol ecule. \Ve sha ll return to co mpounds o f th is typ e whe n mo lecul ar o rb ital th eory is introduced in Sec tio n 9.3. The two n itroge n o xid es , N O a n d N0 2 , are membe rs of a class of che m ica l su bsta nces called fre e radic als . Free radicals are che mica l sp ecies- bo th a to mic a n d mo lecu la r-c-with a n u np aired el ectron . Free radica ls a rc ge n e ra lly quite rcac366
Chapt er
8
I
Bonding and Mo lecular Structure
Case Study
The Importance of an Odd-Electron Molecule, NO
°
Small molecules such as H2, 2• H20. CO, and CO 2 are among the most important molecules commercially, environmentally, and biologically. Imagine the surprise of chemists and biologists when it was discovered a few years ago that nitrogen monoxide (nitric oxide. NO). which was widely considered toxic, also has an important biological rote. Nitric oxide is a colorless, paramagnetic gas that is moderately soluble in water. In the laboratory, it can be synthesized by the reduction of nitrite ion with iodide ion: KNO,(aq) + KI(aq) + H,SO,(aq) -> NO(g) + K,SO,(aq) + H,O(C) + 'I, I,(aq) The formation of NO from the elements is an unfavorable, energetically uphill reaction (j'fW ~ 90.2 kJ/mol). Nevertheless, small quantities of this compound form from nitrogen and oxygen at high temperatures . For example. conditions in an internal combustion engine are favorable for this to happen. Nitric oxide reacts rapidly with O2 to form the reddish-brown gas N0 2• 2 NO(colorless, g) + O, (g) ->
The result is that compounds such as N0 2 a nd HN0 3 arising from reactions of NO with O2 and H20 are among the air pollutants produced by automobiles. A few years ago, chemists learned that NO is synthesized in a biological process bya nimals as diverse as barnacles, fruit flies, horseshoe crabs, chickens, t rout, a nd humans. Even more recently, chemists have found that NOis important in an astonishing range of physiological processes in huma ns and other animals. These include a role in neurotransmission, blood clotti ng, and blood pressure control as well as in t he immune system's ability to kill tumor cells and int racellular parasites. Questions: Oxygen is needed by manyliving
organisms, but some reactions with oxygen can lead to oxidative damage. One species that can produce damage in an organism is the superoxide ion, O2 - , Fortunately, this ion is removed extremely rapidly by reaction with NO to produce the peroxynitrite ion, ONOO -. 1. Draw the Lewis structure for the ion.
The colorless gas NO is bubbled into water from a high-pressure t ank. When the gas emerges into the air, the NO reacts ra pidly with O2 to give brown NO, gas.
2. Are there any multiple bonds in the ion? 3. Are there any resonance structures needed? Answers to these questions are in Appendix O.
2 NO ,(brown, g)
rive. Free ato ms such as II and CI, fo r exa mp le, are free rad ical s an d read ily co mbin e with each o th er to give mo lecu les su ch as 1-1 2 , C12, and H C\. Free rad ica ls are involved in man y reac tio ns in the e nviron me n t. Fo r exa mple, sm all amo un ts o f NO are re leas ed from ve hicle exhausts. The NO ra pid ly form s :t\' O~h wh ich is eve n mo re har m ful to hum an health and to plant s. Exposu re to N0 2 at co nce n tra tio ns of 50- 100 parts per m illio n ca n lead to sig nifica n t infl am ma tion of lu ng tissue. Nitrogen di o xid e is also ge ne rated by na tural p ro cesses. Fo r ex am ple, whe n hay, wh ich has a h igh level of nitrates, is sto re d in silos o n farm s, N0 2 can be ge nera te d as the hay ferm e nts , an d there have bee n re ports o f far m workers dying fro m exposu re to thi s gas in the silo. T he two nitrogen o xides, NO and N0 2 , are un iq ue in th at they ca n be iso la te d an d ne ith er has the ex tre me reac tivity of most free ra dicals. \Vhe n co o led, howeve r, two N0 2 mol ecules j oin o r "d ime rize" to for m co lo rless N20 -t; th e u np aired el ectro ns co m bine to fo rm an N-N bo nd in N20-t (Figu re 8.3) .
8.6
Molecular Shapes
O ne reason fo r d rawing Lewis ele ct ro n do t str uctures is to be able to p red ict the t h ree-di mensio nal geo me try of mol ecul es and ion s. Becau se the p hysica l and chemical p roperties o f co m po un ds are Lied to the ir struc tures, the importan ce of th is su bject ca nn o t be oversta ted . 8.6
I
Molecular Shapes
367
FIGURE 8 .3 Free radical chemistry. When cooled, the brown gas N0 2, a free radical, forms colorless N204 , a molecule with an N-N single bond.The coupling of twofree radicals is a common type of chemical reaction. Because identical free radicals come together, the productis catteda dimer, and the process is calted a dimerization. (Sign in to ChemistryNow, Screen 8.9, to see a video of this reaction.)
When cooled, N0 2 free radicals coupleto form N20 4 molecules.
" ~
c
e L. & _ Aflask of brown N0 2 gas in warm water
• VSEPR Theory The VSEPR theory was devised by Ronald J. Gillespie (1924- ) and Ronald S. Nyholm (1917-1971).
J....._ Aflask of N0 2 gas in ice water
T he valen ce sh ell ele ctron-pair repulsion (VSEPR) model is a reli abl e metho d for predi ct ing th e shapes of cova le n t molecules and pol vato m ic ions. T hi s mo d el is based o n the id ea that bond and lone electron [mi rs in the valence shell oj an element n11el each other and seek to be as Jar ajm,.t as jJOssibl('. T he po sitio ns assu med by the va lenc e electro ns o f an atom thus d e fin e th e ang les between bo nds to su rro u nd ing ato m s. VSEPR is remark abl y successful in p redicting str uctures o f mol ecul es and ion s of m ain gro up elemen ts. H owever; it is less effect ive (a nd sel d o m used ) to pred ict structu res o f co m po u nd s co n ta in ing tr an sition met als. To have a se nse o f how vale nce she ll ele ctron pa irs repel and d e termin e struc ture , blow up sever al balloons to a similar size. Imagin e that each ballo o n represe nts an e lec tro n cloud . Wh en two , three , fo ur, five , o r six ballo ons arc tied together a t a ce n tral po in t (rep res en ting th e nucleus and co re e lec tro ns o f a ce n tral ato m), the ball o on s natu rall y form th e sha pes shown in Fig u re 8.4 . T hese geo metric arra ngc men ts min imi ze int e ra ctions be tween the balloon s.
Central Atoms Surrounded Only by Single-Bond Pairs T he simplest applicatio n of VSEPR th eo ry is to molecul es and ions in which all the electro n pairs aro un d the central ato m are involved in sing le covalen t bo nd s. Figu re 8.5 illustra tes the geo me tries pred icted fo r molecul es o r ions with the genera l fo rmu las AXil' where A is the ce n tral atom and n is the nu mber of X gro u ps bo nded to it.
Linear
368 C h a p t e r 8
Trigonal planar Tet rahedral Trigonal bipyramidal Octahedral FIGURE 8 .4 Balloon models of electron-pair geometries for two to six electron pairs. If two to six balloons of similar size and shape are tied together, they willnaturallyassumethe arrangements shown. Ihese pictures illustrate the predictions of VSEPR.
I
Bonding and Mo lecular Structure
linear
Tetrahedral
'. ®
Trigonal bipyramida l
Octahedral
~
',.' .. ....... :~; ..... "
W 180" 120"
AX,
AX,
AX,
AX,
Example: BeF2
Example: BF3
Example: PFs
Example: SF6
Active Figure 8.5 Various geometries predicted by VSEPR. Geometries predicted byVSEPR for moleculesthat contain only single covalent bonds aroundthe central atom.
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• explore an interactive version of this figure accompanied by an exercise.
T he line ar ge o metry for two bond pairs and th e tr igo nal-planar ge o metry for three bo nd pairs involve a cen tral ato m th at docs not have an o ctet o f electro ns (see Sect io n 8.5). The ce n tra l ato m in a tetrahedral mo lecule o beys the octet rule with four bond pa irs. T he cen tral atoms in tr igo na l-bipyram id al and oc ta hed ra l m olecul es have five and six bond ing pairs, respectively, and are ex pec te d o nly when th e ce ntral ato m is an elem en t in Period 3 or high er of the period ic table (.... page 372). ChemislryJ)"NOw~
•
Sign in at www.thomsonedu.comjlogin and go to Chapter8 Contents to see: Screen 8.13 for an animation of the electron-pair geometries and on identifying geometries • Screen 8.14 for practice predicting molecular geometry
, . Predicting Molecular Shapes Problem Predict the shape of silicon tetrachloride, SiCl4 • Strategy The first step is to draw the Lewis structure. The Lewis structure does not need to be drawn in any particular way because its purpose is only to describe the number of bonds around an atom and to determine if the re are any lone pairs. The number of bond and tone pairs of electrons around the central atom determines t he molecular shape (Figure 8.5).
• Lewis Structures and Molecular Shapes Drawing the Lewis structure is the first step in determining the shape of a molecule or ion.
SoLutionThe Lewis structure of SiCL. has four electron pairs, all of them bond pairs, around the central Si atom. Therefore, a tetrahedral structure is predicted for the SiCL4 molecule, with CL-Si -CL bond angLes of 109.5". This agrees with the actual structure for SiCL, . Lewis structure
Molecular geometry
:[l:
I
.Cl-r-Si-r- El:
..
I
..
:[l:
8 .6
I
MolecuLar Shapes
369
EXERCISE 8 .8
Predicting Molecular Shapes
Whatis the shape of the dichlorometh ane (CH 2Cl 2 ) molecule? Predict the Cl-C-Cl bond angle.
Central Atoms with Single-Bond Pairs and Lone Pairs To see how lon e pairs affect th e geom e try o f th e molecule o r pol vatornic ion , re tu r n 10 the balloon mo de ls in Figu re 8.4. Recall th at the balloon s re prese nted all the electro ll pairs ill th e valence she ll. Th e balloo n model therefor e predi ct"; th e "elec tro n-pai r geome try" rath e r th an th e "molec u lar geome try." T he electron-pair geometry is the geome try take n np by all th e valen ce electr on pa irs aro u nd a cen tral atom , whe reas th e molecular geometry d escribes th e arra ngeme n t in space of th e ce n tra l ato m and th e a to ms directl y attac hed to it. It is important to re cognize th at lone f)airs of electrons on the central a/om Or(ujf} .\jHltial jJOsitions, porn though their location if n ot i nclu ded in thr verbal description oj the slurj)e oJ the molecule or ion. Le t us use th e VSEPR model to predi ct th e m ole cul ar geo me try and bond a ng les in th e N1I3 mol ecul e . O n d rawing the Lewis str uc tu re , we see th e re are fou r pair s of e lectrons in th e nit rogen valence she ll, three bond pai rs, and o ne lon e pair. T h us, th e predict ed elearon-poir geometry is tetrah edral. T he molecular geometry, however, is sai d to he trigonaljJ)'mmidalhecause th at d escrib es the lo cation o f th e ato ms . The n itrogen ato m is at th e apex o f th e pyrami d, and th e three hydrogen a to ms fo rm th e tri gonal base .
H- N-H
I
H l ewisstructure
Etectr on-patr
geometry, tetrahedral
Actual H-N-H angle = 107.5
C
Molecular geometry, trigonal pyramidal
Effect of Lone Pairs on Bond Angles Because the e lec tro n-pa ir geo me tr y in N lI :~ is tetrahedral , we wo u ld ex pect th e H - N-H bond ang le to be 109.5 °. H owever, th e ex perimen ta lly d eterm in ed bo nd ang les in N II 3 are 107 .5 °, and th e 11- 0-11 ang le in wa ter is sma ller still ( 104.5°) (Figu re 8.6) . Th ese a ng les are close to th e tetrah ed ral ang le but not ex ac tly th at value . T his highligh ts th e fact th at VSEP R is no t an accu ra te model; it can o n ly p redict the approxima te geo me tr y. Small vari ations in geo me tr y (e.g., bon d ang le s a few degrees diffe re nt fro m predi ct ed ) are q uite co mm o n and often ar ise becau se th ere is a differen ce betw ee n the spatia l re q uireme n ts of lon e pairs and bond pairs. Lon e pai rs of e lectro ns seem to occ u p y a large r volume th an bonding pairs, an d t he increased vo lu me of lo n e pairs ca uses bon d pai rs to sq ueeze clo ser to gether. In ge neral, th e rel ati ve stre ng t hs of repulsio ns are in t he order Lo ne pair-lo ne pair > lone pair- bond pair > bond pair-bo nd pair T he differen t spa tia l re q uire me n ts of lon e pairs an d bond pairs ca n be used to pred ict vari ati on s in th e bo nd ang les in se ries o f mol ecul es. Fo r ex amp le , th e bond ang les d ecrease in th e series Cl L . N1I3 , and 1120 as th e nu m ber of lo ne pairs 0 11 th e ce ntra l ato m in cr eases (Figu re 8.6) . 370
Cha pt e r 8
I
Bonding and Molecular Structure
FOUR ELECTRONPAIRS Electron PairGeometry = tetrahedral Tetrahedral
FIGURE 8.6 The molecular geometries of methane, ammonia, and water.
Bent
Trigonal pyramidal
All have four electron pairs around the central atom, so all have a tetrahedral
Met hane, CH 4 4 bond pairs no lone pairs
Ammonia, NH) 3 bond pairs 1 lone pair (b)
(a)
Wate r, H20 2 bond pairs 2 tone pairs (c)
electron-pair geometry. (a) Methane has four bond pairs and so has a tetrahedral molecular shape. (b) Ammonia has three bond pairs and one lone pair, so it has a trigonal-pyramidal molecular shape. (c) Water has two bond pairs and two lone pairs. so it has a bent, or angular, molecularshape. The decrease in bond angles in the series can be explained by the fact that the lone pairs have a larger
spatial requirement than the bond pairs.
Finding the Shapes of Molecules Problem What are the shapes of the ions H30 - and ClF 2 - ? Strategy Draw the Lewis structures for each ion. Count the number of lone and bond pairs around the central atom . Use Figure 8.5 to decide On the electron -pair geometry. Finally, the location of the atoms in the ion-which is dete rmined by the bond and lone pairs-gives the geometry of the ion. Solution (a) The Lewis structu re of the hydronium ion, H30 +, shows tha t the oxygen atom is surrounded by four electron pairs, so the electron-pair geometry is tetrahedral.
Lewis structure
Electron-pairgeometry, tetrahedral
Molecular geometry, trigonal pyramidal
Because three of the four pairs are used to bond terminal atoms, the central 0 atom and the three H atoms form a trigon al-pyramidal molecular shape like th at of NH3 • (b) Chlorine is the central atom in ClF 2 - . It is surrounded by four electron pairs, so the electron-pair geometry around chlorine is tetrahedra l. Because only two of t he four pairs are bonding pairs, the ion has a bent geometry.
• " Energized Water" wit h a Bond
Angle of 114°! There are many dubious products sold over the internet, and one of them claims that water is "enerqized" by increasing its bond angle. One advertisement said that in the past water had a bond angle of a healthy 110 but now it is "wimpy" and unhealthy with an angle of only 104 Further, it is claimed that distilled water only has a bond angle of 101 and is biologically dead. To cure this problem, you can buy a costly machine that "energizes" water and causes a bond angle increase to as much as 114 Now, it is also claimed, this water has enough energy to destroy pathogens. Tile old circus master, P. T. Barnum, once said there is a sucker born every minute. 0
,
0
•
Lewis structure
Electron-pair geometry, tetrahedral
Molecular geometry, bent or angular
0
0
•
EXERCISE 8 .9 VSEPR and Molecular Shape Givethe electron-pair geometry and molecular shape for BF3 and BF4 - . What is the effect on the molecular geometry of adding an F- ion to BF3 to give BF4 - ?
8 .6
I
Molecular Shapes
371
CentraL Atoms with More Than Four Valence ELectron Pairs
axial atom
90"
-i.
~rialatom FIGURE 8.7 The trigonal bipyra-
mid showing the axial and equatorial atoms. The angles between atoms in the equator are 120 The angles between equatorial and axial atoms are 90 0 • 0
•
T he situatio n b ecomes mo re co m p lica te d if th e ce n tral ato m h as five o r six electron pairs, so me of whi c h are lone pairs. A tr igo n al-b ip yram id al stru ctu re (Fig u res 8.5 a nd 8.7) h as two se ts of positio n s th at a re n ot eq u iva le n t. T h e positio ns in th e tr igon al pl an e lie in th e eq uator of a n imagin ary sp h e re a ro u n d th e ce n tral atom an d a re ca lled th e equa torial posi tion s. T h e north a nd so u th p ol es in th is re p re se n ta tio n a re ca lle d th e axial p o sitio n s. Eac h equato rial ato m h as two n ei ghb o rin g g ro u ps (t h e ax ial ato m s) at 9 00, a nd eac h axi a l a to m h as t hree grou ps (th e eq u atori al a to ms) a t ~10 0 . Th e resu lt is th a t t h e lon e pai rs, wh ic h re q uire mo re space th an bo ndin g pa irs, p refer to occ u py eq ua torial p o sitio ns rath er th an axial positio n s. The e n tries in th e to p line of Figure 8.8 sh ow sp eci es having a to tal of five valen ce e lec tr o n p airs, with ze ro , o n e, 1\ \' 0 , a n d th ree lon e p airs. In SF.\> with o ne lone p air, th e mol ecul e assu mes a "seesaw" shape wit h th e lon e pair in o n e of th e eq ua to rial p ositio n s. T he C IF3 mol ecul e h as three bon d pairs a nd two lo n e pai rs. T he PN O lo n e p ai rs in C IF3 a re in eq ua to rial p o sition s; two bond p airs are axial, and th e th ird is in th e equ a torial p lane, so th e mol ecul a r geometry is "l-sh a pcd . Th e third mo lecu le shown is XeF2 • H ere, all th ree eq ua to rial po sitio ns a re occu p ied by lone p airs so th e mo lecu lar geo me try is lin ea r. T he ge o metry assu m ed by six electro n pairs is oc ta hed ra l (see Figu re 8.8) , an d all th e a ngles a t adjacen t positio ns a re ~10 0. U n like th e tr igo n al bipyram id , th e octa h edro n h as n o d istinct axia l a n d eq ua to rial po sitio ns; all po sitio ns are th e sam e. T he re fore , if th e mo lec u le has o ne lone pair, as in BrF 5 , it m akes n o diff erence wh ich po sitio n it o ccupies. The lo n e pa ir is o fte n d rawn in the to p o r bottom posi-
FIVE ELECTRONPAIRS Electron-PairGeometry = t rigonal bipyramidat
~
PF, 5 bond pairs No lone pairs
ClF3 3 bond pairs 2 lone pairs
SF, 4 bond pairs 1 lone pair
.. Linear
W ·. XeF 2 2 bond pairs 3 lone pairs
• •/
SIX ELECTRO NPAIRS Electron-Pair Geometry = octahedral Octahedral
Square p}!amidal
BrFs 5 bond pairs 1 tone pair
SF, 6 bond pairs No tone pairs
Sguare planar
XeF4 4 bond pairs 2 lone pairs
FIGURE 8 .8 Electron-pair geometries and molecular shapes for molecules and ions with five or six electron pairs around the central atom.
372
Chapter 8
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Bonding and Mo lecular Structure
Lio n to make it eas ie r to visuali ze th e mol ecul ar ge o met ry, which in this case is square-pyramida l. If two pairs of el ectrons in an oc ta hed ral arrange me n t are lo ne pairs, they seek to be as far a part as possib le. The resul t is a sq uare-planar mo lecul e , as illu strat ed by XeF.,. , , Predicting Molecular Shape Problem Wh at is the shape of the IC ~ - ion? Strategy Draw the Lewis structure, and then decide on the electron-pair geometry. The position of t he atoms givesthe molecular geometry of the ion. (See Example 8.7 and Figure 8.8.) Soluti on A Lewi s structure for the ICl,- ion shows that the central iodine atom has six electron pairs in its valence shell. Two of these are lone pairs. Placing t he lone pairs on opposite sides leaves the four chlorine atoms in a square-planer geometry.
Electron-pair geometry, octahedral
EXERCISE 8.10
Mo leculargeometry, square planar
Predicting Molecular Shape
Draw the Lewis structure for ICV , and then decide on the geometry of the ion,
Multiple Bonds and Molecular Geometry Double and triple bonds invol ve mo re el ectron pa irs tha n sing le bonds, but th is has littl e effec t on th e overall molecu lar shape . All of th e electron pairs in a mu ltip le bond arc share d between the same two nu clei and therefore oc cu py th e same reg io n of space. Because they must rem ain in that regio n. two e lectro n pairs in a d o u ble bond (o r three pair s in a tr ip le bond ) have the sa me effect o n th e struc tu re as o ne electron pair in a sing lc bond. That is, all electron pairs in a multiple bond cou n t as o ne bo nd an d co n tri b u te to mo lec ular ge o me try th e same as a sing le bond does. For exam ple , th e carbo n ato m in CO 2 has no lon e pairs and participat es in two doub le bonds. Each double bond co u n ts as one for the pu rp o se o f predicting geom etry, so th e struc ture o f CO 2 is linear.
Q=C=Q Lewis structure, electron-pair geometry = linear
Molecular structure, linear
\Vhe n re sonance struc tu res are possib le , the ge o me try can be predicted from an y of the Lewis re sonan ce str uctu res o r from th e re sonance h ybrid st ruc tu re. For exam ple, th e geometry of th e CO:\2- ion is predi cted to be trig onal planar because the ca rbo n atom has three se ts of bonds and no lo ne pai rs.
8.6
I
Molecular Shapes 373
:..o-cI=o ..'j
l
2-
:0:
l ewis structure, one resonance structure, electro n-pair
geometry = trigonal planar
Molecular structure, trigonal planar
T he N0 2 - io n a lso h a s a trigo n a l-p la n a r electron-pa ir geometry. Be cau se there is a lo ne pair o n the ce n tra l nitrogen atom , and bo nds in the o the r two positions, th e geo metry o f th e io n is ang ular o r ben t.
O:] ..- N= .. [:O Lewis structure, one resonance structure, electron-pair geometry = trigonal planar
Mo lecular struct ure, angular or bent
T he techn iques just O1 I1 1ined ca n be used to fin d the ge o me tries arou nd the a to m s in more comp licat ed molecu le s. Co ns id e r, for e xa mp le , cystei n e , o n e o f the n atu ral a mi no a cids. H
H-
H
:0:
.. I I II .. S- C- C - C- O-H .. I' I' 1 . . H
:N-
H
I
H
Fou r pairs of electrons oc cu r arou nd th e S, N, C 2, an d C3 atoms, so th e electron-pair geo me try aro und eac h is te trah edral. Thus, th e S- C- I-I and II- N - II ang les are p red icted to be ap p rox imately 109 °. T he 0 a tom in th e gro u ping C- O - II and th e S ato m in th e g rou ping H- S- C are also su rro u nd ed by four pairs, and so th ese ang les are likewise ap p rox imate ly 109 °. Finall y, th e ang le m ad e by O -C1-O is 1200 becau se the elect ro n-pai r geo met ry arou n d C 1 is tri gonal planar. , . Finding the Shapes of Molecules and Ions ProbLem What are the shapes of the nitrate ion, N0 3 - , and XeOF4? Strategy Draw the Lewi s structure, and then decide on the electron pair geometry. The position of the atoms gives the molecu lar geometry of the ion. Follow the procedure used in Examples 8.6-8 .8. Soluti on (a) The NO) - and ( 032- ions are isoelectronic. Thus, like the carbonate ion described in t he text above, t he electron-pair geometry and molecular shape of NO)- are t rigonal planar. (b) The XeOF4 molecu le has a l ewis structure with a total of six electron pairs about t he cent ral Xe atom, one of which is a lone pair. It has a square-pyramidal molecularstructure. Two structures are possible, 374
Chapter
8
I
Bonding and Mo lecular Structure
based on the position occupied by the oxygen, but there is no way to predict which is correct . The actual str uct ure is the one shown, wit h the oxygen in the apex of the square pyramid.
:0:
:0:
}-, I > (
:.F....
f /" "-(
:F ..' :, , .'F .:
l ewis st ruct ure
Ele ct ron- pair geometry, octahedral
Xe
1 ...•(
Xe
Motecula r geometry, square pyramida l
EXERCISE 8 .11 Determining MolecuLar Shapes Use lewi s st ructures and the V5EPR mode Lto determine the eLectron-pair and moLecula r geometries for (a) the phosphate ion, P04 3- ; (b) the sulfit e ion, 5°32 - ; and (e) IF5 •
8.7
Bond Polarity and Electronegativity
Th e model s used to represent cova len t a nd io nic bon ding a re th e ex tre me situatio ns in bondi ng. Pure cova len t bonding, in wh ich at oms sh a re an e lec tro n pair equally, occu rs on l)' whe n two identical atoms a re bonded . when two d issimilar a to ms form a cova le n t bond , th e elec tron pair will b e un equally shared. The res u lt is a polar covalent bond, a bond in wh ich th c two a to ms hav e re sidual or parti al charges (Fig u re H.9). Bond s a re pola r becau se no t all ato ms hold on to th eir vale nce electro ns with the same force, no r d o a to ms take on add itio na l e lec tro ns with eq ua l case. Recall fro m the di scussion o f a to m p roperties th at diffe ren t c lcmen ts h ave d if ferent values of ionization ene rg y an d e lec tron affin ity (Sectio n 7.5) . Th ese d ifferen ces in beh avio r for frce atoms carry over to a to ms in mol ecul es. If a bond pair is not eq u ally sha re d between ato ms , th e bonding electro ns a rc o n ave rage n earer to o ne of th e ato ms. The ato m toward wh ich th e p air is di spl aced h as a larger share of th e el ectro n p air an d th us ac q u ires a part ial n egati ve charge. At th e same time , th e ato m at th e o th e r e n d o f th e bond is d eplet ed in electro ns a n d acq uires a part ial positive charge . The bond bet wee n t he two ato ms h as a p os itive e n d a n d a n egat ive e n d ; tha t is, it has n egati ve a n d po sitive pol es. T he bond is ca lle d a polar bond. In ionic co m pou nds, d isp laceme n t of the bonding p air to o ne o f th e two a to ms is esse n tially co m p le te, an d + a nd - sym bo ls a re writte n alo ngs ide th e ato m symbol s in th e Le wis d rawi ngs. For a pol ar covale n t bond , th e polar ity is indicat ed by writing the sym bo ls 0+ and 8- alongside th e a to m sym bo ls, where 0 (the Gree k letter "del ta" ) sta n ds for a partial c harge . Hydrogen Fluoride , wate r, and am mo nia a re th ree sim p le mol ecules havin g po lar, covale n t bond s.
HF
H20
FIGURE 8 .9 A polar covalent bond. Iodine has a larger share of the bonding electrons, and hydrogen has a smaller share, The result is th at I has a partial negative charge (8- ), and H has a partial positive charge (5 + ).
NH3
Three simple molecules with polar covalent bonds. In each case, F, 0, and N are more electronegative t han H. 8 .7
I
Bond Polarity and Electronegativity
375
H
H
\ \
+ Q
/ HF. polar covalent
H2 • totallycovalent .lX
~
0
li F, ionic .lX = 4.0 - 1.0 = 3.0
.lX = 4.0 - 2.2 = 1.8
Increasing ionic character FIGURE 8 .10 Covalent to ionic bonding. As t he electronegativity difference increases between t he atoms of a
bond. the bond becomes increa singly ionic.
With so man y ato ms 10 use in cova lent bo nd formation, it is not surprising tha t bo nd s between atom s can fa ll anywhere in a co n tinuum from pure cova lent to pu re ionic (Figure 8.10) . T he re is no sharp d ivid ing line be twee n an io n ic bo nd and a covalen t bond . In the 1930s, Linu s Pauli ng proposed a param e ter called ato m e lec tronega tivity that allows us to decid e if a bond is pol ar, wh ich ato m of the bond is negative an d whic h is positive , an d if o ne bo nd is mo re po lar than anot her. T he e le ctronegativity, X, of an atom is de fine d as a measure o f the ability (if an alum in (J molecule to attra ct electrons to itsel]. Values of e lectro negativity are give n in Figure 8. 1I. Several features and per io dic trends are appare nt. T he el ement with the largest electronegativity is flu o rine ; it is assig ne d a value of X = 4.0. Th e clem ent with the sma lle st value is the alkali met al ces ium . Elcctronegativitics generally increase fro m le ft to rig ht across a period and decrease down a gro up . Th is is the op pos ite o f the trend o bse rved for met allic character. Meta ls typically have low values o f c lcc tro ncgativiry, rangin g from slightly less tha n I to about 2. Elec trouegativity vallies for the metalloi ds are aro u nd 2, whereas non me tals have values greate r than 2.
lA
2A
3A
4A
SA
6A
7A
Li 1.0
Be 1. 6
B 2.0
C 2.5
N 3.0
0 3.5
F 4. 0
Na 0.9
M9 1. 3
3B
4B
5B
6B
8B 7B ,----------A--- IB
2B
Al 1.6
Si 1. 9
P 2.2
S 2.6
Cl 3.2
K 0.8
Ca 1.0
Sc 1.4
Ti 1. 5
V 1. 6
CT 1.7
Mn 1. 5
Fe 1.8
Co 1.9
Ni 1. 9
Cu 1.9
Zn 1. 6
Ga 1.8
Ge 2.0
As 2.2
Se 2.6
BT 3.0
Rb 0.8
ST 1.0
Y
ZT 1.3
Nb 1. 6
Mo 2.2
Tc 1.9
Ru 2.2
Rh 2.3
Pd 2.2
A9 1. 9
Cd 1.7
In 1.8
Sn 2.0
Sb
1. 2
1.9
Te 2.1
I 2.7
Cs 0.8
Ba 0.9
La 1.1
Hf 1.3
Ta 1.5
W
Re 1.9
Os 2.2
IT 2.2
Pt 2.3
Au 2.5
H9 2.0
Tl
2.4
1. 6
Pb 2.3
Bi 2.0
Po 2.0
At 2.2
0
< 1.0
01.5- 1. 9
0
2. 5- 2.9
0
1.0-1.4
0 2.0-2 .4
0
3.0- 4.0
FIGURE 8 .11 Electronegativity values for the elements according to Pauling . Trends for electroneqativ ities are the opposite of the trendsdefining metallic character. Nonmetals have high values of electronegativity; the metalloids have intermediate values. and the metals have lowvalues. Values for these elements as wellas for the noble gases and for the lanthanides and actinides are available in the following handbook: Emsley, J., The Elements, 3rd edition, Clarendon Press, Oxford , 1998. 376
Chapter 8
I
Bonding and Molecular Structure
T here is a large rlifferen,ad n elec tro negativity for atoms from the left- an d right-hand side s o f th e periodic table. For cesium fluo rid e, for exam p le, the differen ce in e lectron egativity value s, tlX, is 3.2 [= 4.0 (for F) - 0.8 (for Cs) ]. The ho nd is decidedly ion ic in CsF, therefore, with Cs the calion (Cs") an d F the an io n (F-). In co nt rast, th e ele ctrou cgat ivity difference between H and F in H F is o n ly 1.8 [ = 4.0 (fo r F) - 2.2 (fo r 1I) ] . \ Ve co nclud e that bonding in lIF must be mo re covalen t, as exp ected for a compound form ed from two nonmetals. T he lI - F bond is polar, howeve r, with hydrogen being the po sitive end of the mo lecu le an d fluorine th e negative end (H5+_ FO- ) . Chemislry.o.Now'"
•
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• Estimating Bond Polarities Problem For each of t he following bond pairs, decide which is t he more polar and indicate t he negative and positive poles. (a) B-F and B-Cl (b) 5i-0 and Pr- P Strat egy Locate the elements in the periodic table. Recall that electronegativity generally increases across a period and up a group. Solution (a) Band Flie relatively far apart in the periodic table. B is a metalloid, and F is a nonmetal. Here, X for B = 2.0, and X for F = 4.0. Similarly, Band Cl are relativelyfar apart in t he periodic table, but Clis below Fin the periodic table (X for Cl = 3.2) and is therefore less electronegative than F. The difference in electronegativity for B-F is 2.0, and for B-Clit is 1.2. Bot h bonds are expected to be polar, with B positive and the halide atom negative, but a B-F bond is more polar t han a B-Cl bond.
linu s Pauling (1901-1994) . linus Pauling was born in Portland, Oregon, earned a B.Sc. degree in chemica l engineering from Oregon State College in 1922, and completed his Ph.D. in chemistry at the Ca lifornia Instit ute of Technology in 1925. In chemistry, he is well known for his book The Nature of the Chemical Bond. He also studied protein st ructure and, in the words of Francis Crick, was "one of the founders of molecular biology." It was this work and his study of chemical bonding that were cited in the award of the Nobel Prize in chemistry in 1954. Although chemistry was the focus of his life, at the urging of hiswife, Ava Helen, he wasalso involved in nuclear disarmament issues, and he received the Nobel Peace Prize in 1962 for tile role ile played in advocating for the nuclear test ban treaty.
(b) Because the bond is betweentwo atoms of the same kind, the P- P bond is nonpolar. Silicon is in Group 4A and the t hird period, whereas 0 is in Group 6A and t he second period. Consequently, 0 has a greater electronegativity (3.5) than Si (1.9), so the bond is highly polar Px = 1.6), with 0 the more negativeatom. EXERCISE
8 .12 Bond Polarity
For each of the following pairs of bonds, decide which is the more polar. For each polar bond, indicate the positive and negative poles. First, makeyour prediction from t he relative atom positions in the periodic table; t hen check your prediction by calculating j,X. (a) H-F and H-I
(b) B-C and B-F
(c) C-Si and C- S
Charge Distribution: Combining Formal Charge and Electronegativity The way electrons arc di stributed in a mo lecule o r io n is called its charge distributio n . The c harge d ist ributio n ca n p rofoundly a ffec t th e p rope rti es of a mo lecu le. Examples in clu d e its p hysical p roperti es, suc h as its mel tin g a n d bo ilin g points, a nd its c h e mi cal p ropertie s, suc h as its susc e p tibility to attack by a n anion o r ca tio n o r whe ther it is a n acid or a b ase . '\ Ve saw earl ier (.... page :\59) th at fo rm al c ha rg e ca lcu latio ns can lo cate th e site of a c h argc in a mo lecu le o r an io n . However, th is ca n so me times lead to res u lts th at a re in correct becau se form al c h a rge calcu latio ns assume th at there is equal sha ri ng of e lec tro ns in all bo nds. The io n BF4 - illustra tes this point. Boron has a form al c ha rge of - ] in t his ion , wh e rea s th e form a l c ha rge ca lcu lated for the flu o rin e atoms is O. This is not logical: fluorin e is th e more e lectro ne ga tive ato m so the ne gati ve c harg e should reside 011 F a n d 110t o n B. 8 .7
I
Bond Polarity and Electronegativity
377
A Closer Look
Electronegativity
Electronegativity is a useful, if somewhat vague, concept. It is, how ever, relat ed to the ionic character of bonds. Chemists have found, as illust rated in the figure, that a correlation exists between t he difference in elect ronegativity of bonded atoms and the degree of iorricity expressed as "% ionic characte r." As the difference in electronegativity increases, ionic character increases. Does this t rend allow us to say that one compound is ionic and another is covalent? No, we can say only that one bond is more ionic or more covalent t han another. Electron affinity was introduced in Section 7.5. Atfirst glance, it may appear that electronegativity and electron affinity measure the same property, but t hey do not. Electro negativity is a parameter that applies only to atoms in molecules, whereas electron affinity is a measurable energy quantity for atoms in the gas phase. Although elect ron affinity was introduced earlier as a criterion with whic h to predict the central atom in a molecule, experience indicates that electronegativity is a bett er choice. That is. the central atom is generally the atom of lowes t electroneqotivity.
Formalcharge =
o[~ 7-;) ;](2~) .. I ".
:( r
F
:
Formalcharge = - 1 ~ 3 - [0 + t (8l] Formalcharges for the Band F atoms of the BF4 - anion.
Even compounds with high electronegativity differences are not 100% ionic. ~
100
liF
75
ti ••u
~
u
50
"0
.s
"'
25
2
3
Electronegativity difference
T he \vay to resolve th is d ilemm a is to co nside r cl cctro ncgativi ry in co nj u nc tio n with formal charge. Based o n th e elec tro negativity di ffe ren ce be tween flu ori ne and bo ron (d X = 2.0) , th e B- F bonds are e xpecte d to be pol ar, with flu orin e bein g th e negat ive e nd o f th e bond , BO+- Fo- . So , in th is instan ce , pr edictions base d o n clcctronegati vity and fo rmal charge work in o pposite d irections. The form al charg e calcu lat io n places the negati ve cha rg e on boron , but th e c lcct ro ncg ativi ty di fferen ce leads us to say th e negat ive cha rge o n boro n is distri buted o nto th e fluorin e ato ms, effect ively sp read ing it out over the molecule . Linu s Paul ing poi n ted o u t two basic guide lin es to use whe n d escri bing charge d istr ibu tion s in mol ecul es a nd io ns. T he first is th e clcctroncutrality principle. T h is d ecl ares tha t electro ns will be di strib u ted in suc h a way th at the charges o n all ato m s arc as clo se to zero as po ssible . Seco nd , he noted th a t if a nega tive charge is prcscn t, it shou ld re side o n th e mo st el ec tronega tive ato ms. Simil arly, pos itive charg es are ex pec te d o n th e least e lec tro negative ato ms . The effec t of th ese p rin ciples is cle ar ly see n in th e cas e of BF4 " , where the neg at ive charge is di stributed over th e fo ur fluo r ine ato ms rather than resid ing o n boron . Co ns id ering th e co ncepts o f e lectronegati vity and formal cha rge togethe r ca n help to d ecid e wh ich o f several resonan ce struc tu res is th e more im portan t. For exa mple , Lewis struc tu re A fo r CO 2 is th e log ical o ne to d raw. But what. is ' \-To ng with B, ill which each ato m a lso has an o ctet o f elect rons> Formal charges
000
+1
0
- 1
: 0=( - 0:
Resonance structures A
B
For struc tu re A, each at o m has a fo rmal cha rge of 0 , a favora ble situatio n. In B, howeve r, o ne o xygen atom has a form al charge of + I , an d the ot her has - I. Th is is con trary to the princip le o f e lcc tr o nc u trality. In ad d itio n , B pl aces a positive 37 8
Chapt er
8
I
Bonding and Molecular Structure
charge on th e more elec tro negative 0 a to m. Thus, we ca n con clude th at struc tu re B is a mu ch less satisfac to ry structure th an A. Now use th e log ic ap plied to C0 2 l O decide whic h o f the three possibl e res o na nce struc tu res for the O CN - io n is th e mo st reason abl e. Form al cha rges for each ato m are given a bove t he clement 's sym bol. Formal charges
- 1
Resonance structures
[:
0
0
~-C =N
0
0
- 1
+1
0
-2
J ~ [: o=c=~r ~ [: o= c-~J
0 = 6 - [2 + ('1,)(6)]
~
+1
C = 4 - [0
+ ('1,)(8 )]
~
0
N ~ 5 - [6
+ ('1,)(2 )]
~ - 2
Sum of formal charges = - 1 = charge on the ion.
c
B
A
• Formal Charges in OCNExample of formal charge calculation: For resonance form C for OCN -, we have
Structure C will not co n tribu te sig nifica n tly to th e ove ra ll elect roni c struc tu re o f th e ion . It has a - 2 fo rm al cha rge on th e N a to m and a + I fo rm al charge on th e o a tom . Not o nly is the charge 011 the N a to m high , but 0 is more el ectronegati ve th an N and wo u ld be ex pected to rake on a ne gati ve chargc. Structure A is more sign ifica nt than struct u re B because th e negative ch arge in A is p laced o n th e mo st e lec tr o nega tive ato m (0). \'Ve p redict , th erefore , that struc tu re A is th e best represc n tati ou for thi s ion and that th c carbo n-nitrogc n bond will rese mble a tr ipl e bo nd . The result fo r O CN - also allows us to predict that proto uarion of th e ion will lead to I-JOCN an d not J-I NCO . Th at is, an J-I + ion will ad d to the mo re negarive o xygcn at om . Calculating Formal Charges Problem Boron-containing compounds often have a boron atom with only th ree bonds (and no lone pairs). Why not form a double bond with a terminal atom to complete the boron octet? To answer this, consider possible resonance structures of BF), and calculate the atoms' formal charges. Are the bonds polar in BF)? If so, which is the more negative atom? Strategy Calculatethe formal charges on each atom in the resonance structures. The preferred structure will have atoms with low formal charges. Negativeformal charges should be on the most electronegative atoms. SoLution The two possible structures for BF) are illustrated here with the calculated formal charges on the Band Fatoms. Forma t charge = + 1
Format charge = 0 ~ 7 - [6
+ ; (2))
. /
~ 7 - [4
+ ; (4)]
/ : F:
: F:
II
I
: F-B - F:
: F-B - F:
I
I
Formalcharge = 0 ~ 3 - [0
Forma l charge= - 1
+ l(6))
~ 3 - [0
+ l(8)]
The structure on t he left is strongly preferred because all atoms have a zero formal charge and the very electronegative Fatom does not have a charge of + l , F (X = 4.0) is more electronegative than B (X = 2.0), so the B- F bond is highly polar, the Fatom being partially negative and the B ato m partially positive.
EXERCISE
8 .13 Formal Charge, Bond Polarity, a nd Electronegativity
Consider all possible resonance structures for 502. Wh at are t he formal charges on each atom in each resonance structure? What are t he bond polarities? Do they agree with the formal charges?
8 .7
I
Bond Polarity and Electronegativity
379
Module 13
8 .8
Bond and Molecular Polarity
T he te r m "po la r" was used in Sectio n 8.7 to d esc ribe a bo nd in whic h on e a to m h as a partial p ositive c h a rge a n d th e o the r a partial n eg ative c h a rge. Because mo st
mo lecu les have polar bo nds, mo lec ules as a who le ca n also he pola r. In a polar mol e cul e , el e ctro n d en sity acc u m u late s toward one side of th e mol e cul e , g iving
• Dipole-Dipole Forces The force of att raction between the negative end of one polar molecule and the positive end of another (called a dipole-dipole force and discussed in Section 12.2) affects the properties of polar compounds. Inte rm olecular forces (forces between molecules) influence the temperature at which a liquid freezes or boils, for examp le.
th at sid e a part ial ne gat ive charge, 0 - , a nd leaving th e other side with a partial positive charge of eq ua l valu e , 8 + (Figure 8. 12a) . Before describing th e fac tors th at de term ine whet her a mo lecu le is po lar, let us loo k at th e ex pe rime n ta l measurement of t he po larity of a molec u le. Wh en placed in an elect ric field , polar mo lec ules ex perience a forc e th at lend s to align them with thc field (Fig u re 8.12) . Wh en th e el ec tric field is created by a pair of op pos itely charge d plat es, the po sitive e nd of each mol ecul e is attracted to th e negati ve plate , and th e neg ative end is a ttracted to the positive pla te (Figure 8. 12b). Th e exten t to wh ich th e mol ecul es lin e u p with th e field depe nd s on th eir dipole moment, j.L, which is defined as th e product of th e mag nitude of th e partial chargcs (8 + and 8 -) on the mo lec u le and the dis tance by whic h they are separated . Th e SI u n it o f th e d ipole mome nt is th e co ulo m b-me ter, but dip ole mo me nt s have tr aditionally been g iven usin g a de rived unit called the dpbJP (D; 1 D = 3. :~4 X IO- :JO C . 01) . Experimen ta l values of so me d ipole momen ts are listed in Table 8.7. To pred ict if a mo lecule is polar, we need to consider if th e mol ecul e has po lar bonds and how th ese bon d s a re po sition ed re lative to o ne an o t he r. Diat omic molecules co mpo sed of two at oms with different e lcctro ncgarivi tics a re always po lar (see Table 8.7) ; th ere is one bo nd , and th e mol ecule has a positive and a negati ve en d . But what happen s with a molecul e with three o r more atoms, in which th ere are two or more polar bond s? Co nsider first a linea r tr iatomic mo lecu le suc h as ca rbo n d ioxi d e, CO 2 (Figu re 8. 13) . H ere , each C=O bo nd is pola r, wit h th e ox ygc n ato m the negati ve e nd o f the bond dipol e. Th e terminal atoms are at th e same di sta nce fro m the C ato m; th ey both have th e same 8 - charge, an d they are sym me trica lly arrange d arou n d the ce n tral C atom . T herefore , CO 2 has no mol ecul ar d ipol e , eve n th ough each
(a)
(b)
FIGURE 8 .1 2 Polar molecules in an electric field. (a) Arepresentation of a polar molecule. To indicate the direction of molecular polarity, an arrow is drawn with the head pointing to the negative side and a plus sign placed at t he positive end. (b) When placed in an electric field (between charged plates), polar molecules experience a force that tends to align them with the field. The negative end of the molecules is drawn to the positive plate, and vice versa. The orientation of the polar molecule affects the electrical capacitance of t he plates (their ability to hold a charge), and this provides a way to measure experimentatly the magnitude of the dipole moment. 380
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TABLE 8 .7
Dipole Moments of Selected Molecules
Molecule (AX)
Moment (I'. 0)
Geometry
Molecule (AX,)
Moment (I'. 0)
Geometry
HF
1.7 8
linear
H,O
1.8 5
bent
HCl
1. 07
linear
H,S
0. 95
bent
HB r
0 .79
linear
SO ,
1.62
bent
HI
0 .38
linear
CO,
0
linear
H, Molecule (AX,)
0
linear
Moment (I'. 0)
Geometry
Molecule (A>C.)
Moment (I'. 0)
Geometry
NH,
1. 47
t rigonal pyramidal
CH,
0
tet rahedral
NF,
0 .23
t rigonal pyramidal
CH ,Cl
1.9 2
tet rahedral
BF,
0
t rigonal planar
CH 2C h
1.60
tetrahedral
CH Cl,
1.04
tet rahedral
CCl,
0
tetrahedral
bond is pol a r. T hi s is a n alog o us to a tu g-or-war in wh ich th e peo pl e at o pposite e nds of th e fo p e arc pullin g with eq ua l fo rc e . In co n trast, wa te r is a be n t triato mi c m ol ecul e . Beca use 0 h as a la rge r electrone gativi ty (X = 3.5) th an H (X = 2.2), ea c h of th e O - H bonds is p ol ar, wit h th e H a to m s havin g th e sam e 8 + c h a rg e a n d o xyge n h avin g a n egati ve ch a rge (8 -) (Fig u re 8. 13) . Elec tro n d e n sity accu m u lates OIl tile 0 sid e of tile m ol ecule , m a kin g the m o lecu le e lec trica lly "lo psid ed " a nd th e refo re pol ar (J.l = 1.85 D) . In t rigo na l-plana r BF3 , th e B- F bon d s are h ighly po lar becau se F is much m ore elec tronega tive th an B (X o f B = 2.0 a nd X of F = 4.0) (Figu re 8.14). The molecule is nonpo lar, h oweve r, beca use th e thre e term in al F a to ms have th e sa me 8 - c harge , a rc th e sa m e di stan ce fro m th e bo ro n a to m , a n d are arra nged sym metrica lly a nd in th e sa me pl an e as th e ce n tra l boro n ato m . In co n tras t, th e trigo n al-pla n ar m olecul e p ho sge n e is po lar (Cl, CO, JL = 1.170 ) (Figu re 8.14). He re, th e a ngles a re all a bou t 120 0 , so the 0 a nd Cl a to m s are sym m e tr ically a rra n ged arou nd th e C a to m . T h e
• Peter Debye and Dipoles The common ly used unit of dipole moments is named in honor of Peter Debye (18841966). He was born in The Netherlands, but attended university in Germa ny and Later st udied for his Ph.D . in physics in Munich. He developed a theory on the diffraction of x-rays by solids, a new con- Peter Debye (1884-1966) cept for m a g ~ netic cooling, and (with E. Hackel) a model for interionic att ractions in aqueous solution. As his interests turned more to chemistry, he worked on methods of determining t he shapes of polar molecules. Debye received the Nobel Prize in chemistry in 1936.
No net dipole moment
co, (a)
(b)
Active Figure 8.13 Polarity of triatomic molecules, AX 2• For CO 2, the CO bonds are polar, but t he electron density is distributed evenly over the molecule, and the charges of 0- lie 180 apart. (Charges calculated using advanced molecular modeling software: C = + 0.42 and 0 = - 0.21. ) The mo lecule has no net dipole. In the water molecule, the 0 atom is negative, and the Hatoms are positive. (Calculated charges: H = + 0.19 and 0 = - 0.38.) However, t he positively charged Hatoms lie on one side of the molecule, and the negatively charged 0 atom is on the ot her side. The molecule is polar. (The calculated dipole of 1.86 Dis in good agreement wit h experlment.) 0
TU
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8.8
I
Bond and Molecular Polarity
381
A Closer Look In Chapter 6, you saw atomic orbitals, regions of space within which an electron is most probably found. The boundary surface of these orbitals was created in such a way that the electron wave amplitude at all points of the surface was t he same value (.... page 288). Using advanced molecu lar modeling software, we can generate the same type of pictu res for molecu les, and in Figure Ayou see a surface defining the electron density in the HF molecule. The electron density surface, calculated using software from CAChe, is made up of all ofthe points in space around the HF molecule where the electron density is at least 0.002
e- /A3 (where 1 A= 0.1 nm). You can see that the surface bulges toward the F end of the molecu le. an indication ofthe larger size of t he Fatom. The larger size of the Fatom here is main ly related to the fact that it has more valence electrons than H, and to a lesser extent to the fact that H- F bond is polar and electron density in t hat bond is shifted toward t he Fatom. We can add another layer of information. The electron density surface can be colored according to the electrostatic potential. (Hen ce, this figure is called an electrostatic potentialsurjace .) The comp uter program calculates the electrostatic potential that would be observed by a proton (H+) on the surface. This is the sum of t he attractive and repulsive forces on that proton due to the nuclei and the electrons in the molecule. Regions of the molecule in which there is an attractive potential are colored red. Th at is, this is a region of negative charge on the molecule. Repulsive potentials occur in regions where the molecule is positively charged; these regions are colored blue. As mi ght be expected, the net electrostatic potential will change continuously as one moves from a negative portion of a molecule to a positive portion, and this is indicated by a progression of colors from blue to red (from positive to negative). The electrostatic potential surface for HF shows the Hatom is positive (the Hatom end of the molecule is blue), and t he Fatom is
3 82
Chapter
8
I
Visualizing Charge Distributions and Molecular PolarityElectrostatic Potential Surfaces and Partial Charge
FIGURE A Three views of the electro static potential surface for HF. (left ) The electron density surface around HF. The F atom is at the left . The surface is made up of all of the points in space around the HF molecule where the electron density is 0.002 e " fA' (where 1 A:= 0.1 nm). (middle ) The surface is made more t ranspare nt , so you can see the Hf ato m nuclei inside the surface.
(right) The front of the elect ron density su rface has been "peeled away" for a view of the HF molecule inside. Color sche me : The colors on the elect ron density su rface reflect the cha rge in the different regions of the molecule 384. Colors to the blue end of the spectrum indicate a positive charge, whereas colo rs t o the red end of t he spect rum indicate a negative cha rge.
negative (the Fatom end is red). This is, of course, what we would predict based on electronegativity, Our program also calcu lated that the Fatom has a charge of - 0.29 and Hhas a charge of + 0.29. Finally, the calculated dipole moment for the molecu le is 1.74 D, in good agreement with the experimental value in Table 8.7. Ot her exam ples of electrostatic potential surfaces iUust rate the polarity of water and methylamine, CH3N H2 •
Water
Methylamine
The surface shows the 0 atom of the water molecu le bears a partial negative charge and the Hatoms are positive. The surface for the amine clearly shows the molecule is polar and that the region around t he Natom is also negative. Indeed, we know from experiment that an H+ ion wilt attack the Natom to give the cation CH3 NH3 +. Electrostatic potential surfaces are becoming more widely used, particularlyin organic chemistry and bioc hemistry, to probe the reactive sites of more and more complex molecules. One examp le is the dipeptide glycylglydne.
Bonding and Mo lecular Structure
H
I
H
I
H 1 N - j - il - ~ - j - il- OH
H 0
H H 0
The electrostatic surface for the molecule shows that the 0 atoms of the C=O groups have a partial negative charge as does t he N of t he NH2 group. Positive regions of the molecule include the H atom of the C(O) -N H grouping (the amide grouping) and the H atom of the OH group. Such pictures can help you see quickly the regions of a molecule that may be a proton donor (an acid) or a proton acceptor (a base).
No net dipole moment
Net dipole
Net dipole
r
Jl = 1.47D
" = 1.17D
SF,
fNH,
Active Fi gure 8.14 Polar and nonpolar molecules of t he type AX l • In BF1 , the negative charge on t he F atoms is distributed symmet rically, so the molecular dipole is zero. In contrast, in ([2CO and NH " the negative charge in the molecules is shifted to one side and the positive charge to the other side.
Mo lecule
Cateulated Partial Charges
Ca lculated dipole
B = 0.44, F = - 0.15
o
a = - 0.21, C = 0.23, Cl = - 0.01
1.25
N = - 0.40, H = 0.13
1. 58
The calculated dipoles are in reasonable agreement with experimentalty measured dipoles. (Calculations were done with molecular modeling software from CAChe. ) TU
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• Electrostati c Potenti al Surfaces for BFl and Cl2CO Notice that the charge distribution for BF, (left) is sym met rical whereas that for ([2CO (right) has a partial negative charge on the 0 atom and much less negative charges on the Cl atoms.
clcctroncga tivitics o f th e th ree atOlI1S in the molecul e differ, however: X( O) > X(CI ) > X (C ). There is therefore a net displacem ent of elec tro n de nsity away fr om th e cen te r of the mo lecule, more toward th e 0 ato m than th e CI a to ms. Am mo nia , like I3 F~h has A X ~~ sto ichiome try and pol ar bonds. In co n tras t to I3 F~~, ho weve r, N II 3 is a trigo nal-pyram idal mo lecu le . T he posi tive II at oms are lo ca ted in th e base of the p yramid, a nd th e nega tive N ato m is 0 11 th e apex of the pyramid . As a co nseq ue nce, 0J H :~ is polar (Figu re 8 .14) . In d eed, trigona l-pyramidal mo lecul es are ge nera lly polar. Mol ecul es like carbo n te tr achl o rid e , CCt .. and methane, CI {4, are non polar, owing to th eir sym metrica l, tetra hed ral struc tures. The four atoms bo nded to C have the same partial charge and are the same dis ta nce from th e C ato m. Tetrah edral molecul es with both CI and I I atoms (C I IC1" CI I,CI2 , and C I I"CI) are po lar, however (Figu re 8. 15). T he c1ectronegat ivity for H ato ms (2.2) is less th an th at of Cl ato ms (3.2), an d th e ca rbo n- hyd roge n d istan ce is d ifferent fro m th e ca rho n---eh lorine distan ces. Becaus e CI is more e lectro negative th an I I, the CI ato ms are o n th e more negat ive sid e of the molecu le . T his means th e positive end of th e mo lecular d ipo le is toward th e II at om . To summarize this d iscussio n of molec ular polari ty, loo k again at Fig ure 8.5 (page 3(9) . T hese are ske tches of molecules o f th e type A X" whe re A is th e cen tra l at o m and X is a terminal atom . Yem can pred ict th a t a mo lec ule AXil will not be po lar, regard less o f whet he r the A-X bond s are po lar, if • All the termin al a to ms (o r gro ups), X, are id e n tica l, an d • All th e X ato ms (o r g rou ps) arc a rranged sym me tr ically aro u nd th e ce n tral at om , A. O n th e o the r hand, if o ne o f th e X ato ms (o r grou ps) is d iffe ren t in the str uctures in Figure 8.5 (as in Figures 8. 14 and 8. 15), or if o ne o f th e X po sition s is occ u pied by a lone pair, th e molecule will be pol ar. 8 .8
I
Bond a nd Molecular Polarity
383
J1. = 00
Net dipole
Net dipole
Net dipole
J1. = 00
No net dipole moment
J1. = 1.92 0
J1. = 1.60 D
J1. = 1.04 0
No net dipole moment
0-
.1 +
.l -
JrJ - +
0+
.1 +
.1+ 0+
0+
.1+
.1+
CH,Cl
CH.
a-
J( .1-
0+
-r0-
0-
0+ .I.1CHCl, CC!. FIGURE 8.15 Polarity of tetrahedral molecuLes. The electronegativities of the atoms involved are in the order Cl (3.2) > C(2.5) > H (2.2). Thi s means the C- Hand C-Cl bonds are polar with a net displacement of electron densityaway fromthe Hatoms and toward the Cl atoms [H8+- C8- and C8+-Cl 8- j. Although the electronpair geometry around tile Catom in each molecule is tetrahedral. onlyin CH4 and CCl. are tile polar bonds totally symmetricalin their arrangement. Therefore, CH3Cl , CHzCl2' and CHell are polar molecules, with t he negative end toward the Cl atom s and t he positive end toward the Hatoms .
CH,Cl,
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MolecuLar PoLarity ProbLem Are nitrogen trifluoride (NF 3) and sulfur tetrafluoride (SF,) polar or nonpolar? If polar, indicat e the negative and positive sides of t he molecule. Strategy You cannot decide if a molecule is polar witho ut determining its structu re. Therefore, start with the Lewis st ruct ure, decide on the electron-pair geometry, and then decide on t he molecular geometry. If t he molecular geometry is one of t he highly symmetrical geometries in Figure 8. 5, t he molecu le is not polar. If it does not fit one of t hese cat egories, it will be polar. Solution (a) NFl has t he same t rigonal-pyramidal str ucture as NH 3 • Because Fis more electronegative t han N, each bond is polar, t he more negative end being the Fat om. Because this molecule contains polar bonds and because t he geometry is not symmet rical but has instead t hree positions of t he tet rahedron occupied by bonding groups and one by a lone pair, the NFl molecule as a whole is expected to be polar. You will notice, however, that t he dipole mom ent for NFl is quite small (0.23 0 in Table 8.7) , much smaller than that of NH 3 • This illustrates that lone pairs have aneffect on polarity. For NH 3 , t he N-atom lone pair adds to t he overall polarityof the molecule. (The lone electron pair exte nds into space beyond the Natom and increases the charge separation in NH 3 ; this enhances the dipole.) For NF3, however, the effect of the lone pair on the nitrogen atom of the molecule is counterbalanced by the highly polar N- F bonds on t he other side and the magnitude of the dipole pointing toward t he side with the Fatoms is reduced.
384
Chapter 8
I
Bonding and Molecular Structure
(b) The S- F bonds in sulfur tetrafluoride, SF4 , are highly polar, the bond dipole having f as t he negative end (X for 5 is 2.6 and X for Fis 4.0). The molecule has an electron-pair geometry of a t rigonal bipyramid (see Figure 8.8). Becauset he lone pair occupies one of the positions, the S- Fbonds are not arranged symmetrica lly. The axial S-F bond dipoles cancel each other because t hey point in opposite directions. The equatoriaL S- F bonds, however, both point t o one side of the moLecule.
Comment In qeneral. we do not consider lone pair effects on mo lecular dipoles. Nonetheless, t hey do have an effect, as seen when comparing the dipoLe mome nt for NF 3 (0.23 D) with that for NH] (1.47 D). This is a case in which electrostatic potential surfaces and calculated atom charges are usefuLin showing the difference between these mo LecuLes. The Natom in NH 3 is decidedly negative (-0.40), and t he H atoms are positive (+0.13). In contrast, in NFl the Natom is positively charged, and the Fatoms are negativelycharged (N = +0 .3 and F = - 0.1). The difference in charge between Na nd Fis not as great as between Nand Hin NH l.
NH3
NF3
Electrostatic potential maps.
MolecuLar PoLarity Problem 1,2-DichLoroethyLene can exist in two forms. Is either of t hese pLanar molecuLes poLar?
H
Cl
H
\L
H/
C=C / " Cl
H
"
C= C/
Cl/ A
"
B
Strategy To decideif a molecule is polar, we first sketch the structure and then, using electronegativity values, decide on the bond polarity. Finally, we decide if the eLectron density in the bonds is distributed symmetrically or if it is shifted to one side of the molecule. Solution Here, the Hand Cl atoms are arranged around the C=C double bondswithall bond angles 1200 (and aLL the atomslie in one plane). The eLectronegativities of the atoms invoLved are in the order Cl(3.2) > C( 2.5) > H(2.2). This means the C-H and C-Cl bonds are polar witha net displacement of electron densityaway from the Hatoms and toward the Cl atoms [H s- - C s- and ( li + - (( 6- j. In structure A, the Cl atoms are located on one side of the mo lecule, so electrons in the H-C and C-Cl bonds are displaced toward the side of t he molecule with (1 atoms and away from the side with the Hatoms. Mo Lecule Ais polar. In molecule B, the displacement of electron density toward the Cl atom on one end of the molecu le is counterbalanced by an opposing displacement on the other end. Molecule Bis not polar,
Overall displacement of banding electrons
1
s+
H~
/l
C= C
Cl« B-
s:
s+
\." Cl B-
A, polar, diplacement of bonding electrons to one side of the molecule
s+ H
Cl Displacement of ) bonding electrons
"''.
//
(DiSPlacement of
/.C= C" bondingelectrons H )' \. Cl
s+
s-
B, not polar, no net displacement of bonding electrons to one side of the molecule
8 .8
I
Bond and MolecuLar Po larity
385
Comment The electrostatic potential surfaces reflectthe fact that molecule Ais pola r becausethe elect ron de nsity is shifted to one side of the molecu le. MoLecu le B is not pola r because the electron density
is distributed symmetrica lly.
Molecule A
EXERCISE 8.14
Molecule B
Molecular Polarity
For each of the following molecu les, decide whether the molecule is polar a nd which side is posi-
tive and which negative: BFCl2, NH 2CL and Sel2 _
EXERCISE 8.15
Molecular Polarity
The electrostatic potential surface for OSel2 is pictured here.
(a) Draw a Lewis electron dot picture for the molecule, and give the format charge of each atom. (b) Whatis the motecu lar geometry of OSCl2?
(c) Is the molecu le polar? If so, locate the positiveand negativecharges and the direction of the dipole.
8 .9 Bond Properties: Order, Length, Energy Bond Order
0 = ( =0
Th e o rd er of a bond is th e number of bonding e lectron pairs sha re d by two atoms in a mo lecule (Fig u re 8. 16). You will encount er bond o rd ers of I , 2, and 3, as well as fracti onal bond o rd ers. wh en th e bond orde r is I , th ere is onl y a sing le covalent bond betwee n a pair of atoms. Exa mp les are the bond s in mo lecu les suc h as 112 , NII J , an d Cl L. T he bon d order is 2 when two electron pa irs are shared be twee n atoms, such as the C=O bonds in CO 2 and th e C=C bond in eth ylene, 1I 2C= C1I 2 • T he bond o rd er is 3 when two ato ms are co n nec ted by three bo nd s. Examples includ e the carbon-oxyge n bond ill carbo n mon oxide, CO and th e ni trogen -nitrogen bond in N2 . Fractional bond o rders occu r in molecul es a nd ions having re sonance struct ures . Fo r example , what is the bo nd ord er for ea ch ox yge n-o xygen bon d in 0 3? Each re sonance struc tu re o f 0 3 has one 0 - 0 single bo nd and one 0=0 double bond , fo r a to tal of t hree shared bonding pairs acc ou n ting for 1\\"0 o xygen -oxygen lin ks.
FIGURE 8.16 Bond order. The four
C-H bondsin methane each have a bond order of 1. The two ( =0 bonds of COl
each have a bond order of two, whereas
Bond order = 1
. I 0~ Bond order
: Oy ", O'
the nitrogen-nitrogen bond in N2 has a n
One resonance structure
order of 3.
386
Chapt er
8
I
Bonding and Mo lecular Structure
} Bond orderfor each =
2
(~?rn-oxygen bond -
" or 1.5
V\'e ca n d efin e th e bo n d order between a ny bon d ed p air of a to ms X an d Y as number of shared pairs in all X-V bonds
Bond order =
(8 .2)
number of X-V links in the molecule or ion
Fo r o zone , the re a re three bond p ai rs in volved in two o xyge n -oxygc n lin ks, so th e bo nd o rd er fo r each ox ygen- ox yg e n bon d is :~ , or 1.5.
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Bond Length Bond length is th e di stance bet wee n th e n u cle i of two b o nded atoms. Bond le ng th s a re th e re fore rel ated to th e sizes of th e ato ms (Sec tio n 7.5) , but, for a given p air of atoms, th e orde r o f th e bond also plays a rol e. Tab le 8.8 lists ave rage bo nd le ngt h s fo r a n u m be r o f common chemical bo n d s. It is impo rta n t to rec o gn ize th a t th ese a rc avemgr valu es. Nc ig h bo ring p art s of a mol ecul e ca n affec t th e length o f a parti cul a r bo nd . Fo r exam p le , Ta b le 8.8 spe cifi es th a t the average C- Il bond h as a le ng th o f 110 p m . In met h an e , CII-I, th e measu red bond length is 109.4 pm, whereas th e C- H bo n d is o n ly 105.9 pm long in ace tyle ne, H - C==C-H . Variati ons as g rc at as 10 % fro m th e avc ragc valu es listed in Ta b le 8.8 a re p ossible .
TABLE 8 .8
Some Average Single- and Multip le-Bond Lengths in Picometers (pm}" Single Bond Lengths Group
H
1A
4A
SA
6A
H
(
N
0
74
110
98
94
154
147
143
(
N
140
0
7A
4A
SA
6A
7A
7A
Si
P
S
Cl
Br
7A
92
145
138
132
127
142
161
141
194
187
181
176
191
210
136
134
187
180
174
169
184
203
132
130
183
176
170
165
180
199
128
181
174
168
163
178
197
234
227
221
216
231
250
F Si
P
220
S (I
BT
214
209
224
243
208
203
218
237
200
213
232
228
247 266
Multiple Bond Lengths
*1 p m - 10
12
( ~(
134
(==(
121
( =N
127
(=N
115
(~ O
122
(~O
113
N ~O
115
N==O
108
m.
8 .s
I
Bond Properties: Order, Length, Energy 387
4A
5A
6A
c
N
o
s;
p
s
Relative sizes of some atoms of Groups 4A, 5A, and 6A.
Because ato m sizes var y in a re gular fashio n with the positio n of the element ill the periodic tab le (Figure 7.8) , predi c tions o f tren ds in bo n d len gth ca n be m ad e qu ickly. For ex am ple . the II -X di stance in the hyd rog en halide s incr eases in the orde r predi c ted by the rela tive size s o f the h aloge n s: I-J - F < IJ - Cl < l l -r-Br < H-1. Like wise , bo n d s be tween ca rbo n a nd a n o t he r eleme nt in a given pe ri od decrease going from left to right , in a p red ictable fashion ; fo r e xam ple . C-C > C- N > C- O > C- F. Tr ends for mu ltiple bo n ds are sim ilar. A C= O bond is shorter tha n a C=S bo n d , a n d a C= N bo n d is sh o rt e r th a n a C= C bo n d. T h e effe ct of bond order is evide nt whe n bonds be twee n th e sa m e two atoms a re co m pare d. For ex a m p le, the bo n d s become short er as th e bond orde r in creases in th e se ries C- O , C =O, a nd C==O:
Bond lengths are related to atom sizes.
Bond
(-H
N-H
O-H
Bond Order
110 Si-H 145
98 P-H 138
94 pm S-H 132 pm
Bond Length (pm)
(-0
( =0
143
2
3
122
113
Do u ble bonds a re sh o rte r than sin gle bonds be tween the sa me set of a to m s, a nd tr ip le bond s between those sa me a to m s are sh o rte r still. The carbon at e io n , CO l -, h as three eq u ivalen t resonance stru c tures, Eac h CO bond h as a bo nd order of 1.33 (o r %) because fo u r e lectro n pairs a re used to fo rm th ree ca rbo n- oxyge n lin ks. T he CO bo nd d ista n ce (129 p m ) is inte rmed ia te be[ween a C- O single bond ( 143 pm ) a n d a c=o dou b le bond ( 122 p m ).
:0# 2- , Bond order = II = C
[: ..0 ,(\
Bond order =
2) Av:rage bondorder
1
"". .
..
=
0:
.. Bond order
~ . o r 1. 3 3
s
=
Bond length = 129 pm
1
EXERCISE 8 .16 Bond Order and Bond Length (a) Give the bond order of each of the followi ng bonds, and arrange them in order of decreasing bond distance: C=N , C::=::N, and C-N. (b) Draw resonance structures for N02 - . What is the NO bond order in this ion? Consult Table 8.8 for N-O and N= O bond lengths. Compare these with the NO bond lengt h in N0 2- (124 pm). Accou nt for any differences you observe.
Bond Dissociation Enthalpy T he bond di ssociation e n tha lpy is th e e nthalpy c hange fo r b re ak in g a bo n d in a mo lecu le with th e reactants a n d p rodu ct s in th e gas p hase .
Molecule (g)
Energy supplied =
~H
>0
Energy released =
~H
H, C(g) + CH, (g)
!1,H = + 682 kJ/m ol-rxn
HC=C H(g) --> HC(g) + CH(g)
!1,H
H,[-CH , (g) --> H,C(g)
=
+ 962 kJ/ mol-rxn
Becau se 0.// re p rese n ts th e e n ergy tr an sferred to th e mol e cule from its su rro u n dings, A l l has a po sitive value; th at is, the [nncess oJ breaking bonds in a molecule is alw(1)'s
• Variability in Bond Dissodation Enthal pies The values of ~ rH for eth ane. ethylene, and acetylene in the text are for those molecules in particular. The bond dissociation enthalpies in Table 8.9 are average values for a range of molecules containing the indicated bond.
endothermic. T he e ne rgy su p plie d to brea k ca rbo n-car bo n bonds must he th e sa me as th e e ne rgy re leas ed whe n the sa me bonds form . 77iPforma tion oJ bonds from atoms or radicals in the gas ph ase is alwa)'s exothermic. T h is m ea ns , for exa m p le, th at ArH fo r th e fo r m atio n o f H:\C- CH :'l from two C H ~~( g) rad ical s is - 368 ~J /m ol -rxn .
H, C' (g) + . CH,(g) --> H,[-C H,(g)
!1,H = - 368 kf / mol-rxn
Gen erally, th e bond e nergy fo r a given type o f bo nd (a C-C bond , fo r exa mp le) vari e s so mewha t, dep endin g on th e co mpou nd.ju st as bond length s var y fro m on e molecule to an o th er. T hey are su fficie n tly sim ilar, however, so it is possib le to cre ate a table of average bond dissociation m thnlpies (Ta ble 8.9) . The values in suc h ta bles may bc used to estimate the en t halp y change fo r a reaction , as d escri bed below.
TABLE 8 .9
Some Average Bond Dissociation Enthalpies (kJ/mol)Single Bonds
H C N 0
H
C
N
0
f
Si
P
S
Cl
Br
436
413
391
463
565
328
322
347
432
366
299
346
305
358
485
272
339
285
213
163
201
283
201
201
192
146 155
f
Si
452
335
565
490
222
218 284
253
249
278
293
381
310
234
201
P
S
326 226
CI
255 242
Br
184
216
208
193
175 151
Multi pLe Bonds N ~N
418
C ~C
610
N=N
835
945
C=C
C~N
615
C~O
745
C=N
887
C=O
1046
0=0 (in 0,)
498
" So urce s of di sso cia tio n cn thalpk-s: I. Klo ll an d R. :\1. Ros enberg: Chemical Thermodynamics. 4th Ed. , P. .1')5 . New ) (lrk. J o h n "" i1ey. 1994; an d]. E, H uhccy, E. A. Ke iter, an d R. L. Ke iter: Inorganic Chrmi ,\tT)' 4th Ed" Tab le E. I, New 'ror k, H arp er-Co llin s, 19 93. Sec also Lan g:e 's H alU/book of Chemistr}. j. A. Dea n (cd.}, ~ Ic G raw-Hill lnc. , New )'(n k.
8.9
I
Bond Properties: Orde r, Lengt h, Energy
389
• Bond Energy and Electronegativity Linus Pauling derived electronegativity values from a consideration of bond energies. He recognized that the energy required to break a bond between two different atoms is often greater than expected, based on an assumption that bond electrons are shared equally. He postulated that the "extra energy" arises from the fact that the atoms do not share electrons equally. One atom is slightty positive and the other slightly negative. This means there is a small coulcmbic force of att raction involving oppositely charged ions in addition to the force of att raction arising from the sharing of electrons. This coulombic force enhances the overallforce of attraction.
In reactions be tween mo lecu les, bo nds in re actants are broken ; new bonds are for med as p roduc ts form . If t he total encrb'} rel eased whe n new bo nds form ex ceeds the ene rgy requi red to break the original bo nds, the overall re act ion is exot hermic. If the opposite is true , then the ove ra ll reac tion is e ndothermic. Let us see how this wo rks in p ractice . Le t us use bond di ssociation cru halpt cs to estimate the en thalpy change for the hydrogen ation of p ropen e to p ro pane : H H H
H-
I I I C- C= C-H (g) + I
H
~
H-H (g)
H
H
H
I I I H-C -C -C I I I H
propene
H
H(g)
H
propane
T he first step is to exam ine the reactants an d p rod uct to see what bo nds are bro ke n and what bo nds are fo rmed . In th is case, th e C= C bo nd in p ro pe ne and the H -H bon d in hydrogen are b ro ke n. A C-C bond and two C- II bonds are formed . mol o f C=C bo nds and 1 mol o f H - H bo nd s
Bon ds lnokrn :
H
H H
I I I
H- C- C= C-H (g) + H-H (g)
I
H Energy req uired Bonds formed:
= 610 kJ for C=C bonds + 436 kJ for H- H bonds = 1046 kJ/mol-rxn mo l of C- C bo nd s and 2 mol of C- H bo nd s H
H
H
I I I
H- C- C- C-H (g)
I I I
H
Energy evolved = 346 kJ for C-C bonds
H
H
+ 2 mol X 413 kJ/ mol for C-H bonds
=
1172 kJ/mo l-rxn
• Hydrogenation Reactions Adding hydrogen to a double (or triple) bond is called a hydrogenation reaction. It is commonly done to convert vegetable oils, whose molecules contain C=C double bonds, to solid fats.
By co m bining the e ne rgy re qu ired to break bo n d s and the e nergy evo lved in makin g bo nd s, we ca n estim ate I1J I fo r the hyd rogena tio n o f p ropene and see that the re act io n is ex o thermic. f1,H = 1046 k.f /mol-rxn - 1172 kJ/ mol-rxn = - 126 kf /mo l-rxn
T he exam ple o f the pro pen e- h ydroge n reaction illu strates the fact that the e n thalpy change fo r any reac tio n can be estim ate d using t he eq uatio n • A,H from Enthalpies of Formation Using ')'/1" values for propane and propene, we calculate .).,H for the reaction of - 125.1 kf/mcl-rxn . The bond dissociation enthalpy calculation is in excellent agreement with that from enthalpies of formation. 390
Chapter
8
I
f1,H = :i f1H(b onds broken) - :i f1H (bonds formed)
(8.3)
To use th is eq ua tio n, first ide ntify all th e bon d s in th e reactants th at are b ro ken , and ad d up thei r ho nd dissociation en tha lpies. T hen , identify all the new bonds form ed in the products, and ad d up their bo nd d issociation en thalpies. T he di fferen ce betwee n th e energy re qui red to bre ak bo nd s [= ~ I1 H(bo nds b ro ken ) ] and th e e nergy
Bonding and Mo lecular Structure
evo lved whe n bonds are mad e [= "i!11/(bonds form ed )] gives the estimated en thalpy change for th e reac tion . Such calcu latio ns can give acce ptable result s in ma ny cases.
ChemistrY-9- Now'" Sign in at www.tbomsonedu .com/Ioqtn and go to Chapter 8 Contents to see Screen 8.18 to explore how reactant and product bond energies influence the energy of reaction.
. Using Bond Dissodation Enthalpies Problem Acetone, a common industrial solvent, can be converted to isopropanol, rubbing alcohol, by hydrogenation. Ca lculate the enthalpy change for this reaction using bond energies. O/H
o II
H, C- C- CH, (g)
+ H-
H(g)
~
I
H, C- C- CH, (g)
I
H isopropanol
acetone
Strategy Examine t he reactants and products to determine which bonds are broken and which are formed. Add up the energies required to break bonds in the reactants and the energy evolved to form bonds in the product. The difference in t he sums of bond dissociation enthalpies is an estimate of the enthalpy change of the reaction (Equation 8.3). SoLution Bonds broken: 1 mol of C=O bonds and 1 mol of H-H bonds
o II
H, C- C- CH, (g) ~ ~H(bonds
+ H- H(g)
broken) = 745 kJ for C=O bonds + 436 kJ for H- H bonds = 1181 kl/rnol-rxn
Bonds formed: 1 mol of C- H bonds, 1 mol of c-o bonds, and 1 mol of O-H bonds H
0/
I
H,c - C- CH ,(g )
I
H ~ .l H ( b o n d s
.l ,H ~
formed)
~ .lH(bonds
~
413 kl for C-H + 358 kl for C-O + 463 kl for O-H
broken) -
.l ,H ~ 1181 kJ - 1234 kl
~
~ .l H(bonds
~
1234 kf/mcl-rxn
formed)
- 53 kf/mc l-rxn
Comment The overall reaction is predicted to be exothermic by 53 kJ per mol of product formed. This is in good agreement with t he value calculated from ~fW values ( = - 55.8 kJ/mol-rxn) . EXERCISE 8.1 7
Using Bond Dissodation Enthalpies
Using the bond dissociation enthalpies in Table 8.9, estimate the enthalpy of combustion of gaseous methane, CH 4 • That is, estimate ~ r H for the reaction of methane with O2 to give water vapor and carbon dioxide gas.
8 .9
I
Bond Properties: Order, length, Energy
391
Historical Perspectives
DNA-Watson, Crick, and Franklin
DNA is the substance in every plant and animal t hat carries the exact blueprint of that plant or animal. The st ruct ure of this molecule, the cornerstone of life, was uncovered in 1953, and James D. Watson, Francis Crick, and Maurice Wilkins sha red the 1962 Nobel Prize in medicine and physiology for the work. It was one of the most important scientific discoveries of the 20th cent ury, and the sto ry
has been told by Watson in his book The Double Helix.
When Watson was a graduate st udent at Indiana University, he had an interest in t he gene a nd said he hoped t hat it s biological
role might be solved "without my learning any chemistry." Later, however, he and Crick found out just how useful chemist ry can be when they began to unravel the st ruct ure of DNA. Solving importa nt problems requires teamwork among scientists of many kinds, so Watson went to Cambridge University in England in 1951. There he met Crick, who, Watson said, ta lked louder and faster than ~ anyone else. Crick shared Watson's belief in 1 the fundamental importance of DNA, and the ~ pair soon learned that Ma urice Wilkins and ~ Rosa lind Franklin at King's College in London ~ were using a technique called x-ray crystal~ lography to learn more about DNA's structure. '8 Watson and Cri ck believed that understanding j this structure was crucial to understandi ng genetics. To solve the structural problem, however, they needed experimental data of § the type that could come from t he experiJg ments at King's College. ~ The King's College group was initially
1 i
James D. Watson and Frand s Crick. In a photo taken in 1953, Watson (left) and Crick (right) stand by their model ofthe DNA do uble helix.Together with Maurice Wilkins, Watso n and Crick received the Nobel Prize in medicine and physiology in 1962 . (A Ba rrington Brown/Science/Photo Researchers,Inc.)
reluctant to share thei r data; and. what is more, they did not seem to sha re Watson and Crick 's sense of urgency. There was also an ethical dilemma: Could Watson and Cri ck work on a problem that others had claimed as
s ~
'a
§
•
~
~ 8
j ~
~
....
~
Rosalind Franklin of King's College. london. She died in 1958 at the age of 37. Because Nobel Prizes are never awarded posthumously, shedid notshare in this honor with Watson, Crick, and Witkins. For more on Rosalind Franklin, read Rosolind Franklin: The Dark Lady af DNA by Brenda Maddox.
the irs? "The English sense of fair play would not allow Francis to move in on Maurice's problem," said Watson. Watso n and Crick approached the problem through a technique chemists now use frequently-model building. They built models of the pieces of the DNA chain, and t hey tried various chemica lly reasonable ways of fitting them together. Finally, they discovered that one arrangement was "too pretty not to be t rue." Ultimately, the experimental evidence of Wilkin s and Fra nklin confirmed the "pretty st ructure" to be the real DNA st ruct ure.
8.10 DNA, Revisited This chapter o pe ned with some questions about the struc tu re o f DNA, one of the key molecul es in all bio log ical syste ms. Th e tools are no w in place to say mo re abo ut th e struc tu re of thi s important m olecu le and why it lo o ks th e way it does. As shown in Figure 8. I 7, eac h str and of the do u ble-stran ded DNA mo lecu le co nsists of th ree u n its: a p hos p hate, a d eo xyr ibo se mol ecul e (a sugar molecule with a five-member ring ) , a nd a nit ro gen-contain ing basc o (T he bases in [) ~ A can be o ne of four mo lec ules: adeni ne , guani ne , cytos ine , and thymine; in Figu re 8. 17, th e base is ade n ine.) Two uni ts of t he back bone (without the ade nine on the deoxyribose rin g ) arc also illustra ted in Figure 8. 17. The imp ort an t point here is that th e re peating un it in th e bac kbo ne of DNA co nsists of the ato ms O -P- O - C- C- C. Eac h atom has a te trahed ral elec tro npa ir gco me try. The refore , th e cha in ca n no t be lin ear, In fac t, th e cha in twists as o ne m oves along the back bo ne . T his twisti ng gives DNA its helical sha pe, "'-Th y are th ere two stra nds in DNA with the O -P-O- C- C- C backbone o n th e o utside an d th e nitroge n-co n tai ning bases o n th e in side? T his struc tu re arises fro m th e polari ty o f the bond s in th e base molecules attac hed to the backbon e . 392
Chapter
8
I
Bon ding and Molecular Structure
Five-member deoxyribose ring is slightly puckered owing to tetrahedralgeometry around each Cor 0 atom.
Angles here are all about 120" becauseeach atomis surrounded bythree singleor double bonds p/~ or bytwo single or double bondsand one lone pair. 9.is ~-O
,
;.:S7
P
0-
Repeating unit of DNA backbone: 1 P atom 2 0 atoms 3 Catoms
poe
Sugar (deoxyribose portion) Adenine P- O- Cbond is bent. o atom surrounded bytwo bond pairs and two lone pairs.
Phosphate group, PO/ Electron pairgeometry is tetrahedral.
'lo.-l~d-- B a se
FIGURE 8.17 A portion of the DNA molecule. A repeating unit consists of a phosphate portion, a deoxyribose portion (a sugar molecule with a five-member ring), and a nitrogen-containing base (here adenine) atta ched to the deoxyribose ring.
For exa mp le , t he N-H bo nd s in the ade n ine m ol ecul e are very po lar, whic h leads to a special fo rm of in termolecu la r for ce s-hydrogen bo nding-to th e ba se molecule in th e n ei g hboring ch ain. Mo re abo u t t hi s in Cha p te r 12 whe n we ex plo re inte rm ol e cu la r for ces an d aga in in Th e ChemistlJ of Li]e: Biochemistry (p ag e s 496- ,, 12) .
Chapter Goals Revisited N ow that you have studied this chapter; you shou ld ask mhether you have goals. In panicular; JOu should be able 10:
1IIt'!
the chapter
Understand the difference between Ionic and covalent bonds a. Describe the basic for ms of che m ica l bo nding-ionic a nd cova le n t-an d the di ffe ren ces be twee n th e m , and p red ict from the formu la whe the r a co mpo und has ionic or covalen t bonding, based o n whe ther a met al is part of the fo rmul a (Sectio n 8. 1) . b. Wri te Lewis sym bo ls for ato ms (Sectio n 8.2) . Draw Lewis electron dot structures for small molecuLes and ions a. Draw Lewis struc tu res for mo lecu lar co m po u n ds and ion s (Sectio n 8.2). Study Question(s) assignable in OWl: 6. 8, 10.
b. c.
Understa nd and ap ply t he o ctet ru le; re co gnize exce p tio ns to th e oct et ru le (Secti o ns 8.2-8.5). Study Question(s) assignable in OWL: 6. 8. tc, 12. 56. Writ e reson ance structures , u nd erstand wha t reso nance means, and how and whe n to use this means of rep resen tin g bond ing (Sectio n 8.4). Study auestion(s)
Chemistry.o.Now"" Sign in at www. thom sonedu~com/login to: Assess your understanding with Study Questions in OWL keyed to each goal in the Goals and Homework menu for this chapter For quick review, download Go Chemistry mini-lecture flashcard modules (or purchase them at www.ichapters.com) Check your readiness for an exam by taking t he Pre-Test and exploring the modules recommended in your Personalized Study plan.
Q Access How Do I Solve It? t utorials on how to approach problem solving using concepts in this chapter.
assignable in OWL: 10.
Cha pte r Goats Revisite d
393
Usethe valence shell electron-pair repulsion t heory (VSEPR) to predict t he shapes of simple molecules and ions and to understa nd the st ructu res of more complex molecules. a.
Pred ict the shape or geome try o f molecul es and ion s of main grou p elements using VSEPR theo ry (Sec tio n 8.6) . Table 8. 10 show s a su mmary of the re latio n between valen ce ele ctron pa irs, ele ctron-pa ir an d molecul ar geometry an d molecular p ola rity. Study Question(5) assignable in OWL: 18, 20, 22, 24, 86 , 88 ; GoChemistry ModuLe 12.
Useelectronegativity and formal charge to predict the charge distribution in molecules and ions, to define the polarity of bonds, and to predict the polarity of molecules. a.
Ca lcu late formal ch arges for atoms in a molecule based on th e Lewis structu re
b.
d.
Defi ne elec tro uegativity a nd und erstand ho w it is used to d escrib e the unequ al sharing o f el e ctrons betwe en a to ms in a bond (Sectio n 8.7) . Co mbine form al charge and c lec tr o ncga tivity to gai n a perspecti ve o n the charge dis tri butio n in cova len t mol ecul es an d io ns (Sec tion 8. 7). Study auestion(s) assignabLe in OWL: 28, 29, 31, 32, 34, 71. Understand why so me molecul es are polar whe reas o the rs are no npo lar
e.
Pred ict th e polarity or a molecule (Sec tio n 8.8). Studyauestion(s) assignable in
(Sec tio n 8.3) . Study Question(s) assignabLe in OWL: 14. 16, 36.
c.
(Section OWL:
8.8) .
See Table
8 .7.
Study auestion(s) assignable in OWL: 38.
38, 40, 78, 79, 8t , 86; Go Chemistry ModuLe 13.
Understand the properties of covalent bonds and their influence on molecular structure a. b.
TABLE 8 .10
Defin e an d predict tr ends in bond order, bond le ng th , and bond dissociation e ne rgy (Sec tio n 8.9) . Study auestion(s) assignable in OWL: 27, 42, 44, 45, 48, 58, 81. Use bond di ssociation e nt halpies in ca lcu latio ns (Sectio n 8.9 and Exa m p le 8 . 14). Study auestion(s) and assignable in OWL: 50, 51, 52, 69.
Summary of Molecular Shapes and Molecular Polarity
VaLence ELectron Pairs
ELectron-Pair Geomet ry
2
linear
3
t rigonal planar
4
tet rahedral
5
trigonal bipyramidal
Number of Bond Pairs
Number of lo ne Pai rs
MoLecuLar Geometry
MolecuLar DipoLe?*
2
0
linear
3 2
0 1
trigonal planar bent
no no yes
SnCl,(g)
4 3
0 1
tetrahedral trigonal pyramidal bent
no yes yes
CH 4 • BF4 NHl • PF3 H20. sci,
trigonal bipyramidal seesaw T-shaped linear
no yes yes no no yes no
PF, SF.
2
2
5
0 1
4 3
octahedral
6
3
6
5
0 1
4
2
* Fo r mole cu les o f th e AX n • where th e X a toms a re ide n t ical.
394
Chapter 8
I
Bonding and Molecular Structure
2
2
octahedral square pyramidal square planar
Examples BeCl2 SF). a«,
ur,
Xe F2 • 1)SF6 , PF6 -
nr,
Xe F4
KEY EQUATIONS Equation 8.1 (page 359)
Ca lcu lating the fo rmal c harge on a n atom in a mo lecu le
Formal charge of an atom in a molecule or ion Equation 8.2 (page 387)
Group Num ber - [LPE + 'I,(8E)]
Ca lcu lating bond o rder
Bon d orde r =
Equation 8.3 (page 390) d issociation e n th aIp ies
~
number of shared pairs in all X- V bonds num ber of X-V links in the molecule or ion
Estima ting the en tha lpy cha nge for a react ion using bo nd
tl ,H ~ ~ tlH(bon ds broken) - ~ tlH(bon ds forme d)
STUDY QUESTIONS Online homework for this chapter may be assigned in OWL.
... denotes challenging questions. •
denotes questions assignable in OWL.
Blue-numbered questions have answers in Appe ndix a and fully-worked solutions in the Student Solutions Manual.
Practidng Skills Valence ELectrons and the Octet Ru le (S"" Sa/ion 8,/ an d Chemistrybkno Screen. 8. 2.)
1. Give the pe riodic grou p number and n u mber o f vak nee el ectron s fo r each o f the fo llo wing a tom s. (a) 0 ( 2 O , (g )
D.,H" = - 394 q / m ol-rx n o
Usi ng D., H an d th e bo nd di ssoc iatio n e n tha lpy dat a in Ta ble 8.9, es tima te th e bon d d issociat ion e n t ha lpy fo r th e o xyge n-o xygen bo nd in o zone, 0 :1 • How d o cs yo ur e stim ate co m pare with th e e nergies of an 0 - 0 sing le bo nd a nd an 0 =0 d o u ble bo nd ? Do c s the oxyge no xygen bon d di ssociati on e nthalpy in ozo ne correlate with it" bo nd order?
46 . Co m pare t he ca rhon- oxyge n bon d length s in the formale ion (HCO t " } , in methan ol (CH 3 0 I I) , awl in the carbo na te ion (CO:\2-). In whic h species is th e ca rbono xyge n bon d predi c ted to be lon ge st? In whic h is it p re di ct e d to he sho rtes t? Exp lai n bri efly.
I
H
I
I
45 . • Con sider the nitrogen - o xyge n ho nd le ng ths in NO:/ , NO t - , a nd NO :J - . In whi ch io n is th e bon d pred icted 10 he lo ngest? In whic h is it p re dicted 10 he the sho rtest? Expl ain brie fly.
398
H
I
II
Bond Order a nd Bond Length (8('(' Exercise 8.16 aud Cltnni.s(1)'Now Screen R. / i .)
42 . • Give th e bond o rd er for e ac h bon d in the fo llowi ng molec ule!'; or ion s: (a ) CW (e) S0 " (h) C II ,Cl' (1I) C Il "CII =C Il,
H
I
II - C - C - C = C - II (g ) + 1I, (g )
40. • Wh ich o f th e fo llo win g molecules is (arc ) not pol ar? Whi ch molec ule has bonds with th e larges t po lar ity? (a ) CO (rl) PCI, (h ) RCl, (e) (;"'1, (e ) CF ,
4 1. Give th e bond order fo r e ac h bon d in th e foll owing molecu les o r io ns: (c) N O \!+ (a) CliP (h) SO,z (r1 ) NOCI
H
•
more chaLLenging • in OWt
Blue-numbered questionsanswered in Appendix 0
General Questions on Bonding and Molecular Structure Thew' qu estions are not designated us to ')1)(' or (om/ion in
OIP
dUljJ{n Tlw)' ma)' combine seueml r:m u:t-jJIJ.
53 . • Specify th e n u m ber of vak nee el ectrons for Li , T i, Zn , Si, a n d Cl. 54. In b oron com pou nds, th e R '110 m o fte n is not su rrounded by four valence electron pairs. Illustrate th is wit h RCh . Sh ow h o w th e m o le cu le ca n ac hieve a n octe t co nfigur ation hy formi ng a coo rdin ate covalen t bon d wi th ammo n ia (N H:l) . 55 . W hi ch of th e fo llowi ng com pou n ds Of io n s do n ot h ave a n oc te t o f electrons surrou n d ing the cen tra l a tom; RF 1- , SiF h s-r. , BrF 1 - , Xe F'I?
63 . Wh a t are th e orders of th e N-O b on d s in NO ';! - and NO';! +? T he n itro geu- o xyg en bond leng th in one of th ese ions is 110 pm a n d 124 pm ill th e o ther. \ Vhich b ond le ng th correspon ds to whi ch ion ? Exp la in h ric flv.
64. Wh ich h as th e greater O - N- O hon d a n gle, NO ';! - or NO';! +? Exp lai n briefly. 65 . Co mpare th e F- C1- F an gles in CI F';! + a n d el F';! " . Using Lewis stru ctu res . determine th e approx ima te bon d a n gle in each io n . Dec id e whic h ion h a... the grea ter bo nd a n g le , an d ex p la in yo u r reasoni ng . 66 . Draw a n e lec tron d ot str uc ture for th e cyani d e ion , CN - . In al1ue olls sol u tion , thi s ion in te ra cts with H + to fo r m the acid . Shou ld t h e aci d for mu la be wri tten as H CN or CN H?
56. • In whic h o f th e foll owin g d o cs the central ato m o bey the octet ru le: NO :!, SF t , NlI 3 , SO :h CIO \!. and C I0 2 - ? Arc a llY of these species odd-electron molecu les o r ions?
67 . D raw the elec tron do t stru ctu re for th e su lfi te io n , SO j';! -. In aqueous so lu tio n . the io n inte ract s with H +. Pre dic t whet h e- r a I J+ ion will att a ch to th e S a tom or th e 0 a tom of SO / -.
57 . D raw resonance struc tu re s fo r the form a te ion, I IC0 2 a n d th e n d etermi ne th e C-O b ond order in the io n .
68 . Din itrogen mo noxide , N ';!O, ca n decom pose gen a n d oxygen gas:
58 . • Co ns ider a se rie s o f m olecule s in wh ich carbo n is b o n d e d by single bonds to a toms of secon d -perio d dem e nts: C-O. C -F, C-N, C-C, a nd C- H. Place th ese b onds in orde r o f in cre asin g b o n d le ng th . 59 . To esti ma te the en thal p y change fo r the reaction O, (g )
+
2 H ,(g ) --> 2 H ,O (g)
wh at h ond di sso cia tio n cn thal pics d o you need? Outlin e th e calcu la tio n , bei ng carefu l to show co rrec t algebraic sig ns . 60. What is the p rin ciple of clcc u-oncu tral iry? Use this ru le to ex clude a p o ssible reson an ce str u ctu re o f CO 2 ,
61. Draw Lewis structures (a n d resonance structures where ap propriate ) for th e following mo lec u les a n d io n s. Wh at simi larities a n d differen ces are there in thi s series? (h ) N ,,(c) oew (a) CO,
2 N, O (g) --> 2 N,(g)
10
ni tro-
+ O, (g )
Use bo n d di sso ciat ion cn thal pi cs to es timate th e enth al p y ch a nge tor th is reaction . 69 . .... . T he e quation fo r the co m b ustio n of gaseous m et ha n o l is 2 e ll,OH (g )
+ :,
O,(g) --> 2 e O,( g)
+
4 11,0 (g)
(a) Using th e bond d issocia tion c n th al p ies ill Table 8.9. es timate th e cu th al pv ch ange for this reaction . \ Vh at is th e entha lpy of com b ustion of o ne m o le of g aseo us me thanol? (h) Com pare yo ur an swer in part (a) with a ca lcu lation of !:J.,.H' usin g th ermo ch e mi cal da ta a nd the m e th ods o f C hap ter 5 (sec Eq uation 5.6) .
70. .... Acryloni tri le , C 3 H 3N . is th e bu ild in g blo ck of th e syn the tic fibe r Orion.
62 . Draw res o n a nce struc tu res for th e SO ';! molec u le . and in di ca te th e p a r tial ch a rges o n th e S and () a tom s. Are th e $-0 h o nds pola r, an d is th e molecu le as a whole p ol a r? If so . wha t is th e di rec tion of th e n et d ipole in 5 0 2 ? Is yo u r p re di ct io n con firme d b y th e elec tros ta tic p o tent ial surface ? Explain b riefl y.
rI
H
H-
H
,
I r>. C= C- C= \J
N:
3
Electrostatic potential surface for acrylonit riLe.
(a) (h ) (c) (d )
Electrostati c potential surface for sulfur dioxide.
•
more challenging
• i n OWL
Blue-numbered questions answered in Appendix 0
G ive th e ap proxi ma te values o f a ng le s 1, 2 , a nd 3 . W h ich is th e sh o rt e r carbon-c arbon bo nd? Whi ch is th e stro n ger carbon-carbon bon d? Based on th e elec trostatic p ot ential su rfac e , where are th e po sitive an d nega tive c ha rges loc a te d in th e molecule? (e) Whi ch is tile most p ol a r b on d ? (t) Is th e molecule po lar?
I
399
71. ... . The cyana te io n , NC O - . has the lea st clcctro negaIiV(' atom . C. ill th e cen ter. Th e very u nstable ful mi nat e ion. CNO - . h as the sam e form ula , bill th e N ato m is in th e ce n ter. (a ) Dr aw th e th ree po ssibl e res o na nce str uc tu res o f CN O - . (b) O n th e basis of fo rmal ch a rg es , de cide on th e resonance structure with th e mo st re aso na ble di st ributio n of charge . (c) Me rcu ry fulm in a te is so un stable it is used in b lastin g ca ps . Can yOIl offe r a n exp lan at ion fo r thi s in stabi lity? ( Hi nt: Are th e form al charges in an y resonance struc ture re a sonabl e in view of th e rel at ive c lec tro ncgativitics of th e ato m s?)
75. Hyd ro xyp ro lin e is a less-co m mo n a mi n o ac id.
72. Vani llin is th e flavorin g age n t in van illa ex trac t an d in
76. Am id es are a n im po rta n t class o f organ ic m o lecules.
(a ) • Giv e a p p roximate values for the indicated bo nd
angles. (b ) Whi ch a re th e most pol ar bo nds in th e mol ecule ? Th ey arc usually drawn as ske tc hed here . but anot he r re so nance stru ctu re is possibl e .
vani lla icc cr eam . Its struc ture is shov..'n h ere :
H :0:
I II ..
H- ( - ( -N -H
I
H
(a) Give values fo r th e three bo nd angles indicate d . (h ) Indicat e th e sh o rtes t carbon-oxygen bo nd in th e mo lecul e . (c) Indicate th e mo st polar bo nd in th e mol ecule. 73 . ... Given th at th e spa tia l n -quirenu- n t of a lon e p air is grea ter th an that of a bond pair, e x plai n why (a ) XeF ',;! h as a lin ear m o le cula r struc ture an d no t a bent on e . (h ) CIF:; ha s a 'Dshapcd struc ture an d not a trigo nalplan ar o ne .
I
H
(a ) D raw th at structu re , an d then sugges t wh y it is usua lly no t p ictured . (1)) Sugg est a reason for th e fact th at th e (-I-N-H angit' is close to 1200. 77 . Use the bo nd d isso ciatio n c u th al pics in Ta ble H.9 to calcula te th e e n thalpy ch ange fo r th e d e co mpositio n of u rea (Stud y Question :l O) to hydrazine , H ',;! ;\.'- l'\I-I',;! . an d ca rb on m on o xid e . (Assu m e all co m po u nd s arc in th e gas ph asc.) 78. Th e molecule sho wn here . 2-fu rylme tha ne thi o l. is respo ns ible for th e aroma o f co ffee : H :y--H
74. T he fo rmula fo r nit ryl c h lo rid e is C lNO ',;! . (a) D raw th e Lewi s stru ct ure fo r th e m ol e cul e . includin g a ll res o na nce st ructu res. (h) Wh at is the N- O bond order? (c) De scr ibe th e e le ct ro n-pa ir and m o lec ular geo metries. an d g ive values for all bon d angles . (d) Wh at is th e m o st pol a r bo nd in th e mole cu le ? Is th e m ol ecule po lar? (e) ... The co m pu te r prog ram u sed to ca lcula te electrosta tic potential surfaces gave th e following cha rg es o n ato ms in th e mol ecul e : A = - 0.03. R = - 0.26, and C = + 05 G. Id e nt ify the atoms A. R, and C. Are tlu-se calculated charges in acc o rd wit h YOllr p redi ction s?
1
H
S-
(-
~I
H-
.. I
\(
/
II
~
( -(
(
(
J" /
H/ 2
..
" H
(a ) \Vhat are the fo rm al charges on th e San d o a to ms ? (1)) . Give a p p rox imate value s of an gl es 1.2. a nd 3. (c) Whi ch a re th e shorte r ca rbo n-carbo n bonds in the molecule ? (d) \ Vhich bon d in this m o lec ule is th e m o st po lar? (e) Is (he molec ule as a who le pola r o r no npolar? (f) T he molecul ar mo de l makes it clear that the fou r C ato ms o f the ring ar e a ll in a pla ne . Is th e 0 atom in th at same pl ane (m aking th e five-member rin g pla na r) , o r is th e 0 at o m bent above or below th e pla ne?
Electrostatic potential surface for ClN02• 400
I
•
more chaLLenging
• in OWL Blue-numbered questionsanswered in Appendix 0
79. ... . Dihvdroxva ccton c is a com p o ne n t of qui ck-tann in g lo tio ns . (It reac ts with the am ino aci ds in th e upper laye r o f ski n an d co lors th em b ro wn ill a re action sim ilar to tha t occurrin g when food is browned as it cooks.] (a) Su p pos ing y O ll can m ak e th is co mpound by treating ace ton e with o xyg en . u se bo nd di ssocia tio n c nt halp ie s to es timate th e en thalpy cha nge fo r th e fo llowing reacti on (which is assumed to occur ill the gas p h ase ) , Is th e react ion exotherm ic o r en d o therm ic?
I
I
I
H- ( - ( - ( - H
H
83. l ,2-Di chlo ro et hylen e ca n h e syn thesized hy ad d ing CI2 to th e ca rh o n -carho n tri pl e bo n d o f ace tyle ne .
H :0: H
H :0: H
I II
(h) Whe re d o the posit ive an d n eg at ive cha rges lie in th e m o le cu le ? (c) One molecule foun d in th e 1995 H ale-Bopp co met is I1 C :~ N . Su ggest a struc tu re fo r this mole cule .
+
0,
~
.. I II I .. H-O - ( - (- ( -O- H .. I I " H
H
H
dthydroxyacetone
acetone
(h) Is ace tone p o lar? (e) Po sit ive 1I ato ms ca ll sometimes b e re m ove d (as I-F ) fro m m o le cules wi th stro ng ba ses (wh ich is in p art wh a t h a ppe n s in th e tanning react ion ) . Wh ich II a to ms ar e the m ost po sitive in di lrydroxvaccto nr-P 80. Nitr ic ac id, H N O :I> h as t hre e reso n an ce struc tu res . G ill' of t h e m , h o weve r, co n trib u te s m u ch le ss to the re so na n ce h ybrid than th e oth e r 1\ \' 0 . Ske tch th e three re so n a n ce str uc tures, a n d assign a form al c harge to ea ch a to m . Whi ch one of you r struc tu res is th e leas t im p orta n t?
Us ing bond dissocia tio n c ru h al pi cs , e sti mat e th e e n th alp y change fo r thi s re act io n in the gas p h ase . 84 . T h e molecu le p ict u red below is e p in e p h rin e , a compound used a" a bro nc hodila tor and an tigla uco ma age n t.
H~ H \ \1 1 ( _( H H H H;" ~ !I ~ I I I I 5) H-O - ( ~( -( - ( -N - ( - H
..
\ ( 31 (=
.:1 H- .q
I I
:0:
(
\
I"...'Y I
H
4
H
H H
8 1. ... . Acro lei n is used to ma ke p lasti cs. Su ppos e th is co m po un d can he p re pa re d hy in se rtin g a ca rb o n mou o x ide m ol e cu le into th e C- H b ond of et hylen e .
H
I ..
H ( =0
H H
I
I
I
I
( ~(
H H eth ylene
I
+ :( = 0:
~
I ..
( ~(
I
I
H H acrolein
(a) \ Vh ich is th e stro nger carbon -ca rbo n hon d in ac ro-
(a) • G ive a va lue for each of th e indi cated bond a ngles. (h) Wh a t arc th e m ost pola r bon d s in th e mol ec ule?
ki n? (b) W h ich is th e lo n ger carbon -car bo n bo n d in acrole in ? (c) Is e thylene or acrolei n po lar?
(d ) Is th e reac tion of CO with C 2 11 4 d o the r m ic o r ex o therm ic?
10
gh'l' acrolein e n-
82. Mol ecule s in space:
In the Laboratory 85 . ' th e su bscri p t I s in d icatin g that I s ato m ic o rbitals we re used to create th e mol ecul a r o rb ital. The other mol ecular orbital is co nstruc ted by Jubtrarting o ne ato m ic orbital wave fu nc tio n fro m th e oth er (see Figu re 9 . 16 ) . Wh e n th is h appe ns, th e p robab ility of fin di n g a n el ect ron be tween th e nuclei ill th e mol ecul ar orbital is reduced, an d th e p robability of finding the electron in o ther regi o ns is high e r. \Vith out sig n ifica n t electro n d ensity bet ween th em , the n uclei re pe l o ne a n o th er. Th is type of o rb ital is called a n anti bonding molecular orbital. Becau se it is also a (T o rb ita l, it is lab el ed (T * I\' T he aste r isk sign ifies th at it is a n tibo n d ing. A ntibonrling orbitals hove no counterpart in valence bond thnn )'. A se cond principle of molecular orbital theory is th at the bond iug molecular orbital is lower in ener[!J tha n the !)(lrent orbitals. an d the ant ibondi ng orbital is higher in pnerg} (Figure 9 . lfi) . Th is mean s th at th e ene rgy of a g ro up of atoms is lowe r th an th e e ne rgy of th e se parated a to ms whe n e lectro ns a re assig ne d to bo ndin g mol ecula r orbitals. C he m ists say th e syste m is "stabil ized " by che m ical bo nd fo rmatio n .
• Orbitals and ELectron Waves Orbitals are characterized as electron waves; therefore, a way to view molecular orbital formation is to assume that two electron waves, one from each atom, interfere with each other. The interference can be constructive, giving a bonding MO, or destructive, giving an antibonding MO.
9 .3
I
Mo lecular Orbita l Theory 423
Nodat plane
•
I
+
,---Q---" . ,,
----+
Is
Is
,, o-e-molecutar orbital
> ~
(antibonding)
~ ~
is ;:; ~ ~
;:;
+ Is
-0-< •
Is
----+
Is
"
" ,/
Atomic orbital
a- Is
Molecular orbitals
'
... "
\: ---Q---,
,
''
', ',
• '
..
'' ''
"
,
Is
Atomic orbital
\----{ill-./
c -molecutar orbital
(bonding)
""" (a)
(b) FIGURE 9.16 Molecular orbita ls. (a) Bonding and antibonding a molecular orbitals are formed from two Is atomic orbitals on adjacent atoms. Notice the presence of a node in the antibonding orbital. (The node is a plane on which there is zero probability of finding an electron.) (b) A molecular orbital diagram for H2 • The two electrons are placed in the (f t, orbital. the molecular orbital lower in energy.
Sign in at www.t homsonedu.comj login and go to Chapter 9 Contents to see Screen 9. 11 for an animated version of this figure.
Atomic orbital
Molecular orbitals
Atomic orbital
,.----{j]----., . ,
,
! ,
a
Is
\,
,
': ~
1-0-
-i f
E.D---'-000 2p
2p
2p
7T 2p
,
I
'
\{]D/ 1T2p '1T 2p
:----Q---" . ,
, ,
0 --
~
::; ~
tr nonbondtnq MQ
'" HOMO
7T
bonding MO
F IGURE 9.23. The 7T molecular diagram for ozone. Notice that, as in the other MOdiagramsillustrated (especially Figure 9.21), t he energy of th e molecular
orbitals increases as the number of nodes increases.
Now let 's a p p ly ~fO th eo ry to the pi bonds. A principle of .MO theo ry is t hat th e numbe r of molecular orb itals must equal th e number o f a to m ic o rbita ls. T hus, th e th ree 2/) a to m ic o rb itals must be co m bined in a way th at fo rm s three mol ecul ar o rbitals. One 7Tp ]\10 for o zo ne is a bonding o rbi ta l becau se th e three p orbitals are "in ph ase " across the molecul e . Ano the r 7Tp (\10 is an an tibo nd in g o rb ital becau se th e ato m ic o rbital o n the ce n tra l ato m th at is "o ut o f ph ase " with the term ina l ato m fJ orbitals. T he th ird 7Tp ]\10 is a non bonding o rbital becau se th e middle p orbital does not parti cip at e in the 1\10 . The bonding 7Tp 1\10 is filled by a pair o f e lec tro ns that is de loc alized , or "sp read over," th e molecul e , j ust as th e reso nance hybrid implies. T he non bonding orbital is also o ccu pied , but th e elec tro ns in this orbital are co nce n tra ted near th e two termin al ox yge ns. As th e name implies, e lec tro ns in th is molecul ar o rbital nei th e r help nor hinde r th e bonding in th e mol ecule . Th e 7T bond o rder o f 0 3 is 0.5 (Fig u re 9.23) . Becau se th e if bond o rd er is 1.0 and th e 7T bond o rder is 0.5 , the ne t o xyge n- o xygen bon d o rde r is 1.5-th e same value given by vale nce bon d theo ry. The observati on th at two o f the 7T molecul ar orbitals for ozone extend over th ree ato ms illust rat es an importan t poi n t regardi ng molecular o rbital th eo ry: Orbitals ran extend Ul')'OTU! two atoms. In vale nc e bond theory, in co n tras t, all represe n ta tio ns fo r bonding are based on being a ble to lo cal ize pairs o f electro ns in bonds be tween two atoms. To fu rth e r illu strat e the I\JO a p p roac h, look again a t ben zen e (Figure 9.24 ) . On page 42 1, we n oted th at th e 7T elect ro ns in th is molecul e were sp read o u t ove r a ll six ca rbon ato ms . We ca n no w see how th e sa me case can be made with (\.1 0 th eory. Six /Jo rbitals co n tribute to the 1T syste m . Based on th e premise that the num ber o f m ol ecular orbitals must eq ua l the numbe r o f ato mi c o rbita ls, there must be six 7T molecular orbitals in benzen e. An e ne rgy level d iagram for be nzen e show s that the six 1T elec tro ns re sid e in the th ree lo westene rg y (bo nd ing) molec u lar o r bita ls.
FIGURE 9 .24 Molecular orbital energy level diagram for benzene.
o
Because there are six unhybridized p orbitals, six 7T molecular orbitals can
be formed-three bonding and t hree
antibonding. Thethree bonding molecular orbitals accommodate the six 11 electrons.
'TT
antibonding MOs
00 tr
432
Chapter
9
I
bonding MOs
Bonding and Mo lecular Structure: Orbital Hybridization and Molecu lar Orbitals
Chapter Goals Revisited Now that you have stud fed this chapter; you should ask whether you have met the (!ra/Her goals. In particulm; )'OU should be able to:
Understand the differences between valence bond t heory and moLecular orbita Lthe ory a. b.
c.
d.
Describ e the main features of valen ce bond theo ry and molecul ar o rbi tal th eory, the two co mm o n ly used th eories for cova len t bonding (Section 9. 1) . Re co gnize that th e p remise for va len ce bond th e ory is that bondin g re sults fro m th e ove rlap of atom ic orbitals. By virt ue oft he ove r lap of o rbita ls, elecrron s are con cent rat ed (o r lo calized ) bet ween two ato ms (Sectio n 9.2) . Distin gui sh how sig ma (if) an d pi (7T) bonds arise . For if bonding, orbitals ove rlap in a head-to-h ead fash ion , co nce n tra ting electro ns alo ng the bon d axis. Sid eways overl ap of I) ato m ic or bitals re sul ts ill tt bond fo rm ation , with elect ron density above and below the molecul ar plane (Sec tio n 9.2) . Und ersta nd 11{)\\, m olecul es havin g d ou ble bon d s ca n have iso m eric fo rm s. Stud y Ouestion(s) assignable in OWL: 14 .
Chemislry.o.Now"'" Sign in at www. t homsonedu: com/lo gin to: Assess your understanding with Study Ouestions in OWL keyed to each goal in the Goals and Homework menu for this chapter For quick review, download Go Chemistry mini-lecture flashcard modules (or purchasethem at www.ichapters.com) Check your readiness for an exam by ta king t he Pre-Test and exploring the modules recom mended in your Personalized Study plan. @ Access How Do I Solve It? t utorials on how to approach problem solving using concepts in this chapter.
Identify the hybridization of an atom in a molecuLe or ion a.
Usc th e co ncep t o f hybr idizatio n to ratio nalize mo lecular struc tu re (Sec tion 9.2) . Study Ouestion(s) assig nable in OWL: 2, 3, 4, 5, 6, 8, 9, 11, 12, 21, 22, 24, 27. 32, 35. 36, 38, 44, 45, 47-50,51, 5Z-54; GoChemis1ry Mod ule 14.
Hybrid Orbit als
Atomic Orbitals Used
Number of Hybrid Orbitals
Electron-Pair Geometry
sp
s +p
2
linear
sp'
s +p +p
3
Trigonal-planar
sp'
s +p +p +p
4
Tetrahedral
sp
s+p+p +p +d
5
Trigonal bipyramidal
spld2
s+p+p +p +d+d
6
Octahedral
3d
Understand the differences between bonding and anti bonding molecular orbitals and be able to writ e the molecuLar orbital configurations for simpLe diatomic molecules. a.
b.
Understa nd molecu lar o rbital theory (Sectio n 9.3) , in wh ich atomic o rb ita ls are co m bined to fo rm bon ding o rb itals, non bo nd ing o rbitals, o r an tibo nd ing orbitals th at are d elo calized over sever a l ato ms. In t his d escript ion , the electr ons o f th e mol ecule or ion a re assigned to th e orbitals be g in nin g with th e o ne at lo west ene rgy, acco rd ing to th e Pauli excl usio n prin ciple an d l lu ncl's ru le. Use mo lecu lar orbital th eory to ex pla in t he p roperties o f O 2 and other diatomic mo lec u les. Study Ouestion(s) assignable in OWL: 15-20, 40, 42 , 43 , 57 .
KEY EQUATIONS Equation 9.1 (page 424) Calc u lating th e o rd e r o f a bond fro m the molecular o rbit al e lec tro n con figuratio n
Bond order = 1/ 2 (num ber of electrons in bonding MOs - numbe r of eLectrons in antibonding MOs)
Key Equations
433
STUDY QUESTIONS
.... de notes challeng ing questio ns.
7. Draw the Lewis structu re , a nd th e n specify the elect ron-pair and mol e cul ar geome tries fo r eac h of the foll owing m ol ec ules or ion s. Id e n tity th e hyb rid iza tio n of the ce ntral ato m . (a ) SiF fi ' (b ) Se F, (e) ICI, (d) Xc F,
•
8. • Draw th e Lewis structure, and th e n speci t), th e
Online homework for this chapter may be assigned i n OWL.
denotes questions assignable in OWL.
Blue-numbered questions have answers in Appendix a and fully-worked solutions in the Studen t Solutions Manual.
Practidng Skills VaLence Bond Theory (Sri' Ex amples 9. ]-9.5 find ChemistryNo w Screens 9.2-9.7.) I. Draw the Lewis structure for ch loroform ,
e llch. w hat
arc its electron -pa ir and m o le cul ar geomet ries? w ha t o rbi tal s o n C , H , an d Cl ove rla p to fo rm bonds inv o lving these e le m e nts? 2. • Dr aw t he Lewis structu re for NF3 . Wha t a re its electron-pa ir a nd molecu lar geometries? Wh at is th e hyb ridi zat io n of the nitrogen a tom? Wha t orbita ls 011 N and F ove rla p to fo rm bonds betwee n these c-k-ments?
3. • Specif y the electro n-pair and m olecular geometry for eac h un derli ned ato m in the followin g lis t. Des cribe th e h yb rid orbital set used b y th is a to m in eac h m o lecule o r io n . (h ) ~O, (a) liBr"
4. • Spe cify the electro n-pair an d mo le cul ar geometry for eac h unde rli ned ato m in t he fo llm v·ing list. De scr ibe th e h yb rid orbital set used b y th is a to m in eac h mo lecule or ion . (a) ~Se , (h ) !i0, (e) ~ II ,O
5. • Describe the hybrid orbi tal se t use d by ea ch o f th e ind icat e d a tom s in the mo le cules be low: (a ) th e car bo n a toms and the oxygen a to m in dimethyl ether, H :~C O Cf-1 3 (b ) eac h carbon a to m in p ropene
elect ron-pair and mol e cul ar geome tries fo r eac h of th e fo llowin g m ol e cules or ions. Id e n tity th e h ybrid izatio n of the centra l a tom . (c) cen tra l S in SO F4 (a ) XcOF, (d) cen tral Br in Rr3(h ) RrF, 9. • Draw th e Lewis structures of th e ac id HP0 2F2 an d its anio n P0 2F2" . Wh at is th e molecula r geOIlH'lry and h vb rid izat io n for th e ph o sp horu s a to m in eac h species? (I-I is bonded to an 0 a tom in th e acid .)
10. D raw the Lewis structu res of I1S0 3 F a nd SO:\F-. Wh at is th e m o le cular geometry and h yb rid izat io n fo r th e su lfu r a tom in e ac h specie s? (II is bo nde d t o an 0 a tom in the acid .} II. • Wh at is th e h ybrid izatio n of th e carbon ato m in p h osg e ne , C I2CO ? C ivl' a com p le te de scri pt io n of th e (T ami tr ho nd in g in thi s molec u le . 12. • What is t he h yb ridi zati o n of th e su lfu r a to m in su lfuryl flu o rid e , S02F2? 13 . T he a rrange men t of g ro u ps a ttac hed to th e C a tom s involved in a C=C do u ble bond le a d s to cis a nd tran s isomers, For eac h compound below, draw th e o th e r isom e r.
. "C=C
1I 3C
H/
(a )
113 C
I
I
I
II
CII"
-,
C=C
H/
(a )
(c) th e two carbo n a toms an d the ni trogen atom in the ami no acid g lycin e II II : 0 :
"
(h )
14. • For each co m pou nd below, d e cide whe ther cis and trans isom e rs are po ssib le . If isomerism is pos sible , d raw th e o ther isom e r.
II H 3 C - C = CH ,
H /
H
"C=C
(h) II /
/
H
"C1I,C II ,
C Il 3 /
"
H
CI
"C =C
/ (e) II
C Il ,O Il /
"II
II - N - C - C - O - I I
I
Molecular Orbital Theory
H
(See Examples 9.6-9.8 and ChemistrJNo1l'Sa-PPIlS 9.9-9.1 2.)
6. • Oivc th e h ybrid orbi tal se t u sed by e ach of the und erl in ed atoms in the fo llowing m o le cu les. H H H :0 : H
I
II
I
I
(a ) 1I - 1'I - g - 1\ -1I
II
I
II
I
II
I
..
(h ) lI"g - g = C - g =()
434
I
(e ) H - g = C - g = N :
15. • T he hyd ro ge n m o le cular ion, H 2". can be d etect ed spec trosco pically. \ Vritl' the electro n co nfigu ra tion o f th e io n in mo le cul ar o rbi tal term s. What is the bon d order of the io n ? Is th e hyd rog e n- hyd ro g e n bo nd stron g e r or weaker in 112 + th an ill 11';l?
16. • G ive the elec tron configu ration s fo r the io ns Li2 + a nd Li 2 - in mo le cu la r orbi tal term s. Co m pare tile Li-Li ho nd o rd e r in these io ns with the bon d ord er in
Li 2 • 17. • Calc iu m ca rb ide , CaC 2 , con tai ns the acc rylidc ion , C 22 - . Sket c h th e molecu lar orbi ta l en e rgy level d iagram fo r t he io n . How m an v net (T a nd tr bo nd s docs the io n have? Wh at is th e carbon-carbon bo nd ord e r? How has th e bo nd o rd e r changed o n addi ng el ec tro ns 10 C 2 to o b tai n Ci -? Is th e C 22 - ion param agnetic? 18. • O xyg en , O 2 , ca n acq uire one or two ele ctro ns to g ive O 2 - (xu peroxi de io n ) o r 0 2 2 - (pe ro xid e io n ). Write the elec tron con fig u ra Lion fo r the ion s in m o lec ula r orbital term s, a nd th e n co m pare th e m with th e O 2 m ol ecu le on th e fo llowin g" ba se s. (a) magn e tic character (b ) ne t n u m be r of a and tr bon d s (c) b ond o rde r (d ) o xygen-oxygt'Jl bo nd le n gt h 19. • Assu me th e energy level di agram fo r ho m o uuclea r di ato m ic mole cules (Fig ure 9.2 1) ca n be a p plie d to lictc ron uc lca r dia tomics suc h as CO. (a) \ Vrite th e electron configura tion for ca rbo n mo no xid e , CO , (b) Wh aL is the high est-en e rgy, o ccupied mole cu lar o rbi tal ? (c) Is the molecul e diamagn eti c or paramagn et ic? (d ) WhaL is the n et n u mbe r of (T an d tr bonds? Wh at is the CO bond ord e r?
20 . • T he n itrosyl ion , NO +, has a n interes tin g c he m istry. (a) Is ~ O + di amagne tic o r param ag neti c? If pa ra m agnetic, how m an y u npa ire d e le ct ro ns d oe s it have ? (h) Assume th e molec ular orbital d iagram for a h om on ucl e ar di atomic mo le cu le (Fig u re 9.2 1) ap pl ies to ~ O + . What is th e h igh e sL-('lle rgy mole cul ar o r bita l o ccu pi ed b y el ectron s? (c) Wha Lis the ni troge n- o xygen bon d o rd er? (d) Is the N-O bo nd in :'\0 + stronge r or weaker th an the bond in NO?
23 . • Des cribe the 0 - 5--0 ang le and the hybrid orbital set used by sul fu r in each or th e fo llowing mo le cules or ion s: (a ) SO , (h ) S0 3 (c) S0 3' (rl) SO ,' Do all h ave the sa m e value fo r the 0-5-0 an g le ? Do e s the S a to m in all the se spec ies use th e sam e h ybrid o rbi ta ls? 24 . • Sketch th e Lewis str uc tures of e lF2 + a nd e lF 2 "". Wh at are the electro n-pa ir a nd molecu lar g eometrie s o f e ach ion ? Do bot h h ave.' th e same F-CI-F an gle? \VhaL hvbrid orbita l se t is u sed hv Cl in ea c h io n? 25 . Sket ch th e reso na nce struct ures fo r tilt' n itri te io n , N0 2 -. De scr ibe th e elec tron-pair and mole cu la r ge o met rie s o f th e io n . From the se gvo m etrics . decid e o n th e O-N-O bo nd angle , th e average NO bond orde r, an d the N a to m h yb rid iza tio n . 26. Sketc h th e reson a nce stru ctu res fo r th e ni tr a te ion , N0 3 -. Is the h yb rid iza tio n or the N at o m the sa m e or d ifferent in each st ructure? De scribe th e orbi tals invo lved in bo n d fo rma tio n by th e ce n tra l I\: ato m . 27 . • Sketch th e re sonanc e struc tu res fo r the N 20 mol ecu ll'. Is th e h yb ridi za tio n o f the N a to ms th e sam e or diffe ren t in each structure? Des cri be th e orbit als invo lvcd in bo n d fo rma tio n by th e ce n tra l N a tom . 28. Co m pare the stru cture a nd bondin g in CO 2 an d C0 32 with rega rd Lo the O-C-O bo nd angles, th e CO bon d ord e r, and th e C a to m h yb ridi za tio n . 29. Nume rous molecu les are d en-cu-d ill d ee p spaCt' . T hree o f the m are illu stra ted h e re .
H
I
:0· ,/
( -H
. ' (- H
I
H
H H
Tliesr questions are no! d esignuted as 10 l)'j JP or location. i n the chapter: T hPJ Inri)' combine several concepts [nnn. this and other
H- ( - (=O
2 1. • Draw the Lewis struc tu re fo r AIF4 - . Wh at are its e le ct ro n-pai r a nd m o le cula r ge o m e tries? Wh at orbital s o n Al a nd F ove rla p to fo r m bond s betwee n th e se e lem e n ts? What arc th e form al cha rg es 011 the a to ms ? Is this a rea son ab le c ha rg e d istribu tion ? 22• • Draw the Lewis struc tu re fo r C I F:~. Wh at are its dectron-pair a nd m o le cula r geo me tries? Wh at is th e h yhridi za tio u o f th e chl o rine atom? Wha Lorbi tal s o n Cl a nd F overlap 10 form bon d s be twe e n these e lem ents? morechallenging
. .
H
H H 1
1
Vinyl alcohol ..
H- ( =(-O :
I
dWjJ/fTJ.
•
Acetaldehyde
I I .. 1
General Questions on Valence Bond and Molecular Orbital Theory
Et hylene oxide
1
• i n OWL
Blue-numbered questions answered in Appendix 0
H (a) Co m m en t on the sim ila rit ies or di ffe ren ce s in th e formu las of these com po u n ds. Are Iller isome rs? (b ) In di ca te th e h yb r -idizat io n or eac h C ato m in e ach molec u le . (c) Indi cat e th e value of the H- C-H a ngle in eac h of the th re e m olecu le s. (d) Are any o f the se moh-cu k-s pol a r? (e ) Whi ch molecule sho uld have the stronges t ca rbon-. ca rbo n bond ? T he stro ng es t car bo n- o xyge n bond?
I
435
30. Acrolein , a com po ne n t o f p hotoch em ical smog, has a pun gent od o r an d irritat es eyt·s and mucous mem bra nt's. H
34 . • Lactic acid is a nat ural com pou nd found in so ur mil k. A,...-- IJ
I
H : 0:
I
I , II
I
I '-J ..
II :0"
~
I'.J/J
(a) W hat a re the hvhridi zatio ns of carbon ato ms I a nd 2? (h) What are th e app ro xim a te values o f angles 1\, II,
(a ) I lo w Illa n y 1T bon ds o ccu r ill lactic ac id ? I ro w m a llY a bo nds? (h) De scribe th e h yb ridi zau o u of a to ms 1, 2, a nd S. (c) W hi c h CO bo nd is th e shortest in the m o lec ule? Wh ich CO b o n d is the sLronges t? (d) What are th e approxim at e values of th e bo nd a ng les A, n, an d C?
31 . T h e organic compoun d below is a m em be r o f a cla ss know n as ox imcs. : O -H
I
I
" - C- C= N :
I
35. • Cinnamaldchvdc o ccu rs naturally ill cinna mon oi l.
I
II
C
II
an d C? (c) Is a s-trans iso merism po ssib le h ere ?
II
·· 3
H - C - C - C -O -H
Atl 1:\""\ " H - C = C - C -H I
H :0 :
I
IJ
(a ) Wh at are th e h yb ridi zat io ns of the two C atom s and o f th e N ato m? (h ) What is tilt' app ro xi m a te C- N-O an gle ? 32 . • T he co mpo u nd ske tc he d below is ace tylsa licylic acid , co m mo nl y k n ow n as aspiri n : : 0:
II
..
I
.. ..
n ( C - O - II: 0 : II II ,
r
"",C ,
~
I
/C ~
'--;(
I3
,I,-.Jc
,
/ C,
H \.J C
H
I
A
II
/ l ) T C - C - II
fI
(a) What arc th e ap p roxima te va lu es of the an gle s ma rke d A, H, C, and lJ? (b ) Wha t hybri d orb itals are u sed by ca rbon a toms 1, 2. a nd ~p
(a ) \Vha t is th e m o st po la r bo nd in th e mole cu le ? (b) How ma ny sig m a (rr ) bo nds a nd how many p i ( 1T) bon d s a re th e re? (c) Is cis-trans isom eri sm po ssible ? If so , d raw th e iso-
33 . Ph osph oscriu c is a le ss-comm o n a m in o ac id. A
.>
:() 0
i)- H
""l< ':':";' I
:l
I
H - N - C - II
L'
[j
mer s of th e mo lecule.
II
C
I CH2
I
: 0:
I " ..
: O - p - O:
..
I
(d) G ive t he h ybr idi zat io n of t he C a to ms in t he m o le cu le . (e ) Wh at are th e valu es of th e bo nd a ng les 1, 2 , a nd 3 ?
..
:o :J n I
II
36 . • Io di ne a nd o xygen fo r m a com ple x series o f ion s, am o ng th em IO t- a nd 10 5 3 - . Draw the Lewis stru ct ures for th ese ion s, a nd spe cify t heir el ectro n-pa ir ge ometrie s an d th e sha pes of th e ion s. Wh a t is the h yb ridi zatio n of th e 1 a to m in th ese io ns? 3 7. Antimo ny p cn ta flu o r id c reacts with H F acc ord ing Lo lh e equa tion
(a ) De scr ibe the h yb ridizat io ns of a to ms 1 thro ugh r.. (b ) What are th e a p p rcxirna tc valu es o f the bond angles A , n, C, an d IJ? (c) \Vhat are the mosLpolar bonds in th e mole cu le ?
436
I
2 HF
+ ShF , --> [ H,F ] +[ShF,;]-
(a ) wh at is th e h yb ri d ization o f the 51> a Lom in the re-
actan t and p rod uct? (b) Draw a Lewis stru ctu re for H 2 F+. Wh aL is th e g eo metry of I-I2 F+? W ha t is the h ybrid iza tio n of F in 1I2 F+? •
more chaLLenging
• in OWL
Blue-numbered questions answered in Appendix 0
38. • Xe no n form s well-chara cterize d co m pou nds (.... page 4( 4) . Two xe no n- ox yge n co m pound s are x-o , a n d Xe 0 4' Draw th e Lewis struc tu res of th ese co m poun ds . an d give th eir electro n-pair an d m o le cul ar geo m e tr ies . Wh at are the h yb rid orbita l se ts u sed by xe no n in these two o xide s? 39. T he sim ple vale nce bon d pi cture of O 2 doe s n ot ag ree with th e m o le cul ar orbital view. Co m pa re th ese two th eo rie s with regar d to th e pe ro xid e iOII ,ol - . (a ) Draw a n eh-c tro u d ot struc tur e for O l - . Wh a t is th e bo nd o rd e-r of th e ion ? (h ) Write the m olecul ar orbi tal ele c tro n co nfiguration fo r 0 22 - . Wh at is th e bon d o rd e r based on th is a p proach ? (c) Do th e two th eo rie s of bo nd in g lea d 10 th e sa me m agn e tic characte r an d ho nd o rde r fo r 0 l - ?
44. • Am p he tam ine is a stim u la n t. Replacing on e II a to m o n th e :'\ 1 o r a m in o, gro u p with C1lj gives meth am ph etam ine . a parti cul arl y da nge rous drug com m o nly kn own as "speed .,.
'2.
Amphetamine
40 . • Nitroge n , N 2• ca n ion ize to fo rm N2 + o r ad d a n electro n to g ive N 2 - . Using molec ula r o rb ital th eory, co m pa re th e se species with rega rd to (a) th ei r ma gn e tic chara cte r, (h) ne t num be r of" bo nd s, (c) bond o rd e r, (d ) bon d le ngth . a nd (e) bon d str e ng th . 4 1, Whi ch of th e h om o n uclear, d iatom ic m o le cules
of th e seco n d-pe rio d c lem e nts (fro m Li2 to N(2) a rc pa ra mag netic ? Whi ch h ave a bo nd order o f I ? Whi ch have a bo nd o rde r of 2? Whi ch di at om ic mo le cule has the h igh est bon d orde r?
42 • • Whi ch of th e followin g m o le cule s o r mole cule io n s sho u ld be pa ram ag ne tic? Wha t is the hi gh est occ u p ied mol e cular orbital (H O MO) in eac h o ne ? Assu me the mole cul ar orb ita l d iag ra m in Figure 9 .2 1 applies to all of t he m . (a) N O (e) (),' (e) eN (b)
()F -
(d) N e, +
43. • T h e eN m o le cul e ha s bee n fo u nd in inte rst ell a r space . Assum ing t he e lec tro ni c str u cture of th e mo lecu le can be d e scribed usi ng the mo lecular o rbi tal e ne rgy leve l d iagra m in Fig u re 9 .21 , a nswe r th e foll owin g q uestio n s. (a) Wha t is th e high est e ne rgy o ccupi e d m olecu lar orbita l (H O MO) to 'wh ich an e lectro n (o r electron s) is (a re ) assign ed? (b ) Wh at is th e bo n d order of th e m olecule ? (c) [l ow ma n y net (T bon d s arc th e re ? Ho w ma n y ne t tr bon ds? (d ) Is th e molecu le param ag netic o r diama gnetic?
•
more challenging
• i n OWL Blue-numbered questions answered in Appendix 0
(a ) Wh at a rc th e hybrid o rbita ls u sed by t he C a to ms o f th e eli ri ng , by th e C ato ms of the side ch ai n , a nd by th e N ato m? (b) Give ap p rox ima te values fo r th e bo n d an gle s A, B, a nd C. (c) II ow m an y a bo nd s a nd" bonds are in th e m o lecu le? (d ) Is the molecul e pol a r o r nonpo lar? (e ) Am phe ta mi ne reac ts read ily \eith a pro to n (JI +) ill aq ueo us so lu tio n. \\The re doe s thi s pro to n attac h to the m o lecule ? Doe s th e elec trosta tic pot e n tia l m a p sh own above con fi rm th is po ssibi lity?
I
437
45 . • Me n th o l is use d in so aps . pe r fu m es, a nd foods. It is prese n t in the com m o n herb m in t. an d it can he p repared fr o m tur pen tin e . (a ) What an' the hyh r'idiza tio ns used by th e C a to m s in th e mo lecu le ? (h ) What is the ap proxim a te C- O- li bond a ng le? (c ) Is th e m o le cul e pol ar or n onpol a r? (d ) Is th e six-membe r ca rb o n ring plan a r o r non pla nar? Expl ain why or why no t.
in th e U.S. T he m o lecule h as a three-m em be r ring of two C atom s a nd an 0 a to m .
H
I :0 / I
(-
H
. " ( -H
I
H
Ethylene oxide
(a) Wh at are th e ex pec ted bo nd ang les in the ring ? (h) \Vhat is th e h ybri d iza tion of each ato m in th e ring? (c) Com me n t o n th e relat io n bet we en th e bon d an gles expected based o n hybri d izati o n and th e bond angles ex pe cted for a th ree-nu-mbe r ring. (d) Is th e mol e cu le pol a r? Base d on th e electrostati c pot e nti a l ma p sho wn below, where d o the nega tive a nd positive cha rg es lie in the mol e cu le ? 46. T h e el em en ts o f the seco nd peri o d from bo ro n to o xygcn fo rm co m po u nds of the typ e X"E-EX ", where X ca n be H o r a halo ge n . Sket ch possibl e m o lecu lar stru ctu res fo r B2F" C 2H h "N 2H " a nd 0 2H2' G ive th e hybrid izati o ns o f E in e ac h m ol ecu le an d spec ify a p prox imate X-E-E bo nd a ng les. Electrostatic potential map for ethylene oxide.
In the Laboratory 47 . • Su p pose yo u ca rry o ut th e fo llo wing reac tio n of amm o n ia an d bo ron tr ifluo rid e in the labo rat o ry,
II
F
I I H- N: + B-1" I I H
F
-
II
F
II
F
I I H- N-B - F I I
(a) What is the geometr y of th e boron ato m in BFs? In II :l ~---+ R F:~ ?
(h) Wh at is the h ybridizat io n of th e boron ato m in the two compo u nds? (c) Does th e boro n a to m 's h yb rid izat ion change o n form a tio n o f the co ord ina te covalent bond? (d) Co ns ide ring a to m elect ronega tivincs an d th e bo n d ing in N H s an d BF:h why d o yOll ex pe ct tlu- n itro ge n on NI I3 to don a te an ele ctron pair to th e B ato m of BF3? (e) RF3 also rea cts read ily with wa ter. Rased o n the ammo n ia reactio n above , spec ulate o n how wa ter ca n in terac t with fiF 3 .
49 . • Th e sulfama te io n , H 2NSO j - , ca n be th ought of as h aving been fo rm ed from the amide inn . NI 12 - , a nd sulfur trio xide , S03' (a) Wh at ar e the ge o m et rics of the amide io n and of SO j? Wh at are the hyb ridi za tio ns of th e N a nd S a to ms, re spect ivel y? ( b) Ske tch a stru ctu re fo r th e su lfam a tc ion , a nd cstim a tt' the bond a ng les. (c) Wh at ch a nges in hybrid iza tio n d o p HI ex pec t fo r N a nd S in th e co urse of th e reac tion
1'\ 11, -
(d) Is 50 3 th e d o no r of an ele ctro n pair o r th e acc eptor o f a n el ectron pair in th e re actio n with a m id e io n ? Does the e le ctrostat ic pot e n tial Illap sho wn bel ow co n fi rm r o u r pr ed iction ?
48 . .. . Eth yle ne ox ide is an interme d ia te in th e m an ufactu re of e thyle ne g lyco l (an tifreeze) ami polye ster polyme rs. Mo re than 4 mi llion ton s are p ro duced a n n u ally
438
I
+ S(}, --> II ,N- S0 3 - ?
Electrostatic potential map for suLfurtri oxide.
•
more chaLLenging
• in OWL
Blue-numbered questions answered in Appendix 0
50. A • Th e co m p o u n d whose str uc ture is sh ow n h e re is ac ctvlace to n c . It ex ists i ll two form s: th e en ol fo r m an d th e keto fo r m . H
H
I
H i~ - C = C - C - C H 3
I
: 0:
(
I
l
H , C - C - C - C - CH ,
II
II
:0 :
I
II
: 0: II : 0:
I
II
enol fo rm
keto fo rm
Th e mol ecule re acts wit h O H - to fo rm a n Nic(aq)
+
2 NH, " (aq)
In thi s form , nicotin e is much more read ily absorbe d by the skin an d mucous membran es, so th e co m po u nd is a m uc h mo re potent poi son .
10.4
Nicotine. Two nitrogen atoms in the nicotine molecule can be protonated. which is the form in which nicotine is normally found. The protons can be removed, however, by treating it with a base. This "freebase" form is much more poisonous and addictive. See J . F. Pankow: Environmental Science & Te chnology, Vol 31. p. 2428,
Compounds wit h a Carbonyl Group
Fo rma ld e hyde, ace tic acid, and acetone are amo ng the o rga nic co m po u nds referred to in previou s exa m ples. Th ese co m po u nds have a co mmon st ruc tu ra l feature : Each contains a trigo na l-plan ar ca rbo n ato m doub ly bonded to an o xyge n . Th e C= O gro u p is called th e ca rbonyl gro up, and all o f th ese compounds are membe rs o f a large class of co m po u n ds called carbonyl compounds.
August 1997.
Carbonyl group
Acetone
Formaldehyde
Acetic acid
CH 20
CH3C0 2H
Aldehyde
Carboxylic acid
In thi s sectio n , we will examine five g ro u ps o f ca rbo nyl co m po u nd s (Table 10.6 , pa ge 462) : •
Primary alcohol: ethanol ( H,
H-
•
I (-OH I
•
H
• Second ary alcohol: a-p ropa nol
( H,
I
H- (-OH
I
(H ,
Tertiary alcohol: 2-methyl-2-propanol
( H,
I I
•
Aldeh)'fles (RCII O) have an organic gro u p (- R) and an II ato m attac hed to a carbonyl grou p . Ketones (RCO R') have two - R g ro u ps attac hed to the ca rbo nyl ca rbo n ; th ey may be th e sam e gro u ps , as in ace to ne, or diffe rent groups. Car boxylic acids (RC0 2 H ) have an - R grou p and a n - O H g ro u p a ttac he d to the carbo nyl ca rbo n. Esters (RC0 2 R' ) have - R an d - O R' gro ups attac he d to th e ca rbo nyl carbo n. A mides (RCO N R,' , RCO NHR', and RCO NH , ) have an - R gro u p and an am ino g ro u p (- N I-I2, - N H R, - N R2 ) bonded to the carbo nyl carbo n .
Ald eh ydes, ketones, and carboxylic acids are oxidation products of alco ho ls and, indeed, are co m mo n ly made by thi s route. The product ob ta ine d through o xid ation o f a n alcoho l d epends on th e a lco ho l's struc tu re , whi ch is cla ssified acc o rd ing to the number of carbo n at oms bonded to the C atom be arin g th e - O H g rou p . Pri mary alcohols have one carbo n an d two h yd rogen ato m s attac hed, whereas second ar)' alcohols have two ca rbon ato ms a nd o ne hydrogen ato m attac hed . Tertiavy a tcobols have th ree carbo n at oms a ttached to th e C ato m bearin g th e - 0 1-1 grou p. A jn7mm )' al cohol is o xidi zed in two ste ps. It is first o xidized to an ald e hyde an d th en in a seco nd step to a carbo xylic acid :
H ( - (- OH l
( H,
R-CH , -OH primary alcohol
468
Chapter 10
oxidizing agent
I
Carbon: More Tha n J ust Anot her Ele ment
aldehyde
o II
R-C -OH carboxylic acid
For exam ple , th e air o xid ation o f et hano l in wine produces wine (with excess o xygcn ) vinegar, th e most important in gredient o f wh ich is ace tic ac id. H
H
I I
H- ( - (-OH (€)
I
H
oxidizing agent
I
H
ethanol
acetic acid
Acids have a sou r tas te. The wo rd "vi negar" (from the Fren ch vin aig re) me an s so u r wine. A device to test one 's breath for alcohol reli es on a sim ilar oxidation o f e thano l (Fign res 3.2 1 and 10.10 ) . In co n trast to p rim ary alco ho ls, o xid ation of a snondur» alcohol produces a
FIGURE 10.10 ALcohoL tester. This device for testing a person's breath for the presence of ethanol relies on the oxidation of the alcohol. If present, ethanol is oxidized by potassium dichromate, K2(r207, to acetaldehyde. and then to acetic acid. The yellow-o range dichromate ion is reduced to green Cr3+ (aq), the color change indicating t hat ethanol was present.
k eton e :
OH
I
R- ( -R '
oxidizing agent
o II
R- ( -R '
I
H secondary alcohol
ketone
(- Rand - R' are organic groups. They may be t he same or different.)
Co mmo n o xid izing age nts used fo r th ese reactio ns are reagents suc h as KT\.fn04 and K2 Cr,O, (Table 3.4). Finally, tertia ry alco ho ls d o not react with the usu al oxidizin g agc llts.
{( H,),CO H
oxidizing agent
no reaction
Aldehydes and Ketones Ald ehydes and ketones have plea sant o do rs and are oft en used in frag ran ces. Be nzald e hyde is responsible for the odor of a lm o nd s and cherries ; cin namald e hyde is found in th e bark of th e cin namo n tree; and th e keton e 4· (p-hydrox yph enyl) 2-bu la no ne is resp onsible for th e odo r of rip e raspbe rries (a favo rite o f the au thors of this book ) . Tabl e 10.8 lists several sim ple alde hydes and ket on es.
Benzaldehyde, C6 HsC HO
trans-Cinna maldehyde, C6H sCH=CHC HO
Alde hydes and ket on es are the o xidation produ cts o f primary a nd secondary alco ho ls, res pec tively. Th e reverse reactio ns-reduction of alde hydes to primary alcohols and reduction o f ketones to secondary alc ohols-arc also known . 10 .4
I
Compounds with a Carbonyl Group
469
TABLE 10.8
Simple Aldehydes and Ketones
Structure
BP CC)
(ommon Name
Systematic Name
formaldehyde
methanal
acetaldehyde
ethanal
20
aceto ne
prcp anone
56
methyl et hyl keto ne
butanone
80
diethyl ketone
3-pentanone
0
II
HCH
- 19
0
II
CH ,cH
j
0
II
c
CH ,c CH,
~
~
0 Aldehydes and odors. The odo rs of
almonds and cinnamon are due to alde-
II
CH ,cCH, CH,
0
hydes, but the odo r of fresh raspberries
comesfrom a ketone.
II
CH ,CH,CCH,CH,
102
Co mmo n ly used reage n ts for suc h reduct ions are NaB II _1 and l.i.All L, althoug h II 2 is u se d
0 11
a n ind us tria l sca le.
0
II
R- ( -
OH H
I
NaBH 4 or li AlH 4
R- ( -
H
I H primary alcohol
aldehyde
0
II
R- (-R
OH NaB H4 or LiAlH4
l
R-
I
( -R
I H
ketone
secondary alcohol
EXERCISE 10. 7 Aldehydes and Ketones (a) Draw the structural form ula for 2-pentanone. Draw str uctures for a ketone and two aldehydes th at are isomers of z-penta none. and na me each of t hes e compounds.
(b) What is the product of the reduction of 2-pentanone wit h NaBH 4?
EXERCISE 10.8 Aldehydes and Ketones Draw the structures. and name the aldehyde or ketone form ed upon oxidation of t he following alechols: (a) t -butanol. (b) z-butanol. (c) z-methyl-t-prcpancl. Are these three alcohols structural isomers? Are t he oxidation products structural isomers?
470
Chapter
10
I
Carbon: More Than Just An other Ele ment
Carboxylic Adds Acetic acid is the m ost co m m o n a n d most important carboxylic acid. Fo r m an y years, ace tic acid was made by oxidizing e than o l p rodu ced by fer me nta tio n. 1\0\\', however, ac eti c ac id is ge ne ra lly mad e by co m b in ing ca rbo n mono xid e a n d m ethan ol in th e p rese nce of a cata lyst:
CH,OH(f)
+ CO (g)
catalyst
methanol
CH ,CO,H(f ) acetic acid
Abo u t I billi o n kilograms of ace tic acid are p roduced an n ua lly in th e U nite d Sta tes for use in plas tics, syn the tic fibe rs, an d fun gicides. Man y o rganic ac ids a rc found n aturall y (Ta ble 10.9 ) . Ac ids arc re co gnizabl e by th e ir sour taste (Figure 10.11 ) a nd are found in com mon food s: Citric acid in frui ts, ac etic ac id ill "ill egal', a n d ta rt ari c acid ill gra pes are just three exam p les . So me ca rboxylic acid s h ave co mmo n names d erived from th e source of th e ac id (Tab le 10.9 ) . Becau se fo rmi c acid is found in ants, its n am e co mes fr o m th e Lati n wo rd for a n t (form ica) . Butyric acid gives ra n cid b u tte r its unpl easant odor, a n d th e n ame is rel at ed to th e La tin wo rd fo r bu tter (lmtyrum) . The syste matic nam es o f aci ds (Tab le 10.10 ) are formed by dro pping th e "-e " o n th e n am e o f th e co rrespo nding alka ne a nd ad d ing "coic" (a n d th e wo rd "acid "). Be cau se of the substa n tia l elect ro n egati vity o f oxyge n , th e two () atoms o f the carboxylic aci d group a re slig h tly n eg ativel y c harged, a nd th e II ato m o f th e - 0 11 grou p is positivel y c ha rge d . This c harge d istrib uti on h as several importa nt implicati on s:
FIGURE 10.11 Aceticacid in bread. Acetic acid is produced in bread when leavened with the yeast Saccharomyces exigus. Another group of bacteria, Lactobacillus sanjrandsco, contributes to the flavor of sourdough bread. These bacteria metabolize the sugar maltose, excreting acetic acid and lactic acid, CH ]CH (O H)COzH , thereby giving the bread its unique sour taste.
• T h e pol ar aceti c aci d molecul e di ssol ves readil y in water, wh ich you a lread y kn o w because vin ega r is an aflu eous so lu tion o f acetic aci d. (Acids wit h la rge r organ ic gro u ps are less so lu b le , boweven )
TABLE 10.9
Some Naturally Occurring Carboxylic Acids
Name
Structure
Natural Source
benzoic acid
Q - CO' H
berries
OH
cit ric acid
I
HO,C- CH' - r - CH, - CO, H
citrus fruits
CO, H
lacticacid
H C-CH-CO H ' I a OH
sour milk
malic acid
HO 2C-CH2 - C IH- CO2 H
apples
OH
oleic acid
CH,(CH, ), - CH=CH- (CH,), - CO, H
vegetable oils
oxalic acid
HO,C-CO,H
rhubarb, spinach, cabbage, tomatoes
st earic acid
CH ,(CH,)" - CO,H
animal fats
tartaric acid
HO C-CH-CH -CO H a I I a OH OH
grape juice, wi ne Formic add, HC0 2H . This acid puts the sting in ant bites. 10 .4
I
Compo unds wit h a Car bonyl Gro up
471
H 0
TABLE 10.10
I II I
H- C- C-O-H
Some Simple Carboxylic Adds
Structure
-~
BP CC)
Common Name
Systematic Name
formi c acid
met hanoic acid
101
acetic acid
ethanoic acid
118
propionic acid
propanoic acid
141
butyric acid
butanoic acid
163
valerie acid
pentanoic add
187
0
H
II
HCOH - {;
.)
'-----v-------Ca rboxylic ad d group
Acetic acid. The H atom of the carboxylic ad d group (- C02H) is t he acidic proton
of t his and other carboxylic acids.
0
II
CH ,COH 0
II
CH ,CH,COH 0
II
CH,(CH,),cOH 0
II
CH ,(CH,),COH
• T h e hydrogen of th e - O H grou p is t he acid ic hyd ro gen. As n oted in Cha p te r 3, acetic acid is a wea k acid in water, as are mo st ot her organic acids.
Carbo xylic acids u nd ergo a n um ber of reactions. Among th ese is the red uc tio n of the aci d (with reagents such as LiAIH 4 or Na BH 4) first to an alde hyde and th en to a n alco ho l. For e xampl e , ace tic ac id is reduced first to ace tald eh yd e and the n to ethano l. LiAlH4
LiAlH 4
CH,CO, H ------> CH,CHO ------> CH,CH, OH acetic acid
acetaldehyde
ethanol
Ye t a nothe r im portant as pec t o f ca rb oxy lic acid c he mi stry is th ese aci ds ' reacti on wit h bases to give carbo xyla te a n io ns. Fo r exam p le, acetic acid re acts with so di u m hydroxid e to give sod iu m ace tate (so d iu m e t ha noate} .
CH,CO,H(aq) + OW(aq) ---> CH,CO,-(aq) + H,O(f )
Esters Carbox ylic acids (RCO, H ) reac t with alco ho ls (R'O H) to form es ters (RC0 2 R') in a n esterification reac tion . (T h ese reacti on s arc ge nera lly ru n in th e presence of strong aci ds because acids accelerate th e reac tion .)
o II
RC '------V----' Ca rboxylate group: ~ portion from Portion from ethanol acetic acid
472
o O- H
carboxylic acid
II
CH,COH
CH 3COZCHzCH 3
acetic acid
I
Carbon: More Than Just Another Element
H
II
RC - O- R' + H,O
alcohol
o
Ethyl acetate, an ester
C h a p t e r 10
+ R' -O
ester
o + CH,CH,OH ethanol
II
CH,COCH,CH, ethyl acetate
+ H,O
A Closer Look Having described alcohols and carbonyl compounds, we now pause to look at glucose. the most common, naturally occurring carbohydrate. As their name implies, formulas of carbohydrates can be written as though they are a combination of carbon and water, CA H20)yThus, the formula of glucose. C5H120 6, is equivalent to ( 6(H20)6' This compound is a sugar, or, more accurately. a monosaccharide . Carbohydrates are polyhydroxy aldehydes or ketones. Glucose is an interesting molecule that exists in three different isomeric forms. Two of the isomers contain six-member rings; the third isomer features a chain structure. In solution, the three forms rapidlyinterconvert. Notice that glucose is a chiral molecule. In the chain structure, four of the carbon atoms are bonded to four different groups.
e:tl:':).
Glucose and Sugars CHO , HO
_
HOH!
OH,3 H
-
H
OH4
H
H
OH;
OH
H
HO ~ O OH H
H
Home te st for gLucose.
CH,OH
In nature, glucose occurs in just one of its enantiomeric forms; thus, a solution of glucose rotates polarized Light. Knowing glucose's structure allows one to predict some of its properties. With five polar - OH groups in the molecule, glucose is, not surprisingly, soluble in water. The aldehyde group is susceptible to chemical oxidation to form a carboxy lic acid. Detection of glucose (in urine or blood) takes advantage of t his fact; diagnostic tests for glucose involve oxidation with subsequent detection of the products. Glucose is in a class of sugar molecules called hexoses, molecules having six carbon atoms. 2-Deoxyribose, the sugar in the backbone of the DNA molecule, is a pentose. a molecule with five carbon atoms.
H OH
OH H
H
Open-chain form
e-o-ctoccse
H
HO
I3+Gtucose
Glucose and other monosaccharides serve as the building blocks for larger carbohydrates. Sucrose, a disaccharide, is formed from a motecule of glucose and a molecule of fructose, another monosaccharide. Starch is a polymer composed of many monosaccharide units. H
OH
HO H H
0Q
c -o-Glucose
CH'OOH
HO'" Fructose
.,"
H
OH
H
CH,O H
The str uct ure of su crose. Sucrose is for med from « -c-qtuc ose a nd fructo se. An ethe r linkage is formed by loss of H20 from two - OH groups.
H
deoxyribose. a penrose. part of the ONAbackbone
Wh e n a carboxy lic aci d and an alco ho l reac t to form an este r, the O R gro u p o f the alco ho l ends lip as part o f the es ter (as shown above). T his fact is known because o f isotope labeling experimen ts. If th c reaction is ru n using an alco ho l in whic h the alco ho l oxyge n is I RQ, all of the I RO en ds u p in the es ter molecule. Tab le 10.11 lists a few co m mon es ters an d the acid and alco ho l fro m which th ey arc form ed. T he two-part nam e of an este r is given by ( I) th e na rnc o f the hydrocarbon g ro u p fro m the alco ho l and (2) th e nam e of the ca rboxy late grou p d e rived from the acid name by replacin g "-ic " with "cate." For exa m ple. e thano l (common ly ca lled e thyl alco ho l) and aceti c acid co mbine to give the es te r e thyl ace ta te. An important react io n of es ters is th e ir hydrolys is (literally, reac tio n with water) , a react ion tha t is th e reve rse of th e form at io n o f the es ter. T he react io n , ge nerally
10 .4
I
Compounds wit h a Carbonyl Group
473
TABLE 10.11
Some Acids, Alcohols, and Their Esters
Ester
Alcohol
Add
CH,
Odor of Est er
CH,
0
I
I
II
CH,CO,H
CH,(HCH,(H,OH
CH,COCH,( H,( HCH,
acetic acid
3-methyL-l-buta not
3-methyLbutyl acetate
0
Esters. Many f ruit s such as bananasand
strawberries as weltas consumer products (here, perfu me and oil of wintergreen) contain esters.
banana
II
CH,( H,CH,CO,H
CH,(H,CH,(H,OH
CH,(H,(H,( OCH,(H,(H,(H,
butanoic acid
t-butanol
butyl but anoa t e
pineapple
0
CH,(H,CH,CO,H butanoic acid
• Saponifi cati on Fats and oils are es-
ters of glyceroland long-chain acids.
(O>- CH,OH
CH,(H,(H,i~CH, --,
(
R-C -@) Carboxylic acid
+
\
R'
I
o
Tl
R'
II I
@-N- R' ~ R-C - N-R ' Ami ne
Amide linkage
+ H,O
Amide
This portion from acetic acid
Amides have an o rgani c gro u p an d an am ino gro u p (- NII z, - NII R' , o r - N R' R) attac he d to the carbo nyl gro u p. T he C atom invo lved in the am id e bo nd has three bo nded g ro u ps and no lone pairs aroun d it. vVe wo uld pred ict it sho uld be sj/!' hyhridi zed with trigon al-plan a r geome try and bo nd ang les of ap p roxi matel y 12oo- an d this is what is fo und . H o weve r, the struc tu re o f th e amide grou p o ffe rs a su rp rise. T he N ato m is also observed to have trigonal-planar geo me try with bonds to th ree attached at o ms a t 1200. Becau se the amide n itrogen is su rro u nd ed by four pa irs of electro ns, we wou ld h ave predicted the N a to m would have stJ'~ hybridizatio n and bond a ng les of abo ut 109 °.
10 .4
[J This portion from methylamine
An amide, N-methylacetamide. The N-methylportion of the name derives from the amine portion of the molecule, where the N indicates that the methyl group is attached to the nitrogen atom. The " RCO,R" + R' OH
Recall that fats and oils are esters (page 476), derivatives of glycerol and high- malar-mass organicacids (fatty acids). Their reaction with methanol (in the presence of a catalyst to speed up the reaction) produces a mixture of the methylesters of long chain fatty acids and glycero l.
a II
H C- O- C- R
'I
HC -
~ O-C -
R'
I
+
3 CH OH
)
H C- O- C- R" 2
H C- O- H
II
~
'I
HC- O- H
I H C- O- H e
Hl - O-
°C-II
R
o +
+
R'OH
Irans-esterificatio n:
Biodiesel, a mixture of long-chain esters of fatty adds.
RCO, R'
~ R - ~ -O-H
II
H)C- O- C- R'
o II
Hl - O- C- R"
a Glycerol, a by-product of the reaction. is a va luable commo dity for the health care products indust ry, so it is se parated and sold. The mixture of esters that remains can be used directly as a fuel in existing diesel engines, or it can be blended with pet roleum products. In t he latter
R -~ -10 -R'
0
+ Ht-O - R"
~ R-~ -O -R" + R'OH
In both reactions, the OR' group on the ester combines with hydrogen of second reagent (water or alcohol) as shown. In drawing t his analogy, it is useful to recognize t hat the re are ot her similarities in the chemistry of alcohols and water. Fo r example, bot h ca n be protonated wit h strong acids (giving H30+ and ROH2 +) and deprotonate d by strong bases (giving OH- a nd OR- ).
Questions: 1. Write a balanced chemical equation for the reaction that occurs when methylmyristate, C13HZ7CO zCH3( f ) , is burned, forming CO,(g) and H,O(g).
2. Using enthalpy of formation data , calculate the standard enthalpy change permole in the oxidation of methyl myristate ( :ifW ~ - 771.0 kJ/mal). 3. Which compound, methyl myristate (ClsH300Z) or hexadecane (C1fjH34, one of manyhydrocarbons in petroleum based diesel fuel) is predicted to provide the greater energy per mole? Per liter? ( :ifW f ar C"H" ~ - 456.1 kJ/mal) (d(methyl myristate) ~ 0.86 g/mL, and d(C" H,,)
~
0.77 g/mL]
Answers to these questions are i n Appendix Q.
10 .5
I
Polymers 47 9
(a)
(b) (el FIGURE 10.12 Common polymer-ba sed consumer products. (a) Packaging materials from high-density polyethylene; (b) from polystyrene; and (c) from polyvinyl chloride. Recycling information is provided on most plastics (often molded into the bottom of bottles). High-density polyethylene is designated with a "2" inside a triangular symbol and the letters "HOPE." PVC is designated with a "3" inside a triangular symbol with tile tetter "V" below.
A more chemicall y oriented a p p roac h to polym er cl ass ifica tio n is ba sed on the method of synth esis. Ad dition polymers a re mad e by d irectly ad d ing monom e r u n its togeth er. Condensation polymers a re made by com b in ing mon omer units an d sp litt ing o u t a small mol e cule , often wa te r.
Addition Polymers Po lyeth ylen e , p olystyren e , a n d p ol yvin yl c h lo ride (PVC) are co m mo n ad d itio n polymel's (Figu re 10.12) . They arc built by "ad d in g together" sim p le alkc n cs suc h as e thyle ne (C I-I ,=CI-I 2 ) , styre n e (C6 I-1,CI-I =CI-I 2) , and viny l c h lo rid e (C I-I 2=C H Cl) . Th ese and other ad d itio n polyme rs (Ta b le 10. 12 ) , all derived from alke n es , h ave wid ely va rying propertie s a n d u se s.
ChemislryJ l"Now'"
•
Sign in at www.tho msoned u.com/ Ioqtn and go to Chapter 10 Contents to see Screen 10.9 for an animation of addition polymeri zation.
Polyethylene and Other PoLyolefins Po lyeth ylen e is by fa r the leader in terms o f ad diti 0 11 polymer p roduction. Ethyle ne (C 2H{), the m onomer from wh ich polyeth ylene is made, is a p ro d u c t o f petrole um refin ing a nd on e of th e lop five ch e mica ls produce d in th e United Sta tes. Wh en e thy le ne is h ea ted to between 100 and 250 °C at a pressure o r 1000 to 3000 atm in th e presencc o f a ca ta lyst, po lymers with mol ar masses u p to severa l million a rc fo rmed. The re action ca n be exp ressed as a ba lan ced c h e m ica l eq ua tio n:
n H,C = CH,
~ -(~-~)H
ethylene
480
Chapter 10
I
Carbon : MoreTha n Jus t Another Element
Hn
polyethylene
Ethylene Derivatives That Unde rgo Addition Polymerization
TABLE 10.12
Monomer Common Name
Formula
H H " C= C/
ethylene
"H
H/
U.S. Polymer Polymer Name Uses
polyethylene (polythene)
squeeze bottles, bags.
H H" C= C/
propylene
"CH,
H/
Production
(Trade Names)
polypropylene (Vectra, Herculon)
(Metric tons /year) " 7 million
films, toys and molded objects, electric insulation bottles, films, indoor-
1.2 million
outdo or carpet s
H H " C= C/ H/ " Cl
vinyl chloride
polyvinyl chloride (PVC)
floor tile, raincoats. pipe
H H " C= C/ H/ " CN
acrylonitrile
polyacrylonitrile (Orlan. Acrilan)
TUg S,
styrene
polystyrene (Styrofoam, Styron)
food and drink coolers, building material insulation
0.9 million
vinyl acetate
polyvinyl acetate (PVA)
latex paint, adhesives, textile coatings
200,000
met hyl methacrylate
polymet hyl met hacrylate (Plexiglass, Lucite)
high-quality t ransparent objects, latex paints,
200,000
contact lenses
polytetrafluoroethylene (Teflo n)
gaskets, insulation, bearings, pan coatings
H
1.6
fabrics
million
0.5 million
H
H> = «
O
)
H H " C= C/ H/ O-C -CH II a
"
a H" H/
/ CH, C=C
"-C- O-CH II a
a F
F
F/
" F
" C~ C/
* O m' mente to n =
tet rafluoroethytene
6,000
1000 kg.
Th e ab breviated fo rm ula o f the reaction p ro du ct , f- CII 2 CII z-t-mshows th at pol yethylene is a cha in o f car bo n a to ms, each beari ng two hydro gen s. T he chain len gth for polyeth ylene can be very lo ng. A po lyme r with a molar mass of I milli o n wo u ld co n tain a lmos t 36,000 e thylene mol ecul es link ed toge the r. Pol yethyl en e formcd und er vario us pressurcs and ca talytic co nd itio ns has di fferent p ro perties, as a resu lt of d iffere n t mol ecul ar stru ctu res . Fo r exam ple , whe n chro m iu m o xid e is used as a cata lyst, th e p ro d uct is almo st exclusively a linear chai n (Figu re IO.13a) . If e th yle ne is heat ed to 230 "C at high prcssu rc , ho wever, irregu lar br an ch ing oc curs. Still o ther co n d itio ns lead tel cro ss-lin ke d polyethyle ne , in which d ifferent chains are link ed toge ther (Figu res 10.13b and c) . T he h igh-m olar-mass chains o f lin ear pol yeth ylen e pack closely together and res ult in a mat e rial with a de nsity of 0.97 g/c m :l. Th is mat erial, referred to as
10 .5
I
Polymers 48 1
(a)
(c)
(b)
FIGURE 10.13 Polyethylene. (a) The linear form, hiqh-density polyethylene (HOPE) . (b) Branched chains
occur in low-density polyethylene (LOPE). (c) Cross-linked polyethylene (CLPE).
high -d en sity polyet hylene (IIDP E) , is hard and to ug h, whic h m ak es it suitable for items suc h as m ilk bottle s. If th e p ol yethylene c ha in co n ta ins bran che s, h o weve r, th e c hai ns can not pack as closel y togethe r, a nd a lower-dens ity m ate rial (0.92 g / cm') known as low-d en sity po lyeth yle ne (I..D PE) re su lts. This materia l is so fter an d more flexibl e th an HDP E. It is used in plastic wra p and sandwich bags, among o the r thin gs. Lin king up th e pol ymer chai ns in cross- lin ke d pol yeth ylen e (CLPE) causes the ma teria l to be eve n more rig id an d inflex ible. Plastic bottle caps are ofte n mad e of CLPE . Po lymers formcd fro m su bsti tu te d cthyle ncs (CH 2= CH X) have a ra ngc o f p roperties and uses (see Ta ble 10.12 ) . Sometimes, th e prope rti es are pred ictabl e based on th e mol ecule's structu rc . Pol ymers witho u t pol ar su bstitue n t grou ps , such as polystyre ne, ofte n di ssolve in organi c so lven ts, a property useful fo r so me types o f fabri cat io n (Fig ure 10.14) . Polym ers based on substituted ethytenes. H2C= CHX
t CH,CH \
polyvi nylalcohol
-----t
6CCH,
II a
}n
polyvi nylacetate
polystyrene
Pol yvi n yl alco ho l is a po lymer with littl e affin ity for no n polar so lven ts bu t an affin ity fo r water; which is not su rp rising , based o n the large n u mber of po lar O H g ro ups (Figu re 10. E») . Vinyl alco ho l itself is not a stable co m po u nd (it iso mc rizes to ace ta lde hyde CII 3C II O) , so po lyvin yl alco ho l can no t be mad e fro m thi s co mpoun d . In stcad , it is ma d c by hydrol yzin g th c ester gro u ps in po lyvinyl ace ta te .
LLU+ I
H
I
aCCH,
II
LLU I
H
I
+ n CH,CO, H
aH
a PoLyethylene film. The polymer film is produced by extruding the molten plastic t hroug h a ring-like gap and inflat ing th e
film like a balloon. 482
Chapter 10
So lu bility in water or organ ic so lvents ca n be a liability fo r pol yme rs. Th e many uses of pol ytetrafluoroeth ylen e [Te flo n, f- CF2 CF2 --t fl] ste m fro m th e fact that it d oes not in teract with water or orga nic so lvents, Pol ystyrene , with n = 5700, is a clear, hard, co lo r less so lid that ca n be mo lde d easi ly a t 2:)0 °C. You a rc probably mo re fam iliar with the vcry light, fo am -like marc-
Carbon: More Than Ju st Anot her Element
Ca)
(b)
FIGURE 10.14 Polystyrene. (a) The polymer is a clear, hard, colorless solid. but it may be more familiar as a light , foam -like material called Styrofoam. (b) Styrofoam has no polar groups and thus dissolves well in organic solvents such as acetone. See also Fig ure lO.12b.
ria l known as Styrofoam that is used wide ly fo r foo d and beverage co n ta iners an d for ho me in sul ati on (Figu re 10.14 ) . Styrofoam is p roduced by a p ro cess ca lled "ex pa nsio n molding." Pol ystyren e beads co ntaini ng 4% to 7% of a lo w-bo ilin g liquid like pe n ta ne are placed ill a mo ld and heat ed with steam o r hot air. Heat ca uses th e so lven t to vaporize. creatin g a foam in th e molten pol ym e r th at expa nds to fi ll th e shape of the mo ld .
Natural and Synthetic Rubber Natu ra l ru b ber was fir st in troduced in Euro pe in 1740 , but it remain ed a curiosity u n til 1823 , whe n Cha rles Macintosh in ven ted a way o f us ing it to wate rp roof cotton cloth . T he mac kin tos h, as ra in coa ts are still so me times called, becam e po pu lar d esp ite maj o r p roble ms: Natural ru bber is nota bly weak and is soft a nd tacky whe n war m but britt le at low temperatu res. In 1839 , after 5 yea rs of research o n natu ral rubbe r, th e Ame rica n in vento r C ha rles Goodyear (1800- 1860) di scovered th at heat ing gum rubber with su lfu r produces a m aterial tha t is e lastic, wate r-repe llent , resilie nt, and n o lo nger sticky. Rubber is a naturally occu rring polymer, the mo no me rs of which are mol ecul es of 2-methyl-I,3-bu tad iene , co mmonly called isojnrnf. In natu ral rubber, isop re ne m ono mel's are linked toge ther th ro ug h carbon atoms I and 4-that is, thro ugh the en d car bo n ato ms of th e C.. cha in (Figure 10.16). T his leaves a dou ble bo nd betwee n carbon atoms 2 and 3. In natu ral ru bber, these double bond s have a as co nfigu rat io n. In vulca nized rubber, the m ateri al th at Goodyear di scovered , the po lymer cha ins o f nat u ral rubber are cross-lin ked by short chains of su lfu r ato ms. Cross-lin king hel ps to align th e pol yme r chains, so the mat eri al does not u nd ergo a perman e nt chang e when str e tc hed and it spri ngs back when the stress is removed . Su bstanc es th at beh ave thi s way are ca lled elastomers. With a kn o wledge of th e co m pos itio n and struc tu re o f natural ru b ber, chemis ts began searc hing fo r ways to make syn the tic rubber. Wh en th ey first tr ie d to make th e pol ymer by lin kin g isopren e mon omers togeth er, however, wha t th ey ma d e was sticky and usel ess. T he pro blem was th at syn thesis procedu re s gave a m ixture o f cis and t ra ns pol yisoprcnc. In 1955, however, chem ists at the Goodyear and Fire ston e co mpan ies di scovered special catalysts to prepare rh c all-tis pol ymer. T h is syn the tic ma terial , whic h was structu ra lly iden tical to natu ral ru b ber, is now ma nufactu red cheaply. In fact, more th an 8.0 X lORkg o f syn t he tic pol yisop rene is produced annu ally in th e U ni ted Sta res.
FIGURE 10.15 Slime. When boric add , B(OH)], is added to an aqueous suspension of polyvinyl alcohol, (CH2CHO H)n, the mixture becomes very viscous because boric add reacts with the - OH groups on the polymer chain, causing cross-linking to occur. (The modelshows an idealized structure of a portion of the polymer.)
CH,
I
H, C""C '
I
H
H
I c""C' I
H
H
Isoprene, 2-methyl-1,3-butadiene. 10.5
I
Polymers
483
O th er kinds o f pol yme rs have fu rth er ex pan de d th e repertoire of elasto meric mat eri als no w avai lable. Po lybutadicuc . for exam p le, is cu rre n tly used in th e product io n of tires, ho ses, and helts. So me elastomers, ca lled copolymers, are form ed by pol ymeri za tion of two (or more) diffe rent mon o mers. A co polymer of styre ne and butadi en e , made with a 1: 3 ra tio of th ese raw mat erial s, is th e mo st important synthetic rubbe r no w m ad e ; mo re th an abo ut I billi on kg o f styre ne -b utad ie ne ru b ber (SBR) is produced eac h yea r in th e Un ited Stat es fo r makin g tir es.
HC -- CH II ~ H, C CH,
3n
+
styrene
1 , 3 ~ butad iene
1
FIGURE 10.16 Natural rubber. The
sap that comes from the rubber t ree is a natu ral polymer of isoprene. All the linkages in t he carbon chain are cis. When natu ra! rubber is heated strongly in t he absence of air, it smells of isoprene. This observation provided a d ue that rubber is composed of t his building block.
1
HC = CH
/
H,C
\
HC = CH
/
H,C-CH ,
\
HC=~C/ /
H,C-CH,
\
/
H,C-CH ,
CH
\'
/
CH ),
HC=CH
'
n
styrene-butadiene rubber (SBR)
An d a littl e is left over each ye ar to make bub bl e gum . Th e stretc hi ness o f bubble g u m on ce ca me fr om natural r ubb er, but SBR is n ow used to help yo u blow bubbles.
Chemislryj ;..Now'
•
Sign in at www.t homsonedu.com/ Ioqtn and go to Chapter 10 Contentsto see Screen 10.11 for a self-study module on t he polymer used in bubble gum.
Condensation Polymers A ch emical re action in which two molecu les react by splitt ing o u t, or elim inating, a small m olecul e is ca lled a condensation reaction. The reacti o n of an alco ho l with a carbox ylic ac id to give a ll ester is an exam ple of a co nde nsatio n reactio n. On e way to form a con densa tio n pol ymer uses two different react ant molecules, each co n tai ning two fu nctio nal gro ups. Ano ther route uses a sing le mol ecul e with two different fu n ction al groups. Co m me rc ial pol yesters are mad e usin g bo th types o f reactio ns .
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Sign in at www.t homsonedu.com/ Ioqtn and go to Chapter 10 Contentsto see Screen 10.10 to view an animation of condensatio n polymerization and to watch a video of t he synt hesis of nylon.
Polyesters Copolymer of styrene and butadiene, SBR rubber. The elasticity of bubble gum comes from SBR rubber. 4 84
C h a p t e r 10
Terephthali c aci d co n ta ins two ca rbox ylic ac id gro u ps, and e thyle ne g lyco l co n ta ins two alco hol g ro u ps. Wh en mixe d , th e aci d a nd alco ho l fun ction al gro u ps at both en ds o f th ese molec ules ca n react to form es ter link ages, sp litt ing out wate r. Th e
Carbon: More Than Ju st Anot her Element
res u lt is a p ol ymer ca lle d pol ye th ylen e te re p h thalate (PET ) . T he multi ple es ter lin kages mak e th is su bsta nce a polyester.
o
0
11- 0 - 11
n H OC ~CO H +
n HOCH 2CH 20H
terephthalic ad d
ethylene glycol
+ 2n H20 polyethylene terephthalate (PET),a polyester
Pol yester text ile fib ers mad e from PET are marke te d as Dacron a n d Tcrvlcu c. The in e rt , n on to xic, n o n inflam mato ry, a n d n o n-bl ood-clon in g prop erties o f Dacron p o lymers make Dacron tubin g a n excelle nt su bstitu te fo r hu man blo od vessels in h eart bypass o pe ra tio ns, a nd Dacron shee ts a rc so me times u sed as tempo rary ski n fo r burn victim s. A p olyeste r film , Mylar, has unusu al stre ng th an d can be roll ed in to sh ee ts o ne -th irtie th th e thi ckn ess of a h u man hair. Magn et icall y coate d Mylar film s a re used to make au di o a n d video tapes (Figu re 10. 17). The re is considerab le interest in a nother polyester, polylacric aci d (PLA) . Lactic ac id con ta ins bo th carboxylic aci d an d alco ho l fu nc tio na l gro u ps, so co n de nsatio n between mol ecul es o f th is monomer g ives a polym er.
n
Ho-{-LoH ~ --O-{-~-O-- + n H
CH,
0
\
H
CH,
0
)
H20
n
T h e re is in terest in pol ylacu c ac id for two reasons. First , th e mo no me r used to make this pol ym e r is o b ta ined by b iolo gic al ferme ntat io n of pla nt m at e ri als. (Mos r of the c he m ica ls used in th e manu facture of o the r typ es o f pol ym ers a re d erived fr o m pe tr oleu m, and the re is in cr eased conce rn abou t th e ava ilab ility and cos t o f raw mate rials in th e fu tu re .) Seco nd, this p ol yme r, which is cu rre n tly be ing used in p ackaging mat erial , is biodegrad abl e, wh ich h as th e potential to alleviate land-fill disp osal pro blems.
Polyamides In 19 2H, th e Dul'oru Com pa n y embarke d o n a ba sic researc h p rogram h eaded by Wa llace Ca rothers ( 1896- 1937) . Caro the rs was interested in h igh-mol ar-m ass compo un ds , suc h as rubbers, prot ei ns, a n d re sins. In 1935, his research yiel d ed
FIGURE 10.17 Polyesters. Polyethytene terephthalate is used to make clothing and soda bottles. The two students are wearing jackets made from recycled PET soda bottles. Mylar film, another polyester, is used to make recording tape as weU as balloons. Because t he film has verytiny pores, Mylar can be used for heli um-filled balloons; the atoms of gaseous helium move through the pores in the film very slowly.
10.5
I
Polymers
485
nyI OI I~6 , 6 (Figure 10.18 ) , a pol yamide pr epared fro m ad ipoyl chloride, a d erivat ive of ad ipic aci d , a d iacid , and hcxam c thylcn cdia mi nc . a di am inc :
o n
II
II
ClC(C H,) 4CCl
adipoyl chloride
D
j ..._- - - - - - - -_.... Active Figure 10.18 Nylon6,6. Hexamethyienediami ne is dissolved in water (bottom layer), and adipoyl chloride (a derivative of adipic acid) is dissolved in hexane (top layer). The two compounds react at the interface between the layers to form nylon, which is being wound onto a stirring rod.
ChemislrY..c\Now """ Sign
in at www. thom sonedu:com jlogin a nd go to th e Chapter Contents menu to explore an i nt eractive version of th is figure accompanied by an exercise.
t
0
oII
0II C(C H,) ,C - ~ ( C H , ) , ~
+ 2n H,N(CH ,), NH, hexamethylenediamine
H
t
+ 2n
HCl
H n
amide link in nylon-s .s a polyamide
Nylon ca n be ex trud ed ea sily in to fibe rs that are stro ng er th an na tural fibers and che m ically more inert. T he di scovery of nylo n j o lted the Ame rica n tex tile industry at a cr itica l tim e. Na tura l fibers were no t meeting 20th-ee n tury ne eds. Silk was expe nsive and no t durab le ; woo l was scratc hy; line n crushed eas ily; and co lto n did not have a high -Fash ion image . Perhaps the most iden tifiable usc for the new fiber was in nylon stoc kings . The first public sale of nylo n hosiery took place on O ctobe r 24, 1939, in Wilm ingto n , Delaware (the site o f DuPon t's main office ) . T his use of nylon in co mme rcia l pro ducts curl ed sho rt ly the reafter, however, with the start of World War II. All n ylon was divert ed to makin g parachu tes and o th er military gear. It was not u ntil abou t 19:">2 th a t n ylon reap peared in the co nsu mer mark etp lace . Figu rc 10. I9 illus trates why nylo n makes suc h a go o d fiber, To have goo d ten sile stre ng th (t he abi lity to res ist tearin g ) , the po lyme r chai ns sho uld he able to a ttrac t one ano ther, albeit no t so strongly that th e plastic ca n no t be ini tially ex te nded to for m fibe rs. Ordinary covale n t bo nds betwee n th e chains (cross-lin king) wo u ld be 100 stro ng . Instead , cross-lin king o ccurs by a so mew ha t weaker inte rm o lecul ar for ce ca lled hylmgen bonding ( 110- Section 12.2) between tile hyd rogens of N- II gro u ps o n one chain an d th e ca rbo nvi oxvgc ns o n anothe r chain. T hc po lar itie s o f th e NS- - II S' + gro u p and th e CS+ ' O S::" (h) CH,C H,C H= CHC H, + H, -->
20. Draw the stru ctu re . ami g ive the systematic na me for th e products of tlu- fo llo wing reac tio ns: 1l 3C , / CIl ,CIl , C= C + Hz / (a) 1l 3C II
,
(h) CH 3C = CCII,CIl 3
+
2 Br,
~
2 1. • T he co m pou nd 2-bromoh ut an e is a prod uc t o f addition of HRr to an alke ne . Id e nt i fy th e alke ne an d give its n am e .
seve n-carb on chain a nd on e e th yl su bstitue n t}. Draw tlu- stru ctu res, and name th ese co m po un ds . Is e ith er isom e r chira l?
22. T he com pound 2,3-dibromo-2-me thylhexa ne is fo rme d by ad d itio n of Rrz to an alke n e. Id en tity th e alke n e , a nd write an eq ua t ion fo r thi s reacti o n .
12• • Amon g th e 18 structura l isom e rs with th e fo rm ula C ~J I I H arc two with a five-carbo n ch ain h aving o ne e thy l a nd one m ethyl su bs titu en t g ro u p. Draw the ir stru clu res, a n d na m e these two isome rs.
23. Draw stru ctu res for alke nc s t hat have th e fo rmula C 31I:,Cl, an d nam e e ach com po un d . (T hese a re de rivatives of p ro pe n e in whi c h a chlori ne a tom repl ace s on e hvdrogen ato m.)
13. List several typ ical physical properties of G,H I(J. Predi ct th e followin g p hysica l pro pe rties of dodcc anc , C 1zH z6: co lo r. state (s. t , g ). so lub ility ill \va tc r. so lu bi lity ill a non polar so lve n t.
24 • • Ek-nu-mal a nalysis of a colorless liqu id h as given its fo rmul a as CsH 10. You recogni ze th at th is could be e ith er a cyc loalkan c o r a n alken e . A che mic al te st to detcrminc the cla ss to wh ich thi s co m pound he lo ngs invo lves ad d ing bromin e . Expla in ho w thi s would allo w yOll to d isting u ish be twee n the two cla sses.
•
morechallenging
• in OWL
Blue-numbered questions answered in Appendix 0
I
489
Aromatic Compounds (See f :xa mjJ!f 10.5, Exercise 10.5, and Chon;sl1)'Now S("r('('1/ 10.3.)
25. Draw struc tu ral fo r mulas fo r th e fo llo wing co m p o u nds: (a) 1,3-d ichlo rohe llze n e (a lte r n a tive ly ca lled m-d iclllo roh e nz c lle) (b) l -brom o-l -mcth vlb cn ze u c (a lte rn a tive ly ca lled /r b rom o to lue ne)
26.
( ~ i \'l'
th e syste m a tic n am e fo r each o f th e fo llow ing co m p o u nd s: CI (a)
&"" ¢
(C) Yt
(O(
C, H ,.
NO ,
27. \Vrit e a n e q uatio n fo r the p re p aration o f c th ylb c nzcn c fro m b e n ze n e an d an ap p ro p ria te co mpou nd co n ta in in g a n e t hyl g rou p.
28. • Write an e q u a tio n for th e pre paration o f h cxvlh c nzcnc fr om be n zen e a n d o ther ap p rop riate reagen ts.
29. A single co m p o u n d is form ed hy a lkyla tion of 1,4-rlime thylbe n zen e . Write th e eq ua tion fo r th e re act ion o f th is co m p o u n d with C Il:;C l an d AlCh . \Vh at is th e struc tu re an d na m e o f th e p roduct ?
34 . • Na m e th e foll owin g a rni ncs: (a) C I 13 C112 C112 NIl 2 (b ) (CIl ,),N (c) (C Il , ) (C, Il , ) N Il (d ) CoH " N H, 35 . D raw stru ctu ra l fo r mulas for all th e alcoho ls with th e form ula l.4H WO. Oivc th e syste matic na me o f each . 36 . Draw structu ra l form u las for all p ri mary amin c s with th e formul a C 4 1I ~N II 2' 37 . Co m p le te a n d balan ce th e foll owin g equa tio ns : (a) C,J [,N H ,(C) + H CI (aq ) -> (h ) (C llj),N (aq ) + Il ,SO , (a'l ) -> 38. • Ald e h yd es a n d ca rboxylic a cids are fo rmed hy oxidatio n of prim a ry alco ho ls, an d ketones are fo r med wh e n sec on dary alco ho l.s arc o x id ized. (~ i v e th e nam e an d for m u la fo r th e alco h o l th at , when o xid ized , g ives th e fo llowin g p ro d uc ts: (a ) C Il "CI I, C I I, CI 10 (h ) 2-h exanon e
Compounds with a Carbo nyl Grou p (.\'(,P Exercises 10. i - l 0. 10 and Clu'm iJll}N ow Scnrn 10.5.)
39 . • Draw struc tural fo rmulas fo r (a ) 2-pen ta non e , (b ) h cx an al , an d (c ) p e n tan o ic acid.
30 . N itra tio n o f toluen e gives a m ix tu re o f two p roducts. o n e with the ni tro gro u p ( - I'\ () ~J in th e ortho p o sit io n a n d o ne with the nitro grou p in th e jHlra po sitio n . Draw str uc tu res o f th e two prod ucts.
40 . • Id e ntify th e cla ss o f e ach o f th e following co mp ou nds, an d g ive th e syste m a tic na m e f or eac h : (a) 0
II C I LlCC II "
II
(See Exam/Ill' 10.6 a nd Chem iJll)'Now Screen 10.5. )
CH 3 C H,C H,CH
31. • Give th e syste ma tic n a m e fo r each o f th e foll owin g alc o hols, a n d tell if ea ch is a prim ary, sec o n dary, o r tertia rv alco b o l: (a) C H,C H,C H,O H (b) C II ,C II,C II ,C II , C)If CH, (c) Cl I , (rl )
I .
H , C - C - OH
I C II ,
I I
(c)
I
CI I,Cl I, C I IC I I, CO , f1
. (b )
OH
0
II
CI I, CI I, COCI r,
a n d te ll if e ach is p ri m ary, seco n d a ry, o r te rti ary: (a) l -b ut a n o l (h) 2-b u ta n o l (c ) 3.3-d im c t hyl-2-hu tan o l (d ) 3.3-d ime t hyl-t -h u tano l
I
I
41. • Id entify th e cla ss of each of th e foll owi n g co mp o u nd s. a n d g ive th e syste ma tic n am e for eac h : (a) CH,
H , C - C - CII , CII ,
3 3. \Vrite th e fo rm ula , an d draw th e stru ct u re fo r each o f th e foll owin g ami n cs: (a) cthyl amin c (b) d ipropylamine (c) bu tyld im eth ylamiu e (d ) triethyla mine
o CIl "CC Il,C fI ,C Il ,
32. • Draw structu ral fo r m ul as for th e foll owin g a lco h o ls,
490
o
(b)
Alcohols, Ethers, and Amines
(c)
0
II
C H 3C O C H ,C H,C H,C H ., (d)
0
Br-Q-AoH 42 . Dr aw structu ra l formulas fo r th e foll o win g ac ids an d es te rs: (a ) 2-me t hylhexan o ic aci d (b) penryl butanoat c (wh ic h h as th e o d or o f ap rico ts ) (c ) c etyl aceta te (wh ich h as th e odo r of o ra n g es) •
more chaLLenging
•
in OWL Blue-numbered questions answered in Appendix 0
43 . • Give th e stru ct ural for m ula and syste m atic name for th e product, if an y, from ea ch o f ti lt' fo llowing reaction s: (a ) ponta nal and KMnnl (h) pent an al and LiAIl L. (c) 2-octano ne and LiAlH j (d ) 2-octa no ne and KM no.l 44. Describe h ow to pre pare z-pcu tano l beg inn ing with the app ro priate ketone .
45. Describe h ow to prepare propyl propa noa te begin n ing with I-p ro pa nol as th e only ca rb on- co n tain ing reagent . 46 . • Give th e name a nd st ructure of th e p ro duct of th e reac tion o f benzoic ac id an d 2-p ropanol.
47. Draw str u ctu ral form ul as, and give th e na mes fo r th e p ro ducts o f t he followin g reacti on :
(h) Th e re are four O Il groups in thi s structure . Estimat e th e C-O- I I bon d a n g le s fo r t!Je se grou ps. Will they he th e same va lue (m o re or le ss ) , or should th ere h e significan t diffe ren ces in th ese bond angles? (c) ls the molec ule chiral> H ow m an y chira l carbon a tom s ca n he id en tifie d in th is structure? (d ) Id en tity th e shortest bo nd in th is molecule. (e) Wha t arc t he func tiona l g ro u ps of th e molecule?
Functional Groups (Ser Exa m/llr / 0. i and Chrm;J/r)Now Screen. 10.5.) 5 1. • Id enti fy the fun cti o n al g rou ps in th e fo llowin g mole cules .
(a) CIl 3C Il ,C II ,OIl (h )
°II CH,COCH,CH,CH,C H, + NaO Il
°II
H,CCNH CH , (c)
48. Draw structu ral formu la s, an d g ive th e names fo r th e
CH 3C
p ro d u ct s of th e fo llow ing reacti on :
°
(d )
CH 3
o-~- O-{H
+ NaOH
~
CII ,C Il ,COCII "
?
CH,
52. Con sider th e follo win g molecules: I
49 . • T h e Lewi s stru ctu re of p henyla lani ne , o ne of the 20 ami no aci ds th a t ma ke lip pro te ins , is drawn below (wit h o u t lon e pairs o f electrons) . T h e carbon at om s are nu mbe red fo r the p urpo se o f th is q u e stio n . (a) Wh at is th e g eo m etry of C:J? (b ) Wh at is th e O -C-O bon d angle ? (c) Is thi s molecule ch iral? If so, whic h car b o n ato m is chi ral ? (d) Wh ic h hydro gen a tom in th is com poun d is acidic?
H
H
I
N- H
1,1 ·,
~
r O I I , c= o ~
C -C =-- Il II
I
O- H 50. • T he Lewis stru ctu re of vita min C , whose chemical nam e is ascorbi c aci d . is drawn be low (with o u t 10lH' p a irs of elec trons ) .
II
O il
I
I
HO - C - C -H /O"
I
II
\(' / '
" C/
°
I( " c= c/ HO'
' OH
(a) Wh a t is th e ap proximate va lue for th e O - C- O bond a ngle ?
•
more challenging
• i n OWL
Blue-numbered questions answered in Appendix 0
°II H,CO H °II
.
°II °II
CH,C H,CC H, 2.
CH 3C H,CO H 3. H,C = CH CH ,DH 4.
OH
I
CH 3C Il ,C IlC II , (a) Wh at is th e result of tre atin g com poun d 1 ,...it h NaRII.? Wh at is th e fu n cti on al grou p ill the produc t? Name th e product. (b) Draw th e structu re of th e reaction prod uc t fro m co m p o u n d s 2 a n d 4. Wh at is the fu nction al group in the p ro du ct? (c) Wh at com poun d results fro m adding H 2 10 compou nd 3? Name th e reacti on prod uct. (d ) Wh at compound results fro m ad d in g Na O Il to co m p o u n d 2?
Polymers (See Exmn/llr /0.8, Exercise /0. /0, and Chrm;Jl1JNo'll' Sr:ri'i'Jl.51O.9 and lO./O. )
53 . Pol yvin yl aceta te is the h in d e r ill \v'a tcr-hase d pain ts. (a ) Wri te an equa tio n fo r its form ation fro m vinyl ac e ta te . (b ) Show a p o rt io n of thi s p ol ym er with t h ree m o n omer u n its. (c) Describe h ow to make p ol yvi n yl alcoho l from poly. viny l ace tate .
I
491
54. Neoprene (polvchl oroprcru- , a kind of rubber) is a po lym e r form ed fro m till' chlorin at ed butadie-n e II , C= CII CCI= CII, . (a) Write an eq ua tion sh o wing the form ation of pol ychloro p rcn c fro m th e m onomer. (h) Show a po rtion of thi s polym er with three m onom e r u nits. 55 . Saran is a co po lym e r of L l-dichloroe thcnc and ch loroetln-n e (vinyl ch lo ride ). Draw a possibl e struc ture for th is polyme r. 56. T he stru ctu re of m e thyl m eth acr ylate is g-iven in Table 10.12. Dr aw th e structure of a polymct h yl met hacryla te (P,\1,\1A) po lymer that ha s fo ur m ono mer units. (P M,\lA has excell en t o p tica l pro perties an d is used to m ake ha rd con tac t lc nscs .)
General Questions on Organic Chemistry These questions art' not designated (1.5 to tJjJe or location in the chaptn: T hl)' nUl)' combine seueml concepts.
57 . T h ree d iffe rent co m pou nds with the fo rm ula C~ H~C l~ are kn own . (a) Two of these co m po u nds a rc geo me tric isom e rs. Draw th eir st ruc tu res. (b) T he thi rd co m po u nd is a struc tu ral isom e r of the o th er two . Draw its str uc ture . 58 . • Draw th e struc ture of z-bu rau ol. Iden tity th e ch iral carbon a to m in thi s co m pou nd . Draw the m irr o r im age of the struc ture you fi rst drew. Ar e th e two m o lec u les su pcri m posable? 59 . Draw Lewis stru ctures , an d name three stru ctu ra l isom ers with th e form ula Cl;H I2 • Are an y of th ese isomers ch ira l? 60 . Draw str uc tu res , and na m e th e four alk cucs th a t have th e fo rm u la C 1 11['.. 61 . Write eq ua tions for the reac tions of d.....2-hute ne with th e foll owing reagen ts, re presentin g th e reactants a nd produ cts u sing struc tura l formulas. (a) '1 , 0 (b) H Rr (c) CI,
62 . Draw t he struc tu re , a nd n am e t he produc t formed if th e foll owin g alco ho ls a re o xidi zed . Assume a n exc ess of th e o xid izin g age n t is used. If th e alc o h ol is not expect ed to react with a ch e m ica l o xid izing ag-en t, write N R (no re action ) . (a) CH 3 CH 2 CH,C H, O H (h) 2-bu tall o l (c) 2-met hyl-2-p ro panol (d) 2-me tlwi-l -pro pa no l
64. • Write equation s fo r the fo llowing reaction s, rep re senting th e reactan ts a nd products usin g struc tura l fo rm ulas. (a) Th e fo rm a tio n of et hyl ace ta te from ace tic aci d and et hanol (b) T he hyd rol ysis of g lyce ryl tristearate (the t ric stcr of g lyce ro l with stea ric ac id , a fatty acid )
6 a' . Write a n e q uatio n fo r the fo rm atio n of the fo llow ing po lym e rs. (a) Pol ystyrene , from styren e (C r;l Ir,CIl =C II 2 ) (b) PET (po lye thyle n e tc rc pht halatc }, from et hylen e g lyco l and tc rc pl u ha lic aci d
66. Writ e eq ua tio ns fo r th e fo llowing re ac tion s, rep resen ting the reacta nt s a nd pro d ucts using stru ct ura l form u las. (a) T h e h yd rol ysis of th e amide Ci j ll ~,C ON IIC I h to fo rm ben zoi c acid and meth yla m in e (h) Th e hyd rolysis f-C O(C II 2),C ON II(CII,) "NII -J-m (n ylo n-fi, 6, a polyamide ) to g ive a ca rboxylic acid an d a n am ine 67 . • Draw th e struc tu re of each of th e follo wing co mpound s: (a) 2,2-dimet h ylpentane (b) :~ ,;\-di e t hy l pe n tane (c) 3-e thyl-2-me thy lpe n ta ne (d) :i-e th ylhexa ne 68 . .... Struc tu ral isom ers. (a) Draw a ll of the iso me rs possible fo r C3 1l.S0 . Give th e syste m a tic nam e of each , a nd n-Il in to whic h class of co m po u nd it fits. (h) Draw tilt' struc tural formu las for an aldehyd e an d a keto ne with the mole cu lar for mula GI H ~ O . G ive the syste m a tic n ame of each . 69 . .... . Draw structural form ulas for possib le isomers of the dic hlorinated pro pa ne , C3 H j jCh. Na me eac h co mpound.
70. • Draw struc tural fo rmulas fo r possible isom ers with the fo rmula C 3II hC lRr, an d nam e eac h iso m er.
7 1. Give stru ctu ra l formu las and syste ma tic names for the th ree stru ct ur al isomers of trimetlrylbcnzcn c, Ctj1I3(C II 3h 72 . Give st ruct u ra l fo rmul as a nd syste ma tic names fo r possible isome rs of di c h lorobenzene , C 6 H 4C 12 • 73. Voodoo lilies de pend on carrion bee tles for po llination . Carri on beetles are a ttra cted to de ad a nimals, a nd he ca use dead and putrefying a n ima ls give off the ho rr iblesmell ing amine cadaveri ne. the lily likeveise re leases cadaveri ne (and the clo sely rela ted co mpound putre scine ) ( page 466). A hi o logi cal cata lyst, an e nzym e , co nve rts th e nat urally o ccurring am ino acid lysine to cadaverine. 'I
I I CII
11 2 NCII , CI1 2CII 2CII 2 - C - Nil,
63 . Wri u- eq ua tio ns for the follo win g reactions, re p resen tin g th e re actants and products usin g struc tu ra l formulas. (a) The reacti o n of ac et ic ac id and sodi um h yd roxide (b) T he reactio n of me t hylamin e with HC l 492
I
lysine •
more chaLLenging • in OWL
OH
°
Blue-numbered questions answered in Appendixa
Wh at g-rou p of a to m s m ust be re placed in lysine to m ak e cadaverine ? (Lysin e is esse nti al to h u m an nutritio n bu t is n ot syn thesiz e d in th e h um an hod y.) 74. Be nzo ic acid oc curs in m an y be rries. Wh en hu m an s c a t berrie s, ben zoi c ac id is co nve rted to hip pu ric aci d in the body by re ac tion with th e am ino acid g lycin e H 2l':CH 2C0 2 H . Draw the stru ctu re of h ip pu ric ac id , knowi ng it is a n a m ide fo rmed by reac tion o f the ca rbo xylic aci d gro u p of he n zo ic acid an d th e a m ino g ro u p of g lycine . Wh y is h ip pur ic ac id referred to as a n acid ? 75. • Co n sider th e re acti on o f cLo;-2-h u te ile with 11 2 (in the pre se nce of a catalyst) . (a ) Draw th e struc tu re , a nd g ive th e n am e of th e reactio n pro d u ct. Is th is re ac tio n prod uc t chi ral ? (h) Dr aw a n isom e r of the re act io n product.
76. • Give the name of each co m po u n d bel ow, an d name th e fun ctional g ro u p in vo lved. 0 11 (a ) H , C -
I I
C - CH , CH , CH ,
80 . \ '\Trite a chemi cal equa tion describin g the reac tio n hetwee n glyce rol and stea ric acid to give glycery l n-isu-ara u-, 8 1. • T he produ ct of a n ad d ition re ac tio n o f an alkene is ofte n pred icted by Ma rko vn ikov's rul e. (a) Dra w the st r uc tu re of the product o f adding HB r to propcnc, a nd give the name of t he produ ct. (h) Draw the str uc ture, a nd g ive th e n am e of t he co mpound that re su lts fro m add ing H 2 0 to 2-m c t hyl-
l-bun-n e. (c) If yo u add H 20 to 2· me thy l-2-h u te ne , is the product the sam e o r diffe ren t th an th e p ro duct fr om th e IT · ac t io n in part (b)? 8 2. An unknown co lorle ss liqu id has the formul a G 1HIOO.
Draw the stru ctu res fo r the fou r alco ho l co m po un d s th at have th is fo rmula .
In the Laboratory 83 . • Whi ch of th e fo llowi ng co m po u nds produces ace tic acid whe n treated with a n o xid izing agen t su ch as KMnO_l?
0 11
H
o II
(a)
H , C - CH,
(e) H , C -
"
(b) II"C - CCII,ClI, CII, II
o II
0
I II I
(e) II "C - C - C - II
,
o II
C - OH
77. Draw the str uctu re o f g lyce ryl u- ilaurat c . When thi s tri estel' is sapo nified, what arc th e products? (Se e pag-e 476.) 78 . ... A we ll-known co m pa ny se lling o utdoo r clo thi ng h as recent ly introdu ced j acke ts m ad e of recycled pol yct hylcue u-rc pln halat c (PET) , th e prin cipal ma te rial in man y soft d rin k bott les. An o ther co m pany m a ke s PET fib ers by treatin g recycled bottles with me th ano l to g ive th e d ieste r dim ct hvlt c rcphth a lat c an d et hyle ne g lyco l a nd th en rcpolym e rizcs th ese com po un ds to g ive n ew PET. \Vl'ite a che m ical eq uatio n to sh ow h ow the reactio n of PET with meth an ol ca n g ive d imct hylu- rcp hth alate a nd et h yle ne glyco l. 79 . • Iden tify the re ac tion products, and write an equa tio n fo r the fo llowing re act ion s of C H 2 =C HC H 2 0 H . (a) [1 2 (hyd ro ge na tio n , in the presen ce of a catalyst) (b) O xid a tion (e xcess o xidizin g age n t ) (c) Additio n po lyme riza tio n (d ) Este r fo rma tio n , usin g acet ic acid
•
more challenging
• i n OWL
II"C - C - II
( 6 (O,(g) f}. ,W
= - 2803
+6
H,O(f)
kJ j mol-rxn
Rath er than ca n ) ' o ut thi s reaction in o ne ra pi d an d exoth ermic ste p, a ce ll ca rr ies out a more co n tro lled o x idatio n ill a se r ies of ste ps so that it can o b tain th e e nerb')' in sm all in cre m ents. In add itio n , it wo uld he in effi cien t if eve ry part o f a ce ll need ed to have all the mecha nisms nec essa ry to car ry o u t the ox ida tio n of eve ry type o f m ol ecule use d for e nergy. In ste ad , it ca rr ies o ut the o xid a tion of co mpoun d s suc h as gluco se ill one lo cati o n a nd stores th e energy in a sm a ll set o f co m po u nds th at ca n he used almost anywhere in th e cel l. T h e principal co m po u nd used to perfo rm th is fu nc tion is ad enosine 5 '-t ri p hosp ha te (AT P). T his ribo n ucleo tide co nsists o f a r ibo se molec ule 10which th e ni n-og eu uus base ad eni ne is con necte d a t the I ' po sitio n a nd a tri phosphate g ro u p is co n nec te d at th e 5' po sition (Figu re 22 ). In ae ro h ic res piratio n , th e eq uivale nt o f 30-32 m ol e s of AT P is typically pro duced pe r mole o f g lucose oxidi zed . Based on th e f:j,H va lues fo r th e p ro ce sses, a g rea te r p roduction of AT P mi g ht he ex pected , h ut the p ro ce ss is not co m plet el y effic ien t.
0-
0-
N'" :):
~ I
QN
- o - pL I o - p-L I o - P-L I O- CH
I
I
I
0-
0-
0-
N
1
'>
N
0
H
H
HO
H
H
OH
Figure 22 Adenosine-S'-triphosphate (ATP).
Th e h ydrolysis o f AT P to ad e no sine 5'·d ip hos p ha te (ADP) an d inorgan ic ph o spha te (Pi) is a ll exo the rm ic p ro cess (Fig u re 23) . ATP
+
H,D --> ADP + Pi
f}. ,H = - 24 kJ jmol-rxn
\Vhy is thi s reaction exotherm ic? We ca n assess thi s hy evaluating bo nd e n thal pics (pag e 388) . In thi s re acti on , we mu st b rea k two bonds, a P- O bon d ill AT P and a ll H - O bond ill water. But we also fo r m two new bo nd s: a P- O bond bet wee n th e ph osphat e grou p being cle aved o ff th e ATP and th e O I l of the orig ina l water and an 1-1- 0 bond be tween the h ydrogen from th e water and th e po r tio n o f th e ATP th at fo rms ADP. In the overall pr o cess, m o re energy is rel eased in fo rm ing th ese new bo nds in th e p ro ducts than is required to b reak the necessar y bond s in th e re actan ts. Thus, th e overall react ion is ex o the rm ic. In ce lls, Illan y chemical processes th at wo uld he en doth ermi c 0 11 th eir own are lin ked with th e h yd rolysis of ATP.
N~N 0-
0-
I
I
~.J'> ~N N
0-
I
- O- PL O- P-L O- P-L O- C q
I
I
I
1
0-
0-
0-
H
0 H
H
H
HO
HPO ,' -
"
+
HO-
0-
0-
I PL I
I PL I
0-
O-
0-
OH
ATP Adenosine-5'-tn phosphate
~;;-) ~NJ-N
q
O-C
1
H
H
HO
0
H
H
OH
AOP Adenosine-5'-diphosphate
Figure 23 Ihe exothermic conversion of adenosine-S '-tripho sphate (AlP) to adenosine-5 '-diphosphate (ADP) .
Metaboli sm
Th e co mb in atio n o f a n e nerge tica lly unfavorabl e p ro cess with th e en erget ica lly favorable h ydr o lysis o f AT P ca n yield a proc ess that is energetica lly favo ra b le. For exa m p le, m o st cell s h ave a greater co ncen tra tio n o f pot assium ions an d a sm aller co ncen tra tio n of so d iu m ion s in sid e the m th an ar e present outsid e th em . Th e n atural tenden cy, th erefo re , is for sod iu m io ns to flow in to the cell a nd for pot assium ions to flow out. To m aintai n th e correct co nce n tra tions , the cell m ust co u n teract thi s movem ent and use active tran sport to pUlllp sod iu m ion s ou t o f th e ce ll and po tassiu m ions into th e cel l. This requires e nergy. To acco m p lish thi s fe at , th e cell lin ks thi s pump ing process to th e hydrolysis o f ATP to ADP. Th e ene rb'l' released from th e h ydrolysis react ion provides the ene rb,}' to ru n a m olecular pump (a tran sport p rotein ) th at moves the ions ill th e d irectio n th e ce ll n eeds.
Oxidation-Reduction and NAD H Ce lls also need co m po u n ds th at ca n h e use d to ca rry o u t ox id ati o n-red uc tio n reac tio ns. Just as AT P is a co m poun d used in m an y bioch emical reacti ons when e nerb')' is n eeded, so nature uses an ot her sm all set o f co m po u n ds to run Illan y redo x reactions. An important exam ple is nicotinamid e adeni ne dinucleotide (NAnH). This co m po u n d co ns ists o f two ribonucleotides joined at th eir 5' po sitions via a diph o sph a te lin kage . One of th e nucl e otides ha s ad enin e as its n itro gen ous base , whereas th e o th er h as a n ico tin ami d e rin g (Figu re 24). Wh en NA DH is ox idized, changes occ u r in the ni cotinamide rin g, suc h that th e eq u ivalen t of a hyd ride ion (H -) is lo st. Because th is h ydride io n h as two e lec tro n s associated with it, th e n icotinamide ri ng loses two elec tro ns in the proc ess. Th e resu lting species, referred to as NAD +, is sh own O il th e right in Figure 2,1. In m an y bi o chemi cal reaction s, whe n a particular species n eeds to be redu ced , it reac ts with NADH . Th e NADII is
er H.,
H
I
OI- ~2 0 N H H I
H
-o-rO-
nicotinamide
H
HO 0
C-N Hz
I
-O -pL O-
oI
o II
(=C
OH
- CH z
0
N
k( u~
HHH HO
Oxidation NH 2 N
N)
adenine
OH
NAD H Figure 24 The str uctures of NADH and NAO+.
Reduction
I
511
ox idized to NA D +-, losing two elec tr ons in th e process, a nd th e spe cies o f in te res t is reduced by ga ini ng th ese electrons. If a spec ies must h e ox id ized , th e o p p osite process ofte n occ u rs: th at is, it reacts with NAD+. Th e NAD+ is reduced to NADH , a nd th e species of in terest is o xid ize d .
Respiration and Photosynthesis In th e process o f respiration , a ce ll breaks d own mol ecul es such as g lucose, ox idizi n g th em to C O~ a nd H ~O .
( .H 12 0. (s)
+ 6 02(g) ---> 6 (02( g) + 6 H20(f)
Th e c ne rb'l' rele ase d ill th e se re actions is used to ge ne ra te th e AT P n eeded by th e ce ll. Th e sugars empl oyed in th is pro ( . H,20.(S)
+ 6 0 2(g)
Green plants h ave found a \.. . ay to use light to p ro vid e th e enerb,)" n eeded to ru n th is e ndo ther mic reactio n. Th e key mo lecul e in volved in trappin g th e ene rgy from ligh t ill ph ot o syn th esis is ch loroph yll. Grec n pl an ts co n lain two type s o f ch loro p hyll: ch lorop hyll a an d ch lo ro p hyll b (Figu re 2.'')), Th e a bso rbance spec tra o f ch lo ro p hyll a and ch lo ro p hyll b are also shoev-n ill Figure 25 . No tice that th ese mol e cul es ab sorb best in the b lu e-viol et a n d red-oran ge re gi ons pf th e visib le sp ectru m , Not much lig h t is absorbed in th e g ree n n.~g i o ll . wh en wh ite ligh t shi nes o n cht o rophvll . red-o ra nge an d blue-vio let light a re abs o rbed b y th e ch lo ro p hyll; g ree n light is not abso r bed hut rather is re fle cted. \Ye see th e re flected light, so pl an ts a p pear g ree n to tis. Th e ligh t e ne rgy abso rbe d b y th e ch lorop hyll is used to d rive th e proce ss o f ph oto synth esis.
512
I
The Chemistry of Life-Biochemistry
R
Chlo rophyll a - CH3 Chlorophyll b - CHO
80
R H
60 H , C ~ CH
~
v
c
ro
-e ~
~
40
-« 20
Hydrophobic phytyl side chain 500
600
700
Wavelength (om)
Figure 25 The stru cture of chlorophyll and the visib le absorbance spectra of chlorophyll a and b.
Concluding Remarks In this brief ove rview of bioch emistry, we h ave e xa m ine d p rot eins a n d th ei r str u ctu re s, n ucle ic ac id s, protein syn thesis, lipid s, a n d m e taboli sm . As yo u h ave seen, th e prin ciples o f chem istry yo u h ave been le a rn in g in gen era l c h e m istry ca n he a p plie d to unde rstandin g b iol ogi cal pro cesses. \\'e hope yo u h ave be gun to re cogni ze th e m a rvel o u s com p lexity of life as wel l as so m e o f the underlying p a ttern s th a t e xist with in th is com ple xity. Thi s di scussion h as, h owever. m e rely scra tched th e su r fa ce o f th is fascin at in g an d im po rtan t field o f stu d y. Vast a reas o f biochem istry rem a in to be stu d ie d, a nd Illany questio n s persist for whic h th e a nswe rs a re cu rre n tly u n known . Pe rhaps yo u will p u rsu e a ca re e r d oin g re searc h ill th is a rea . At the \"el')' le ast, we hope you h ave gain e d a n a p p reciatio n o f th e im portan ce an d sco pe o f th is area of sc ie n ce.
SUGGESTED READINGS 1.
J. E. Ba rrick an d
R. R. Bre ak er: 'T he Power of Rib o switch c s.' Scientific A merican, Vol. 296, No. I , pp .
50-57. 2007. 2. M. K. Ca m pbe ll a nd S. O . Farrell: Bio dIf'1J1 L~ l1)l , 5t h cd. , Belmont, Ca lifo rn ia: Th om so n Bro ok s/ Co le , 2006. 3. T. R. Cec h: "RNA as an Enzyme." Sa entijic A mn i uUl, Vol. 255, No . 5. pp . 64-75, 1986.
4. \V. C. Galley: "Exo the rm ic Bond Bre aking : A Persistent Miscon cept ion ." Journal of Chemical Education, Vol. 8 1, pp . 523-525, 2004. 5. R. II. Garren and C. ~1. Gr isha m: Biochemistry; Srd ed ., Belmon t, Californ ia: Th om so n Brook s/ Cole , 2007. 6. D. C. Ph illips: "T h e Thre e-dim e nsion al Structure of an En zyme Molecu le." Scientific A nmican, Vol. 2 15, No .5 , pp. 78-90, 1966. 7. J. D. Wal./Jou bll' Hl'lix: A Personal A a:ou nl ofthl' / >L~wvn)' of the Stru ctu re of!l N A, New Yo rk : M entor, 1968.
STUDY QUESTIONS Bhu--numbr-n-d questions h ave a nswe- rs in Ap pend ix r a nd full y-wo rked solut io ns in th e Student Solutions Ma nu al.
1. (a ) Draw the Lewis stru ctu re fo r th e ami no ac id valine , showin g th e am ino gro u p a nd th e carboxylic acid g ro u p in th e ir unio nized for ms. (h) Draw th e Lewis structure for th e zwiu c rio uic for m of va line . (c) Wh ich of th ese struc tu res will he th e p redo mina nt fo rm a t ph ysiological pH ? 2. Co nside r th e a mino ac ids ala nine, leu cine, serine , ph enylalan in e, Ivsiuc. a nd aspa rt ic aci d. Wh ich h ave pol ar R gro u ps, an d whic h have non polar R grou ps? 3. Using Lewis struc tures, show two d iffe re n t ways th at alan ine ami glycine may he com bined in a peptid e ho nd .
Study Questions
4. Wh en listin g th e seq uenc e o f am in o ac ids in a pol vpeptid e Of pro tein , the seq uen ce always be gin s with th e a m in o ac id that h a s th e fre e am in o g ro u p a n d e n d s wit h th e a m in o acid th at h as th e free carboxylic acid grou p . Draw th e Lewis structure for the trip e p tide : se r inc-l cu cin c-val ine . 5, Dr aw two l .cvv.is st ruc tu res fo r th e d ipe p tid e ala n ine iso leu cin e t hat sh ow th e reso n a n ce stru ctu res o f th e a m ide li n kage .
6. Id e n tify th e typ e o f struc ture ( p rim a ry, secon dary. t ertiary, o r q ua ternary) th a t co rre sp onds to th e fo llowin g sta tem ents. (a) T hi s typ e o f structu re is th e a m in o ac id seq ue n ce in th e p rote in . (b) Th is type o f structu re ind icates h o w di ffe rent p e ptide c h a in s in th e overal l p rotein a rc arranged with re sp e ct to o n e a n ot he r. (c) Th is typ e o f structu re refers to h ow th e pol ypep tid e c hain is fo ld ed , in clud ing h ow a m in o acids th at are fa r apa rt in th e seque n ce e n d up in th e overall molecu le. (d) T h is type of stru cture d eals with how am ino aci ds ncar o n e an o ther in th e sequ en ce a rrange the m selves.
7. (a) Draw th e Lewis stru ctu re (b) Draw th e Lewis st ruc tu re sine (it consist s o f rib os e (c) D ra w th e Lewi s struc tu re sin e 5 '-m o n o p h os p ha te .
fo r th e sligar ribo se , fo r th e n ucleosid e ade noa n d adenine} . for th e n u cleo tid e aden o-
8, A DNA or RNA seq m 'lKe is usually written from the cu d wi th a fre e 5 '-O H to the end with a fre e 3 '-O H . Dr aw the Lewis stru ct ure fo r th e tctranu clcot idc AUec. 9, Do th e D ~ A se qu e n ce s ATGC a nd CG rA f ep rese nt th e sa me molecu le ?
10. (a) Wh at typ e of in te racti on h ol d s DNA's d o u bl ehelic al stra n ds to ge tlle r? (h ) Wh y wo u ld it n o t he go o d fo r DN A's d o uble-h elic al stra n d s to b e h el d tog e th e r b y r oval cu t b o n d s? II. Compl e m e n tary stra n d s of n u cle ic acids run in opposite d ire ctions. T ha t is, th e 5 ' e n d of on e st ran d will he lin ed u p with th e 3' e n d o f th e o ther. Give n th e fo llowin g n u cleo tide se q u e n ce in DNA: 5 ' - AC (;ce ;ATTC- 3 ' : (a) De termine th e seq u ence of th e co m ple m en ta ry s tra n d o f DNA. Repo rt thi s s('q ue nCl' b y writing it fro m its 5' end to its 3 ' e n d (th e u su al wa y o f rcport ing n u cle ic ac id seq u e n ces ). (b) Writ e th e seq uence (5 '-3 ') for th e stra n d o f m RNA th at wo ul d b e com pl e me n tary to the o rigi n al stra n d o f DNA . (c) Assu m ing t ha t th is se qu e n ce is p art o f th e co d in g seque n ce fo r a p rotei n an d th at it is prop e rly lin ed up so th at th e first co d o n of th is se q ue nce h egins with the 5 ' nucleo tide of th e m RNA, write the seq u e nc es for th e th ree arui codons th at wou ld be com ple m entary to thi s stran d o f mRN A in th is regio n.
I
513
(d ) Wh at se q u e n ce of amin o aci d s is co ded fo r hy th is mRNA?
12. (a) Acc o rd ing to th e gene tic code in Table 1, wh ich a m in o acid is co d e d fo r by the m RNA co d o n G AA? (b) What is th e seq u e nc e in th e o rigin al DNA th at le d to th is co don be ing presen t in th e mRNA ? (c) If a m u tat ion o cc u rs in th e DNA in whi c h a (; is su bstit u ted fo r t he n ucle otide a t th e seco n d p o sitio n o f t h is co d ing re gio n in t h e DNA, wh ich am in o acid will n ow be selec te d ? 13. (a) Descri b e wha t o ccurs in the p ro cess of tr an scri pt io n. (h) Desc rib e wh at o ccurs in the p ro cess o f tr an slatio n . 14. Sketch a sec tio n or a p hosp ho lip id b ilayer in wh ic h yo u let a circle rep re se n t th e pol a r head grou p and cu rvy lin es rep resent th e h yd ro ca r b o n ta ils. Labe l th e regions o f th e bi layer as bei n g pol ar or no n p ol ar.
15. \ Vha t stru ctu re d o all ste ro ids h ave in co m mon?
16. Th e secti on abo u t m e tab olism p rovid ed a val ue for ~ r Ho
fo r th e o xida tio n of on e m ole o r g lu cose. Using valu es at 25 °C, ver ify tha t th is is th e correct valu e fo r th e eq uatio n ~lr
C"H " O , (s)
+6
O , (g ) -> 6 CO ,(g)
t.j H" [C,;f{ "CM s) ]
~
+6
H, () (t)
- 1273.3 kf/ mo l
17, Wh ich of th e fo llowin g sta te me n ts a re tru e? (a) Bre aking th e P-O b on d in AT P is an e xoth ermic p roce ss. (h) Ma ki n g a new bo nd between th e ph o sph o rus a to m in th e ph o sph ate g rou p b ei ng cl eaved o ff ATP an d th e 01 I g ro u p o f wa ter is an ex o therm ic proce ss. (c) Breaking b o n d s is an e n d oth e r mi c p ro ce ss. (d ) T h e e n ergy rel ease d in th e h yd rolysis of AT P m ay b e u sed to run e n do th e r mi c reac tio n s in a ce ll.
18. Co ns ider th e fo llow ing reactio n : NA D II + II + + ~ 0 ';1 ---4 NA D + + II 20 (a ) Wh ich sp ec ies is bei n g oxid ized (NAD H, H +, o r ( 2)? (b ) Whi ch sp ec ies is b ei n g red u ced ? (c) Wh ich sp ecies is th e ox idi zin g agen t? (d ) W h ich spec ies is t h e re d u ci ng ag en t?
19. (a) Ca lc ula te th e e n thalpy cha ng e fo r th e p ro d u ct ion o f OlHC mole of glu co se by th e process o f p ho tosyn th esis at 25 "C. t./ r [glucose fs) ] = - 1273.3 ~J / mo l 6 CO , (g)
+ 6 1l, O(() -> C,;f I" O ,,(s) + 6 O , (g )
(h) Wh at is th e en th alp y c h a ng e in volved in p ro d u cin g
1 molecule o f g lu co se by thi s process? (c ) C h loro p h yll mole cules abso rb ligh t o f vario u s wawlength s. O n e wave le n g th a bsorb e d is 650 n m . Calcu la te th e e ll e q,,~' of a p hoton o f ligh t h avi n g th is wavel e ng th . (d) Assu m ing tha t all of th is e nergy goes toward p roviding th e e ne rgy requ ired fo r th e p h otosyn th eti c rcactio n , can the ab so r pti o n o f on e ph oton at 650 l llll le ad to th e p ro d uct io n of on e mole cule of gl uco se , o r must multip le ph o to ns be ab so rbed?
STATES OF MATTER
11
Gases and Their Properties
The Atmosphere and Altitude Sickness Some of you may have dreamed of climbing to t he summits of the world's tallest mountains, or you
may be an avid skier and visit hig h-mountain ski areas. In either case, "acut e mountain sickness"
(AMS ) is a possibility. AMS is common at higher
i"
altitudes and is characterized by a headache, nau-
-e
'"
sea, insomnia, dizziness, lassitude. and fatigue. It .:; can be prevented by a slow ascent, and its symp- ~ toms can be relieved by a mild pain reliever. AMS and more serious form s of high altitude sickness are gener-
P(O ,)(mm Hg)
Approximate Percent Saturation
ally due to hypoxia or oxygen deprivation. The oxygen concentration
90
95%
in Earth's atmosphere is 21%. As you go higher into t he atmosphere,
80
92%
the concentration remains 21%, but the at mospheric pressure drops.
70
90%
When you reach 3000 m (the altitude of some ski resorts), t he barometric pressure is about 70% of that at sea level. At 5000 rn , barometric pressure is only 50% of sea level, and on t he summit of Mt. Everest, it is only 29% of the sea level pressure. At sea level, your
60
85%
SO
80%
40
72%
For m o re
0 11
rln- a tmosp here , see pa ge
5~ 1 .
blood is nearly saturated with oxygen, but as t he partial pressure of oxygen drops, the percent sat uration drops as well. At P(O ,) of 50 mm Hg, hemoglobin in t he red blood cells is about 80% sat urated. Other saturation levels are given in t he table (for a pH of 7.4).
Questions: 1. Assume a sea level pressure of 1 atm (760 mm Hg). What are t he O2 partial pressures at a 3000-m ski resort and on Mt. Everest? 2. What are the approximate blood saturation levels under these conditions? Answers to these questions are in Appendix Q.
514
Chapter Goals Set' Chap In Goals Revisited (/Hlgf' 54 4) lor Stud), Questions kf')'f'd to these f,.,rtJa l \ fi nd a.uigllahlf' i n OH'L
Chapter Outline 11.1
Gas Pressure
11.2
Gas Laws: The Experimental Basis
Understa nd the basis of the gas laws and know how to use those laws (Boyle's law, Charles's law, Avogadro's hypothesis, Da lton's law).
11.3 The Ideal Gas Law
Use the ideal gas law.
11.4 Gas La ws and Chemica l Reactions
App lyt he gas lawsto stoichiometric calculations.
11.5
Gas Mixtures and Partial Pressures
Understa nd kinetic-mo lecular t heory as it is app lied to gases , especially
11.6 The Kinetic-Molecular Theo ryof Gases
the distribution of molecular speeds (energies).
11.7
Recognize why gases do not behave like ideal gases under some conditions.
11.8 Some App tications of the Gas Laws and KineticMolecular Theory
Diffusion and Effusion
11.9 Nonideal Behavior: Real Ga ses
o u n tain climbe rs, hot ai r ballo ons, SCU BA d iving, and au to mobile a ir bags (Fig ure 11.1) d ep end o n the p ro pe rt ies o f gases. Aside from understan d ing how these work , there are at least three reason s for studying gases. First, so me com mo n elements and com pounds (suc h as oxygen , nitro gen , and m ethane ) exist in th e gaseous slate u nd er normal cond itio ns o f p ressu re an d temperature . Furthe rmore , llla llYliqui d s suc h as water can be vaporized, and the p hysica l p roperties of th ese vapors are impo rta n t. Second, o u r ga seo us atmos p here provid es o ne mea ns o f tr ansferri ng energy and material throug hout the g lobe, and it is the sou rce o f life-sustai nin g chemicals. T he third re ason for stud ying gases is also co mpe lling. Of th e three states of ma tter, ga ses are reaso nably simple whe n viewed at the mo lecular level , and, as a res u lt, gas be hav ior is well u nderstood . It is po ssibl e to descri be the pro perties of gases qualitatively in terms of th e be havio r of th e mo lec u les th at make lip the gas . Even more imp ressive , it is possib le to describe the properties of ga ses fjlum titat ivf'l)' using simp le math em at ical mode ls. One objec tive of scie n tists is to d evel op p recise mathemati cal and conce p tua l models of natural p he nomena. a nd a stud y of gas behavio r will introduce yOll to this approach. To descri be gases , chem ists have learn ed that on ly fo u r q uant ities are needed: the p ressu re ( P ), volu me (\1) , an d tempe ratu re ( 'I: kelvins) of th e gas , an d amou n t ( n, mol) .
M
ChemiSlrY-9-NOw~ Th roughout t he text this icon introduces an opportunity for self-study or to explore interactive tuto rials by signing in at www.t homsonedu.com/ login.
FIGURE 11.1. Automobile air bags. Most automobiles are now equipped with airbags to protect the driver and passengers in the event of a head-on or side crash. Such bags are inflated with nitrogen gas, which is generated by the explosive decomposition of sodium azide:
2 NaN3( s) -l- 2 Na (s) + 3 Nz(g) The airbag is fullyinflated in about 0.050 s. This is important because the typical automobile collision lasts about 0.125 s. See Chemist ryNow Screen 11.1 for questions about automobile air bags. 515
,/ Vacuum
,/ Column of mercury
760 mm Hg for standard atmosphere
Atmospheric pressure
11 .1 Gas Pressure Pressure is the force exe r ted on an o bj ect d ivide d by the area ove r which It IS exerted , and a barom et er depends o n th is to m easu re atm osp her ic p re ssu re . A baro mete r can be ma d e by filling a tu be with a liquid , o fte n mercury, and inverting th e tube in a di sh co n ta in ing th e sam e liquid (Fig u re 11.2) . If th e air has bee n removed co m ple tely from th e ve rt ica l tube, th e liquid in th e tube assum es a level suc h th at the pressure ex erte d by th e mass o f th e co lu m n o f liquid in th e tub e is bala nced by the p ressu re of th e atmo sp he re pressing d own 0 11 the su rface o f the liqu id in th e d ish . Pressure is ofte n reported in units o f millimeters of m ercur y (m m Hg), th e height (in mm ) of th e m ercury co lu m n in a m ercury baro me te r above til e su r face o f th e mercury in the d ish. At sea level, this height is abou t 760 111 m . Pressures are also rep orted as standard atmospheres (atm ), a unit d efined as follows: 1 standard atmosphere (1 atm) = 760 mm Hg (exactly)
FIGURE 11.2 A barometer. The pressure of the atmosphere on the surface of the mercury in the dish is balanced by the downwa rd pressure exerted by the column of mercury. The barometerwas invented in 1643 by Evangelista Torricelti (1608- 1647). Aunit of pressure called the torr in his honor is equiva lent to 1 mm Hg .
Th e SI un it o f pressu re is the pascal (Pa). 1 pascal (Pa)
= 1 newton/ meter'
(T he newton is th e SI uni t o f for ce .) Because th e pascal is a ve ry sm all u nit co mpa red with o rd ina ry p ressu res, the uni t kilopascal (kPa) is more o fte n used . Ano ther u ni t used fo r gas pressures is th e bar, where I bar = 100 ,000 Pa. To su mmarize, th e units used in science fo r pressu re are 1 atm = 760 mm Hg (exactly) = 101.325 kilopascals (kPa) = 1.01325 bar
• Hectopascals Meteorologists have long measured atmospheric pressure in millibars. However, after the 51 system of units became more widespread, they began to use the unit "hectopascal," which is equivalent to t he millibar. 1 hectcpascal (hPa ) = 100 Pa = 1 mbar 1 kilopascat (kPa ) = 1000 Pa = 10 hPa
or 1 bar
~
1 X 10 5 Pa (exactly)
~
1 X 10' kPa
~
0.9872 at m
Pressure Unit Conversions ProbLem Convert a pressure of 635 mm Hg into its corresponding value in units of atmospheres (atm), bars, and kilopascals (kPa). Strategy Use the relationships between millimeters of Hg, atmospheres , bars, and pascals described earLier in t he text . SoLution The relationship between millimeters of mercuryand atmospheres is 1 atm
... 635 mm Hg X -
1 atm 76 0
mmHg/ =
The relationship between atmospheres and bars is 1 atm
0.836 atm
= 1.013 bar.
0.836 attn X 1.013 bar 1 atm
=
0.846 bar
The relationship between millimeters of mercury and kilopascals is 101.325 kPa
gX 635 91m H
516
Chap t er
11
I
Gases and Their Properties
= 760 mm Hg.
101.3 kPa 760 _~m Hg
~
84.6 kPa
= 760
mm Hg.
A Closer Look
Measuring Gas Pressure
Pressure is the force exerted on an object divided by the area over which the force is exerted :
Vacuum
Pressure = force/a rea This book, for exam ple, weig hs more tha n 4 lb and has an area of 82 tn", so it exerts a pressure of about 0.05 lbji n2 when it lies flat on a surface. (In metric units, the pressure is about J g/cm2 . ) Now consider the pressure that the colum n of mercury exerts on the mercury in the dish in the barometer shown in Figure 11.2. This pressure exactly balances the pressure of the atmosphere. Thus. the pressure of the atmosphere (or of any othe r gas) can be measured by relating it to the height of the column of mercury (or any other liquid) the gas can support. Mercury is the liquid of choice for barometers because of its high density. A barometer filled with water would be over 10 m in height. [The water column is about 13.6 times as high as a colum n of mercury because mercury's density (13.53 g/cm3) is 13.6 times t hat of water (density ~ 0.997 g/cm', at 25 O() .]
EXERCISE 11,1
add gas
Vacu um (no gas present) -No pressure
exerted on Hg
In the laboratory, we often use aU-tu be manometer. which is a mercury-filled, If -sha ped glass tube. The closed side of the tube has been evacuated so that no gas remains to exert pressure on the mercury on that side. The other side is open to the gas whose pressure we want to measure. When the gas presses on the mercury in the open side, the gas pressure is read directly (in mm Hg) as the difference in mercury levels on the closed and open sides.
Yo u may have used a tire gauge to check the pressure in your car or bike tires. In the U.S., such gauges usually indicate the pressure in pounds per square inch (psi) where 1 atm = 14.7 psi. Some newer gauges give the pressure in kilopascals as well. Be sure to recognize that the reading on t he scale refers to the pressure in excess of atmospheric pressure. (A flat ti re is not a vacuum ; it contains air at atmospheric pressure.) For example, if the gauge reads 35 psi (2.4 atm}, the pressure in the tire is actually about 50 psi or 3.4 atm.
Pressure Unit Conversions
Rank the following pressures in decreasing order of magnitude (from largest to smallest) : 75 kPa, 250 mm Hg, 0.83 bar, and 0.63 atm.
11.2
Gas Laws: The Experimental Basis
Boyle's Law: The Compressibility of Gases Wh en you pump u p th e tir es of you r bicycle , th e p um p squeez es th e air II1to a smalle r vo lum e (Figu re 11.3) . T his p ro per ty o f a gas is ca lled its compressibility. \Vhile study ing th e co m pressibility of gases, Ro ber t Boyle ( 1627- 169 1) o bserve d th at the volu me o f a fixed amou n t o f gas at a g iven temperat u re is inversel y propo rtional to th e p ressu re e xe rted hy th e gas . All gas es beh ave in this mann er, an d we now refer to th is rel ationshi p as Boyle's law. Boyle 's law can be d emon st rat ed in m an y ways. In Figu re 11 .4, a hypo d erm ic syring e is filled with air a nd sealed. Whe n pr essure is ap plied to th e movabl e plun ger of the syringe . the a ir in side is co m presse d . As the pressure (P ) in creases on the syring e, th e gas vo lu me in th e syr ing e ( V) de cr eases. Wh e n 1/ V o f th e gas in th e syring e is plotted as a fu nctio n of P, a straigh t line results. T hi s type o f plo t d e mo nstrat es th at t he pressure and volu me of the gas are inve rsel y pro po rtio nal; that is, th ey change in opposite di recti on s.
11.2
FIGURE 11.3 A bicycle pump-
Boyle's Law in action. Th is works by compressing air into a smaller volume. You experience Boyte's law because you can feelthe increasing pressure ofthe gas as you press down on the plunger.
I
Gas Laws: The Experimental Basis
517
0.150
0.100
~[
I
j
0.050
•1
'l'
oL---------------o
0
500
jJ
1000
1500
2000
Mass of lead on syringe
~
Active Fi gure 11.4 An experiment to demonstrate Boyle's law. Asyringefilled with air was seated. Pressure was applied by adding lead shot to the beaker on top of the syringe. As the mass of lead increased. the pressure on the air in the seated syringe increased, and the gas was compressed. Aplot of (t/volume of air in the syringe) versus P (as measured by the mass of lead) is a straight line.
Chemislry.o.Now'" Sign in
at www.thomsonedu.comjlogin and go to the Chapter Contents menu to explore an i~teraetive version of this figure accompanied by an exercise.
Math emati call y, we can write Bo yle 's law as: 1
P o: -
V
when nand T are constant
whe re th e sym bo l ex mean s "proportional to." Wh en two quantities are proportion al to each oth er, th ey can be eq ua ted if a pmp ortio na lit)1 cons tant, here ca lled Ci\, is introduced . p
=(,
1 V
X -
or
PV
= (8
whe n nand T are constant
T his form o f Boyle 's law ex p re sses th e fact th at the produ ct of tlie /Jressu re a nd volume oj a WUi sample is a constan t at a J.,riven tetn/wra ture, whe re th e co nsta n t CB is determined by the amo u n t o f ga s (in moles ) and its tempe rature ( i ll kelvin s) . It foll o ws from thi s th at , if th e pr essu re-volum e product is kn own for a gas sam ple un d er o ne se t of co nd itio ns (PI an d Vi), th en it is kn own for an other se t of co nd itio ns ( P':!. an d V2 ). Und e r eithe r set o f con d itio ns, th e P Vp rodu ct is eq ua l to CB, so (11.1 )
This fo nn of Boyle 's law is useful whe n we wan t to know, fo r exa m p le, what happen s to th e volu me of a g iven amo u n t o f ga s whe n th e pressure changes at a co nstan t temperature .
I
Boyle's Law
ProbLem A sample of gaseous nitrogen in a 65.0-L automobile air bag has a pressure of 745 mm Hg. 1f this samp le is tra nsferred to a 25.0-l bag at the same t emperatur e, what is th e pressure of the gas in the 25.0-L bag?
518
C h a p t e r 11
I
Gases a nd Their Properties
Stra tegy Here, we use Boyle's law, Equation 11. 1. The original pressure and volume (PI and V) ) and the new volume (V2 ) are known . Soluti on It is often useful to make a tab le of the information provided. Initial Conditi ons PI
Final Co nditio ns
= 745 mm Hg
V1 =65 .0 L
Pz
~
P,V, -1'
20
Absolute zero
T (OC)
T (K)
Vol. H, (mL)
Vol. 0, (mL)
300 200 100 0 - 100 200
573 473 373 273 173 73
47.0 38.8 30.6 22.4 14.2 6.00
21.1
17.5 13.8 10.1 6.39
-
- 273.15°(
10
.'.......
~
- 300
- 200
o
- 100
100
200
300
Temperature (0C)
Active Figure 11.6 Charles's law. The solid lines represent the volumes of the samplesof hydrogen and oxygen at different temperatures. The volumes decrease as the temperature is lowered (at constant pressure). These lines, if extended, intersect the temperature axis at approximately - 273 0(. lU
Chemistry.c).Now Sign in at www.t homsonedu.comj login and go to the Chapter Content s menu to exptcre an i~teractive version of this figure accompanied by an exercise.
• Boyle's and Charle s's la ws Neither
Boyle's law nor Charles's law depends on the identity of the gas being studied. These
laws describe the behavior of any gaseous substance, regardless of its identity.
are ex te nde d to lower tempe rat ures, they all reach ze ro vo lu me at th e sam e temperature , - 273. 15 "C. (Of co u rse, gases will no t actually re ach zero vo lu me; th ey liq uefy above that re m pera tu re .) Th is te mpe ra ture is sign ifica nt, howeve r. \Villia m T h omso n ( 1824-1907), also known as Lo rd Kelvin , propo sed a temperature sca lenow known as the Kelvin scale- for which th e zero po in t is - 273.15 °C (.... pa ge 27). Whe n Kelvin temperatu res are u sed with volu me measu reme nts , th e vo lu metempe rat u re relatio ns hip is
whe re Cc is a proportion ality constant (which de pends o n the a mo u n t of gas an d its p re ssu re ) . T hi s is Charles's law, wh ich sta tes th at if a g iven qu an tity o f gas is held at a co nsta n t p ressu re , its volum e is di rec tly p ropo rtion al to the Kelvin tem peratu re . Writ in g Charles's law ano ther way, we have V/T = (~ ; th at is, the volu me o fa gas d ivided by the temperatu re of th e gas (in kelvins) is co ns ta n t for a given sam ple of gas at a specified p ressu re . T herefore, if we know the volu me an d tempe rature of a g iven quant ity o f ga s ( ~ and T J ), we can find the volum e , V2 , at so me o ther tem pe rat u re , 12, usi ng the equa tio n VI T1
= ~ at constant nand T2
P
(11 .2)
Ca lcu latio ns usin g C harles's law a re illustr at ed by th e followi ng example and exercise. Be su re to no tice th at the te mpe ratllre T must always be expressed in kelvins. Charles 's Law ProbLem Asample of CO 2 in a gas-tight syringe (as in Figure 11.4) has a volume of 25.0 mLat room temperature (20.0 cC) . Whatis the final volume of the gas if you hold the syringe in your hand to raise its temperature to 37 °C? Stra te gy Because a given quantity of gas is heated (at a constant pressure), Charles's taw applies. Because we know the original Vand T, and want to calculate a new volume at a new, but known , temperature, use Equation 11.2. 520
Chapter 11
I
Gasesand Their Properties
Solution Organize the information in a table. Remember the temperature must be converted to kelvins. Initial Conditions VI T,
= 25.0 mL 20.0 + 273.2
~
Final Conditions
,- .
V - , ~
293 .2 K
T,
~
37
+ 273
~
310. K
Substitute the known quantities into Equation 11.2, and solve for V2: V2
,) = = T2 (-V T,
25.0 310. K( -mL) 293.2 K
=
26.5 mL
Comment As expected, the volume of the gas increased with a temperature increase. The new volume (V2 ) must equal the original volume (VI) multiplied by a temperature fraction that is greater than 1 to reflect the effect of the temperature increase. That is,
EXERCISE 11.3
Charles's Law
A balloon is inflated wit h helium to a volume of 45 Lat room temperature (25 "C). If the balloon is cooled to -10 0(, what is the new volume of the balloon? Assume that the pressure does not change.
Combining Boyle's and Charles's Laws: The General Gas Law Th e vo lume of a given amo u n t of gas is inversely proportional to its p ressu re at co nstant te mperatu re (Boyle's law) and directly pro po rti on al to the Kelvin temperature a t co ns tan t pressure (Charles 's law) . Hut what if we need to know what h appen s to th e ga s whe n two of the three parame ters (F: \~ and T ) ch an ge ? For exa mp le , \.. . hat wou ld hap pen to the pressure of a samp le of n itrog en in an au tomob ile air bag if the same amo u n t o f gas were placed in a smaller bag and hea ted to a hig her tem perature? You can deal with thi s situation by com bining th e two equat ions th a t ex p ress Boyle' s and Cha rles's laws. PV ----.-L.1T1
PV = -----.L.£ T,
.
for a qiven amount of gas, n
(11.3)
T h is equation is so me time s called the general gas law or combined gas law. It a pp lies spec ifica lly to situation s in whic h th e nnunin t (if gas does not c!lange. Aweat her balloon is filled wit h helium. As it ascends into the troposphere, does the volume increase or decrease?
. General Gas Law Problem Helium-filled balloons are used to carry scientific inst ruments high into the atmosphere. Suppose a baLtoon is launched when the temperature is 22.5 O( and the barometric pressure is 754 mm Hg. If t he baLloon's volume is 4.19 X 103 l (and no helium escapes from the balloon}, what will t he volume be at a height of 20 miles, wherethe pressure is 76.0 mm Hg and the temperature is - 33.0 O( ? Strategy Here we know the initial volume, temperature, and pressure of the gas. We want to know the volume of t he same amount of gas at a new pressure and temperature. It is most convenient to use Equation 11.3, the general gas law, Solution Begi n by setti ng out the information given in a table. Initial Conditions
Final Conditions
= 4. 19 X 103 L P1 = 754 mm Hg
V2
VI
T,
~
22.5 O[ (295 .7 K)
=?l P2 = 76.0 mm Hg
T,
~
- 33.0 O[ (240 .2 K)
11.2
I
Gas l aws: The Experimental Basis
521
We can rearra nge the genera l gas law to calculate the new volume V2 :
v, ~ (~) X (P,V,) ~ V,X 'l X ~ P2
~
T1
P2
TI
240.2 4. 19 X 10 ' L( 754 mm Hg )( --K) 76.0 mm Hg
= 3.38
295.7 K
X 104 l
Comment The press ure decreas ed by almost a facto r of 10, which should lead to abou t a te n-fold volume incre ase . This increase is partly offset by a drop in te mpe ratu re t hat leads to a volume decrease . On balan ce, the volume increases becaus e t he press ure has dro pped so subs tantially. Notice t hat t he solutio n was t o multi ply t he original volume (VI) by a pressu re factor larger th an 1 ( beca use t he volume in creases wit h a lower pressure) and a temperat ure factor smaller th an 1 (because volume decreas es wit h a decre ase in te mpe ratu re).
EXERCISE 11.4
The General Gas Law
You have a 22.-l cylinder of helium at a press ure of 150 atm and at 31 °C. How many balloons ca n you fill, each with a volume of 5.0 L on a day when t he atmos pheric pressure is 75 5 mm Hg a nd the t emperature is 22 o( ?
The ge ne ra l ga s law lead s to o the r, useful p redictio ns of ga s beh avior, For example , if a give n am ou n t of gas is held in a clo sed co n ta iner, the pressure of th e gas will inc re ase with increasing temperatu re . T = VZ and 5 0 P2 = R1 X ~ T 1
• Gay-Lussac's Law Gay-lussac's law states that, at consta nt volume, the pressure of a given mass of gas is proportional to the absolute te mperat ure. In 1779 Jos eph la mbert proposed a definition of absolute zero of temperatu re based on this relationship.
Th at is, whe n 7"2 is greater th an Ti . P2 will be great e r th an PI. In fact, thi s is the rea son tir e ma n ufa cturers rec ommend checking tire pressu res whe n th e tires are co ld . Afte r d riving for some d istan ce , frictio n war ms a tir e and inc reases t he interna l pressure. Filling a warm tire to the reco m me nded p re ssure may lead to an u nd er-inflat ed tir e.
Avogadro's Hypothesis Fron t and side air bags are now co mmon in automobiles. III the event of an accident, a bag is rapidly infl at ed wit h nitro gcn gas ge nerated by a chem ica l rea ctio n . T he air hag uni t has a sen so r that is se nsitive 10 sud d en d ecelera tio n of th e vehicl e an d will send an electr ical sig nal th at will tr igger th e reactio n (Figu res 11.1 and 11.7). In man )' typcs o f air bags, th e ex p losio n of sodium azid e ge ne ra tes nitro gen ga s. 2 NaN 3( s) ---> 2 Na(s) + 3 N,(g)
Driver-sid e air hags in flat e to a volu me of abo ut 35-70 L, and passe nger air bags infl at e lo about 60- 160 L. The fin al vo lu me o f the bag will d e pe nd o n the am o u n t o f n itr ogen ga s ge nera ted. The re la tionship between vo lu me an d amou n t of ga s was firs t not ed by Amedeo Avogadro . In 1811 , he used work o n gases by th e chem ist (and ear ly ex perime nter with hot air balloons) Joscph Cay-Lussac ( 1778- 185 0 ) to propo se th at equal volu mes oj ga.~es u nder the same conditions oj temperature an d {nessure turae equa l numbers oj j)(i rtitles (either molecules or ato ms, d epe nd ing on th e composition of th e ga s.) This idea ca me to be known as Avogadro's hypothesis. Sta te d an othe r way, th e vo lume
522
C h a p t e r 11
I
Ga ses and Their Pro pertie s
-
I
o ~ ~-. "'II.
!!_ .......;.;;;;0 When a car deceleratesin a collision, an electricalcontact is made in the sensor unit.
Driver-side air bags inflate with 35-70 l of Nz gas, whereas passengerair bags hold about
The propellant (green solid) detonates.
60-160 l.
The bag deflates within 0,2 s, the gas escaping through holes in the bottom of tile bag.
releasing nitrogen gas, and the folded nylon bag explodes out of the plastic housing. FIGURE 11.7 Automobile air bags. See ChemistryNow Screen 11.1 for more on air bags.
of a gas a t a g ive n tempe rature a n d p re ssure is d irec tly p ro portio nal to th e a m o u n t of gas in m ol es: V :x n at constant T and P
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• Avogadro's Hypothesis Problem Ammo nia can be made directly from t he elements: N,(g) + 3 H,(g) -> 2 NH,(g) If you begin with 15.0 L of H2 (g). what volume of N2(g) is required for complete reaction (both gases being at the same Tand P)? What is t he theoretical yield of NH], in liters, under the same conditions?
Strat egy From Avogadro's law. we know that gas volume is proportional to the amount of gas. Therefore. we can substitute gas volumes for moles in this stoichiometry problem. Solution Calcu late the volumes of N2 required and NH] produced (in liters) by multiplying the volume of
H2 available by a stoichiometric factor (also in units of liters) obtained from the chemicalequation:
.
V (N2 requi red}
= (15.0
.
reqUiredr: ) = 1-) ( 1 l N,.avai
L H2_3vallao e
-
V (NH] produced) = (15.0 l
3 L !:'1 3v31tal: i le
. .-(2l
H1__available)
-
.
5.00 L N2 required
'r-
NH produced ) 3 = 10.0 LNH 3 produced 3 l !:l2_3vallab e
EXERCISE 11 .5 Avogadro's Hypot hesis Methane burns in oxygen to give CO2 and H20, according to the balanced equation CH, (g) + 2 O,(g) -> CO,(g) + 2 H,O(g) If 22.4 Lof gaseous CH4 is burned, what volume of O2 is required for complete combustion? Wh at
volumes of CO 2 and H20 are produced? Assume all gases have t he same temperature and pressure.
11.2
I
Gas laws: The Experimental Basis
523
11.3
The Ideal Gas Law
Four in terrel ated q uantities ca n be used to d escri be a gas: pressure , vo lum e, te mperature, and amo u n t (mo lcs). \Ve know from ex per ime n ts th at three gas laws can be used to d escrib e the rel a tio nsh ip of these properties (Sectio n 11.2 ). Boyle' s Law
Charles ' s Law
Avogadro ' s Hypothesis
(const ant P, n)
(constant T, P)
V " (l iP)
(consta nt T, n)
If all t hree laws are co mbined, th e resu lt is V
'X
~ P
• Properties of an Ideal Gas For ideal gases , it is assumed that there are no forces of attraction between molecules and that th e molecules themse lves occupy no volume.
This can be m ade in to a mathematical equation by intro d uci ng a p roportiona lity co nstan t, now labeled R. T hi s constan t, called th e gas constan t, is a u niversal constan t, a num be r you can use to in terre late th e pro perties of an y gas:
or
(11.4)
PV = nRT
• STP-What Is It? A gas is at STP, or standard te mperature and pressure, when its temperature is 0 °C or 273.15 K and its pressure is 1 atm. Under t hese conditions, exactly 1 mol of a gas occupies 22.414 l.
T he eq ua tio n P\l = n R T is called th e id eal gas law. It describes th e be havio r of a so-called id eal gas. As you will learn in Sec tio n 11.0 , however, th ere is no suc h thin g as an "id eal" ga s. No net he less, real gases a t pressures aro u nd one at mosp here o r less a nd temperatures arou nd ro om te mperatu re usua lly behave clo se enoug h to th e idea l th at P\l = nRT adeq uatel y describes th e ir be havi or. To use the equation P\l = n R 7~ we need a value for U. This is readi ly det e rm in ed ex pe rimen tally. By care fu lly measuring P, V, n, and T for a samp le of ga s, we ca n ca lcu late th e value of R from th ese values using the id eal gas law eq ua tio n. Fo r ex a mple, under co nd itio ns of s tand ard te m perature and p res sure (STP) (a gas tem pera tu re of 0 °C or 273. 15 K and a pressu re of I an n), 1 mol of gas occ up ies 22.4 14 L, a qu antity called th e standard molar volume. Su bstitu ting these value s into the ideal gas law g ives a value fo r R: R = PV = (1.0000 at m)(22.414 L) nT (1.0000 mol)(273.15)
L . atm
= 0.082057--
K ' mol
With a valu e fo r /-1., we can now use the idea l gas law in calcula tio ns.
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• Ideal Gas Law ProbLem The nitrogen gas in an aut omobile air bag , with a volume of 65 L, exerts a pressure of 829 mm Hg at 25 °C. What amo unt of N2 gas (in moles ) is in t he air bag ?
r 524
Chapter 11
I
Strategy You are give n P, V, a nd T and want to calculate t he amou nt of gas (n). Use the ide al gas law, Equati on 11.4.
Gases and Thei r Pro pe rti es
Solution First. list the information provided. P = 829 mm Hg
V = 65 l
T = 25 °C
n
=?
To use the ideal gas law with R having units of (L' at m/ K' mol). the pressure must be expressed in atmospheres and the temperature in kelvins. Therefore, P
= 829 mm Hg [ ~
T ~ 25
1 atm ) 760 ~'!1 m H g
= 1.09 at m
+ 273 ~ 298 K
Now substitute the values of P, V, T, and R into the ideal gas law, and solve for the amount of gas, n:
n ~ PV ~ ~=~ ( lcc·0,-,9-,a-,tm-'),:,(6,-,5--,L,,;)= :-::-cc- ~ 2.9 mol RT
(0.082057 L· atm/ K· mol)(298 K)
Notice that units of atmospheres, liters. and kelvins cancel to leave the answer in units of moles. EXERCISE 11. 6
Ideal Gas Law
The balloon used by Jacques Ch arles in his historic balloon flight in 1783 (see page 533) was filled with about 1300 mol of H ~ . If the temperature ofthe gas was 23 O( and its pressure was 750 mm Hq. what was the volume of the baUoon?
The Density of Gases The d ensity o f a gas at a g iven te mp e rature and pressu re (Figure I I .R) is a useful quant ity. Beca use the amou nt (n, mol) of any co m pound is given by its mass (rn) divided by its molar mass (AI), we ca n su bstitu te mliH fo r n in th e ideal gas eq uation. PV
~ [:
) RT
Density (d) is defin ed as mass divided by vo lu m e (mit). \Ve ca n rearra nge th e form of the gas law above to give th e foll owing eq ua tio n, which has the term (mIV) o n the left. T h is is the d ensity of the ga s. m PM d = - = V RT
(11.5)
FIGURE 11.8 Gas density. (a) The balloons are filled with nearly equal amounts of gas at the same temperature and pressure. One yellow balloon contains helium, a low-density gas (d = 0.179 giL at STP). The other balloons containair, a higl1er density gas (d = 1.2 gil at STP). (b) A hot-air balloon rises because the heated air has a lower density than the surrounding air.
(a)
(b) 11.3
I
The Ideal Gas l aw 525
Gas d e nsity is d ire ct ly proportional to the pressure an d molar mass and inve rsely pro portional to th e te mperat u re . Equa tion 11.5 is useful becau se gas d ensity ca n be ca lcu lated from th e mol ar mass, or the mol ar mass can be fo und fro m a measure men t o f gas d ensity a t a given p ressu re and te mperatu re .
Density and Molar Mass ProbLem Calculate the density of CO2 at STP. Is CO2 more or less dense t han air? Strategy Use Equation 11.5, the equation relating gas density and molar mass. Here, we know the molar mass (44.0 q/rnol}. t he pressure (P = 1.00 atm), the t emperature (T = 273.15 K), and the gas constant (R) . On ly t he density (d) is unknown. FIGURE 11.9 Gas density. Because
carbon dioxide from fire extinguishers is denser than air, it settles on top of a fire and smothers it. (When CO2 gas is released from the tank, it expands and cools significantly. The white cloud is condensed moisture from the air.)
SoLution The known values are substituted into Eq uation 11.5, which is t hen solved for molar mass (M) : d
~ PM ~ c::-c=(1c:.o:::o:::a_lm~)cc(4-c4~.0~g~j:::m;-;o:,:I)::::::~ ~ 1.96 gj L RT
(0.082057 L· at mjK ' mol)(273 K)
The density of CO 2 is considerably greater tha n t hat of dry air at STP (1.2 q/ L} . EXERCISE 11.7
Gas Density and Molar Mass
The density of an unknown gas is 5.02 q/Lat 15.0 °Cand 745 mm Hg. Calcu late its mo lar mass.
Gas d en sity has p ractical implications. From the equal ion d = Pivf/R7: we re cognize th at th e density of a gas is d irectly pro porti o nal to its mo lar mass. Dry air, whic h has an aycrage molar mass of abo u t 29 g l mo l, has a density of abo u t 1.2 giL at I arm and 25 "C. Gases or vapo rs with mo lar masses greater than 29 g lmol have d en sities larger than 1.2 giL u nde r these same cond ition s (1 atm and 25 "C) : Gase s suc h as CO 2 , SO~:!o and gaso line va po r settle alo ng the gro un d if re lease d in to th e atm osphere (Figure 11.9) . Co nvers ely, gase s suc h as 1I 2, l Ie , CO, ClI.} (methane ) , and N II .~ rise if released in to the atmosp here. T he sign ifica nce of gas d en sity has be en re vealed in seve ra l tragic even ts. One occurred in the Afri can country of Cameroon in 1984 when Lake Nyos ex pe lled a huge b ubbl e of CO 2 in to the atmosp he re . Because CO 2 is d e nser than a ir, th e CO 2 clo ud hu gged t he g ro u nd, killing 1700 peop le nearby (pag e ( 30) .
Calculating the Molar Mass of a Gas from P, V, and T Data Wh en a new co mpou nd is iso lated in th e lab oratory, o ne o f the first thi ngs to be d o ne is to d et ermine its molar mass. If th e co m po u nd is in th e gas p hase , a classica l meth o d o f d eterm ining th e mola r mass is to measu re th e pressure a nd vo lu me exe r -ted by a given mass of th e ga s at a given tem pe ra tu re .
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, Calculating the Molar Mass of a Gas from P, V, and T Data ProbLem You are trying to determine, by experiment, the formu la of a gaseous compound to replace chlorofluorocarbons in air conditioners. You have determined t he empirical formula is CHF2 , but now you want t o know the molecu lar formula. To do t his, you need the molar mass of t he com pound. You therefore do another experiment and find t hat a 0.100-g sample of t he compound exerts a pressure of 70.5 mm Hg in a 256-mLcontai ner at 22.3 0(, Wh at is the molar mass of the com pound? What is its molecu lar formu la? 526
Chapt er
11
I
Gases and Their Properties
Stra tegy He re, you know the mass of a gas in a given volume (V), so you can calculate its density, d. Then, knowi ng the gas pressure and temperature, you can use Equation 11.5 to calculate t he molar mass. Solution Begin by organizing the data:
= mass of gas = 0.100 g P = 70.5 mm Hg , or 0.0928 atm
m
V = 256 mL, or 0.256 l
T ~ 22.3 "C. or 295.5 K The density of the gas is the mass of the gas divided by the volume: d ~ 0.100 g ~ 0.39 1 gil 0.256 L Usethis value of density along with t he values of pressure and temperat ure in Equation 11.5 = PM/RT), and solve for the molar mass (M).
(d
M ~ dRT ~ (0.3g1 g( L)(0.082057 L· atm(K' mol)(2g5.5 K) ~ 102 g(mol P 0.0928 atm
With this result, you can compare the experimentally determined molar mass with the mass of a mole of gas having t he empirical formu la CHF 2• Experimental molar mass Mass of 1 mol CH F2
102 g(mol 51.0 q/fc rmula unit
Therefore, t he formu la of t he compound is C2 Hl4 • Comment Alternatively, you can use the ideal gas law. Here, you knowt he P and T of a gas in a given volume (V) , so you can calculate the amount of gas (n).
n ~ PV ~ RT
(0.0928 at m)(0.256 L) ~ 9.80 x 1 0~ mol (0.082057 L· atm( K' mol)(295.5 K)
You now know t hat 0.100 9 of gas is equivalent to 9.80 Mo lar mass
=
X
10- 4 mol. Therefore,
0.100 9 9.80 X 10-4 mol
= 102 g/mol
EXERCISE 11 .8 MoLar Mass from P, V, and T Data A0.105-g sample of a gaseous compound has a pressure of 561 mm Hg in a volume of 125 mLat 23.0 "c. Whatis its molar mass?
11.4 Gas Laws and Chemical Reactions Man y industriall y im po rt an t reactio ns involve gas es . Two examp les are the combina tion of n itroge n and hydro gen to p ro d uce am mo nia ,
N2(g ) + 3 H2(g ) --> 2 NHJ(g) and th e elect ro lysis of afJ ueous NaC l to produce hydro gen and chlo rine ,
2 NaCl(aq) + 2 H20 (C) --> 2 NaO H(aq) + H2(g ) + Cl2(g ) If we want to un d erstand th e quant ita tive aspects of suc h re actions, we need to ca n)' out sto ichio metry ca lcu latio ns . T he sche m e in Figure I 1.10 con nects these ca lculatio ns for gas reactio ns with the stoi chiometry calculat ions in Cha p te r 4. Chemislry.c'i.Now'"
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Sign in at www.t homsonedu.comj login and go to Chapter 11 Contents to see Screen 11.7 for a tutorial on gas laws and chemical reactions: st oichiometry.
11.4
I
Gas Laws and Chemical Reactions
527
Mass of A(9)
Mass of B (9)
multiply by
stoichiometric factor
Concentration A 1--
- - - ..1
x Volume A
FIGURE 11.10 A scheme for stoichiomet ry calculatio ns. Here, Aand Bmay be either reactants or products . The amount of A (mol) can be calculated from its mass in grams and its molar mass,from the concentration and volume of a solution, or from P. V. and T data by using the ideal gas law. Once the amount of Bis determined, this value can be converted to a mass or solution concentration or volume, or to a volume of gas at a given pressure and temperature.
. Gas Laws and Stoichiometry ProbLem Yo u are asked to design an air bag for a car. You know t hat t he bag should be filled wi th gas with a pressu re higher t han atmospheric pressure, say 829 mm Hg, at a tempera t ure of 22.0 "C. The bag has a volume of 45.5 L. What quantity of sodium azide, NaN ], should be used to generate t he required quantity of gas? The qas-producinq reaction is 2 NaN,(s) --> 2 Na(s) + 3 N,(g) Strategy The general logic to be used here follows a pat hway in Fi gure 11.10. Us e PV = nRT with gas data ---7 Amount of N2 required ---7 Use stoichiometric factor t o calculate amount of NaN] required ---7 Use molar mass to calculate mass of NaN ] required SoLution The first ste p is t o find t he amount (mol) of gas required so t hat t his can be relat ed t o the quantity of sodium azide required:
P = 829 mm Hg (1 at mj 760 mm Hg) = 1.09 atm V ~ 45.5 L T = 22.0
-c. or 295.2 K
PV n = N2 required (mol) = -
RT
(1.09 atm)(4 5.5 L) (0.082057 L . at m/ K . mol)(295.2 K) Now th at th e required amount of nit rogen has been calculat ed, we can calculate t he quantity of sodium azide that will produce 2.05 mol of N2 gas. Mass of NaN]
=
2.05 mol N2 2 mol NaN ] ) ( 65.01 9 ) ( 3 mol N 1 mol Na N] 2
=
88.8 g Na N]
• Gas Laws and Stoichiometry ProbLem You wish to prepare some deuterium gas, 02, for use in an experiment . One way to do this is to react heavy water, 0 20 , with an active metal such as lit hium. 2 Li (s)
+
2 0, 0(1:) --> 2 LiOO(aq)
+ O,( g)
What amount of O2 (in moles) can be prepared from 0.125 g of li metal in 15.0 mLof 020 (d = 1.11 gjm L) . If dry O2 gas is capt ured in a 1450-mL flask at 22.0 "C. what is the pressure of t he gas in mm Hg? (Deuterium has an atomic weight of 2.0147 gj mol.)
528
Chapter 11
I
Gases and Their Prope rties
Strategy You are combining two reactants with no guarantee that they are in the correct stoichiometric ratio. This exam ple must therefore be approached as a limiting reactant problem. You have to find the amount of each substance and t hen see if one of t hem is present in a limited amo unt. Once the limiting reactant is known, t he amount of O2 produced and its pressure underthe conditions givencan be calculated. Masses of Li and ° 20 Step 1
x (l j molar mass) Step 2
Moles of Li and ° 20
Decide on .. llmltlng reactant
•
5tore . hitometn c
factor
Motes of 02 produced
Step 3
Use ideal gas law
Lithium metal (in the spoon) reacts wit h drops of water, H20, to produce LiOH and hydrogen gas, H2• If heavy water, 020, is used , deuterium gas, D2, can be produced.
Step 4
Pressure of 02 SoLution Step 1. Calculate the amount (mol) of Li and of Ol D: 0.125 9 L·I ( 1 mol Li . ) 6.94 1 9 U 15.0 mL 020 ~
= 0.0180 moI L·1
gO'O)( 1 mol 0,0
1.11 . ( 1 !IlL-02D
2 NH,(g)
Assume that 355 Lof H2 gas at 25.0 °Cand 542 mm Hg is combined wit h excess N2 gas. Wh at amount of NH 3 gas, in moles, can be produced? If t his amount of NH 3 gas is stored in a 125-Lta nk at 25.0 0(, what is t he pressure of the gas?
11.4
I
Gas La ws and Chemica l Rea ctions
529
TABLE 11.1
Components of Atmospheric Dry Air MoLar Mass·
Mole Percent
Partial Pressure at 5TP (atm)
N2
28.01
78.08
0.7808
O2 CO 2
32.00
20.95
0.2095
44.01
0.0385
0.00033
Ar
39.95
0.934
0.00934
Constituent
*T h e averag e m ol ar ma ss o f dr y air = 28 .9fiO g / m ol.
11.5
Gas Mixtures and Partial Pressures
The air yo u b reath e is a mi xture o f n itro gen , oxyge n , argo n , carbo n di o xid e , water va po r, a nd sma ll a m o u n ts o f othe r ga se s Cra ble 11.1 ) . Eac h of th e se gase s exe r ts its own p ress ure, and atmos p he ric p ressu re is th e sum o f th e p ressures exerted by eac h gas. T he pressure of eac h gas in the mi xture is ca lled its partial pressure. J ohn Dalton (1766- 1844) was th e fir st to observe th at th e pre ssu re o r a mi xture of ideal ga ses is th e su m of the parti al p ressures o f th e different gases in th e m ixture . T his o bserva tio n is now kn O\\'11 as Dalton's law of partial pressures (Figu re 11.11 ) . Ma the ma tically, we ca n wri te Dalton 's law of partial p ressu re s as (11.6)
whe re PI, P'l. ' a nd P:~ are the pressures of the diffe ren t ga ses in a mixture , an d ~ 01 a l is the total pressu re . In a m ixture of gases, each ga s beh ave s ind e pe nd en tly o f a ll o t he rs in th e In ixlu re . Th erefo re , we can consid e r th e be havior o f each ga s in a m ixtu re se para te ly. As a n examp le , let us tak e a m ixture of three ideal g ases, label ed A, R, and C. T he re are TlA moles o f A, HI'. mole s of B, a nd. rlc mol es o f C. Assu me th at th e mi xtu re (u lo ta ! = U /\ + nil + nc) is co ntained in a give n volu me ( V) at a given tempera ture ( 7) . \Ve ca n ca lcu late th e p ress u re ex e rted by each ga s from the idea l ga s law equat ion : P, V = n,RT
1.0-liter flas ks
0.010 mol Nz
0.0050 mol 0 2
25 O(
25 O(
0.0 10 mol Nz 0.0050 0 z 25 -c
mix P = 186 mm H9
P = 93 mm H9
(P=279 mm H9 )
FIGURE 11 .11 Dalton's law. In a 1.0- Lflask at 25 "(, 0.010 mol of Ne exerts a pressure of 186 mm Hg, and 0.0050 mol of O2 in a 1.0-l flask at 25 "C exerts a pressure of 93 mm Hg (left and middle) . The Na and O ~ samples are mixed in a 1.0- l flask at 25 "C (right). The total pressure, 279 mm Hg, is the sum of t he pressures that each gas alone exerts in the flask. 530
Chapter 11
I
Gases a nd Thei r Pro pe rti es
whe re each gas (A, B, and C) is in the same vo lu me Vand is at th e same temperature T According to Dalton 's law, the total pressure ex e rted by the m ix tu re is th e su m of the p ressu res exe rted by eac h co m po ne n t:
P" "l =
P+ P,+ Pc=
nA(
A
P" "l = (nA+ n, + nc)(
R: ) + n,(R: ) + nc(R: )
R:) (11.7)
Fo r m ixtu res of gase s, it is co nvenie n t to introduce a quantity ca lled the mol e fractio n, X, whic h is d efi ned as the number o f moles o f a pa rti cular su bsta nce in a mi xture di vided by th e total numbe r of mo les of all su bstances p resen t. Mathem at ically, the mole fraction of a substa nce A in a mi xtu re with B and C is exp ressed as nA
XA
nA +
nB
_
+ nc
Now we ca n co m bine thi s equation (written as for 1~.I" and I~ ota h and derive the eq uatio n
nA n total
Utota l
= riA l X.I,,) with th e equation s
(11.8)
This eq uat io n is useful because it tell s us th at the [nessure oJ a gas in a mixture oJ gases is the !n nd ucl oJ its mole[taction and the total pressure oJ the mix ture. For exam p le. the mol e fraction o f N 2 in a ir is 0.78 , so, a t ST P, its partia l pressure is 0.78 a t m or .,9 0 rnrn II g. ChemislrY~J.NowN Sign in at www.t homsonedu.comj login and go to Chapter 11 Contents to see Screen 11.8 for tutorials on gas mixture s and partial pressure s.
Partial Pressures of Gases Problem Halothane, C2H BrctF3, is a nonflammable, nonexplosive, and nonirritating gas that is commonly used as an inhalation anesthetic.
F
Br
I I F- (- ( I I F
H
Cl
t . L i -trifluorobromochloroethane, halothane
The total pressure of a mixture of 15.0 9 of halothane vapor and 23.5 g of oxygen gas is 855 mm Hg. What is the partial pressure of each gas? St rategy One way to solve this problem is to recog nize th at the partial pressure of a gas is given by the total pressure of the mixture multiplied by the mole fraction of the gas. 11.5
I
Gas Mixtures and Partial Pressures
531
SoLuti on l et us first calculate t he mole fractions of halotha ne and of O2 • Step 1. Calculate mole fractions: Amount of C2HBrC lF3
=
mo l ) 15.0 g -1 ( 197. 49
1 mOl Amount of O2 = 23.5 g -- ) ( 32.00 9
Total amount of gas
= 0.0760 mol
= 0.734
mol
= 0.0760 mol C2HBrCLF2 + 0.734 mol O2 = 0.810 mol
Mole fraction of ( 2HB rClF3 = 0.0760 mol ( 2HB rClF3 0.810 t otal moles
= 0.0938
Because the sum of the mole fraction of halot hane and of O2 must equal 1.0000, t his means that t he mole fraction of oxygen is 0.906. Xh a loU1 a o~
+ Xoxyg~o = 1.0000
+ Xoryg~o = 1.0000 = 0.906
0.0938
Xoryge"
Step 2. Calculate partial pressures: Partial pressure of halothane =
FIGURE 11.12 A molecular viewof
gases and liquids. The fact t hat a large volume of N2 gas can be condensed to a smallvolume of liquid indicates that the distance between molecules in the gas phase is very large as compared with t he distances between molecules in liquids.
Phalothaoe
= 0.0938 '
P haloth aoe
=
P tow l
Pha lotha,,~ = Xhalothaoe ' P tow l
= 0.0938 (855 mm Hg)
80.2 mm Hg
The total pressure of the mixture is the sum of t he partial pressures of the gases in the mixture. Ph a loth a o~
+
p o. ygeo
= 855 mm Hg
and so P oxygeo Poxygeo
= 855 mm = 855 mm
Hg -
Phalothane
Hg - 80.2 mm Hg
=
775 mm Hg
EXERCISE 11.10 Partial Pressures The halothane-oxygen mixt ure described in Example 11.11 is placed in a 5.00-l tank at 25.0 "C. What is the total pressure (in mm Hg) of the gas mixture in t he tank? What are the partial pressures (in mm Hg) of the gases?
•
Module 16
11.6
The Kinetic-Molecular Theory of Gases
So far, we have dis cussed the macroscopic p ro pert ies of ga ses, p ro pe rties suc h as p ress ure and volume th at result from the be hav ior of a syste m with a large n u m ber of pa rt icles. Now we tu rn to the kin etic-mo lec u lar th eory (.... pagc 7) for a description of the beh avio r o f ma ile r at th e mo lecu lar or atomic level. H un d red s of experimc n ta l o bservat ion s have led to the fo llowin g postu lates re gard ing t he behavio r of gases. • Gases cons ist of particles (mo lecules o r ato ms) who se separatio n IS mu ch greate r than the size of th e pa rticles themselves (see Figure 11.12) . • T he particle s of a gas are in co n tin ual, random , and rap id motion . As they move, th ey co llide with one ano ther and wit h th e walls o f th eir co n ta iner, bu t the y do so without lo ss o f e ne rgy. • T he average kin etic e nergy of gas parti cles is pro portional to the gas ternperatu rc. All gases, regardless of their molecular mass, haue the same average kinetic
enerb..rv at
til t> same
temperature.
Let us dis cus s the beh avior o f ga ses fr o m this point of view. 532
Sig n in at www.thomsonedu.comjlogin to download the Go Chemistry modu lefor this section or goto www.ichepters.comtc purchase modules.
Historical Perspectives
Studies on Gases: Robert Boyle and Jacques Charles
Ro bert Boyle (1627-1691) was born in Ireland as the 14th and last child of the first Earl of Cork. In his book Uncle Tungsten, Oliver Sacks tells us that "Chemistryas a true science made its first emergence with the work of Robert Boyle in the midd le of the seventeenth century. Twenty years [Isaac] Newto n's senior, Boyle was born at a time when the practice of alchemy still held sway, and he still mai ntained a varietyof alchemical beliefs and ~ practices, side by ~ side with his seen-
2 NO , (g)
(a) If yo u m ix NO an d O 2 in th e correc t sto ich io m etric rati o a n d N O h as a vo lu m e o f 150 ml.. wha t vo lum c o f O :! is re q u ired (a t th e sa me p ressu re an d te mpe ratu re )? (h) After rea ct io n is co mplet e b e twe en 150 mL of NO a n d th e sto ich io m e t ric vo lume o f O 2 , wh at is th e volu m e o f N O~ (a t the sa me p ressu re a n d tem p e ratu re )? 16. Ethan e, c'2H 6, b urns in air accord ing to th e e q u a tio n 2 C, H 6 (g) + 7 O, (g) -----> 4 CO ,( g) + Ii H ,o (g) Wh at volu m e o f O :! (L) is re q u ired for com p le te rene lio n with 5 .2 L o f C 2 H 6 ? Wh a t vo lu me of H 2 0 va por (L) is p rod uced? Assu me a ll gases are m ea su re d a t th e sa m e tem p e rature a n d p ressu re .
Ideal Gaw l aw (See Fxamjl!" 11.6 and Ch"mislr)'Now
&1"('('1/
11.4.)
17. A 1. 2 5~g sam ple of CO';! is con tai ned in a 75 0 .~m l. fl ask at 22 .5 "C. Wh at is th e p r~ ssure of th e gas? 18. • A ball oo n ho lds 30.0 kg of hel iu m . What is the \'01HUH' of the halloo n if th e fin al pre ssure is 1.20 a un an d th e te m peratu re is 22 "C? 19. A flask is fi rst evac uated so t hat it co n ta ins no gas at a ll. T he n , 2.2 g of CO ';! is intro duce d in to th e flask. O n '»armi ng to 22 "C. the gas exerts a pr essure o f 3 18 mm I lg. Wh a t is the volum e of the flask? 20. A ste el cylinde r h ol ds 1.50 g of e thanol , C,;! I1"O II. Wh a t is th e p ressu re of th e e tha no l va po r if th e cylin der h as a vo lume of 25 J em" an d the te mpe ra ture is 250 "C ? (Assu m e all of the et ha nol is in th e va por ph ase at th is tc rnpc-ra tu rc. ) 2 1. A balloon fo r lo ng-d ista nce !lyin g co n ta ins 1.2 X 107 L of hel ium . If th e helium p ressu re is 737 m m Hg at 25 "C, what m ass of he-lium (i n grams) doc s th e balloon co n ta in? (See Study Question 14.) 22. • What m ass of h eli um , in grams, is req uired to fill a !J .O-1. ball oon 10 a pre ssure of 1.1 a nn a t 25 "C?
Gas Density (St't' Exrunple 11.8 and Chnnisl ryNow Screen 11.5.)
23. Forty m ile s above Eart h 's su rfa ce . th e te mperatu re is 250 K, an d th e p ressu re is o n ly 0.20 m m II g. 'Vh at is th e d en sity of a ir (in g ra ms per lite r) a t thi s altitu de ? (Assu m e th e mola r mass of air is 28. 96 g / mol.) 24 • • Dic th yl eth er; (C';! I l"h O , vapo rizes easily a t roo m te m pe rat u re . If th e vapor exerts a pressu re o f 23:3 m m I1g in a fla sk a t 25 "C, wha t is th e den sity of the vapo r? 25. A gascous o rg a uofluo rine co m pound has a densi tv of 0. 355 gi l. a t 17 "C a nd 189 mm Hg. Wh at is the molar m ass of the co m pou nd ? 26. • Ch lo rofo rm is a co m mon liq uid used in th e labo ratory. It va po rizes read ily. If th e pressu re of c hlo rofor m vapor in a flask is 195 m m I1g a r 25. 0 "C an d th e d en sity o f the va por is 1.25 gi L. what is the mola r ma ss of chl oroform ?
Ideal Gas laws and Determining Molar Mass (Sft' Ex amples fl. i and 1 f .S and C!wmis(1)'Now Sawn 11.6.) 27. A 1.00 7-g sa m p le of an unkn own gas exe rts a pre ssu re of 7 15 mm fi g in a 452-m l. co n ta ine r a t 23 "c. wh at is th e m ol ar ma ss of th e gas? 28. • A 0.0 I25-g sa m p le of a gas wit h a n e m p irica l form u la of C IIF 2 is pl aced in a 165-m l. flask. II h as a pressur e o f 13.7 mm Hg a t 22.5 "C. What is the molec ular fo rmul a of th e co m po u nd?
•
morechallenging
• i n OWL
Blue-numbered questions answered in Appendix 0
29 . A n ew ho ro n hydrid e. RJ I)', h as bee n isol at ed . To fi n d its m ol a r m ass. yo u m easu re th e pr essu re of th e gas ill a kn own vo lu me a t a known te m pera ture. T h e fo llowing ex peri me ntal da ta are co llec ted: Mass of gas = 12.5 m g
Pre ssure of gas = 24.S mm Hg
Tempe ra ture = 25 °C
Vo lum e of flask = 12':) mL
,"Vhich fo rmu la co rre sp o nds to the calc u la te d molar m ass? (a ) B, H G (rl) BGH IO (h) R.;HlIJ (e) RIOH 1.; (e) I\, H" 30. • Acet a ld e hyde is a co m mo n liqu id comp ou nd th at vapo rizes re ad ily. Det ermin e th e mo lar m ass of acetaldeh yd e from th e followin g da ta: Sa m ple m ass = 0.10 7 g Te mperat u re
= 0.0
"C
Volume of gas = 125 m l . Pres su re = 33 1 m m II g
Gas laws and Stoichiometry (5;"" Examples 11. 9 and 11./0 and Chnnistl)'Now Screen II . i .) 3 1. Iro n react s with h yd ro ch lo ric aci d to produce ironf l l ) ch loride a nd hydr o gen gas: Fe (s)
+
2 II CI (a'! ) -> Fe CI, (a'! )
+
[l , (g )
Th e l Iz gas from t he re actio n of 2.2 g of iro n with excess acid is co llec ted in a 10.0-L flask a t 25 "C. \Vh a l is the pre ssu re of th e 11 2 gas in th is flask? 32. • Silane. Sil L , reacts with O 2 an d wa te r:
10
give silicon di o xide
SilI ,(g) + 2 O , (g ) -> SiO , (s) + 2 1I,o (f) A 5.2 0-1. sam ple of sm, gas a t 35 6 mill II g pre ssu re a nd 25 °C is allowe d to react with O 2 ga s. What vo lu m e of O ';! gas . in lite rs. is req u ired fo r co m ple te react ion if the o xyge ll h as a pre ssu re of 425 m m Hg a t 25 °C? 33 . Sodium az ide , the ex p losive co m po un d in au to mobil e air bags, de co m poses accord ing to th e foll owin g eq ua tio n : 2 NaN,(s ) -> 2 Na (s)
+ 3 N,(g)
Wh at m ass of sod iu m azid e is re q u ire d to pro vid e the n itro gen need ed to infl a te a 75.0-L hag to a p ressu re of 1.3 atm a t 25 °C? 34. • T he hyd ro ca rbon o ctane (C,J -I I ~ ) h urn s to give CO ';! an d wate r vapor:
If a 0.048-g sam ple o f o cta ne h u ms co m ple tely in 0 2. wh at will he th e pre ssure of wat er vapor in a 4. 75-1. flask a t 30.0 "C? If th e O ';! gas n ee ded for comp le te com bustio n was co n ta ined in a 4. 75-1. flask a t 22 "C, what wo u ld its p ressu re he ?
I
547
35 . H vd ra zinc reacts with O 2 ac cordi ng to the followin g eq ua tion : N, H 4(g )
+
O , (g ) ---> N,(g)
+
2 H, O (f )
Assume the O 2 ne eded for th e reaction is in a 45 0-1. ta n k a t ~ 3 "c. Wh at mu st th e o xyge n pressure be in the ta n k to have e no ugh oxyge n to co nsu me 1.00 kg of hyd ra ziu e cc m p k-u-ly? 36 . A self-con tained u nderwa ter brea thing ap para tus uses cani ste rs co n ta ining pot assiu m su pcrox idc . The su pe rox ide co ns u mes th e CO 2 ex h aled hy a person an d re places it with ox yge n . 4 KO , (s)
+
2
co, (g)
---> 2 K,CO, (s)
+
3 O ,(g )
\Vh at m ass o f K0 2 , in g rams , is re quired to re act with 8 .9 0 I. of CO 2 at ~~ . O "C a nd 767 mm lI g ?
Gas Mixtures and Dalton's Law (See Example 11. /1 and ClwmiJtr)Now
SUf'f'1I
11.8.)
37 . Wh at is the to ta l p re ssu re in atm o sph e re s of a gas m ixtu re th at co n tains 1.0 g of 1I2 a nd 8.0 g o f AI' in a 3.0-L co n ta in e r a t 27 DC? Wh at are th e parti al pressu re s o f th e two g ases?
38. A cylin d er of co m p ressed gas is labeled "Co m pos itio n (m o le %) : 4.5 % 112S, :~. O % CO 2 , bal an ce N 2 . " T he p res·
42 . Eq ual masses of ga seou s N 2 an d Ar are p la ced in se para te flasks o f eq ual vo lu m e a t the same te mpe ratu re . Tell wh e the r each of th e foll owin g sta te m e n ts is true o r false. Briefl y e x plai n you r a ns we r in e ach case. (a ) The re a re more m o le cule s o f N 2 p re se n t th a n a to ms o f AI'. (h ) Th e p re ssu re is g rea te r in the Ar flask. (c) Th e AI' a to m s h ave a g rea te r average spee d th a n the 1'\2 m o le cule s. (d) Th e N 2 molecules coll ide m o re fre que ntl y with th e walls of th e flask tha n d o th e AI' a to ms . 43. If the spe ed of an ox ygen mol ecul e is 4.2H X 10.1 cm /s at 25 DC, wha t is th e spe ed of a CO 2 m olecule a t th e sa m e tem pe rature? 44 . Calc u la te th e rms speed fo r C O m o le cules at 25 DC. Wh at is the ra tio of th is spe-ed to th at o f AI' a to ms at th e sam e te mpe rat u re ? 45 . • Plan ' th e foll o wing g ases in o rde r of increasing average mo le cul ar spee-d a t ~ 5 °C: Ar, C1I 4 , N 2 , CII 2Fz. 46 . Th e re act ion o f S02 with CI2 gives di chlo rinc o x ide, wh ich is used to bleach wo o d pulp an d to tr ea t waste'wa te r: SO, (g )
ga ug e a ttac h ed to th e cylin der reads 4fi aun. Calcu la te the par tial p ressure of eac h gas , in a tmosp he res, in th e cylinder. SUf{'
39 . • A h al oth an e- o xyge n mix ture (C 2 1IRrCIF3 + O 2 ) ca n be used as a n a nesth et ic. A ta n k co n ta in in g suc h a m ixtu re h as th e fo llo win g part ial pressures: P (haloth a ne ) = 170 rn m Hg an d P (0 2 ) = 5 70 nun Hg. (a ) What is the ra tio of th e n u m be r o f m o les o f haloth an e to the n um b er of m o les o f 0 2? (b) If th e la n k co n ta ins 160 g of 0 2, what mass o f C 211RrCIF3 is p re sen t? 40 . • A co lla psed ball o on is fill ed with He to a vo lu me o f 12.5 L at a p ressu re of 1.00 a tm. O xyg en , 0 2, is th en ad ded so th at th e fina l vo lu me of th e halloo n is 26 L 'with a to tal pressure o f 1.00 a tm . Th e tem pe rature , whic h re m ain s co nstan t throu ghou t, is 2 1.5 DC. (a ) \ Vha t m ass o f H e d oc s the hall o o n co n tain? (h) Wh a t is the fi na l partial p re ssu re of H e in th e bal lo o n ? (c) Wh a t is the part ial p re ssu re of O 2 in the hallo on ? (d) \ Vha t is the m o le fraction o f ea ch ga s?
Kinetic-Molecular Theory (See Section 11.6, Example 11.1 2, and Chnn iJl/),N ow Sorms 11. 9- 11.12.)
41 . • Yo u have two flasks o f e q ual vo lu m e . Flask A co n ta in s H 2 a t 0 DC and 1 a tm p re ssu re . Flask B con tains CO 2 ga s a t 25 DC an d 2 a tm p re ssu re . Compare th e se two gases with respect to eac h of the foll owin g: (a) average kinetic en ergy pe r m o le cule (b ) ave ra ge molecu la r velo ci ty (c) nu mbe r of m o le cul e s (d ) mass of gas 548
I
+
2 Cl,( g ) ---> O SCI, (g )
+
C I, O (g)
All o f th e co m po u n d s involved in the re a ctio n arc gas es . List th e m in o rde r of increasing averagc spe e d .
Diffusion and Effusion (See Example 11. 13 and ChemiJfr;Now Screen 11.12.)
47 . • In each pair o f gases bel ow, tell wh ich will effu se faste r: (a) CO 2 or F2 (b ) O 2 o r N 2 (c) C 2H.1 or C 2 H 6 (d) lW O ch lo rofluorocarbo ns: CFCl3 o r C 2 CI2F1 48 . Arg o n gas is 10 time s d e nse r t ha n hel iu m g as a t the sa m e tem pe rature a nd pressu re . \ Vhich gas is pre dicted to e ffuse faster? How m uch faste r? 49 . A ga s whos e m ol ar m ass yo u wish to k now effuses th rough an o pe n ing a t a ra te o n e th ird as fast as th at o f heliu m g as. \ V11 at is the mo lar m ass o f th e u nkn own ga s? 50 . ... A sa m p le o f u ran ium fluo rid e is fou nd to effu se at th e rate of 17.7 mg / h . U nder co m pa rable co nd itio ns, gaseou s 12 effu ses a t th e ra te of 15.0 mg / h . Wh at is th e m ol a r mass of th e u ran iu m fluo rid e ? ( Hi nt: Rat e s m ust be co nve rted to units o f m o les per tim e .)
NonideaL Gases (See Section 11. 9.)
51. • In th e text, it is sta te d th at the pressure of 4.00 m o l o f C I2 in a 4. 00-1. ta n k a t 100.0 DC sho u ld be 26. 0 atm if ca lcu la ted usin g the van d e l" Waals eq ua tio n . Ve ri fy thi s re sult, and co m pa re it with the p re ssu re p redi cted by th e ideal gas law. •
more chaLLenging
•
in OWL
Blue-numbered questions answered in Appendix 0
52 . • \ ( 111 want to sto re 165 g o f C O t ga s in a 12.5-1, lan k a t ro om te m pera ture (25 °C ). Calcu la te the p re ssure the gas woul d haw using (a) th e id ea l g as law a nd ( h) the van dcr Waals equal io n . ( Fo r CO ';.!. a = :t 59 atm : L2/ moI2 a nd b = 0.042 7 L/ mol. )
25 °C . wha t m ass of d im cth vlclichl o ro.. .ilan e. (C I f 3h SiCIA g ). ca n he fOnJ H'(I? S i(s )
Tb ese qu estions art' 1/01 designated as
10 I),p t'
ortocation j" the 61.
53 . Comph-n- the foll owin g ta ble : mm atm
Hg
kP a
bar
2 C II ,CI(g ) --> (C II , l,SiCI,(g)
\\1Ia t p ressu re of (C I 13hSiC lt (g) wo ul d yo u ex pec t in t hi s same flask a t 95 °C 011 com ple tio n of th e reactio n ? (Dim c thyld ich lo rosilaru- is o ne ...tarring ma terial used to make sil ico nes. polymeric substan ces used as lu h ricants. antistick agents. a nd wa ter- p roofing ca u lk.)
General Questions dUl!lln: Tlw)' mflJ m mbin r several COIl Cf'!I/'{ .
+
~ i ( CO ) 4 ca n be made hy re acti ng fin e ly divided nickel with g ao;eolls CO . If yo u han- C O in a 1..; 0-1. flask a t a pre5.o;ure of 418 mill Hg a t '2;).0 "C. a lo ng wit h 0.450 g of Xi powder. wh at is the theoret ical yield of :\' i (CO) 4?
62. T h e gas R.-J l ti burns in air to giH' lit O and B:!O j .
Standard atm o sp h e re Partial p re ssu H' o f :'\';.! in th e a tmosp here T~U1k
of co m p ressed H 2
Atm osph e ric pressure a t till' lo p of ~I o u n t Everest
:1:1.7
54. On combust io n. 1.0 L o f a gasl'ous co m pou nd o f hydrogcn . carbo n , an d ni trogen g ives 2.0 L o f C O t, :1.5 I. o f H 20 va por. a nd 0.50 1. or N';.! a t STP. Wh at is the empirical formu la of the co m pound:'
55. .&
\ (HI h an- a sa m ple o f heliu m gas a t -:1:\ °C . an d yo u wan t 10 in crease the average speed o f hel iu m atoms hy 10.0 % . To wh a t temperatu re sh o u ld th e g as he ln'an-d 10 accom pli sh Ihi s?
56. If 12.0 g o f 0 t is re quired to in fla te a ha llo on ta in size at 27 °C , wh at m ass o f O ';.! is re q uired it to th e sa me size (a n d p ressure ) at 5.0 °C?
10 10
a ce rin fla te
57 . But yl merca pt an . C 4H!!S H , h as a w ry had o dor a nd is a mong th e co m pou n ds ad ded to na tura l gas to lu-l p d etec t a lea k o f ot herwise o d orle ss na tural gas . In a u l'Xpl'ri ITI(·I1 I. you b u rn 95 .0 m g o f C.~ H 9'-" 1 1 a nd co lle ct the p ro duct gasl's (5 0 2 • Ca t. an d 11 2 0 ) in a 5 .2;') I. fla k at 25 "C . \\11at is th e total gao; p rl'ssu re ill tlufla k, and wh a t i... the partial p re ssure of eac h of thl' product ga ...es? 58, A hicvclc tin ' Ita... an in tern al volume of 1.52 I. and co n ta ins OA06 mol o f air. The lin ' will hurst if its internal p ressure reac hes 7.25 a tm . To wh at Il'mpl'ratttH'. in degn'es Ce lsius, does the air in the tire need to 1)(' In-an-d to ca use a blowou t? 59 . • The lem pl' ra tu re of th e a tmosphere Oil xtars ca n be as h ig h as 27 "c at the equator at no o n . an d rlu- atmosp heric p n ·ssure is a bo ut 8 m m Hg . If a spacec raft co u ld collec t 10. m :J of this a u n o sp he n -. co m p ress it to a sma ll volu m e , and se nd it ba c k to Earth . how m an y moles would the sa m ple co n ta in? 60. If you plan' 2.25 g o f so lid silico n in a 6.5()·1. flask that cot u ai us C1I3C1 with a pfessuH' of 5H:j m m 1Ig- a t ... more challenging • in OWL Blue-numbered questions answered in Appendix 0
(a) Th re e g-ases are invo lved in thi s reac tion . Place them in order of in creasing nn s speed. (Assu m e all ar c a t the same n-mpera mrc.} (b) A 3.26-1, fla sk co n ta ins R.! lI ti ' l( a p n·5.o;u re o f 256 mm II g a nd a le m pe ra lure o f 25 °C. SUPPOSl>O e gas is added to the flask u n til R.! H 1i a nd O ';.! are ill the correct sto ichio me tric ra tio fo r the co mbus tio n reaction. At th is po int. wha t is the par tial p ressu re o f Ot? 63 . • You have four g as sa m ples: I . 1.0 L o f II ';.! a t STP 2. 1.0 L o f Ar at ST P 3. 1.0 L of II ';.! a t 2 7 °C a nd 760 nun H g 4. 1.0 L of Il l' 'II 0 -c a nd 9 00 nun JIg (a) Which sa m ple has th e la rge st numbe r o f gas part icle s (a to ms o r m ol ecul es ) ? (1)) Which sa m ple co n tains th e sm all ;-.. . l nu m ber of particl e s? (c) Wh ich sam ple re p U'sl' ll ts th e la rg-est m a ss? 64, Propa ne reacts wit h oxyge n to giyc ca rbon d io xid e a m i wa ter ,·apo r. CJH, (g )
+ 5 O , (g ) --> :1 CO , (g) + 4 H, O (g )
If yo u mi x Csl-I" a nd O t ill the correc t sto ich io me t ric rati o. a nd if th e tot al p n ·.. .sure of the m ixt u re is 288 m m Hg. what are the parti a l p rt·SSU rt·s of CsH l'l a nd O t? If th e tem perature a nd vol u me do not c hange. what is the prl'Ss ure of the water ' mm Hg . Wh at a mo un t of heliu m is in th e ballo on? (Re m e m be r that gauge pre ssure = total p ressu re bar o m e tr ic pressure. Sec page 5 17.) 74. If yo u h ave a sample of wat er in a closed contain e r, so m e of th e wa ter will eva porat e unt il the pressure of th e water vapor, at 25 -c , is 2 :t~ m m Hg. Ho w many m olecule s of water per cu bic ce n tim e ter exi st in the vapor ph ase? Thermal decomposition of (NH.) 2Cr207'
69 . T h e density of air 20 k m above th e ea rth 's su rfac e is 92 g / m 3 . T he p re ssu re of th e atmosphere is 42 mm Hg, a nd t he te mperature is - 63 -c . (a) What is th e averag(~ mo lar ma ss of the a tmo sp here at th is alti tu de? (h ) If th e a tm os phere a t thi s alt itude co nsis ts of on ly 0 t and Ne. wh at is th e m o le fractio n of e ach gas? 70 . • A 3.o-L bu lb co nta in ing He at 145 mm Hg is co nnec ted hy a va lve to a 2 .0~ L bulb co ntain ing AI' at 355 m m lI g . (Se e th e accompan ying fig urc.) Ca lcula te th e parti al pre ssure of e ac h ga s an d the tot al p ressu re after the valve between th e flasks is opened.
He V = 3.0 L P = 145 mm Hg
He + Ar
550
I
M V = 2.0 L P = 355 mm Hg
75 . ' ( HI a rt' g iven 1.5G g of a mi x ture of KC10 3 a n d KC\. When he at ed , th e KCI() :~ deco m poses to KCI a nd ()2, 2 KCIO ,(s) ---> 2 KCI(s)
+ 30, (g )
and 327 ml , of ()2 with a p re ssu re of 735 mm Hg is col lcctcd at 19 "C. What is the weig ht percentage of KC10 3 in th e sa m ple? 76. ... . A study of climbers wh o rea ched the sum m it of Mo unt Everest witho u t supplementa l oxygen showed th at th e partial pressures of O 2 and COt in their lun gs wer e 35 mm I Ig and 7.5 m ill I Ig, re spectively. Th e barometric pressure at th e sum m it was 253 m m Hg. Assume th e lung gases a rc satu ra ted with m o istu re a t a body te m pe ratu re of 37 °C [whi ch mean s th e partial pressu re of wa ter vapo r in the lungs is P (lI t O ) = 47. 1 mill fi g] . If yo u assume the lu ng gases consist of o n ly 0 2, N 2 , CO 2, and H 20 , wha t is the partial pr essure of N 2? 77 . Ni trogen monoxide reac ts with o xyge n to g ive ni trogen dioxide ; 2 NO (g)
+ O , (g ) ---> 2 NO, (g)
(a) Place the three ga se s in order of inc re asing rm s speed a t 298 K. (b) If you mix NO and O 2 in th e correct sto ich io me tric ra tio a nd N O has a partial pre ssu re of 150 m ill fi g, what is th e parti al pr essu re of 0 2? (c) Afte r reaction between 1':0 and O a is comple te , what is th e pressure of N0 2 if th e xo origina lly h ad a pr essur e of 150 m m lIg and O e was ad ded ill the co rrec t stoichiomet ric a mo u n t?
He + Ar
•
more chaLLenging •
in OWL
Blue-numbered questions answered in Appendix 0
78. ... . Am moni a gas is syn th esized by combini ng hyd roge n a nd ni trogen :
3
+
Il , ( ~ )
N,(~)
-> 2
N Il 3(~)
(a) If yo u wa nt to produce 5 62 g of N H 3 , wha t vo lu me of H 2 gas , at 56 °C and 745 m m Hg, is required ? (b) To p ro duce 5 62 g of N H 3 , wha t vo lu m e of air {th e so u rce of N 2) is requ ir ed if the ai r is intro d u ced at 29 °C a n d 745 m m Hg ? (Assu m e the ai r sa m ple has 78. 1 mo le % N 2.) 79 . Ni troge n tr ifluor ide is prepare d hy the rea ctio n of a mmo n ia and fluo rin e .
4
N H 3 (~ )
+ :,
F, ( ~ )
-> 3 N H ,F (s) +
NF3 (~)
If yo u mi x NIl s with F2 in the COIH'ct sto ichi om et ric ra tio,
and if th e tota l p ressu re of th e m ixtur e is 120 nun II g, wha t arc the part ial pr essures of I\ I hand F2? Wh en the re actan ts have bee n com ple tel y r o usu rru-d , wh at is the to tal p n..'ssu re in the flask? (Assum e T is co ns tan t.] 80. Chlorine trifluoride, ClF3 , is a valuable reagent because it can be used to co nvert m et al ox ides to metal fluori de s:
6 NiO (s) + 4
C 1 F, ( ~)
-> 6 NiF, (s) + 2
C I,(~)
+ 3 O, ( ~)
(a) WhaL mass of N iO will react with e lFj gas if the gas has a pre ssure of 2:>0 m m Hg at 20 °C in a 2.5-L flask? (b) If t he CIF 3 d escribed in pa rt (a) is com ple tel y con su me d, wha t a rc th e pa rtial pr e ssure s o f C l2 a nd o f O 2 in th e 2.5 -1. flask a t 20 °C (in m m Hg ) ? Wh at is the tota l pressure in the llask? 81. ... . Rela tive h um idi ty is th e ra tio of th e pa rt ial pressure of water in air at a g iven temperatur e to th e vapo r pres-sure of water a t th at tem perature . Calc u late th e m ass of wate r pe r liter o r air un de r th e foll owin g co ndit io ns: (a ) a t 20 °C a nd 4.» % rel ative h umi d ity (h ) a t () °C an d 95 % relative hum id ity Under wh ich circ u ms ta nces is th e m ass of 1120 per lite r g rea Le r? (Se e Ap pen di x G for th e vapo r p re ssu re of water.) 82• • How mu ch wa te r vapor is p resent in a dormito ry ro o m whe n th e rela tive hum id ity is 55% a nd th e temperatu re is 23 °C? Th e d ime n sion s of the ro o m a rc 4.5 m 2 floor a re a a nd 3.5 m ce iling h e igh t. (See Study Que stion HI fo r a d efin itio n of re la tive hu m id ity a nd Ap pend ix G fo r the vapo r p re ssu re of wa tc r.)
In the Laboratory 83. ... . " iHl have a 550 .·mI. ta n k of gas with a pre ssure of 1.56 a un at 24 "C. \ tlll thought the gas was pure ca rb o n mon o xide gas, CO , h ut yOll late r fo und it was cou tamina n-d by sma ll quan tities of ga seo us CO 2 an d O 2. Ana lysis shows th at th e: ta nk pressure is 1.34 an n (a t 24 °C) if th e CO 2 is removed . Ano ther experim en t shows th at 0.0870 g of O 2 call he remove-d che mically. Wh at an ' th e m asses of CO and CO 2 in the ta nk, and what is the p,-u ·· rial p f{' SSU H ' of ea ch of th e th ree gases a t 25 °C? •
morechallenging
• in OWL
Blue-numbered questions answered in Appendix 0
84 . 1/;. . Meth ane is burned in a labo ratory Bun sen burn e r to give CO 2 a nd wa ter vapo r. Meth a ne gas is su p plied to th e burner at th e ra te of 5.0 L/ m in (a t a tem peratu re of 28 °C a nd a pn ' ssu re: of 773 m m I 19 ) . At what rate m ust o xygen be su pplied Lo th e burner (a t a pr essure of 742 m ill I-I g a nd a te m perature of 26 °C )? 85 . ... . Iro n form s a series o f co m po u nds of th e type Fe ACO ))" In air, th ey are o xid ized to Fe20 j and C O 2 ga s. After he ating a O.142-g sam ple of Fe .t (CO ) .~ in ai r, yo u isol at e th e CO 2 in a 1.50-1. flask a t 25 °C. T he pressure of th e ga s is 44 .9 m m Hg . wh ar is the em pirical fo rmul a of Fe x(CO )/ 86 . ... . Group 2A me tal ca rbo na tes a re deco m pose d to the metal oxide a nd CO 2 o n heati ng: ~I C0 3 ( s )
-> MO (s) + CO,(g)
You h e at 0 .158 g o f a wh ite , so lid ca r bo nate of a Group 21\ met al ( M) an d find th at th e evo lve d CO 2 h as a p n.'SSU H' of 69 .8 mm Il g in a 285· m L flask at 25 °C. Id en tify M. 87 . O ne way to syn th es ize di horan c , R2II h , is th e re act io n
2 Na RIl ,(s)
+ 2 ll,rO , (aq ) -> R , H 6 ( ~ ) + 2 Na H, r O ,(aq) + 2 H, (g )
(a) If yo u have 0.136 g of Na BH,j a nd ex cess H jP O j, a nd yO Il co llect the R211h in a 2.75 L flask a t 25 °C, what is the pre ssure of th e B2Hti in th e flask? (h ) A by-produc t. of th e reactio n is 112 gas. lfbo th R211 li a nd H 2 gas co me fro m thi s react io n , what is the lolal pre ssu re in th e 2. 75-L flask (after re ac tion of 0.136 g of Na RH ,j wit h ex ces s H 3P () ~ ) a t 25 PC? 88. • You a re give n a so lid mix tu re of 1':a1\0 2 a nd NaCI a nd are asked to an alyze it fo r the: amoun t of NaN0 2 pres· cu t. To d o so , yo u allow th e mix ture to reac t with sulfam ic acid , HS0 3 NII 2, ill wa ter ac co rd ing to th e equation
Na NO ,(aq)
+ I ISO , N Il , (aq ) -> Na HSO , (aq)
+ H,O (f ) +
N ,(~ )
What is th e we igh t perce ntage of Na N0 2 in 1.232 g of the so lid mi xtu re if reaction with sulfa m ic ac id p ro d uce s 295 m L o f N 2 gas with a pressu re of 7 13 m ill Hg a t 2 1.0 °C? 89 . ... You h ave 1.249 g of a m ixture of Na Il C0 3 and Na2CO :J. \ Na CI(aq)
Na,CO , (aq)
+
Il , O (f )
+
C O ,(~ )
+
H, O (f )
+
C() , ( ~ )
+ 2 HCI (aq ) -> 2 NaCl( aq)
\ Vha t vo lum e of CO 2 is evolve d a t 74 5 m m II g and 25 °C?
I
551
90 . ... . A m ix ture of Na HCO:'1 and Na tC O J h as a ma ss o f 2.50 g. When treat ed with II CI(aq ). 6G5 m l , of CO t gas is libe rated with a pre ssu re of 7:~5 mm I1g a t 25 "C. \Vh al is lIlt' \v·c igh t percen t of ~ a 1lC 0 3 and Na tC O :l in th e m ixture ? (Se c Stu dy Que stio n 89 for th e react ion s that occu r.) 91. ... Ma ny n itra te sail" ca n he d ecomposed by h ea tin g. Fo r exam ple , blu e , an hydrou s co pper (ll ) nitrat e produces n itro ge n dioxide and oxygen when heated. In th e laborat ory, yo u fin d th at a sa m ple o f thi s sa lt prod uced 0.195 g of mix ture of NO t a nd O 2 with a pressu re o f 725 m m Hg a t 35 "C in a 125-m L flask (a nd bl ack, so lid CuO was left as a resid ue) . Wh a t is the average molar mass of the gas mixtu re? What are th e mol e frac t io ns of NO t an d 0 t? Wha t amou n t of eac h ga s is in the mix ture ? Do these amouuu.. re flect th e rc la tivc amou nts of N0 2 and 0 t ex pec te d based o n th e bala nced eq ua tion ? Is it possi ble th a t th e fa ct that so me NOt m o lecu les co mbine to give N 2 o.1 play s a ro le?
Summary and Conceptual Questions The[otlot nng questions
Wi)'
use corurptsfnnn thr [nrmous dlfljJlp1S.
9 3. A 1.0-1, flask co n tains 1o.n g eac h of O t an d COt a t 25 °C, (a ) Whi ch gas has tlu- greater partial press ure, O a o r CO t , o r a re they th e sa me ? (h ) Wh ich m olecules have the grea ter average speed , or arc they th e sa m e? (c) Which mole cu les have th e g rea te r average kin e tic ene rgy, o r are th ey the sa me ? 94 . If e q ual m asses of 0 t a nd Nt arc p laced in separate co nta ine rs of eq ual volu me at the same te mperat ure , whic h of th e followi ng sta te m en ts is true? If false , te ll wh y it is fa lse . (a ) T he P H ' SSU H ' in th e flask co nt aini ng Ne is grea ter t han that in the flask con tai nin g 0 t . (h ) The re are more m o lecule s in th e flask co n tai ning O , than in th e flask co ntaining l'\t . 95 . You have two pre ssu re-p ro o f ste el cylin ders of eq ual volume , one co n tain ing 1.n kg o f CO an d th e o th er containing 1.0 kg of ace tylene , Ct H t . (a) In which cylinder is the pressure grea ter a t 25 °C? (h ) Wh ich cylinder co nt ains th e g re a te r n u m ber of mol ec u les ? 96 . • Two flasks, e ach with a vo lu m e of 1.00 L, co n ta in 0t ga s with a pr essure of :~ 8 0 m m Hg. Fla sk A is a t 25 "C, aud flask R is a t 0 "C. w hich flask con tain s th e grea ter n umber of () t molecules?
"
~ 1I;;;;;._--11......t..._
....
Heati ng copper{II) nitrate produces nit rogen dioxide and oxygen gas and leaves a resid ue of copper( II) oxide.
92 . ... . A co m pound contai ni ng C, II , N, a nd 0 is burn e d in excess oxygen . T he gaM' S prod uced hy bu rn ing 0.11 52 g are fir st tr eate d to co nve rt th e nitro ge nco n ta ining product gase s in to N 2 , ami then th e re sulting mix ture of CO t , H 2 0 , Nt , an d excess 0 t is passed th rough a be d of CaCl t to absorb th e water. T he CaC!2 increases in mass hy 0.099 12 g. T he remai n ing gases are bubbled into wate r to fo rm II t CO j • and th is so lu tio n is titrate d with 0 .3283 M NaO H : 28.8 1 mL is requ ir ed to ac hi eve th e secon d eq u ivale nce point. The ex cess O e gas is removed by reactio n with co p pe r m eta ! (to giYC CuO ) . Fin a lly, the Na ga s is collec ted in a 225 .0-m L flask, where it h as a p re ssu re of 65. 12 m m Hg a t 25 "C. In a se pa ra te ex pe rime nt, the u n known compound is fou nd 10 have a molar mass of 150 g /mol. Wh at a re th e empi rical an d mol e cul a r form ula s of th e unknown co mpo un d ?
552
I
97 . ... Sta te whe t her e ac h of th e fo llowing sa m ple s of matu-r is a gas . If th e re is Bo t enough in forma tion for you to dec ide , write "in su fficie n t informa tio n ." (a ) A ma teri al is in a steel ta n k a t 100 ann p ressure . Wh e n th e ta nk is ope n ed to the a tmosphere, th e mat e rial sud denly ex pands, in cr e asing its vo lu me bv 10%. (h ) A 1.0-mL sa m ple of ma te rial weig hs 8.2 g. (c) T he m an- rial is tran sp are n t a nd pa le green ill col or. (d ) One cuhi c m et e r of m ateria l co n tains as ma n y m o lecules as 1.0 Il l"~ of air a t th e same te mpe rat u re and pre ssu re . 98 . Eac h of th e fou r tires of a ca r is fill ed wit h a diffe re n t gas . Eac h tir e h as th e sa m e vo lume, an d eac h is fill ed to the sam e pressure, 3.0 a tm , a t 25 "C. O ne tire co ntains 116 g of a il', an o ther tire has 80 .7 g of neon . a no ther tire has 1G.O g of helium , an d th e fo urth tire ha s 160. g of an u n kn own gas. (a ) Do a ll fou r tires co n tain th e sa me n um ber of gas molecu les? If not, whic h o ne has the greatest n u mhe r of m ok-c u k-s? (h ) How ma ny ti mes he avie r is a m o lec ule of th e u nknown gas th a n an atom of hel ium ? (c) In whic h tire d o th e molec u les have the largest kinetic enagy? T he h igh est ave rage speed?
•
more chaLLengi ng •
in OWL
Blue-numbered questions answered in Appendix 0
99. You haw two ga s-filled ballo on s, one co n ta in ing He a nd th e ot he r co n taining 112' T he 112 balloo n is twice th e size o f th e H e ballo o n . T h e pressure of gas in th e I12 balloon is 1 arm . a nd th at ill th e He balloo n is 2 a tm. T he H 2 balloon is o u tsid e in th e snow ( -5 °C), and the fi e ha lloon is in side a war m b uild in g ( 23 °C) . (a) Wh ich ball o on con tai ns th e g reate r number of m olec ules? (h) Whic h balloon contains the greater ma ss of g as?
100. T h e so d iu m azid e required for automobile air bag s is m ad e by the reac tion of sodiu m m e tal with di ni tro gen o xide in liqu id am mon ia: 3 N, O (g )
+
4 Na (s)
+
N H 3 (t) -> NaN,,(s) + 3 NaO Il (s)
+
2 N, (g )
(a) Vou haw 65 .0 g of so d iu m and a :~ 5 . 0· L flask containin g N 2 0 gas with a p re ssure of 2. 12 a tm a t 2:1 0(:. What is th e theoret ical yiel d (in grams) of NaN 3 ? (h ) Dr aw a Lewi s struc tu re tor the azide io n . In cl ud e all po ssible re son an ce structures . \ Vhich re sonance stru ct ure is m o st like ly? (c) What is the sha pe of the az ide io n ?
102. Ii. Ch lo rin e gas (C I2 ) is used as a di sinfect ant in muni cipal wat er su p plies , al th o ug h ch lorin e d ioxid e (CI0 2 ) a nd oz o ne a re becomin g more widely used . Cl0 2 is a be n et" ch oi ce th a n C I2 in th is a p plica tion heca use it leads to fe we r c hlo rina ted by-products, whi ch a re th em selves pollut ants. (a) How many va le n ce e le ctro ns a re in ClO t ? (h) T h e ch lori te ion , CI0 2 "" , is o b tai ned h y re d ucing CI0 2 • Draw a po ssible electro n dot structu re for CI0 2 - . (Cl is the ce n tra l ato m. ) (c) Wh at is th e h ybridizat ion of th e cen tral CI a to m in CI0 2 -? What is the sha pe o f th e io n ? (d) Which spe cies h as the larg e r bon d a ngle , 0 3 or CI0 2 -? Ex plain briefl y. (e) Ch lo rine d io xid e , ClO t, a yellow-gre e n g-as, ca n be m ad e by th e reac tion o f ch lori ne with so d iu m chl orit e :
2 Na C10, (s)
+ Cl, (g) -> 2 1':aC l (s) + 2 C10 ,(g )
Assu me yo u react 15 .6 g o f NaC 10 2 with chlori ne ga s, wh ic h has a p re ssure o f 1050 mm fi g in a 1.45-1. flask a t 22 °C. Wha t mass of ClO t ca n he produced ?
10 1. • If the ab solute te mpera tu re of a gas doubles, by
ho w m uc h does the ave ra g e speed of the gaseous molc o de s in crease? (Sf'f' ClwmiMrJ-Now Sarr n 11. 9.)
•
morechallenging
• in OWL
Blue-numbered questions answered in Appendix 0
I
553
STATES OF MATTER
12
Intermolecular Forces and Liquids
Antarctica Scene-Icebergs, Penguins, Snow, Ice, and Fog Anta rctica is a place of un iqu e won der and beauty. Aside from the
livi ng creatures-many species of penguins, whales, and bi rds-one mostly thinks of ice and icebergs. Some iceberg s are as large as a
small state, while ot hers are only the size of a foo t ball field. We all kn ow ice floats on water, but what is the reason for this?
554
Questions: 1. Given the density of seawater (d = 1.026 q/ rnl.) and of ice (d = 0.917 q/c m"), is most of the volume of an iceberg above or below the waterline? 2. What volume is above or below the waterline? Answers to these questions are in Appendix Q.
Chapter Goals
Chapter Outline
Set' Chap In Goals Revisited (/Hlgf' 580) lor Stu d)' Questions kf')'f'd to these f,.,rtJa l \ fi nd a.uigllahlf' i n
OH'L
Describe intermolecula r forces and their effects. Understa nd the importance of hydroge n bonding.
Understa nd the properties of liquids.
12.1 States of Matter and Intermo lecular Forces 12.2 Intermolecular Forces Involving Polar Molecules
12.3 Intermolecular Forces Involving Nonpolar Molecules ...-, 12 .4 Properties of Li quids
u nd red s of co mm o n com po u nd s can ex ist ill the liq uid and vapo r sta tes at o r n c ar r o om tem p erature an d u nder ordinal)' p r ess ures , a nd th e on e th at comes
H
to mind im med iate ly is water. Wat er mo lecules in the atmosp here ca n in te ract a t lower tem peratures and co me to geth er to form clo uds o r fog , and still larger cluste rs of m olecul es eventually can fall as rai n drops. At onl y slig h tly lower tem peratures, th e molecules asse mble into a cr ysta lline la ttice , whic h you see as snow flakes and ice. T he pri ma ry o bjec tive s of this chap te r arc to exa mine the in te rmo lecular forces that allow mole cules to interact and then to lo ok at liqu id s, a res ult of suc h in te ract ions. You will find this a useful chapter because it explains, among other thin g s, why your body is cooled whe n yo u sweat and why ice ca n float on liq u id water, a property shared by a lmos t no ot her su bstance in its liqu id and so lid states.
12.1
Chemislry.o.Now'"
•
Throughout t he te xt t his icon int roduces an opport unity for self-study or to explore inte ractive t utoria ls by signi ng in at www.thomsc nedu.com/ Iogt n.
States of Matter and Intermolecular Forces
T he kine tic-mo lecu lar th eory o f gases (.... Sect ion I 1.6) assu mes th at gas molecules or ato ms are wide ly se parated and th at these particles ca n be co nsid e red to he ind epe nde nt of o ne ano ther. Conseq uent ly, we can re late the pro perties of gases u nd e r most cond itio ns by a simp le math emat ical equatio n . PV = nR '}: know n as th e ideal gas law eq uation (Equatio n 11.4) . In real gase s, howe-ve r, the re are fo rces be tween molecules-intermolecular forces-and these re qu ire a more co mp lex ana lysis o f ga s be havior (... page 543 ) . If th ese intermolecular for ces become strong enough , the substance can co nd ense to a liq uid and eventually to a so lid . For liqu id s in part icular, the exist ence of in termolec ular forces makes the picture more co mp lex, and it is not possible to create a sim ple "id ea l liqu id equatio n. " lI ow di fferent are th e states of ma tter at the particulate level? vVe can get a se nse of this by comparing the volumes occupied by equal n um bers of molecul es of a ma terial in d ifferent sta te s. Figu re 12.1a shows a flask con tai nin g about 300 rnL of liquid ni trogen . Ifall of the liqu id were allo wed to evaporate , th e gaseous n itr ogen , at 1 ann and room temperatu re , wo uld fill a large ballo on (m o re th an 200 L) . A grcat amou nt of s p a(~ c ex ists bet wee n mol ecul es in a gas, whereas in liqui d s th e molec u les are close together. T he increase in volume whe n co nve rt ing liq uid s to ga ses is str ikingly large. In co n t ras t, no dra ma tic cha nge in vo lume occurs whe n a so lid is co nve rted to a liq u id . Figure 12.1b shows the same a mo u n t of liqu id and so lid benzene , C1jlI jj , sid e by sid e. As yo u see , th ey are Hot ap p reciably differe nt in volume. T his means that the at oms in th e liq uid arc pa cked to gether abou t as tightly as the ato ms in the so lid p hase.
12.1
I
Early morning fog over Lake Champl ain in upstate New York.
• The Lotus Effect The photograph on the cover of t his book illustrates t he importa nce of intermolecular forces. You can read about the lot us effect on the back of th e title page and on the back cover of the book.
States of Matte r a nd Intermolecula r Forces
555
FIGURE 12.1 Contrasting gases, liquids, and solids. (a) When a 300 ~ml sampleof liquid nitrogen evaporates, it will produce more than 200 l of gas at 25 °Cand 1.0 atm. In the liquid phase, the molecules of N2 are do se together; in the gas phase, they are far apart. (b) The same volume of liquid benzene, CoHo, is placed in two test tubes, and one tube (right) is cooled, freezing the liquid. The solid and liquid states havealmost the same volume, showing that the molecules are packed together almost as tightlyin the liquid state as they are in the solid state.
li quid benzene
Solid benzene
liquid nitrogen
(e}
(b)
\f e know th at gases ca n be co mp resse d easily, a process th at in volves for cing the gas mo lecule s much closer togethe r. In co n tras t, th e mo lecu les, ion s, o r atoms in liq ui d or so lid ph ase s stro ngly res ist for ce s that wo uld push them even clo se r together. Lack of compressibility is a cha rac teristic property of liq uid s an d so lid s. For exa mp le , th e volume of liqu id water changes o n ly by 0.005 % pe r at mosp here o f p re ssure ap plied. In term o lecul ar for ce s in flu e nce chem istry in man y ways: • T hey a re d ire ct ly rel ated to pro pe rt ies su ch as m e lting point , boili ng poi n t, and the energy ne ed ed to co nve rt a so lid to a liquid or a liquid to a vapor. • T hey are important in determ ining the so lu bility of gases, liq uid s, and so lid s in var ious so lve n ts. • T hey are cruci al in d e termin in g the struc tu res of bio logica lly im po rtan t mo lecu les su ch as DNA and proteins . Bond ing in ionic compound s d epe nds o n th e electrostat ic for ce s of attractio n be tween o p positely charged io ns. Simi larly, th e in termolecular forces attract ing one mo lecu le to ano ther are e lectrostatic . By co m par iso n , th e attractive force s between the ion s in io ni c co m po u n d s are usu ally ill the ran ge o f 700 to 1100 kJ /rno l, an d mo st cova le nt bond en erg ies are in th e rang e of 100 to 400 kJ /m o l Cra ble 8.9) . As a ro ug h guide line, inte rmolecular forces are ge ne rally less th an about 15 % of the values o f bon d ene rg ies. No ne thel ess, these in teractio ns ca n have a p ro found effect on molecul ar pro pert ies an d are the su bject of t his sectio n . T he sections that follow are o rgani zed around the po larity of the mo lec ules in volved . We shall first d escri be fo rces in vol ving polar molecu les an d then those involving non polar mo lecu les. In Chap ter 13, we shall d escri be ionic and m et all ic so lids and the bond ing in those su bstance s.
ChemistrY.{J.Now'" Sign in at www.t homsonedu.com/ Ioqtn and go to Chapter 12 Contents to see: Screen 12.2 to view an animation of gases, liquids, and solids at the molecular level • Screen 12.3 for an outline of the important int ermolecular forces
556
Chapter 12
I
Intermolecular Forces and li quids
12.2
Intermolecular Forces Involving Polar Molecules
Interactions Between Ions and Molecules with a Permanent Dipole T he d istri bu tion of bo ndin g elect ron s in a m ol e cul e often re su lts in a permanent dipo le moment (.... Section 8.7) . Becau se polar mo lecu les have positi ve and negati ve ends, if a pola r mo lecu le and an io nic co m po u nd are mix ed , the negat ive e nd of t he dipo le will be attracted to a positive cation (Fig u re 12.2) . Simi la rly, the po sitive end of th e dipol e will be attracted 10 a negati ve an ion . Fo rces of attraction bet ween
Water surrounding a cation
a positive o r nega tive ion and po lar mo lecu les-io n-d ipo le fo rces- are less t han those fo r jo n-ion attractio ns (wh ich ca n be o n the order of 500 kJ/mol ) , but th ey are grea te r than o the r types of forces betw een mo lecu les, whether polar o r no np olar. lou -dipole attractio ns can be evalua ted based o n Coulomb's law (.... Equ ation 2.3), whic h informs us that th e fo rc e of attractio n be tween two charged obj ects depends on the p roduct of thei r charges di vid ed by the sq ua re of the distan ce bet ween them (.... Section 2.7) . T herefore, when a po lar molecule e nc ou n ters an io n, the attractive fo rces d ep end on three factors:
Water surrounding an anion
• The di stance bet ween th e io n and th e d ipo le . The clo ser th e io n and dipole , the stro nger the attract io n . • T hc charge o n t he ion . The h ighe r t he ion charge , t he st ro ngc r t he attrac tio n . • T he magnitud e of th e d ipo le. T he greater the magn itude of the d ipole , the stro ng e r the attrac tio n . The formatio n o f hyd rated ion s in aqucous so lut io n is one o f t he most important ex am ples of th e in teract ion betw een an ion and a pol ar mo lecul e (Figure 12.3) T he e n tha lpy cha nge asso cia ted with the h yd ration of io ns- whi ch is generally ca lled th e en tha lpy o f so lvatio n or, for ion s in wate r, t he en thalpy o f hydration-is su bsta n tia l. T he so lvatio n e n thalpy for an individual io n ca n no t be measured direc tly, but values ca n be estima ted . Fo r exa mp le, the hydratio n of sodiu m ion s is d escr-ibed by the fo llowing reactio n : Na ' (g)
+ x H20 (C) -->
[Na(H2 0),] ' (aq) (x probably
~
6)
:",w
~
- 405 kJ/ mol
Th e en thalpy of hyd ra tio n d epends on l i d, where d is the di stan ce he twee n the ce nte r of the ion and the op pos itely charg ed "pole " o f the d ipole . As the ion rad ius becomes larger, d incr eases, and th e e n tha lpy o f hydration beco mes less ex o the rmic. T his trend is illustrated by th e en thalpies of h ydratio n of th e alkali me tal ca tio ns (Table 12.1 ) and by those for Mg 2+ , Li" , an d K + (Fig ure 12.3) . It is in teresting to co mpare th ese value s with th e en thalpy o f hydratio n of th e
••••• d
d K+,r = 133 pm ~H
li +, r = 78 pm j, H = - 515 kJjmol
d Mg2+ , r = 79 pm ~ H = ~ 1922 kJjmol
Active Figure 12 .2 Iondipole inte racti ons. When an ionic compound such as NaCl is placed in water, the polar water molecules surround the cations and anions.
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• Coulomb's Law Th e force of att raction between oppositely charged particles depends directly on t he product of t heir charges and inversely on the square of the distance (d) between t he ions (ljd 2) (Equation 2.3, page 78). TIle energy of the attraction is also proportional to the charge product, but it is inversely proportional to the distance between t hem (ljd) .
FIGURE 12.3 Enthalpy of hydration. The energyevolved when an ion is hydrated depends on the dipole moment of water, the ion charge, and the distance d between centers of t he ion and the polar wate r molecule. The distance d increases as ion size increases.
= - 321 kJj mol
Increasing force of attraction; more exothermic enthalpyof hydration 12 . 2
I
Inte rmolecular Forces Involving Polar Molecules
557
TABLE 12 .1
Radii and Enthalpies of Hydration of Alkali Metal Ions Ion Radius (pm)
Enthalpy of Hydration (kJjmol)
Li+
78
- 515
Na"
98
- 405
Cation
K'
133
- 321
Rbo
149
- 296
Cs +
165
- 263
II + ion , estima ted to be - 1090 kJ/ mol. T his ex trao rd inari ly large value is d ue to th e tin y size of the H + ion.
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Hydration Energy ProbLem Exp lain whyt he enthalpy of hydration of Na " (- 405 kJ/mol) is somewhat more exothermic than th at of cs' (-263 kl/mol). whereas that of Mg 2 " is much moreexothermic (-1922 kJ/m ol) than t hat of either Na " or Cs". Strategy The strength of ion-dipole attractions depends directly on the size of the ion charge and the magnit ude of the dipole, and inverselyon the distance between them. To judge the ion-dip ole distance, we need ion sizes from Figure 7.12. SoLution The relevant ion sizes are Na " = 98 pm. Cs" = 165 pm. and Mg 2+ = 79 pm . From t hese values, we can predictt hat the distances between the center of the positivecharge on the metal ion and the negative side of the water dipole witt varyin this order: Mg2 - < Na " < Cs ". The hydration energyvaries in the reverse order (with the hydration energy of Mg 2+ being t he most negative value). Notice also t hat Mg2 + has a 2+ charge, whereasthe other ions are 1+. Thegreater charge on Mg2- leads to a greater force of ion-dipole attraction than for the other two ions, which have only a 1 + charge. As a result. the hydration energy for Mg2 + is much more negative tha n for the other two ions. Weak attraction
EXERCISE 12 .1 Hydranon Energy
Which should have the more negative hydration energy, F- or Cl- ? Explain briefly.
Interactions Between Molecules with Permanent Dipoles wh en a po lar mo lecul e en co unt e rs an o ther po lar molecu le , of th e same or a d ifferent kind, the positive end of one mo lecule is a ttrac te d to the negati ve e nd o f the o the r pol ar mol ecul e.
Man y mo lecu les have di pol es, an d th ei r int e ract io ns o ccu r by dipole-dipole a ttraction. 558
Chapt er
12
I
Intermolecular Forces and Liquids
A Closer Look
Hydrated Salts
Solid salts with waters of hydration are common . The formulas of these compounds are given by appending a specific number of water mole cules to the end of the formula, as in BaCl l ' 2 H20 . Sometimes, the water mo te-
cules simply fill in empty spaces in a crystalline latti ce, but often the cation in the se salts
is directly associated with water molecules. Fo r example, the compound (r([3' 6 H20 is better written as [Cr(H 20)4ct2]Cl ' 2 H20. Four
Compound
Com mon Name
Uses
NazCO] • 10 H2O
Washing soda
Wa ter softener
Na ZS,03 • 5 H2O
Hypo
Photography
Epsom salt
Ca thartic, dyeing and ta nning
CaSOIi • 2 H2O
Gypsum
Wa llboard
(u SO• • 5 H2O
Blue vit riol
Biocide
MgS04
•
7 H2O
Hydrated cobatt (II) chloride, ( oel2 ' 6 H20. In the solid stat e, the compound is best described by the formula [Co (H20)4CI2J • 2 H20. The cobalt (lI) ion is surrounded by four water molecules and two chloride ions in an octahedral arrangement. In water, the ion is completely hydrated, now being surrounded by six water molecules. Cobalt (II) ions and water molecules interact by ion- dipole forces. This is an example of a coordination compound, a class of compounds discussed in detail in Chapte r 22.
of the six wate r molecu les are asso ciated with the CrH ion by ion- dipole attra ctive forces; t he remaining two water molecules are in the lattice. Common examples of hydrated sa lts are listed in the table.
J-
- - - -
Fo r polar mo lecul es, di pole-dipole attractio ns in flue nce, amo ng other th ings, the evaporatio n ofa liq u id an d th e co ndensation o f a gas (Figure 12.4) . An energy cha nge occu rs in both pro cesses. Evapora tio n re q u ire s th e inp u t o f e nergy, specifically th e e ntha lpy ofvaporlzauon (8.,""p l r ) [see Section 5.3 an d Se ction 12.4J. T he value fo r th e e n thalpy of vaporizat ion has a positive sig n, in di cating th at eva p~ oratio n is an e ndot he rm ic p rocess. T he en thalpy cha nge for th e co nde nsa tio n process- t he reverse of evapo ration - has a negative value . T he g reate r the fo rces of attraction between mo lec ules ill a liqu id , the greater the e ne r6'1' th at must be supplied to se parate th em . T hus, we ex pect polar co mpoun ds to h ave a high er value fo r their e n thalpy of vaporiza tio n th an no npol ar co m po un ds with simi lar molar masses. For ex am ple , no tice t hat 8.\'a p l JO for po lar FIGURE 12.4 Eva poration at th e molecular level. Energy must be supplied to separate molecules in t he liquid stat e against intermolecular forces of attraction.
c
-(
Liquid
12 .2
I
Inte rmolecular Forces Involving Pola r Molecules
559
TABLE 12 .2 Molar Masses , Boiling Points, and t>"p W of Nonpolar and Polar Substances
Nonpolar
Polar
M
BP
!i vapH o
M
BP
j,vapH o
(g/ mol)
(0C)
(kl/mol)
(g/ mol)
(0C)
(kl/mol)
N,
28
- 196
5.5 7
CO
28
- 192
6.04
SiH4
32
- 11 2
12.10
PH,
34
- 88
14.06
Ge H4
77
- 90
14.06
AsH]
78
- 62
16.69
160
59
29.96
ICI
162
97
Br,
mol ecul es is g rea ter th an fo r nonpo la r m o lecu les of a p p rox imatel y th e sa me size and m ass (Table 12.2). The bo ilin g poi n t o fa liq u id also d epen ds o n in term olecu lar forces of a ttraction. As th e temperatu re of a su bsta nce is rai sed , its mo lecu le s gain kine tic encq ,T),. Even tually. when the boi ling po int is reached, the molecu les have suffi cien t kinet ic c nc rb"1' to esca pe the fo rces of att ractio n of th ei r ne igh bo rs. For m o lec u les o f simila r mo lar mass, th e greate r th e po larity, th e h ig he r th e te m pe ratu re req u ired fo r th e liquid to boi l. In Table 12.2, yOll see that the bo ilin g point fo r po lar l e i is greater th an th at for nonpol ar Br-, fo r exam p le. In term o lecular forces also influ en ce so lubi lity. A qu al itative obse rvat ion o n so lubil ity is th at "like di sso lves like ." In ot her words, po lar mol ecu les are likely to d isso lve in a po lar so lven t, and no n po lar mo lecu les arc likely to di ssol ve in a non po lar so lven t (Figu re 12.5 ) (..... Chap te r 8). T hc co nverse is a lso tru e : that is, it is u nli ke ly th at pola r mo lecu les will d issolve in nonpol ar so lvents o r that no npo lar molecu les will d issolve in po lar so lven ts. For ex am ple, water an d e thano l (C zH;jO H) can be m ixed in any rati o to give a hom ogen eous m ixture. In co ntrast, water d o es no t d issolve in gasoline to an a p-
Hydrocarbon
(a) Ethylene glycol (HOC H2CH20H), a polar com pound used as antifreeze in automobiles, dissolves in water.
(b) Nonpolar motor oil (a hyd roca rbon) dissolves in nonpolar solvents such as gasoline or CCl 4• It will not dissolve in a polar solvent such as water, however. Com mercialspot removers use nonpolar solvents to dissolve oiland grease from fabrics.
FIGURE 12.5 "li ke dissolves like." 560
Chapter 12
I
Inte rmolecular Forces and Liquids
prcciablc ex tent. T he d ifferen ce in the se two situa tio ns is th at e thano l and water are pol ar molecul es, whe reas t he hydrocarbon molecul es in gasoline (e .g., oc ta nc, C8 II HI ) are no np ol ar. The water--ethano l int e raction s are stro ng enoug h th at the energy ex pended in p ush in g water m olecul es apart to make roo m fo r ethano l molecules is co m pe nsa te d for by th e energy of a ttrac tio n be tween th e two kin d s of polar molecul es. In cont ras t, water- hydrocarbon a ttra ctio n s are weak . Th e h ydrocarbon m olecul es ca n not d isru pt the stro nge r wate r-water a ttrac tio ns . Chemislry.o..Now'"
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Hydrogen Bonding Hydrogen fluo ride , wate r, am monia and many o the r co m po u nds with O -H and N- H bonds have exce ptio na l p roperties. Co nside r, for exam ple. the boiling po ints fo r hydr ogen co mpo u nd s of elem e nts in Groups 41\ th ro ug h 71\ (Figure 12.6) . Ge nerally, th e bo iling points of rel ated co mpou nds increase with molar mass. T his trend is see n in th e boiling poin ts o f th e hydro gen compou nds of Gro u p 4A clem ents, for example (Cl l, < Sil I, < Ce l I, < Sn l L ) . The same effect is also operating fo r the heavie r molecules o f th e hydrogen co mpounds of c lemen ts o f Gro ups IlA, 6A, and 7A. T he boiling poin ts o f NH 3 , H ~O, and H F, however, deviate sign ifica n tly fro m wha t migh t he expected based o n molar mass alo ne. If we extrapolate the curve for the boiling points of H 2Te , H 2Se , and H 2S, the bo ilin g point o f water is predicted to be arou nd - 90 "C. T he bo iling po int of water is almos t 200 "C higher th an thi s value! Sim ilarl y, th e bo ilin g po in ts ofNl l, and I1F are muc h high er than would he ex pected based on molar mass. Because th e te m pera ture at which a su bstan ce boils d e pends on th e attrac tive forces between mo lecu les, th e ex traordina ri ly high bo iling po ints of 1120 , I IF, and NI I :~ in dica te stro ng int ermolecul ar attractio ns.
C
Active Figure 12.6 The boiling points of some simple hydrogen compounds. The effect of hydrogen bonding is apparent in the unusuaUy high boiling points of H20, HF. and NH ]. (Also, notice that t he boiling point of HCl is somewhat higher t han expected based on the data for HBrand HI. It is apparent that some degree of hydrogen bonding also occurs in liquid HCl.)
~
HzTe SbH3
~
HI
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SnH4
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0
0
B
•
~
E
'" - 100
o
2
4
3
5
Period
12 .2
I
Int ermolecular Forces Involving Pola r Molecules
561
Negatively Charged Region Positively Charged Region
Hydrogen Bond
Hydrogen bonding between HF molecules. The partially negative F atom of one HF molecule interacts through hydrogen bonding with a neighboring HF molecule. (Red regions of t he molecule are negatively charged, whereas blue regions are positively charged. For more on electrostatic potential surfaces, see page 382.)
T he clcc tro nc gativitics of N (3.0) , a (35), and F (4 .0) are amo ng the h igh esl of all th e c lem en ts, whe reas th e e lcc tro nega tivity o f hydrogen is mu ch lower (2.2) . This large d ifferen ce in elec tr o negativity m ean s th at N - I 1, 0 -1 I , and F- I I bo nd s are ,'ery pol ar. In bonds bet ween II and N, 0 , or F, the more elect ro negati ve elerncn t tak es on a sign ifica n t negati ve charg e (sec Figu re 8.11), an d th e hyd rogen ato m acq uires a sign ifica n t positive cha rg e . T here is an unusu all y stro ng attractio n be tween all e lec tro nega tive a to m with a lon e pai r of el ectrons (m os t o fte n , an N', 0 , or F ato m in ano the r molecul e o r even in th e sa me mol ecul e ) and th e hydroge n a to m of th e N - II, 0 -11 , or F-II bon d . Thi s type o f in teraction is kn own as a hydrogen bond. Hydro gen bo nd s are an extre me fo rm o f di pol e-dipol e interaction whe re on e ato m invo lved is a lways H and th e other a to m is h ighl y e lectro negative, most o ften 0 , N, o r F. A h yd rogen bo nd ca n be re p rese n ted as
X- H
Y- H
T he hydrogen ato m beco mes a bri dge bet ween th e two e lectron egative ato ms X an d ' : an d the das hed line re presents th e hydrogen bon d . T he mo st pron oun ced effec ts of hydro gen bonding occu r where both X a nd Y are N, 0 , or F. En ergies associate d with mos t hydrog en bond s involvin g these el em ents arc in th e ran ge o f " to 30 q /m ol. Types of Hydrogen Bonds [X-H - - - :Y] N-H---,N-
O- H - - - ,N-
F-H---, N-
N-H - - - ,0-
O- H - - - ,0-
F-H - - - ,0-
N-H---,F-
O- H - - - ,F-
F-H - - - ,F-
Hyd ro gen bo ndi ng has important im pl ication s fo r an y property of a co mpo u nd th a t is in flu en ced by intermolecular force s of a ttracti o n . Fo r e xam p le, h ydrog e n bonding aff ects the str uc tu res o f m ol e cul ar so lids . In so lid ace tic aci d , CH 3C0 2 H , for e xa m p le, two mol ecules arc j o in ed to one ano ther by hydro gen bondin g (Figu re 12.7) .
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The Effect of Hydrogen Bonding FIGURE 12.7 Hydrogen bonding. Two acetic add molecules can interact through hydrogen bonds. This photo shows partly solid glacial acetic add. Notice that the solid is denser than the liquid, a property shared byvirtually all substances, the notable exception being water. 562
Chapter 12
I
ProbLem Ethanol, CH 3CH 20H, a nd dimethyl ether, CH 30CH 3 , have the same formula but a different arrangement of atoms. Predict which of these compounds has the higher boiling point.
Inte rmolecula r Forces and Liquids
Ethanol, CH1CH,OH
Strategy Inspect t he st ructure of each molecu le to decide whether each is pola r a nd, if pola r, whether hydroge n bonding is possib le. Solution ALt hough these two compou nds have identi caLmasses, they have differe nt structures. Etha nol posse sses an O- H group, and a n electrostatic potential surface i n t he ma rgi n shows it to be polar with a partia lly neg ativ e 0 atom and a parti ally negati ve H atom. The resu lt is th at hydrogen bond ing between etha nol molecules is possible an d makes an importa nt contrib ution t o its i ntermo lecular forces .
( H,(H2 - O:. -H- O:
I
H
I
( H2(H ,
hydrogen bonding i n ethanol, CH3CH 20H
In cont rast. dimethyl eth er, altho ugh a polar molecule, prese nts no o pportu nity for hydrogen bo nding because there is no O-H bond. The H at oms a re atta ched to much less electronegative ( atoms . We ca n predict, th erefo re, t hat intermo lecula r forces will be larger in et hanol tha n in dimethyl et he r and th at et han ol will have t he high er boiling point . Indeed, et hanol boils at 78.3 " C. whereas dimethyl ether has a boilin g point of - 24.8° (, more th a n 100 O( lower. Unde r sta nda rd condi tio ns, dimethy l et her is a
Polar O-H Bond
Electrostatic potential surfaces for ethanol (Left) and dimethyl ether (right) . The surface for eth anol clearly shows the polar O-H bond. The 0 atom in dimethyl ether has a partia l negative charge, but the re is no H ato m attached. [Color coding: Red indicates a region of largest negative charge. Colors from yellow to green to t urquoise indicate increasing positive charge (or decreasing negative charge). Blue indicates a region of partial positive charqe.]
gas, whereas et han ol is a liquid.
EXERCISE 12 . 2
Hydrogen Bonding
Using struc tural formu las, describe the hydroge n bo ndi ng between met hanol (CH3 0H) molecules. What physical properties of metha nol are likely to be affected by hydrogen bonding?
Hydrogen Bonding and the Unusual Properties of Water On e o f th e mo st str iking d ifferences bet ween o u r pl an et and others in o u r so lar syste m is th e presen ce of lar ge amou n ts of wate r on Ea rt h . T h ree fourths of th e pla net is cove re d by o ceans; th e polar reg ions are vast ice fields; an d eve n so il an d roc ks hold large amounts of water. Altho ug h we tend to tak e wate r for gra nted , almost no o t hc r su bsta nc e behaves in a sim ila r man ne r. Wat er's uniqu e featu res reflec t th e ability ofl 12 0 mol ecul es to cli ng ten aciously to one anot her b y h ydro gen bo ndin g. One re ason for ice 's unusual stru ctu re and wate r 's unusu al p rop erties is that eac h h ydr ogen ato m o f a wate r mo lecu le can fo rm a h ydr ogen bond 10 a lon e pai r of elec trons on the o xygen ato m of an adjacen t water molecule. In ad d itio n , because th e ox ygen ato m in wate r has two lon e pairs o f el ectrons, it ca n form two more hydrogen bonds with h ydrogen ato ms from adjacen t molecul es (Figu re 12.841 ) . The resu lt, see n part icul arl y in ice , is a tet rah ed ral arra ng em en t fo r th e hydrogen at o ms aro u nd eac h o xygen , in vol ving two cova len tly bonded hydrogen ato ms and two hydrogen-bonded hydroge n ato ms. As a co nse q ue nce of th e regul ar arrange ment of wate r molecul es link ed by hydroge n bond ing, ice has an open-cage structu re with lots o f empty space (Figu rc 12.8b ) . The result is th at ice has a d en sity abo u t 10% less th an that of liqu id wate r, whic h explains why ice floats. (In co n tras t, virt ua lly all other so lid s sin k in the ir liquid phase. ) \Ve can also see in thi s structu re th at th e oxygc n ato ms are arra nged at th e co rne rs of pucke red , he xago nal ri ngs . Snow flake s are always based on six-side d figu res (... pag e 69 ), a reflectio n o f th is int ern al mol ecul ar struc tu re of ice . \Vhen ice melts at 0 °C, th c regul ar structu re imposed o n th e so lid state by hyd rogen bon din g break s down , and a relatively large increase in d en sity occu rs (Fig u re 12.9). Anot her su rp rising thing occu rs whe n t he te m peratu re of liquid water is ra ised fro m 0 °C to 4 °C: T he d en sity of water increases. For almo st c\'e ry o the r su bsta nce known , d ensity d ecr eases as the temperature is ra ised . O nce aga in , hydrogen bond ing is th e reaso n fo r wate r' s see m ing ly odd beh avior. At a tempera12 .2
I
• :• 01• • •
Hydrogen bonding in wat er. (above) Water readily forms hydrogen bonds. (below) An electrostatic potential surface for two water molecules shows the hydrogen bond invoLving t he negatively charged a atom of one molecule and the positively charged H ato m of a neighboring molecule.
Intermolecula r Forces In volving Pola r Motecu les
563
FIGURE 12 .8 The str uct ure of ice. (a) The oxygen at om of a water molecule attaches its elf to two ot her water molecules by hydrogen bonds. Noti ce th at the four grou ps t hat surrou nd an oxygen ato m are a rrang ed as a distorted tetra hedron. Each oxygen ato m is covalently bonded to two hydrogen ato ms a nd hydrogen bonded to hydrogen atoms from two other molecules. The hydrogen bonds a re longer t han the covalent bonds. (b) In ice, the structural unit shown in part (a) is repeated in t he crystalline lattice. This compute rgenerated st ructure shows a small portion of t he exte nsive lattice . Notice t he sixmember, hexagonal rings. The corners of each hexagon are 0 atoms, and each side is composed of a normal O-H bond and a slightly longer hydrogen bond.
1.0000 0.9999
E
0.9998
~ 0.999 7 1 - - - - - + - - - - - -, .~ Q
0.9180
Ice
0.9170
L--;'---;6---:'---;'-:O---::'-':-6:-8:-::1O:-Temperat ure (0C)
Active Figure 12.9 The tem perature dependence of the den sities of ice and water.
ChemistryANowT~ Sign in at www. thom sonedu~comjlogin a nd go to th e Chapter Contents men u to explore an interactive version of this figure accompanied by an exerdse . 564
Chapter 12
I
o H
:/
Hydrogen bond
••
(al
ture just above th e me lting poi n t, some of th e water mol ecules co ntinue to cluster in ice-like arra ngements, whic h req uire extra spa("C. As the te m pe rature is raised from 0 "C to 4 "C. the final vestige s of the ice str ucture d isappear, and th e volum e co n tracts fu rt her, giving rise to th e increase in d en sity. Wat er 's d ensity reac hes a max imum at abo ut 4 "C. From th is poi nt , the de nsity d ecl in es with increas ing tem perature in th e normal fashio n . Becau se of th e way th at water's density changes as the tem peratu re app roache s the freezing point , la kes do not freeze so lid ly fro m the bo ttom u p in th e win ter. \ Vhen lak e water co o ls with the ap proach ofwinter; its density increases, the cooler water sinks, and th e warmer water rises. T his "tu rn oye r " process con tinues u ntil all the water reac hes 4 "C, th e maxiruuru d en sity. (T his is th e way ox yge n-r ich water moves 10 the lake bo tto m to restore the oxyge n us ed d uring the summer an d n ut rients are b rought to the to p laye rs of the lake .) As the tem peratu re decreases furt her, th e co lder wate r stays o n the to p of the lak e , be cause wate r cooler than 4 °C is less de nse th an water at 4 "C. \ Vi th further heat lo ss, ice can then beg in to for m on th e su r face , float ing there and p rotec ting the un derlying water and aq uatic life from fu rt her heat lo ss. Extensive hydrogen bo n d ing is also t he origin of the extraordi narily high heat ca pa city o f water. Altho ug h liq ui d wate r d o es 110 t have the reg ula r struc tu re of icc, h yd ro ge n bondi ng sti ll occu rs. Wit h a rise in tem peratu re, th e extent of hydrogen bon ding di minish es . Disrupting hydrogen bo nd s re q uires energy. T h e high heat ca pacity of water is, in large pa rt , why oceans a nd lake s have such a n enormous effect on weat he r. In au tum n , whe n the te mpe ratu re of the air is lower than th e tem perat u re of t he ocean or lak e , the water transfers energy as hea t 10 the atmosp he re , moderating the d rop in air tempe ratu re . Fu rthe rmore , so mu ch energy is avai la ble to be tra nsferred for each degre e d rop in te mpe ratu re th at the d ecline in water tem pera ture is grad ual. For this reason , th e tem pe rature of the ocean or of a large lake is ge nerall y hi gh er than the average ai r te m perat ure unt il lat e in t he autu m n .
ChemistrY-9- Now'" Sign in at www.t homsonedu.comjlogln and go to Chapter 12 Contents to see Screen 12.7 to view an animation of the transformation of ice to water and for a table listing all of the unusual properties of water.
Inte rmolecula r Forces an d Liquids
A Closer Look
Hydrogen Bonding in Biochemistry
It is arguable that our world is what it is because of hydrogen bonding in water and in biochemical systems. Pe rhaps the mostimportant occurrenceis in DNA and RNA where t he
organicbases adenine, cytosine, guanine, and thymine (in DNA) or uraci l (in RNA) are attached to sugar-phosp hate chains (Figure A). The chains in DNA are joined by the pairing of bases, adenine with thymine and guanine with cytosi ne. Figure B illust rates the hydrogen bonding between adenine and thymi ne. These models show that t he molecules naturally fit together to form a six-sided ring, where two of the six sides involve hydrogen bonds. On e side consists of aN· · · H- N grouping, and the other side is N-H . . . O. Here, t he electrostatic potential surfaces show t hat the Natoms of adenine and the 0 atoms of thymine bear partial negative charges, and the H atoms of the N- H groups bear a positive charge. These charges and the geometry of t he bases lead to t hese very specific interactions. Th e fact that base pairing t hrough hyd rogen bonding leads to t he joining of the sugarphosphate chains of DNA, and to the double helical form of DNA, was first recognized by James Watso n and Francis (rick on t he basis of experimental work by Rosalind Franklin and Maurice Wilkins in the 1950s. It was this development that was so important in the molecu lar biology revolution in the last part of the 20th century. See page 392 for more on t hese scientists.
Cytosine
FIGURE A Hydrogen bonding in DNA. With the four bases in DNA, the usual pairings are adenine with thymine and guanine with cytosine. This pairing is promoted by hydrogen bonding.
FIGURE B Hydrogen bonding between adenine and thymine. Electrostatic potentialsurfaces showthat the
polarN-H bond on one molecule can hydrogen bondto an electronegative Natom in a neighboring molecule.
12.3
Intermolecular Forces Involving Nonpolar Molecules
Ma rr y im porta nt m olecul es such as O ~h N 2, an d th e halo gen s are not pol ar. \Vhy, the n , does O 2 d isso lve in polar water? \Vhy ca n the N 2 of the atmos p he re be liquefied (see Fig ure 12.1)? So me inter molecul ar fo rces m ust be ac ting be twee n O 2 and water and bet ween N 2 mo lecu les, but what is th eir nature?
Module 17
• Van der Waals Forces The name "van der Waals forces" is a general term applied to attractive intermolecular interactions. (P. W. Atkins: Quanta: A Handbook of Concepts, 2nd ed.• p. 187. Oxford. Oxford University Press, 2000.)
DipoLe/Induced DipoLe Forces Polar m olecul es suc h as water can ind uce , or cr eat e , a di pole in mol ecul es that d o n o t have a pe rm an en t dip ole. To see how th is situ at ion ca n o cc u r, picture a polar water mol ecul e ap p roach ing a nonpolar mo lecu le suc h as O 2 (Figu re 12.10 ) . T he Sign in at www.thomsonedu.comj login to download the Go Chemistry module for this section or goto www.ichapters.com to purchase modules.
565
FIGURE 12 .10 Dipole/ induced dipoLe int eraction. (a) A polar molecule such as water can induce a dipole in nonpolar Oe by distorting t he molecule's electron cloud. (b) Nonpolar 12 dissolves in polar ethanol (CzH,OH). The intermolecular forc e involved is a dipole/ induced dipole force.
(a) The dipole of water induces a dipole in O2 by distorting the O2 electron cloud.
(b)
Polar ethanol (C2HsOH) induces a dipole in nonpolar 12 ,
The Solubility
TABLE 12.3
of Some Gases in Water* Molar Mass (g/mol )
Solubility at 20 DC (g gas/100 9 water)'
2.01
0.000160
H, N,
28.0
0.00190
0,
32.0
0.00434
* Data
ta ken fro m]. Dean: L (/ llge :~ I/rl1ulhoo!l oJ Otem i.l/ry. 141h Ed., pp . 5 .3-5 .8 , :-.J ew York, .'.I e-C raw-H ill,
1992.
t
Me asu re d under co n ditio ns whe re pressure o f gas + p re ssure o f water vapo r = 760 mill H g.
electron cloud of an isolat ed (gase o us) O 2 mol ecule is sym metrically di stributed between the two o xygen a to ms. As the neg at ive end of th e polar 1120 molecule approache s. however, th e O 2 electron cloud be co mes di sto rted. In thi s process, the O 2 mol ecul e itse lf be comes polar; that is, a dipo le is induced in the oth erwise no npo lar O 2 molecul e . The resu lt is that H 20 and O 2 molecu les are now att racte d to one an ot her, albe it on ly weak ly. O xygen can d isso lve in wat er bec ause a force o f att raction exists between water's pe rman ent d ipole a nd the induced dipole in O 2 • Che m ists refer to such interactions as dipol e/induced dipole intera ctions. The p ro cess o f induci ng a dipo le is callcd polarization, and the d egree to whic h the electron cloud of an ato m or a molecule can be di storted d epends o n the polari zabiIity of th at ato m or mo lecu le. The electron cloud of all ato m o r molecule with a large, ex tended electro n clo ud, suc h as 12> can be polarized more read ily th an the elec tr o n clo ud in a much sm aller a to m o r m o lecul e, suc h as lie o r 112, in which the valen ce e lec tro ns are close to the n ucle us and more tig ht ly hel d. In general, for an ana logous series of su bsta nces, say t he halogen s o r alka ne s (su ch as CJI .j" C 2H jj , C 3I1R, and so o n ) , the higher the molar mass, the gu:,ater the jHJlanwbility oj the molecule.
Enthalpies of Vaporization and Boiling Points of Some Nonpolar Substances TABLE 12.4
J. vapH
Element/ Compound 8P (0C)
o
(kJ/mol)
N,
5.57
- 196
0,
6.82
- 183
8.2
- 161.5
29.96
+ 58.8
(benzene)
30.7
+ 80 . 1
I,
41.95
+ 185
CH,
(methane) 8r,
H
(6 6
566 C h a p t e r 12
I
T he so lu bilities of co mmo n gas es in water illus trate the effec t of interactio ns between a dipole and an ind uc ed dipole. In Table 12.3, yo u see a tren d to hig her so lu bility with in creasing mass of the no n pol ar ga s. As the m olar mass of th e gas increases, th e polarizabi lity of the electro n cloud incr eases and th e stre ng th o f the dipo le/induced dipole in te raction increases.
London Dispersion Forces: Induced Dipole/Induced Dipole Forces Iod in e , Iz, is a so lid and no t a ga s aro u nd room te mperat ure a nd pressure , illu strating that no n polar mo lecu les mu st also ex perie nce intermo lecular for ces. An estimate of these fo rces is provided by the e n thalpy of vapo riza tio n of the sub stance at its bo ilin g point. The d at a in Table 12.4 suggest that these for ces ca n range from vcry weak (N z, 0 z, and CH 4 with low c n thalpics of vapo riza tio n and very low boiling points) to more su bsta n tial (12 and benzene ). To understand ho w two nonpol ar mo lecu les ca n att rac t eac h ot her, re call that the el ectro ns in atoms o r mo lecul es are in a sta te o f con sta nt mo tion . \Vhe n 1:\\'0
Intermolecular Forces and Li quids
Two nonpolar atoms or molecules (depicted as having an electron cloud that has a time-averaged spherical shape).
Momentary attractions and repulsions between nuclei and electrons in neighboring molecules leadto induced dipoles.
Correlation of the electron motions between the two atoms or molecules (whichare now dipola r) leads to a lower energy and stabilizesthe system.
FIGURE 12.11 Induced dipole interactions. Momentary attractions and repulsions between nuclei and electrons create induced dipoles and lead to a net stabilization due to attractive forces.
at oms or nonpol a r mole cules approac h each o ther, attractions o r repul sion s between the ir elec trons a nd nucl e i ca n lead to disto rt io ns in th e ir elec tron clouds (Figu rc 12.11 ) . T ha t is, dipol es ca n be induced momenta rily in n eighboring a to ms o r mol ecules, a n d these in d uc ed dipol es lead to intcrmolecul ar a tt ract io ns. T he inte rm ol ecular fo rce of a ttraction in liquids a n d so lids co m p osed of n o npol ar m olec u les is a n induced dipole /induced dipole force . C he m ists o fte n ca ll th cm London dispersion forces. London di sp ersion fo rces ac tu a lly a rise between all m ol ecules, both n onpola r an d p ol a r, but London dispersion forces are the only intermolecular forces that allow nonpolar molecules to interact.
A Closer Look Hydrogen bonds involving water are also responsible for the structure and properties of one of the strangest substances on earth (Figure). When methane (CH4 ) is mixed with water at high pressures and lowtemperatures, solid methane hydrate forms. Although the substance has been known for years, vast deposits of methane hydrate were only recently discovered deep within sediments on the floor of Earth's oceans. How these were formed is a mystery, but what is important is their size. It is estimated that global methane hydratedeposits contain approximately 1013 tons of carbon, or about twicethe combined amount in all known reserves of coal, oil, and natural gas. Methane hydrate is also an efficient energy storehouse; a liter of methane hydrate releases about 160 liters of methane gas. But of course there are problems to be solved. One significant problem is how to bring commercially useful quantities to the surface from deep in the ocean. Yet another is the possibility of a large, uncontrolled release of methane. Methane is a very effective greenhouse gas, so the release of a significant quantity into the atmosphere coutd damage the earth' s climate. Among the many sources of information is: E. Suess, G. Bohrmann, J. Steinert. and E. lausch, Scientific American, November 1999, pp. 76-83.
Methane Hydrates: An Answer to World Fuel Supplies?
(a) Methane hydrate bumsas methane gas escapes from the solid hydrate.
(b) Methane hydrate consists ofa
lattice of watermolecules with methane molecules trapped in the cavity. Methane hydrate. (a) When a sample is broughtto the surface from the depths of t he ocean, the methane oozes out of the solid, and the gas readily burns. (b) The structure of t he solid methane hydrate consists of methane molecules trapped in a lattice of water molecules. Each point of the lattice shown here is an 0 atom of an H20 molecule. The edgesare O-H -O hyd rogen bonds. Such structures are often called "clathrates." (For more on methane hydrates , see pages 259-260. )
12 .3
I
Intermolecular Forces Involving Nonpolar Molecules
567
Intermolecular Forces Problem Suppose you have a mixt ure of solid iodine, 12, and the liquids wate r and carbon tet rachloride (CCL.) . Whatintermolecular forces exist between each possible pair of compounds? Describe what you might see when t hese compounds are mixed. Strategy Fi rst , decide whether each substanc e is polar or nonpolar. Second, determine the types of intermolecular forces t hat could exist between the different pairs. Finally, use t he "like dissolves like" guideline to decide whether iodine will dissolve in water or CCl and whether CCl4 will dissolve in water.
I c
"
~
~
Br2
Induced dipole/induced dipole forces. Br2 (left) and 12 (right) both consist of nonpolar molecules. They are a liquid and a solid, respectively, implyi ng that there are forces between the molecules sufficient to cause them to be in a condensed phase. These forces between nonpolar substances are known as London dispersion forces or induced dipole/i nduced dipole forces.
SoLution Iodine, 12< is nonpolar. As a molecule composed of large iodine at oms, it has an extensive electron cloud. Thus, the molecule is easily polarized, and iodine could inte ract wit h water, a polar molecule. by dipole/ induced dipole forces. Carbon te trachloride. a tet rahedral molecule, is not polar (see Figure 8.15). As a consequence. it can inte ract with iodine only by dispersion forces. Water and CCl4 could int eract by dipole/ induced dipole forces. but t he interaction is expected t o be weak. The photo here shows t he result of mixing t hese t hree compounds. Iodine does dissolve to a small exte nt in water t o give a brown solution. Whe n thi s brown solution is added t o a test t ube containing C([4, the liquid layers do not mix. (Polar water does not dissolve in nonpolar C([4') (Notice the more dense CCl4 layer [d = 1.58 q/m l.] is underneath the less dense water layer.} Whe n the tes t tub e is shaken. however, nonpolar Ia dissolves preferentially in nonpolar CCL., as evidenced by the disappear ance of t he color of 12 in t he water layer (top ) and t he ap pearance of the purple 12 color in t he CCl4 layer (bottom).
Non polar 12 Polar H20
Nonpolar CCl4 - __~
EXERCISE 12 .3
Shakethe test tube
IntermoLecular Forces
You mix water, ( ( l4' and hexane (CH1CH2CH2CH2C H2( H,) . What type of intermolecular forces can exist between each pair of t hese compounds?
Summary of IntermoLecuLar Forces Geckos use intermolecular forces! A little gecko can climb vertically 1 m up a polished glass surface in 1 s. Geckos have millions of tiny hairs or setae on their feet, and each setae ends in 1000 or more even tinier hairs at the tip. Recent research has found that qeckos are unique in that they adhere to a surface through van der Waals forces of attraction between the hairs and the surface. (K. Autumn , "How gecko toes stick." American Scientist Vol. 94, pages 124- 132, 2006.) 568
C h a p t e r 12
I
Interm olecular forces involve molecules th at are polar o r tho se in whic h polarity call be induced (Ta ble 12.5) . Fu rt he rmore , seve ra l type s of int erm olecul ar forces ca n be at work in a sing le type of mo lecu le (Figu re 12.12 ). Also no te in Figure 12.12 th at whi le eac h ind ividual ind uced dipole/induced d ipo le force is usua lly quite small, th e su m o f th ese fo rce s oyer th e e n tire structu re o f a mo lecu le can act ua lly be q uite great, even in polar molecul es.
Sign in at www.thomsonedu .com/Ioqtn and go to Chapter 12 Contents to see Screen 12.5 to view an animation of induced dipole forces and for an exercise and tutorial on intermolecular forces.
Intermolecular Forces and Liquids
Summary of IntermoLecular Forces
TABLE 12.5
Approximate
Energy (kJ/mol)
Example
Type of Interaction
Factors Responsible for Interaction
Ion-d ipoLe
Ion change, magnitude of dipole
40-600
Na 1"
H20
Di pole moment (depends on ato m electronegativities and molecular
20-30
H20
CH,OH
Dipole-dipo le
st ructu re ) Hydrogen bonding,
Ve ry polar X-H bond (where X = F, N, 0) and atom Ywith lone pair
X- H . . . :Y
of e lectro ns
Dipole/ induced dipole
Dipole moment of polar molecule and polarizability of nonpolar molecule
Induced dipole/ induced dipole (London dispersion forces)
Polariza bility
5-30 2- 10 0.05-40
. Intermolecular Forces Problem Decide which are the most important intermolecular forces involved in each of the followi ng, and place t hemin order of increasing st rength of interaction: (a) liquid met hane. CH 4 ; (b) a mixture of water and methanol (CH 30H) ; and (c) a solution of bromine in water. Strategy For each molecule, we consider its structure and decid e whether it is polar. If polar. we consider the possibility of hydrogen bonding . SoLution (a) Methane is a covalently bonded molecule. Based on the Lewis structure, we can conclude t hat it mu st be a tetrahedral molecuLe and t hat it cannot be polar. The onLy way methane molecules can interact with one another is t hrough induced dipole/i nduced dipole forces.
50
45
1-;::=============:: :;-- - - - ---, • Dipole-dipole force o Induced dipole/ induced dipoleforce
o Di pole/ induced dipoleforce o Total force
40 35
25 -
20
-
15 10
- -
,- -
5 I-
0
Ar
CO
HI
HBr
HCl
NH,
H20
FIGURE 12.12 Energies associated wit h intermolecular forces.
12 .3
I
Inter molecular Forces Involving No npolar Molecules
569
(b) Both water and methanol are covalently bonded molecules; bot h are polar; and both have an O- H bond. They therefore interact t hrough the special dipole-d ipole force called hydrogen bonding, as well as by dipole-dipole and London forces.
' -H
<s0, :
" HI
"H-
,O
s and
I CH,
H'-' -
I
0
"H-
H, C
,O
I
H"
(c) No npolar molecules of bromi ne, Br2, interact by induced dipole forces, whereas water is a polar molecule. Therefore, dipole/ind uced dipole forces (and London forces) are involved when Br2molecules interact with water. (This is similar to the I2- et hanol interaction in Figure 12.10.) In order of increasing strength, t he likely order of interactions is liquid CH 4 < H20 and Br2 < H20 and CH 30H
EXERCISE 12.4 IntermoLecular Forces Hydrogen bonding involving methanol (CH3 0H) and water.
Decide whi ch type of intermolecular force is involved in (a) liquid 02; (b) liquid CH,OH ; and (c) N2 dissolved in H20. Place the interactions in order of increasing strengt h.
12.4
Properties of Liquids
o r the th ree states of matte r, liquids are th e most diffi cul t to d escribe precisely. T he mo lecu les in a gas un d e r normal con di tio ns are fa r apa rt and m ay be co ns idered mo re o r less inde pe nde n t of o ne another. T he stru c tu res o f so lids ca n be d escr ibed read ily because the part icles tha t m ake up so lids-ato ms , molecules, o r ion s-are clo se together a nd are usu ally in a n o rde rly arrangeme n t. T he particles of a liqu id in te ra ct with th ei r neighbo rs, like th e part icles in a so lid, but , u n like in so lids , there is littl e lo ng-range order in their arra nge me n t. In spite of a lack of precision in d escri bin g liquids, we ca n still co ns id er the be havio r of liquids at th e mo lecul ar level . In th e fo llowi ng sec tio ns, we will lo o k furthe r a t th e p ro cess of vapo rizat io n , at th e vapor p ressu re o f liq uid s, at their bo iling po in ts and cr itica l p ro perties, an d at the behavio r th at results in th eir su rfac e te nsio n , capilla ry act io n , and viscos ity. Vaporization and Condensation Vaporizati on or eva poratio n is the process ill wh ich a su bsta nce in th e liquid sta te beco mes a gas. In thi s process, molecul es escape from th e liqu id su r face and e n ter the gase o us sla te. To u nd erstand eva po ra tio n , we have to look at molecular e ne rg ies. Molecu les in a liqui d have a ran ge o f e ne rg ies (Figu re 12.13) that clo sely resembles th e distr ibutio n of e nergies for molecules o f a gas (see Figure 11.14) . As with gases, th e average ene rgy for molecu les in a liqui d d ep end s o nly O il tem pe rature : T he high er the temperature , th e high er th e ayerage energy and the greate r th e relative num ber o f mo lecu les with high kin etic ene rgy. In a sample of a liqu id, at least a few mo lecules have more kinet ic eller6'!' than the pote nt ial energy of the in termolecul ar attrac tive fo rces hold ing th e liquid mo lecu les to o ne ano ther. If th ese high-en ergy mo lecu les are at th e su r face o f the liqu id and if the y are moving in the rig ht d ire ction , they ca n b reak free o f thei r ne ighb o rs and en te r the gas p hase (Figu re 12.14) . Vap o rization is an endothe rmic process becau se ene rgy mu st be added to the system to overco me the intermolecu lar fo rces of attractio n hold ing the molecul es togeth er. T he en er6l)' requi red to vaporize a sam ple is often given as the stand ard 570
C h a p t e r 12
I
Intermolecular Forces and Liquids
D Numberof molecules havingenough energy to evaporate at lower temperature, T1
FIGURE 12.13 The distribution of energy among molecules in a liquid sample. T1 is a higher temperature than Ti. and at the higher temperature. there are more molecules with an energy greater than E.
D Number of molecules +
havingenough energy
D to evaporate at higher temperature, T2
Energy _
molar en thalpy of vaporization, A,.apW (in units of kilojoules per mole; see Tabl es 12.4 and 12.6 and Figure 12.4).
Liquid
vaporization heat energyabsorbed by liquid
Vapor
a va pW = mola r heat of va porizati on
A m olecu le in the g as phase ca n transfe r so me o f its kin e tic energy by co llid ing with slower gaseous molecul es and so lid obj ects. If thi s molecu le loses su llicie n t en e rgy a nd co mes in co n tac t with the su rface o f the liquid. it ca n re e n ter the liqu id ph ase in tile p rocess ca lled condensation .
Vapor
condensation heat energy released byvapor
Liquid
Co nd e nsatio n is th e reverse of vapo rizatio n . Con de nsa tio n is exo the rm ic, so e nc rgy is transfe rred to the su rr o u nd ings. 17u~ entha ljJ)' change for condensation is equa l but ojJjJOsite in sign to the enthaljJ)' of oapmization. For examp le , t he enth alp y change for the vapo riza tio n of 1.0 0 mol of wate r at 100 °C is + 40. 7 1-:.1. O n co nde nsing 1.00 mol of wa te r vapo r to liqu id water at 10 0 °C, the e n thalpy change is - 40 .7 1-:.1. In tile d iscussio n of int erm olecul ar fo rces, we pointed o u t the relati onsh ip between the AvapH o values fo r various su bstanc es and th e tc mperaturcs a t whi ch th ey boil (Ta ble 12.6) . BOLli prope rti es reflect the attrac tive fo rces be tween particl es in the liq uid . Til e boiling poin ts o f non po lar liquids (e.g., th e hyd ro carbons, atm osp he ric g ases, and the halogcn s) in crease with increasing ato m ic o r mo lecu lar mass,
Vapor
FIGURE 12.14 Evaporation . Some molecules at the surface of a liquid have enough energy to escape the att ractions of their neighbors and enter the gaseous state. Atthe same time, some molecules in the gaseous state can reenter the liquid.
Liquid
12 .4
I
Properties of liquids
57 1
TABLE 12.6 Molar Enthalpies of Vaporization and Boiling Points for Common
Substances" Compound
Polar Compounds HF HCl HBr HI NH, H,O SO,
o
Boiling Point (0C) (Vapor
Molar Mass (g/mol)
(kJ/mol)t
pressure = 760 mm Hg)
20.0 36 .5 80 .9 127 .9 17.0 18.0 64 .1
25.2 16.2 19.3 19.8 23.3 40.7 24.9
19.7 - 84.8 - 66.4 - 35.6 - 33.3 100.0 - 10.0
16.0 30.1 44 .1 58 .1
8.2 14.7 19.0 22.4
- 161. 5 - 88.6 - 42.1 - 0.5
0.08 1.7 6.4 12.6
-
:lvapH
Nonpolar Compounds
CH, (methane) C, H. (et hane) C3 Hij (propane) C4 HlO (butane)
Monatomic Elements He Ne Ar Xe Diatomic Elements H, N,
4 .0 20.2 39 .9 131.3
268.9 246.1 185.9 108.0
252.9 195.8 183.0 0, 188.1 F, ct, - 34.0 Br, 58.8 * Da ta ta ke n from D. R. Litle : Em ir La!Jom lOTJ a mi l ndustrinl Clwmirals, Bo ca Raton , FL, eRe 2.0 28.0 32 .0 38.0 70 .9 159 .8
0.90 5.6 6.8 6.6 20.4 30.0
-
I' n~ s s ,
199:l . t !i,'al-' ! JO is me-asur ed at th e norma l bo ilin g p oi nt o f th e liqui d .
a reflecti on o f increased in te rmo lecu lar di spersion fo rces. Th e alka nes (suc h as methan e ) listed in Ta ble 12.6 show th is tr end clearl y. Simil arl y, th e boi ling points and en th ai pies of vapo rizu tio n o f th e heavie r hyd rogen halid es (HX , whe re X = Cl, Br, and I) increase with in cr easing molecul ar ma ss. Fo r these m olecul es, h yd ro gen bonding is not as impo rt ant as it is in HF, so d isp ersio n fo rces and o rdi nary d ipo le-dipo le forces accou n t for th eir intermolecul a r a ttrac tio ns (see Figure 12.12 ) . Because d ispersion forces becom e in creasingl y impo rtant with increasing mass, the boil ing po ints are in th e order HCl < H Br < HL Also noti ce in Table 12.6 th e very high e n tha lpies of vapo rizatio n of wate r and hydrogen flu oride th at result fro m ex te nsive hydrogen bo ndi ng. ChemislryJJ.Now~
•
Sign in at www.thomsonedu.com/Ioq tn and go to Chapter 12 Contents to see Screen 12.8 to view an animation of the vaporization process and for a table of j,v~p Ho values.
Enthalpy of Vaporization
r. I 572
Chapter 12
I
Problem You put 925 mLof water (about 4 cupsfu l) in a pan at 100 ' C, and the water slowly evaporates. How much energy must have been transfe rred as nea t to vapori ze t he wate r?
Inte rmole cula r Forces a nd Liquids
Stra tegy Three pieces of information are needed to solve this problem: 1. J.V3p W for water
= +40.7 kJ(m ol at 100°C.
2. The density of water at 100 °C = 0.958 g( cm3 • (This is needed because J. V3p W has units of kilojoules per mole, so you first mu st find t he mass of water and then t he amount.) 3. Molar mass of water
= 18.02 g(mol.
SoLution Avolume of 925 mL (or 9.25 X 102 ern') is equivalent to 886 g, and this mass is in turn equivalent to 49.2 mol of water.
g)[
0.958 925 mL -[
1 mL
mOl ) -1 18.02 9
= 49.2 mot H20
Therefore, t he amount of energy required is
40.7 kJ ) 49.2 mol H20 -[ mol
~
2.00 X 10' kJ
2000 kJ is equivalent to about one quarter of t he energy in your daily food intake.
EXERCISE 12.5 Enthalpy of Vaporization The molar enthalpy of vaporization of methanol, CH 30H, is 35.2 kJ(mol at 64.6 "C. How much energy is required to evaporate 1.00 kg of this alcohol at 64.6 °C?
Wat e r is exc ep tio na l among the liq u id s listed in Ta ble 12.6 in t hat an enormous am ount of heat is requir ed to co nve rt liquid water to water vapor. T h is fac t is important to yo u r own p hysical wel l-being. Wh e n yo u exe rcise vigo ro usly, yo u r body responds by sweating to rid itsel f o f the excess heat. Energy fro m yo u r body is tr an sferred to sweat in the p ro cess of evaporatio n , and your body is cooled . Enth alpies of vapo rizatio n and co nde nsation o f wate r also play a role in wea ther (Figu re 12.15) . For exam ple, if enoug h water co nde nses fro m th e air to fall as an inc h of rain on an acre o f grou nd , th e heat released exceeds 2.0 X IOR~J ! T his is equivalen t to abo u t 50 tons of ex plode d dynamite, th e ene rgy released by a small bomb.
Vapor Pressure If yo u put so me wate r in an ope n beak er, it will eventually eva po ra te co m ple te ly. Air movemen t an d gas di ffusio n re m ove the water \'ap or from th e vicin ity of th e liq uid su rface, so many water m o lecu les are no t ab le 10 re turn to the liquid . If yo u pu t wate r in a sealed flask (Figu re 12.16) , ho wever, th e wate r vapor can no t escape, and so me will reco nde nse to form liq uid wate r. Even tua lly, the masses of liq uid and of vapor in the flask remain co ns ta nt. T hi s is an o the r exam ple of a dynami c equi librium (.... page 1\ 9 ) . Liquid
Va por
Mo lecul es still move co n tin uo usly fro m th e liq uid p hase to the vapo r p hase an d fro m th e vapor ph ase back to t he liq uid phase . T he ra te at which mo lecules rno ve fr0111 liq u id to vapo r is th e same as th e rate at whic h they I1Hwe from vapo r to liq uid ; th us, th ere is no ne t cha nge in th e masses o f the two p hases. Wh en a liq uid- vapor equi librium has been es ta blishe d, th e eq u ilibr iu m vapo r p ressu re (o fte n just ca lled the vapo r pressure ) can be measured . The equili brium vapor pressure o f a su bsta nce is the p ressure exe rte d by th e vapo r in eq uilibrium with th e liq ui d phase . Co ncep tually, the vapor p ressu re o f a liq uid is a measure of th e tende ncy o f its mo lecu les to escape from the liq uid p hase an d en ter th e vapor
FIGURE 12.15 Rainstorms release an enormous quantity of energy. When wa ter vapor condenses, energy is evolved to the surroundinqs. The enthalpy of condensation of water is large, so a large quantity of ener9Yis released in a rainstorm.
12.4
I
Properties of Liquids
573
Vapor pressure. Avolatile liquid is placedin an evacuated flask (left). At the beginning, no molecules of the liquid are in the vapor phase. Afte r a short time, howeve r, some of the liquid evaporates, and the molecules now in the vapor phase exert a pressure. The pressure of the vapor measured when the liquid and the vapor are in equilibrium is called the equilibrium vapor pressure (right). Active Figure 12.16
ChemislryJJ.Now~ Sign in at www. thomscnedu•.cormloqtn and go to th e Ch apt er Cont ent s menu to explore an inter active version of t his figure accompanied by an exercise.
• EquHibrium Vapor Pressure At the conditions of T and P given by any point on a curve in Figure 12.17, t he pure liquid and its vapor are in dynamic equilibrium . If rand P define a point not on the curve. the system is not at equilibrium. See Appendix Gfor the equilibri um vapor pressures of water at various temperatures.
( EQUILIBRIU M
Time
• Ptot al
Volatile liquid
= Pvapor
Vapor pressure at t emperat ure of measurement
Hg in t ube open t o flask
ph ase a t a given temperature . This tendency is referred to qu alita rivel y as th e vo la tility of the co m po u nd . The hig he r the eq uilib riu m vapor pressure a t a g iven tempe ra ture , th e more vo latile th e su bstance. As d escr ibed p reviously (see Figure 12. 13) , the d istr ibution of mo lec u lar e ne rg ies in th e liq uid p hase is a functio n of tem peratu re . At a hig he r tempe rature, more mol ecu les have sufficie n t ene rgy to esca pe the sur face of t he liquid . The eq uilibriu m vapo r p ressu re must, th e re fo re , in crease with tempe ratu re . It is useful to re presen t vapo r pressu re as a fun ctio n o f te mpe rature. Figure 12.17 shows the vapor pressure curves for severa l liquids as a fun ction o f te m pe rat ure. All j){)ints along the vajHJr {nessure versus temperature furves represen: conditions oj {nessure and tempemture at nihirh. liquid and vaj){)r are in equilibrium. For exam ple , at 60 °C th e vapo r pressure o f wat er is 149 mm H g (Ap pe nd ix G ) . If wate r is placed in an evac ua te d flask th at is mai ntain ed at 60 "C. liq u id wate r will eva po rat e u nt il the p ressure exe rt ed by the wate r vapor is 149 mm JI g (assu m ing en o ug h wat er is in the flask so th at so me liq uid remains when eq u ilibr iu m is reach ed ). ChemislrY.{J.Now~ Sign in at www.t homsonedu.com/ Ioqtn and go to Chapter 12 Contents to see Screen 12.9 to view an animation on equilibrium vapor pressure and for a simulation of vapor pressure curves.
Vapor pressure curves for diethyl et her [(C 2 Hsh O], et hanoL(C2 HsOH), and wate r. Each curve represents conditions of T and P at which the two phases, liquid and vapor, are in equilibrium. These compounds exist as liquids for temperatures and pressures to the left of the curve and as gases under conditions to the right of the curve. Active Figure 12.17
Chemislry..o..Now"" Sign in at www. thom sonedu~ com/Login and go to th e Chapter Contents menu to explore an inter active version of th is figure accompanied by an exercise.
1000
800
I
Normal BP 34.6 O(
760 mm Hg
Normal BP 78.3 O(
or 600
I
O(
E
oS.
•
"•
;J EthanOl! 400
/
~
/
200
~: - 20°
00
20 0
/' 40 0
Temperature (0C)
Chapter 12
100
Ii
0;
0
574
Norma l BP
Intermolecula r Forces and Liquids
60 0
80 0
100 °
120°
· Vapor Pressure Problem You place 2.00 Lof water in an open container in your dormitory room; t he room has a volume of 4.2 5 X l OAL. Yo u seal the room and wait for the water to evaporate. Wi ll all of the water evaporate at 25 ° C? (At 25 O( th e de nsity of wat er is 0.997 gJrn L, and it s va por pressu re is 23.8 mm Hg .) Strategy One a pproach to so lving th is problem is to calculat e th e quantit y of water t hat must evap orate
to exert a pressure of 23.8 mm Hg in a volume of 4.25 X 104 Lat 25°C. We use the ideal gas law for this calculation.
Solution Calculate the amount and then mass and volume of water that fulfills the following conditions: P ~ 13.8 mm H9, V ~ 4.15 X 10' L, T ~ 25 vC (198 K).
P = 23.8 mm H9(
1 atm
760 mm Hg
) = 0.0313 atm
n ~ PV ~ (0 .0313 atm)(4.15 X 10' L) RT
54.4 mol
( 0 .0820 57 L' at m ) (298 K) Kvmo l
54.4 mol H20 980, 9 H,o (
18. 02 9 ) ( 1 mol H 20
1 mL ) 0 ,9979 H,O
= 980. 9 H20
~ 983 mL
Only about half of t he available wat er needs to evaporate to achieve the equilibrium water vapor pressure of 23.8 mm Hg at 25 °C.
EXERCISE 12.6 Vapor Pressure Curves Examine t he vapor pressure curve for etha nol in Figure 12.17. (a) What is the approximate vapor pressure of et hanol at 40 o(? (b) Are liquid and vapor in equilibrium when t he temperat ure is 60 "Cand the pressure is 600 mm Hg? If not does liquid evaporate to form more vapor, or does vapor condense to form more liquid?
EXERCISE 12.7 Vapor Pressure If 0.50 9 of pure water is sealed in an evacuat ed 5.0- Lflask and the whole assembly is heat ed to 60 0(, willt he pressure be equal to or less tha n t he equilibrium vapor pressure of water at this temperature? What if you use 2.0 9 of water? Under either set of conditions, is any liquid water left in the flask, or does all ofthe water evaporate?
Vapor Pressure, Enthalpy of Vaporization, and the Clausius-Clapeyron Equation Plo tti ng the vapor pressure for a liq uid at a se ries of te m pe rat ures re su lts in a curved lin e (Figu re 12.17) . Howeve r, th e German p hysicist R. Clausius (1822- 1888) and the French m an B. P. E. Clapeyro n ( 1799- 1804 ) showed th at, fo r a p ure liquid , a linea r rela tio ns h ip ex ists between th e reciprocal of th e Kelvin tem perature ( 1/ T ) and th e na tural logarithm of vap o r pressu re (In P ) (Figu re 12.18) .
In P = - p."pH"/ RT) + C
(12.1)
H e re , D. \'ap'JO is the e n tha lpy of vapo riza tio n of the liq uid; R is the id eal g as co nstan t (8.3 14472 .Il K · mo l) ; an d C is a consta n t chara cte ristic of th e liquid in qu estion . T hi s equat ion , now called the Clausius-Clapeyron equation, p rovid es a method of o btaining values fo r D.vapl JO. T he eq uilibr iu m vapo r pressu re of a liqu id can be measured at severa l d iffere nt tc m pcraturcs, and the logarithm o f th ese prcssu rcs
OL0.00 25
_ 0.0030
0 .0035
0.0040
l/ rl' - ') FIGURE 12.18 Clausius- Clapeyron equation. When the natural logarithm of the vapor pressure (In P) of water at various temperatures (7) is plotted against liT, a straight line is obtained. The slope of the line equals - j. .. pHoI R. Va tues of T and P are from Appendix G.
12.4
I
Properties of Li quids
575
is p lotted versus 1/7: T he result is a straight Iine with a slo p e of - !1l'ap IlOj R For example , p lotti n g data for water (Figu re 12.18 ), we find the slo pe of th e line is - 4.90 X 10'" which give s "'""pliO= 40 .7 kJ/ mo l. As a ll a lte rnative to p lotting In P vc rslIs 1/ '1 : we ca n writ e th e fo llowin g equation t ha t allows us to calcu late d ,"apH o kn owing th e vapo r p re ssu re of a liquid a t two different temperatures. In P2 - In Pl
=
[
- Ll. yapW RT,
+ c] -
[ - Ll. yapW RT,
+ c]
This can be sim p lified to
lnll
P,
= _ Ll." ,W [~ _ ~] R
T,
T,
(12 .2)
For ex a mp le , et hylen e g lyco l h as a vap o r pressure of 14.9 nun Hg ( PI) at 373 K ( T,) , a nd a vapo r pressnre of 49.1 mru JI g ( P,) at 398 K (7; ) . l (49 .1 mm H9 ) n 14.9 mm Hg 1.192
Ll." pH'
~
~
Ll." pW [ 1 1 ] 0.00 83145 kJ/K' mol 398 K - 373 K
Ll." pW ( 0.0083145 kJ/K ' mol
= 59.0
0.000168) K
kJ/ mol
ChemistrY-9-Now~ Sign in at www.t homsonedu.com/ Ioqtn and go to Chapter 12 Contents to see Screen 12.9 for th ree tutorials on using the Clausius-Clapeyron equation .
~
-
~ .r
~ ; ,," ..
"
..J
. -\....
FIGURE 12.19 Vapor pressure and boiling. When the vapor pressure of t he liquid equals the atmo spheric pressure, bubble s of vap o r begin to fo rm wit hin t he body of liquid, a nd t he liquid boils.
576
12.8 Clausius-Clapeyron Equation
, '.
~
a a
"" ji:;;' . . -I:: ~. ~, : ' ..:.;;{.
EXERCISE
Ca lculate the enthalpy of va porization of diethyl et her, (C2H sh O (see Figure 12.17). This compound has va por pressures of 57.0 mm Hg and 534 mm Hg at - 22.8 "Cand 25.0 "c. respectively.
C h a p t e r 12
I
Boiling Point If you have a beaker of wat er open to the atmosp he re , the atmosp here pr esses d own on th e surface . If e n ough e ne rgy is add ed , a tem perature is eve n tu ally rea ched a t wh ich the vapor pressure of the liq uid equals the atmospheric p ressu re. At th is tem pe ra ture , bu bbl es o f th e liq u id 's vapor will not be cr ushed by th e atmosp heric pressu re . T he bubb les ca n rise to th e su rface , a n d th e liq u id bo ils (Figu re 12.19 ). T h e boiling point of a liqui d is the te mpe ra tu re at wh ich its vap o r pressure is equal to th e ex te rna l pressure. If th e extern al pressu re is 760 rum Hg , this te mperatu re is raile d th e normal b oiling poin t. This point is highligh ted on th e vapor p ressure curves for the su bs ta n ces in Figure 12.1 7. T h e normal bo ilin g point o f wate r is 100 "C, and in a great man y p laces in th e Uni te d Stat es, wat e r bo ils at or near thi s tempera ture . Ifyo u live at hi gh er altitudes , however, suc h as in Salt Lake City, U ta h, wh e re the barome tric pressure is ab o ut 65 0 mm Hg , water will boi l at a n o ticea bly lowe r temperat u re . T he cu rve in Figu re 12.19 shows th at a pressure of 650 mm II g co rrespo n ds to a boiling tempe ra mre of a bo u t 9 5 "C. Food , th e refo re , h as to he cooked a lillie longe r in Salt Lake Cit)' to ac hi eve the sa me re su lt as in Ne w 'ro rk Cit)' at sea level .
Intermolecular Forces and Li quids
ChemiSlry.{)"Now~ Sign in at www.thomsonedu.com/Ioqtn and go to Chapter 12 Contentsto see Screen 12.10 to watch a video of the boiling of a liquid in a partial vacuum and to explore the relationship of molecular composition and structure and boiling point in two simulations.
TABLE 12.7 Critical Temperatures and Pressures for
Common Compounds* Compound
T, ( "e)
P, (atm)
CH, (metha ne)
- 82.6
45.4
Critical Temperature and Pressure
C, H, (eth ane)
32.3
49.1
O n fi rst th ought, it m ig h t see m th at yap o r pre ssure-temperature cu rves (s uc h as shown in Fig u re 12.1 7) sh o u ld co n tin ue upward wit h o u t lim it, hut th is is n ot so . In ste ad , whe n a sp eci fic te mpe ra ture a nd p r essu r e are re ac h ed, th e interface be twee n t he liquid and th e va po r d isa ppea rs. T h is p oint is ca lled th e critical point. T h e te mp erature a t wh ic h th is p he n o me no n occ u rs is the cri tical temp erature, Tn a n d th e corres po ndi ng pressu re is the critical pressure , P; (Figu re 12.20 ) . T he su bst an cc th at ex ists und e r th ese co n d itio n s is ca lle d a supercritical fluid . It is lik e a gas und er suc h a h igh pre ssure th at its d ensity resem bl es th at of a liq u id , wh ile its viscosity (a b il ity to flo w) remai n s close 10 th at of a gas ( ~ page 580) . Consider wha t th e substa nce mig h t lo o k like at the mol ecular level unde r th ese co nditions. T he m ol ecules have been fo rced almost as close togethe r as the y a re in th e liquid sta te, but eac h m ol ecule has e nough kinetic e ne rhy)' to exceed th e forces h ol ding m olecules together. As a result, th e su pe rc ritica l fluid h as a tightly packe d mol ecul a r arra nge ment like a liq u id , but th e in termolec u lar fo rces o f attract io n th at c harac terize th e liqu id stat e are less th an th e kine tic e ner~n' o f th e p articles. Fo r m o st su bstances, t he critical point is a t a very h igh te mp e ra tu re and p ressu re Cra ble 12.7) . Wat er, for ins ta nce, h as a cr itica l tempe rature of 374 °C a n d a cr itica l p ressure of2 17.7 a tm . Su perc ritical fluids can h ave u n e x p ec ted pro perties , suc h as t he a b ili ty 10 d issol ve nor m all y in so lu b le m at erial s. Su pe rc r iti cal CO 2 is es pecia lly u se ful. Car bo n di o xid e is wide ly availabl e , esse n tia lly no nto x ic , n onflam mable , a n d in ex pensive . It is re la tive ly easy to reac h its cri tica l te mp e r atu re o f 30 .99 °C an d critica l p re ssu re of 72 .8 a tm. O ne u sc of su pc rc ri tica l CO 2 is to ex trac t caffeine fro m coffe e . T he co ffee bea ns a rc tre at ed with stea m to bring th e caffei ne to t he sur face . Th e be an s are then im m e rse d in su pc rc ri tica l CO 2 , whi c h se lectively d issol ve s th e caffe ine but le ave s in tact th e co mpou n ds tha t g ive fla vor to coffee. (Dec a ffei n a ted coffee co n ta ins le ss than 3 % of th e orig inal caffeine.) T he so lutio n of ca ffei ne in su p e rc ri tical CO 2 is po ured off, an d the CO 2 is evaporate d , tra p ped, a n d re u se d .
C3 Ha (propane)
96.7
41.9
C4 HlO (but ane)
152.0
37.3
CCl,f, (CfC-12l
111.8
40.9
NH,
132.4
112.0
H2O
374.0
217.7
P, = 217.7 atm
CO,
30.99
50,
157.7
71.8 77.8
*Da ta ta ke n fro m D. R. Lid e : Basic 1-alloI"(1I01)" and I ndustri al Chemicals,
Boca Rato n , FL,
eRe
Press, 1993.
• Green Chemistry and Supercritical CO 2 It is not surprising t hat other uses are being sought for supercriticalCO 2, One application being investigated is its use as a dry cleaning solvent. More than 10 billion kilograms of organic and halogenated solvents are used worldwide every year in cleaning applications. These cleaning agents can have deleterious effects on the environment. so it is hoped that many can be replaced by supercritical CO 2, (for more about supercritical CO 2 see page 609.)
FIGURE 12.20 Criticaltemperature and pressure for water. The curve representing equilibrium conditions for liquid and gaseous water ends at the critica l point; above that temperature and pressure, water becomes a supercritical fluid.
Super critical fluid
--------------------Gas
li quid
: T, = 374.0
O(
Tem perature -
12.4
I
Properties of Liqui ds
577
FIGURE 12.21 Intermolecular forces in a Liquid. Forces acting on a molecule at the surfaceof a liquid are different than t hose acting on a molecule in the interior of a liquid.
Water molecules on the surface are not completely surrounded by other water molecules.
Water molecules under the surface are completely surrounded byother water molecu les.
Surface Tension, Capillary Action, and Viscosity
• Viscosity Long chains of atoms, such as those present in oils, are floppy and become entangled with one another in the liquid; the longer the chain, the greater the tangling and the greater the viscosity.
Mol ecules in th e in ter ior of a liq uid in teract with mol e cul e s all a ro u nd th em (Fig u re 12.2 1) . In cont ras t, m ol e cul e s on the surfac e of a liq uid are affected on ly by th o se m ol e cul e s lo cated a t o r bel o w th e surface layer. T his lead s to a n et in wa rd fo rce of a tt ra ctio n o n th e su r fa ce mol ecule s, co n trac ting th e su rf ace a rea and m ak ing thc liq uid be have as t houg h it had a skin. T he toughness of thi s skin is measu re d by its su r fac e tension- th e e ne rgy requ ire d to b reak th rough the su r face o r to d isru p t a liqu id dro p an d sp read th e m at erial o u t as a fil m . Su rfac e te nsi on causes wate r drops to be sp he res an d n o t littl e cu bes, for ex am p le (Fig u re 12.22a ) , becau se a sp he re has a smalle r su r fac e area th an a ny o t he r sh ape o f th e sa me vo lu me . Capillary acti on is closel y re late d to su r face te nsion . Wh en a small-d iamete r g lass tu be is place d in water, the wate r rises in th e tube , just as water rises in a piece of pa pc r in water (Figu re 12.22b ) . Because polar Si-O bond s are p resent o n th e
FIGURE 12.22
Adhesive and cohesive forces.
, (a) Aseries of photographsshowing the differentstages when a water drop falls. The drop wasilluminated by a strobe light of 5-ms duration. (The total time forthis sequence was0.05 s.) Water droplets take a sphericalshape because of surface tension.
578
Chapt er
12
I
Intermolecular Forces and Li quids
(b) Capillaryaction. Polar water molecules are attracted to the - OH bonds in paper fibers, and water rises in the paper. If a line of ink is placedin the path of the rising water, the differentcomponents of the ink are attracted differently to the water and paper and are separated in a process called chromatography.
(c) Water (top layer) forms a concave meniscus, whilemercury (bottom layer) forms a convex meniscus. Tile different shapes are determined bytile adhesive forces of the molecules of the liquid with the wallsof the tube and the cohesive forces between molecules of the liquid.
Case Study
The Mystery of the Disappearing Fingerprints
The events of September 11, 2001, are etched in everyone's memory. The possibility of domestic terrorism, however, began almost two years before when a man was appre-
hended in late December 1999 at the Canadian border with bomb materials and a map of the los Angeles International Airport. Although he claimed innocence. his fingerprints were on the bomb materials, and he was convicted of an attempt to bomb the airport. Each of us has a unique fingerprint patte rn, as first described by John Purkinji in 1823. Not long after, the English in India began using fingerprints on contracts because they believed it made the contract appear more binding. It was not until late in the 19th century, however, that fingerprinting was used as an identifier. Sir Francis Galton, a British anthro pologist and cousin of Charles Darwin, established that a person's fingerprints do not change over the course of a lifetime and that no two prints are exactly the same. Fingerprinting has since become an accepted tool in forensic science. In 1993 in Knoxville, Tennessee, detective Art Bohanan tho ught he could use it to solve the case of t he kidnapping of a young girl. Th e girl had been taken from her home and driven away in a green car. The girl soon managed to escape from her attacker and was able to describe the car to the police. After four days, the police found the car and arrested its owner. But had the girl been in that car? Art Bohanan inspected the car for her fingerprints and even used the latest technique, fum ing with superglue. No prints were found. Th e abductor of t he girl was eventually convicted on other evidence, but Bo hanan wondered why he had never found her prints in the ca r. He decided to test the permanence of children's fingerprints compared wit h adults. To his amazement, he found that chil-
Taking a fingerprint at the LocaL poLice st ati on.
dren's prints disappear in a few hours, whereas an adult's prints can last for days. Bohanan said, "It sounded like the compounds in children's fingerprints might simply be evaporating faster than adult's." To answer this, you should know about the nature of fingerprints. The residue deposited by fingerprints is 99% water. The othe r 1% contains oils, fatty acids, esters, salts, urea [(NHzhCO], and amino acids. An example of a fatty acid is myristic acid: CH3(CHJ1ZCO ZH. An este r is t he combination of an acid with an alcohol such as et hano l (C H3CH 20H), so an este r of myristic acid would be: CH ,(CH, j"CO,CH,CH, Scientists at Oa k Ridge National Laboratory studied the fingerprints of 50 child and adult volunteers, identifying the compounds present by such techniques as mass spectrometry (~~I
Cr Mn Fe
I- i-:
W Re Os
Primitive
Co
Ir
Ni Cu
C
N
0
~ Si
P
5
Zn Ga In
Ne CI
Ar
Ge As Se Br Kr Sn Sb Te
Pt Au Hg Tl Pb Bi
I
Xe
BA t
Rn
Cubic close packing (Face-cente red cubic)
Body-cente red cubic
identic al ato ms , molecu les, o r io ns at th e corners of th e cu bic u nit cell. Th e bcc and fcc ar ra nge me n ts, however, diffe r from the primitive cube in th at they have additional particles at o the r loc ations. T he bee structu re is ca lled "bo dy-ce n tered " because it has an additional particle , of th e same type as th o se at th e co rne rs, at th e ce n ter o f th e cu be . The fcc arrange men t is ca lled "fac e-cen tered " be cause it has a particle, of the same type as th e co rner atoms, in the ce n te r of eac h of the six faces of th e cu be. Exa m ples of each struc tu re are found amo ng th e crystal lattices of the metals (Figure 13.5). The alkali metals, for exam p le , are body-centered cu bic, whe reas nickel, co p per, and a lu m in u m are face-centered cu bic. No tice that only on e me tal , polon ium , has a pri miti ve cu bic latti ce . Wh en th e cu be s pack together to mak e a th ree-dimensio n al crysta l of a met al , th e atom a t each corner is shared am ong eigh t cu bes (Fig u res 13.3, 13.4, a nd 13.6a ) . Because o f thi s, o nly o ne eig h th o f each corner atom is ac tua lly within a given un it ce ll. Furthermore , be cause a cu be has eigh t co rne rs, and be cau se one eigh th of th e ato m at each co rner "belo ngs to" a particular unit ce ll, th e corner ato ms co ntribute a net of one ato m to a given uni t ce ll. Thus, the [nimitiue cubic arrangement has one net atom within the unit cell.
(8 corners of a cube)(% of each corner atom within a unit cell) = 1 net atom per unit cell for the primitive cubic unit cell
FIGURE 13.6 Atom sharing at cube corne rs and faces. (a) In any cubic lattice, each com er particle is shared equally among eight cubes, so one eighth of t he particle is wit hin a particular cubic unit cell. (b) In a face-centered latt ice, each particle on a cube face is shared equally bet ween t wo unit cells. One half of each particle of thi s type is within a given unit cell. 592
Chapter 13
I
The Chemist ry of Solids
(aJ
(b)
FIGURE 13.7 X-ray crystallography. In t he x-ray diffraction experiment, a beam of x-reys is directed at a crystalline solid. The photons of t he x-ray beam are scatt ered by the atoms of t he solid. The scattered x-rays are detected by a photographic film or an elect ronic detector, and t he patt ern of scatt ered x-rays is related to t he locations of th e atoms or ions in the crysta l.
In co nt ras t to the p rimit ive cu bic lattice , a body-cente red cu be has an ad dit io na l atom who lly wit hin th e unit ce ll at the cu be 's ce n te r. T he cen te r particle is prcsent in ad d itio n to th o se at th e cu be co rners, so the body-centered cubic arrangement has a net of two atoms within the un it rell. In a face -ce n te re d cu bic arra ngemen t, th ere is an a to m o n ea ch of th e six faces of the cu be in additio n to th o se at the cu be co r ne rs. O ne half of eac h ato m o n a fac e be lon gs to a given u nit ce ll (Fig u re 13.6b ) . T h ree net par ticles ar e th erefore co n tr ib u te d by th e particles on th e faces of th e cu be:
(6 faces of a cube)(V, of an atom within a unit cell) = 3 net face-centered atoms within a face-centered cubic unit cell Thus, the f ace-centered cubic arrangemen t has a net offour atoms wilhin the unit cell, o ne co n tr ib u te d by th e cor ner ato ms and anothe r th ree co ntri b u te d by the ato ms ce ntered in th e six faces. An ex pe rime nta l tec h n iq ue , x-ra y crysta llograp hy, can be use d to determin e th e struc tu re of a crystalline su bsta nce (Fig u re 13.7) . Once th e str uc ture is known , th e in form ation ca n be co mbine d with other ex pe ri me ntal inform ation to calcu late suc h use fu l param eters as th e rad ius o f an ato m (Stud y Question s 13.7- 13.10 ) . Chemislry.o.Nowm
•
Sign in at www.thomsonedu.comjlogin and go to Chapter 13 Conten ts to see Screen 13.2 for a self-study module on crysta l lattices.
Determining an Atom Radius from Lattice Dimensions Problem Aluminum has a den sity of 2.699 g/cml , an d t he atoms are packed in a face- centered cubic crystal latti ce. What is the radius of an alu minum at om? Strategy Our st rategy for solvi ng t his problem is as follows: 1. Find the mass of a unit cell from t he knowledge t hat it is face-ce ntered cubic. 2. Combine the density of a lumi num with the mass of t he unit cell t o find t he celt volume.
Aluminum meta l The metal has a facecentered cubic unit cell with a net of four Al atoms in each unit cell.
3. Find the lengt h of a side of the unit cell from its volume. 4. Calculate t he at om radiu s fro m the edge dime nsion. 13 .1
I
Crystal l attices and Unit Cells
593
SoLuti on
1. Calculate the moss cf tne unit cell.
g)(
26.98 Massof 1 Al atom = -( 1 mo l
1 mol
)
6.02 2 X 1023 atoms
= 4.480 X
. cell = ( 4.480 X 10- " 9 ) ( 4 Alato ms) = 1.792 Mass of unit 1 Al atom 1 unit celt
10- 23 gJatom
. X 10- 22 gJumt
cell
2. Calculate the volume of the unit cell. Volumeof untt. cell One face of a face-centered cubic unit cell. This shows the cellface diagonal, J2 x edge, is equalto four times the radius of the atoms in the lattice.
= ( 1.792 X 10- " unit cell
g)(-1 em'--g ) 2.699
= 6.640 X 10- 23
em' J umt. cell
3. Calculate the length of a unit cell edge. The lengt h of t he unit celt edge is t he cube root of t he celt volume. Lengt h of unit cell edge
= V6.640 X
10
23
em!
= 4.049 X
10- 8 ern
4. Calculate the atom radius. Notice in the model of aluminum in the margin (and in Figure 13.4) that t he At atoms at the celt corners do not touch each other. Rat her, t he four corner atoms touch the facecentered atom. Thus, t he diagonal distance across the face of the cell is equal to four times the Al atom radius. Cell face diagonal = 4 X (Alatom radius) The cell diagonal is t he hypotenuse of a right isosceles t riangle, so, using t he Pythagorean theorem, (Diagonal distance)!
=
2 X (edge)2
Taking t he square root of both sides, we have Diagonal distance = ./2 X (cell edge) = ,fi X (4.049 X 10- 8 em)
= 5 .72 7 X 10- 8 em
We divide the diagonal dista nce by 4 to obtain t he Al atom radius in em. .
At atom rad ius =
5. 727 X 10- 8 em
4
= 1.4 32 X 10- 8 cm
Atomic dimensions are often expressed in picometers, so we convert the radius to that unit. 1.4 32 X 10- 8
cm (~) ( 1 X110pm12 m) = 143 .2 pm 100 ern
This is in excellent agree with t he radius in Figure 7.8. EXERCISE 13.1
Determining an Atom Radius from Lattice Dimensions
Go ld has a face-centered unit cell, and its density is 19.3 2 gJcm3• Calcu late t he radius of a gold atom.
EXERCISE 13.2
The Structure of Solid Iron
Iron has a density of 7.8740 gJcm3, and the radius of an iron atom is 126 pm. Verify that solid iron has a body-centered cubic unit cell. (Besure to note t hat the atoms in a body-ce ntered cubic unit cell touch along the diagonal across the cell. They do not touch along the edges of the cell.) (Hint: t he diagonal distance across t he unit cell is edge X 13.)
594
Chapter
13
I
TheChemistry of Solids
A Closer Look
Packing Oranges
It is a "rule" that nature does things as efficiently as possible. You know this if you have ever tried to stack some oranges into a pile that doesn't fall over and that takes up as little space as possibte. How did you do it ? Clearly, the pyramid arrangement below on the right works. whereas the cubic one on t he left does not.
ers repeat t heir pattern in the manner ABA BA B... . Atoms in each A layer are directly above the ones in another A layer; the same holds t rue for the B layers. In the ccp a rra ngement, the atoms of t he "to p" layer (A) rest in depressions in the middle layer (B), and those of t he "bottom" layer (C) are oriented opposite to t hose in t he top layer. In a crystal, the pattern is repeated ABCABCABC. . . . By t urning the whole crystal, you can see that t he ccp arrangement is t he face-cente red cubic structure (Figure 1).
(b) Cubic close-packing =- face-centered cubic
(a) Hexaqonal close-packing (hcp)
Top
layer
If you could look inside the pile, you would find that less open space is left in t he pyramid stacking than in the cube stacking. On ly 52% of the space is filled in t he cubic packing arrangement. (If you could stack oranges as a body-centered cube, that would be slightly better; 68% of the space is used.) However, the best method is the pyramid stack, which is really a face- centered cubic arrangement. Oranges, atoms, or ions packed this way occupy 74% of the available space. To fill t hree-dimensional space, the most efficient way to pack oranges or atoms is to begin with a hexagonal a rrangement of spheres, as in this arrangement of marbles.
Succeeding layers of atoms or ions are then stacked one on to p of t he other in two different ways . De pending on the stacking patt ern (Figure 1), you will get either a cubic close-packed (ccp) or hexagonal close-p acked (hcp) arrangement. In t he hcp arrangement, additiona l layers of pa rticles are placed above and below a given layer, fitting into t he same depressions on either side of t he middle layer. In a three-dimensiona l crystal, t he lay-
Middle layer
(fcc)
A
B
Bottom A layer
c
FIGURE 1 Effident packing. The mostefficient ways to pack atoms or ionsin crystalline materials are hexagona l close-packing (hcp) and cubic close packing (ccp).
NaCl(s)
Latt ice e neq,T)' is a measu re of th e stre ng th of ionic bondi ng. Often , however, che mists use lattice ent halpy, 111alt iceH rather than lattic e energy because o f the di ffic u lty of est ima ting so me ene rg y q ua ntities. Th e same tren ds are see n in bo th , th o ugh , and , becau se we are d ealin g with a co ndensed ph ase, the nu meri cal values are nearly identical. We shall fo cu s here on t he d e pend en ce of la tt ice e n tha lp y o n io n cha rges an d sizes. As g iven by Co u lom b's law, t he highe r th e io n c ha rges, th e g rea te r th e a ttraction between o p positely charged ions, an d so !:J.lanin.ll has a larg er negative value for more h ighly charg ed ion s. Th is is illustrat ed by t he latt ice e n tha ipies of MgO and NaF. T he value o f dl attin.H for :\lg O ( -40IiO kJ /mo l) is abou t four tim es more nega tive than th e value for NaF ( - 926 kJ/mol ) because th e charges o n the ,Mg2+ and 0 2 - ion s [(2 +) X (2 -)] are twice as large as th ose o n Na t an d F- ions. Because the attractio n between ions is inversely p ro portio nal to the d ista nce between th em , th e e ffect of io n size on latti ce en tha lpy is also pred icta ble : A la ttice bu ilt fro m sm alle r io ns ge nera lly lead s to a m ore negativ e valu e for t he lattice e n thalpy (Ta hle 13.2 an d Figure 13.11 ) . Fo r alkali me ta l hal id es, for exam ple , t he latt ice e n thalpy fo r lithi um compoun d s is gene ra lly m o re n egat ive th a n th at fo r potassiu m co mpo u n ds because th e Li" ion is mu ch sma lle r tha n th e K + ca tio n . Sim ilarly, fluori des are more strongly bonded than are iodides with the same ca tion . • Born-Haber Cycles Ca lculation of lattice energies by t his procedure is named for Max Born (1882-19 70) and Fritz Haber ( 1868- 1934), German scien-
tists who played prominent roles in thermodynamic research.
600
C h a p t e r 13
Calculating a Lattice Enthalpy from Thermodynamic Data Lattice en tha lpies can be calculated using a th erm odynam ic rel at ion sh ip known as a Born-Haber cycle. T his calculation is an application of H ess's law (-4 page 233 ) . Suc h a cycle is illust rated in Figure 13.12 fo r solid sod iu m ch loride .
The Chemistry of Solids
FIGURE 13 .11 Lattice e nergy. J.l' tticeU is illust rat ed for the formatio n of
the alkali metal halides. MX(s), from the ions M+(g)
- 1100
-0 E
~
c
- 800
~
e
"B ~ ~
- 700
~
:
F
Br
+ X- (g).
li
- - - - - - Na
-~
- 600
Cl
Steps 1 and 2 in Figure 13. 12 in volve form ati on ofNa +(g ) a n d CI-(g ) ions from th e elements; th e enthalpy c h a nge for eac h of the se ste ps is known (Ap p e n d ice s F a n d L) . Step 3 in Figure 13.12 gives the la ttice e n tha lpy, ~ l a tticJ{. is th e stan da rd m o lar e n tha lpy of formation of NaC l(s) (Ap pe n d ix L) . The e n thalpy values for e ac h ste p arc rel a te d by th e follo win g e q u a tio n:
air
ti.JHO [NaCl(s)]
= d HSt e p 1a + tiH St e p tb + ti.HSt e p 23 + d HSte p 2b + ti. HSte p 3
Because th e va lue s for a ll o f th ese qu antit ies a re kn own exce pt for a / r"'t~'r :\ (dlalticc/ l). th e va lue fo r thi s ste p ca n be ca lcu lated. Step Step Step Step
la. lb. 2a. 2b.
+ 121. 3
~J /rnol (Ap pe n d ix L) - 349 kl /m ol (Ap p e n d ix F) + 107 .3 kJ/mo l (Ap pe n d ix L) + 496 kJ/lI1o l (Ap p e n d ix F)
En tha lpy of forma tio n of CI(g)
!1 H for Cl (g ) + e "
---?
Cl - (g)
Ent halp y of form ation of Na (g ) ---? Na +(g) + e "
!1H fo r Na (g )
T he sta n d ard e n tha lpy o f form ation o fNaC I(s) , o.JHo, is - 4 11. 12 kJ/mo l. Co m bin ing th is with the known valu es o f Steps I and 2, we ca n ca lcu la te 6.Hs lt: p3 , which is th e lattice e n thalpy, D. 1a lt ic J l. FIGURE 13.12 Born- Haber cycle fo r
Cl(g)
t he formation of NaCl(s) from t he elemen ts. The calculation in the text uses ent ha lpy values, and t he value obta ined is
STEP 18
Cl- (g)
the lattice ent halpy, ~ 1"ttk.H . The difference between J.liIttic. U and ,llatbceH is qenerally not significant and can be corrected
+ Na +(g)
for, if desired . (Note t hat th e energy dia-
STEP 28
STEP lA »,
gram is not to scale.)
g>
Na(g)
~
co
STEP 2A
,....-..-.-..., 1 ., CI,(g)
~
+
STEP 3
.:1latticeH
Na(s)
!l.f H" NaCI( s)
13 .3
I
Bonding in Ionic Com pounds: Lattice En ergy 6 0 1
Ste p 3.
Formation o f NaCl(s) from the io ns in th e gas phase =
~fls« . p:l
~f/,tqJ3 = ~IH o [NaC I(s)] - ~H."(t.p la - ~f/Slt:p Ih - ~HS(q' 2;l - ~f/Sl °C) by exe rt ing a mo d erate p re ssure 011 t he gas. In fact, CO 2 is reg u larly sh ip ped in tan ks as a liq uid to laborat ories an d ind ustr ial co m pan ies. Fina lly, the critical p ressu re an d te mperatu re for CO 2 are 73 atrn and 3 1 OC t respec tively. Because th c critica l te m pe ra tu rc and p rcssu re arc ea sily atta ine d in th e labora to ry, it is po ssible to o bse rve t he tr an sfo rm at ion to su pe rc ritica l CO 2 (Fig u re 13.20 ) .
ChemislryANow
N
•
Sign in at www.t homsonedu.com/ login and go to Chapter 13 Contents to see Screen 13.8 to view animations of phase changes and to do an exercise on phase diagrams.
The separate phases of CO 2 are seen through the window in a highpressurevessel.
As the sample warms and the pressureincreases,the meniscus becomes less distinct.
As the temperature continues to increase. it is moredifficult to distinguish the liquid and vapor phases.
Once the critical Tand P are reached, distinct liquidand vapor phases are no longer in evidence. This homogeneous phase is "supercrtticet CO 2, '"
FIGURE 13.20 Transformation to supercritical CO 2 •
13 .6
I
Phase Diagrams
609
Chapter Goals Revisited
Chemislry.o.Now"· Sign in at www. thom sonedu~com/Login to : Assess your understanding with Study Questions in OWL keyed to each goal in the Goals and Homework menu for this chapter For quick review. download Go Chemistry mini-lecture flashcard modules (or purchase t hem at www.ichapters.co m) Check your readiness for an exam by t aking the Pre-Test and exploring t he modules recommended in your Personalized St udy plan.
fj Access
How Do I Solve It ? tutorials on how to approach problem solving using concepts in t his chapte r.
N ow that you have studied this cnapter; you should ask u ihether )'Oll have met the chapter gonL\', In part icula r, you should be able to:
Underst and cubic unit cells a. Describe the three type s o f cu bic un it ce lls: primitive cu bic (pc), bodyce n te re d cubic (bee) , and face-cen tered cu bic (fcc) (Sect ion \ 3. \ ) . b. Relat e ato m size and u ni t ce ll dimensions. Study Question(s) assignable in OWL: 7, 8, 10,26,29,32,34,36,43; GoChemistry Module 18.
Relate unit cells for ionic compounds to formuLas a. Unde rstand the relatio n of u nit ce ll structu re and fo rmu la fo r io ni c co mpounds. (Sectio n 13.2 ) Study Questi on(s) assignable in OWL: 4, S, 6, 8; Go Chemistry Modu le 18.
Describe the properties of soLids a. Unde rsta nd lattice energy and how it is ca lculate d (Se ctio n 13.3 ) . Study Question(s) assignable in OWL: 11,13 ,14, 16, 38. b. Charac terize differen t types of so lid s: metalli c (c .g., co p per) , io nic (e.g., NaC I and CaF2 ) , mo lecu lar (e. g., wate r an d 12 ) , network (e.g., d iam ond ), and am o rphous (c .g., glass a nd mall Ysyn the tic pol yme rs) (Ta ble 13. 1) . Study Question(s) assignable i n OWL: 17. c. Den ne th e processes of mel ting, freezing, and su blim atio n and thei r ent halpies (Sectio ns 13.4 and 13.5) . Study Question(s) assignable in OWL: 20. Understand the nature of phase diagrams a. Id en tify the d ifferent poin ts (tr iple po int, normal boil ing poi nt , freez ing po in t) and regio ns (so lid , liquid , vapor) o f a p hase d iag ram , and lise th e d iag ra m to eva lua te th e vapo r p ressure o f a liq uid a nd th e rel ative d en sit ies o f a liquid and a so lid (Sec tio n 13.5) . Study Question(s) assignable in OWL: 21, 22, 23, 24.
STUDY QUESTIONS OnLine homewo rk for this chapter may be assigned in OWL. .&. denotes challenging questions.
•
denotes questions assignable in OW L.
Blue-numbered questions have answers in Appendix 0 and fully-worked solutions in t he Student Solu tions Manual.
Practidng Skills Metallic and Ionic Solids (S('f' Examples IJ. I- I J.3 and Clwmi.\'f/}N ow Screens 13.2
and 13.3.) l. O ut line a two-d imen sio nal uni t cell fo r the p att ern
show n here . If the bl ack sq u ares a re labele d A a nd the
white squan.'s are R. what is th e sim plest fo rmu la fo r a "co m po u nd " based o n thi s patte rn ?
I I
• • • I I
f--
•• •
I I
~
I I I I
610
l-
~-
•
2. Ou tlin e a two-di m en sion al uni t ce ll fo r th e pa ttern shown h e re . If the bl ack sq ua res are la bele d A an d the wh ite sq ua res are R. wha t is th e sim plest fo rmul a fo r a "co m po u n d " based on th is patt e rn ?
6. • T he mi ne ra l flu o rite , wh ich is co m pose d of ca lciu m io n s an d fluor id e ion s, h as th e un it cell show n he re . (a) Wh at typ e of u n it cell is de scribe d hy th e Ca2+ io ns? (b) Wh e re a re th e F" io ns locat ed , in octa he dra l ho le s o r tetrahedral ho le s? (c) Based 0 11 thi s u nit ce ll, wha t is th e fo rmula of fluo ri te?
3. One way o f viewing th e unit ce ll of perovskite was illustrat ed in Exa m p le 13.2. Anot he r way is show n he re . Pro ve th a t th is view also lead s to a fo rmul a o f CaT i0 3 . 7. • Ca lciu m me tal cr ystallize s in a face-cen te red cu bic uni t ce ll. T he den sity of th e so lid is 1.54 g /cm 3 . Wh a t is th e ra d ius of a calc ium a to m ? 8• • T Ill' den sity of co ppn me tal is 8.9:1 g/C1T1 3. If th e rad iu s of a co pp e r a to m is 127.8 p m , is th e co p per u n it ce ll prim it ive , bod y-cen tere d cu b ic, o r face-ce ntered cubic?
o'-- + .3lIi;;:;
4• • Rutile , T iO i?, cr ystallizes in a struc ture charac teris tic of man y oth e r ion ic co m poun ds . How m an y fo rm ula units of T iO i? are in th e un it ce ll illu str ated h ere ? (T he ox ide io ns marked by a n x a rc wholly with in th e cel l: the o thers are in the cel l faccs.)
9. Po tassiu m iodide has a Face-ce nte red cubic unit cell of iodi d e ions '.. . ith pot assium ion s in octah e d ra l holes. T he de nsity of KI is 3. 12 g / ern ". Wha t is the kngt h of o ne side of th e unit ce ll? (Ion sizes a rc fou nd in Tahle 7,12,)
10. ... . A u n it ce ll of cesiu m chloride is shown on page 596. Th e den sity of th e so lid is 3.99 g/ Cl n 3, a nd th e rad iu s of th e CI- io n is 181 p ill. Wh at is th e radius of th e Cs" ion in the ce n ter of th e cell ? (Assu me t ha t th e Cs+ io n touch es a ll of th e co rn er CI- ion s.) Ionic Bonding and Lattice Energy (See Clwmisll )1Now Srnen 13.4.) 11. • List the foll owing co mpo und s in o rd e r of increasin g latti ce elH.>rbT)-· (fr o m le as I n ega tive to most n egative ) : LiI, LiF, Ca O , Rhi.
5• • Cu prite is a se m icon d uc tor. O xid e io ns are a t the cu be co rne rs and in the cube ce n te r. Cop pe r ion s a rc wh o lly with in the un it cel l. (a ) \ Vhat is th e fo rmula of cuprite? (h) Wh at is th e o xid a tion n um ber of co p pe r? Copper
12. Exa m in e the trend s in latt ice e nergy in Tahle I :t 2. Th e value of the la ttice e ne rgy In-co m es somewh a t more nega tive on go ing fro m Na I to Na lsr to NaCI , and a ll a re in the ra nge of - 700 to - SOO ~J / m ol. Suggest a reaso n for the obse rvat ion th at th e latti ce e nergy o f N"aF (8 b ll in .U = - 926 kJ/m ol ) is m uch more nega tive th an th ose of th e o the r so d iu m h alid es. 13. • To m elt a n io ni c so lid , e ne rgy must be su p p lied 10 d isrup t th e force s between io n s so th e regular array of io ns co lla pses. If th e d istance be tween th e anion a mi the ca tio n in a crys talline so lid decreases (bu t ion charges rem a in th e sam e ), sho uld tilt.' me ltin g p o int de cr ease or in cre ase ? Expla in .
•
more challenging
• i n OWL Blue-numbered questions answered in Appendix 0
I
6 11
14. • Whi ch co m po u nd in e ach of th e fol lowin g pa irs sh o uld re q u ire th e h igh e r tem perat u re to melt? (Se e Study Question 13.) (a) Na CI or RhCI (h) lIaO or MgO (c ) Na CI o r MgS 15. Calc ul a te th e m o la r e n thalpy of form a tion , /j,/H o, of so lid lithi um flu o rid e m ing th e approach o u tlined on pages 599-GO ~ . /j,Jl-fO [Li (g )] = 159 .37 kJ/ mol , and o the r req ui red dat a ca n he fo un d in Ap pe nd ice s F and L. (Se e a lso Exe rcise 13.5.) 16. • Ca lc ula te the la ttic e e n thalpy fo r RhCl . In ad d itio n to d ata in Ap pe ndi ces F and L, yo u will need th e fo llowing informa tio n: ~
b.JH
O
[Rh (g) ]
b.JH
o
[RhCI (s) ]
20. • Th e speci fic h ea t ca pac ity of silver is 0.2:35 Ji g ' K. Its meltin g poi n t is 96 2 °C, an d it." e n tha lpy of fusio n is 11.3 ~J /m o l. wh at q ua n tity of e ne rgy as he a t. in joules, is requi red to chang e 5 .00 g of silver from a so lid a t 25 °C to a liqui d at 962 °C?
Phase Diagrams and Phase Changes (See Chf1l1islryNow Screen 13.8.)
21. • Co ns ider th e phase d iag ram of COt in Fig ure 13. 19 . (a) Is th e de nsity o f liq uid C O t g re a te r or less than rhat of so lid COt? (b) In wh at ph ase d o yo u fi nd C O t at 5 atrn a nd 0 °C ? (c) Can CO t be liq uefied at 45 °C? 22 • • Use the phase di agram give n here to answe r th e fol lowin g quest ion s:
HO.9 k./llllo! ~
Normal MP
- 435 .4 kJ/ llloJ
Normal BP
1.0 ------ - --------- ------------
Other Types of Solids (See Chf1l1isl1)Now Sa vens 13. 6 a nd I J . i .)
I
•~
17. • A di amond u nit ce ll is shown h e re . (a) II m\' many ca rb on a to m s an ' in on e unit ce ll? (b) Th e u nit ce ll ca n be co ns id ered as a cu bic u nit cell of C a to ms with o th e r C ato ms in ho le s in th e lattice . Wh at typ e of unit ce ll is thi s (p c, be e, fcc )? In wh at h o les are o ther C ato m s lo cat ed, octahe d ral or tetrahedral ho les ?
18. T he stru cture o f g ra p h ite is give n in Figure 2.7. (a) What type of intermo lec ul ar ho n d ing fo rc es exist bet we e n the laye rs of six-m em ber ca rbo n rin gs? (h) Accoun t for th e lu brica ting ability of gra p h ite . T h at is, why d oes gra p hite feel slip pe ry? Wh y d oes pen cil lead (wh ich is re all y g ra p h ite in clay) lea ve bla ck ma rks on paper?
Physical Properties of SoLids 19. Ben ze n e , C1iH h , is an o rga n ic liq uid th at fr ee ze s at 5. 5 °C (se e Figure 12.1 ) to form bea u tiful , fea ther-l ike crys ta ls. [l ow mu ch e neq.,'Y as he at is evolved whe n 15.5 g of ben zen e freezes a t 5.5 OCt (T h e e n thalpy of fu sio n of ben zen e is 9.95 ~ J I m o l.) If the 15.5-g sa m ple is n- me hcd. agai n a t 5 .5 °C, wha t q ua n tity of c n e rgy as he at is re qu ired to con ve rt it to a liq u id ?
612
I
•
Solid Gas
0.5
e,
- 120
- 115
- 110
- 105
Temperat ure (0C)
(a) In wha t ph ase is th e substa nce fo u nd at room temperatu re and 1.0 at m p re ssu re ? (h) If the pn.'s su re exerte d on a sa m ple is 0. 75 a tm an d th e te m peratu re is - 1 14 °C, in what ph ase doc s th e su bstanc e ex ist? (c) If yo u measu re th e vapor pre ssu re of a liquid sa mple an d fi nd it to he 38 0 mm II g, what is th e rem perat u re of th e liq uid ph ase ? (d) What is the vapor pressur e of tile so lid at - 122 °C? (e) Whi c h is th e de n se r ph ase-so lid o r liq uid ? Exp lai n briefly. 23. • Liqui d am mo nia, NII 3 (f ) , was o nce used in ho m e refr igerators as th e h ea t tra nsfer flu id , T he spec ific he a t ca pacity of th e liq uid is 4.7 Ji g · K and th a t of th e va po r is 2.2.1 /g . K. T he en t ha lpy of vaporiza tio n is 23.33 ~J/m o l a t the hoiling poi n t. If yo u heat 12 kg of liquid am moni a fro m - 50.0 °C to its ho iling point of - 33.3 °C, allow it to eva po ra te , and th en co n tinue war ming to 0.0 °C, how much e nergy must yo u su pply? 24 . • If yo u r a ir con dit ioner is mo re th a n several years old , it ma y use the chlorofluorocarbo n CClt Ft as the he at tra nsfer flu id . T he no rm al ho ili ng poin t of CC ltF 2 is - 29.8 °C, a nd th e e n tha lpy of vapo riza tion is 20.1 1 kJ / m o1. T Ilt' gas a nd till' liquid have spe cific h ea t ca pa cities of 117 .2 .I/m ol . K an d 72. 3 J / mol . K, re spec tively. H ow mu ch e ne q.,'Y as h ea t is evo lved when 20.0 g of CCl tF t is coo led fro m + 40 °C to - 40 °C?
•
more challenging • in OWL
Blue-numbered questionsanswered in Appendix0
General Questions T Ill's" questions an' not drsignntrd as 10 [J/Jl' or location ill the clutpter: They ilia)' combine SfV('Ta/ WI/(1)!J,
25 . Co ns truct a ph ase di ag ram for
O~
from th e fo llowi ng
iutcrma tio u: norm al ho iling po int , 90.18 K; norma l m el ti ng poi n t, 54.8 K; a nd trip le po in t, 54.34 K at a p re ssu re o f 2 m ill II g . Very ro ug h ly estimate th e va po r p re ssu re of liq u id {)2 a t - }96 -c, th e lowe st Lc m p c ra~ lu re ea sily reach ed in th e laboratory. Is th e d en sity o f liq uid O ':.! g reater o r le ss than th at of so lid O ':.! ? 26. .A. • Tu n gsten crys ta llizes in the u n it cell shown he re .
1316.5pm
(a) Wha t typ e of unit ce ll is th is? (b) How ma n y tu ng ste n a tom s oc cu r per un it ce ll? (c) If the e dge o f th e unit cel l is 3 16.5 pm , wh at is th e radi us of a tu ngsten a to m ? ( Hi nt: T he W a to ms touch e ac h o ther alo ng the di ag o na l lin e from o n e co rne r of th e un it ce ll to th e o p pos ite co rne r o f the un it ccll.)
27 . Silver crys tallizes in a facc-ccn u-rc d cubic un it cel l. Eac h side of th e u n it ce ll h as a le ngt h o f 409 p m . Wh at is the ra d ius o f a silver a to m? 28 . ... . T h e u n it ce ll shown here is fo r calc iu m carb ide. How m an y calc iu m ato ms a nd how m an y ca r bo n atom s arc in eac h u n it cell? Wh a t is the form ula o f calci um ca r bide? (Calciu m io ns are silver in co lor an d ca r bo n a to ms are g ray.)
31. ... Calci um fl uo ride is the we ll-known m ine ral fluo rite . It is known th at eac h un it cell co n tains fou r Ca 2 + ion s a nd e igh t F- io ns a nd th a t th e Ca 2 + io ns are arranged in an fcc la ttice . Th e F- ions till a ll th e te trahed ra l ho les in a face-ce n te re d cu b ic latti ce o f Ca 2 + ion s. T he edge of the Ca F2 un it ce ll is 5.4629 5 X IO- H ern in le n gth . T he d e n sity o f th e so lid is 3. 1805 g /c m 3 , Use th is inform a tion to calc ula te Avogadro 's num be r. 32. ... . Iro n has a body-cent e red cu bi c un it cell wit h a ce ll d im e nsio n o f 28 6.65 pm . T he d en sity of iron is 7.874 g /cm 3. Use th is info rm at ion to calcu la te Avogadro 's nu m be r. 33 . ... Vou ca n ge t so me id e a o f how effi cie n tly sp he rical a to ms or ion s are p acke d in a th ree-d im ensional so lid by se e ing how we ll ci rcular a to ms pack in t\VO dim en sions. Using th e d rawin gs shown he re , p rove th at B is a mo re e fficie n t way to pack circu lar a toms than A. A unit ce ll of A con tains po rti o ns o f four circles a nd o ne h o le . In R, packing coyc rage ca n he calcu la ted hy lo o kin g at a tri a ngle that con tains po rti o ns of th ree circles an d on e hol e . Show that A fill s abo u t 80 % of tilt' avail abl e space . whereas B fill s clo ser to 90 % o f the ava ilable space.
A
B
34. ... . Assuming th a t in a p rim itive cu bic u n it cell llle sp herica l a to ms o r io ns j us t touch a lo ng the cu be's edge s, cal cula te th e pe rce n ta g e o f em p ty spac e withi n th e u ni t ce ll. ( Rec a ll that th e vo lu me of a sp he re is (4/ 3) rrr:\ whe re r is th e radiu s o f th e sp hc re. ) 35 . ... T he sol id sta te struc tu re of silico n is
29 . • T he very d en se metal iridiu m h as a face-cen te red cu bic un it ce ll a nd a d e n sity of 22.56 g / cm 3. Use this in for mation to calcula te the radius o f a n atom of the ele men t. 30. Van adium m et al has a de ns ity o f 6. 11 g/cm :1. Assu m ing th e van ad iu m a to m ic ra d ius is 132 pm , is th e vanad iu m unit ce ll primitive cu bic, body-ce n te red cu bic, or fac ecentere d cu bic? •
morechallenging
• in OWL
Blue-numbered questions answered in Appendix 0
(a) Describe thi s crystal as pc , bee. or fcc. Wh at typ e o f ho le s are occ u pie d in th e latt ice ? (b) Calcu la te th e d e n sity o f silico n in g/cm 3 (g ive n th at the cu be ed ge has a len g th o f .543.1 pm ) , a n d es tim a te the ra d ius of the silico n a to m. (N o te: th e Si ato ms o n th e e dges do no t tou ch o ne ano th c r.)
I
613
36 . ... . T h e so lid sta te stru ctu re of silico n carbide , SiC , is sh o wn bel ow, Kn ov.... illg that the Si-C bond le ng th is 188.8 pm (a nd th e Si-C-Si bond an gle is 109.5°), cal cul ate th e density of SiC.
38 . • Using the therm o chem ical d at a bel ow, an d an es tim at ed value of - 248 1 ~ I / m o l fo r the la tti ce elll'rh'Y for Na 20, calcu la te the value fo r th e second ele ct ron affin ity of ox yge n [O - (g ) + e " ---7 0 2- (g ) ] . Quantity
Numerical Value ( kJj mol)
Enthalpy of atomization of Na
107.3
Ionization Energy of Na
495.9
- 418.0
Enthalpy of formation of solid Na20 Enthalpy of formatio n of O(g) from 0,
249.1
- 141.0
First electron affinity of 0
Unit cell of SiC.
In the Laboratory 39. Lead su lfide , PbS (co m m o nly called ga len a) , has th e sa m e formula as Zn S.
Sample of silico n carbide. Unit cell of PbS.
37 . Spinels a re so lids with the general fo rmula i\ B ,;l)~ (where A2+ a nd B3+ are metal catio ns of th e same o r different me tals. T he be st-known e xam ple is co m m o n ma g netite , Fc:\O, [which you ca n fo rm ula te as (Fe 2 +) (Fe3+hn d. An ot her ex am ple is the m ine ra l ofte n referred to as sp inel, '\l gAI 2 0 ,.
.,..-~
'. '. q
.
~
.
I"R.;
-
l·: .~
• '; ',\
I~ I
.
-. (
J.
'.. '. >:: .:\~.~.~~. '. Sample of galen a.
Does PbS h ave the sa m e solid stru ctu re as ZIlS? If diffcrc n t. how a re they d ifferen t? How is the uni t ce ll of PhS rel ated to it" formul a? A crysta l of the spinel MgAI20 4 on a mar ble chip.
T h e ox ide io ns of spin els form a face-cent e red cu b ic lattice . In a normal spinel, cation s oc cu py ~ of the tetrahe d ra l sites and ~ of th e octahedral sites. (a) In MgA12 n " in what typ e of h o le s arc t he magnesium an d alu m in um ions found ? (b) T he m ineral chromite h as th e form u la FeCr2 o.,. What ion s a re involved, and in what type of ho les are they found?
614
I
•
more challenging • in OWL
Blue-numbered questionsanswered in Appendix0
40. Ca T iO j , a pcrovskitc , has till' structu re below. (a ) If th e d ensity of the so lid is 4.10 g/cm 3 , what is Ilnlen gt h of a sid e o f the uni t cell? (b) Calcula te th e rad ius o f the Ti -l+ ion in the ce n ter of the uni t cel l. How we ll d oe s you r ca lcu la tio n agree with a lite rat ure value of 75 pili?
Summary and Conceptual Questions T he fo llowinf; qu estions
nUl)'
use concepts from this nnd inruious
r!lajllen.
41. ... Boron p hosp hide , BI>, is a se m ico n d uc tor a n d a hard ,
ab rasio n-resistan t ma te rial. It is made by re ac ting boro n u-ibromid c and p hosp horus tribromidc in a hyd rogen a tm o sp h e re a t hi gh tempera tu re (> 750 DC ) . (a) Wri te a hal anc ed che m ica l equation for th e svn th csis o f BP. (Hi nt: Hyd mgen is a reducing agen t.) (h) Boro n p hosp hi d e crysta llizes in a zin c hlcndc structure , formed fr o m boron atom s in a face -centered cu b ic lau.icc a nd phosph orus a to m s in tet ra h edra l ho les. How many tet ra hedral ho les arc filled with P a to m s ill each u n it ce ll? (c) T he len g th of a u nit ce ll of BP is 4 78 p m . Wha t is the den sity of th e so lid in g /cm 3 . (d ) Calcu la te th e clo.se st di stan ce between a R a nd a I> a tom in t he u nit cel l. (Assu me t he B a to m s d o not touch alo ng th e ce ll e dge . T he R ato m s in th e f~I Cl' S touch the B a to ms at th e corners of the unit cel l. Set' page 594.)
Unit cell of the perovskite CaTi0 3 •
4 2. ... Why is it no t po ssible for a salt with the formu la iv13X CNa3Po... for e xample ) to have a face-ce n te red cubi c la ttice of X anions ,...irh ~1 ca tio ns in o ctahedral h oles? 43 ,
Asample of perovskite CaTi0 3 •
.at. .
"1'\"'0 identical swim m ing pools arc filled with unifor m sp he res o f icc packed as clo sely as po ssible . T he sp heres in th e first pool a re th e size o f g ra ins of san d; tho se in the sec o n d poo l a re the size of o ra nges. T h e ice in bot h pool s melts. In whic h poo l, if ei ther, will th e watc r level he hig her? (Ignore a n y differe nc es in filling sp ace a t the plan e s nex t to the walls an d hottom .)
44. Sp inel s art' d es cri bed in St udy Question ~1 7. Con side r two norm a l sp ine ls, Co ALll j and SnCo zo. 1• Wh at m etal io n s are invo lved in each ? wh at are th e ir electro n co nfigurations? Are the metal ion s para magnetic, and if so h ow m an y u n paired el ect ron s are invo lved ?
•
morechallenging
• in OWL
Blue-numbered questions answered in Appendix 0
I
615
STATES OF MATTER
14
Solutions and Their Behavior
Safe Flying You are sitting in an airport in the wi nter. Outside t he weather is bad,
Questions: 1. Why does ethylene glycol (HOC H,CH,O H) dissolve so we ll in
wa ter? 2. Why use a solution of ethylene glycol and water as an antifi rst the plane has to be de-iced. Ice and snow on t he wings severely freeze? impairs the ability of wi ngs to provide lift, so it is unsafeto try to ta ke 3. If you mix 100. g of ethylene glycol with 500. g of water, what off unless the plane is de-iced. The fluid that is sprayed on the plane is t he freezing point of the mixture? with blowi ng snow and ice. You want to get on the plane home, but
is a mixture of ethylene or propylene glycol and other substances. These are the same chemicals used in t he antifreeze in your car's radiator.
616
Answers to these questions are in Appendix Q.
Chapter Goals
Chapter Outline
Set' Cha/lln Goals Revisited (/H1gf' 646) lor Stud), Questions kp)'f'd to these bJ'(}al\ ruul a.nigllahlf' i n OH'L
14.1 Units of Concentration 14.2 The Solution Process
Ca lculate and use the solution concentration units mo lality, mo te fractions, and weight percent.
Understand the solution process. Understand and use the coUigative properties of solutions.
14.3 Factors Affecting Solubility: Pressure and Temperature
14.4 Colligative Propertie s
",.~"",~""'l>
14 .5 Colloids
e come in to contact with solu tio ns e' "ery d ay: 3C( ueous so lu tio ns of ionic salts , ga so line with add itives to im p rove its properties. and househ ol d cleaners such as a m monia in wate r. \Ve p urposely ma ke so lutio n s. Ad di ng su ga r, flavorin g . an d sometimes CO ;! to water p roduces a palatable so ft drink . Ath le tes drink co m me rcial beverages with d issolved sa lts to m at ch salt concent rations in body fluid s p recisel y, thus allowing t he flu id to be taken in to the bo d y mo re rapi dl y. In medicin e , saline so lutions (aqueo us so lutions co n tain ing NaCI and other so lu ble salts) are infuse d in to th e body to replace lo st fluid s. A solution is a homogeneous mix tu re of two or more substances in a sing le p hase . By convention , th e component present in largest amount is identified as the solvent and th e other co mpo nent (s) as th e so lu te (s) (Figure 14.1) . Althoug h o the r types of so lu tio ns ex ist (such as alloys, so lid so lu tio n of met al s), the o bjective in th is chap te r is to develop an un d e rstand ing of ga ses , liquids, an d so lids disso lve d in liq uid so lve n ts. Experience tells you tha t ad d ing a so lu te to a pu re liqu id will change th e p ro perties of the liq uid . Ind eed , that is the reaso n some solu tion s are made. For instance , add ing antifreeze to the water in you r car's radia tor prevents the coolant fro m bo iling in the su mmer and freez ing in th e winter. T he cha nges that occur in the freezing and boili ng points when a su bstance is disso lved in a pure liquid are two observations we shall examine in d e tail. These prope rt ies, as well a" th e osmotic pressu re o f a so lution and ch anges in vapor pressu re , are exam ples of co lligat ivc p ro pe rt ies.
W
ChemistrY.(J"Now'" Throughout t he text th is icon intr oduces an opp ortunity for self-st udy or to explore interactive tutorials by signing in at www.thomscnedu .com /Iogtn .
FIGURE 14.1 Making a solution of copper(II) chloride (the solute) in water (the solvent). When ionic compounds dissolve in water, each ion is surrounded by water molecules. The number of water molecules is usually six, but fewer are possible.
(a) Copper(II) chloride, t he solute, is added to water, t he solvent.
(b) Inte ractions between water molecules and Cu2+ and Ct" ions allow the solid to dissolve. The ions are now sheat hed with wate r molecules.
617
Co lligative properties a re p roperti es of so lu tio n s th at d epend on ly o n th e numbe r of so lu te part icles per so lven t mo lecu le a n d no t o n th e id entity o f th e so lut e.
14.1
VsoLn > 1.00 L VH20 added = 1.00 L 0.100 molal solution
Vso1n = 1.00 L VH/ O added < 1.00 L 0.100 molar solution
FIGURE 14.2 Preparing 0.100 moLal and 0. 100 moLar solutions. In the flask on the right, 0.100 mol (19.4 g) of KZCr0 4 was mixed with enough water to make 1. 000 l of solution. (The volumetric flask was filled to t he mark on its neck, indicating that the volume is 1. 000 L. Slightly less than 1.00 l of wate r was added.) If 1.00 kg of wate r was added to 0.100 mol of KlC r04 in t he flask on the left, t he volume of solution is greater than 1.000 L. (The small pile of yellow solid in front of the flasks is 0.100 mol of K2Cr04 . )
Units of Concentration
To an a lyze the co lligat ive properties o f so lutio n s, we ne ed ways o f m easuring so lu te co nc e n tra tio ns th at refl ect t he number of m ol ecu les o r ions of so lu te per m o lecule of so lvc n t. Mo la ri ty, a co nc e n tr a tio n unit u seful in sto ichi ometry ca lcu la tio ns , is not u sefu l whe n de aling wit h m o st co lliga tivc properties. Recall th at molarity ( M) is defined as the n umber of mo les o f so lu te pCI' liter o f so lu tio n (... page 174 ) , so u sing m o larity does not allow u s to id e n tify th e e xac t a m o u n t of so lven t u sed to make th e so lutio n . T his fact is illu str at ed in Figu re 14.2 . T h e flask o n the righ t co n ta ins a 0.100 ~1 aq u cous so lu tio n of po tassiu m ch ro mate. It was made by addin g e n oug h wate r to 0 .100 m ol of K;!CrO" to m ake 1.000 L of so lutio n . Th ere is no way to id entify the amou nt of so lve n t (v..ra te r ) that was ac tua lly ad ded. If 1.000 L of wate r h ad been added to 0 .100 mol o f K2CrO ", as illu strated with the flask on the le ft in fi g ure 14 .2, th e vo lu me of so lu tio n would b e g re a ter th an 1.000 L. T h ree co nc e n tra tio n units a re d escri bed h e re that reflect the nu m ber of m olccu lcs or io ns of so lu te pe r so lve n t m olecule : mo lality, m o le fracti on , a nd wc igh t percent. The molality, m, o f a so lu tio n is d efin ed as the a m o u n t of so lu te ( m o l) pe r ki logra m o f solve » r.
Concentration (c, mol/ kg)
= molality of solute =
amount of solute (mo l) mass of solvent (kg)
(14.1)
T he m o lal ity of K2C rO" in the flask o n the left sid e of Figu rc 14 .2 is 0 .100 m o l/kg . It was prepared from 0 .100 mo l ( 19.4 g) of K2CrO, a n d 1.00 kg ( 1.000 L X 1.00 kg/L ) o f wa te r. No tice th at differe nt quan titie s o f water we re u sed to m a ke th e 0 .100 :vl (0. 100 mo la r) a n d 0 .1OO m (0 .100 mo lal) so lu tio n s of K2Cr04. T hi s me an s the molarity and the molality C!f a J"rlvl'n solu tion cannot be the same (a lt hough th e d ifference ma y be ne gligib ly sma ll whe n the so lu tio n is qui te dil ute ) . The m ol e fraction , X, of a so lu tio n co m po ne n t is defined as the a m o un t of th at component (n..\,) divi de d by the to ta l a m o u n t o f a ll o f th e co m p o ne n ts of t he m ixrurc ( n..\, + Ull + tit; + ...). Mathematical ly it is re presen ted as
Mole fraction of A (X, )
~
n,
- - -"'----nA + nB + nc + ...
(14.2)
Co n sider a so lu tio n th ai co n ta in s 1.00 mol (46. 1 g) of e tha n o l, C;!lI",O H , in 9.00 m ol ( 162 g) of wa ter. T he m o le fract io n of alco hol is 0. 100 , a n d that of wa te r is 0.900 . • Molarity and Molality The use of the ter ms "molar" (symbol M) and "molal" (sym bol m) is common practice among chemists. Recently, however, NISTsuggested t hat use of these ter ms and symbols to represent concentrations should be discontinued and replaced with the formal units of concentration (moljLand
mol/kg). 618 C h a p t e r 14
1.00 mol ethanol Xe t hilnol
1.00 mol ethanol
+ 9.00 mol water
9.00 mol water X wate r
1. 00 mol ethanol
+ 9.00 mol water
No tice th a t the slim of the mo le fra ctio n s o f the com p o n e n ts in the so lu tio n eq u a ls 1.000 , a relati o n ship th at is true for all so lu tio ns.
Solutions and Their Behavior
Weight percent is th e mass of one co m ponen t divided by th e total mass of th e mi xture , multiplied by 100%:
.
weiq ht % A =
mass of A X 100% mass of A + mass of B + mass of C + ...
3inf
Cat Spray • Kil l,
(14 .3)
t'~. ~nd TlI:h rlll'1l("l1;n day. • K I II . H(BfW ·W~
-
/CllIl.....mt
The a lco ho l-w a ter mi xture has 46 . 1 g of eth anol a nd 162 g of water. so the tot al mass of so lu tio n is 208 g , and th e we ight % of alco ho l is 46. 1 9 ethanol 46. 1 9 ethano l
,
+ 162 9 water
\Vei ght pe rcent is a common unit in co nsu m er products (Figu re 14.3 ) . Vinegar, for exa mp le, is an aq ue o us so lu tio n co n tain ing ap p rox im ately 5 % ace tic aci d and 95 % wat er. The lab el o n a co m mon household bleach lists its ac tive ingredie nt as 6.00% so d iu m hypochlorite (N a'GCl) and 94 .00% inert ing re dients. Na tu ra lly occu rrin g so lu tio ns are often very dilut e . En vironmen tal che m ists, biologi sts, ge o log ists, oc eanog ra phe rs. and o the rs frequent ly use parts per million (p pm) to ex p res s th eir co nce n tra tio ns . The unit ppm refers to relative quantities by mass; 1.0 pp m represents 1.0 g of a su bsta nc e in a sam ple with a tot al m ass of 1.0 mill ion g . Because wa te r a t 25 °C has a d en sity of 1.0 g / m L, a co nc e n tra tio n of 1.0 mg /L is eq u ivale n t to 1.0 mg of so lu te in 1000 g of water o r to 1.0 g o f so lu te in 1,000 ,000 g o f wate r: th at is, units of ppm and mg /L are approxim at ely eq uivalen t.
0lIIlI~
II IlL
r;..;_.ol ,
It ....
'''' '"
'111M
"i"i\
!.'!':!!'.'f.""" . "'"
FIGURE 14.3 Weight percen t . The composition of many common products is often given in ter ms of weight percent. Here, t he label on t he cat spray indicates it contains 1.15% active ingredients.
m
Chemislry.o.Now
•
Sign in at www.t homsonedu.comj logi n and go to Chapter 14 Contents to see Screen 14.2 for an exercise on calculati ng soluti on concent ratio ns in variou s units.
Calculating Mole Fractions, Molality, and Weight Percent Problem Assume you add 1.2 kg of ethy lene glycol, HOCH 2CH 20H , as a n a ntifreeze t o 4.0 kg of wat er i n the radiator of your car. What are t he mole fraction, molality, an d weight perce nt of t he ethylene glycol? St rategy Calculate th e amount of ethy lene glycol and water, an d the n use Equations 14 .1- 14. 3. Soluti on The 1. 2 kg of ethy lene glycol (mo lar mass water repres ents 220 mol.
= 62. 1 gjmo l) is equi valent to 19 mot
and 4.0 kg of
Mole fraction :
Xg 1y, ol
=
19 mol eth ylene glycol 19 mo l et hylene g lyeo l + 220 mo l water
0.0 8 0
Molality:
' glycol
=
g _m_o:-l~e~ t h~y_le_n~e~g~ly~,~ ol _l_ 4.0 kg water
=
4.8 mol/kg
=
4.8 m
Weight perce nt: Weight %
1.2 X 10 3 g et hylene glycol
= c-:cc:-=-=~~--;--""-;-----,~----,:---:-:-;:;---:1.2 X 10 3g ethylene glycol + 4.0 X 10 3 g wat er
X 100%
=
23%
Commercial an tifreeze . This solution contains ethylene glycol, HOCH ,CH2 0H, an organic alcohol t hat is readily soluble in water. Regulations specify t hat t he weight percent of ethylene glycol in ethylene glycol- based antif reeze must be at least 75%. (The remainder of t he solutio n can be other glycols and water.) 14.1
I
Units of Conce nt rat ion
6 19
EXERCISE 14.1 MoLe Fraction , Molality, Weight Percent, and Parts per Million (a) If you dissolve 10.0 g (about one heapingteaspoonful) of sugar (sucrose, CI2H n0 1t ) , in a cup of water (250. g). what are t he mole fraction . molality. and weight percent of sugar? (b) Sea water has a sodium ion concentration of 1.08 X 104 ppm . If the sodium is present in the form of dissolved sodium chloride, what mass of NaCl is in each liter of sea water? Sea water is denser than pure water because of dissolved salts. Its densityis 1.05 q/ml..
14.2 • Unsaturated The term unsaturated is used when referring to solutions with concentrations of solutethat are less than that of a saturated solution.
If so lid CuC I2 is ad ded to a beak er of wat er, the salt will begin to dissolve (see Figure 14.1) . Th e amoun t o f so lid d im in ish es, an d th e co nce n trations o f Cu 2 +(a q) an d CI- (aq) in th e so lu tio n incr ease . If we con tin ue to ad d CuCI2 , however, we will eve n tua lly rea ch a point whe n no ad d itio na l CuC I2 seems to d isso lve . T he co nce ntrati on s of Cu 2 +(aq) and CI- (aq) \ \-'iII not in crease furth e r, an d any add itio nal so lid CuC I2 ad ded after th is po int will re main as a so lid at th e bottom of t he beak er. \Ve say t hat suc h a so lu tio n is sa turated . Altho ug h no cha ng e is obse rved o n the m acro scopic level , it is a d ifferen t ma tter o n the particu lat e level. The process o f d isso lving con tin ues , with Cu H an d CI- ion s leaving the so lid state and e n te ring so lu tio n. Co ncu rre n tly, a seco nd process is o ccurrin g: the fo rm at io n o f so lid CuC I2(s) from Cu2+(a '1) an d CI-(aq), The ra tes at which C uC l2 is d isso lving and re p recipita ting are eq ua l in a satura ted so lution , so th at no net cha ng e is observed o n the macr osco pic level .
A Closer Look
Supersaturated Solutions
Although at first glance it may seem a contradiction, it is possible for a solution to hold more dissolved solute than the amount in a satu rated solution. Such solutions are referred to as supersaturated solutions. Supersatu rated solutions are unstable, and the excess solid eventually crystallizes from the solution until the equilibrium concentration of the solute is reached, The solubility of substances often decreases if the temperature is lowered. Supersatu rated solutions are usually made by preparing a sat urated solution at a given temperature and the n carefully cooling it . If the rate of crystallization is slow, the solid may not precipitate when t he solubility is exceeded. Going to still lower temperatures results in a solution that has more solute than the amount defined by equilibrium conditions; it is supersaturated . When disturbed in some manner, a supersaturated solution moves toward equilibrium by precipitating solute, This change can occur ra pidly. often with the evolution of thermal energy. In fact, supersatu rated solutions are used in "heat packs" to apply heat to injured
620
Chapter 14
I
The Solution Process
.. THE-H EA T ,
SOLUTION'
Supersaturated solutions. When a supersaturated solution is disturbed. the dissolved salt (here sodium acetate, NaCH 3C0 2) rapidly crystallizes. (See ChemistryNow Screen 14.2 to watch a video ofthis process.) muscles. Whe n crystallization of sodium acetate (NaC H3C02) from a supersaturated solution in a heat pack is initiated, the tempera-
Solutions and Their Behavior
Heat of crystallization. Aheat pack relies on the heat evolved by the crystallization of sodium acetate. (See ChemistryNow Screen 14.6 to watch a video ofa heat pack.}
ture of the heat pack rises to about 50 0(, and crystals of solid sodium acetate are detectable inside the bag,
(a)
(b)
Less dense layer of nonpolar octane, CgH 18'~ Solution of CuS04 in -
.
~
~------
water.
----0""1
-
•
This process is ano the r exa m ple o f a dynam ic eq u ilibriu m ("4 page 119 ), a nd we ca n d escrib e the situatio n in terms o f an eq uatio n with su bstan ces linked by a se t of d oubl e a rrows (
):
A satu rated so lu tio n gives us a way 10 d efin e preci se ly the so lu bility o f a so lid in a liquid. Solubility is the co nce n tr atio n o f so lu te in equilibrium with u nd isso lved so lu te in a satu rated so lut io n . T he so lu bility ofCuC l:b for exa m ple , is 70.6 g in 100 mL ofware r at 0 "C. If we ad d 100.0 g ofC uC I2 10100 m L of water at 0 "C, we ca n expect 70.6 g to di ssol ve, and 29.4 g o f so lid to remain .
Liquids Dissolving in Liquids If tw o liquids mi x to a n a p p rec ia ble exten t to fo rm a so lu tio n , th ey are said to be miscible. In co n tra st, immiscible liquids d o no t mi x 10 form a so lution ; th ey ex ist in co n tac t with each othe r as separat e layers (see Figures 12.5 and 14.4). T he pol a r co m po un d s e tha n o l (C 2 H 50 H) an d wa ter are m iscible in a ll propo rtio ns as a re the non polar liqu ids oc ta ne (C gIl ul ) an d carbo n te tra ch loride (Ce Lt) . O n the othe r hand, neith er CRII IR no r Ce Lj is m iscible wit h wa te r. Observati o ns like th ese have led to a fa m ilia r rul e o f thu mb: Like dissolves like. T ha t is, two o r more nonpol ar liquids fre q ue nt ly are mi scible , j ust as are two or m o re pola r liquids. Wh at is th e molecul ar basis for th e "like d issolves like " g uid el ine? In pure water and pure e thano l, th e maj or fo rce be twee n mol ecul es is hydroge n bo nd in g involving 0 - 1I gro u ps. Wh en tile two liquids are mi xed , hydrogen bo nd in g betwe en e than o l a nd water mo lecu les also occurs a nd assists in the so lu tio n process. In contrast, m olecul es of pu re oc tane or pu re CCI-t, both of whic h are non pol ar, are held toget her in the liq uid ph ase by d ispe rsio n fo rces (.... Sectio n 12.3) . Th e energy associate d wit h th ese fo rce s o f attrac tio n is sim ilar in valu e to th e e ne rgy due to the fo rces of attract io n be tween octane a nd CCLI m olecul es whe n th ese no npol ar liquid s are mi xed . Thus, littl e o r no e nergy change occ urs when o ctane-o ctane and CCl r CCI4 a ttractive fo rces arc replaced with octane- Cf .l, fo rces. T he so lu tio n
OH group like dissolves like. Glucose has five - OH groups on each molecule, groups
that allow it to form hydrogen bonds with water molecules. As a result, glucose dissolves readily in water. 14. 2
I
Th e Solution Process
621
~
+
~. - r..")
-. ..( ,')'I
H20
Separate liquids
CH 30H
Mixture
FIGURE 14.5 Driving the solution process-entropy. When two similar liquids- here water and methanolare mixed, t he molecules intermingle, and t he energy of t he system is more dispersed than in the two, separ ate
pure liquids. A measure of t his energy dispersal is entropy, a ther modynamic function described in more detail in Chapter 19.
• Entropy and the Solution Process Although the energetics of solutio n formation are important, it is generally accepted t hat entropy is a more important contributor to the solution process. See Chapte r 19 and T. P. Silverstein: "The real reason why oil and water don't mix." Journal of Chemical Education, Vol. 75, pp. 116- 118, 1998.
p ro cess is ex pecte d to be nearl y energy neut ral. So , wh y do th e liquid s mi x? The an swer lies d eep er in th ermodynami cs. As yo u shall see in Chap ter 19 , spo n taneo us change s, suc h as th e mi xing ofliquids, are accompanied by an increase in en tro py, a th erm odynam ic fun ction that is a measure of the di spersal of the energy o f th e particles in th e m ixtu re relative to th e pure liquids (Figu re 14.5) . In contrast, polar and nonpolar liquids usually d o not mi x to an ap preciable d egree ; whe n placed to gethe r in a co n tainer, th ey se parate in to two d istin ct layers (Figu re 14.4 ) . T he expl an ati on is complex and involves th e int erplay of th e en tha lpy of mix ing and e n tro py. T he en thalpy of mi xin g is ze ro or ne arl y so, but mixin g d issim ilar liquids leads to a decrease in en tr o py. As e xpl ain ed in Cha p te r 19, this mean s th at mi xin g di ssimilar liquids is not thermod ynami call y favorabl e .
Solids Dissolving in Water
Active Figure 14.6 SoLubility of nonpolar iodine in polar water and nonpoLar carbon tetrachLoride. When a solution of nonpolar 12 in water (t he brown layer on top in the left test t ube) is shaken with nonpola r ecl. (the colorless botto m layer in the left test tube), the 12 t ransfers preferentially to th e nonpolar solvent. Evidence for this is t he purple color of the bottom eeL. layer in the test t ube on the right.
The "like d issolves like " gu id el inc also holds for molecul ar solids dissolving in liquids. No n polar solids such as naph th ale ne , ClU II~h d issolve re adil y in nonpolar so lve n ts such as benzene , C1iH r" and he xane, Cr,H 14. Iodine, [2, a nonpola r ino rgani c so lid, dissolves in water to so me exten t, but, g iven a cho ice, it di ssolves to a largcr exte n t in a nonpol ar liquid such as CCLt (Figu re 14.6 ). Sucrose (suga r) , a polar mol ecul ar so lid , is no t 'Try so lu ble in nonpolar solven ts but is readily solu ble in water, a fact that we kn ow well becau se o f its usc to swee te n beverage s. The presen ce o f O -H g ro u ps in th e str uctu re of sugar and ot her su bstances suc h as glucose allows these mol ecules to interact with pol ar water mol ecules through hydrogen bond ing. "Like di ssolves like " is a so mew ha t less effe ctive but still useful gu ide line whe n co nsid ering the so lu bility of io nic so lids . Thus, we can reason abl y predict th at io nic com po un ds , whic h ca n be co nsid ered ex tre me exa m ples of polar compounds, will
ChemistryANowT~ Sign in at www. thomsonedu~com/login and go to the Chapter Contents menu to explore an interactive version of this figure accompanied by an exerdse. 622
C h a p t e r 14
I
Solu tio ns an d Thei r Beh avior
No npolar 12 Polar H20
Nonpolar CCl 4 -.~
~+---- Polar H20
Shake the test t ube
Nonpolar
~1--- CCI4 and 12
FIGURE 14.7 Dissolving ionic solids and entha lpy of solution. (a) Dissolving NaOH in water is a strongly exothermic process. (b) A"cold pack" contains solid ammonium nitrate, NH 4 NO" and a package of water. When tile water and NH 4N03 are mixed and t he salt dissolves, the temperature of the system drops, owing to the endothermic enthalpy of solution of ammonium nitrate P ',olnW = + 25.7 kJjmol).
(b)
not dissol ve in n onpo la r so lve n ts. T h is fact is amp ly borne out by o bservation . So d iu m ch lo ride, for exa m p le , will n ot dissolve in liquid s su ch as h exan e or CC I4 , hu t th e salt d o es h ave a sign ifica n t so lu bility in water, Man y io nic co m po u n d s are so lu b le in water; oU I , acco rd in g to th e so lu b ility guide lines 0 11 page 126, th e re are m all )' o th e r ionic so lids th at a re n o t. Pred ict in g th e so lu b ility of io nic co mpoun ds in water is complica ted. As mentioned ea r lier, 1\\'0 factors-ent halpy and en tro py- toge ther de te rmine the exten t to wh ich o ne substa nce di ssol ves in a not h e r. Fo r io n ic co m po un ds d isso lving in water, en trop y usu all y (but no t always) favors so lut ion . A favo rable en tha lpy facto r (n egative Doll) ge nera lly leads to a co m pou nd bein g soluble . Fo r exam p le , whe n sod ium hydro xid e di ssol ves in water, the so lu tio n warms up (Figure 14.7a) , an d sod ium h ydro xide d issol ves read ily in wa ter. An unfa vo ra bl e en thalpy factor, h owever, do es n ot g ua rantee that an ion ic com po u nd will no t be so lu b le. Wh en ammonium nitrat e d issolves in wate r, the so lu tio n becomes co lder (Figu re 14.7b ) , but ammo ni u m nitrate is still very so lu b le in water. Netwo rk so lid s, includ ing gra p h ite , di am ond, a n d q ua rt z san d (Si0 2 ) , do n ot dissolve in wate r. In deed, wh e re wou ld all th e beach es be if sa n d di ssol ved in wate r? T h e covalent chemical bo n d in g in n e two rk solids is sim p ly too stro n g to b e b ro ken; t he lattice rem ains intact whe n in co n tac t with water.
Enthalpy of Solution To understan d th e e n e rge tics of th e so lu tion process, let us view th is process at th e mol ecul a r level. \Ve will use th e process of d issolvin g p ot assiu m flu o rid e , KF, in water to illu strate what occ u rs, a nd th e e nergy-leve l di agram in Figu re 14.8 will assist us in following th e cha nges. So lid potassiu m flu o rid e h as a n io n ic crysta l la tt ice with a lternating K+ a n d F- ion s held in p lace by att ractive force s due to th ei r o pposite c harges. In water, th e se ion s arc se p a rated fr om eac h o t h er a n d hvdrated: th at is, th e y are su rro u nde d b y water mo lec u les (Figu re 14.1 ) . Io n-d ip ol e fo rce s o f attrac tio n h ind wate r mo lec u les stro n g ly to eac h ion. The e nergy c ha nge to go fro III th e reacta n t, KF (s) , to th e products, K+(aC\ ) a nd F-(aC\) , ca n b e co n sidere d to take pl ace in two stages:
1. Energy must be su pp lied to separate the io ns in th e lattice against th eir a ttractive forces. This is th e reve rse of the process defi n ing th e lattice e ntha lpy of a n io nic co m pou n d wit h a n e n tha lpy equal to - ~ la ll in:H ( ... page fiOO). 14.2
I
The Solution Process
623
FIGURE 14.8 Model for energy changes on dissolving KF. An estimate of the magnitude of the energychange on dissolving an ionic compound in water is achieved by imagining it as occurring in two steps at the particulate level. Here, KF is first separated into cations and anions in the gas phase with an expenditure of 821 kJ per molof KF. These ions are then hydrated, with j, ~d'" tion H estimated to be - 8 37 kJ. Thus, t he net energychangeis - 16 kl, a slightly exothermic enthalpy of solution.
~H =
- lattice enthalpy kJjmot
j,Hhydrduon = - 837
kJjmol
j. H = +82 1
(See ChemistryNow Screen 14.4 for exercises on t he energetics of solution form ation.)
- 16
kJ/m1
Se pa rating th e ions fr om on e an o the r is h ighly e nd o th e rmic bec a use the a ttrac tive fo rces between ions a re stro ng. 2. En e rgy is evo lved whe n th e in d ividual io ns a re transferred into wa ter, where eac h ion becom es su rroun de d by wate r mo lecules. Agai n, stro ng fo rces o f attrac tio n (io n- d ip o le forces) are involved . T his p rocess, re fe rr ed to as hydration wh en wate r is th e so lve n t, is stro ngly exoth e rmic. \Ve ca n th erefore re p rese n t the process o f di sso lvin g KF in term s of th ese ch emica l eq ua tio n s: Ste p I KF (s) -----> K +(g ) + F- (g) Ste p 2 K +(g) + F- (g) -----> K+ (aq)
- A lau icell
+ F- (aq)
A h yd ra tio nll
T he overal l re action is th e su m of th ese two ste ps. T h e e nt h alp y of th e o verall reaction, ca lled the e n th alpy of solution ( Aso11I / /), is th e su m of the two e n th alp ics.
Overall \\" e ca n use th is to estimate the value of A hyd ratio llll. For exa m p le, we estimate the lattice e n e rb'T fo r KF to be - 821 ~J/ mo l usin g a Born -Habe t cycle calculatio n (... page fiOO ) , a n d we measu re the valu e o f A,oIlJI in a ca lo rimetry ex pe rim e n t to be - l fiA kJ/ mol. From these two values , we ca ll d etermine A hyd ratiOlJ I to be - 837 kJ/mo l. As a ge n e ral ru le , to be so lu b le, a salt will have an en tha lpy of solu tio n th at is exot herm ic or o n ly slightly e n d o the rmic (Figu re 14.9 ). In th e latte r insta nce, it is assu me d th at the e nthalpy-d isfavo red so lu tio n p ro cess will be balan ced by a favora ble e ntro py o f so lu tio n. If th e e n tha lpy o f so lut io n is ve ry e ndothermic-beca use of a low hyd ra tio n e nergy, fo r exam p le- th e n the co mpound is unlikely to be so lu ble. \Ve ca n reaso n abl y specu late th at no npol ar so lven ts wo u ld 1I0 t so lvate io ns stro ngly, an d th at so lu tio n fo rm ati on wou ld th us be e n e rge tically un favorabl e. \re the refore predict th at a n ion ic co mpoun d , su ch as co p per(II) su lfate, is n ot very so lu ble in no n p o lar so lve n ts suc h as ca rbon tetrachl o rid e a nd octa ne (Figu re 14.4) . It is also useful to recog n ize th at th e e n thal py of so lu tio n is th e diffe re nce be twee n two ve ry lar ge num bers. Small variatio ns in e ither lattice e nergy o r hyd ratio n e nt hal pies ca n d et erm in e whe th e r a salt d isso lves e n doth erm ically or exo the rm ically. Fin ally, not ice th at th e two cn c rgy qu an tities, A lilttin:H a n d A hYllratiollH, a rc both affected by io n sizes an d io n cha rges (... pages 55 7 an d 600 ) . A salt co m posed of smalle r ions is expected to have a gre ater (mo re n egative) lattice e nt halpy beca use th e ions ca n be clo ser togethcr a n d expe rie n ce h igh er attractive force s. H owever, th e small 624
Chapter 14
I
Solutions and Their Behavior
Dipole-dipole attraction
andhYdr] :ondin9 8 - 0
8 +
• Ion-dipole attraction; ::--------J~=~ defined by entha lpy of :
hydration,
il hydrati(mH
000
Act ive Fi gure 14.9 Dissolving an ionic solid in water. This process is a balance of forces. There are intermolecular forces between water molecules. and ion- ion forces are at work in th e ionic crystal lattice. To dissolve, the ion-dipole forces between water and the ions (as measured by .l 1l~dra tion H) must overcome the ion- ion forces (as measured by il liltticeH) and the intermolecular forces in wate r.
Chemislry) ).Now'" Sign in at
Na+
www. • thomsonedu.ccm/Ioqtn and go to the Chapter Contents menu to explore an interactive version of this figure accompanied by an exercise.
Cl1000
Ion-i on att raction; defined by th e lattice ent halpy, illattk eH
size will also allow a clo ser a p proac h of so lve n t mole cul es a nd a greate r so lvatio n e ntha lpy. The net resu lt is th at sim p le co rre lations of so lu bility with stru ctu re (io n ic rad ii) o r th ermodynami c param et ers (dlallice H) arc ge ne ra lly n ot successfu l.
Enthalpy of Solution: Thermodynamic Data
Form ation of NaCl(s):
Na (s) + 'I, Cl,(g) -----> NaCl( s)
Data for Calculating Enthalpy of Soluti on
TABLE 14.1
As mention ed ea rlier, th e enthalpy of so lu tio n for a sa lt ca n be measured usin g a ca lori me te r. Thi s is usu ally d on e in a n o pe n syste m suc h as th e co ffee-cu p calori me te r d escr ibed in Section 5 .6. For a n ex pe rime n t run und er sta n da rd con d itio ns, th e resulting measurem ent produces a value fo r th e sta n da rd e n t halpy of so lu tio n , AsolJ I" , whe re sta n d a rd co n d itio ns refe r to a co nc e n tra tio n o f I mol al. Ta b les o f th e rm odyn ami c values o fte n include valu es fo r th e e n th aI pies o f formation o f aCJ ueous so lu tio ns of salts. Fo r exam p le, a value of dIH o fo r NaC I(aq ) o f - 407.3 ~J /m ol is listed in Tabl e 14.1 a n d Appe ndi x L. Th is value re fers to the formati on of a 1 m so lu tio n o f NaCI fro m the cleme n ts. It may be co nside red to in volve th e c n th alp ics of t\\TO ste ps: ( 1) th e form ati on of NaC l( s) from th e clements Na(s) a n d CI2 (g ) in thei r sta n dard states, a nd (2) th e form ation o f a I m so lut io n by di ssol ving so lid Na CI in wate r: D.jH'
~
- 411.1 kJ/mol
Dissolving Na Cl:
NaCl(s) -----> NaCl(aq, 1 m)
6.solnHo = + 3.9 kJjmo l
Net process:
Na(s) + 'I, Cl,(g) -----> Na Cl(aq, 1 m)
D.jH'
~
- 407.3 kJ/ mol
t>tH'(s) t>tH'(aq, t m)
(kl/mol)
(kl/mot)
liF
- 616.9
- 611.1
Nar
- 573.6
- 572. 8
KF
- 568.6
- 585.0
RbF
- 557.7
- 583.8
LiCl
- 408.7
- 445.6
NaCl
- 411.1
- 407.3
KC l
- 436.7
- 419.5
RbCl
- 435.4
- 418.3
NaD H
- 425.9
- 469.2
NH 4N0 3
- 365.6
- 339.9
Compound
Chemislryj l"Now'"
•
Sign in at www.thomsonedu .com/ Ioqtn and go to Chapter 14 Contents to see Screen 14.3 for a visualization of the process and for a problem on t he solution process • Screen 14.4 for an analysis of the dissolution of KF and an exercise on solution energetics
Calculating an Enthalpy of Solution
r
Problem Use the dat a give n in Table 14.1 to determ ine the entha lpy of solutio n for NH,N03 , the compou nd used in cold packs. St rategy Use Equation 5.6 and data from Table 14 .1 for reacta nts a nd products. 14 .2
I
The Solution Process
625
Henry's Constants ( 25 °C)' TABLE 14.2
SoLution The solution process for NH 4N 0 3 is represented by the equation La w
NH,NO,(s) --> NH ,NO,(aq) The entha lpy change for thi s process is calculat ed using entba tples of formation given in Ta ble 14.1:
Gas
kH (mol/ kg · bar)
N,
6.0 X 10- 4
0,
1.3 X 10 - 3
CO,
0.0 34
" From hltp :/ / webhook .ni st.g-ov/ cheuiistry/ . Nou-: I ba r = 0.9869 atm .
A" , H' ~ L [At H' (product)J - L[At H' (reactant)J ~ At H' [ NH, NO,(aq)] - At H' [ NH,NO,(sl] ~
- 339.9 kJjmo l - (-36 5.6 kJjm ol)
~
+25. 7 kJjm ol
The process is endot hermic, as indicated by the fact that ;lwlnW has a positive value and as verified by t he experime nt in Figure 14.7b. EXERCISE 14.2
Calculating an Enthalpy of Solution
Use t he data in Table 14.1 t o calculate the entha lpy of solutio n for NaOH .
14.3
Factors Affecting Solubility: Pressure and Temperature
Pr essu re a n d temperature a re two extern al factors th at influence so lu b ility. Both affect th e so lu bility of gases in liq u id s, whe reas o n ly temperature is an important factor in th e so lu b ility of so lid s in liq u ids.
Dissolving Gases in Liquids: Henry's Law The so lu b ility of a gas in a liq u id is direct ly proportio n al to the gas pressu re . T h is is a state me n t of H enry's law, (14.4)
FIGURE 14.10 Gas solubility and
pressure. Ca rbonated beverages are bott led under CO2 pressure. When t he bott le is opened, the pressure is released, and bubbles of CO" form within the liquid and rise to t he surface. After some time, an equilibrium between dissolved COl and atmospheric COl is reached. Because CO 2 provides some of the taste in the beverage, the beverage tastes flat when most of its dissolved CO 2 is lost. 626
C h a p t e r 14
I
where ,~ is th e gas so lu bility, Pg is th e p artial pressu re of the gas eo us so lu te , a nd kll is H enry' s law co ns ta n t Cra b le 14.2) , a constant c ha rac te r istic of the so lu te and so lvent. Ca rbo nated so ft d rin ks illu strate how He nry's law works. These beverages a re pa cked u nder pressure in a c ha m be r filled with carbo n dioxide gas , so me o f whic h dissolves in the beverag e . \ Vh e n th e ca n or bottle is ope ne d , th e parti al p ressu re of CO 2 a bove the so lu tio n drops, which ca use s the so lu bility o f CO 2 to d rop. Gas bubbles out of the sol ution (Fig u re 14.10 ). H e n ry's la w h as import ant co nsefJ11ences in SCUBA d iving. \ Vh e n you dive , th e p ressu re of th e air you breat h e must be balan ced agains t th e ex tern al p ressure of the water. In dee p e r d ives, the pressu re o f the gases in th e SCUBA gea r must be several atmos p heres and, as a result, mo re gas dis so lves in the blood. This ca n lead to a problem. If yo u ascend too rapid ly, yo u ca n ex p e rie n ce a p ainfu l a n d pote nti ally leth al condition referred to as "th e be nds," in which nitrogen gas bubbles form in th e blood as th e so lu bility of nitrogen d ecreases with d ecreasin g pressu re, In an effort to p re vent the bend s, divers may use a hel ium-oxyge n mi xture (ra th e r th an nitrogen-oxygen ) because h el iu m is not as so lu ble in blood as nitrogc n . \ Ve ca n bette r u n d e rsta n d th e effec t of p ressu re on so lu b ility by exam in ing th e syste m a t the part icul at e le vel. T h e so lu bility of a ga s is defi ned as th e conce n tration o f the di sso lved gas in eq u ilibriu m wit h th e su bs ta nce in the gas eous state . At eq u ilib riu m , th e ra te at wh ich so lu te gas molecu les esca pe th e so lu tio n a n d e n ter th e ga seo us sta te equa ls th e ra te at wh ic h ga s molecules ree n te r the so lu tio n . An in crease in p rcssure resu lts in more mo lec u les of ga s striking th e su rface of the liquid and entering so lu tio n in a give n tim e . The so lu tio n even tu ally reach es a n e w equilibrium whe n the conce n tratio n of gas dis so lved in the so lve n t is hig h e noug h
Solutions and Thei r Behavior
tha t the ra te of gas molecules esc aping the so lu tio n again equals the rate of gas mo lecu les e n teri ng th e so lu tio n. ChemistrY-9-Now~ Sign in at www.thomsonedu.com/Ioqtn and go to Chapter 14 Contents to see Screen 14.5 for an exercise and tutorial on Henry's law.
Using Henry's Law Problem Whatis t he concentratio n of O2 in a fresh water stream in equilibrium with air at 25 °Cand 1.0 bar? Express the answer in grams of O2 per kg of solvent. Strategy To use Henry's law to calculate t he molar solubility of oxygen, t he partial pressure of O2 in air must first be calculated. SoLution The mole fraction of O2 in air is 0.21, and, assuming the total pressure is 1.0 bar, the partial pressure of O2 is 0.21 bar, Using this pressure for Pg in He nry's law, we have:
""
Solubility of O2
= kfjPg =
(1.3
X IO-
l
kg . bar
mOl ) (0.2 1 bar) = 2.7 X 10- 4 mol/ l
This concentration, in grams per liter, can t hen be calculated using t he molar mass of O2: SoIuhili 1 tty 0 f O2
g)
= ( 2.7 X 10-' mO I) (31"O -- = kg
mol
0.0087 9/k 9
This concentration of O2 (8.7 mgjkg) is quite low, but it is sufficient to provide the oxygen required by aquatic life. EXERCISE 14.3
Hyperbari c chamber, an illustration of Henry's law. Aperson is placed inside the chamber, and the partial pressure of oxygen is raised to severa l times the norma l, atmospheric level. This raisesthe amount of oxygen dissolved in bloodand tissues. This techniqueis used to treat, among other things, decompression sickness, carbon monoxide poisoning, severe anemia, and certain nonhealing wounds.
• Limitations of Henry's Law Henry's law holds quantitatively onlyfor gases that do not interact chemically with the solvent. It does not accurately predict the solubility of NH] in water, for example, because this compound gives small concentrations of NH 4 + and OH- in water.
Using Henry's Law
Wh at is the concentration of CO 2 in water at 25 O( when the partial pressure is 0.33 bar? (A lthough CO 2 reacts with water to give traces of H+ and HCO]-, t he reaction occurs to such a small extent that He nry's law is obeyed at low ( 0 2 partial pressures.)
Temperature Effects on Solubility: Le Chatelier's Principle Th e so lu bility o f all gases in water d ecreases with in cr easing tem pe ratu re . You may re alize thi s from eve ryday o bse rva tio ns suc h as th e a ppcarance of bubbles of air as water is heated be low the boili ng poi nt. To u nd erstan d th e e ffec t o f tem peratu re 0 11 the so lu bility of gases, let us reexami ne th e en thalpy o f so lu tio n . Ga ses th at di ssolve to an ap p re cia ble ex te n t in water usu all y d o so in an e xothe rmic pro cess
T he reverse p ro cess, loss o f di ssol ved gas molecu les fro m a so lu tio n, requires c nergy as heat. T he se two p ro cesses can reach eq uilibriu m even tua lly. To und erstand how tem peratu re affec ts so lu bility, we tu rn to Le Chate lier' s principl e , wh ich states that a change in any of the facto rs dcter min ing a n equi libr iu m ca uses the system to adj ust by shifting in the d irec tion that reduces or co u n terac ts the effect o f the chang e. If a solu tio n of a ga s in a liqu id is he-ated, for ex am ple, th e eq uilibriu m will shift to absorb so mc of the added cnergy. T hat is, the re act ion Gas + liquid solve nt
Exothermic process is negative.
.i. solnH
saturated solution
+ energy
Add energy. Equilibrium shifts left. 14.3
I
Factors Affecting Solubility: Pressure and Temperature 627
esCl
180
.
~.
r
)Ijl-
160 140
RbCI
120
liCl
I
-SOl)
ml
• 51.
~
II
-'00
q, 100 r ~
o ~
~ 60
~~
C'l
KCl
40 ~,..l!i!!!~===== N aCl ~ li S0 2
20
'00 4
"--20
40
60
80
100
Temperat ure (0C)
(a) Temperature dependence of the solubility
(b) NH 4C l dissolved in water.
of some ionic compounds.
(e) NH 4Cl precipitates when the solutionis cooled in ice.
FIGURE 14.11 The temperature dependence of the solubility of some ionic compounds in water. Most compounds, such as NH4Cl,increase in solubility with increasing temperat ure.
FIGURE 14.12 Giant crystals of
potassium dihydrogen phosphate . TI1e crystal being measured by this researcher at Lawrence Livermore Laboratory in California weighs 318 kg and measures 66 x 53 X 58 em. The crystals were grown by suspending a th umbnail-sized seed crysta l in a 6-foot tan k of satu rated KH 2 P04 • The temperature of the solution was gradually reduced from 65 QC over a period of about 50 days. The crystals are sliced into thin plates. which are used to convert light from a giant laser from infrared to ultraviolet.
Module 19
shi fts to the le ft if the temperature is ra ised be cause ene rgy is abso rbed in the p ro cess that p roduces fre e gas mo lecules and p u re so lve nt. T his shift co rresponds to less gas d isso lved a n d a lowe r so lu bility a t h igh e r te mp e ra ture-the o bse rved res ult . The so lu bility o f so lid s in water is also affected by temperature , but, un like th e situatio n in volving so lu tio ns o f gas cs, no genera l pattern of be havio r is o bserved . In Figure 14.1 1, the so lu biliti es of several salts are plotted \'erSHS tem pe ra tu re . T he so lu bility o f man y sa lts in creases with inc reasing temperat ure , but there are notab le excep tio ns . Pre di ctio ns based on whet her th e en tha lpy o f so lutio n is po sitive o r nega tive wo rk most of the time , bu t exce p tio ns do oc cu r. Che m ists take adva ntage of the var iatio n of so lu bility with tem perature to pu rify compoun ds. If a co m po u nd is mo re so lu ble in a g iven so lven t at highc r temperatures, an impure sa mple of th e co m po u nd is di sso lved in the so lven t at a hig h tem perat u re . T he so lu tio n is coo led to d ecre ase the so lu bility (Fig u re l 4. l l c) . \Vhen the lim it of so lu bility is reached at the lower tcm pe rature, crystals o f th e pu re co m po u nd fo rm . If th e p ro cess is d one slowly and ca re fully, it is so me times po ssib le to obtain very large crysta ls (Fig u re 14.12) . Chemislry.o.Now'"
•
Sign in at www.thomsonedu .comjlogin and go to Chapter 14 Content s to see Screen 14.6 for a tut orial on temperature and Le Chatelter's principle.
14.4
Colligative Properties
If yOll d isso lve so me salt in wate r, the vapo r pressure o f the wate r over the so lu tio n will decrease , and th e water will evapo ra te less rapidly under the same con d itions. In add itio n , t he so lut io n will freeze below 0 °C and bo il above 100 "C. T hese observatio ns refer to th e co lliga tivc properties o f the so lu tio n, properties th at d e pe nd 628
Sign in at www.thomsonedu.comj togin to download the Go Chemistrymodule for this section or go to www.ichepte rs.comtc purchase modules.
o n the relati ve numbers o f so lute and so lvent particles in a so lut ion and no t o n th e ir iden tity
Changes in Vapor Pressure: RaouLt's Law Th e e qui liln-iu rn vapor p re ssure a t a pa rti cul ar te mperatu re is th e pressure of the vapo r when the liq ui d and th e vapor are in equilibrium (.... page 573) . \Vhen the vapo r prcssu re of th e so lve n t ove r a so lu tion is measured at a give n te mpe ratu r e , it is ex perimentally observed that • T he vapo r pressure o f the so lven t over the so lu tio n is lowe r than the vapor p ressu re o f th e p ure so lvent. • T he vapor pressu re of t he so lven t, P'O!WI1 I, is pro po rtio na l to th e rel ati ve numher of so lve n t molecules in the so lu tion ; that is, the so lve n t vapor pressure is proportion al to th e so lve n t mol e frac tio n, /~"J !wn l oc X." ln 'IlI ' Because so lve n t vapor pressure is p ropo rti onal to the re lative nu mber of so lve n t molecules, we can wri te th e fo llowing eq ua tion for th e eq uilib riu m \'apo r prcssure of the so lvent ove r a so lution : P solvent
= Xsolvent posolvent
(14.5)
This e q ua tio n, ca lle d Raoult's law , te lls u s that th e vapo r p ressu re of so lvent over a so lu tio n (P,o!l'en,) is so m e fract ion of t he pure so lve n t e quil ib ri u m vapor pressure ( P '\ olveJll)' For exam p le , if 95 % of the m ol ecul es in a so lution a re so lven t mo lec u le s ( X,o!n'rH = 0.95 ) , the n the va po r pressure of tile so lvent ( P:,,,II'l'nt) is 95 % of /J osolvent Like the id eal ga s law, Raoul t's law d escrib es a sim plified m o d el of a so lution . \Ve say that an ideal solu tion is one that obl)'J Raou lt 's law. No so lution is id eal , ho weve r, just as no gas is tr u ly id e al. Neve rt hel ess, Raoul t's law is a good approx imatio n of so lu tio n be havior in m an y ins tances, especially a t low so lute co nce n tra tio n . For Raoult 's law to hol d , th e forces of attraction be tween so lu te and so lvent m ol ecul es m ust be the sam e as those between so lve n t mo lecules in the p u re so lvent. T h is is fr eque ntl y the case whe n m ol ecul es with sim ilar st ructu res are invol ved . Sol u tion s of o ne hyd ro ca rbon in an o the r ( hexa n e , C 6 H J.j., d issol ved in octane , C gII I8 , for examp le ) fo llow Raoult 's law qu ite close ly. If so lvent-so lute inte raction s are st ro nge r than so lvent-so lven t int eractio n s, th e actual vapor pressure will be lower th a n calcu la te d by Rao ult's law. If th e so lven t- so lu te in te ractio ns are wea ke r than so lve n t- so lven t int eractions, th e vapor pressure will he high e r.
• Raoult's l aw Raoult's law is named for Francois M. Raoult (1830-1901), a professor of chemistry at the University of Grenoble in France, who did the pioneering studies in this area.
Chemislrv.ci.Nowm
•
Sign in at www.t homsonedu.comj login and go to Chapter 14 Contents to see Screen 14.7 for an exercise and a tutorial on Raoult's law.
. . Using Raoult's Law
f Problem Suppose 651 g of et hylene glycol, HOCH,CH ,OH, is disso lved in 1.50 kg of water. What is t he
I
va por pressu re of t he water over th e solution at 90 ' C? Assume ideal behavio r for the solution .
Strategy To use Ra oult's Law (Equation 14.5). we first must calcuLate t he moLe fraction of the solvent (water). We aLso need the vapor pressure of pure wat er at 90 °C (= 525.8 mm Hg, Appendix G) . 14 .4
I
Co LLigative Properties
629
Chemical Perspectives
Henry's Law and the Killer Lakes of Cameroon
It was evening on Thursday, August 21, 1986. Suddenly, people and animals around l ake Nyos in Cameroon, a small nation on the west coast of Africa, collapsed and died. By the next morning, 1700 people and hundreds of animals were dead. The calamity had no apparent cause-no fire. no earthquake, no storm. What had brought on this disaster? Some weeks later, the mystery was solved. lake Nyos and nearby lake Monoun are crater lakes, which formed when cooled volcanic craters filled with water. Lake Nyos is lethal because it contains an enormous amount of dissolved carbon dioxide. The CO 2 in the lake was generated as a result of volcanic activity deep in the earth. Underthe high pressures found at the bottom of the lake, a very large amount of CO 2 dissolved in the water. On that fateful evening in 1986, something happened to disturb the lake. The (02saturated water at the bottom of the lake was carried to the surface, where, under lower pressure, the gas was much less soluble. Approximately one cubic kilometer of carbon dioxide was released into the atmosphere, much like the explosive release of CO 2 from a can of carbonated beverage that has been shaken. The CO 2 shot up about 260 feet; then , because this gas is more dense than air, it hugged the ground and began to move with the prevailing breeze at about 45 miles per hour. When it reached the viltages 12 miles away, vital oxygen was displaced. The result was that both people and animals were asphyxiated. In most lakes, this situation would not occur because lake water "turns over" as the seasons change. In the autumn, the top layer of water in a lake cools; its density increases; and the water sinks. This process continues, with warmer water coming to the surface and cooler water sinking. Dissolved CO 2 at the bottom of a lake would normally be expelled in this turnover process, but geologists found that the lakes in Cameroon are different. The chemocline, the boundary between deep water, rich in gas and minerals, and the upper
630
Chapter 14
I
lake Nyos in Cameroon (western Africa), the site of a natural disaster. In 1986, a huge bubble of COl escaped from the lake and asphyxiated more than 1700 people.
layer, full of fresh water, stays intact. As carbon dioxide continues to enter the lake through vents in the bottom of the lake, the water becomes saturated with this gas. It is presumed that a minor disturbance-perhaps a small earthquake, a strong wind, or an underwater landslide-caused the lake water to turn over and led to the explosive and deadly release of CO 2, Lake Nyos remains potentially deadly. Geologists estimate that the lake contains 10.6 to 14.1 billion cubic feet (300-400 million cubic meters) of carbon dioxide. This is
Solutions and Their Behavior
about 16,000 times the amount found in an average lake that size. Ateam of geologists from France and the United States has been working to resoLve this potentiaL threat . In early 2001, scientists lowered a pipe, about 200 meters long, into the lake. Now the pressure of escaping carbon dioxide causes a jet of water to rise as high as 165 feet in the air. Over the course of a year, about 20 million cubic meters of gas willbe released. While this has been a successfulfirst step, more gas must be removed to make the lake entirely safe, so additional vents are planned.
Solution We first calculate the amounts of water and ethylene glycol and, from t hese, t he mole fraction of water. Amou nt of water
= 1.50 X 103 g
= 651 g
Amount of ethylene glycol
=
X
(~) = 83.2 mol water 18. 02 g (~) 62.0 7 g
83.2 mol water 83.2 mol water + 10.5 mot glycol
water
= 10.5 mol glycol
= 0.888
Next. we apply Ra oult's law, calculating t he vapor pressure from t he mole fraction of water and t he vapor pressure of pure water:
=
P"'~te r
X wate lOwater
= (0.888)(525.8 mm Hg) = 467 mm Hg
The dissolved solute decreases t he vapor pressure by 59 mm Hg, or about 11%: .lP"'ater
= Pwate,
-
powater
= 467 mm Hg - 525.8 mm Hg = - 59 mm Hg
Comment Et hylene glyco l dissolves easily in water, is noncorrosive, and is relatively inexpensive. Because of its high boiling point. it witl not evaporate readily. These features make it ideal for use as antifreeze. It is, however, toxic to animals, so it is being replaced by less toxic propylene glycol for this application. EXERCISE 14.4
Using Raoult's Law
Assume you dissolve 10.0 g of sucrose (C 12 H220 ll ) in 225 ml (225 g) of water and warm t he water to 60 °C. Wh at is the vapor pressure of t he wate r over t his solution? (Appendix G lists PO(H zO ) at various temperatures.)
Ad d ing a no nvol at ile so lute to a so lve nt lowers th e vapo r p ressure o f the so lven t (Exam ple 14.4) . Raoul t's law can be mod ified to ca lculate d irectl y th e loweri ng o f th e vapo r p ressu re , .iP""lvt'llt, as a fu nction of the mo le fra ction of the so lu te. ~ Psolvent = P solvent -
Substitu tin g Raoul t's law fo r ilP solvent
P~oh'r l1l'
posolvent
we have
= (Xsolvent P" solvent)
-
P" solvent = -
(1 -
Xsolvent)
pOsolvent
In a so lut io n th a t has o n ly th e volatil e so lve n t and one nonvolatil e so lu te, the su m of the mol e fra ction of so lve n t an d so lu te mus t be I: Xsolvent
Th erefo re , 1 -
Xs " ll'clll
+ Xsolute
=1
= X"',hun and the equatio n for d,P~"hTIll ca n be rewritten as d, P solvent
= - XsoLu te posolvent
(14 .6)
Thus, th e change ill the vapor p ressure o f th e so lven t is proport ion al to th e mol e fra ct io n (the re lative numbe r o f panicles) o f so lu te .
14.4
I
Colligative Properties
631
Boiling Point Elevation Suppose you have a so lu tio n of a nonvolatile solu te in the vo latile so lve n t benzene. If th e so lu te co nce n tra tio n is 0.200 mol in 100. g of benzen e (C 6 116 ) ( = 2.00 mo ll kg ) , thi s mean s th at Xh l• l1/ t>tlt: = 0 .865. Using Xh t' ll / t>t l t> a n d applyin g Raoult 's law, we ca n calcu late th at th e va p or pressure of the so lven t at 60 "C will drop from 400. nun II g for th e pu re so lve n t to 346 mm II g for th e so lutio n: Pbenlene = Xbenlene
P\ enlene = (0. 865)(400. mm Hg) = 346 mm Hg
This point is marked on th e vapor pressure g ra p h in Figure 14.13. Now, what is th e va por pressu re whe n the temperature of the so lu tio n is rai sed an other 10 "C? Th e vapo r pressure of pure benzene , P\ c: tlll'Ill" b ecomes larger with in creasing temperature , so l~w tl Jt> ll t' for th e so lu tion mu st also become larg e r. T h is n ew point, and additional o n es ca lcu lated in th e sam e \,,'ay for other temperatures, define the va po r p ressu re curve for th e so lu tio n (the lower cu rve in Figure 14. 13 ) . An important o bse rvatio n we ca n make in Figure 14.1 3 is th at th e va por pressure lowerin g ca use d by th e n onvolatile so lu te leads to an in crease in th e boiling point. Th e n ormal boilin g point of a liquid is th e temperatu re at wh ich its va por p ressure is eq ua l to I atm o r 760 mm Hg (.... p age 576) . In Figure 14.13 , we see th at th e no rmal boiling point of pure benzene (at 760 mm Hg ) is a bou t 80 "C. Traci ng the vapo r pressure cu rve for th e so lut io n , we also see that the vapor pr essure reaches 760 1010 Hg at a temperature a bo u t 5 "C higher th an thi s valu e . The vap o r pressure curve and in crease in th e bo ilin g point shown in Figure 14.1:-1 re fer specifically to a 2.00 m so lu tio n. \Ve mig h t wo nder h ow th e b o iling point of th e so lu tion wou ld va ry with so lu te co nc e n tra tion. In fact , a sim p le rel ati onship ex ists
FIGURE 14.13 Lowering the vapor
pressure of benzene by addition of a nonvolatile solute. Tile curvedrawn in red represents tile vapor pressureof pure benzene, and the curvein blue represents the vapor pressure of a solution containing 0.200 molof a solute dissolved in 0.100 kg of solvent (2.00 m). This graph was created using a series of calculations such as those shown in the text. As an alternative, the graph could be created by measuring variousvapor pressures for the solution in a laboratory experiment. (See ChemistryNow Screen 14.8 Colligative Properties, to view an animation of t his vapor pressure lowering.)
800
700
600
E S
500
&
," ~
~
~p
LIQUID
400
at 60
:;
for
~
•
>
Xsolute = 0.135
300
VAPOR
Pure benzene 200 Benzene + solute 100 0 20
632
C h a p t e r 14
I
Solutions and Their Behavior
O(
} = 54 mm
I
BP pure benzene 30
40
50 60 Temperature (0C)
, , , , ,
70
st
',
15.1°1 BP
, , , solution , , 80
Some Boiling Point Elevation and Freezing Point Depression Constants
TABLE 14.3
NormaL Boiling Point
NormaL Freezing Point
('e) Pure Solvent
SoLvent
100 .00
Water Be nze ne
80.10 207.4
Camphor
('e) Pure Solvent
KIp ('(1m)
+ 0.5121
0.0
- 1.86
+ 2.53
5.50
- 5. 12
179. 75
- 39.7
+ 5.611
61.7 0
Chlo roform
Kbp ('(1m)
+ 3.63
(( HCl,)
be tween boilin g po int elevati on a n d m ol al co nc e n tration : T he boi ling point elevatio n , A7 hp, is d irectly pro portio nal 10 the m olalit y of th e so lu te .
Elevation in boiling point = ATbp =
Kbpm solute
(14.6)
In thi s eq ua tio n , l\bp is a pr oportionality co nsta nt ca lled th e molal boiling point elevation constant.lt has th e units of deg recs/ mo lal (OC/rn) . Values for kbp a re dc tc rmi ne d ex pe rime ntally, a n d d iffe rent so lve n ts have di ffe re nt value s (Table 14.3) . Fo rmally, th e valu e co rres po nd s to th e e levation in boili ng point fo r a I m so lut ion.
Chemislry.o.Now'
•
Sign in at www.thomsonedu .com/Ioqtn and go to Chapter 14 Contents to see Screen 14.8 for an exercise and a tutorial on the effect of a solute on the solution freezing point.
Boiling Point Elevation Problem EugenoL the active ingredie nt in cloves, has t he formula ClO Hu 0 2 • What is the boiling point of a solutio n conta ining 0,144 g of th is compound disso lved in 10.0 g of benzene? Strategy We can use Equation 14.6 to calculate the change in boiling point . This va lue is the n added to t he boiling point of pure benzene t o provide the answer. To use Eq uation 14.6, you need a value of Kbp a nd the molality of the solutio n. The Kbp value for be nzene is given in Table 14.3, but you need to calculat e th e molality, m. SoLution 0.144 g eugeno l
c
-
euge
~ P u re wa te r
.....:'\-_ _. -- 5% sugar 95%water
~ Se m i permeable membrane
Height of solution column
1\.
FIGURE 14.15 The process of osmosis. (a) The bag attached to the tube containsa solution that is 5% sugar and 95% water. The beaker contains pure water. The bag is made of a material that is semipermeable, meaning that it allows water, but not sugar molecules, to pass through. (b) Over time,water flows from the region of low solute concentration (pure water) to the region of higher solute concentration (the sugar solution). Flow continues untilthe pressure exerted bythe column of solution in the tube abovethe water level in the beaker is great enough to resultin equal rates of passage of water molecules in both directions. The heightof the column of solution (b) is a measure of the osmotic pressure.
(See ChemistryNow Screen 14.9 Colligative Properties, for an animation of osmosis.)
14 ,4
I
Colligative Properties
635
which o n ly certain types of molecules can pass. H ere , water m o lecul es ca n pass th ro ug h the memb ran e , but large r suga r mo lecu les (o r hydrat ed io ns) can no t (Figu re 14.16) . Wh e n th e ex pe riment is begu n , the liqnid level s in th e beaker and the tub e are the same. Ove r tim e, howeve r, th e level of t he sugar so lu tion in side th e tub e rises, th e level of pure water in th e beak er falls, and th e sugar so lu tion becomes mo re dilu te . Event ua lly, no fu rth e r net cha nge occu rs; equi libri um is reac hed. From a molecul ar point of view, th e sem ipe rmeable m embran e d oes not p resent a barri er to the move ment o f water mol ecul es, so the y move th rough the m em b ran e in both directions. Over time, mo re wate r molecul es pass t h ro ug h th e m embran e fro m the pure wate r side to th e so lu tio n side th an in th e o p pos ite d irec tion . In effect, wate r molecul es ten d to move fro m regions of low so lu te concentration to regions o f h igh so lu te concentra tio n . T he same is true fo r an y so lven t, as lo ng as th e membran e allows so lven t mol ecul es but not so lu te mo lecul es or io ns to pass th rough . "'-Thy does th e syste m eve nt ually reac h eq uilibrium? Clea rly, the so lutio n in th e tube in Figu re 14.15 ca n never reach ze ro sugar o r salt concen tratio n, whic h wou ld be re q uire d to eq ua lize th e nu mber of water molecul es m ovin g th rough the memb ra ne in eac h d irectio n in a given time. Th e answe r lies in the fact th at the so lu tio n moves high e r a nd higher in th e tube as os mosis co n tin ues and water mo ves into th e sugar solu tio n . Eventually, th e p ressu re exerted by th is co lu m n of so lu tion cou n te rba lances the pressure exe rted by th e wate r moving th rough the m embran e fro m th e p ure wate r side, and no furth e r ne t move me n t of water o ccurs. An eq uilib rium of forces is ac hieved. Th e p ress ure created by th e co lu m n of so lutio n fo r th e syste m at eq ui librium is called th e osmotic pressure, A measu re o f this p ressure is th e diffe ren ce betwee n th e height of th e so lu tio n in th e tube and th e level of pure water in th e beaker.
n.
FIGURE 14.16 Osmosis atthe particulate level. Osmotic flow th rough a membrane th at is selectively permeable (semipe rmeable) to wat er. Dissolved substances such as hydrated ions or large sugar molecules cannot diffuse.
636
Chapter 14
I
Soluti ons and Their Beha vio r
O sm o tic pressu re is a co lliga tive prope rly. From ex pe rime ntal measu rem en ts o n dilute so lu tio ns, it is known t ha t os mo tic p ressure and co nce n tratio n ( r) are rel ate d by the equa tio n
n=
cRT
(14.8)
In th is eq uatio n, c is th e molar co nce n tra tio n (in moles per liter ) ; R is th e gas co ns ta n t; a nd Tis th e abso lu te tem perature (in kelvin s) . Using a value fo r th e gas law co nstan t of 0.082057 L . at m/ K . mol allo ws calc ula tio n of th c os mo tic prcssu rc in atmo sp he res. Th is eq ua tio n is analogo us to th e ideal gas law ( PV= uR T ), with takin g th e place o f P and c be ing equiva len t to ulE Beca use pressu rc s o n th e order o f 10 - 3 atm are easily measu re d , co ncc n tra tio ns of very dilute so lu tio ns (as low as a bo u t 10- 4 .M) ca n be d et ermin ed th rough measu re me n ts of o smo tic p ressu re . O ther exa mp le s of os mos is a rc sh own in Figure 14.17. In this ca se , the cgg's m embran e serves as the se m ipermea ble membr an e . O smosi s occurs in o n e di rection if th e co nce n tratio n o f so lu te is g reate r in side th e egg th an in th e e xte rio r so lu tio n an d oc cu rs in th e o t he r d ir ec tio n if t he co nc e n tr a tio n so lu tio n is less in sid e t he egg tha n it is in th e exte rior so lu tio n . In bo th cases, so lven t flo ws fro m th e re gio n o f lo w so lu te co nce n tra tio n to the regi o n of h igh so lu te co nce n tra tio n.
n
n
Colligative Properties and Molar Mass Determination Early in th is boo k, yo u learn ed how to ca lcu late a molecu lar form ula fro m an e mpirical formula whe n give n th e mol ar mass. Bu t how do yo u know th e mol ar m ass of an unkn own co m pou nd ? An ex pe rim e nt must be carried o u t to find thi s cruc ial pi ece of informa tio n , and o ne way to do so is to usc a co lliga tivc p ro perty of a
(a) Afresh egg is placed in dilute acetic acid. The acid reacts with the CaC0 3 of the shell but leaves the egg membrane intact.
(b) If th e egg, with its shell removed, is placed in pure water, t he egg swells.
(c) If t he egg, wit h its shell removed, is placed in a concentrated sugar so lution , t he egg shrivels.
FIGURE 14.17 An experiment to observe osmosis. You can try th is experiment in your kitchen. In the first step, use vinegar as a source of acetic acid. (See Chemistry Now Screen 14.1, Puzzler, and Screen 14.9, Colligative Properties, for a video of this expertment.)
14 .4
I
Colligative Prope rties
637
so lu tio n o f the co m po u nd. T he same basic lo gic is used fo r each o f the co lliga tive p ro pert ies stud ied: Change in vapor pressure, boiling point elevation, freezing point depression, or osmotic pressure
Solution concentratio n
Determining Molar Mass from Boiling Point Elevation Problem A solution prepared from 1.25 g of oil of wintergreen (methylsalicylate) in 99.0 g of benzene has a boiling point of 80.31 "C. Determine the molarmass of this compound. Strategy Ca lculations using colligative propertiesto determine a molar mass always follow the pattern outlined in the text.
Solution We first use the boiling point elevation to calculate the solution concentration: Boiling point elevation (..H bp) = 80.31
=~=
c solute
K
bp
0.21 °( 2.53 O(jm
O( -
80.10
O(
= 0.21
O(
0.083 m
The amount of solute in the solution is calculated from the solution concentration: 0.083 mOl )
Amount of solute = ( 1.00 kg
(0.099 kg solvent)
= 0.0082 mol solute
Now we can combine the amount of solute with its mass to obtain its mo lar mass: 1.25 9 0.008 2 mol
-:-:'c=--"---c ~ 150 gjmol
Comment Methyl salicylate has the formula
( sHsD} and
a molarmass of 152.14 gjmol.
. , Osmotic Pressure and MoLar Mass Problem Beta-carotene is the most important of the A vitamins. Its mola r mass can be determined by measuri ng the osmotic pressuregenerated by a given mass of l3-carotene dissolved in the solvent chloroform. Calculate the molar mass of l3·carotene if 10.0 mLof a solution containing 7.68 mg of l3·carotene has an osmotic pressure of 26.57 mm Hg at 25.0 "C. Strategy First, use Equation 14.8 to calculate the solution concentration from the osmotic pressure. Then. use the volume and concentration of the solution to calculate the amount of solute. Finally, find the molar mass of the solute from its mass and amou nt.
Solution The osmotic pressurecan be used to calculate the concentration of l3-carotene:
Now the amount of l3-carotene dissolved in 10.0 mLof solvent can be calculated: (1.429 x 10- ' moVL)(0.0100 L)
~
1.43 X 10- 5 mol
We can combine the amount of solute with its mass to calculateits molarmass: 7.68 X 10-' 9 ~ 538 gjmol 1.43 x 10- s mol
Comment Beta-carotene is a hydrocarbon with the formu la C4oHs6 (molar mass = 536.9 gjmol). 638
Chapter 14
I
Solutions and Their Behavior
EXERCISE 14.7 Osmotic Pressure and MoLar Mass A 1.40-9 samp le of polyethyle ne, a common plastic, is dissolved i n enough benzene to give exactly 100 ml of solution. The measured osmotic pressure of the solution is 1.86 mm Hg at 25 °C. Calculate the average molar mass of the polymer.
Ji ;'·"''''1'1(~
CoLligative Properties of Solutions Containing Ions
c
In th e n o r th ern U n ite d Sta te s, it is commo n prac tice to sca tte r sa lt o n snowy o r icy roads o r sidewalks. Wh e n th e su n sh ines O il the sn ow or patc h of ice , a sma ll am o u n t melts, and so m e sa lt di ssol ves in th e wa ter. As a re sul t o f the disso lved so lu te , the fr eez ing po int of the so lu tio n is lowe r th a n 0 "C. T h e so lu tio n "ca ts" its way th ro ug h the ice , brea king it u p, a n d th e icy pat c h is no longe r d an ge ro us fo r drivers o r for people walkin g. Salt (N aC l) is th e m o st co m mon su bs ta nc e u sed o n ro ad s be cause it is in e xpensive a n d di ssol ves read ily in wat e r. Its rel atively low mo lar m ass mea n s th at th e effec t
A Closer Look
Osmosis and Medicine
Osmosis is of practical significance for people in the health professions. Patients who become dehydrated through illness often need to be given water and nutrients intravenously. Water cannot simply be dripped into a patient's vein, however. Rat her. the intravenous solution must have t he same overalLsolute concentration as the patient's blood: t he solution must be Isoosmotic or isotonic (Figures Aand B. part a). If pure water was used, the inside of a blood cell would have a higher solute concentration
(lower water concentration), and water would flow into the cell. This hypotonic sit uation would cause the red blood cells to burst (lyse) (Figure B, part c). The opposite situation, hypertonicity, occurs if the intravenous solution is more concentrated than the contents of the
~" Putting salt on ice assist s in melting the ice.
blood cell (Figure B, part b). In t his case, t he cell would lose water and shrivel up (crenate). To combat this, a dehydrated patient is rehydrated in t he hospita l with a steri le saline solution that is 0.16 MNaCl, a solution t hat is isotonic with the cells of t he body.
.:
-,•.....(......../
:./ '\ . •
/"':
, c ..,.'
'
;1
. . JI'. ••.
•
~1011 . . . .....
•
'u
-~ :
-_Il . - ~~
•
. . ...
='":-::='!.t:i!
! I
:~~ii:
FIGURE A An isotonic saline solution. This solution has the same molality as body fluids.
(a) Isotonic solution (b) Hypertonic solution (c) Hypotonicsotution FIGURE B Osmosis and living cells. (a) Acellplaced in an isotonic solution. The net movement of waterinto and out of the cellis zero because the concentration ofsolutesinside and outside the cellis the same. (b) In a hypertonicsolution, the concentration of solutes outside the cellis greater than that inside. There is a net flow of waterout of the cell,causing the cellto dehydrate, shrink, and perhaps die. (c) In a hypotonic solution, the concentration of solutes outside the cellis less than that inside. Th ere is a net flow of waterinto the cell,causing the cell to swell and perhaps to burst (or lyse).
14 .4
I
Colligative Properties 639
per gram is large . III ad d ition, salt is especially effective be cau se it is an el ectrol yte . T hat is, it d issolves to give ions in so lu tio n: NaCl(s)
---?
Na +(aq)
+ Cl-(aq)
Recall that co lligative p roperti es depe nd no t o n what is d isso lved b ut onl)' on the number of particles of solute per solvent particle. When I mo l o f NaC I d isso lves, 2 mo l of io ns form, whic h mean s th at th e effect on th e freez ing po int of wate r sho uld be twice as large as that ex pected for a mole of sugar. A 0. 100 m sol u tio n o f NaCI really co n ta ins two so lutes, 0. 100 m Na ' and 0. 100 m CI- . Wh at we sho uld use to estimate th e freezing po in t d ep ression is th e total mo lality o f so lu te pa r ticles:
m"", ~ m(Na +) + m(Cl-)
~ (0 .100
+ 0 .100 ) mol/kg ~
0 .200 mol/kg
M ,p = ( - 1.86 ' Cj m )(0.200 m) = - 0 .37 2 ' C
To estim ate th e freez ing point d e pressio n for an io nic co m po u nd . first find the mo lality of so lu te fro m the mass and molar m ass of the co m po u nd an d the m ass of the so lvent. T he n, mu ltiply the m o lality by the num be r of io ns in the fo rmu la : two for NaC l, three for Na 2S0 ,1, four fo r LaCI;3, five fo r AI2 (S0 4h , and so on . Tab le 14.4 shows that as the co nc e n tra tio n of NaC I d ecr eases, A 7lp fo r NaC l ap p roaches but does no t qu ite reach a value that is two tim es larger than the value d etermined assu m ing no d isso ciat ion . Likewise , A'I,p fo r Na2S0 4 approaches bu t does not reach a value that is th ree time s larger. T he rati o o f the ex perime ntally observed value o f AJ lp to the value ca lcula ted, assu m ing no d issoc iat io n , is ca lled th e van't Hoff fa cto r after J aco bus H curi ku s van't Ho ff (1852-1 9 11) , who was invo lved in stud ying this p hen o men o n. T he van 't lIoff fac tor is re presen te d by i.
i=
ilTfp . measured ~T~ , calculated
Of
o.Tfp measured
= Kfp
XmXi
(14.9)
Th e nu m be rs in the last co lu m n of Tab le 14.4 are van ' t Ho ff factors. These values ca n be used in ca lcu latio ns o f any co lliga tivc property. Vapo r prcssu re lowerin g,
TABLE 14.4 Mass %
Freezing Point Depressions of Some Ionic Solutions m
.l Tf p
.;lTf p
.lTtp, measured
(mol/ k9)
(measu red, 0c)
(calculaled, 0c)
J.1fp, calculated
NaCi
0.007 00 0.500 1.00 2.00
0.0120 0.0860 0 .173 0.349
-
0.0433 0. 299 0. 593 1.186
-
0.0223 0.160 0 .322 0.649
1.94 1.87 1.84 1.83
0.00 493 0.0354 0.071 1 0.144
-
0. 0257 0. 165 0.320 0.606
-
0.00917 0 .0658 0.132 0. 268
2.80 2.51 2.42 2.26
Na2S0..
0.00700 0.5 00 1.00 2.00
640
Chapter 14
I
Solutions and Their Behavior
Case Study
Henry's Law in a Soda Bottle
By now, everyone has seen it on YouTu be or has tried it. Drop a Mentos 1M into a bottl e of soda (preferably diet soda), and a geyser of
It turns out this chapter, as well as the chapters on gases (Cha pter 11), kinetics (Cha pte r 15), and equilibrium (Chapter 16) can help us explain what is happening. Carbonated sodas are bottled under a high pressure of CO2 , Some of the gas dissolves in the soda, but some also remains in t he small space above t he liquid (called the "heads pace"). The pressure of the CO 2 in t he headspace is between 2 and 4 at m. Whe n the bottle cap is removed, the CO 2 in t he headspace escapes rapidly. Some of the dissolved CO 2 also comes out of solution, and you observe this as bubbles of gas rising to the surface (Figure 14.10). If the bottle remains open, t his continues until equilibrium is established wit h COz in t he atmosphere (where t he partial pressure of CO2 is 3.75 X 10-' atm),
soda erupts from the bottle.
CO ,(soiution)
A Mentos was dropped into a large bottle of Diet Coke. For moreinformation, see J. E. Baur and M. B. Baur, Journal o/Chemical Education,
CO,(g)
and the soda goes "flat." If the newly opened soda bottle is undistu rbed, however, the loss of CO 2 from solution is rather slow because bubble form atio n is not rapid, and your soda keeps its fizz. But why is bubble formation slow? The reason for this is explained by the physics of
bubble formation. Fo r a bubble to form, nucleation sites must be available. These can be impurities in the water or the rough surface of an ice cube or bottle or drinking glass. Th e more nucleation sites t here are available, the more rapid the bubble formation. The surface of a Mentos a pparently has many such sites and promotes very rapid bubble formation.
Questions: 1. If the headspace of a soda is 25 mL and the pressure of CO2 in the space is 4.0 atm ( = 4.0 bar) at 25 of, whatamountof CO 2
is contained in the headspace? 2. If the CO 2 in the headspace escap es into the atmosphere where the partial pressure of CO 2 is 3.7 X 10- 4 otm, whatvolume would the CO 2 occupy (at 25 °C)? By what amount did the CO2 expand when it was released? 3. CO 2 obeys Henry's law to about 5 bar. What is the solubility of CO 2 in water at 25 °C when the pressure of the gas is 4.0 bar ( = 4.0 atm)? What amountof CO 2 is dissolved in 710 g of diet soda? 4. What is the solubility of CO 2 in water at 25 °C when the pressure of the gas is 3.7 X 10- 4 bar? Answers to these questions are in Appendix O.
Vol. 83, pages 577-580, 2006.
boiling po in t eleva tio n , freezin g poin t d ep ression , and osmo tic p ressure are all larger fo r e lectro lytes th an for no nclcct rol ytcs o f th e sam e mol ality. The varr 't H off factor ap p roaches a who le num be r (2 ,3, and so o n ) o n ly with very dilu te so lu tio ns. In more co nc en tra te d so lut io ns, the ex pe rimen ta l freezing po in t d epressio ns ind icate th at th ere are fewer io ns in so lu tio n th an ex pect ed . T h is beh avio r; which is typ ical of all io nic co m po u nds, is a co nseq uence of the str o ng attrac tio ns be tween io ns. T he result is as if so me o f the positive and negative ion s arc paired , d ecreasing th e tot al mo la lity o f particles. Indeed , in m ore co ncen tra ted so lu tions , and es pecially in so lve n ts less po lar than wate r, io ns are ex te ns ively associated in io n pairs and in even larg er clu sters.
. , Freezing Point and Ionic Solutions Problem A 0.00200 m aqueous solution of an ionic compou nd, Co(NH lh(NOz)Ct. freezes at - 0.00732 How many moles of ions does 1.0 mol of the salt produce on being dissolved in water?
r
-c.
Strategy First, calcu late .:lTfp of the solution assuming no ions are produced. Compare this value with the actual value of .:lTfp' The ratio wi ll reflectthe number of ions produced. 14.4
I
Co lligative Properties 641
SoLution The freezing-point depression expected for a 0.00200 m solution assuming that the salt does not dissociate into ions is -l Tfp calculated
= Kfpm = ( -1.86
O( )( 0.0200 m)
=
- 3.72 X 10- 3
O(
Now compare the calculated freezing point depression with the measured depression. This gives us t he van't Hoff factor:
i
=
-IT,p, measured ~Tfp '
calculated
- 7.32 X 10- 3 - 3. 72 X 10 3
O(
---'-=---'-~~ ~
O(
1. 97 = 2
It appears that 1 mol of this compou nd gives 2 mo l of ions. In t his case, the ions are [(o(NH3h(N02 ) ] +
and (1 -. EXERCISE 14.8
Freezing Point and Ionic Compounds
Ca lculate t he freezing point of 525 g of water t hat contai ns 25 .0 g of NaCl. Assume i. the van't Hoff factor, is 1.85 for NaCl.
14.5
Colloids
Earl ier in thi s cha p te r, we d efined a so lu tio n br oadl y as a homogeneous mi xture of two o r more su bsta nces in a single ph ase ( pa ge ( 17) . To th is d efinition we sh o u ld add th at , in a tr ue so lu tion , n o se ttling of th e so lu te sh o u ld be obse rved a n d the so lu te particles sh o u ld be in th e fo rm o f io n s o r rel atively sm all mol ecules. T h us , NaCI a n d suga r form true so lu tio ns in water. 'rou a re also familiar wit h suspe ns io ns, wh ich resu lt, for exa mp le, if a h an dful o f fine sa n d is ad de d to water a n d sha ke n vigo ro usly. Sand p a rticles a re still visib le a n d g ra d u ally sett le to th e bottom o f th e beak er o r b ottle. Colloidal dispersions, also ca lled colloids, represent a sta te intermedi at e between a so lu tio n a n d a sus pe ns io n . Co llo id s incl ude many o f th e foods yo u ea t and th e mat erials a ro u n d yo u; a mon g th em a rc .JELL-O ®, milk, fog , and porcel ain (see Table 14.5 ) . Aro u n d 1860 , the British c h e m ist T ho mas Gra ham ( 1805- 1869) fou nd th at substa nces suc h as sta rc h , g ela tin, g lue , a n d a lb u mi n from eggs diffuse d o n ly very slowly whe n placed in wate r, co m pa re d with su gar o r salt. In addition , th e fo rme r su bsta nc es differ sign ifica n tly ill th eir a bility to d iffuse t h ro ugh a th in m embran e : Su ga r mol ecul es can diffuse through man y membran es, but th e very large mol ecul es th at ma ke up sta rch , ge latin , g lu e , a nd albu m in do no t. Mo reover, Gra ham fou n d th at
Gold colloid. Awater-soluble salt of ( A u C ~ l - is reduced to give colloidal gold metal. The colloidal gold gives the dispersion its red color. (Simila rly, colloidal gold is used to give a beautiful red color to glass.) Since the days of alchemy, some have claimed that drinking a colloidal gold solution "cleared the mind, increased intelligence and will power, and balanced the emotions."
642
Chapter
14
I
TABLE 14.5
Types of Colloids
Type
Dispersing Medium
Dispersed Phase
Examples
Aerosol
Gas
li quid
Fog, clouds, aerosol sprays
Aerosol
Gas
Solid
Smoke, airborne viruses, auto mobile exhaust
Foam
Liquid
Gas
Shaving cream, whipped cream
Foam
Solid
Gas
Styrofoam, marshmallow
Emulsion
li quid
liq uid
Mayonnaise, milk, face cream
Gel
Solid
li quid
J elly, J ELL-O'!l, cheese, butter
Sol
Liq uid
Solid
Gold in wate r, milk of magnesia, mud
Solid sol
Solid
Solid
Milkglass
Solutions and Their Behavior
FIGURE 14.18 The Tyndall effect. Colloidal dispersions scatter light. a phenomenon known as the Tyndall effect. (a) Dust in the air scatters the light coming t hrough t he t rees in a forest along the Oregon coast. (b) Anarrow beam of light from a laser is passed through an NaCl solution (left) and then a colloidal mixture of gelatin and water (right).
(a)
(b)
h e co u ld not crys ta llize th e se su bsta nce s, wh ereas he cou ld c rysta llize su ga r, sa lt , an d other ma teri a ls that form tru e so lutio n s. Gra h a m co ined th e word "co lloid " (fro m th e G ree k, m e ani n g "glue") to describe th is class of su b stances th at are d istin ctl y differen t fro m tru e so lu tio n s a n d su sp e ns io ns . We now know that it is possible to crystallize so me collo id al su bstances, albeit with di fficu lty, so there reall y is no sha r p d ividing lin e be tween the se classes based 011 thi s p ro perty. Collo ids d o, howeve r, h ave two di sting ui shi ng c ha rac te ristics . First , co llo ids generally have high mo lar masses; this is tr ue of p ro tei ns suc h as hem og lob in tha t h ave m ol a r masses in the thousands. Second , the particles o f a co llo id a re rel ati vel y la rge (say, 1000 n rn in di am et er ) . As a co n seq ue nce, t hey ex h ib it th e Tyndall effect; they scatter visib le light wh e n disperse d in a so lve n t, m a kin g th e m ixt u re ap pea r clou d y (Fig u re 14.18 ) . T h ird, e ve n thoug h co lloidal particles a re lar ge , th ey are no t so larg e th at they se tt le ou t. Graham a lso gave u s the words so l for a co lloida l dispersio n of a so lid su bsta nce in a flui d m edium a n d gel fo r a co llo idal d isp ersion that h as a structu re that preve n ts it from be ing mobile . J ELL-O ® is a so l when th e so lid is first mi xe d with boiling water, bu t it be co mes a gel whe n co o le d . Ot her ex am p le s of ge ls are the
gelatino us precipitates of AI(O Il )" Fe (OIl )" and Cu (O H ), (Fig ure 14.19 ) . Collo idal di sp e rsio n s co ns ist of fin el y di vid e d part icles th at, as a re sult, have a very hi gh su rface area. Fo r e xa m p le , if yo u h ave one million th of a m ol e of co lloid al p a rti cle s, eac h assu med to be a sp h e re wit h a di am e ter of 200 nm , th e total surface area of th e p a rticl e s wo u ld be on th e o rde r of 20 0 m ill ion cm 2 , or th e size of several footba ll fie lds. It is not su rprising, therefore, tha t m an y of the properties of co llo ids d e pen d on the p roperties of surfaces.
FIGURE 14.19 Gelati nous predpftates . (left ) AI(OH),. (center) Fe(OH),.
and (right) [u (OH),.
Types of CoLLoids Colloi d s a rc classified accord in g to th e sta te of the d ispersed p hase a n d th e d isp e rsing m e d iu m . Ta b le 14 .5 lists seve ra l typ e s of co lloi d s a nd gives examples of eac h . Collo ids wit h water a s the di spe rsing med iu m ca n be classified a s hydrophobic (fr o m th e Greek, mea ni ng "wa te r-fe a ring" ) o r hydrophilic (rwa tc r-lo ving") . A hyd rophobic co llo id is one in whic h o n ly weak attractive fo rce s e xist between the wa te r and the surfaces of th e co llo idal p arti cle s. Exam p les in clud e d ispe rs ions of metal s a nd of n e arly inso lub le sa lts in water. Wh en co m pou nds like Agel pre cipitate , the re sult is often a co llo id al di spersio n . T h e pre cipitation rea ctio n oc curs too rapidl y for ion s to gather from lo ng d ista n ce s a nd mak e large crystals, so the ion s aggregate to fo r m small p articl e s th a t re mai n su sp e n d e d in th e liqu id . 14.5
I
Colloids
643
FIGURE 14.20 Hydrophobic eelIotds. Ahydrophobic colloid is stabilized by positive ions absorbed onto each particle and a secondary layer of negative ions. Because the particles bear similar charges, they repel one another, and precipitation is prevented.
I ~t-"'::-- Su rrounded by positive ions Sheathed in negative ions
FIGURE 14.21 Formation of silt . Silt forms at a river delta as colloidal soil particles come in contact with salt wate r in t he ocean. Here, the Ashley and Cooper Rivers empty into t he Atlantic Ocean at Charleston, South Ca rolina. The high concentration of ions in sea water causes the colloidal soil particles to coagulate. 644 Chapter 14
I
",Thy don 't the par ticles co me toge ther (co agulate) an d form large r pa rt icles? T he answer is th at th e co lloida l particles ca rry electric cha rges. An AgC I particle, fo r example , will absorb Ag + io ns if the ions are present in su bstantial co nce n tratio n; an attraction occurs between Ag" io ns in so lutio n and Cl" ions 011 th e su r face o f th e particle. In thi s way, th e co lloidal particles be come pos itively charg ed, allowing the m to at trac t a seco ndary layer of anio ns . Th e part icles, now su rro u nd ed by layers of io ns, rep el o ne another an d are prevented from co ming togeth e r to for m a precipi tat e (Fig u re 14.20) . A stable hyd ro pho bic co lloid can be mad e to coagula te b y in tro d ucing ion s into the dis pe rsing med ium . Milk con ta ins a co lloidal suspension of p rotein-ri ch casei n micelles with a hyd rophobic core. Wh en milk fe rm en ts, lact ose (milk sugar) is co nverted to lactic aci d , whic h forms lactate io ns and hyd rogen ions. T he p rotective charges o n the su rfaces of the co llo idal particles are overcome, and th e mil k coag u lates; th e milk so lids co me tog ether in clum ps ca lled "cu rd s." Soi l panicles are often carried by water in rivers and streams as h yd ro pho bic co llo ids. Wh en river wate r carryi ng large amou nts of co llo idal particl es mee ts sea wate r wit h its high co nce n tra tio n of salts, th e pa rticl es coagu late to form t he silt see n at the m outh o f th e river (Figu re 14.21 ) . Municipal water treatmen t pla n ts often add salts such as AI2(SO -th to clarify wate r. In aq ueous so lutio n , alu m in u m ion s ex ist as [AI( 1-I 20 ) fiP + cations, whic h neutrali ze the cha rge o n the hyd ro phobic co lloidal so il pa rticles, causing these particles to aggrega te an d settle o ut. l lydropliilic colloids are strong ly a ttrac ted to water mo lecules. T he y often have groups suc h as - 0 1-1 and - N1-I 2 o n th eir su r faces. T hese gro u ps fo rm strong hyd rogen bonds to water, there by stabilizing the co lloid . Pro tei ns an d starch are im po rtan t examples o f hydrophilic co llo id s, and ho mogeni zed milk is the mo st fami liar exam ple. Emulsions are co llo ida l dis persions of one liqu id in ano the r, such as oi l o r fa t in water. Familiar examples in clud e salad d ress ing, m ayo n naise, and mi lk. If vegeta ble oi l and vinegar arc m ixed to make a salad dressin g, th e mi xture qui ckly separates in to two laye rs because the no n polar oi l molecu les do no t interac t with the polar water and acetic acid (CI13 C0 211) mo lecules. So why are m ilk and mayo n naise ap pa re n tly homogen eous mi xtures th at do not se pa ra te into layers? T he answer is th at they co n ta in an emulsifying agen t suc h as soap or a protein . Leci thi n is a p hosp ho lipid fou nd in egg yo lks, so mixing egg yo lks with oil and vinegar sta bilizes th e co llo idal dis pe rsio n known as mayonn aise . To understand thi s pro cess
Solutions and Their Behavior
fu rth e r, le t u s lo ok into th e fu n ct ion in g of soa ps a n d d ete rg e n ts, su b sta nces kn own as su rfacta u ts.
Surfactants Soaps a n d de te rgen ts are e m u lsifyin g age n ts. Soap is m ade by heati ng a fat with sod iu m or potassium hyd ro xid e (... page 476), whic h pr oduces the a n io n of a fatty acid.
o U H, C(CH 2h, - C- 0- Na+
'--~_..J' '--v----' Hydrocarbon tail Polar head Soluble in oil Solubte in water sodiumstearate, a soap
Th e fatty ac id a n io n h as a sp lit p e rson ality: It has a n o npol a r, hydroph obi c hyd ro carbon tail tha t is so lu b le in o th e r sim ila r hydro carbon s a n d a pol a r, hydrophilic he ad th at is so lu b le in wate r, Oil ca n n o t be read ily washe d a\vay fro m di sh es o r clothing with wate r because oil is n onpol ar an d thus inso lu b le in water. In stead, we add soa p to th e wate r to clean away th e o il. The n onpol ar m olecul es of th e o il interact with th e non polar hydro ca rbon tail s o f th e soa p mol ecules, leaving th e p olar heads of th e so ap to int eract with su rrou nd in g wate r mo lecu les. T he o il a n d wate r th en mi x (Fig ure 14.22) . If th e oily mat eri al o n a piece o f clothing or a d ish also co n ta ins so m e dirt particles, th at d irt ca n n ow be was hed away. Substa n ces suc h as soaps th at affec t th e properties o f su rfaces, a n d th erefore affect the interacti on be twee n two phases, a rc called su rf ace-ac tive age n ts, o r surfa ctants, for sh o rt. A su r fac ta n t u sed fo r clean in g is ca lled a detergent. On e fu n c tio n of a su r facta n t is to lower th e su r fac e tension of wa te r, whic h e n h a n ces th e clean sin g actio n of th e detergent (Figu re 14.23) . Man y deterge n ts used in th e h om e a nd ind ustry are syn th e tic. One exam p le is sod iu m laurylbcn zcn csulfon at e , a biodegrad a bl e co m pou nd .
• Soaps and Surfactants A sodium soap is a solid at room temperature, whereas potassium soaps are usually liquids. About 30 million tons of household and toilet soap, and synthetic and soap-based Laundry detergents, are produced annually worldwide.
CH,CH2CH2CH2C H2CH2CH2CH2C H2CH2CH2CH2- Q - SO, - Na + sodium tauryLbenzenesulfonate
/
-~ /~
-7r\
Detergent molecules
Water
Hydrophobic nonpola r tail FIGURE 14.22 The cleaning action of soap. Soap molecules interact with water through the charged, hydrophilicend of the molecule. The long, hydrocarbon end of t he molecule is hydrophobic, but it can bind through dispersion forces with hydrocarbons and other nonpolar substances. 14.5
I
Colloids
645
FIGURE 14.23 Effect of a det ergent on th e surface ten sion of wat er. Sulfur (density = 2.1 q/crn' } is carefully placed on the surface of water (density, 1.0 gjcm3) (left). The surface tension of t he water keeps the denser sulfur afloat. Several drops of detergent are t hen placed on the surface of t he water (right). The surface tension of the water is reduced, and the sulfur sinks to the bottom of the beaker.
add
surfactant
•
In ge nera l. syn th e tic d ete rg e n ts usc th e su lfo na te gro up, - S O:~ - , as the pola r head instea d o f the carbox ylate gro u p, - CO ;!'". The carboxylate anions fo rm an in so luble precip itate with a n y Ca 2+ o r ~lg2 + io ns p rese nt in water. Becau se hard wate r is c ha racte rized by hig h co n ce n tratio ns of th ese io ns, using soaps co nta ini ng carboxylates prod u ces bat htu b rings a n d tell-tale gray cloth ing. Th e synt he tic su lfo n ate d e te rge n ts h ave the ad vantage th at they do n o t for m suc h p recipit ates beca use the ir ca lciu m salts are more so lu ble in wate r.
Chapter Goals Revisited
Chemislry.o.Now"· Sign in at www. thom sonedu~com/Login to: Assess your understanding with Study Questions in OWL keyed to each goal in t he Goals and Homework menu for this chapter For quick review. download Go Chemistry mini-lecture flashcard modules (or purchase them at www.ichapters.com) Check your readiness for an exam by taking the Pre-Test and exploring t he modules recommended in your Personalized St udy plan.
Yl Access How Do I Solve It ? tutorials on how to approach problem solving using concepts in t his chapte r.
N ow that you have studied this chapter, you should ask whether J Olt have met the chapter goals. III puvticulrn; )'ou should be able to:
Calculate and use the solution concentration units moLality, mole fraction, and weight percent a. Defi ne th e te rm s so lut io n , so lve n t, so lu te, and co lliga tive p ro perties (Sec tio n \ 4.\ ) . b. Usc th e fo llowin g co n centration un its: mol al ity, mol e fractio n , a n d we igh t pe rce n t. (Sectio n 14 .1 ) . Study Question(s) assignable in OWL: 6, 9, 10, 12, 56. c. Un dersta nd th e di stinc tio ns betwee n satura ted , u nsa tu ra ted, a n d su p e rsa tura te d so lu tio ns (Sectio n 14 .2 ). d. Defi ne a n d illu strate t he terms m iscib le a n d imm iscible (Sec tio n 14 .2 ) . Understand the solution process a. Describe the process o f di sso lving a so lute in a so lve n t, in clu di n g th e e n e rgy chan ges th a t may occu r (Sectio n 14 .2 ) . Study Ouestion(s) assignable in OWL: 13, 16, 88 .93.
b.
Understa n d the rela tio ns h ip o f la ttice e nthalpy an d en thalp y of h yd ration to th e e n th alpy of so lu tio n for an ionic so lu te (Sec tio n 14 .2 ) . Study Ouestion(s) assignable in OWL: 77, 88.
c.
646 Chapter 14
I
Describe the effect of pressure and temperatu re on th e so lu bility of a so lu te (Sectio n \ 4.2) .
Solutions and Their Behavior
d.
Use H en ry's law to calcu late th e so lu bility o f a gas in a so lven t (Sectio n 14.2). 21, 22, 67 . Ap ply Le Chatelier's pri nci ple to th e change in so lu bility of gases with te mpe ra tu re changes (Section 14.2) . Study Quesnon(s) assignable in OWL:
e.
Understand and use the coLligative properties of soLutions a. Calc u late th e mole fraction of a so lven t ("Oh'O:111) and th e effec t ofa solu te o n solven t vapo r pressu re ( P,oln~l1 l) using Raoult's law (Sec tio n 14.4) . Study Question(s) assignable in OWL: 24. 74. 96 . b. Calc u la te th e boil ing point el evati on o r freez ing po int d e p ression ca used by a so lute in a so lven t (Sec tio n 14.4) . Study Question(s) assignable in OWL: 28, 30, 32, 45, 53,60; Go Chemistry Module 19. C. Ca lcu late th e os mo tic pressure (n) for so lu tio ns (Sec tio n 14.4) . Study Ouestion (s) assignable in OWL: 47,78,84,97. d. Use co lliga tive prope rties to de te rm ine th e mol ar m ass o f a so lute (Sectio n 14.4) . Study auestion(s) assignable in OWL: 35, 39, 61, 64, B4, B5. e. Charac terize th e effect of io nic so lu tes on co lligative pr ope rt ies (Sec tion 14.4) . Study Ouestion(s) assignable in OWL: 43 .
f.
Use th e van 't H off fa ctor, i, in calc ula tions invo lving co lliga tive properties (Sectio n 14.4) . Study Question(s) ass ignab le in OWL: 41,79 ,80.
KEY EQUATIONS Equation 14.1 (page 618) of so lven t.
Mol ali ty is d efin ed as th e a mou n t o f so lu te per kilo gr am
. . amount of solute (moL) Concentration (c, mol/kg) = molality of solute = - - -c--:----'--:-'mass of soLvent (kg) Equation 14.2 (page 618) T he mole fraction , X, o f a so lu tio n co m po nent is d efined as th e n um ber of moles o f a given componen t of a mixture ( nA ' mol) di vid ed by th e total num ber o f moles o f all of th e co m po ne n ts o f th e mi xtu re . Mole fraction of A (X,) =
nA
n,
+ nB + n( + ...
Equation 14.3 (page 619) \Veigh t percen t is the mass o f on e co mpo nen t d ivid ed by the total mass of the mix ture (m ultiplied by 100 % ) . Weig ht % A =
mass of A X 100% mass of A + massofB + massofC + ...
Equation 14 .4 (page 626) Hen ry's law: th e solu bi lity of a gas. Sg. is eq ua l to th e p roduct o f th e part ial p ressu re o f th e gaseous so lu te (Pg- ) an d a constant (HII) characteristic o f th e so lu te and so lve n t.
Equation 14 .5 (page 629) Raoult's law: th e eq ui lib riu m vapo r pressure of a so lve n t over a so lu tio n at a given te m perature, P ' OI\'t'll l> is th e prod uct of th e m ole frac tio n of th e so lven t ( X,olvt'llI) and th e vapor pressu re o f the p ure so lven t ( PO soln~llI ) ' Psolvent
= Xsolvent posolvent I
64 7
Equation 14.6 (page 631) The eleva tio n ill boiling poin t of th e so lven t in a soluti on , d'lill» is th e product of th e mol ality of th e so lu te, fll,;" lu k' and a co ns ta n t characteristic of th e so lvent, l\qp' Elevat ion in boiling point
= 6.TbP = Kbpm solute
Equation 14.7 (page 634) Th e d epression o f th e freezing po int of th e solve n t in a so luti o n , d'l;·p, is th e product of th e molalit y o f th e so lu te , m ' o lut e> and a co ns ta n t characte ristic o f th e so lvent, Kfp ' Freezing point dep ression = 6.Tf p =
Kfpmsolute
Equation 14.8 (page 637) The osm o tic pressure , [], is the product of th e so lu te co nce nt ration c (in mo l/L), th e universal gas co nstan t R (0.082057 L . atm yK . mol ), and the tem peratu re T (in kelvins) .
11 =
cRT
Equation 14.9 (page 639) Th is mod ified eq ua tion fo r fre ezing point d ep ression acc o u n ts for th e possibl e di ssociation of a so lu te. The van't Ho ff factor, i, th e rati o of th e measu red fr eezin g poi n t d ep ression and th e freezin g point d epression ca lculat ed assu ming no so lu te di ssoc iat ion , is rel at ed to th e relative number of particles p roduced by a dissolved so lu te . 6.Tfp measured
STUDY QUESTIONS OnLine homework for this chapter may be assigned in OWL. ... denotes challenging questions. •
denotes questions assignable in OW L.
Blue-numbered questions have answers in Appendix 0 and fully-worked solutions in the Student Solu tions Manual.
= Kfp
XmXj
3. Fill in th e bl an ks in th e ta ble . Aq ueous so lu tio ns are assu med . Compound
Molality
NaI
0.1 5
C2H 5OH C12H uO u
Weight Percent
Mole Fraction
5.0 0.15
4. Fill in the hl ank s in the ta ble . Aq u eo u s so lu tio ns are assu med .
Practicing Skills
Compound
Concentration (81'f' Exm np lrs 14.1 rind 14. 2 a nd CllI'mi.\l ry'\/ow Screen 14.2.)
KN0 3
l. Su p pose yo u di ssolve 2..e,G g of succ ini c acid ,
C 2H I (C 0 2H h , in 500. mL of wa ter. Assumi ng that th e de nsity o f wa te r is l.OO g / cm :l , calcula te th e m ol ality, mole fracti on , a nd we igh t percen tag e of acid in the so lu tio n.
2. • Assume yo u dissolve 45.0 g o f cam p hor, C lOII J1jO, ill 425 mL o f e than ol , C2H.'i O H . Calcu la te the molal ity, mole fra ct io n , a nd wei g h t perce nt o f ca m p ho r in th is so lu tio n . (T h e de nsity o f eth anol is 0.785 g / m L.) 648
CH 3C0 2H HOC H2CH 2OH
Molality
Weight Percent
Mole Fraction
10.0 0.0 183 18.0
5. Wh at mass of Na2C0 3 must yo u add to 125 g o f wa ter to p repare 0 .200 m Na2CO ~\ ? Wh at is the mo le fractio n o f Na 2CO :I in th e re su lting so lu tio n? 6. • You want to p repare a so lu tio n th at is 0 .05 12 111 ill NaNO J. wh ar m ass of NaN O j m ust he added 10 500. g of wa te r? What is th e mol e frac t io n o f NaNO :} in th e so lu tio n?
7 . You wish to pre pa re a n aq u eous so lu tio n o f glyc ero l, C 311:,( O Ilh. in wh ich the mole fractio n of th e so lu te is 0.093 . W ha t mass of glyc e rol m ust yo u ad d to 425 g of wa ter to make thi s sol u tio n? Wh at is th e molality of th e sol utio n?
8. You wan t to p repa re a n a q ueous so lu tio n of e thy le ne glycol, H () C H 2CH ~.P H , in whic h t he m ol e fra ction of sol u te is 0 .125. Wh at mass o f e thyle ne g lyco l, in g rams , should yo u co mbine with 955 g of wa ter? \ Vhat is the molality of t h e so lu tion? 9 . • H yd rochl ori c aci d is so ld as a co nce n tra te d aq u eo us sol utio n . If th e mol a rity o f co m mercial II CI is 12.0 a n d its d e ns ity is 1.18 g/ cm 3 • ca lcu la te th e followi ng: (a) th e m o la lity o f th e so lu tio n (b) th e we ig h t pe rce nt o f H CI in th e so lu tio n
10. • Co n ce n tra te d su lfuric aci d has a d e nsity o f 1.84 g/cm :'l and is 95 .0 % hy we ight 112S0 4• Wh a t is tile molality o f th is aci d ? Wh at is its mo larity? 11. T h e ave rag e lith ium io n con centrat ion in sea wa ter is 0 .18 p pm . Wh a t is th l' mo lal ity of Li " in sea water? 12. • Silver ion h as a n average conce n tratio n of 28 ppb (p a r ts pCI' billi on ) ill U .S. wa ter su p p lies . (a) W ha t is th e m o lal ity of th e silver io n ? (b) If yo u wa n te d 1.0 X 10 2 g o f silver a n d co u ld re cover it ch e m icall y from water su p p lies, wha t volu me of wa te r in lite rs, wo u ld yo u h ave to treat ? (Assu me th e den sity o f wa ter is 1.0 g/c m 3. )
The SoLution Process (See Exa m/Ill' 14.3 and Clwmislr),Now Screens /4.3 and /4.4.) 13 . • W hi c h p airs o f liqu id s will be misci b le ? (a) Il ,O an d CII "CIl,C Il ,C Il , (b ) C(iH (i (b e nzene) an d CCI4 (c) H 20 a n d C H:'lC0 2 H
15 . Use th e da ta o f Ta ble 14.1 to calcu late th e e n tha lpy o f so lutio n of LiC !. 16 . • Use th e followi ng data to ca lc u la te th e e n t hal py o f sol utio n o f so d iu m p e rch lo rate , NaCI0 4 : ~
- 3H2.9 k.J/n lOl an d d jH O(aq , 1 m) = - 3 69 .5
~J / mo l
17. Y-. (~
W
Figure 13 Representation of glass structure. (a) Silica glass (Si02) may have some order over a short distance but much less order over a larger portion of the solid (b). (c) The Sia 2 structure can be modified by adding metal oxides, which leads to a lower melting temperature and other desirable properties. (In t his simple representation, the graySi atoms are shown at t he center of a planar t riangle of red 0 atoms; in reality. each Si atom is surrounded tetrahedrally by0 atoms. The structure is not planar but is three dirnenslonat.)
Ceramics
Incident light
i = angle of incidence r = angle of refraction
i>( Refracted light
o
(aJ {b] Figure 14 Refraction of light. (a) When light enters a different medium, its velocity changes. This causes the path of a photon to change direction in the material. (b) Observing an object in a glass of water illustrates the effect of light refraction.
and the index of refrac tio n will affec t how mu ch o f th e light is reflec ted a nd the an gl e at wh ich it bends in th e second m at erial. You ca n observe this e ffec t hy putting an objec t in a g lass o f wate r a nd lo ok in g at th e ap pa ren t be nd th at resul ts in th e object (Figu re 14b ). T h is co m bina tio n of th e tran sm ission an d refl ection charac te r istics of glass has allowe d scien tists and e ng inee rs to d evelop optical fibe rs (Figu re 15) . Optical fibe rs are d esign ed to have a p rope rty ca lle d total intern al refl ection , whe reby all th e light th at e n ters at o ne e nd o f th e fib er stays withi n the fib er th ro ug h refl ect io ns with th e inte r io r su rface as the light tr avel s fr o m o ne en d of th e fibe r to
Figure 15 Optical fibers. (left) Glass fibers t ransmit light along the axis of the fiber. (right) Bell Laboratory scientist Joanna Aizenberg recently discovered that a deep-sea sponge, made chiefly of silica (Si02 ) , has a framework that has the characteristics of optical fibers. (For mo re about the structure of the sponge, see Figure 2 on page 27.)
I
665
th e o ther. Tot al in te rnal re flec tion in these fib e rs is ac hieved by con tro lling th e ra tio o f the indi ce s o f refraclion be twee n the fib e r's co re and its outside su rface. Chem ically. the index of' re fra cti o n is con tr o lle d by ad just ing th e quantity and type o f ca tio n ic network m odifie rs that are ad de d 10 th e g lass. T he index of re fr action o f a g lass fib er ca n he co n tro lle d so th a t it h as o ne value at th e co re o f th e fiber hut ch anges smoo thly across th e radius o f th e fibe r to a different valu e at th e sur face. This is acco m plished by a n lo u-exch ang e pro ce ss durin g fibe r production in wh ich , fo r exam ple , K+ io ns a re replaced by Tl ' ions. O p tical fib ers are tra ns form ing the co m m u n icatio ns ind ustry in an am azing fashi o n . In ste ad o f transmi tting inferIllation usin g e lec tro ns travelin g through m et all ic wires, o p tical fibers allow co m m un icatio n to occu r by tr an sm itting ph otons through g lass fib er bund les. Sign al tr an smi ssion by optical fibers, known as photonics, is much faster and more ec on o m ical than tr an smi ssion usin g co pper wires and cahies. For exam ple , th e qu antity o f co p per re q uire d to carry the eq u ivale n t am ount of information transm itted by o p tical libel' would we igh 300,000 times mo re than the o p tical fiber material!
Fired Ceramics for Special Purpos es: Ce ments, CLays. and Refractories O th er classes of ce ra m ics include ce me n ts, clays, a nd re fractories. Un like glasses, these ce ra m ics are proce ssed by sha ping, dryin g , an d then firin g, witho u t ever m elting th e solid. Cemen ts a re ex tre m ely stro ng and are co m mo nly use d as struc tur al mat e rials. They ca n be fo rmed into a lmost an y sha pe . Wh en mi xed with water, th ey produce a paste th at ca n he poured into m ol d s a nd allowe d to dry a nd harden . C lays a re ge ne ra lly m ixt ures of h ydrated alu m ina (AI20 j ) an d silica (Si0 2 ), b u t m ay also co nt a in o ther ingrcd le nts. suc h as tri calcium silica te , (3 Ca O . SiO :!) , d ica lciu m silica te , (2 Ca O . Si0 2 ) , a nd !\lgO. Th ei r co m positio n is irregular, an d , be cause th ey are powd ers, th ei r crys ta llin ity ex te nds for o nly short di stan ces. C lays ha ve the useful pro perty o f becomi ng \'e ry plastic when wa ter is ad de d. a characteristic re fe rred to as hydropl asticity, This pl asti city, and clay's abi lity to h o ld its sha pe darin g firin g, are \'e ry importan t for the forming p ro cesses used to crea te various objec ts. The laye red m olecul ar struc ture o f clays res ults in micr o scopic p latel et s that can slide ove r eac h o the r easily when wet. Th e layers co ns ist o f Si0 4 tetrah edra join ed with AlO t.i octahedra (see Sec tio n 21 .7) . In ad d itio n to th ese ba sic silico n- and alum inum-based str uctu res, diffe re nt ca tio ns ca n he su bs titu te d into th e fra mewo rk to change th e prope rti es of the clay. Com mo n su bs titue n ts includ e Ca 2+ , Fe2+. an d Mg2+. Differen t cia)' materials ca n th en be created h y \'aryin g th e co mb ina tio ns of layers and th e su bstitue nt ca tio ns.
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Th e Ch emistry of Mod ern Mat erials
n ow calle d p iezo electricity, is th e p rope r ty th at a llov»s a m ech ani cal di stortion (su ch as a slig h t bendin g ) to induce an e lec trical cu rre n t a n d, co nversely, a n el ectri cal curren t to ca use a d istortion in th e m at e ri al. No t all cry stalline ce ra m ics exh ibi t piezoel ectricity. T h ose th a t d o have a specific un it ce ll str u ctu re (Section 13.2) th at ca n lo osely trap an im pu r ity ca tio n. Th e ion's p osition sh ifts whe n th e un it ce ll is d e for med hy m ech ani cal stre ss. T h is sh in ca use s an induced d ipo le (see Sectio n 12.3) an d, th erefore, a p o tential differen ce across th e m ateri al th at ca n h e co nve rt ed
~
" ~
Figure 16 Aerogel, a networked matrix of Si02• (left) NASA's Peter Isou holds a piece of aerogel. It is 99.8% air, is 39 times moreinsulating than the best fiberglass insulation, and is 1000 times less dense than glass. (right) Aerogel was used on a NASA mission to collectt he particles in comet dust. The partides entered the gel at a very high vetodty, but were slowed gradually. Scientists studied the t racks made by the particles and later retrieved the particles and studied their composition. Refractories co n stit u te a class of ce ra m ics tha t ar e capabl e of withs ta n di ng \'ery hi gh te mpe ratu re s with o u t d e formin g , in so m e cas e s up to 1650 °C (3 000 OF ) , a n d th at are th e r m all y in su lat ing . Because of th e se properties, r efr acto r y b r ic ks ar e used in ap p lica tio ns su ch as furn ace lin in gs an d in m etallurgi cal o pe r ation s. Th ese m at erials a re the r m ally in sul a tin g larg el y b ecau se of th e porosit y of th ei r str uc ture ; that is, h ol e s (o r po res) are di sp ersed evc nlyv•.ithin th e so lid. However, whil e p o ro sity will m ak e a m at eri al more th erm all y in su la ting, it , v.ill also weake n it. As a co n se q uence , re frac tories are n o t as strong as ce m ents. An amazin g exa m p le of th e use o f p oro sity to in crease th e in sul atin g ca paci ties o f a ce ra mi c is found in a m at erial d eveloped at NASA called arrow>! (Fig u re 16 ; see Case Study: Th e world's Ligh test Sol id , page 607) . Aerogel is m ore tha n 99 % a ir, with th e rem ai nd er co n sisting of a n etwo rked m at ri x o f Si0:l' Th is m akes aerogel about 1000 tim es less d e n se th a n glass h u t g ives th e m at eri al ex trao rd ina ry th erm al in sula tin g a bi lities. NASA used aeroge l o n a mi ssion in wh ich a sp acecraft flev..· through th e tail of the co m e t \Vild 2 an d re tu r n ed to Ea rth with spac e p art icles e m bed de d in the aeroge l.
to an el ec tri cal signal. In ad di tio n to th e min erals orig inally tested h y th e Cu r ie brothe rs, materials k n oven to exh ih it th e p iezoelec tr ic effect in cl ude titan ium co mpo u n d s of barium a n d le ad , lead zircona te ( Pb Zr0 3), an d am mon iu m d ihyd ro g en ph o sph ate ( N H ., H ~ P O .d. Ma te r ial s th at e x h ib it p ie zoe lec tr icit y h ave a g re a t m arry a p plica tions. r an gin g fr om h orn e gadge ts to so p hi sticate d m ed ical a n d sc ientific a p p licat io n s. O n e use \.. . ith wh ic h yo u lIlay b e familiar is th e a u to ma tic igniti on system s o n so me bar be cu e g r ills and lighters (Fig u re 17) . All digital wa tc h b ee p e rs a re b ased o n pi e zo cerami cs, as a re sm o ke d etector alarms. A less famili ar ap p licatio n is foun d in th e se ns ing lever of so me ato mic force micro sco pes (Af' Ms) a n d sc a n n in g-tu n n e lin g mi cr o scopes (ST Ms), in strume nts th at co n vert m ech an ical vib ra tio ns to e lec tr ical sig n a ls. Scientists an d engi neers are always search ing for materi als with n ew a nd useful p roperti es. Pe rh aps th e m ost dramati c property that h as been obse rved in n ewly devel oped ceramics is su p erco n d uc tivity at rel atively high temperatures.
Modern Ceramics with Exceptional Properties In 1880 , Pierre Cu r ie an d hi s b ro the r J acque s work ed in a sm a ll labora tory in Paris to exam in e th e el ectrical p rope r ties o f certa in cr ysta lline su bs ta nces. Usin g nothin g m ore th an till foil , glu e , wir e, a n d m agn e ts, th ey wer e ab le to con fir m th e prese n ce of su r face charges on sa m p les o f m at erials su ch as tou rma line , quartz, an d to paz wh e n they were su bj ected to m ech ani cal stre sses . This ph en om en on ,
Figure17 Devices that depend on the piezoelectriceffect. These devices work by using a mechanical stress to produce an electric current. Piezoelectric devices are widely usedin ignitors and in devices that convert electricimpulses to vibrations, such as in the timing circuit of a wristwatch.
Biomaterials: Learninl'{ from Nature
I
667
Once agai n, we sec th at co m bi n ing a to ms into so metimes complex ch e m ica l co m p osi tio ns a llows scien tists to d evelop m ateri als with particular propertie s. In cera m ics , wh ich arc n o rm all y electrica lly in sul atin g , thi s incl udes even th e ab ilit y to conduct electricity.
Biomaterials: Learning from Nature
Superconductivity is a phen omen on in wh ich th e e lec trical resistivity o f a mat erial drops to nearl y zero a t a particular temperature re ferre d to as the critical temperature, T; (Figu re 18) . .Mo st m etals n aturally h ave resistivities th at d e... crease with tem peratu re in a constan t m ann er b u t still have sign ifica n t resis tivity eve n at temperatures ncar 0 K. A few metals an d metal alloys h ave b ee n fo un d to ex hi bit superco n d uctiv ity. For m etals, h owever, th e cri tical temperatures are ex tre mely low, be tween 0 an d 20 K. Th ese ternpcraturcs arc costly to ac hieve an d d ifficult to maintain . Rec en t scie ntific a tten tio n h as, th e refore , fo cu sed on a class of cera mics with su perco nd uctive critical tcmperatures ncar 100 K. Th csc m aterials in clude YB a~Cu 30 7 , with 1 ~ = 92 K (Figu re 19), a nd HgBa:lC a:lCU:lO:1, with 7: = 153 K.
Mo st of th e m at e rial s d e scribed so far in thi s cha pte r co me from nonlivin g so u rces a n d , ill m an y ca ses, arc th e re sult of laborat ory syn theses. However, an important bran ch o f materials re sea rch d e als with ex am in ing, understanding , an d eve n co pying m a teri als produced by livin g syste ms. The stu d y of n aturall y occu rr ing mat erials h as led to tile d evelopment of syn theti c mate rials that po sse ss impo rtan t properties. A good exam p le is rubbe r (C h a p ter 10, pagc 483) . Th e pol ym er we k n ow as rubber was initi all y obtain ed fro m certa in tr ees an d ch e mically modified to co nver t it to a useful mate ri al. Natu ral ru bbe r was found to b e so u se ful th at che mists eve n tual ly ac h ieve d th e Syll· th esis o f a str uc tu rally id enti cal materi al. Research on ruhhe r, whi ch sp an ned more th an 200 years , h as h ad imp o rtan t co nsefJ lle lKcs for human s as evid e nced by th c myri ad ap plica tio ns of rubber today. Today, scie n tis ts co n tin ue to look to n atur e to provide n ew m at e rial s a n d to provid e clues to improve th e m at erlals we alread y use . T he sea ur ch in a n d its cera m ic sp ines (Figu re 12) a n d th e spo nge whosc skele to n h as th e charac te ris tics o f o p tica l fibers (Fig u re 15) ar e just two exampl e s wh ere bioru ateri a ls res ea rc h h as fo cused on se a life ill a sea rc h fo r Il C W m at erials. Scientists h ave also exam ined co nc h sh e lls to understand the ir In credible fracture stre ngth . Th ey used sca n n ing ele ct ro n mi cro scopy (SEI\t ) 10 scrut in ize th e str uc ture o f th e shell whe n it was fractu red. Wh at th ey di sco vered was a cri ss-cro sse d. layered str uc ture that is the equivalent of a "ce ra m ic pl ywood " (Fig u re 20) . T h is mi croarchitecuu-e prevents fra ctures th at o ccur on th c outside su rface of th e she ll from
Figure 19 The lattice ofYBa2Cu30" a superconductor. Yttrium ions are yellow; barium ions are red; copper ions are green; and oxygen ions are blue. (Reprinted with permi ssion of Dr. Klaus Hermann of the Fritz Ha ber Institution.)
Figure 20 A scanning electron microscope picture of the shell of the conch. Photos from S. Ka mat , X. Su, R. Ballarini, and A. H. Heuer. Structural basis for the fracture toughness of the shell of the conch Strombus gigas. Nature. VoL 405 , pp. 1036- 1040, 2000.
Figure 18 Superconductivity.When a superconducting material is cooled to a low temperatu re, say in liquid nitrogen (boiling point is 77 K), it generates a very strong magnetic field. In this photo, a i -pound magnet is levitate d in the field created by t he cooled superconductor.
668
I
Th e Ch emistry of Mod ern Mat erials
t5:i:
b e ing tr a n sferred into the in ne r layers. Figure 21 Strong mussels. (left) A common blue mussel can cling to "T?" H3 T h e d iscove ry h as in sp ired ma terials en gi- almostany surface, including this Tefton sheet, even underwater. (right) The HN n ee rs to create mat e rials that are sign ifi- adhesive precu rsor is a protein interlinked with iron(III) ions. Side chains o NH ca n tly srrc ugthc n cd b y in co rpo ratin g a on the protem are dthydroxyphenylalanioe (DOPA), and an ironflll) 0 ~ HN OH fibr ous ceram ic matri x, suc h as SiC (silico n 10n binds to the hydroxyl groups (-OH) 10 three Side chams. ca t-hi de) wh iskcr s. CH3 CH 3 I In anoth er are a of re search fo cu sing o n sea cr eatures , th e connecti ve tissues of sea O ~? ~O O ~ cu cu mbers and o th e r ec h inoder m s (ma r ine I Fe 0 inverteb ra tes with tube feet a nd calc iteHN "'" 0/ "0 covere d , radi ally sym me trica l bodi e-s) h ave ~ b ee n stu d ied in an a tte m p t to di scover h ow =>~ I th ese ani mals ca n reve rsib ly co n tro l the stiff- ~ \ 00 ')-< ness of th eir o u te r skill. Th e co n n ec tive tis- if ~ N0N , ~ CH3 sues o f th ese ani mals in clu d e th e protein ~ 0 ~H co llage n in a cross-lin ked fibe r struc tu re , H3C OH sim ilar to th e denn is, an inner layer o f th e skin co ns isting of sen sitive conn ective tissue Sci e n tists who h ave re sea rch ed mussel ad h e sives h ave o f Illan y mamm als. At th e sa me tim e, o the r proteins a nd so lub le molecu les in the ec h ino d e rm syste m allow th e an ib een ab le to d e termin e th at th e amin o acid 3,4 -d ihydroxymals to change th e characteristics of th e co n n ec tive tissue ph en ylalani n e (DO PA) is th e agen t p rimari ly r es p o ns ible in response to th eir n ervo us system. As a res u lt, crea tu res fo r th e strength o f th e ad hesio n. But DOPA alo ne ca n no t suc h as sea cuc u m be rs ca n mov e ab out an d , iII so me cases, ex p lain th e in cred ib le strength o f th e mu ssel glues. Th e se cret lies in the co m h in a tion of a n Fe 3+ io n with DO PA can d efe n d th e m sel ves by hardenin g th ei r skin to all almost sh ell-like co ns iste ncy. Th e ensu ing lab orat o ry research h as to form a cr o ss-link ed m at rix o f th e m ussel's protein (Figu re 2 1). T he cu r ing process, or h ardenin g of th e n atfo cu sed 0 11 th e for mulatio n o f a syn thetic coll agen-based ural protein aceous liqu id p roduced by th e mussel , is a p olym er co m p osite m at er ial in wh ich the stiffness ca n be cha nged repeat edly through a ser ies of oxidation an d reducresu lt of the iron- p ro te in interact io n th at occurs to form tion react ion s. Scienti st... arc 1l0\\' d evelo p ing mode ls fo r Fe( DO PA)3 cross-lin ks. synt he tic skin and muscle ba sed 011 th eir find ings. Research o n ad hes ive m at erials re p rese n ts anothe r a rea in wh ic h sea cr ea tu re s can provid e so me clu es, Ge tting th in gs to stick to gether is im p o rt an t ill a m u lti tu d e of apThe Future of Materials p licat ions. Th e lo ss of th e space sh u ttle Colu m b ia in early 2003, ca use d hy lo ss o f so m e ce ra m ic tiles whe n a pi ece of T he moder n tools an d tech n iqu es o f che m istry are mak in g in sul a ting foam th at fell off durin g laun ch hit th em , ofit p ossibl e for scie n tists not o n ly to d evelop n ovel m ateri als, but also to proceed in n cv.., an d u nforese e n d irection s. T h e fe re d a so be ring lesso n in ad hes ive failu re u n d er extre me co n di tio ns o f temperatu re an d h u m id ity. If yo u lo ok fie ld of nano te chnology is a n exam p le. In nanotech no lo gy, around, yo u will probably find so me lh ing wit h all ad h esive str uc tu res with d im ensio n s o n th e orde r of n an o m eters are u sed to ca r ry out specific funct ions. For ex am ple, sc ilabel , so me thi ng with an attached plastic p art, so mc th ing with a rubber sea l, or p erhap s so m ethi ng taped together. e ntists ca n now crea te a tu be o f ca rbo n a to ms embed ded Ad hesives h ave also p rc)\'en usefu l fo r m edical ap p lica tio ns, in a sligh tly lar g er ca rbo n tube to act as a ba ll b eari ng at where spec ialize d glucs h el p d octors sea l tissues with in th e th e mo lecul ar leve l (Figu re 22) . human b CH Iy. Fo r e\'er y type of sticking a p p lica tio n, differNa n oscience has pro dded p rofoundl y im po rtan t a p p lie n t properties are n eeded for the ad h esive mat eri al. ca tio ns fo r m edici n e , co m p u ting , a n d e ncq,']' consumpNa tu re p r ovid e s num erous exam p les of ad hes io n. tio n , Fo r exa m p le, sc ie n tists h ave d eve lo ped qu antu m dots Geckos and Flies th a t ca n walk o n glass wh ile co m ple tely (Figu re 23 ) , n an o m e te r-scale crys tals o f different m aterials in vert ed hold dues to the kin d of b io logica lly b ased adh eth at can em it light an d ca n even b e m ade to Fun ction as sio n th at could b e th e b asis o f syn thet ic analogs. Mat-inc lasers. Quantum do ts h ave bee n used as hio log icalm a rkers mussels, wh ich ca n stick quite well to wood . metal, a nd by attachin g th em to vari o us ce lls, By sh in ing ligh t 0 11 ro ck , also hold gre at interest fo r scie n tists stu dyi ng adheth em , th e quantu lIl d ots will flu oresce in diff e rent co lo rs, sio n (Figure 21 ) . all owin g the ce lls to he im aged.
XC
n.
'( Y ,r HOyO 0 "'" 07' r > 0x:D H,r
b
H~'Co
t
Y
~"'" 0 HN\
,, enerf,lJ' outpu t pe r minute? 4. Using th e result of the calcu latio n in Q ues tio n 3, cs tima ll' th e n umber of so lar ce lls th at would he needed to power a 700-\\7 m icrowave oven . If th e so lar ce lls were assem bled int o a p an el. wha t wo uld he the a pp ro xim ate area of th e pan el ? 5. Descri be how you co uld calc ula te th e den sity of pewkr from th e de nsitie s of th e co m po ne n t clem ents, assuming th at pewte r is a su bstitu tiona l allo y, Look up th e de n sities o f the consti tuen t ele men ts an d ca rry o ut th is ca lcu latio n . Den sities can he foun d on a websi te suc h as www.webcl em c n ts.co m . or go to Ch e mis tryN O \V a nd click 011 th e period ic tabl e tool. Click o n the sym bo l of each of the ele me nts in pewte r. A ta ble of a tom ic prop erties includes th e cle men t's d e ns ity. 6. Ca lcu la te a n a p prox ima te value fo r th e d e nsit y of aeroge l using the fact th at it is 99 % a ir, hy volume , an d tlure main der is Si0 2 (Density = 2,3 g/ cm 3) . Wh at is th e mass of a I ,O-cm:\ p ie ce of aerogel?
CONTROL OF REACTIONS
15
Chemica l Ki netics: The Rates of Chemical Reactions
,.
\ Fading of t he color of phenotphthalein with time
(etapsed time about 3 minutes)
Where Did the Indicator Go? The indicator phenolphthalein is often used for t he tit ration of a
This chapter is about one of t he fundamental areas of chemistry: the rate of reactions and how they occur. In Study Question 71. you
weak acid using a strong base. Achange from colorless to paLe pink will see some data t hat will allow you to discover howthe rate of the indicates t hat the equivalence point in the reaction has been reached phenolphthalein reaction depends on the hydroxide ion concentra( [N,O,]
~~----------- ~=---~~~
1.0
2.0
3.0
4.0
5.0
6.0
Time (r). hours
672
Chapter 15
I
0.22 mol/L - 0.42 rnol/L (6.3 h 4.0 h) (60 min/h)
!>t
Chemica l Kineti cs: The Rate s of Chemica l Reacti ons
7.0
B.O
9.0 10
11
T h e rat e of this reac tio n fo r any in te rval of tim e ca n be e xpressed as th e c ha nge in conc e n tra t io n of N 2 0 :; d ivid ed by t he c ha nge in ti m e :
Rate of reaction
=
• Representing Concentration Recall that square brackets around a formula indicate its concent rations in mol/L (Section 4.5).
change in [ N20 S] change in time
Th e minus sig n is re q u ire d b ecause th e co nce ntratio n o fN 20 3 d ec reases wit h time (tha t is, d [1':,O,,] = [1':, 0,] (final) - [1':,0 ,] (in itia l) is n ega tive ) , an d rat e is always ex pressed as a po sitive qu a nti ty. Using data fro m Figu re 15.2, t h e ra te o f di sa p peara nce of N zO,·) be tween 40 m in a n d 5S m in is give n by
Rate ~
d [ N,Os] ~ (1.1 0 mol/ l) - (1.22 mol/ l ) ~ il t 55 min - 40 min mo l N20S consumed Rate = 0 .00 80 -~~--;-- l vmi n
• Calculating Changes RecaUthat when we calculate a change in a quantity, we always do so by subtracting the initial quantity from the final quantity:
+ 0,12 mOl/l 15 min
.i c
= Cfinal -
Cinitial'
No te the units for reaction rat es; if co nce ntrat io n is ex p resse d in m o l/L, th e u nits for ra te will be m o l/L . tim e . Du ri ng a c he m ica l react io n , am oun ts o f reac ta nts d ecrease with tim e, and amou n ts of p roducts in crease . Fo r th e d ecomposition of N z0 3 , we cou ld a lso ex press the ra te e it h er as 6.[N 0 2]/ 6.101' as 6.[02] / 6.1. Rat es based o n changes in concentrations o f prod u cts will have a positive sign be ca u se th e co ncen tra tio n is increasin g . Fu rt he rm ore , th e num e rica l values o f ra tes d efi ned in these ways will b e diffe rent fro m value o f 6.[N 20 i;] / at. No te th at th e ra te of d ecompositio n o f N zO :, is o ne h alf the rat e of fo rm ati on of N0 2 a n d twice th e rate o f form ati on o f O z. T he rel ationshi p be twee n th ese rate ex pressio n s is d ete rmi n ed from th e coeffic ie n ts in th e chem ica l equat io n .
Rate of reaction
= -
d[N,Os]
M
=
1 d [ NO,] +- - - -
2
M
=
d [O, ] + 2- - -
M
Fo r th e l f i-min u te int erval be twee n 40 and 55 minut es. th e ra tes fo r th e formation of N 0 2 an d 0 2 a re
Rate = d [NO, ] = 0.0080 mo l N,Os consumed X 2 mol NO, formed M l · min 1 mol N20S consumed = 0.01 6 mol NO, form ed l mi n v
Rate
=
d [O, ] M
=
0.00 80 mol N,Os consumed X g mol O2 formed l. vrnin 1 mol N20S consumed
= 0.0040 mal O,
formed
l min v
T h e gra p h of [N zO ,;] versus tim e in Figu re 15.2 does n o t give a straig h t lin e becau se th e ra te o f the reac tio n c ha nges d u ring th e co u rse o f th e reac tio n. T h e co nce ntrati o n o fN zO j d ecr eases ra p id ly a t th e be gin n in g of th e reactio n but m ore slowly n ea r th e en d . \ Ve ca n ve rify th is by co m paring th e ra te of d isappearan ce o f N zO :, calculate d previously (the concen tra tion d ecreased by 0.12 m ol / L in 15 mi n ) to th e ra te of reac tio n ca lcu late d fo r th e tim e in terval fro m 65 h to 9.0 h (whe n
15.1
I
Rates of Chemical Reactions
673
the co ncen tratio n d rops by 0. 12 mol/L ill 150 m ill) . The rat e in th is later stage of thi s re ac tion is o n ly one tenth o f th e previous valu e . (0.10 rnol/L) - (0.22 rn ot/L) 540 min - 390 min =
• The SLope of a Line The instantaneous rate in Figure 15.2 can be determined from an analysis of the slope of the tine. See pages 39-41 for more on findi ng the slope of a line.
~ + ,,0 .,1,-2,-,m ,-=02 V~ L
150 min
0.00080~ L' min
T he procedure we have used to calcu lat e the reaction ra te gives th e ave rage rat e ove r th e cho se n time in terval. 'fe mig ht a lso ask wha t the ins tan taneous rate is at a single poin t in time. In an au to mobile , the in stantan eo us rate can be re ad fro m th e speed omet er. For a che mical reactio n , we ca n e xtrac t th e ins tantan eo us ra te fro m th e co nce n tration - time graph hy drawi ng a line tangent to the co nce n tra tion-tim e cu rve a t a part icul ar time (see Fig u re 15.2) . Th e in sta nt an eous rate is o bt ain ed fro m th e slo pe o f thi s lin e. For exa m ple . whe n IN20 5] = 0. 34 mo l/L and t = 5.0 h , th e ra te is t. [N20, ] M
Rate when [ N20,] is 0.34 M
= 1.4
X
= + 0.20 moVL 140 min
1O-1~
L'm in
At that particula r mom ent in time , (t = 5.0 h ) , N 20 :" is bein g con su med at a rat e of 0.001 4 mol /L . mill . ChemislrY-9-NDW~ Sign in at www.t homsonedu.com/ Ioqtn and go to Chapter 15 Contents to see Screen 15.2 for a visualization of ways to express reaction rates.
Relative Rates and Stoichiometry ProbLem Relate the rat es for t he disappearance of reactants and formation of products for the followi ng reaction:
• Reaction Rates and Stoichiometry Fo r the general reaction a A + b B ~ c ( + d D. the international convention defines the reaction rate as
4 PH ,( g) --> P,(g)
Strategy In t his reaction. PH] disappears, and P 4 and H2 are formed . Consequently, t he value of .:i.[PH]]f .:i.t will be negative, whereas .:i. [P4 J/ .:i.t and .:i. [H2 J/ .:i.t will be positive. To relate the rates to each other, we divide ~ [ rea g e n t]f j, t by its stoichiomet ric coefficientin t he balanced equation.
1-"'[AJ -_ - -1-"'[-8J Rate -_ - ~ a ell
b ell
~ +~ "'[eJ ~ + ~ "'[D J c
"'t
d
+ 6 H, (g)
Solution Because four moles of PH ] disappear for every one mole of P4 formed , t he numericaLvalue of the rate of form ation of P4 is one fourth of the rate of disappearance of PH]. Similarly, P4 is formed at only one sixth of the rate t hat H2 is formed .
s:
Rate of Reaction
r 674
Chapter 15
I
ProbLem Data collected on the concentration of dye as a function of time (see Figure 15.1) are given in the graph below. What is t he average rate of change of t he dye concentration over the first 2 min? What is t he average rate of change during t he fifth minute (from t = 4 min to t = 5 min)? Estimate t he instantaneous rate at 4 min.
Chemica l Kinetics: The Rates of Chemical Reactions
Strategy To find the average rate. calculate the difference in concentration at the beginning and end of a time period (.1.c = Clin.l - Cinitia l) and divide by the elapsed time. To find the instantaneous rate at 4 minutes. draw a line tangent to the graph at the specified time. The negative of the slope of the line is the instantaneous rate. 3.50
z:
3.00
.s.
2.50
~ 0
0
M
2.00
X c
." 0
1.50
bc •uc
1.00
0
u
•
0.50
~
0
0.00
0
1
2
4
3
6
5
7
8
Time (min) SoLution The concentration of dye decreases from 3.4 x 10 - 5 molJL at t = 0 min to 1.7 x 10 - s molJL at t = 2.0 min. The average rate of the reaction in this interval of time is Average rate
= _ j, [ Dye1=
(1.7 X 10 - ' moVl) - (3.4 X 10- ' moVl)
lit
2 .0 moL 6
Average rate
= +
8 .5 X 10- mol l . min
The concentration of dye decreases from 0.90 X 10 - 5 moljl at t = 4.0 min to 0.60 t = 5.0 min. The average rate of the reaction in this interval of time is Average rate
X 10 - s
moljLat
(0. 60 X 10- ' moVl) - (0. 90 X 10- ' moVl)
=
1.0 mol
Average rate =
+ 3 .0
X 10-
6
mol
L' min
From the slope of the line tangent to the curve. the instantaneous rate at 4 min is found to be + 3.5 X 10- 6 mcl/t ' min. EXERCISE 15 .1
Reaction Rates and Stoichiometry
Whatare the relative rates of appearance or disappearance of each product and reactant, respectiveLy, in the decomposition of nitrosyl chloride, NOel? 2 NOCI(g) -> 2 NO(g)
+ CI,(g) :J 0.04 c,
EXERCISE 15 .2
!
Rate of Reaction
Sucrose decomposes to fructose and glucosein acid solution. A plot of the concentration of sucrose as a function of time is given here. Whatis the rate of change of the sucrose concentration over the first 2 h? Whatis the rate of change over the last 2 h? Estimate the instantaneous rate at 4 h.
0.03
."~o
0.02
~ ..3
0.01
0.000'-----=-----,-----,----,-----, 4
6
8
Time (hours) Sign in at www.thomsonedu .com/Ioqtn and go to Chapter 15 Contents to see Screen 15.2 for a self-study module on rate of reaction.
Concentration versus time for the decomposition of sucrose. (See Exercise 15.2.) 15.1
I
Rates of Chemical Reactions
675
15.2
Reaction Conditions and Rate
Seve ra l facto rs- reac ta n t concentra tio ns, te m pe rature , a n d prese nce of ca ta lystsa ffe ct th e rate o f a r ea cti o n . If the reacta nt is a so lid, th e su rface area ava ila b le fo r
reactio n is also a factor. T he "io d in e cloc k re a cti on " ( Fig u re 15 .3 ) illus tra te s th e effec t of co nce n tra tio n
and tem pe ra tu re . T he reac tion mi xtu re co n ta ins h ydroge n pero xide ( H 20 2 ) , io di de io n (1 - ), vitamin C (asco rbic acid), and starc h (which is an ind icato r of the presence of iod in e , 12) , A seq u e n ce o f reactio ns begins with th e slow o xid a tion of iod id e ion to 12 by H 20 :!. H,O,(aq)
+
2 r-(a q)
+
2 H,O+(a q) -> 4 H,O(C)
+ I,(aq)
As so o n as 12 is form ed in th e so lu tio n , vitam in C rap idl y reduces it to 1- .
• Effect of Temperature on Reaction Rate Cooking involves chemical reactions, and a higher te mperature results in foods cooking faster. In t he laboratory, reaction mixtures are often heated to make reactions occur faster.
Wh en a ll of th e vitam in C h as bee n co ns ume d , 12 remains in so lu tio n a n d forms a bl u e-black com p lex wit h starch . Th e time measu red represen ts h ow lo ng it has tak e n for th e give n a mou n t of io d ide ion to react. Fo r the first ex perime n t (A in Figure 15 .3) th e tim e required is 5 1 seco n ds . when th e co nce n tratio n of io di de io n is smaller (8) , th e tim e re quired fo r the vitamin C to be consu med is lo nge r, 1 min ute and 33 seco n ds . Fina lly, whe n the co nce n tra tions a re again th e same as in ex perime n t 8 bu t th e react ion mix tu re is h ea ted, t he reac tio n oc curs m ore ra pidl y (56 secon ds) .
(a) Initial Experiment. Th e blue color of the starchiodine com plex develops in 51 seconds.
(b) ChangeConcentration. Th e blue color of starch-iodine complex develops in 1 minute. 33 seconds when the solution is less concentrated than A.
Smallerconcent ration of 1- t han in Experiment A.
Solutions containing vitamin C, H20 2 , r , and starch are mixed.
(c) ChangeTemperature. The blue color of the starchiodine complex develops in 56 seconds when t he solution is the same concentration as in B but at a higher temperature.
Same concentrations as in Experiment B, but at a higher temperature.
FIGURE 15.3 Theiodine clock reaction. This reaction illustrates t he effects of concentration and temperature on reaction rate. (You can do these experiments yourself with reagents available in the supermarket. For details, see S. W. Wrig ht: "lhe vitamin Cdock reaction." Journal of Cnemico! Education, Vol. 79, p. 41, 2002.) See ChemistryNow Screen 15.11 for a video of the iodine clock reaction. 676
Chapter 15
I
Chemical Kinetics: The Rate s of Chemical Reacti ons
00
00
(~
FIGURE 15.4 Catalyzed decomp osition of HzOz" (a) The rate of decomposition of hydrogen pe roxide is
increased by the catalyst MnOz. Here, HzO z (as a 30%aqueous solution) is poured onto the black solid Mn Oz and rapidly decomposes to O2 and H20. Steam forms because of the high heat of reaction. (b) Abombardier beetle uses the catalyzed decomposition of HzOz as a defense mechanism. The heat of the reaction lets the insect eject hot water and ot her irrita ti ng chemicals with explosive force. (c) A nat urally occurring catalyst, called an enzyme,
decomposes hydrogen peroxide. Here, the enzyme found in a potato is used to catalyze HzOz decomposition. and bubbles of O2 gas are seen rising in the solution.
Catalysts arc su bstances th at accel era te chem ica l reactions but are not th em selves co nsu m ed. Co ns ide r the effect of a cata lyst o n th e decomposition o f hyd rogen pero xide, H 20 2• to form wate r and oxygen . 2 H,O, (a q) --> O, (g)
+ 2 H,O(f)
This d ecomposit ion is very slow; a so lu tio n o f H 20 2 call be sto re d for man y mo n ths with on ly min imal chang e in co ncen tra tio n. Ad d ing a man gan ese salt, an io di d eco n ta ini ng salt, or a bio log ica l su bstanc e called an enzyme causes th is react io n to oc cur rapidly. as shown by vigorous bu bb ling as ga seous o xygen escapes fro m the so lu tio n (Fig u re 15.4) . T he su r face area o f a so lid reactan t ca n a lso affe ct th e react ion rat e . O n ly mol ecul es at the surface of a so lid ca n co m e in co nt act with ot her reactants. Th e sm aller th e pa n icles o f a so lid, th e more m olecul es are fo u nd o n th e so lid 's su r face . With ve ry sma ll particles, th e effect o f su r face a rea o n ra te can be quir e d ra matic (Fig u re 15.5) . Fan ners know th at ex p los io ns of fin e d ust particles (suspe nd ed in th e air in an e nclosed. silo or at a feed mi ll) re presen t a majo r hazard.
(a)
ChemistrY-9- Now'" Sign in at www.thomsonedu .com /Ioqtn and go to Cha pt e r 15 Contents to see: Screen 15.3 for a visualization of the factors controlling rates • Screen 15.4 for a simulation of t he effect of concentration on rate
e
J...
...
~
15.3
Effect of Concentration on Reaction Rate
(b)
On e importan t go al in studying the kineti cs of a reactio n is to de termin e its mech anism ; th at is, how th e reacti on o ccu rs at the molecul ar leve l. T he place to beg in is 10 learn ho w co nc e n tratio ns of reac tan ts affect the reaction rat e . The effect of co ncent rat ion ca n be determined by evalua ting the rate of a reaction using d iffe re nt co nce ntra tio ns of eac h reactan t (with th e temperature held co nstant) . 15 .3
I
FIGURE 15.5 The combustion of lycopodium powder. (a) The spores of
th is common fern burn only with difficulty when piled in a dish. (b) If the spores are ground to a fine powder a nd sprayed int o a flame, combustion is ra pid.
Effec t of Conce ntr at ion on Reaction Rate
677
Co nsid er, for exam ple, th e decomposition o f N20 ,·, to 1\0 2 and 0 2' Figu re 15.2 presen te d d at a o n th e co ncen tra tio n of N 20 5 as a fu nc tion o f time . \Ve previou sly ca lcu late d tha t, when [N 20 ",] = 0.34 0101/1" th e instan ta ne o us rat e o f disappearan ce of N20 ,j is 0.0014 mol/L . min . An evaluatio n of th e instantane o us rate of the reaction when [N 20 :-J = 0.68 mol/L reveals a rate of 0.0028 mo l/L . min . That is, d oubling th e co nce n tration of N2 0 j doubles th e reactio n ra te. A sim ilar exe rcise shows th at if [N 2 0 ,.J is 0,17 mol / L (half of 0.34 mo l/L) , the reac tio n ra te is also halved . From th ese resu lts, we know that th e reaction rat e must be directly p roportion al to the reac tant co nce n tratio n for this reactio n:
Rate of reaction = [N20sl where th e sym bo l 'X m ean s "p ropo rtion al to ." Different rel ation sh ips bet wee n reaction rate and reac tan t co nce n tra tio n are enco u n tered in o ther reactions. Fo r exa m ple. the reaction ra te co u ld be indep end ent o f co nce ntr atio n , o r it m ay depend on th e reac ta n t concen tratio n raised to so me powe r (tha t is, [reacta n t] 1/) . If th e reaction in vol ves seve ra l reac ta n ts, th e reaction rat e may depend on th e concentrations o f eac h of th em o r o n only one of th em . Finally, if a catalyst is involved , its co nce n tratio n may also affect th e ra te , as ca n th e co nce n tra tio ns o f product s.
Rate Equations Th e rel ati onship bet ween react ant co nce n tra tio n and. reaction rat e is ex p ress ed by an eq ua tio n calle d a rate equatio n, o r rate law. For th e d eco mposition o f N"20 j th e rat e eq ua tio n is N20S(g) --> 2 N0 2( g) + 'I, O2 Rate of reaction
= k[N 20sl
whe re th e proportionality co ns tan t, k, is called th e rate constant. T his rate eq ua tio n tell s us that thi s rea ction ra te is proportional to th e co ncen tratio n of th e reactant. Based on th is equatio n, we ca n d et ermin e th at when [N 20 :j ] is doubl ed , th e reaction rate d oubles. Generall y, for a rea ction suc h as a A+b B -->xX • Exponents on Reactant Concentrations and Reaction Stoic hiometry It is important to recognize that t he exponents m and n are not necessarily the stoichiometric coefficients (a and b) for the balanced chemical equation.
th e rate equation has the form Rate of reaction = k[A]m[B]' The rat e equation expresses th e fact th at the rate of rea ction is propo rtional to th e rea ctant concentrations, each co ncen tra tio n being rai sed to so me po wer. T he export ents in thi s equa tio n are often positive who le numbers, but they can a lso be ne gative numbers, fra ction s, o r ze ro and they are d et ermined by ex pe rime n t. If a homogen eous catalyst is present, its co nce n tra tio n migh t also be included in the ra te equati on , eve n though th e ca ta lytic spec ies in not a pro duct o r reac ta n t in th e eq uatio n for the reacti on . Co ns ide r, for exam ple , th e d eco mpo sition o f hydrogen peroxid e in th e pr esen ce o f a ca ta lyst suc h as iodide ion. H20 2(a q)
1- (" )
H20(t ) + Y, 02(g)
Expe rime n ts show th at thi s rea ction has th e fol lowin g rat e eq ua tio n : Reaction rate 6 78
Chapter 15
I
Chemica l Kinetics: The Rate s of Che mical Reacti ons
~
k[H2 0,] Wl
Here , th e co n ce n tratio n o f l " a p pears in the ra te law, even th ough it is n o t in vol ved in th e balan ced eq u atio n .
The Order of a Reaction Th e order o f a reac tio n with resp ect to a p articu la r react ant is th e exp o ne n t of its co nce ntrat io n te rm ill the rat e e xp ressio n, a nd th e overall reactio n order is th e su m of the ex po n e n ts o n all co nce n tra tio n term s. Co ns id e r, for ex am p le, th e reaction of N O a nd C12 :
• The Nature of Catalysts A catalyst does not appear as a reactant in the balanced, overall equation for the reaction, but it may appear in t he rate expression . A common practice is to identify catalysts by name or symbol above the reaction arrow. as shown in the example. A homogeneous catalyst is one in t he same phase as the reactants. For example, both H20 2 and 1- are dissolved in water.
2 NO(g) + Cl,(g) -> 2 NOCl(g) Th e ex peri men ta lly det ermin ed rate eq uatio n for th is reaction is Rat e
= k[NO]' [Cl,l
This reactio n is sec o nd o rd er in NO , first orde r in Cl 2 , a n d th ird order overall. How is thi s rela te d to th e ex pe rime n ta l d at a for th e rate of di sappea ran ce o f NO ? Experi ment
[NO] mOl/L
[t l,l mOl/ L
0.250
0.250
J.
X2
• Overall Reaction Order The overall reaction order is the sum of the reaction orders of t he different reactants.
Rate moljL • s 1.43
J.
J.. no change
x 10- 6 X 4
x 10 - 6
2
0.500
0.250
5.72
3
0.250
0.500
4
0.500
0.500
x 10- 6 11.4 x 10- 6 2.86
Compare Experim ents 1 and 2: Here, [Clz] is held co ns tan t, a n d [NO ] is doub led . The c ha nge in [NO ] lead s to a reaction ra te in crease by a fac to r of 4; th at is, th e rate is p roportional to th e square o f th e NO conce n trati o n. • Compare Experiments 1 an d 3: In experim en ts I a n d 3, [NO ] is h e ld con sta nt, a nd [Cl z] is doubled, ca using the rat e to d ouble. T ha t is, th e rat e is proportion al to [el, ] . • Compare Experim ents 1 and 4: Both [NO ] a n d [Clz] are d oubled from 0 .250 ~ I to 0.5 00 1\1. From previo us experiments, we kn ow that d o ublin g [NO ] sh o u ld ca use a fo u r-fo ld increase , a n d douhlin g [CI2 ] causes a two-fold increase . Therefo re , doublin g both co nce n tra tio ns sh o u ld ca use a n eight -fo ld increase , as is observe d ( 1.43 X 10 - 6 X 8 = 11.4 X 10- " rIl o l/L · s) . •
The de compo sition o f am mon ia on a p lat inum su r face a t 856 °C is a ze ro order reactio n . NH,(g) -> If, N,(g)
+ %H,(g)
Thi s mean s th at the reacti on rate is independent o f
N II :~
co nc e n tratio n.
Rate = k[NH, l' = k React ion o rde r is impo rt ant b ecause it gives so me in sig h t int o th e mo st int e resting que stion of all- how th e reactio n o ccu rs . T hi s is described fu rth er ill Se ction 15 .6.
The Rate Constant, k Th e rat e co nsta n t, k, is a proportion ality constan t th ai rel at es ra te an d concent ration at a g iven temperature . It is an im po rtan t qua ntity becau se it e n a b les yo u to li nd the react ion ra te fix a new se t o f co nc e n tratio ns. To see how to lise k, co ns id e r 15. 3
I
Effect of Concent ratio n on Reacti on Rat e
679
• Time and Rate Constants The time in a rate constant can be seconds, minute s, hours, days, years, or whatever time unit is appropriate . The fraction f/ti me can also be written as time: ". For example, 1fy is equivalent to y- l, and 1/s is equivalent to S-l .
th e su bst itu tio n o f CI- io n by water in t he cancer chemot herapy age n t cisplatin, I't (NH .,),CI, . Pt(N H, ), Cl,(aq)
+
H,O(C)
+
+
T he rate law for thi s reaction
IS
Rate
= k[Pt(NH )), CI,l
an d th e rate co nsta n t, h, is 0.27/h at 25 "C. Knowing k allows y O ll to calcul at e th e ra te at a pa rtic ular reactan t co nce ntratio n-for example , when [Pt(N II 3 h CI2 ] = 0.018 mol/L: Ra te = (0.27/h)(0.018 mcl/L) = 0.0049 rnol/ L: h • Some Rate Constant s First Order k( l /s) 2 N20 5(g) --. 4 NO,(g)
+ O,(g)
3.38 X 10- 5 at 25 °C
C,H. (g) --. 2 CH, (g)
5.36 X 10- 4 at 700 -c
Sucrose(aq, H30 +) ~ fruetose(aq ) + gtucose(aq)
6.0 X 10- 5 at
Second Order
k( L/mo l' s)
2 N06' (g) --. 2 NO(g)
0.80 at 10 "C
25
As no te d , earl ier, reac tio n rates have units o f mol/L . time whe n co nce ntra tio ns are g iven as mo les per liter. Rat e constants mu st have units co nsiste n t with th e units fo r the other terms in th e ra te equati on . • First-o rder reactio ns: the un its of If are I/t im e . • Second-orde r reactions: th e uni ts o f k are Llm o l . time. • Ze ro-order react ion : th e units of k are mo l/L . tim e.
-c
Determining a Rate Equation + 6,,( g)
H,(g) + I,(g) --. 2 HI(g)
0.0242 at 400 "C
O n e way to d et e rm in e a rate eq uat ion is by usin g the "met hod o f init ial ra tes." T he initial rat e is th e in stantan eous reac tio n ra te a t th e sta r t of the reac tio n (th e ra te at t = 0) . An ap p roxima te va lue o f the initial rate can be o btaine d by m ixing th e react ants a nd d e te rmin in g d [p rod uct] /d l or - d [ reacta n t]/ d l after 1% to 2% o f th e lim iting re act a nt has bee n co ns umed. Me asuring th e ra te du rin g the ini tia l stage o f a reaction is co nvenie n t becau se ini tia l concen tratio ns are kn own . As an exa mp le o f th e dete rm in ati on of a reaction rat e by th e me thod of initial rates, let us co ns ider the reactio n of so d iu m h ydro xid e with me th yl acetate to prod uce ace ta te ion an d m ethanol. CH ,CO,CH, (aq)
+
+
•
-
CH ,CO, - (aq)
+ CH ,OH(aq) +
Reactant co ncentrations and initial rates fo r th is reactio n for several ex perime n ts at 25 "C are co llec ted in the ta ble be low.
680
C h a p t e r 15
I
Chemica l Kinetics : The Rates of Chemical Reactio ns
Initial Concentrati ons (mol/L)
Experiment
[OW]
(mol/ l · s) at 25 · C
0.050
0.050 .l x 2
0.000 34 .l x 2
0.05 0
0.10
0.00069
I.x 2
1- no change
I x 2
0.10
0.10
0.00137
J. no change 2
3
Initi al Reaction Rat e
[CH,CO,CH,1
As no ted in the tab le , whe n th e ini tia l concen tra tio n o f o n e reac ta n t (e ith er CIIj CO~ CII 3 or O II - ) is doubled wh ile the concen tra tio n of the other reac ta n t is
held co ns ta n t, the in itial re action rate doub les. T h is rate doub ling shows th at the reaction ra te is d irectly p ropo rtion al to th e co nc e n tratio ns of both CH :\CO:!C H 3 and O I r ; thus, the reaction is first order in eac h of th ese reac ta n ts and seco n d orde r overall. T he ra te law th at re flect s th ese ex p e rime n ta l o bse rvat ions is Rate
~
k[CH,CO,CH,][OW l
Using thi s equat ion, we ca n predict that do ub ling bo th co nce n tra tio ns at the same time sho uld cause th e rat e to go up by a fac to r o f 4. What hap pens, however, if on e co nc e n trat io n is d oubled and the othe r is halved ? T he rat e eq ua tio n tells us the rat e sho uld not cha nge! If the rate e q uatio n is know n , the valu e o f k, the ra te con sta nt , can be fou nd by su bstitu ting values for the ra te a nd co nce n tra tion into the rat e eq ua tio n . Usin g the d at a for th e methyl ace ta te / hyd ro xid e ion re act io n fro m the fir st e xperi me nt. we h ave
Rate = 0.00034 moljl . s = k(0.050 molj l )(0.050 molj l ) k~
0.00034 moljl . s ~ 0.14 l/ mol . s (0.050 moljl)(0.050 moljl)
Sign in at www.t homsoned u.com/ Ioqtn and go to Chapter 15 Contents to see: Screen 15.4 for a self-study module on control of reaction rates Screen 15.5 for a simulation, a tutorial, and an exercise on determining rat e equati ons from a stud y of the effect of concentration on reaction rat e
Determining a Rate Equation Problem The rate of the reactio n betwee n CO an d N0 2 at 540 K
CO( g) + NO,(g ) -> CO,( g) + NO (g) was meas ured sta rting with vario us concentratio ns of CO a nd N0 2• Determine th e rate equation a nd the value of t he rate consta nt . Initi al Concentrations Experiment
Initial Rate
[CO] , mol/ l
[NO,]. molfl
(molf l ' h)
5.10 X 10- 4
0.350 X 10- 4
3.4 X 10- 8
5.10 X 10- 4
0.700 X 10- 4
6.8 X 10- 8
5.10 X 10- 4
0.175 X 10- 4
1.7 X 10- 8
0.350 X 10-
6.8 X 10- 8
10- 3
4
1.02 X
5
1.53 X 10- 3
4
0.350 X 10- 4
10.2 X 10- a
15 .3
I
Effect of Concent ration on Reaction Rat e
681
Strategy For a reaction involving several reactants, the general approach is to keep t he concent ration of one reactant constant and t hen decide how the rate of reaction changes as t he concent ration of t he other reagent is varied. Because the rate is proportionalto t he concentration of a reactant. say A, raised to some power n (the reaction order) Rate oc [AJ'
we can write the general equation Rate in experiment 2 Rate in experiment 1
[A,]' ~ ( [A, ] ), [A,], [A,]
If [A] is doubled and the rate doubles from experiment 1 to experiment 2, t hen n t he rate goes up by 4, then n = 2.
= 1. If [A] doubles and
SoLution In the first three experiments, t he concentration of ( 0 is held constant. In t he second expert-
ment, the N02 concentration has been doubled relative to Experiment 1, leading to a twofold increase in t he rate. Thus, n = 1 and t he reaction is first order in N0 2• Rate in experiment 2 Rate in experiment 1 2 ~ (2)" and so n
6.8 X 10-' moVL' h 3.4 X 10- ' mol/ L: h
0.700 X 10- ' ) " ( 0.350 X 10- 4
= 1.
This finding is confirmed by experim ent 3. Decreasing [N0 2] to half its original value in experiment 3 causes the rate to decrease by half. The data in experiments 1 and 4 (wit h constant {N0 2l) show that doubling [CO] doubles t he rate, and t he data from experiments 1 and 5 show that tripling t he concent ration of CO triples the rate. These resuLts mean that the reaction is first order in [CO]. We now know the rate equation is Rate
~
k[COj[NO,]
The rate constant. k, can be found byinserting data for one of the experime nts into the rate equation. Using data from experime nt 1, for example, Rate
~
3.4 X 10-' moVL· h
~
k
k(5 .10 X 10-' moVL)(0.350 X 10- ' mol/L)
~
1. 9 L/ mol . h
• . Using a Rate Equation to Determine Rates ProbLem Using the rate equation and rate constant dete rmined for t he reaction of CO and N0 2 at 540 Kin
Example 15,3, determine the initiaL rate of the reaction when [CO ] [NO ,]
~
= 3.8 X 10- 4 molfLand
0.650 X 10- ' mof/L.
Strategy A rate equation consists of three parts: a rate, a rate constant (k), and the concentration terms. If two of these parts are known (here k and t he concentrations), the third can be calculated. SoLution Substitute k (= 1.9 Lj mot · h) and t he concentration of each reactant into the rate law deter-
mined in Exam ple 15.3. Rate
~
Rate
= 4.7 X 10- 8 molfL· h
k[COj[NO, ]
~
(1. 9 ljmo l ' h)(3.8 X 10- ' moVL)(0.650 X 10- ' moVL)
Comment As a check on the calcuLated result, it is sometimes useful to make a n educat ed guess at t he answer before carrying out the mathematical solution. We know that t he reaction here is first order in both reactants. Com paring t he concentration vaLues given in t his probLem with t he concentratio n values in found experiment 1 in Example 15.3, we notice th at [CO] is about t hree fourths of t he concentration value, whereas [N0 2] is almost twice t he value. The effects do not precisely offset each other, but we might predict that the difference in rates between this experiment and experiment 1 will be fairly small, wi th t he rate of t his experiment being just a little greater. Th e calculated value bears t his out.
682
Chapter 15
I
Chemica l Kinetics: The Rates of Chemical Reactions
EXERCISE 15 .3
Determ ining a Rat e Equati on
The initial rate of the reaction of nitrogen monoxide and oxygen
2 NO(g)
+ O, (g) --> 2 NO,(g)
°
was measured at 25 O( for various initial concentratio ns of NO and 2 _ Determine the rate equatio n from these data. What is the value of the rate constant, k, and wha t are its units? Initial Concentrati ons (molj L)
Experiment
[NO]
[0,]
Initial Rate
(molj L• s)
0.020
0.010
0.028
2
0.020
0.020
0.057
3
0.020
0.040
0.114
4
0.040
0.020
0.227
5
0.010
0.020
0.0 14
EXERCISE 15 .4
Using Rate Laws
The rate constant, k, at 25 O( is O.27/h for the reactio n Pt(N H,),Cl,(aq) + H,O(f) --> [Pt(NH ,),(H ,O)Cll'(aq) + CI -(aq)
and the rate equation is Rate
~
k[Pt(NH ,),CI,]
Calcu late the rate of reaction when the concentration of Pt(NH 3 hCl 2 is 0.020 M.
Concentration-Time Relationships: Integrated Rate Laws 15.4
~.::::..g. Module 20
It is often im portant for a che m ist to know how lon g a re act io n m ust pro ceed to re ac h a predet erm in ed co nce n tra tio n of so m e reactan t or p ro d uct , o r what the reacta n t and prod uct co nc e n tra tio ns will be after so me time has elapsed . A ma thc ma tica l eq uatio n that re lates time and co nce n tratio n- that is, an eq uatio n that d escr ibes co ncen tra tio n- time cu rves like the on e shown in Figure 15.2- can be use d to d et erm in e this info rma tion . With suc h an eq ua tio n, we co uld ca lcu late the co ncen tra tio n at an y give n tim e o r th e length o f tim e need ed fo r a g iven am o u n t of react ant to react.
First-Order Reactions Su p pose the reaction "R -? prod ucts" is first o rd er. T his mean s the reactio n rat e is directly p ro po rtio na l to th e co nc e n tr atio n of R ra ised to th e first power, o r, ma th e mat ically, - t. [R] = k[R] M
Using calculus, th is rel atio nsh ip ca n be tran sformcd into a vc ry usefu l eq uatio n callcd an in tegrated ra te eq uation (be cause in tegra l calculus is used in its d erivat ion ) . ln iRk [R]o
=-
kt
(15.1)
Signin at www.t homsonedu.comjtogin to download the Go Chemistry module for this sectionor go to www.ichapters.com to purchasemodules.
683
A Closer Look
Rate Laws, Rate Constants, and Reaction Stoichiometry
Qualitatively. the rat e of a reaction is easy to
concentrations of reactants and products. Based on the reaction stoichiometry, we know that the rate of appearance of Oz is one half the rate of disappearance of N20 S and one fourth the rate of N02 appearance. Second, this equation specifically defines what we mean by rate; it provides a single, numericalvalues for this parameter; and it tetls us how to calculate the value of reaction rate from experimental data. If we were to follow the disappearance of NzOs as a measure of reaction rate and base our definition of rate on the stoichiometry above, we should write the followi ng differential form of the rate law.
understand: it represents the change in conce nt rati on of the reactants a nd products.
When we deal with rates of reaction quantitatively, however, we need to be specific about the reaction stoichiometry. Co nsider the first order decomposition of NzOs. a reaction that we mentioned earlier 2 N20 ,
-->
4 N0 2 + 0 2( g)
The rate of the reaction can be expressed (and measured in lab) as the change in conce nt rat ion of either reactants or products as a
function of time. The numerical values of the rates of formation of reactants and products a re related . but t hey a re different because of t he reactio n stoichiometry. If we equate the ra te of reaction to the rate of app ea ra nce of
O2 , we would write Rate
~
-l [ 02J -l t
~ _ ~ ~ ~ + ~ ~N02J 2
-l t
4
-l t
This relation is written by dividing each rate by its stoichiometric coefficient (.... page 674). There are two different pieces of information in the equation above. Fi rst it gives the relationship between the rates of change of
• Initial and Final Time, t The time 0 does not need to correspond to the actual beginning of the experi ment. It can be the time when instrument readings were started, for examp le, even though the reaction may have already begun. t =
and the half-life equation is
t ,/ 2 =
N20 ,
~t
a
-l [ N0 ] Rate ~ - - - '-'- ~ 2k [N 0 ] ~t
a s
the integrated rate equation is (see page 683),
2 N0 2 + '12 0 2(g)
Oifferential rate equation: - .1 [N,O ,] ~t
~
°]
k'[ N
a s
Integrated rate equation:
IJN20 ,} ~
[N20,]o
- i«
Half-life equation:
5
From t his definition, it follows that the differential rate equation is
-->
Followi ng the reasoning above, the rate laws for the reaction written this way wo uld be:
Rate = - -1 --l [-N20, - ] = k [ NO ] 2
0 .69 3/ 2k
We can also write t he equation for NzOs decom position as follows:
t'/2 ~ 0.693/k'
Note that t he differentia l and integrated rate laws derived based on the two different chemical equations have the same form, but k and k' do not have the same values. In this case, 2k
~
k' ,
For more on this issue, see K. T. Quisenberry and J. Tellinghuisen, Journal of Chemical Education, Vol. 83, pp. 510-512, 2006.
In [N20,} ~ - 2kt
[N20,l
Here , [R}« and [R] / are concen tra tio ns of the react an t at tim e l = 0 and at a lat er tim e , t, respect ively Th e ratio of co nce n tratio ns , [R]/[ RJo, is th e frac tio n of reacta nt th at rem ain s after a give n time has elapse d . In word s, th e equatio n says
.h ( Nat ura I Iogant m
concentration of R after some time ) conce ntration of R at sta rt of experiment
= In(fraction ~
of R remaining at time, t) - (rate constant)(elapsed time)
No tice th e negative sign in th e equa tion . The ratio [R] /[ R]o is less th an I because [RL is always less th an [R] c: th e re actant R is co ns u med durin g th e reaction . This mean s th e logarithm o f [RJ/[ R]o is negative. so th e othe r side of th e eq ua tio n must also be ar a negative sign . Eq ua tio n 15.1 can be used to ca rry out th e follo wing calculatio ns: • If [RL/[ Rl o is measu red in th e labo ratory after so m e amo u n t of time has elap sed , th en k can be ca lcu lated. • If [Rl u and k are known , th en th e con centrati on of mat erial remain ing after a given am o u n t of tim e ([ R] /) ca n be ca lcu late d . • If k is known , th en th e tim e elapsed until a specific fraction ([RL/ [Rl o) re mains can he calcu lated . Fina lly, not ice that k for a fir st-order reacti on is inde pe nde nt of con ce ntration : k has units of time - \ (J,-I or \"- \ for exam ple ) . This mean s we can choose any 684
Chapter
15
I
Chemical Kinetics: The Rates of Chemical Reactio ns
co nvenie n t uni t fo r [RL and [R]o: mo les pe r liter, m oles, gram s, nu mber of ato m s, n u m ber o f mol ecules, or ga s p ressure . ChemistrY-9-Now~ Sign in at www.thomsonedu .com/Ioqtn and go to Chapter 15 Contents to see Screen 15.6for a tutorial on the use of the integrated first-order rate equation.
The First-Order Rate Equation Problem In t he past, cyclopropa ne, C3H 6 , was used in a mixture with oxygen as an anesthetic. (This practice has almost ceased today, because the compound is flammable.) When heated, cyclopropane rearranges to propene in a first-order process. H I
H
H I
CH ,CH
C" I I C I ~ C' I
H
I H
= CH,
H
cyclopropane
propene
Rate = k[cyclopropane]
k
~
2.42 h- ' at 500 °C
If the initial concentration of cyclopropa ne is 0.050 mo f/t. how much time (in hours) must elapse for its concentration to drop to 0.010 molfl?
Strategy The reaction is first orderin cyclopropane. You know t he rate constant, k, and t he initialand final concentrations. Use Equation 15.1 to calculate the time (t) elapsed to reach a concent ration of 0.010 molfl. Solution Values for [cyclopropane]; [cydcpropanej -. and k are substituted into Equation 15.1; t (time) is t he unknown: [ 0.010J In- [ 0.050J
~
(2.42 h-')t
t ~ - In(0.20) ~ - (- 1.61) ~ 0.665 h
2.42 h
12.42h l
Comment Cycloalkanes with fewer than five carbon atoms are strained because the C-C-C bond angles cannot match the preferred 109.5°. Because of ring strain, the cyclopropane ring opens readily to form propene.
• . Using the First-Order Rate Equation Problem Hydrogen peroxide decomposes in a dilute sodium hydroxide solution at 20 °Cin a first-order reaction:
+ 'I, O,(g) Rate = k[HzOzl with k = 1.06 X 10- 3 min' H,O,(aq)
--7
H,O(f)
?
Whatis the fraction remai ning after 100. min if the initial concentration of HZ0 2 is 0.020 molfL? Whatis the concentration of the peroxide after 100. min? Strategy Because the reaction is first order in HZ0 2, we use Equation 15.1. Here, [HzOz]o, k, and tare known, and we are asked to find the value of [HzOz]t and the fraction rema ining. Recall that
1 = fraction remaining
[R
[RJ,
Once this value is known, and knowing [H 20 2 Jo. we can calculate [H20d t. 15 .4
I
Concent ration-Time Re lationships: Integrated Rate Laws
685
SoLuti on Substitute the known values into Equation 15.1. In [H,O,], ~ - kt ~ - (1.06 X 10- ' min- ' )(100. mi n) [H,O,j, In [H,O,j, ~ - 0.106 [H,O,], Taking the antilogarit hm of - 0.106 [i.e., the inverse of In(- 0.106) or e- O•106 j, wefind the fraction remai ning to be 0.90. Fra ction remaining
= IH20Jr = 0.90 [H,O,],
Because [H 202] O = 0.020 moljL, this gives [H 20 2] r = 0.018 moljl.
EXERCISE 15 .5
Using the First-Order Rate Equatio n
Sucrose, a sugar, decomposes in acid solution to give glucose and fructose. The reaction is first order in sucrose, and the rate constant at 25 °Cis k = 0.21 h- 1 • If the initial concentration of sucrose is 0.010 moljL, wh at is its concentration after 5.0 h?
EXERCISE 15 .6
Using th e First-Order Rate Equati on
Gaseous azomet hane (CH jN 2CH j ) decom poses to ethane and nit rogen when heated:
+ N,(g) The disappearance of azomethane is a first-order reaction with k = 3.6 X 10- 4 S - 1 at 600 K. CH,N,CH,g)
--7
CH ,CH,(g)
(a) Asample of gaseous CHjN 2CH j is placed in a flask and heated at 600 Kfor 150 s. Wh at fraction of t he initial sample remains after t his time? (b) How tong must a sample be heated so that 99% of the sample has decomposed?
Chemislryj J"Now'"
•
Sign in at www.t homsonedu.com/ Ioqtn and go to Chapter 15 Co ntents to see Screen 15.6 for a selfstudy module on the first-order rate equation .
Second-Order Reactions Su p pose th e re actio n "R
---7
products" is seco nd orde r. Th e rat e equation is
_ u[R] ~ k[R], M Using calculus, thi s relationsh ip ca n be tran sform e d into th e fo ll owin g equa tio n th at relates reacta nt co nce n tra tio n and tim e: 1
1
[R],
[Rl,
~ -~ = kt
(15.2)
T he sa me symbolism used with first-order react ions ap plies: [R] o is the co nc e n t ratio n of reacta n t at th e tim e t = 0; [RL is th e co nce n trat io n a t a lat er tim e ; and k is th e sec o nd-o rd er rat e co nstant which has the units o f L/ mo l . tim e .
686
Chapt er
15
I
Chemica l Kinetics: The Rates of Chemical Reactions
Using the Second-Order Integrated Rate Equation Problem The gas-phase decomposition of HI
HI(g) --> 'I, H,(g)
+ 'I, I,(g)
has the rate equation
_ !>[HI] ~ k[HII' jJ
where k = 30. L/m ol' min at 443 °C. How much time does it ta ke for the concent ration of HI to drop from 0 .010 molf l t o 0.0050 molfL at 443 °C? Strategy Substit ute t he values of [HIlo, [HIl t, and k into Equation 15.2, and solve for the unknown, t. SoLution Here, [HIl o = 0.010 molfl and {HIlt = 0.0050 molfL. Using Equati on 15.2, we have
1 1 0.0050 molfl 0.010 molfl (1.0 X 10' l/ mol) - (1.0 X 10' l/ mol) t
EXERCISE 15.7
~ (30. l/mol. min)t ~
(30. l/ mol · min)t
=
3.3 min
Using the Second-Order Integrated Rate Law Equation
Usi ng the rate consta nt for HI decomposition given in ExampLe 15.7. calculate the concent ration of HI after 11 min if (HI!o ~ 0.010 mcl/L.
Zero-Order Reactions If a reactio n (R ---7 product s) is zero order, th e rate eq uation is _ a[R] = k[R]O
at
T hi s e q uatio n leads to th e integrated ra te e q uatio n (15.3)
[Rlo - [R], = kt where the un its o f k a re mo l/L . s.
Graphical Methods for Determining Reaction Order and the Rate Constant \ Vc ca n de rive a co nvenie n t way to determin e th e o rde r of a reac tio n a n d its rat e co ns ta n t u sin g gra p h ica l m et hods. Eq u a tio ns 15.1 , 15 .2, and 1:'").:1, if rearran ged slig h tly, h ave th e form j' = rnx + b. T h is is th e e q ua tio n fo r a stra igh t lin e , whe re m is th e slo pe of th e lin e a n d b is th e j- in rc rcc p t. In th e se e q ua tio ns , x = I in eac h cas e. Zero order
(RI,
~
- kt
First order
+ [R]o
1 11 y
mx
b
In [R] ,
~
Second order
- kt + In [RIo
1 1 1 y
mx
b
1 [R] ,
+ kt +
1 [R]o
11
1
mx
y
15 .4
I
• Finding th e Slope of a Line See pages 39-41 for a description of methods for finding the slope of a line.
b
Concentrati on-Time Relation ship s: Int eg rated Rate Laws
687
Active Figure 15.6 The decomposition of azomet hane, CH3N 2C H3 • If data for the decompositio n of azomethane.
CH]N2C H](g)
-1
CH]CH ](g)
+ N2 (g)
are plotted as the natural logarithm of the CH ]N 2CH 3 pressure versus time. the result is a straight line with a negative slope. This indicates a first-order reaction. The rate constant k = - slope.
P X 10 2
I(s)
In P
- 2
aIm 0 1000 2000 3000 4000
8.20 5.72 3.99 2.78 1.94
- 2.50 - 2.86 - 3.22
-r
k ~ (-3.94) - (-2.86) (4000 1000)s k = 3.6 X 1O- 4(s - 1)
- 3
~ 3 . 58
- 3.94
Sio e ~ P
- 2.5
~
5
- 3.5 - 4
Chemislry..0.....Now"'" Sign in at www. t h om sonedu~com/logi n and go to t he
- 4.5
0
1000
2000
Chapter Conte nts menu to explore an interactive version of t his figure accompanied by an exercise.
3000
4000
5000
Time (s)
As an exam p le o f th e grap hical metho d fo r d etermi ni ng reactio n o rde r, co nside r th e deco m posit io n of azo mct ha nc. CH,N,CH,(g) -> CH,CH,(g) + N,(g) T he decom pos itio n o f azo mc rha nc was fo llowe d at 60 0 K by observin g th e decre ase in its partial p re ssu re with lime (Figu re l 5.fi) . (Rec a ll fro m Cha p te r I I th a t pressure is proportion al 10 concen tra tio n at a g ive n te mpe rature an d vo lu m c .) T he t hi rd co lu m n lists valu es o f In P( CH :~ N 2 CH 3 ). A p lo t o f p ressure vs. tim e for a fir st-order reaction is a cu rved line (see Fig ure 15.2 ) . As shown in Figure 15.6, ho weve r, a grap h of In P( CII .~ N 2 CII 3 ) vers us time produces a st raig h t line , showing th at th e r eacti on is first o rd e r in C H :~ N 2 C H 3' T he slo pe of th e lin e can be measured , an d the n eg ative of the slo pe eq ua ls the ra te co nsta n t fo r th e re ac tio n, 3.6 X 10- 4 S - 1. The d eco mpo sition of N0 2 is a second-orde r pro cess. NO,(g) -> NO(g) Rate
+ 'I, O, (g)
= k[NO,J'
T his fac t ca n be veri fied by showi ng th at a plot of 1/ [ N 0 2 ] versus time is a straig h t lin e (Figu re 15.7). He re , the slo pe of the line is eq ual to k. Fo r a zero-order reaction (Fig u re 15.8 ) , a plot of co nce ntrat io n vs. tim e gives a straig h t lin e with a slo pe eq ua l to th e negative o f th e rat e co ns ta n t.
FIGURE 15 .7 Asecond-order reaction. A plot of 1/ [N02] versus time for the decomposition of N02• NO, (9) --. NO (9) + 'I, 0, (9) results in a straight line. This confirms that this is a second-order reaction.The slope of the line equals the rate constant for this reaction.
Time
[NO,l
I/[NO, l
(min)
( mol/ L)
( L/ mol)
0 0.50 1.0 1.5 2.0
0.0 20 0.015 0.012 0.010 0.0087
50 67 83 100 115
120
100
i" E
2:
M I~
80
0.5
1.0
1.5
Time (min)
688 Chapter 15
I
Chemical Kinetics: The Rates of Chemical Reacti ons
2.0
[NH,l,
'Z
0.540 mmolj l - 1.29 mmoljl
Slope =
,
1000 s - 500 s
0
2.00
= - 1.5
•
"0 E E
:e-
[NH,lo - kt
2.5
"E ~
~
FIGURE 15.8 Plot of a zero-order reaction. Agraph of t he concentration of ammonia. [NH 3]!. against time for t he decom position of NH ~. 2 NH ,(g) --7 N, (g) + 3 H, (g) on a metal surface at 856 O( is a straight line. indicating that t his is a zero-order reaction. The rate constant, k. for this reaction is found from the slope of the line; k = < slope. (The points chosen to calculate the slope are given in red.)
1.50
x 10 - 3 mmol
=- k
L. 5
•
~
::0-
J
0
E E
--' 1.00
£:
s:
•
mmol
k = 1.5 X 10- 3
L·
••
5
•
J
0.50
0 200
400
600
1000
800
Ti me (t) , seconds
Table 15.1 su m m a r izes th e rela tio n sh ips between conce nt ra tio n a nd tim e for first-, se cond-, a n d zero-order processes.
ChemiSlrY.{J"Now'" Sign in at www.thomsonedu.com/Ioqtn and go to Cha pter 15 Contents to see Screen 15.7 for a tutorial on graphical methods.
EXERCISE
15.8 Using Graphical Methods
Data for the decom position of N2 0 S in a particular solvent at 45 O( are as follows: t, min
[N,Osl. moljL 2.08
3.07
1.67
8.77
1.36
14.45
0.72
31.2 8
Plot {N20 S], In [N ~ 0 5] ' and 1/[ N20 s] versus time, t. What is t he order of the reaction? What is th e
rate constant for t he reaction?
TABLE
15.1 Characteristic Properties of Reactions of the Type "R - > Products" Rate Equation
Integrated Rate Equation
0
- ~ [ R ] /~ t ~
k[RJo
[R] o - [Rl,
1
- ~ [ R l/ M ~
k[Rl'
In ([R1J[RJol
2
- ~ [ R J /~ t ~
k[R],
(l/[ R],l - (l/[R]o)
Order
~
kt ~
- kt ~
kt
Straight-Line PLot
Slope k Units
[Rl,
- k
mol/L . time time- t
VS. t
In [Rl,
V5.
t
- k
l/[R],
VS.
t
k
15 .4
I
lj mol . time
Concent rati on-Time Relationsh ips: Int egrate d Rate La ws
689
Half-Life and First-Order Reactions
• Half-Life and Radioactivity Half-life is a term often encountered when dealing wit h radioactive elements. Radioactive decay is a first-order process, and halflife is commonly used to describe how rapidly a radioactive element decays. See Chapter 23 and Example 15.9.
T he half-life, t ,~ , of a reactio n is th e time re quired fo r th e co nce n tratio n of a re ac ta n t to dec rease to o ne h alf its in it ial value . J I alf-life is a co nve nien t \\Tay to descr ibe th e rate a t wh ich a reac ta n t is co ns u me d in a c he m ica l re acti on : T he longe r th e hal f-life , th e slo we r th e reaction . H alf..life is used p rim a ril y whe n dealin g wit h fir st-o rd e r processes. T he h alf-life , t ljl> is th c ti me when th e frac tion of th c reac ta n t R rem ain ing is
[Rl t
= = X [R], or @k [RJo
1/ /1
J Je re , [R](J is th e initial co ncen tration , and [R] I is the co ncen tration after th e reac tion is half co m p leted. To evaluat e l lh. fo r a first-ord er reaction, we substit u te [Rl / [R]o = ~ an d t = t lh. in to th e in regrated first-order rate equat ion (Eq ua tion 15 .1) ,
In (X)
~
- kty,
or
In 2
~
kt Y,
Rea rran g in g th is equatio n (a n d calcu lat ing th a t In 2 = 0.( 9 3) p rovides a u seful eq ua tio n th at relat es h alf-life a n d the first-orde r rate co ns ta n t:
tx = -0.693 k-
(15.4)
T h is eq ua tio n id en tifies an impo rta nt fea ture of first-order reac tio n s: t il l is iruiependent oj concentration. To illustra te the concept of h alf-life , cons ider again th e fir st-o rd e r d ecompositi on of azornc tha nc, C H :3N 2C H 3.
CH,N,CH,(g) -> CH,CH,(g) + N,(g) Rate = k[CH, N,CH,J with k • Half-Life Equations for Other Reaction Orders For a zero-order reaction, R ~ products
Give n a ra te co n sta nt o f 3.6 X 10 - 4 32 min . t y" =
For a second-order reaction, R ~ products I. , h
1
k[R] ,
Note t hat in both cases the half-life depends on the initial concentration.
= 3.6
S- l , \\TC
0.693 3.6 X 10
4 S 1
X 10- 4
, - 1
at 600 K
ca lcu la te a half-life o f 1.9 X 10:3 s or
=
1.9 X la' s
The pa rti al pressu re of azo met hane h as been p lo tted as a fun ct io n of tim e in Figu re 15 .9 , a n d thi s gra p h shows th at P (azomethan e ) d ecr eases by h alf eyc ry 32 min ut es. The in itial p ressu re was 82 0 m m fI g , b ut it dropped to 4 10 rum II g in 32 minutes, and then d ro pped to 205 nuu JIg in a not her 32 m inu tes. T hat is, after two h alf-lives (64 min ) , th e p ressure is (1;2) X ( Y:1) = ( ~) 2 = lit or 25 % o f th e in it ial pressu re . After three h alf-lives, the p re ssure h as d rop ped furt her to 102 nnn II g Of 125 % of th e in itia l value and eq ua l to ('1,) X ( '1,) X ('1,) = (Y2 )3 = Y8 o f th e initial value . It ca n be h a rd 10 visualize whet her a reacti o n is fast or slow from th e ra te co nsta n t value. Ca ll yo u tell fro m th e rate constan t, k = 3.6 X 10 - 4 s", whe ther th e azo m ethan e d ecompo sit ion will ta ke sec o nds, h ours, o r days to reach co mple tio n? Proba bly n o t, but th is is easily assesse d fr o m th e value of th e h alf-life for th e reaction (32 min ) . Now we know th at we wo uld o n ly h ave to wait a few h ours fo r th e react ant to b e esse n tia lly co nsu med .
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690
C h a p t e r 15
I
Chemical Kinetics: The Rate s of Chemical Reactions
Act ive Figure 15 .9 Half-life of a first- order reaction. The curveis a plot of the pressure of CH3N 2CH 3 as a function of time. (The compound decomposes to CH 3CH 3 and N2 with k = 3.6 X 10- 4 S-l) . The pressure of CH .NlH 3 is halved every 1900 s (32 min). (This plot of pressureversus time is similar in shape to plots of concentration versus time for all other first-order reactions.)
9
I'0
C
7
~ ~ ~
5
~
6
E 0
N
~
'0
5
1 half-life, 1900 s P ~ '/1 (820 mm Hg)
C; I
E
4
Chemislry) ).Now'" Sign in at www.
-S ~
~
2 half-lives, 3800 s P ~ '/4 (820 mm Hg)
3
~
~
;;;
."
I."
• jlogin and go to the thomsonedu.com Chapter Contents menu to explore an interactive version of this figure accompanied by an exercise.
2
1 0 0
500
1000
1500
2000
2500
3000
3500
4000
4500
Time (5)
• , HaLf-Life and a First-Order Process Problem Sucrose, ( t, HnOl1 , decomposes to fructose and glucose in add solution wit h the rate law Rate = k[C 12H nOIl ]
k = 0.216 h- t at 25°C
(a) What is the half-life of C12 HnOll at this temperature? (b) What amount of time is required for 87.5% of the initial concentration of C12H n0 11 to decompose? Strategy (a) The decomposition of ( 12HnO tt is first order in this compound, so Equation 15.4 can be used to calculate the half-Life. (b) After 87.5% of t he C12H u 0 I1 has decomposed, 12.5% (or one-eighth ofthe sample) remains. To reach this point, three half-lives are required. Half-Life
Fraction Remaining 0.5
2
0.25 0.125
Therefore, we calculate the half-life from Equation 15.4 and then multiply by 3. SoLution (a) The half-life for t he reaction is t 'l, ~ 0 .693/k ~ 0.693/(0 .216 h- ') ~ 3.21 h
(b) Three half-lives must elapse before the fraction remaining is 0. 125, so Time elapsed
= 3 X 3.21 h =
9.63 h
•. Half-Life and First-Order Processes
r
ProbLem Radioactive radon-222 gas (WRn) from natural sources can see p into the basement of a home. The half-life of w Rn is 3.8 days. Assume that the radon gas is trapped in the basement and cannot escape. If a basement has 4.0 X 1013 atoms of m Rn per liter of air initially. how many atoms of w Rn per liter will remain after one month (30 days)?
15 .4
I
Concentrati on- Ti me Relationships: Integrated Rate Laws
691
St rat egy Using Eq uation 15.1. and knowing t he number of atoms at t he beginning «(Rlo), t he elapsed time (30 days), and the rate constant, we can caLculate the number of atoms remaining ([Rl!). First, the rate constant, k, must be found from t he half-life using Equation 15.4. SoLution The rate constant, k, is k
= 0.693 = 0.693 = 0.18 d- 1
ty,
3.8 d
Now use Eq uation 15.1 to calculate the number of atoms remaining after 30. days.
In
[Rn], 4.0 x 10n atom/L [Rn]t 4.0 X lOll atom/ L
~ - (0.18 d-')(30. d) ~ - 5.5 = e-S.5 = 0.0042
[Rn] t = 1.7
X
1011 atc m/L
Comment Thirty days is approximately 8 haLf-lives for this element. This means that the concentration at the end of the month is approximately (1/2)8 or 1/m t h of the original concentratio n.
EXERCISE 15.9 Half-Life and a First-Order Process Americium is used in smoke detectors and in medici ne for t he treatment of certain malignancies. One isotope of americium, w Am, has a rate constant, k, for radioactive decay of 0.0016 y-1, In contrast, radioactive iodine-125, which is used for studies of t hyroid functioning, has a rate constant for decay of 0.011 d- 1• (a) Whatare the half-lives of these isotopes? (b) Which isotope decays faster? (c) If you are given a dose of iodine-125 containing 1.6 X 1015 ato ms, how many atoms remain after 2.0 days?
15.5
A Microscopic View of Reaction Rates
Th ro ugh o u t this book, we have turned to the particu late level of c he m istry to u nd erstand c h e m ica l phe nom ena. Rate s of reacti o n are n o e xcep tio n. Loo king at th e way rea ct io ns oc cu r at the a to mi c an d mo lec u la r levels ca n provide some in sight into the vario us in flu e n ces o n rat es of react ions. Le t u s review t he macroscopic o bservation s wc h ave m ad e so far concern ing reactio n ra tes . We know that the re a re wide differences in rates of reactio n s- fro m ve ry fast reactio ns lik e th e e xp lo sio n th at o ccurs wh e n h ydrogen a nd o xygen a re e xposed to a spark o r flam e (Figu re ] . ]8 ) , to slow reactio ns like th e formation o f rust that occ ur over days, weeks, or yea rs. Fo r a specific reactio n , factors tha t in fluence reactio n rate in cl ud e th e concentratio n s of the reacta n ts, the te m perature of the reac tio n syste m , a n d t he p re se n ce of catalysts. Le t us lo o k a t eac h of these influences in more dept h .
Concentration, Reaction Rate, and Collision Theory Conside r the gas-p hase reaction of n itr ic oxide a nd ozo ne :
NO (g) + O,(g) -> NO,(g) + O,(g) The rate law fo r this p rodu ct-favo red reaction is first order Ra te = k[I\'O] [O :d. How can thi s reac tio n ha ve thi s rate law? 692
C h a p t e r 15
I
Chemica l Kinetics: The Rates of Chemical Reactions
III
each reactant:
Le t us consider the react ion at th e particula te level and im ag ine a flas k co n taining a mi xtu re o f 0J0 and 0 :\ mol ecul es in th e gas phase. Bo th kinds o f molecul es are in rapi d and ra ndo m mo tion withi n the flask. T hey strike th e walls of the vessel and co llide with o ther m olecul es. For this o r any othe r reactio n to occ u r, the co llision theory of reaction rates states th at th re e co nd itio ns mu st be me t: L T he reacting mol ecul es must co llide with o ne another.
2. The reacti ng mo lec ules must co llide with sufficien t energy to initi ate the process o f breaking and for m ing bonds. 3. T he mol ecul es must co llide in an o rien ta tio n th at ca n lead to rcar ran gement o f th e a toms and th e forma tion of prod uc ts. \ Ve shall d iscu ss each o f th ese co nditions within th e con text o f the effec ts of co n-
ce n trat ion and temperature o n reaction ra te . To reac t, mol ecul es m ust co llide with one anot he r. It is reason able to propo se thai the rare of th e ir reaction be re late d to the n u mb er of co llisions, whic h is in tu rn related to thei r co ncen tratio ns (Figure 15.10). Do ub ling the co ncentra tio n o f on e reagent in th e NO + 0 3 reaction , say NO , will lead to twice th e number of mol ecul ar co llisions. Figu re 15. IOa shows a sing le molecul e o f o ne o f the reactan ts (0J0 ) moving random ly amo ng sixteen 0 3 mo lec ules. In a give n tim e perio d , it mi ght co llid e wit h two O J mol ecules. T he number of N O - O j co llisio ns will double , however, if the co ncentration of NO mo lec u les is d ou bled (to 2, as shown in Figure 15.I Ob) or if tile n u m ber o f 0 3 m olecul es is dou bled (to 32, as in Figure 15.1Oc) . T h us, we ca n explain th e d ependen ce o f reaction rate o n co nce n tra tion : The n umbe r of co llisio ns bet ween the two reactant mo lec u les is direc tly pro portional to the co nce ntration o f each reactan t, a nd th e rate of th e reac tio n shows a first-ord er d epend en ce o n each react ant. ChemistrY-9-NDW~ Sign in at www.thomsonedu.comjlogin and go to Chapter 15 Contentsto see Screen 15.9 for a visualization of collision theory and for tutorials on using half-life.
Temperature, Reaction Rate, and Activation Energy In a lab orat ory or in the che mical indus try, a che mical reaction is often carried out at e levated temperat ure hecau se th is allows the react io n to occur mo re rapi d ly. Co nverse ly, it is so me times desirable to lower th e tcmperature to slow d own a che m ical reaction (to avoid an un controll abl e reac tio n o r a potentiall y da ng c ro us explosion) . Che m ists are very aware of the effe ct o f te m pe ra tu re o n th e rate o f a reac tio n .
(a) 1 NO: 16 0 3 - 2 hits/second
(b) 2 NO: 16 0 3 ~ 4 hits/second
(c) 1 NO: 32 0 3 ~ 4 hits/second 15.5
FIGURE 15.10 The effect of concentration on the frequency of molecular collisions. (a) Asingle NO molecule, moving amo ng sixteen 0] molecules. is shown colliding with two of them per second. (b) If two NO molecules move among 16 OJ molecules, we would predict that four NO-O ] collisions would occur per second. (c) If the number of OJ molecules is doubled (to 32), the frequency of No-O] collisions is also doubled, to four per second.
A Mi croscopic View of Reaction Rates
693
FIGURE 15.11 Energy distribution curve. The vertical axis gives the relative number of molecules possessing the energy indicated on the horizontal axis. The graph indicates the minimum energy required for an arbitrary reaction. At a higher temperature, a larger fraction of the molecules have sufficient energy to react. (Recall Fig ure 11.14, the Boltzmann distribution function, for a collection of gas molecules.)
lower te mperature
-, Higher temperature •
Minimum energy
\:7·~' . ~
..
~ . ~
Energy
A d iscu ssion of the effec t of tem perature o n reactio n ra te begin s with refe re nce to d istrib ut ion of e nergies for mo lec ules in a samp le of a gas or liq u id . Recall fro m stud ying gases and liqui d s tha t th e m olecul es in a sam ple have a wide range of ene rg ies, describe d ea rlie r as a Bo ltzmann d istr ibution o f e ne rgies (... Figu re 11.14 ) . In any sa mp le o f a gas or liquid, so me mo lec ules have very low energies; others have very high energies ; but most have so me int e rmediate e nergy. As the tempe ra ture in creases, th e average energy of th e m olecul es in th e sample inc reases , as does th e fractio n havin g h igh e r energies (Figu re 15.11 ) .
Activation Energy Molecules re q u ire so m e muu mum energy to react. Chemists visualize th is as an ene rbT)' barrier th at must be su r mo u n ted by th e reac ta n ts for a reaction to occ ur (Figu re 15.12) . T he energy re qu ired to su r mo u nt the barri er is ca lled the acti vat ion energy, Ea. If the barrier is low, t he e nergy re q uired is low, an d a high p ro portio n o f th e m olecul es in a sample may have sufficie n t energy to reac t. In suc h a case , th e react ion will be fast. If th e barr ier is h igh , th e activation energy is h igh , and o n ly a few reactan t m olecul es in a sam ple m ay have suffic ient energy. In this case, the reacti o n will be slow. To illustra te an activa tio n energy harrier, co nsider the co nversio n of N0 2 an d CO to NO and CO z o r the reve rse reaction (Fig u re 15.13 ) . At th e mo lec ular level , atom fro m an NO z we im agi ne th at the reac tio n involves t he tr an sfe r of an at o m mo lecu le to a CO mo lec u le (or, in the reverse re acti on , the tra nsfe r of an from CO 2 to NO) .
°
NO,(g) + CO(g)
FIGURE 15.12 An analogy to chemical activation energy. For t he volleyball to go over the net, the player mu st give it suffic ient energy. 694
C h a p t e r 15
°
NO(g) + CO,(g)
\fe ca n descri be this process by using an energy d iagram or reaction coordinate d iag ram . T he ho rizontal ax is describes the reactio n p rogress as th e reac tio n proceed s, and the vertical ax is represen ts the pot e n tial energy of the system during the reactio n. Whe» N0 2 and CO ap p roac h an d 0 atom transfer begins, an N-O bon d is bein g broken , and a C=O bo nd is forming. Energy inp u t (the activatio n ene rgy) is required fo r this to occur. The energy of the system reach es a maximum at th e transitio n state. At th e tra nsitio n state, sufficien t ene rgy has been co ncentrated in th e approp riate bo nds; bond s in the reacta nts ca n now br eak, and new bonds ca n form
Chemical Kinetics: The Rate s of Chemical Reacti ons
Reactants
Transition state
Products
Active Figure 15.13 Activation Energy. The reactionof N0 2 and CO (to give NO and CO 2) has an activation energybarrier of 132 kJjmol-rxn. The reverse reaction (NO + CO 2 ~ N0 2 + CO) requires 358 kJjmol-rxn. The net energy change for the reaction of N02 and CO is - 226 kJjmol-rxn.
--------r---Ea = 132 kJjmol-rxn
Reactants N02 + CO
e, =
ChemislrY;J.NDW~ Sign in at www. thomsonedu:com/login and go to the Chapter Content s menu to explore an interactive version of this figure accompanied by an exerdse.
358 kJj mol-rxn
ilH = - 226 kJj mol-rxn
Products NO + CO 2 Reaction progress
to give p ro d ucts. The system is poised to go on to products. Alternatively, it can re tu rn to th e reac tan ts. Because th c tran sition state is at a maximum in pote n tial eneq,ry, it ca n no t be isolated. Using computer molecul ar modeling tec h n iques, however, chemists ca n desc ribe wha t th e transitio n state m ust look like. In th e N0 2 + CO rcacti on , 132 kJ/nlOl is required to reac h th c tra nsiti on sta te; th at is, th e top of th e e nergy barrier. As th e reac tio n co nti n ues towa rd th e p rod ucts- as t he N-O bond is fin all y b ro ken an d a C=O bo nd fo rm s- th e re acti on evo lves e ncrgy, 358 kJ/m ol-rxn . T he nct e ne rgy cha nge inv ol ved in this exot hermic re act ion is - 226 kJ/m ol-rx n . !i U= + 132 kJ/ mol-rxn + (-358 kJ/ mol-rxn)
=-
226 kJ/mo l-rxn
What happ en s if NO and CO 2 are mi xed to form N0 2 and CO? Now the re action requi res 358 kJ/m ol-rxn to reach t he tr an sition state, and 132 J.s.J/ mo l-rx ll is evo lved o n pro ceeding to th e p rod uc t, N0 2 and CO. T he reaction in th is d irec tio n is endo thermic, re q uiring a n in put of + 226 kJ/m ol-rxn . ChemislrY~J.NDWN Sign in at www.thomsonedu .com/Ioqtn and go to Chapter 15 Contents to see Screen 15.10 for a simulation of reaction coordinate diagrams.
Effect of a Temperature Increase Th e conversion of N0 2 an d CO to p ro ducts at ro om tempe ra tu re is slow because o n ly a small fraction o f th e mo lecules have eno ug h ene rgy to reach th e transition sta te. T he rate ca n be inc reased by heatin g the sam p le. Raisin g th e tempe ra tu re increases the reac tio n rate by increasing the fraction of mol ecul es with e noug h ene rgy to surmo unt th e act ivation energy barrier (Fig u re 15.11 ) .
Effect of Molecular Orientation on Reaction Rate Havin g a sufficie n tly high ene rgy is ne cessary. but it is not suffici e n t to e nsure that reactan ts will for m p roduct s. T he reactan t mo lec u les must also come toge ther in the correct o rie n tation. For the reactio n o f N0 2 and CO, we can imag in e th at the transition sta te struc tu re has o ne o f the 0 ato ms of N0 2 beg inn ing to hind to th e 15.5
I
A Mi croscopic View of Reaction Rates
695
C ato m of CO in prepara tion for 0 atom transfer (Fig u re 15.13) . The lowe r the pro bability o f ac h ieving the p ro pe r a lignme n t, the smalle r the value of h, and th e slowe r the reactio n. Im agi ne what happens whe n two o r more com plicated mol ecules co llide. In on ly a small frac tion of t he co llisions will th e mol ecul es co me tog ether in exactly th e rig h t orientatio n . T hus, o n ly a tin y fraction of the collisio ns can be effe ctive . No wo nd er so me re actions are slo w. Converse ly, it is am azi ng th at so man y are fast!
Th e Arrhenius Equation T he o bse rvation t ha t reactio n rates d epe nd o n the en erbTY and frequency o f co llisio ns be twee n re acti ng molecu les, on the tem peratu re , and O il whet her the co llisio ns have th e co rrect g eo metry is su m m arized by the Arrhenius equation: k = rate consta nt = Ae- f,lRT
~action
Frequency factor /
• Interpreting the Arrheniu s Equation (a) Til e exponential ter m. This gives t he fraction of molecules having sufficient ene rgy for reaction and is a function of r. Value of e- £i/ Itr Temperature (K)
for E. = 40 kJjmol-rxn
298
9. 7 X 10 - 8
400
5.9 X 10- 6
600
3.3 X 10- 4
(b) Significance of A. Alt hough a complete understanding of A goes beyond t he level of t his text, it can be noted t hat A becomes smaller as t he reacta nts become larger. It reflects t he fact t hat reacti ng molecules must come toget her in t he app ropriate geometry.
(15.5)
of molecules with minimum energy for reaction
[ 11 t h is eq ua tio n , h is the rate co nsta n t, whe re R is the ga s constan t with a value o f 8.3 145 10 X 10 - 3 ~J /K · mol , and T is t he kelvi n te m peratu re . T he paramet er A is ca lled the frequen cy fa ctor, and it has u nits of L/m ol . s. It is re lated to the n umbe r of co llisio ns and to t he fraction of co llisio ns that have t he co rrec t ge ome try; A is spec ific to each reactio n and is tempe ratu re de pe ndent. T he facto r p-/'?tI R T re presen ts theJmction oj molecu les having the minimu m em!TgJ requiredJor reaction; its value is a lways less th an I. As th e tab le in th e margi n shows, th is fractio n cha ng es signi ficantly with te mperatu re . T he Arr he ni us equa tio n has signi fica nt uses.
• It can he used to calculate f~ from the temperatu re , d epen d ence of th e rare co ns tant. • It can be u sed to calcu late the rat e co nsta n t for a g ive n te m peratu re , if 1~1 and A a re kn own .
If rate co nsta nts of a given reaction arc measured a t severa l te mpera tu re s then o ne can apply grap hical techn iq ues to determine thc ac tivatio n c nergy of a reactio n . Taki ng th e na tural logarithm o f ea ch side of Eq uatio n I!l.5, we have in k = lnA +
(- :~ )
Rearrang ing th is expression slig ht ly shows that In h and l iT are rel ated linearl y. lnk
(1)
E, T = - /I
+ lnA
1, mx
1,
f-
Arrhenius equation (15.6)
1, y
+b
(- Equation for st raight line
7:
T his mean s that, if th e na tural logarithm of h (In h) is plo tted versus I I the re su lt is a d ownward-slo pi ng straig h t lin e with a slo pe of ( - f~1 R) . T he ac tivatio n e nergy, E;ll ca n be o btained fro m the slo pe of th is lin e .
ChemistrY-9- Now' Sign in at www.thom sonedu .com /Ioqtn and go to Chapter 15 Conte nts to see Screen 15.11 for a simulation and tutorials on the temperature dependence of rea ction rates and t he Arrheniu s equation .
696
C h a p t e r 15
I
Chemica l Kinetics : The Rates of Chemical Reacti ons
A Closer Look
Reaction Coordinate Diagrams
Reaction coordinate diagrams (Figure 15.13) convey a great deal of information. Areaction t hat would have an energy diagram like that
Energy of
intermediate
in Fi gure 15.13 is t he substitutio n of a halogen atom of CH 3 Cl by an ion such as F- . Here,
the F- ion attacks the molecule from t he side opposite the Cl substituent. As F- begins to form a bond to carbon. t he ( - Cl bond weak-
+
ens, and the CH3 portion of the molecule changes shape. As time progresses, the products CH3F and Cl - are formed .
,
H20
FIGURE A A reaction coordinate diagram for a two-step reaction , a process involving an intermediate.
+
[F"
Reaction progress
The diagram in Figure Adescribes a reaction that occurs in two steps. An example of such a reaction is the substitution of the - OH group on methanol by a halide ion in t he presence of acid. In the first step, an H+ ion attaches to the 0 of the ( - O-H group in a rapid, reversible reaction. The energy of this protonated species. CH30H2 " . a reaction intermediate, is higher than the energies of the reactants and is represented by the dip in the curve shown in Figure A. In the second step, a halideion, say sr', attacks the intermediate to produce methyl bromide, CH 3Br, and water. There is an activation energy barrier in both the first step and second step.
' CH, " ' ClJ -
+ Cl-
+
->
CH ,OH 2 +
Br-
+ H20 Notice in Figure A, as in Figu re 15.13. that the energy of t he products is lower tha n t he
energy of the reactants. The reaction is exothermic.
• Determination of E. from the Arrhenius Equation Problem Using the experimental data shown in the table, calculate the activatio n energy fa for the reaction 2 N,O(g) --> 2 N,(g) Experiment
Temperature (K) 1125
4
+ O,(g)
k (L/mol' s) 11.59
105 3
1.67
100 1
0.380
838
0. 00 11
15.5
I
AMi croscopic View of Reaction Rates
697
4,------. 2
St rategy To solvethis problem graphica lly, wefirst need to calculate In k and 1fTfor each data point. These data are th en plotted , and f a is ca lculated from the resu lting st raight li ne (stope = - EJ R). Solution First, calculate 1f T a nd In k.
o
l iT (K- ' )
Experiment
- 2
In k
8.889 x 10-
~
4
2.4501
.5 - 4
9.497 X 10 - 4
0.513
- 6
9.990 X 10- 4
- 0.968
11.9 x 10- 4
- 6.81
4
-8
Plotting these data gives the graph shown in Figu re 15.14. Choosing t he large blue points on the graph, the slope is found to be Slope
FIGURE 15.14 Arrhenius plot.
Aplot of ln k versus liT for the reaction 2 NzO(g) - t 2 N2(g) + 02(g). The slope of
.l In k .l (l /T)
=
2.0 - (-5 .6)
- 3.0 X 10 4 K
(9.0 - 11.5)(10 ' )K '
The activatio n energy is evaluated from the slope.
the line gives E•. See Example 15.10.
=
- fa
Slope
E,
- 3.0 X 10 4 K
8.3 1 X 10 l kJ/K· mol
R 250 kf/mc l-rxn
E,
T he ac tivation ene rbT)', l~. , for a reactio n ca n be o bta ined a lgebraica lly if k is known at two different tem peratures. \Ve can wri te an eq uati o n for each set of these co nd itio ns :
In kj
_(io-) + In A or RT
=
In k1
=
_(io-) + In A RT 1
j
If one of the se eq uation s is su b tracted from the ot he r, we ha ve
In k1
-
[1 1]
In kj = In -k1 = - -E, - - kj RT1 Tj
(15.7)
Exa m ple 15. 11 d e mon str a tes th e use of th is equatio n .
Calculati ng E, Numerically • E. , Reaction Rates, and Temperature
A often-used rule of thumb is that reaction rates double for every 10 O( rise in temperature in the vicinity of room temperature.
ProbLem Use values of k determined at two different temperatu res to calculate the value of f a for the decomposition of HI: 2 HI(g)
-7
H,(g)
+ I, (g)
k,
~
2.15 X 10-' L/ (mol · s) at 6.50 X 10' K(T,)
k2
=
2.39 X 10- 7 L/ (mol' s) at 7.00 X 10 2 K(T2 )
Strategy Use Equation 15.7. Solution In 2.39 X 10- ' lj(mo l · s) 2.15 X 10- ' lj(mol · s)
8.31 5 X
10~~ kJ/K . mol
X [ 7.00 : 10 2 K - 6.50: 102 K]
Solving this equation gives f a = 180 kl/ mc l-rxn . Comment Anot her way to write the difference in fractions in brackets is
This expression is often easier to use. 698
Chapter
15
I
Chemical Kinetics: The Rates of Chemica l Reacti ons
Calculating E, from the Temperature Dependence of k
EXERCISE 15 .10
The colorless gas N20 4 decomposes to the brown gas N0 2 in a first-order reaction. N,O,(g) --> 2 NO,(g) The rate const ant k tion energy. f a?
= 4.5
X 10 3 5 - 1 at 274 Ka nd k = 1.00 X 10 4 5 - 1 at 283 K. What is the act iva-
Effect of Catalysts on Reaction Rate Ca talysts a rc substa nc es th at sp eed u p th e rat e o f a che m ica l react ion . \ Vc have see n seve ra l ex am p les of ca ta lysts in ea rli e r di scussio n s in thi s chap te r: l\l nO ~ , io di de ion , an e nzym e in a po ta to, an a hydr o xid e ion all ca talyze th e decom position of hyd rogen peroxid e (Fig u re 15.4) . In biol ogi cal syste ms , ca talysts called e nzymes influence th e ra tes of mo st reac tio ns (page 702 ) . Catalysts are n o t consu med in a c he m ica l reactio n . T he y a re, h owever, in tim a tel y in vol ved in th e d et ail s o f th e reacti on a t th e particul at e level. T hei r func tion is to p rovid e a d iffe ren t pa th way with a lo we r activatio n energy fo r th e reactio n . To illustr at e how a ca talyst p a rti cip a tes ill a re action , le t us co ns id e r the isome ri zation of cis-2-bute ne, to th e slig h tly more sta ble isom er, trans-2-bu te n e.
H,\ I
H
/ H, \
I
H
cis-2-buten e
l
H,\
(g)
C= C
H Transition state
C=C
\
(g)
CH,
• Enzymes: Biological Catalysts Catalase is an enzyme whose function is to speed up the decomposition of hydrogen peroxide. This enzyme ensures that hydrogen peroxide. which is highly toxic. does not build up in the body.
• Catalysts and t he Economy "One third of [the] material gross national product in the U.S. involves a catalytic process somewhere in the production chain." (Quoted in A. Bell, Science. Vol. 299, page 1688, 2003.)
trans- 2-bute ne
oJ Endrotates
V
'If
bond breaks
T he activ ation e n e rgy for t he u n ca ta lyze d co nve rs io n is re la tive ly la rge26 4 kJ / m o l-rxn- becau se the 7T b on d must be b ro ke n to a llow on e e n d o f the m ol ecul e to rotat e in to a n ew positio n . Be cause o f t he h igh ac tivatio n e n e rgy, this is a slow reactio n , a nd ra th er h igh te m peratu res a re req uired for it to oc cur at a reason abl e rate . T he (is- to tm ns-2-bu te n e reactio n is g reatly acc ele rated by a cata lyst, iodine. In the prese nc e of io din e , this react ion ca n he ca rried o ut at a tempe ratu re seve ral hun dred degrees lower th an fo r th e un ca ta lyzed react ion . Io di n e is n ot co nsu m ed (n o r is it a product ), a n d it docs n ot a p pear in th e ove rall balan ced eq ua tio n . It d oes a p pe ar in th e re ac tio n rat e law, h oweve r; the ra te of the react io n de pe nds on th e sq u a re ro o t of th e io din e co nce n tra tio n : Rat e ~ k[cis-Z-butene ][ I, ],/'
The presen ce o f 12 c hanges t he \vay th e reactio n occu rs; th at is, it changes th e mechan ism o f t he re acti on (Figu re 15.15) . T he best hypoth esis is th at iod ine molecu les first d issociat e to fo rm io dine ato m s (Ste p I ) . An iodi n e a to m th e n ad ds to o n e of the C a to m s of th e C= C do ubl e b on d (Ste p 2) . Th is con ver ts th e d o ubl e bo nd be twee n th e ca rbo n atoms to a sing le bond (the 7T bon d is b rok e n ) a nd allows th e e nds o f th e mo lecule to twist freely rela tive to each o th er (Ste p 3). If th e iodin e ato m th en d issoci ate s fr o m th e inte rm edi at e , th e d o u ble bo n d ca n re-fo rm 15.5
I
• Butene Isomerization Isomerization of os-z-butene is a first order process with the rate law"Rate = k[cis-2-butene]." It is suggested to occur by rotation around the carbon-carbon double bond.The rate at which a molecule will isomerize is related to the fractio n of molecules that have a high enough energy. (See ChemistryNow Screen 15.8 to view an animation of the interconversicn of butene isomers and the energy barrier for the process.)
A Microscopic View of Reaction Rates
699
t......
FIGURE 15.15 The mechanism of t he iodine-catalyzed i somerization of d s-2-butene. Cis-2-butene is converte d to trans-2-b utene in the presence of a
Step 5
catalytic amount of iodine. Catalyzed reactions are often pictured in such dia-
rj
Step 1
grams to emphasize what chemists refer to as a "catalytic cycle."
in th e trans con figu ra tio n (Step 4) . T he iodine ato m ca ta lyzing th e rotation is now fre e to add 10 ano ther molecu le o f d s-2-bu lene, The res u lt is a kind of chain reaction, as o ne molecule of ci,'i-2-b ute ne after an o ther is co nver ted to the trans iso mer. T he cha in is br oken if th e iodine a to m recombines with another iodine a to m to re-fo rm m o lecul ar iod ine, An ene rb'1' profil e fo r th e ca ta lyzed reactio n (Figu re 15.16 ) shows th at the overall energy barr ier is mu ch lower th an for th e un cat alyzcd reaction . Five se para te steps are ide n tified fo r th e mech anism in the e ne rgy profile . This proposed mechanism also in cludes a series of che m ica l spec ies called reaction in ter m ed iates , species fo rmed in one ste p o f th e rea ction and co nsu m ed in a lat er step. Iodin e ato ms are int ermed iates, as are the free rad ical species fo rm ed whe n an iodine a to m ad ds to ci,'i-2-bu tene.
FIGURE 15.16 Energy profile for t he iodi ne-catalyzed reaction of d 5-2-
300 '- - - - - - - - - - - - - - - - - - - - - - - - - "
butene. A catalyst accelerates a reaction by altering the mechanis m so that the
Uncata lyzed rea ction
activatio n e nergy is lowered. With a smaller barrier to overcome, more reacting molecules have sufficient energy to surmount the barrier, and t he reaction occurs more rapidly. The energy profile for the uncatalyzed conve rsion of cis-cbutene to trans-2-butene is shown by t he black curve, and th at for the iodine catalyzed reaction is represented by the red curve. Notice t hat the shape of t he barrier has changed because the mechanism has changed.
700
Chapter 15
200
Ea
';5 E
2 BrNO(g)
A single-ste p rea ctio n wo u ld require th at th re e reactant mol ecules co llid e simu ltancously inj ust the rig ht. o rie ntatio n . T he p robabili ty o f thi s o ccu rring is small; thus, it would be reasonabl e to loo k fo r a m ech anism that o ccu rs in a series of steps, with each step invo lving o n ly o nc or two mol ecul es. In o ne possib le mech anism , Br2 and N O might co mbine in a n ini tia l ste p to produce an in termed iat e species.
• Rate Laws and Mechanisms Rate taws are derived by expe riment ; t hey are macroscopic observatio ns. Mechanisms are schemes we propose t hat speculate on how reactions occur at t he particulate level. 15 .6
I
Reaction Mechanism s
701
Case Study
Enzymes: Nature's Catalysts
Withi n any living organism, there are untold numbers of chemical reactions occurring, many of them extremely rapidly. In many cases, enzymes, natu ral cata lysts, speed up reactions that would normally move at a
in the blood as hemoglobin loses O2, The resulting HC0 3- ions are transported backto the lungs. When hemog lobin again takes on O2, it releases H30 + ions. These ions and HC0 3- re-form H2C03, from which CO 2 is liberated and exhaled. You can do an experiment that illustrates t he effect of carbonic anhydrase. Fi rst, add a small amount of NaOHto a cold, aqueous solution of CO 2, The solution becomes basicimmediate ly because there is not enough H2C0 3 in the solution to use up t he NaOH . After some seconds, howeve r, dissolved CO2 slowly produces more H2C03, which consumes Na OH, and t he solution is again acidic. Now t ry the experiment again, this time adding a few drops of blood to the soluti on (Fi gure A). Carbonic anhydrase in blood speeds up reactions 1 and 2 by a factor of about 107 , as evidenced by the more ra pid reaction under these conditions. In 1913, Leonor Micha elis and Ma ud l. Me nten proposed a general t heory of enzyme action based on kinetic observatio ns. They assumed t hat t he substrate,S (t he reactant), and the enzyme, E, form a complex, ES. This complex th en breaks down, releasing t he enzyme and t he product, P.
snail's pace from reactants to products. Typically, enzymes give reaction rates 10 7 to
10 14 times faster than t he uncatalyzed rate. Enzymes a re typically large proteins, ofte n containing meta l ions such as Zn2+ . They are thought to function by bringing t he reactants together in just the right orie ntation in a site
where specific bonds can be broken and/or made. Carbonic anhyd rase is one of many enzymes important in biological processes (Figure A). Carbon dioxide dissolves in water to a small extent to produce carbonic acid, which ionizes to give H30 + and HC0 3- ions. CO,(g) P
"'!
I (a)
1 =0
(b)
r e a sec
(e)
t
e
rs sec
(d)
t = 17see
(e)
r e zr sec
FIGURE A CO 2 in water. (a) Afew drops of blood are added to a cold solution of CO 2 in water. (b) Afew dropsofa dye (bromthymol blue) are added to the solution, the yellow cotor indicating an acidic solution. (c and d) Aless-than-stoichiometric amount ofsodium hydroxide is added, converting the HlO to HC0 3 (andCOJ2- ). The blue color ofthe dye indicates a basic solution. (e) The blue color begins to fade after some seconds as CO 2 forms more H2CO The amount of H2CO formed is finally sufficient to consume the added NaO H, andthe solution is again acidic. Blood is a source of the enzyme carbonic anhydrase, so the last stepsare noticeably more rapid than the reaction in the absence of blood. j
j •
702
Chapter
15
I
Rate
Rate max
- - - - -; -:.:-.-...
"-------~
[5]
FIGURE B Rate of enzyme catalyzed reaction. This plot of substrateconcentration [5] versus reaction velocity is typical of reactions catalyzed byenzymes and following the Michaelis-Menten model.
When the subst rate concentra tion is low, the rate of t he reaction is first order in 5 (Figure B). As [5] increases, however, the active sites in t he enzyme become saturated wit h subst rate, and t he rate reaches its maximum value. Now the kinetics are zero order in subst rate.
Questions: 1. Catalase can decompose hydrogen peroxide to O2 and water about 107 timesfaster than the uncatalyzed reaction. If the latter requires one year, howmuch time is required by the enzyme-catalyzed reaction ? 2. According to the Michaelis-Menten model, if l/R ate is plotted versus 1/ [5}, the intercept a/the plot (when 1/(5] ~ 0) is l/ Ratemax• Find Ratemax for a reaction involving carbonic anhydrase. [5], mol{L
Rate {mtltimoles/rntn}
2.500
0.588
1.00
0.500
0.714
0.417
0.526
0.370
0.250
0.256
j
Chemica l Kinetics: The Rates of Chemica l Reactions
Answers to these questions are in Appendix
Q.
+
Step 1
Br,
FIGURE 15.17 A reaction mecha"i sm. A representation of the proposed
~ ~
NO
Br.2NO
two-step mechanis m by which NO and BT2
are converted to NOBr.
+
Step 2
NO
+
~
Br2NO
BrNO
BrNO
Br2NO (Fig ure 15. 17) . This in ter mediat e would the n react with an o ther NO mo lecu le to give the reac tio n p roduct s. Th e eq uatio n for t he overa ll reaction is obtained by ad d ing th e e q uatio ns for th e se two ste ps : Step 1Step 2. Overall Reaction:
Br, (g) + NO (g) Br, NO (g) Br, NO(g) + NO(g) --> 2 BrNO(g) Br, (g) + 2 NO(g) --> 2 BrNO(g)
Each ste p in a mul tistep react ion seq ue nce is an el ementary step , d efi ned by a ch e m ica l e q uatio n th a t d e scrib e s a sing le m o le cu la r even t suc h as th e for ma tio n
o r ru p tu re of a che mical bo nd resulting from a mo lec u lar co llisio n . Eac h ste p has its 0\\' 1l ac tivatio n e nergy, f :", and ra te co ns ta n t, h. Add ing th e eq ua tio ns fo r each step must g ive th e balanced eq uation for the overall reac tio n, and the time required to com ple te a ll o f the ste ps defin es th e ove rall re actio n rate. T he se ries o f ste ps co nstitu tes a possible reac tio n mechan ism . Mech anism s o f reactions are usually po stula te d sta rting with ex perimen ta l d at a . To see how thi s is do ne , we fir st d escribe th ree types of eleme n tary ste ps in term s of th e conce p t of moleculaiity.
Molecularity of Elementary Steps Eleme n ta ry ste ps are classified by th e nu m ber o f re acta n t mol ecul es (o r ions, a to ms, o r fr ee radicals) that co me togeth e r. T h is who le, po sitive number is ca lled the molecularity of the e lemen ta ry ste p. When o ne m olecul e is the o n ly rea ct ant in an e leme n ta ry ste p, the reac tio n is a unimolecul ar p ro cess. A bimolecul ar eleme n tary pro cess invo lves t\'I'O mo lec u les, whic h may be identical (A + A -? prod uct s) o r d iffere n t (A + B -? products) . T he mechan ism pro po sed fo r th e d eco mpositio n of ozo ne in th e stratosp he re illustrate s th e use of th ese term s. Step 1Step 2. Overall Reaction:
Unimolecular Bi molecular
O, (g) --> O, (g) + O(g) O,(g ) + O(g) --> 2 O, (g) 2 O, (g) --> 3 O,( g)
A te rm olecular e le me ntary ste p invol ves three mo lecules, wh ich co u ld be the same o r di ffere nt (3 A -e p ro d uc ts: 2 A + B -? pro du ct s; o r A + B + C -? prod ucts). Be aware , however, the simu ltaneous co llisio n of th ree mo lecu les has a low p ro bability, u n less o ne of the m olecul es in vol ved is in high co nce n tra tio n , suc h as a so lven t molecul e. In fac t, most tc rmo lcc ular pro cesses invo lve th e co llisio n o f two reac ta n t mo lec u les and a th ir d , inert mo lec u le. T he fu nction of the ine rt molecu le is to abso rb t he ex cess energy prod uced when a new che mica l bo nd is fo rm ed by th e first two mo lecu les. For exam ple , N2 is un ch an ged in a termol ecul ar reaction be tween o xygen mo lecu les and o xyge n a toms th at produces ozo ne in th e u p per atmo sp here : O(g)
+ O,( g) + N,(g ) --> O,( g) + ene rgetic N, (g) 15 .6
I
Reaction Mechanisms
703
T he p ro bability that four o r more mo lecules will sim ultaneo usly co llide with suffici en t kine tic enerbT)' and proper o r ien ta tio n to reac t is so small th at reactio n mol ecul a rities gr eater th an three are never pr oposed .
Rate Equations for Elementary Steps Th e expe rime n ta lly d eterm ined ra te equation for a reactio n can no t be predicte d fro m its ove ra ll sto ichiom et ry. In co n trast, th e rat e eq uatio n for any eleme n tary ste p is d efin ed by th e reac tio n sto ichio me try. The rate eq uat io n of an e leme n tary step is g ive n by th e p ro duc t o f th e rat e co ns ta n t a nd the concen tr at ions o f the reac ta n ts in th at ste p. We ca n therefore write th e rat e eq uatio n for a ny el eme n tary step , as shown by ex a mp les in th e follo win g tabl e : Elementary Step
Molecularity
Rate Equation
A ~ product
unirnolecutar
Rate = k[A]
A + B -7 product
bimolecular
Rate = k[AIIB]
A + A ~ product
bimolecular
Rate = k[A F
2 A + B ~ product
termolecular
Rate
= k[A]'[B]
For e xa m ple , th e rate laws fo r each o f th e two ste ps in th e d eco mp o sition o f o zone are Ra te for (unimolecular) Step 1
~
k[O, ]
Rate for (bimolecular) Step 2 = k'[O,][ O]
When a reaction mech anism co ns ists o f two ele menta ry ste ps, th e two steps will likel y occu r at different rat es. T he two ra te co nstan ts (k and k' in th is exa m ple ) are no t ex pec ted to have th e same value (no r the same un its, if the two ste ps have different rnol ccul aritics) .
Molecularity and Reaction Order A un imolecul ar el em en tar y step must be first o rder; a bimo lecu lar elemen tar y ste p mu st be seco nd order; and a term olecular eleme n tary step must be third order. Such a d irect rel ati on between molccul arity and o rde r is emphatically not tru e for a multiste p reaction . If yo u learn fro m an ex pe ri me n t th at a reaction is fi rst o rde r, yo u can no t co nclud e th at it occurs in a single , u nimolecul ar eleme n tary step. Sim ilarly, a seco nd-o rd er rat e eq ua tio n does not impl y th at th e reactio n oc cu rs in a sing le , bim olecul ar elem entary ste p. An illust ration of thi s is the de composition of N2 0 5 : 2 N,O,( g) -> 4 NO,( g) + O,(g)
Here , th e rat e law is "Ra te = k[N 20 ii ], " but che m ists arc fairly ce rtain th e mechan ism invo lves a se ries o f u nimol ecul ar and bimolecul ar steps . To see how th e expe rim en ta lly o bserved rate equati on fo r th e overall reacti on is co n nected with a po ssible mech anism or seq uenc e o f eleme ntary ste ps req u ire s so me chem ica l intui tion . \Ve will p rovide o n ly a g lim pse o f the su bject in tb e ne xt sectio n. N20S Decomposition. The first step in the decomposition of N20 S is thought to be t he cleavage of one of the N-O bonds in the N-O- N link to give the odd-electron molecules N0 2 and NO]. These react further to give the final products. 704
Chapt er
15
I
ChemislryJ :>"Now'"
•
Sign in at www.thomsonedu.comjlogin and go to Chapter 15 Contents to see: Screen 15.12 for exercises on reaction mechanisms • Screen 15.13 for exerciseson reaction mechanisms
Chemical Kinetics: The Rates of Chemical Reacti ons
Elementary Steps Problem The hypochlorite ion undergoes self-oxidation-reduction to give chlorate, ([° 3 - , and chloride ions.
3 ClO-(aq) --> ClO ,-(aq)
+ 2 Cl-(aq)
This reaction is thought to occur in two steps: Step 1:
CIO -(aq) + CIO -(aq) --> CIO,-(aq) + Cl-(aq)
Step 2:
CIO,-(aq) + CIO -(aq) --> CIO ,-(aq) + Cl-(aq)
What is the molecularity of each step? Write the rate equation for each reaction step. Show that the sum of these reactions gives the equation for the net reaction.
Strategy The molecularity is the number of ions or molecules involved in a reaction step. The rate equation involves the concentration of each ion or molecule in an elementary step, raised to the power of its stoichiometric coefficient.
SoLution Because two ions are involved in each elementarystep. each step is bimolecular. The rate equation for any elementarystep involves the product of t he concentrations of the reactants. Thus, in this case, the rate equations are Step 1:
Rate ~ k[CIO -]'
Step 2:
Rate ~ k'[CIO,-j[CIO-)
From the equations for the two elementarysteps, we see that the (l0 2- ion is an intermediate, a product of the first step and a reactant in the second step. It therefore cancels out. and we are left with the stetchiometric equation for the overall reaction: Step 1:
CIO- (aq) + CIO -(aq) --> CIO,-(aq) + Cl-(aq)
Step 2:
CIO, - (aq) + ClO-(aq) --> ClO ,-(aq) + Cl-(aq)
Sum of ste ps:
3 ClO-(aq ) --> CIO,- (aq)
+ 2 Cl-(aq)
EXERCISE 15.11 Elementary Steps Nitrogen mo noxide is reduced by hydrogen to give nitrogen and water: 2 NO(g) + 2 H,(g) --> N,(g) + 2 H,O(g)
One possible mechanism for this reaction is 2 NO(g) --> N,O,(g) N,O,(g)
+ H,(g) --> N,O(g) + H,O(g)
N,O(g) + H,(g) --> N,(g) + H,O(g)
What is the molecularity of each of the t hree steps? What is t he rate equation for t he third step? Identify the intermediates in this reaction; how manydifferent intermediates are there? Show that the sum of these elementary steps gives the equation for the overall reaction.
Reaction Mechanisms and Rate Equations Th e d ependen ce of ra te on co nce n tra tio n is an ex peri me n ta l fact. Mechan isms, by co n tras t, are co nstruc ts of our imagin ation , in tu ition , and goo d "che m ica l sens e." To de scribe a mech anism , we need to mak e a guess (a go od g uess, we hope ) ab o ut how th e react ion o ccu rs at the part icu lat e level. Several mechan ism s can always be proposed that correspo nd to the obse rved ra te equa tion, and a po stu lated m ech an ism is o ften \\-To ng. A go od mecha nism is a wor thy goal beca use it allows u s to understand th e chem istr y bett e r. A p ract ica l co nseq ue nce o f a good m ech an ism is tha t it allows us to predict, for exa mp le, how to co n tro l a re actio n better and ho w to d esign ne w experim en ts. One o f th e im po rtant guide lines o f kin etics is that products (if a reaction mn nruer be produced at a ra te Jas ter than till' rate oj the slouwst stejl. If one ste p in a m ultistep reac tio n is slowe r than the othe rs, the n the rate oj the overall reaction is limited If)' the combined rates oj all elementary StP/IS utI thmugh the slowest stefl in the mechanism. O ften 15 .6
I
Reaction Mechanisms
705
the overall reac tio n ra te a nd th e rate of th e slow ste p are ne ar ly the same. If the slow ste p d et e rm ines th e rate of th e reaction , it is ca lled th e r ate -de te r m ining step, o r ra te-limi ting ste p. Imagi ne that a reac tio n takes place with a m ech an ism invo lving two seq ue n tia l steps , and assu me th at we know th e rat es of both ste ps . T he first ste p is slow an d the second is fast: • (an You Derive a Mechanism? At this introductory level, you cannot be expected to derive reaction mechanisms. Given a mechanism, however, you can decide whether it agrees with experimenta l rate laws.
Elementary Step 1
A +B
Elementary Step 2
M+A
Overall Reaction
2A + B
k,
Slow, E. large k, Fast, E. small
) X+ M , Y
' X+ Y
III the first step, A and B co me toget her and slowly react to for m o ne o f the p ro ducts (X) p lus ano the r react ive spec ies, !\L Almos t as soo n as !\.I is fo rmed, however, it is rapidly co nsu me d by reacting with a no ther mol ecu le o f A to fo rm th e seco nd p roduct \: T he rat e-deterrn in iug ele me n tary ste p in this exam ple is the first ste p . Th at is, the rat e of th e first ste p is eq ua l to the rat e o f the ove ra ll react io n . T hi s ste p is bimolecu lar and so has the rat e eq uati o n
where kl is the ra te co ns ta nt for th at ste p. T he overall reac tio n is ex pec ted to have this same seco nd-o rde r rate eq uation . Le t us ap ply t hese ideas to the mech anism o f a rea l react io n . Co nside r the reactio n of ni tr ogen dioxid e with fluo rin e whic h has a seco nd-order rat e eq uatio n :
Overall Reaction
2 NO,( g)
Rate
+ F, (g) -> 2 FNO, (g)
= k[NO ,][ F, J
Th e rat e eq ua tio n immediately ru les o u t the possibility th at th e reactio n occu rs in a sing le ste p . If the eq uatio n fo r the reac tio n represented an eleme n ta ry ste p , the rate law would have a sec o nd-o rd er de pend e nce o n [N 0 2 ] . Because a sing le-step re ac tio n is ru led o u t, the mech anism must include at least two ste ps . \Ve can also co nclude from the rat e law Ihat the ra te-d e ter m ining e leme n tary step m ust in vol ve N0 2 a nd F2 in a 1:I rat io . O ne po ssible mecha nism propo ses th at mo lecul es o f i'\ 0 2 an d F2 first react to produce on e mo lec u le of th e prod uct (FN0 2) plus o ne F ato m . In a sec o nd ste p, the fluo rine ato m pro d uced in the first step reacts with ad d itio na l N0 2 to g ive a seco nd m olecul e of p rod uct. If the first, bimo lec ular step is ra re d ete rminin g, th e ra te equati o n wou ld he "Rat e = kl [N0 2] [ F2 ] , " the same as the ex perime n ta lly o bserved rat e eq ua tio n . T he ex perimen ta l rate co ns ta n t wo u ld be th e same as hI'
15
I
S(g)
~
NO ,(g)
Elementary Step 2
Fast
NO ,(g) +
OveraLL Reaction Chapter
~
Stow
jt 706
+ F, (g)
Elementary Step 1
Chem ical Kinetics: The Rates of Chemical Reactio ns
+ •
2 NO ,(g) + F, (g)
k,
~
~
FNO, (g) +
FNO,(g)
r
2 FNO ,(g)
S(g)
T he fluo ri n e a to m formed in th e first ste p o f the N0 2 / F2 react ion is a reac tio n in termediate . It d oes not ap pear in the eq ua tio n descri bin g t he overall reactio n. Reactio n int ermedi ates usu ally have o nly a flee ting ex iste nce, but o ccasio nally th ey have long en o ug h life times to be o bserved . T Ile d et ectio n and identificatio n o f an int ermed iate are stro ng evid e nc e for the p roposed mechan ism . Elementary Steps and Reaction Mechanisms Problem Oxygen atom transfer from N0 2 to CO produces nitrogen monoxide and carbon dioxide (Figu re 15.13) :
NO,(g) + (O(g)
-?
NO(g) + (O,(g)
The rate equation for this reaction at temperatures less than 500 K is:
Rate
~
k[NO,] '
(an this reaction occur in one bimolecular step? Strategy Write the rate law based on the equation for the N0 2 + CO reaction occurring as if it were an elementary step. If this rate taw corresponds to the observed rate law, then a one-step mechanism is possible. Solution If the reaction occurs by the collision of one N0 2 molecu le with one CO molecule, t he rate equation would be
Ra te
~
k[NO,][CO]
This does not agree with experiment, so the mechanism must involve more than a single step. In one possible mechanism, the reaction occurs in two, bimolecu lar steps, the first one slow and the second one fast:
Elementary Step 1
Slow, rate -
- .No,(g) + NO(g)
2 NO,(g)
determining
Elementary Step 2
Fast
.No,(g) + (O(g)
-
+ (O, (g)
.t. +•• - ... +...
NO, (g) + (O(g) -
Overall Reaction
NO,(g)
NO (g) + (O, (g)
The first (rate-determining) step has a rate equation that agrees with experiment, so this is a possible mechanism . EXERCISE 15 .1 2
Elementary Steps and Reaction Mechanisms
The Raschig reaction produces hydrazine. N ~H 4' an industrially important reducing agent, from NH 3 and oet- in basic, aqueous solution. A proposed mechanism is
Step 1 Fast
NH,(aq) + OCl-(aq) -? NH ,Cl(aq) + OH-(aq)
Step 2 Slow
NH ,Cl(aq) + NH ,(aq) -? N,H,(aq) + Cl -(aq)
Step 3 Fast
N,H,(aq) + OH-(aq) -? N,H.(aq) + H,O(f)
(a) What is the overallstoichiometric equation? (b) Which step of the three is rate determining? (c) Write the rate equation for the rate-determining elementary step. (d) What reaction intermediates are involved?
15 .6
I
Reaction Mechanisms
707
A co m mo n two-step reaction mech anism in volves an initial fast reaction th at produces an inte rmediate, followed by a slower seco nd step in wh ich the intermediate is co nve rted to the fin al product. The rat e o f th e re action is d etermin ed by th e second ste p, for whi ch a rat e law ca n be wr itte n. The ra te of th at ste p, ho wever, depends o n th e co ncentration o f the int ermediate . Remembe r, t hough , th at th e ra te law must be writt en with re spect to the react ants on ly. An int e rm ediat e , who se co nce n tra tio n will probabl y not be measu rabl e , ca n no t ap pear as a term in th e overall rate equation. The reacti o n of nitrogen monoxide and oxyge n is an exam ple of a two-step reac tio n where th e first ste p is fast and the seco nd step is rat e d et ermining. 2 NO( g) + O,(g ) --> 2 NO,(g ) Rat e
~
k[NOJ'[O,l
The expe rime n ta lly d et erm in ed rat e law shows sec o nd -o rd er dependen ce on NO and first-orde r dependence on O 2, Althoug h this ra tc law wou ld be co rrect for a termolecu lar reaction , ex perime n tal evide nce in d ica tes th at an int ermediat e is form ed ill this reaction . A possib le two-step mechanism that proce eds through all int e rmed iate is Elementary Step 1:
NO(g) + O,(g)
Fast, Equilibrium
k,
k- 1
OO NO (g) intermediate
Elementary Step 2: Slow, Rate-determining
k, NO(g) + OONO(g) ---> 2 NO,(g)
Overall Reactio n
2 NO(g) + O, (g) ---> 2 NO, (g)
Th e seco nd ste p of thi s reaction is th e slow ste p, and the overa ll ra te depends o n it. \Ve ca n write a rat e law fo r th e second step: Rate = k,[NOJ(OONOl
Th is rate law can no t he co m pared directly with th e ex pe rime n tal rate law becau se it contains th e co nc en tra tio n of an in te rmed iate . OO NO . To eliminat e th e intermediate front thi s rat e exp ressio n, we look at th e rapid first ste p in thi s reac tio n seq ue nc e th at invo lves an equilibriu m bet ween the in te rmed iate spe cies and the reactants. At th e beginn ing of th e reaction, NO and O 2 reac t rapidly and produce th e int ermediate OO NO . T he rate of formation ca n be defined by a rate law with a ra te co ns tan t k1: Rate of production of OONO = k,[NOl [O,l
Because the inte rm ediate is co nsu me d o nly very slowly in th e sec o nd ste p, it is po ssibl e for th e OONO to revert to NO an d O 2 before it re acts further: Rate of reverse reaction (OONO---> NO + 0,)
~
L , [OONOl
As NO an d O 2 form OONO, their co nce n tra tio ns drop, so th e rat e of th e for ward rea ction d ecreases. At the same tim e, th e co nce n trati o n o f OO NO builds up , so th e ra te of the reverse react ion in creases. At equ ilib riu m . th e ra tes of th e for ward and reverse re actions become th e same . Rat e of forward reaction
k, [NOJ(O,l
= ~
rate of reverse reaction
L , [OONOl
Rearran gin g thi s eq uatio n , we find -""- =
L, 708
Chapter
15
I
Chemical Kinetics: The Rates of Chemical Reactio ns
[N,O, l = K [NOJ(O, ]
Problem Solving Tip 15.1
Relating Rate Equations and Reaction Mechanisms
The connection between an experimental rate equation and the proposed reaction mechanism is important in chemistry.
and molecular structure and bonding, general chemical experience, and intuition. 3. The proposed reaction mechanism is used to derive a rate equation. This rate equation can contain only those species pres-
1. Expe riments mu st first be performed to
determine the experimental rate equation. 2. A mechanism for the reaction is proposed
ent in the overall chemical reaction. If the derived and experimental rate equations
on the basis of the experimental rate
are the same, the postulated mechanism may be a reasonable hypothesis of the reaction sequence. 4. If more than one mechanism can be proposed, and they all predict derived rate equations in agreemen t with experiment. then more experiments must be done.
equation, the principles of stoichiometry
Both k] a nd k_1 a re co ns ta n ts (they will ch a nge o n ly if th e temperature changes). \Ve ca n d efin e a n e w co nstan t K equa l 10 the ratio of th ese two co ns ta n ts a nd ca lle d t he equ ilibrium constant, whic h is equ a l to th e quotient [O O N'O ] / [N O ] [0 2 ] , From th is, we ca n d e rive a n ex p ress io n for the co n ce nt ra tio n o f OO NO : [OONO l
= K[NO][O,l
If A.1 N O ] [0 2 ] is su bs tit uted fo r [O ONO] in th e ra te la w for th e rate-d et ermining e leme ntary ste p , we h ave Rat e = k,[NO][OONO] = k,[NO](K[NOJ(O,ll ~
k,K[NOl' [O, ]
Because bo th k2 a nd K a re co n sta n ts, the ir product is a not he r co n sta n t k' , a n d we have Rate
• Equilibrium Const ant The important concept of chemical equilibrium was introduced in Chapte r 3 and will be described in more detail in Cha pters 16-19.
= k' [NO]' [O,]
• Mechanisms with an Initi al Equilibrium In this mechanism, the forward and reverse reactions in the first elementary step are so much faster than the second elementary step t hat equilibrium is established before any significant amount of OONO is consumed by NO to give N0 2 • The state of equilibrium for the first step remains throughout the lifetime of the overall reaction.
This is exac tly th e ra te law derive d fro m e xperime n t. T hus, th e se q ucn ce of reaction s o n whic h th e rat e law is based may be a re ason a bl e m ech an ism fo r th is reaction . It is n o t th e o n ly po ssibl e m ech an ism , howeve r. T h is rat e eq uat io n is also co ns iste n t with th e reac tio n occurrin g in a sing le tcrrn ol ccu la r ste p. Ano the r possib le me ch anism is illu strated in Exa m p le 15.14.
. Reaction Mechanism Involving an Equilibrium Step Problem The NO
+ O2 reaction described in t he t ext could also occur by t he followin g mechanism:
Element ary Step 1: Fa st. equilibrium k,
NO(g)
+ NO(g )
N,O,(g) L 1 intermediate
Element ary Step 2: Slow, rate-determining
N,O, (g)
+ O,(g )
k, --> 2 NO , (g)
Overall Reaction: 2 NO(g) + O,(g) --> 2 NO,(g) Show that this mechanism leads t o the followi ng experimental rate law: Rate
= k[N0j2(02]'
St rat egy The rate law for the rate-determining elementary step is
Rate
~
k, [N,O,][O,]
Theintermediate N20 2 cannot appear in t he final derived rate law. To obtain the rate law, we use t he equilibrium constant expression for t he first step. 15 .6
I
Rea ction Mechanisms
709
SoLution [N 20d and [NO] are related by the equilibrium constant . ~ ~ [N,O,] ~ K k_, [NOJ'
Solving this equation for [N20 2} gives [N20 2 ]
= K[NOF. When this is substituted into the
Rate
~
derived rate law
k,[ K[NO,1' }[O,)
the resulting equation is identical wit h the experimental rate law where k2K
= k.
Comment Three mechanisms have been proposed for the NO + O2 reaction. The challenge for chemists is t o decide which is correct. In t his case, further experimentation detected t he species OONO as a shortlived int ermediate, confirming the mechanism involving this intermediate. EXERCISE 15 .13
Reaction Mechanism Involving a Fast Initial Step
One possible mechanism for the decomposition of nitryl chloride, N0 2Cl, is k,
NO,(g) + (L(g)
Elementary Step 1: Fast. Equilibrium
NO,(l(g)
Elementary Step 2: Slow
k, NO ,( L(g) + Cl(g) ---> NO, (g) + Cl,(g)
L,
What is the overall reaction? What rate law would be derived from this mechanism? What effect does increasing the concentration of the product N0 2 have on the reaction rate?
Chapter Goals Revisited
Chemistry.c\NoW™Sign in at www.
thomsonedu•.comjlogin to: Assess your understanding withStudy Questions in OWLkeyed to each goal in the Goals and Homework menu for this chapter For quick review, download Go Chemistry mini-lecture flashcard modules (or purchase them at www.ichapters.com) Check your readiness for an exam by taking the Pre-Test and exploring the modules recommended in your Personalized Study plan.
@ Access How Do I Solve It? tutorials on how to approach problem solving using concepts in this chapter.
Now that you have studied this chapter; you should ash whether JOlt hm se met the chapter goals. In particular; you should be able (0 :
Understand rates of reaction and the conditions affecting rates a. Explain the co nce p t of react io n ra te (Sec tio n 15.] ) . b. Derive the avcrage and instantane ous rates ofa reaction fro m con cen tratio ntime da ta (Sec tio n 15 .1 ) , Study auestion(s) assignabLein OWL: 5. c. Describe factors th at affec t re actio n rat e (i.e ., react an t co nce nt ra tions, ternperatu re . prcscnce o f a ca ta lyst, an d t he sta te o f th e re ac ta n ts) (Sec tio n ]5.2 ) . Study auestion(s) assignable in OWL: 8, 10, 62, 76, 77, 81.
Derive the rate equation, rate constant, and reaction order from experimental data a. Defi ne th e vario us part s of a rat e eq ua tio n (the ra te co nstan t an d o rder of rea c tion }, a nd un d ersta nd th e ir sign ificance (Sec tion ] 5 .3 ) . StudyOuestion(s) assignabLe in OWL: 12, 14, 68. b.
De rive a rat e eq ua tion from ex perime n ta l in fo rmatio n (Sec tio n 15 .3 ) . Study auestion(s) assignable in OWL: 12, 14, 48, 56, 58, 71.
Use integrated rate laws a. Describe and use the re latio nships be tween reactan t co nce n tra tion and time for zero-ord er, first-o rd er, and secon d-order reaction s (Section 15 .4 and Tab le 15 .1 ) . Study Ouestion(s) assignabLe i n OWL: 16, 18, 19, 20, 24; GoChemi stry ModuLe 20.
b.
Ap ply gra p hica l met hods fo r d ete rm ining reactio n orde r and the rate co ns ta n t from ex perime n ta l d ata (Sectio n 15 .4 a nd Table 15 . 1) . Study Ouestion(s) assignable in OWL: 28, 30.
c.
Use the co nce p t o f h al f-life (t l},) , es pecially fo r fir st-orde r r eact ion s (Section 15 .4 ) . Study Question(s) assignable in OWL: 22. 24. 26, 60, 71.
710
Chapt er
15
I
Chemical Kinetics : The Ra tes of Chemical Reactions
Understand the collision theory of reaction rat es and the role of activation energy a. De scribe th e co llisio n theory of re acti on rates (Sec tio n 1:)5 ) . b. Relate ac tivatio n ene rgy U:a) to the ra te and th ermodynamics of a reac tion c.
d. e.
f.
(Sec tio n 15. 5 ) . Study Question(s) assignabtein OWL: 36. 69. 83. Usc collision th eory to de scribe th e effec t o f re actant conce n tr ation o n re ac-
tio n rate (Sec tio n l.' .5 ) . Understa nd the effect of mo lecu lar o rien tation 0 11 re action ra te (Sec tion 15.5) . De scri be th e effect o f temperatu re 011 reaction rate u sing th e co llisio n theory o f reac tio n ra tes and the Arr he ni us eq ua tio n (Equa tio n 15.7 and Section 155 ) . Use Eq ua tio ns 15.5 , 15.6, and 15.7 to calcu late the ac tivatio n en ergy from rate co ns tan ts at di ffere n t te m pe ra tu re s (Sec tio n 15.5 ) .
Relate reaction mechan isms and rate laws a. Describ e th e fun ction in g o fa catalyst an d its effect o n the ac tivati o n energy and mec ha nis m of a reactio n (Sec tio n 15.5). b. Understand reaction coo rd inate d iag ram s (Sec tio n 1:'").5) . c. Understand th e co nce p t o f a reaction mechani sm (a proposed seq ue nce of bo nd -mak ing and bo nd-breakin g ste ps that occurs durin g th e conversio n o f react ant'; to p roduct s) a nd thc rel ati on of th e mechanism to th e overall . sto ichi o me tric eq ua tio n fo r a reactio n (Sec tio n 15.G) . d. Describe th e eleme ntary steps of a mechan ism , a nd give th eir mo lecularity (Sec tio n 15.6) . Study Ouestion(s) assignable in OWL: 40, 42, 44, 74. e. Defin e th e rare-de term in ing ste p in a mech an ism , and identify any re action intermed iat es (Sec tio n 15.6) . Study Ouestion(s) assignable in OWL: 44. 69, 70, 78. 80.
KEY EQUATIONS Equation 15.1 (page 683) In teg rated rat e eq uatio n fo r a first-order reactio n (in whi ch - t. [Rl/t.t= k[ R]) .
lJRJ, =
- kt
[ RJo Here , [R] o and [R] I are co nce n tra tio ns of the reactant at tim e t = 0 and at a later time, t. T he rati o of co nce nt ratio ns . [R] j[ R]o. is the fr act ion of reac ta n t tha t re ma ins afte r a g iven time has e lapse d. Equation 15.2 (page 686) In tegra ted rate eq uatio n fo r a second-orde r reac tio n (in which - t. [Rl/t.t= h[R ]') . 1
1
- - = kt [R], [R]o
Equation 15 .3 (page 687) In tegrated rat e eq uat io n for a ze ro-o rde r reac tio n (in whi ch - t. [Rl/t.t= k[ R]O).
[R]o - [R], ~ kt Equation 15.4 (page 690) T he re lation be tween the half-life co nstan t ( k) fo r a first-order re act ion .
(t ~)
and th e rate
0.6 93
ty, - - k-
I Key Equations 711
Equation 15.5 (page 696) k
Ar rhe ni us equatio n in expo nential form
= rate constant = Ae- f / frequency f actor
RT
/~fraction.
motecules with wt mu .nmum . 0 f molecues energy for reaction
A is the fre q uen cy fac tor; f;" is the activa tio n ene rgy; T is th e temperat ure (in ke lvins) ; a nd R is the gas constan t ( = 8.3 145 10 X 10- " I;J/K · mo l) .
Equation 15.6 (page 696)
In k =
Expanded Arrhenius equat ion in logari thm ic fo rm . -
E,(1) Ii T + In A
! y
~
Arrhenius equation
! +b
mx
"'-
-
- Eq uation for straight line
Equation 15 .7 (page 698) A versio n of the Arrhen ius equation used to calcula te the acti vation en ergy for a re acti on when yo u know th e values of th e rat e constan t at two temperatures (in ke lvins ) :
In k2
STUDY QUESTIONS Onli ne homework for this chapter may be assigned in OWL. .. denotes challenging questions. •
denotes questions assignable in OWL.
Blue-numbered questions have answe rs in Appendix a and fully-worked solutions in the Student Solutions Manual.
Practidng Skills
-
In k,
~
k2 In ~ k,
[1 1]
E, R
- - -
~ -~
T2 T,
4. In the synt hesi s of am m onia, if - 6,[112 ]/6,1 = 4.5 X 10- -1 mo l/I.·mi n , what is ~[ NH :I]/~l ? N, (g)
+
5. • Experi mental data are listed here for the re act ion A
~
2R.
Tim e
[B]
(s)
(molj L)
0.00
0.000
10.0
0.326
10.0
0.571
Reaction Rates (8('(' Ex(u n!J/I'J / 5.1- /5.2, Exerd srs / 5./-/5.2, awl ClmlliJl I)Now
30.0
0.750
Sm'e11 / 5.2. )
40.0
0.890
l. Give the re lative rates o f di sappearan ce of reactants
and form atio n of products fo r eac h of the fo llowin g re action s. (a) 2 O , (g ) -----? :l O , (g ) (b) 2 1I0 F (g ) -----? 2 II F (g ) + O , (g )
2. Give the re lative rates o f di sappearan ce of reactants and fo rm ation of products for ea ch of th e foll owing re action s. (a) 2 NO (g) + IIr, (g ) -----? 2 NOllr (g ) (b ) N, (g) + 3 H, (g ) -----? 2 NH,,(g)
3. In th e react ion 20:I (g ) -----7 3 0 2(g ). th e ra te of form ati on o f O 2 is 1.5 X 10- 3 m o l/ I. . s. Wh at is th e rate of d eco m pos itio n of 0 3? 712
3 H , (g ) --> 2 N H,(g)
(a) Prepare a gra p h fr om th ese d a ta ; co n nec t th e points wit h a sm o o th lin e ; and calcul ate the rat e o f ch a nge of [R] fo r e ac h l{)..s interval from 0.0 to 40 .0 s. Do es the ra te of cha nge decrease from one tim e in te rval to th e IH.'Xt? Suggest a reaso n for th is re su lt. (h) How is the ra te of change of [A ] rel a te d to the ra te of chan ge of [R] in e ac h time interval ? Calcu lat e the ra il' of change of [A ] for the tim e in te rval fro m lO .O to 20, 0 s. (c) 'Vh at is the in stantaneous ra te , ~ [R ]/6,t , when [ II] ~ 0.750 mol/L?
6 . P h enyl ac e ta te, an es te r, re acts with wa te r ac co rdi ng to til l' e q ua tio n
°II
CII 3CO C,;lI,
+
H 20
ph eny l acetate
acetic acid
phenol
T h e d ata in the table were co ll ecte d fo r th is reaction a t 5 -c . Time
[Phe nyl acetate]
(s)
(molfL)
t
NO(g)
+
Ilr, (g ) -->
t
NO llr(g)
Experi men ts sh ow th a t th is reactio n is sec o n d o rder in N O a nd first orde r in Br2' (a) Wri te th e ra te equa tio n for th e reacti on . (b) H ow d ocs the in it ial re actio n ra te ch a n ge if th e rou ccnu- a tion o f Rr2 is changed fro m 0 .0022 m o l/ I. to O.006 fi m o l/I.? (c) Wh at is th e cha nge in th e initi al ra te if the concen tra tio n of NO is changed from 0 .00 24 mol/I. to 0. 00 I t mol/I..?
0
0.55
15.0
0.42
30.0
0.31
45.0
0.23
60.0
0.17
75.0
0. 12
0.010
0.010
2.5 X 10 - '
90. 0
0.085
0.020
0.010
1.0 X 10 - 4
0.010
0.020
5.0 X 10 - '
11. Th e d at a in the tahle a re fo r the re act io n o f Nf) and O 2 at 660 K.
t
Concentration and Rate Equations (Sp(' Ex a mples / 5.] - /5.4, Exenises / 5.3-/5.4, and Chnni,\'t,}Nm.ll Screens 15.4 an d 15.5 .) 7. Usin g th e rat e eq u atio n "Ra te = k[ A ]2[ H], " defin e th e o rder of th e re actio n with resp ect to A an d B. wh at is th e to tal o rder o f th e re act ion? 8• • A reactio n h as th e ex perimen ta l ra te equa tion Rate = k{Ap. lI ow will th e rat e c ha nge if the conce n tra tio n o f A is trip le d ? If th e concen tra tio n of A is h alve d ? 9 . T h e reaction bcu»ccn o zo ne a n d nitro ge n dioxi de at 23 1 K is fir st order in b oth {N0 :l] a n d (O J].
t
NO ,(g)
+
O A g ) --> N, O,(s)
+
O , (g )
(a) Wri te th e ra te eq u a tion fo r th e re actio n . (b) If the con cen tration o f N0 2 is tripl ed (a n d [O J] is n o t c h angc rl), wh at is th e c h a ng e in th e re action ra te? (c) \ Vhat is th e effec t on rea ct ion rate if the co nce n tration o f 0 3 is h alve d ( n o ch a nge in [N 0 2])?
more challenging
NO(g )
+
O , (g ) --> t NO ,(g )
Reactant Concentration (moLJ L) [NO] [0 ,]
(a ) Plo t th e ph e n yl ac e ta te concentration ve rsus time , a n d d escrib e th e sha pe of th e cu rve observe d . (h) Calcu la te th e ra te of ch a ng e o f th e ph e n yl ace ta te co uccru ra tio n d u rin g th e p e rio d 15 .0 s to :-lO.O s a n d a lso d uri n g th e period 75.0 s to 90 .0 s. Wh y is one valu e sm aller than th e o th e r? (c) Wha t is th e rate of cha nge o f t h e p heno l co nce ntr atio n d u rin g th e time pe riod 60 .0 s to 75 .0 s? (d ) W ha t is th e instan tan eous rate a t 1.").0 s?
•
10. • Nitrosy l bromide , NO Br, is fo rmed fr o m KO an d Br2:
• i n OWL Blue-numbered questions answered in Appendix 0
Rate of Disappearance of NO
(molfL . s)
(a) De te rm in e th e ord er of th e re actio n for eac h reac ta n t. (h ) Write th e ra te eq u a tio n for th e re act ion . (c) Ca lcu la te th e ra te co ns ta n t. (d ) Calcu la te th e ra te ( in m ol y l . . s) a t tln - in stant wh en [NO ] = 0.0 15 rnol y l . an d [ 0 , ) = 0 .005 0 m o l/I.. (e ) A t th e in sta n t wh en N O is re actin g a t th e rat e 1.0 X 10 - 1 m o l/I. . s, wh a t is th e ra te a t wh ich O 2 is re acting an d N 0 2 is form in g? 12. • T h e react ion
t
NO(g)
+t
1I, (g ) --> N,(g)
+t
Il,o (g )
was studie d a t 904 °C , an d th e d ata in th e tabl e we re col lec ted . Reactant Concentration (mo LJL) [NO) [H,)
Rate of Appearance of N2
(mol/ L . s)
0.420
0. 122
0.136
0.210
0.122
0.0339
0.210
0.244
0.0678
0.105
0.4 88
0.0339
(a ) Determ in e th e order of th e rea ction for each reac ta nt. (b) Write th e ra te eq uatio n for th e react ion . (c) Ca lcu la te th e ra te cons ta n t fo r th e re act io n . (d ) Fin d the ra te of ap pearann' o f N 2 at th e instan t whe n [N O ] ~ 0 .350 mof/L an d [II ,] ~ 0$)5 mo l/I..
I
713
13. Da ta fo r the re action 2 N O (g ) arc g iven in th e tabl e.
0 2(g )
Concentrati on (moljl) [NO] [0,]
Experiment
---?
2 N0 2(g )
Initial Rate
(mol/ L · h)
3.6 X 10- 4
5.2 X 10 - 3
3.4 X 10- 8
3.6 X 10- 4
1.04 X 10- 2
6.8 X 10- 8
4
2
1.7 X 10- 8
1.8 X 10-
1.04 X 10-
1.8 X 10- 4
4
+
5.2 X 10 - 3
19. • H yd ro gen peroxid e , H 2 0 Aaq ) , d eco mpo se s to H 20 (O an d 0 2(g ) in a reacti on that is first o rd e r in H 20 2 an d has a ra te co nsta n t k = 1.0 6 X 10- 3 min - I a t a g iven te m pe ra tu re . (a ) How lo ng will it take for 15 % o f a sa m p le o f H 2 0 :.! to d ec o m pos e? (b) How lo ng will it tak e fo r g5 % o f t he sa m ple to decom pose? 20, • T h e de composition of n itrog e n di oxi de a t a h igh tem pe ra tu re
(a) Wh at is the rate law fo r th is reacti o n ? (h ) What is t he rate constan t for th e rea cti on ?
;-': O , (g ) --> NO(g)
(c) Wh at is th e ini tial ra te o f th e reac tion in cx pcrimont 4?
Experiment
4
+ NO , (g)
--> C O ,(g)
Concentration (moljl) [CO] [N0 2l
+ ;-': O(g) (molfL . h)
5.0 X 10- 4
0.36 X 10- 4
3.4 X 10- 8
5.0 X 10- 4
0.18 X 10- 4
1.7 X 10- 8
1.0 X 10- 3
0.36 X 10 - 4
6.8 X 10- 8
1.5 X 10- 3
0.72 X 10 - 4
(a) What is th e rate \;:1\'/ for thi s reacti on ? (h ) What is th e rate constan t fo r t he reac t io n? (c) What is t he ini tia l ra te of th e re action in ex periment 4?
(See Exam!Jlf's /5.8 and /5 .9, Exercise /5.9, am! Cllfmist ryNo1lJ Screen 15.8.) 21 . The ra te equa tion fo r rlu- decom positio n of N20 ~, (g iving N0 2 a nd ( 2) is Rat e = k[N 20 :;J, T he value of k is 6.7 X 10 - :' s- I fo r th e reac tio n a t a pa rti cu la r te mperatu re . (a) Calc ula te th e half-life o f N 20 :;. (b) How long does it tak e fo r th e N 20 5 concent rat io n to d ro p to on e tenth of its origi na l value? 22 • •
+
fruc-
H,O (f ) -> 2 C, H I2 0 6 (a q )
is - d [su cro se]/d t = k[C I2H 220I d. Afte r 27 m in a t 27 °C, the suc rose conce n tra tio n d ecr eased from 0.01 46 M to 0 .0 132 M. Find the rate co ns ta n t, k. 16. • T he d e co mpo sition o fN 20 :-' in CC14 is a fir st-o rder reaction . If 2.56 mg o f N 20 " is presen t in itia lly, a nd 2.50 mg is pre sent afte r 4. 26 m in at 55 °C, wha t is th e val ue of th e ra te co ns ta n t, h? 17. T he d ec ompositio n of S02Cl2 is a fi rst-order reactio n : SO,CI,(g) -> SO, (g )
+ Cl,(g)
T he ra te co ns ta n t fo r the re action is 2.8 X 10 - 3 mi n - I at 600 K. If th e ini tial co nce ntratio n of S02CI2 is 1.24 X 10 - 3 mot/ L, how lo ng will it ta ke for th e co nce n tra tio n 10 d rop to 0.31 X 10 - 3 mol/I.? 18. • Th e co nversio n of cyclo pro pane to p ropene (see Exam ple 15.5) occu rs wit h a first-o rde r ra te co nsta n t of 2.42 X 10 - 2 h - I. H ow long will it take fo r th e co nce n t ra tion of cyclo p ro pa ne to decrease fro m a n in itia l co nce n tra tio n 0.080 m o l/ I. to 0.0 20 m ol/ I.? 714
I
+ Cl,(g)
is first orde r in S0 2Cb , an d the reac tio n ha s a hal f-life o f 245 m in a t 600 K. If you heg in with :t 6 X 10- 3 mol o f S0 2C12 in a 1.0-1, flask, how lo ng \v·ill it la ke fo r th e am o un t of S02CI2 to d e cr e ase to 2.00 X 10 - 4 mo l?
Screen 15.6,) 10
T he d e co mpo sition of SO:.! CI2 SO,C1,(g) -> SO, (g)
Concent rati on- Time Relationships (See Exa mples 15.5- /5. 7, EXa n Jf S 15.5- /5 .7, ami Chfmistf)Noll'
C 12H" OIl( aq )
O , (g)
Half-Life
Initial Rate
15. Th e ra te equation for th e h yd ro lysis of sucrose to se an d g luc ose
!.;;
is secon d order in thi s reacta nt. T h e ra te co nsunu fo r th is reactio n is 3.4 0 Llmol . mi n . De termin e the time neede d fo r th e concentrati on o f !':0 2 to d e cr e ase fro m 2.00 m o f/ L to 1.50 mo l/ I..
14. • Data fo r th e followin g react ion are g iven in th e tabl e below. CO(g)
+
23. C ase ous az om ctha nc , C H 3N =NC H 3 , deco mposes in a fi rst-o rder reac tio n whe n h eat ed : C H,N=NC H,,(g) -> N, (g)
+ C, Hfi(g)
T he ra te co ns tant fo r thi s reacti on a t 60 0 K is 0 ,02 16 miu "' . If the in itial q uant ity o f azo rru-thaue ill th e flas k is 2.00 g , how much re main s after 0.0 500 m in? What q ua n tity of N 2 is fo rmed in this time ? 24 . • T he compound Xe(CF 3h d e co mp o se s in a fi rst-orde r reactio n to e le men tal Xc with a hal f-life of 30. m in . If yo u p lace 7.5 0 m g o f Xe( CF 3 h in a flask , how lo ng must yo u wa it until o n ly 0 ,25 mg of Xe(CF3 h rem ains? 25 . T he ra d ioactive iso tope li·J CU is used in the fo r m o f co p pcrf l l ) ace ta te to study Wilson's d isease . T he isotope has a ha lf-life of 12.70 h . Wh at fr action of ra d ioactive co p pcrf l l ) ace ta te re ma in s a fter 64 h? 26. • Rad io act ive go ld- 19g is u sed in th e di ag nosis o f liver pro blem s. T he h al f-life o f th is iso to pe is 2.7 days . If you beg in with a 5.6- mg sam ple of th e iso tope , h ow m uc h of t hi s sam ple re mai ns after 1.0 d ay? •
more challenging • in OWL
Blue-numbered questionsanswered in Appendix 0
Graphical Analysis: Rate Eq uations and k (SI'(' Exercis e 15.8 find ChemiJ(,}Nuw Screen 15.7.) 27. Dat a for th e d ecomposition o f d in itroge n o xid e
+
2 N , O (g) -> 2 N ,(g)
32. Bu tadiene , C.jII h(g ) , di mcrizcs when h ea te d , formin g
O , (g)
a gold surface at 900 °C are given bel o w. Veri ty th at th e reac tio n is fir st o rder by p repa ri n g a grap h o f In [N 2 0 j versus tim e . De rive th e ra te co ns ta n t fr o m th e slo pe o f th e line in th is g ra p h. Usi ng th e ra te lav.... a nd valu e of 11, d ete rmine the d eco mposit ion rat e a t 900 °C whe n [ N 20 J = 4 N O, (g)
2 N O, (g) -> 2 N O (g)
+ O, (g)
Th e concen tration o f NO:! was measured as a fun ct ion o f time. A gra p h of I / [ N 0 2 ] vers us time gives a straig h t line with a slo pe o f 1.1 L/m ol . s. Wh at is the ra te law fo r th is reac tio n ? wh at is the rate co nstan t? 30. • T he decom position of IIOF occ urs a t 25 2 HOF (g) -> 2 H F(g)
+
0(:.
O, (g)
Using th e data in the ta ble bel ow, de termine th e ra te law, a nd th c u calculate the ra te constan t, [HOF] (molfL)
o
•
2.00
0.754
5.00
0.5 26
20.0
0.24 3
50.0
morechallenging
• in OWL
+ O, (g)
fro m t he o bserved ra te con sta n ts: Ie a t 25 °C = 3.46 X 10 - 5 S- I ami Ie at 55 °C = 1.5 X 10 - 3 S -
I.
34 . If the rat e co n stan t fo r a re actio n triples \4..' he n th e tern perature rises from 3.00 X 102 K to :t 10 X 102 K, what is th e ac tiva tion ('uergy of th e reactio n?
29. Gaseo us 1'\0 2 d e co mposes a t 573 K.
0.81 0
II
Kineti cs and Energy
Pl ot In [N II :d versus time an d 1 / [ NII ~d versus time . Wh at is th e order o f th is reac t ion with res pect to N I-I :~? Fin d the rate cons ta n t for the reactio n fro m the slo pe .
0.850
CH
15 -cyc to o c lad ic n c
28. • Ammonia d e co mposes whe n he at ed according to th e eq ua tion
o
HC
/ 2 11 2C - C1I
8.7 X 10- 3
N H ,,(g) -> N H , (g)
CII
II
"
1,3-hutadic nc
[N,O) (molfL)
15.0
Time (h)
~
II C
Blue-numbered questions answered in Appendix 0
35 . Whe n heat ed to a h igh te mp eratu re, cyclo b u tane , C 4 II x, d e comp o se s to eth yle ne:
Th e act ivatio n ene rgy, f~:1> for this react io n is 26 0 ~ l /mo l- rxn. A t 800 K, th e rat e constant 11 = 0.0:'\15 s- I. Dete rm in e ti ll' va lue of 11 a t H50 K. 36 • • Wh e n heated , cyclopro pane is co n vert ed to p rope n e (se e Exam ple 15.5 ) . Rat e co ns tan ts for thi s reaction at 47 0 °C and 51 0 °C are 11 = 1. 10 X ]0 - 4 s- I aud Ie = I,(J2 X 10 - 3 S- I, res pective ly. Dete rmi n e the activa tion e nergy, f~" from th e se data.
I
715
37. Th e react ion o f lit molecules with F a to ms 1I, (g )
42 . • Th e reactio n of N O:z(g) an d CO (g) is th ought to o ccu r in two ste ps:
+ F (g ) ---> II F(g ) + II (g)
h as an ac tiva tio n e ne rgy of 8 kJ / mnl-rxn an d a n e ne q:.,IY chan ge of - 133 kJ / mo l-rxn. Draw a d iag ra m simi la r to Figu re 1 5 . 1 :~ for th is p roce ss. In d ica te th e a ct ivatio n e ne rgy and e n thalpy ch ange o n thi s di agram . 38. Answe r th e foll owing questions based o n th e d iagram below. (a) Is the rea ct io n exo th erm ic or e n d o ther mic? (b) Do es the re a ct ion oc cu r ill mo re th an o ne ste p? If so , how m any?
Step I
Slow
NO,(g)
+
NO,(g) ---> NO(g)
Step 2
Fast
N0 3(g)
+
CO (g ) ---> NO ,(g)
+
NO ,,(g ) C O , (g )
(a) Show th a t th e e le me n ta ry ste ps ad d up to g ive th e overall , sto ich io me tr ic e q uatio n. (b) Wh a t is the m ol ccu la ri ry o f eac h ste p? (c) Fo r th is mech ani sm 10 he consiste n t wit h kin e tic d at a , wha t m ust be th e ex pe rimen ta l rate eq u a tion ? (d) Iden tify a ny in u-rm edia rcs in th is re actio n . 43. A p ro posed m e ch an ism fo r th e reactio n of NOt an d CO is Step 1 Slo w, e nd o therm ic 2 NO, (g) ---> J"O(g)
Products Reactants
+
N0 3(g)
Step 2 Fast, exo therm ic N0 3 (g ) + CO (g ) ---> J"O, (g)
+
CO , (g)
Ove rall Re action Exo therm ic NO ,(g) + C O(g) ---> NO(g)
+
CO,(g)
(a) Id entify e ach o f th e fo llowin g as a reac ta n t, p ro duc t, or in te rm ed iate: N O t(g) , CO(g) , N O :\(g-), CO,(g) . N O(g). (h) D raw a reaction coo rd inat e diagram fo r th is reactio n . In di cat e on th is d rawin g th e ac tivatio n e nergy for eac h ste p a nd th e overal l e n tha lpy ch an ge .
Reaction progress Reaction Mechanisms (See Examples 15.12- 15 . /4 , Exercises 15.1 1-15. 13, and Chl'flli.\byNow Screens 15. 12 awl 15 . J3.)
39 . Wh at is th e ra te law for each o f the fo llowing e le mentary reaction s? (a) NO(g) + N0 3(g) ---> 2 NO,(g) (h) C I(g ) + H, (g ) ---> HCI (g ) + H (g ) (c) (C H,),Cll r (a q) ---> (C H 3 ),C'(aq) + llr- (aq )
+
44 . • Til e mec han ism fo r the reac tion o f e lI 30 H a nd lI Br is beli eved to invol ve tw o ste ps . Th e overa ll reactio n is exoth e rmic. Step 1 Fast. e nd o ther m ic C H 30 H + II " Step 2 Slow CII :l>l l t +
40 . • Wh at is the ra te law tin eac h of th e foll owing e lcm e ntar y reacti on s? (a) CI(g) + ICI (g ) ---> I(g ) + Cl,(g) (b) O (g ) + O ,,(g ) ---> 2 O , (g ) (c) 2 J"O,(g) ---> N, O .,(g )
+
C H 3 0 H, +
Br" ~ ClI j Rr
+
I I..!O
(a) Wr ite an eq uatio n fo r the overal l reacti on. (h) D raw a re acti on co ord ina te d iag-ram fo r thi s rea ctio n. (c) Sh ow th at th e ra te law fo r th is reaction is Rate = k[C H,D Il][ I I +][ llr - ].
41. O zo ne , 0 3 , in the earth 's u p pe r a tm os p here de compo ses accordi ng to th e eq ua tio n 2 0 3(g ) ---> 3 O , (g ) Th e me chani sm o f th e re acti on is th o ugh t lo pro ceed th rough a n init ia l fast , rever sibl e ste p followed by a slow, secon d ste p. St ep 1
Fa st , reversible
0 3(g )
St ep 2
Slow
0 3(g )
o :z( g )
+
+
716
I
These questions are not designated
(is
to fJIJI' or location in the this find other
dlajJfn: Thl)' I/Ul)' combine seueml concepts from
rhnpters.
O (g )
O (g ) ---> 2 O , (g )
(a) Wh ic h of till' ste ps is ra te-d et ermin ing? (h) Wri te the rat e eq uation for th e ra te d et ermi ni n g
su-p.
General Questions
45 . A reac tio n h as the foll owin g experim en tal ra te e quatio n: Rat e = k[A]:l[ B] , If the conce n tratio n o f A is doubled a nd th e co nce n tra tio n of B is ha lve d , what hap pen s to the rea ct ion ra te ? 46 . For a fi rst-orde r re a ctio n , what fractio n of re ac tant m ai n s aft er five h al f-live s h ave el apsed ?
•
more challenging • in OWL
1"1>
Blue-numbered questionsanswered in Appendix0
47 . To de term ine the co n ce n tr a tio n d epe nde nce o f th e r ate of th e re a ction H,I'O ,,- (aq )
+
O l-l r f aq ) ---> HI'O ,,' - (aq )
+
H , O (( )
yo u mi g ht m e asu r e [O H -] as a fun ction o f ti m e u sin g a pl I me te r. (To do so , yo u would se t u p co n di tio n s u n d e r whi ch [I-{ 2P O :~ -] rem ain s co ns ta n t by usi ng a larg e e xcess o f thi s rcac tant .) 1I0w would yo u prove a se co nd-order rat e depende nce fo r [O ll - ]?
48. • Da ta for the fo llowing reaction are given in th e tah ie . 2 NO (g ) Experiment
+
5 1. W hen h e ate d , tet rafl u oroeth vle n e dimerizes to form octafluorocvclo h u tan e.
Rr,(g ) ---> 2 NO Rr (g )
2 C,F , (g) ---> C,F , (g) To de termine th e ra te of th is reac tion a t 488 K, th e data in th e ta ble were co lle cted. Analysis wa s do ne g r a p h ically, as sh own below:
Time (s)
[C,F,1 (M) 0. 100
0
0.080
56
0.060
150.
[NO]
[Br2]
Initial Rate
0.040
335
(M)
(M)
(mol/L . s)
0. 030
520.
1.0 X 10- 2
2.0 X 10 - 2
10- 2
10 - 2
4.0 X
1.0 X 10- 2
2.0 X
2.4 X 10- 2 0.384
5.0 X 10- 2
6.0 X 10- 2
W h at is the o rder o f th e re action wit h respect to [1\0 1 and [Br2], and wh at is t he overall order of the reaction?
(a ) Wh at is th e- ra te law for th is re action ? (b) Wh at is th e va lu e of th e ra te co n stant? (c) Wh at is th e concent ratio n of C .lF4 after 600 s? (d ) H ow lo n g will it ta ke u n til t he reaction is 90 % co m p le te?
49 . Formic acid decompose s a t 550 °C acco rd in g to th e equatio n
II CO ,II (g) --->
co, (g)
+ 1I, (g )
T h e re ac tio n foll ows fi rst-o rd e r kine tics. In an experim e nt, it is d etermined th at 75 % of a sam p le of HC0 2H h as d ecomposed in 7'2 sec onds. Dete r m in e I"" for thi s reaction.
50 40
~ E
2:
30
~: 20
~
---~
10
50. Iso merization of C H 3 NC occurs slowly wh en CH 31\:C is lu-a u-d .
0
CII "NC (g ) ---> CII "CN (g) To stu dy th e rate of th is reactio n at 4SS K, d a ta on [C H 3 NC] were co llec te d at variou s tim e s. Analysis le d 10 th e g ra p h be low, (a) \ Vhal is th e ra te law for thi s reactio n ? (b ) Wha t is th e eq uat io n for th e s traight lin e in th is g ra p h? (c) Calc u la te th e rate consta n t fo r thi s reaction . (d) H ow lo n g docs it take fo r h al f o f the sa m p le to isome rize ? (e ) \ Vh a t is th e co n ce n tr a ti o n of C H:lNC after 1.0 X 10'1 s?
0
100
300
CO (g) + NO, (g ) ---> CO , (g ) + NO (g) (a) De r ive th e ra te equatio n . (h) De te r m in e th e r eactio n o r d e r with respect to each reactan t. (c) Calcu late th e r ate constan t, giving th e correct un its for h. [CO] 5.1 X 10-
- 5.0
4
z
:£' - 6,0 ~
..5
0 ' --
4000
0.70
4
x 10 - ' 10 - 4
3.4 X 10- 8 6.8
x 10 - 8
1.7 X 10 - 8
5.1 X 10- '
0. 18 X
1.0 X 10- 3
0.35 X 10- 4
6.8
10- 3
0.35 X 10- '
10.2 X 10- 8
1.5 X
- 7.0
0.35 X 10 -
Initial Rate
(molfL . h)
[NO,]
5.1 X 10 - '
tr
700
52. Data in the tab le were co llected at 540 K for th e following react ion :
Initial Concentration (moIj L)
- 4.0
500
Time, seconds
x 10- 8
-8000
12,000
Time, seconds •
more challenging
• i n OWL
Blue-numbered questions answered in Appendix 0
I
717
53. Am mo ni u m cya na te , N lI ..NCO, rea rra nges in water to give u rea, (N lfzhCO .
56 , • A Nitryl flu o rid e can be made b y tre at ing n it ro gen di o xid e with fluo rin e:
N H,NCO(aq) -> (N H, ),CO(aq ) [NH , NCO] (molfl)
Time (min) 0 4.50 x 10
0. 370
1.07 x 102
0.292
2
0.212
2.30 X 10
102
Initial Concentrati ons (moIj L)
0.114
54 . Ni trogen o xides, NO ., (a m ixture or NO a n d N0 2 co llec tively designat ed as N O x ), play a n essen tia l rol e in tlu- product ion of polluta nt.s found in ph ot och emi cal s mog. T he N O x in th e atmosph ere is slow ly b roke n down to N2 and () 2 in a first-ord er rea ctio n . Th e averag e hal f-life of NO ., in the smo kes tac k emi ssion s in a la rg e city du rin g dayligh t is 3.9 h . (a) Starting ,...ith 1.50 mg in an expe rim en t, what quan tity of NO ., re mai ns a fte r 5.25 h ? (h) How many ho u rs of da ylight must have elapse d to decrease 1.50 m g of NO x to 2.50 X lO - li mg?
55 . At tempe rat ures bel ow 500 K, th e reac tio n be twee n carb ou mon ox id e an d ni tr oge n d io xid e
[NO, Fj
(mol/l . s)
0.001
0.005
0.001
2.0 x 10- 4
2
0.002
0.005
0.001
4.0 X 10- 4
3
0.006
0.002
0.001
4.8 x 10- 4
4
0.006
0.004
0.001
9.6 x 10- 4
5
0.00 1
0.00 1
0.001
4.0 x 10- 5
6
0.001
0.00 1
0.002
4.0
2 N, O ,,(g ) -> 4 NO ,(g)
T{K)
k (,- ')
338
4.87 x 10- ' 1.50 x 10- 3
318
4.98 x 10- 4
308
1.35 x 10- 4
Mechanism 1
298
3.46 x 10- 5
273
7.87 x 10- 7
Single , elem entary ste p
NO z + N O z ---7 N0 3
Fast
1\'0 , + CO -> NO , + CO ,
Mecha nism 3
I
+ CO -> CO , + 1\'0
Slow
+ NO
:\'0, -> 1\'0
Fast
CO
+ o ,(g )
59. T he d ecompo siti on o f g aseo us d im eth yl e ther a t o rd inary pressu res is fi rst order. Its half-life is 25 .0 min a t 500 ' C:
CH , OC H,,(g) -> CH ,(g ) + CO (g) + H, (g )
Two ste ps
Slo w
10- 5
58 . • Th e data in th e tabl e give th e tempera ture depe nd e n ce o f th e ra te constant fo r th e reac t io n N 2 0 ,,(g ) ---7 2 NOAg ) + 112 O Ag ) . Plot these data in th e a p p rop ria te way 10 d erive th e activa tio n energy fo r th e reaction .
328
Two ste ps
x
has th e following rate eq uation: Rat e = k[N 20 ,,] . It ha s be e n found ex perime n ta lly tha t th e d ecompo sition is 20 5 % co mple te in 13.0 h at 2HH K. Ca lculate th e ra te co ns tan t an d th e half-life at 298 K.
h as th e foll o win g ra te e q ua tion : Rat e = k[N0 2 F. Whi ch of the th ree mec ha n isms suggested h ere best ag rees wit h the ex pe rimen tally o bse rve d rat e eq ua tion?
Mechanis m 2
718
[F,]
57 . The d ecompo sition of din itro geu pcnt ao xidc
+ NO ,(g) -> CO , (g ) + 1\'0 (g )
NO,
Initi al Rat e
[NO,]
Experiment
Using the d at a in th e tahl c : (a) Decide wh e ther th e re acti on is fi rst o rd e r or sec o n d order. (b) Ca lc u la te k for thi s reacti on . (c) Ca lcu late th e ha lf-lift.. of a m mo nium cya nate un d er these co nd itio ns. (d) Calcu late th e co ncen tration of NJ-I 4NCO after 12.0 h .
CO (g)
+ F, (g) -> 2 :\'O , F(g )
Use th e rat e d a ta in th e tab le to do th e fo llowin g: (a) ,,,rrite th e ra te e quati o n fo r th e rea cti on . (b) Ind icate th e ord e r o f re actio n wit h res pec t to eac h co rn p o m-n t o f th e reaction . (c) Fin d th e num e ri cal valu e o f th e rate consta n t, k.
0.458 1
6.00 x
2 :\'O ,(g)
+
°
(a) Starting with ROO g o f dimethyl ether, vvha t mass rcmain s (in grams) after 125 m ill and after 145 min ? (b) Calcu la te th e tim e in m in u tes required to decrease 7 .60 n g (nanograms) to 2.25 ng . (c) Wh at frac tio n o f th e o riginal dim e th yl e th er rc ma in s a fter 150 min ?
+ 0 -> CO ,
•
more challenging • in OWt
Blue-numbered questionsanswered in Appendix 0
60• • Th e d e compo sitio n of ph o sphin e, PH :> . p ro ce ed s accord ing to th e eq ua tio n 4 PH , (g )
P, (g )
-7
+
6 H, (g )
It is fo u n d th at the reac tion has th e foll owin g ra te eq uatio n: Rate = k[ PII 3 1. Th e h alf-lif e o f PH 3 is 3 7.9 s a t 120 -c . (a ) 1I0w m ilch tim e is re qu ired fo r th ree- fo urth s o f th e PH 3 to d ecompo se ? (b) Wh at fraction o f th e o rigin al sa m ple of PII J re mains afu-r 1.00 m in ? 61. T he o zo ne in th e earth 's o zo ne layer d ecomp o se s acco rd in g to th e e q ua tio n
T he me ch a ni sm of th e rea c tio n is though t to p ro cee d th rough a n in itia l fast eq u ilib riu m an d a slow ste p: Step 1
Fast , reversible
O J(g )
St ep 2
Slow
O , (g )
O :.! (g )
+ O (g )
-7
+
O (g )
2 0 ,(g)
Sh ow that th e mech an ism agrees with th is ex peri mcn tal rate law:
- Il[ O,l! d l
~
k [0 , )'/[ 0, ].
62. • l l u udn..ds of different reac tio ns can oc cu r in th e stratos p he re, among the m reac tions th at d estroy th e eart h .... ozone layer. Th e table bel ow lists severa l (sec o ndord e r ) reaction s of CI ato ms with ozo n e an d organ ic co mpou nds; each is give n with its rat e co ns ta n t. Reaction
Rate Constant (298 K, cm3jmolecule . s)
(a) Cl + OJ -,\0 ClO + O2
1.2 X 10- 11
(b) Cl + CH 4 -,\0 HC[ + CH 3
1.0 X 10 - 13
(c) Cl + CJH 8 -,\0 HC[ + C3H 7
1.4 X 10 - 10
(d) Cl + CH ,fCl ~ HCl + CHfCl
3.0 X 10 - 18
Fo r e q ua l co nc e n tra t io ns o f CI a n d th e o t he r re a ct a n t, whic h is th e slowest reac tio n? Wh ich is th e fastest re actio n ? 63 . Da ta for the reactio n
[Mn (CO h(CIl ,CN )l ' + NC;l I , ----> [Mn (COl, (NC, I I"W + CH ,CN a re given in the table . Ca lcu la te f.;l fro m a plot o f In II. vers us I I T. T(K)
•
298
0.0409
308
0.0818
318
0.157
more challenging
• i n OWL Blue-numbered questions answered in Appendix 0
64 . • T he ga s-p ha se re a ct ion
2 N, O,(g)
-7
4 NO,(g) + O, (g)
h as an act iva tion e ne rgy of 103 kJ/ m ol ·rxn, a nd th e ra te co ns ta n t is 0.0900 m in - I a t 328 .0 K. Find the rate co ns ta nt at :j 18.0 K.
65. . ... Egg p rotein albumi n is p re cip ita te d when a n egg is co o ked in boili ng ( 100 °C) water; Eli for th is fir storder reac tion is 52. 0 kJ / nlOl. Estima te the tim e to p repa re a 3-m in egg a t a n altitud e at which wa ter bo ils a t 90 °C. 66. . ... Two mole..cules o f 1,3-h u tad ie lle (C-I116 ) form 1,E>-cyc!o octad ien e , CllH l Z a t h igh er tem peratu res.
Use th e fo llo wing da ta to determi ne th e o rde-r of th e reactio n a nd th e ra te co ns ta n t, k. (Note th at th e total p re ssu re is th e p re ssu re o f th e unrca ct ed C ~ I I I; at an y tim e a nd till' pressu re of th e C!,\ 1llz ,) Time (min)
Total Pressure (mm Hg)
0
436
3.5
428
11.5
413
18.3
401
25.0
391
32.0
382
41.2
371
67. . ... l lyp ofluo ro us ac id, HO F, is wry u nstabl e, dec o m po sin g in a first-o rd e r react ion to g ive H F a nd O z. wi th a half-life of 30. min a t room te m perature :
11 0 F(g)
-7
IlF (g) +
Y, O, (g)
If th e partial p ressure o f HO F in a l.OO-L flask is in itially 1.00 X 10'.! mm Hg a t 25 °C , wha t a re th e to tal p re ssu n.· in th e flask and the pa rti al p re sslllT o f I10 F afte r ex actly :j O mi n ? Afte r 45 m in ? 68 . . ... 'Ve kn ow th at th e d e compo sition o f SO zCl'.! is first order in SO :.! Cl:.! . SO ,CI, (g )
-7
SO, (g )
+ CI,(g)
wit h a half-life of 24':) min a t 600 K. If yo u beg in with a pa rt ial p rt' sSUH' o f SO '.! C1:.! o f 25 mm II g ill a J.(J. J. flask . what is th e part ial p ressu re of' each re actant am i prod uct after 245 min ? Wh at is th e partia l pressure o f eac h reactan t after 12 h ?
I
719
(Fo r more details o n th is re action se e L N ich olson ,
69 . • A Nitramid e , N0 2K1I 2, decom poses slow ly in aqueo us so lu tion accord in g to th e fo llowin g reaction :
[oum al of Cliemiml Ed ucation; Vol. 66, p age 725, 191'39.) Concentration of Phenolphthalein {molJl}
+ H,O (t )
NO ,N H, (a q ) ---> N, O(g)
Th e reac tion fo llow s th e ex pe ri men ta l ra te law
0.0050
kINO,NII , ]
Rail' =
[11,0' ] (a ) Wh a t is th e app a ren t o rde r o f th e reaction in a buffered so lu tio n? (h) Which o f the fo llo wing m echanisms is th e m o st a pp ro priate for th e in te r pretat ion o f thi s ra te law? Expla in.
Mechan ism 1 k,
Mechan ism 2
+ H,O (ratf'limiting .\(fjJ)
Mechan ism 3
+
H;lO +
(rapid eouililniu m) N0
2N II -
k"
l
N 20
11,0 ' + 011 - iI,;
I
+
O Il -
2 11,0
(mte
limiting .\(fjJ)
(Vf1)'
[ast mution}
(c) Sh ow the rela tionship b et wee n til t, experime nt a lly o bse rved rate cons ta n t, k, a n d th e rat e co ns tan ts in th e selected m ech ani sm . (d) Sh ow th at hyd ro xyl ion s ca ta lyze th e d e co mpositio n o f nitram ide .
0.00
0.0045
10.5
0.0040
22.3
0.003 5
35.7
0.0030
51. 1
0.0025
69.3
0.0020
91.6
0.00 15
120 .4
0.00 10
160 .9
0.00050
230.3
0.00025
299.6
(a) Plot th e d ata a bove as [p henolp h thale in ] versu s tim e. a n d d e term ine th e ave ragt' ra te from t = 0 to t = 15 s an d fro m 1 = IOO s to t = 125 s. Do cs th e rate chan g e? If so, why? (h) Wh al is the ins ta n ta n e o us ra re a t 50 s? (c) Use a g ra p hical m eth od to d ctcrrniu c th e o rd e r of th e reactio n with re spect to ph cu olph th ak-in . Wriuth e rate law, and de te r mine the rate con stan t. (d) wh a t i... th e half-life fo r the reaction ?
(rajJid equilibrium}
k,
Time (s)
72 . A \Ve w an t to stu d y th e h yd rol ysis o f the b eaut iful gree n , co ba lt-ba sed co m p le x ca lled tmwrd ich lo ro -h is (e lhyk ned ia m in e)co h alt( l l l) ion ,
70 . • Ma ny b ioc hemi cal re ac tio ns are ca ta lyzed by acids. A typi cal me ch an ism con sisten t with the ex perime nta l resul ts (in wh ic h lIA is tlu - acid a n d X is th e reactant) is
Step I
Fa.-'it, reversib le
II A
Step 2
Fast, reversible
X
Step 3
Slow
XH+ 4
+
l
[-P
11 + +A XH +
+
produ cts
Wh at rate law is d e rived from th is mech an ism ? Wh at is th e o rder o r th e re action with re sp ec t to HA? How would doub lin g the conce ntra tion of HA affec t th e react ion ?
In the Laboratory 71. • The co lor ch ange accom panyin g th e reactio n o f phen olp h th ale in wit h stro n g ba se is illu strate d o n page 670 . T h e c ha ng e in co n cen tra tion of the d ye can he fo llowed b y spe ctro p h o to mv trv (page 190) , a n d som e d ata col lected by th at a ppro ac h a rc g iven bel ow. T he in itia l conce n tra tio ns we re [p henolp ht halei n ] = 0.0050 mol y L an d [OIl - ] = 0.61 mol / L . (Da ta a re ta ken fr om re view ma terials fo r kin e tics at chemcd .chcm. purd uc .cdu .}
7 20
I
In th is h yd rol ysis re ac tion, the g re e n co m p lex ion tnUl.\O- [Co( e nhC12]+ fo r m s th e red co m p le x ion [Co (en h (H 20 ) Cl ] 2+ as a Ct: ion is re p lac ed with a wa ter m ole cul e o n th e Co:1+ ion (e n = 112NCI hCII 2NII ~ ) . tnms- [Co( e n )2C12] +( aq) + 1120 (£) -? gre en [C o( l'II),( II,O) CI]"(aq ) r ed •
more chaLLenging
• in OWL
+
CI- (a q)
Blue-numbered questions answered in Appendix 0
T he reac tio n p rogress is foll owed b y observi ng th e co lo r of th e so lu tion. T he original so lu tion is gree n , a n d th e fin al so lu tio n is re d , b u t at so me int e-rmediate stage when bo th th e n..aCLa 11l a nd p roduct a n ' p re se nt , the solu tio n is gray.
Original solution
Intermediate solution
(c) Th e a ct ivatio n ene q.,T)' for a re act ion ca n be fo u nd hy p lott in g In k versus l i T. Ho wever, he re we do no t need to measure k di re c tly. In ste ad , bec aus e k = - ( l / f) l n ( [ R] / [ RL.) , the lim e need ed In ac h ieve th e g ray co lor is a m e asu re o f k. Use th e d a ta bel ow to fin d the acti va tio n en ergy. Temperature °C
Time Needed to Achieve Gray Colors (for the sam e initial concentration)
56
156 s
60
114 s
65
88 5
75
47 5
73. T he e n zyme chymo trypsi n ca talyzes th e hydro lysis o f a pe p tid e co n tain ing phe n yla lan in e , Using the d at a below a t a g iven tem pe ra Lure , calc u la te the maxi m um ra te of th e rea ction , Rat e rnax ' (Fo r more inform at ion on en zyme ca talysis and th e Michacl ix-Mcntcn m ode l, sec
page 70t.) Peptide Concentration (mol jL)
2.2 X 10- 6
5.0 X 10-
4
3.8 X 10 - 6
10.0 X 10- 4
5.9 X 10- 6
4
7.1 X 10- 6
15.0 X 10-
Final solution
Re action s such as thi s h ave be e n stu died ex te nsively, and ex perirru-u ts suggest that tile ini tial , slow ste p in th e re ac tio n is th e brea king of th e c' o-c'l ho nd to give a five-co o rdi nat e in te rm ed ia te . T he intcrrn e dia u- is th e n a uackvd ra p id ly by wa ter. Slow:
Fast:
fnlll.r[c'o( e n hCI2] + (aq ) ~ [Co(c n),CI]" (aq )
[C o (c n),CI]" (a '1)
+
+
Cqaq )
1I, 0 (a'1) --. [C o(cn ),( H, O )C l]" (aq )
(a) Rased o n the re ac tio n mec han ism , wha t is th e predic «..d ra te law? (b) As th e reac tion p ro cee d s, th e co lo r changes fro m gree n to re d with an in te rmedia te stage where the co lor is gray. T he g ray co lo r is reach ed a t the same tim e , n o m att e r wha t the co nce n tra tio n of th e g ree n sta r ting m a terial (at th e same tem pe ratu re ) . [l ow d oe s th is show th e rea ct ion is fi rst ord er in the g reen fo rm ? Ex p la in ,
•
more challenging
• in OWL
Blue-numbered questions answered in Appendix 0
Reaction Rate (mol jL ' min)
2.5 X 10- 4
74. • T he su bstitu tio n of CO in N i( CO ) j by a no ther molecuh - L [wh e re L is an elec tron-pair d o nor suc h as P ( c' H :~ h ] was stud ie d so me years ago a nd led to an u nd ersta ndi n g" o f some o f th e ge n e ra l p rinciples th a t g overn the ch em istry of co m poun ds havin g mcral- Ct) bo nds. (See J. P. Day, F. Basolo, an d R. G. Pe arso n : j ou rnal oI tllf' Amnicall Chemicnl Society, Vol. 90 , p. 692 7, 19 68 .) A deta iled stu dy of t he kinetics o f the re actio n led to th e fo llo win g" m e cha ni sm : Slow Fast
Ni(CO ). -> Ni(C O ), Ni(CO ),
+ CO
+ L -> Ni(C O ),L
(a ) Wha Lis th e m o lccul arity of eac h o f t he ele me n ta ry reac tio ns ? (b ) Dou bl in g the co ncen tra tio n o f Ni(CO) 4 incr e ase d the rea ct ion rat e by a factor o f 2. Do uhli n g th e co nce n tratio n o f L had no effe ct Oil the reac tion ra te. Based on thi s info rm ation , w r -i te th e rate equatio n for t he re actio n. Doe s th is agree with the m e chan ism de scr ibed? (c) T he (..xpc rin u-n tal ra re co ns ta n t fo r the reacti o n , when L = P (C,(iH "h is 9 .;l X 10- 3 S-l at 20 °C If the ini tial couccn ua tion of Ni(COh is 0.025 1\1, what is t he co nce n tra t io n of th e product after 5.0 mi n?
I
721
Summary and Conceptual Questions TIU' Iollouri ng questions
!!lft) ' 11-\'(' C01/ujJ[J
[rom this an d
jn1'OiU1H
r1wl)ler:~.
75 . Hyd ro g e n ation re ac tions, processe s wh erei n H 2 is
added to a molecule, art' usually ca talyzed. An excel le nt ca talyst is a very finely d ivided m e ta l sus p e n d e d in th e reacti on so lve n t. Te ll why fill ely di vid ed rh odium , fo r exam p le . is a much mo re effi cie nt ca ta lyst th a n a sm all b lo ck of th e m eta l.
76. • ... Su ppose yo u h ave 1000 blocks, eac h of wh ich is
79 . De scrib e e ac h of th e fo llowin g sta te m en ts as true o r fa lse . If fa lse , re write th e se n te nc e to m ak e it co r rec t. (a) Th e rat e-d e term in in g elem e ntary ste p in a re ac tio n is the slowes t ste p ill a m ech a n ism . (h) It is p ossible to ch ange th e ra te co ns ta n t by ch an gin g th e tempe ratu re . (c) As a reactio n proce ed s a t co ns ta n t temperatu re, th e ra te rem ai n s co ns tan t. (d) A re actio n th at is t h ird o rd e r overall must in vo lve mo re th an one ste p .
80 . • ld entifv which o f th e fo llowin g sta tem en ts arc in cor-
1.0 em 0 11 a sid e. If all 1000 o f th ese b locks a rc stacke d to g ive a cu b e th at is 10. em o n a sid e , wh at fr acti on o f th e 1000 bl o ck s h ave a t le ast on e su rface on th e ou tsid e su rfa ce o f the cu be ? Ne xt, d ivid e the 1000 block s in to eig ht equal piles of blo cks a nd fo rm th e m in to e ig h t cu be s, 5 .0 em on a sid e . Wh at fr actio n of the b lo cks I1 m \ ' h ave a t le ast on e su rfa ce on t h e o u tsid e o f th e cu be s? H ow d oe s th is mat he m atica l m o d el perta in to Stu dy Q ues tio n 75?
rec t. If tl; e sta te m e n t is in co rr ect , rewr ite it to h e correct. (a) Reac tions a re fa ster at a h igh e r tempe rature b eca us e activa tio n e ne rgi es are lower. (h) Ra tes in crease with incre asin g co n cen tra tion o f re acta n ts becau se th e re a rc m o re co llisio ns betwe en reac ta n t mo le cu le s. (c) At h ig h e r temp e rature s, a la rger fracti on o f m olecu les h aw e nough e n e rgy to g 2
1 1I ( ~ )
Ra te = k[ H, ][ I, l
Determin e wh ic h o f th e foll owin g sta te me n ts are tr u e . If a sta te m e n t is fal se , indi cate why it is in co rre ct. (a) Th e reaction m u st o ccur in a sing le ste p. (h) T hi s is a secon d-o rd er re ac tion ove ra ll. (c) Rai sin g th e temperatu re will caus e t h e va lue o f k to d e cre ase . (d) Rai sin g the temperatu re lowe rs th e ac tiva tion e n ergy fo r th is reac tio n. (e ) If th e co nc en tra tion s of bo th re actants are d oubled , th e ra te will d o u ble. (f) Ad din g a ca ta lyst in th e reactio n will ca us e th e in itia l rate to in cre ase .
81. • Th e re ac tio n cycl o p ro p an e ~ propene oc cu rs o n a p la tin u m m e ta l sur fac e a t 20 0 °C. (T h e p la tinum is a ca talvs t.) T h e re ac tion is firs t order in cyclo propa ne . I n d i ~a te h ow the fo llowing q u ant itie s c1;a n ge (i n crease , d ecre ase , o r 110 change) as t h is reac tion progresses, as~ Slim ing co n stan t te m p e ratu re . (a) [cyclo p ro pane] (b) [ pro pen e] (c) [cat alyst] (d) th e ra te co n sta nt, k (c-) th e o rd e r' o f the reac tio n (f) th e h al f..life o f cyclo pro pan e
82. Iso to p es are o fte n used as "tra ce rs" to follow an a tom
78. • C h lo ri ne a to m s co n trib u te to the d e struc tion o f th e
thro ugh a chemical reac tio n, a n d th e fo llo wing is an ex am ple . Ace tic aci d re acts with meth an ol (Ch ap te r II ) .
e a rth's o zone laye r hy th e fo llowing seque n ce of re acti o n s:
CI + 0 , ---> C1 0 + 0 , CIO
+
+ 0 ---> CI + 0 ,
wh ere th e 0 ato m s in th e sec o n d stc,' p co m e fro m th e d ecompo sit io n of o zone by su n lig h t:
O , (g) ---> O (g)
CH,C02H
+
CH ,OH
+ O , (g )
\Vh at is lI lt' n e t eq u a tio n o n su m m ing th ese th ree e q u a tio ns? \Vh y d ocs th is lead to oz o ne lo ss in th e stra tos p h e re? W h a t is th e role p layed hy Cl in this scquen ce of re ac tio n s? \ Vh a t n a m e is g ive n to sp ec ies suc h as C lO ?
CH,CO,CH,
+
•
+
H20
Ex p lain h ow yo u co u ld u se the iso to pe 1HO to show wh eth er th e o xyg CaCO,(s) + CO,(g) + H,O(C) If yo u next add pieces of dry ice to th e sus pension of CaCO J (o r if yo u bubb le gaseo us CO 2 into the mi xtu re ) , you will observe that th e so lid CaCO j d issolves (Figure 16.l b) . T his hap pen s because a reactio n occ urs th at is th e reve rse of th e re action th at led to preci pita tion o f CaCO j ; th at is: CaCO,(s) + CO,(aq) + H,O(C) -> Ca"(aq) + 2 HCO, -(aq) Now imagin e what will hap pen if th e so lutio n of Ca 2 + and I-ICO:I - io ns IS III a d osed co ntainer (u nlike th e reac tio n in Figu re 16.1, whic h was do ne in an o pe n co n tainer). As the reactio n begins , Ca 2 + and HCO:\- react to give p ro d uc ts at so me rate. As th e reactants are used u p, th e rate of th is reaction slows. Att he sa me time , howeve r, th e reactio n pro d uct s (CaCO:!> CO 2 , an d 11 20 ) begin to co mbine to re form Ca 2 + and H CO :\- , at a rat e th at increases as t he amou nts of CaCO:~ a nd CO 2 increase. Even tually, the ra te of the fo rward reactio n , the fo rm atio n of CaCO:1> and t he ra te of t he reverse react ion , th e redi ssol vin g of CaCO:h become eq ua l. \Vi th CaCO :~ bei ng formed and redisso lving at th e same ra te, no fu rt her macroscopic change is observed. \Ve have reached equilibrium , and no fu rther net cha nge is observed. \Ve describe an eq u ilib riu m syste m with an eq uatio n that co n ne cts reacta nts an d products with double arr o ws. T he do ub le arrows, indi ca te th at the reaction f
• Cave Chemistry This same chemistry accounts for stalactites and stalag mites in caves. See page 119.
,
16 .1
I
Ch emical Equili brium: A Review 725
(a)
• Reversibility of Reactions All chemi-
cal reactions are reversible, in theory. You may see some of t hese in your laboratory. Afew examples include NH;(g) + HCl(g) ==' NH,Cl(s) CuS04 ' SHzO(s)
~
Cu50,( s) + 5 H,O(g) (NH, j,CO,(s) ==' 2 NH,(g) + H,O(g) + CO, (g)
Practically speaking, some reactions cannot be reversed. Frying an egg , for example, is not a reversible process in practicat te rms.
(b) (e) FIGURE 16.1 Equilibria in the C02/Ca2+jH 20 system. (a) Combining solution of NaHCO J and CaCl2 produces solid (a CO], (b) If excess dry ice (the white solid) is added to the Ca CO] precipitated in (a), the calcium carbonate dissolves to give Cat' {aq) and HCO l "(aq ) (c). (See also Figure 3.6.)
is reve rsible an d tha t th e reaction will be studi ed us ing the co ncepts o f che mical equ ilib ria. Ca2+(aq) + 2 HCO, - (aq) > [N O] [0,,] . If sto ich io me tric a mo u n ts of NO a n d 0 3 a re m ixed and a llowe d to come to e qu ilib rium , virtua lly none of th e react an ts will be fo u nd (Figu re 16.4a ) . Esse n t ially, a ll will h ave be e n co nve rted to N0 2 a nd 0 2. A che m ist wou ld say that "the re ac tio n h as go ne to co m p le tio n. " Co n verse ly, a sma ll va lue o f K m ea n s th at ve ry little o f thc p rodu cts e x ist whe n e quilibriu m h as bee n ach ieve d (Figu rc 16.4b ) . T hat is, thc reac tants are favore d ove r th e prod u cts a t e q u ilib riu m.
K < 1: Reaction is reactant-favored at equilibrium. Concent rations of reactants are great er th a n concentra tions of products at equilib rium. Th is is tru e fo r th e fo rm atio n o f ozo n e fro m o xygen . 3/2 O, (g) , K=
730
C h a p t e r 16
I
Principles of Reactivity: Chemical Equilibria
~ [0,]
'/2
O,(g )
= 2.5 X 10- 29 at 25 °C
SeLected Equilibrium Constant VaLues
TABLE 16.1
Equilibrium Constant, K (at 25 °C)
Product - or Reactant-Favored
4 .2 X 10 52
K > 1; product -favored
10 81
K > 1; product -favored
Reaction
Combination Reaction of Nonmetals 5( s)
+ O, (g) + O, (g)
2 H, (g ) N, (g) N,(g)
50, (g)
+ 3 H, (g) + O, (g)
2 H,O(g)
3.2 X
2 NH, (g)
3.5 X 10 8
2 NO(g)
1.7
X
K > 1; product -favored
10 - ' (at 2300 K)
K < 1; reactan t-favored
Ionization of Wea k Acids and Bases
+
HCO,H(aq)
+
H, O(C)
[1,]
738 Ch apter 16
I
Principles of Rea ctivity: Chemical Equilibria
Assu m ing t he ini tial 12 co nce n tratio n is 0.45 M, and setting up the ICE tabl e in the usu al manne r, we have 2 [(g)
Reaction
[,(g)
Initial (M )
0.45
0
Change (M)
-x
+ 2x
Equili brium (M )
0.45 - x
=='
2x
Fo r th e equ ilibr ium co nsta nt expression . we aga in arr ive at a quad ratic eq ua tio n . K
~
5.6 X 10 - 1>
~
(2x)'
c-'--'-c-
(0.45 - x)
All houg h we could so lve thi s e q uation using the quad ra tic fo r mula , t here is a sim p ler way to reach an a nswer. No tice th at th e value of K is ve ry sma ll, indi ca ting th at th e a mo u n t of 12 th at will be di sso cia ted (= x) is vc ry sma ll. In fac t, K is so small that su btracting x from the origi na l reacta nt co ncentratio n (0.45 mol /L ) in th e d en omi nato r of t he eq u ilibriu m co nsta nt ex p ress io n will leave th e d eno m in at or es se n tial ly un ch an ge d . T ha t is, (0.45 - x ) is esse n tia lly eq ua l to 0.45 . Thus, we drop x in th e d en omin at o r and have a simpler eq uatio n to so lve. K
(2x)' = 5.6 X 10- 1> = (0.45)
Th e so lutio n to this eq uatio n gives x = 7.9 X 10 - 7. From thi s value , we ca n determine th at [12 J = 0.45 - x = 0.45 mo l/L and [ I] = 2x = 1.6 X 10 - (; mo l/L. No tice that th e answer 10 th e 12 di ssociation p robl em co nfirms th e assum p tion th e d issoelat ion o f 12 is so small th at [12] at eq u ilibriu m is esse nt ially eq ual to the in itial co nce n tra tio n . \Vh en is it possible to sim p lify a q uadratic equal io n? The d ecision de pe nd s on both the value o f th e init ial co ncentratio n of th e react ant and th e value o f x, which is in tu rn re lated to th e value of K. Co ns ide r th e gene ra l reac tio n A ~- B + C
whe re K = [ B][C] /[A]. Assu me we know Ka nd th e initial co nce n tra tio n of A ( = [A]o) an d wish to find the eq uilibriu m co nce ntratio ns o f R and C ( = x). The equilib riu m constan t expressio n now is K ~ [B][eJ ~ (x)(x) [AJ, - x [AJ
\Vhen K is very sma ll, th e value o f x will be m uc h less tha n [A] ", so l.A] o - x = [A] o. T herefore , we ca n wri te th e following ex pressio n. (16.3)
If 100 X K
-c. (a) Wh at is th e value o f Kp? (h ) Su p pose yo u m ix 0.45 m ol of 112 a nd OA5 mol of 12 in a 10.0-1. flask a t 425 °C. What is th e tot al p re ssu re of th e mi xtu re before a nd after eq u ilibrium is achi eved? (c) What is th e parti al pressu re of eac h ga s a t eq uili bri um ?
•
more chaLLenging
•
in OWL Blue-numbered questions answered in Appendix 0
54 . • Sulfuryl ch loride , 501C11, is a co mp o u nd with \'ery irrit ating va po rs; it is used as a reage n t in th e syn th es is of org ani c co mpounds. Whcu heat ed to a su ffic ien tly hi gh te mperatu re, it de co mpo ses to S01 an d C11. SO, Cl, (g )
SO, (g )
K
+ C l, (g)
~
0.045 a t 375 "C
(a) A 1.00-1. flask con tai ning 6.70 g o f S0 1CI2 is h eated to 375 °C. Wh at is th e co nc en tra tio n of each of the co m po u nd s in th e syste m whe n eq ui libri u m is achi eved? Wh a t fra ctio n o f 5 0 1C11 has di sso ciat ed? (b) What are th e co nc e n tra tio ns o f 50 1C12 , S0 1, and Cl1 at equilib riu m in the 1.00-1. flask at 375 cC if yo u be g in with a m ix tu re of S0 1C12 (6.70 g ) an d C12 ( 1.00 a tm )? Wh at fra ction IlfS0 2C12 1lao,; dissocia u-d ? (c) Com pa re the frac tion s o f 5 0 1C1',;! ill p arts (a ) an d (1)) . Do they ag re e with your e xpe cta tio ns based on Lc C hau-licr's p ri n ciple? 55 . .... Il em ogl ohi n ( Hb} call form a com plex with both 0 1 a nd CO . For th e reacti o n IIbO ,( a q )
+
C O (g )
IIbCO (aq )
+
O , (g )
a t bo d y tempe ratu re , K is a bo u t 200. If the ra tio [ H bC O]/ [ H b0 2 ] comes dose to I, d eath is probabl e . Wh at parti al p re ssure o f CO in th e ai r is likely to be fatal? Assu me th e parti al pressure of 0 1 is 0.20 a tm .
56 . . .... Lim estom- d ec o m poses at h igh temperatu res. CaCO , (s)
+
Ca O (s)
C O ,(g)
At 1000 -c, K Il = :l 8 7. If pure CaCO j is placed in a 5.0 0-1. fla sk a nd he at ed 10 1000 °C, wha t q u antity o f CaC O :1 m ust d e compo se to ac h ieve th e eq uili b ri u m p re ssu re of CO 2?
57 . At 1800 K, o xyge n d issoci at es very sligh tly into its at om s. O , (g ) ,
2 0 (g)
«; =
1.2 X 1O- I U
60 . • A. Th e d issociatio n o f calciu m carbo na te h as a n equilibrium co nsta n t of Kp = 1.1() at 800 0(:. CaCO, (s)
CaO (s)
+
C O , (g )
(a) What is K fo r th e re actio n ? (b ) If yo u pl ace 22.5 g of CaC 0 3 in a 9.56.. L co ntainer a t 800 "C, what is th e pr essure of CO 2 in th e con ta ine r? (c ) Wh at percentage of th e origi na l 22.5.. g sam ple of CaCO :\ re ma in s undccompo scd at eq u ilib riu m ? 61. A. A sa m ple o f N 1o.j g as with a pressure of 1.00 a tm is pl aced ill a flask. Wh e n equi libri um is achiev ed. 20.0 % of th e 'N2 0 t has been co nve rted to N0 1 ga s. (a) • Calcu la te Kp(h) If th e orig inal p re ssu re of N 2 o.j is 0.10 aun , wh at is th e pe rcent di sso ciation o f th e gas? Is th e re sult in agree men t with l .c Cha tcl iur's pri ncipl e ?
62. . .... A re a ction impo rta nt in sm og form a tion is O , (g )
+ NO (g )
O, (g )
+ NO ,(g)
K
~
s.o X
1031
(a) If th e initial co nce n tra tio ns are [0 3 ] = 1.0 X IO-..jj M, [N O ] ~ 1.0 X 10--> M, [N O , ] ~ 2.5 X IO-l M, and [0 2 ] = 8. 2 X 10-3 M, is the system at eq u ilibriu m ? If not , in wh ich di rectio n d oes th e re acti on proceed ? (b) If th e tempe ratu re is inc rease d , as on a very war m day, will th e con ce n tratio ns of th e products in cre ase o r decrease ? ( H int: Yi.Hl m ay ha ve to cal cula te the e n thal py ch ange for th e reac tion to find o u t if it is e xo therm ic o r end oth r-nn ic.)
In the Laboratory 63. A. The ammonia co m plex of tr imc thyl borau c, (NH 3 ) B(C H lh di ssociat es at 100 "C to its co m po nen ts with KI' = 4.62 (whe n th e p re ssu res are in atmos p he res) .
(N H,)B(C H,),(g )
B(CH ,), (g)
+
NH , (g)
If yo u pla ce 1.0 m o l o f 0 1 in a 10.-1. vesse l an d he at it to 1800 K, how many 0 atom s are present in th e flask?
+v~
58. . .... Nitr os yl b romide , NO Rr, is prepar ed by th e direct reacti on o f NO an d Br',;!. 2 NO (g )
+ Br,(g ) -> 2 NO ll r (g)
T he com poun d di sso ciat e s re ad ily a t ro o m tempe rarure. ho weve r. N O Br( g )
NO (g )
+ 'Ii Br, (g )
Som e NO Rr is pl ace d in a flask a t 25 "C a n d allow e d to d isso ci at e . Th e tot al pressure a t e q ui libriu m is 190 mm II g a nd the com po u n d is foun d to be 34% d isso ciated . wh at is the val ue o f Kp? 59 . . .... Bo ric aci d an d g lycerin fo r m a com plex II(OH ), (aq )
+
g lycer in (aq}
B (OH ), ' g lycerin (aq )
wit h an equilibrium co nstan t of 0.90 . If th e co nccut ratio n of bo ric a cid is 0.10 M, how m uch g lycerin sho uld be adde d , per lite r, so th at 60. % of th e bo ric ac id is in the form o f th e co m plex ? •
morechallenging
• in OWL
Blue-numbered questions answered in Appendix 0
If N II J is cha nged to so me o ther mole cule. the cq ui libr ium co nstan t is d iffer e nt Fo r [ (C H ,) , P I II(C H,) ,
KI'
~
Fo r [ (C Il , ),N ] B (C Il , ),
K I'
= 0.472
0 .1 28
(a ) If yo u be gin a n ex perime n t hy pl acin g 0.010 mol of e ach com p lex in a flask , wh ic h wo u ld have th e larg.. es t p artial p re ssure o f R (C113 h a t 100 °C? (h ) 11'0. 7:) g (0. 0 10 m ol ) o f (N H,) B( C H ,) , is placed in a 100 ... m L flask an d heat ed to 100 °C , what is th e pa rti al p ressure o f eac h gas in th e eq u ilib riu m mi xture . a nd what is th e tot al p re ssu re ? Wh at is th e pe rce nt d isso ciat ion of (N II 3 H\ (CII 3h ?
I
757
64 . T h e ph o tog raph s below show wha t occu rs whe n a sol utio l\ o f p otassiu m ch rom a te is tr eat ed with a few d rops o f co nc e ntra ted h ydroch lo ric aci d. So m e o f th e h rig h t yellow chroma te io n is co n ver ted to th e orange di ch ro mate io n .
(a) Adding KSeN
,
(a ) As m ore K...IOlCN is ad ded to the so lu tio n, th e colo r be co me s eve n mo re red . Exp la in thi s o bse rva tio n . (h ) Silver io ns fo r m a whi te precipit ate with SCN - io ns. \Vha t wo uld yo u observe o n add ing a few d ro ps of aque ous silver n it rat e to a re d so lu tion of [Fe (1120 )"SCN] + ion s? Expla in yo ur o bserva tio n .
-
(a) Expla in thi s ex pe rime n ta l o bse rva tio n in term s of Le C hatcli e r's pri ncipl e . (b) What wo uld yOll observe if you tre at ed t he o ra nge so lu tio n with so d iu m h yd ro x id e? Ex pla in yo u r o bse rva tio n .
66 . • A. T he ph oto g ra ph s at th e bottom o f the page show wha t o cc urs wh e n yo u ad d a mm o n ia to aq u eo us ni ckel ( lI ) ni tra te an d th e n ad d e t hyle ned ia m ine (N H 2CH 2 CH 2 N H 2 ) to the inte rm edi at e blue-p u r ple solu tion . [Ni (H , O )"JH (aq ) + {) NH,,(aq) g ree n [N i (N II "lo)" (aq ) + blu e-p u r ple
6S . T h e ph o tog ra ph s in (a) show wh at o ccu rs ..•vh e ll a so lutio n of iron (HI) n it rat e is tre at ed with a fe w d ro ps o f aqu eo us pot assiu m thi o cyan at e . (Se c Cl u -mi strvlvo w, Scree n 16.4.) T he nearly colorl ess iron ( IH) io n is conve rt ed to a re d [Fe (l h O L;SCN ]2+ io n. (T h is is a class ic te st fo r the p re sen ce of iron (H I) ions in sol u tio n. ) [Fe( H ,o )"l" (aq )
+
SCN - (a q) [Fe( II , O ),SCN] H (a'l )
(b) Adding Ag'
{) 11, 0 (1')
[N i (N lI ,,)cl H (aq ) + 3 N lI ,ClI ,ClI ,N lI , (a'l) blu e-pu rple [N i (N H ,C H, C H ,NH ,) , JH (aq ) + {) NH 3 (aq ) K, vio le t
+ 11, 0 (1')
Wh ich equ ilibrium co nstan t is g rea ter, K. or K2? Exp lain .
Add ethylenediamine NH 2CH2CH2N H2
758
I
•
more chaLLenging
•
in OWL
Blue-numbered questions answered in Appendi x 0
Summary and Conceptual Questions T Ill' [ollounng questions
Wi)'
use concepts from this fwd tnrnious
clutpters.
2 C(g)
67 . De cid e whe t her each of th e fo llow ing sta te men ts is true or fal se . If false, change th e wording to m ake it true . (a ) Th e magnitude of th e e q u ilibriu m con stant is always independent o f temperature . (h ) Wh en two che mical equa tio ns a rc ad d ed to give a n et equatio n , the e q u ilibriu m consta n t for th e n et e q uatio n is th e p roduc t o f th e equilibrium co nstants o f the su m med eq u a tio ns.
(c) Th e eq u ilibrium constan t fo r a reac tio n h as th e sa me val ue as K for th e rev e rse re acti on . (d ) Only the concen tra tion of CO t ap pears in the e q uilib rium consta n t ex p ress io n for th e re act io n CaC0 3(s) Ca O(s) + co, (g) . (c) Fo r th e react io n CaCO:1(s ) ~ CaO (s ) + C O 2 (g) , th e va lue of Kis nu m ericall y th e sa me , whe ther th e a m o u n t o f CO 2 is exp resse d as mo les/lite r or as gas pressure.
68 . • Neit her Ph C12 n or PbF2 is ap prec ia b ly so lu b le in water. If so lid Ph Cl 2 a n d so lid Ph F2 a re pl aced in eq ual amo u n ts o f wa ter in se pa rate beakers, in wh ich be aker is th e co ncen tra tio n o f Pb N greater> Equil ib rium con stan ts for th ese so lids d issol vin g in wa te r are as foll o ws: PhCl , (s)
Ph" (aq)
+
2 Cl " (aq )
PhF, (s)
Ph" (aq )
+
2 F- (ag )
K ~ 1.7 X I O:"
K
~
3.7 X 10-"
69 . Charac terize ea ch o f th e fo llowing as p rodu ct- o r rcactan t-favorcd . (a) CO(g) + \', O, (g ) CO ,(g) K p ~ 1.2 X 1O,,; (h) Il,O (g ) (c) C O(g)
•
morechallenging
70 . • Cons ider a gas- p hase reac tio n where a co lorless compound C produces a b lue co m p o un d R.
, H , (g )
+ C I,(g )
• in OWL
+ \', O ,(g) COCI, (g )
«; ~ !I. I «; ~ 65
B (g )
Afte r reaching equ ilib riu m , th e size o f th e flask is hal ved. (a) What color ch an ge (if a ny) is o bserved im mediately up on h a lving th e flask size? (h ) What colo r cha nge (i f an y) is o bs erved a fter cquilibri um h as b een ree stablish ed in th e flask?
71 . An ice cu be is p la ced ill a b eaker of wa ter a t 20 °C . Th e ice cu be p artiall y m elts, a n d th e tempe rature o f th e wa te r is lo wered to 0 "C . A t this point, both icc and water arc at 0 °C , and no fu rt her change is appa re nt. Is th e system a t eq u ilib riu m? Is th is a d yn ami c eq u ilibri um ? T ha t is, arc eve n ts still occurri ng a t th e mol e cular level ? Suggest an ex peri m e n t to te st wh e th er this is so. ( Hint : Consider usin g D';.!O.) 72. Se c th e simula tion on Che mis tryblow, Scree n 1GA. (a) Se t the conce n tra tion o f Fe3+ at 0.0050 M a n d that o f SCN - a t 0 .0070 ~L Cl ick th e "Re act" b u tton . Does th e concen tra tion of Fe 3 + g o to zero? \ Vhen e q ui libriu m is reached , wha t are th e conce n tra tio ns o f th e re actants a nd th e produ ct s? Wh at is the equ ilib rium constan t? (h ) Begin with [Fe3+] = [SCN - ] = 0 .0 M a m i [FeSCNN ] = 0.0080 M. Whcu equilibriu m is reac hed , whic h io n h as th e larges t concentra tio n in so lu tio n ? (c) Beg in with [Fe H ] ~ 0 .001 0 M. [SCN - ] ~ O.O()2 0 ~l, and [FeSCN"2+] = a.OOgO M. De scribe th e resu lt o f all owin g thi s syste m to come to equil ib ri um .
X IO-l l X lO "
Blue-numbered questions answered in Appendix 0
I
759
CONTROL OF REACTIONS
17
The Chemistry of Acids and Bases
~
~ ~
e
Aspirin Is Over 100 Years Old! Aspirin is one of t he most successful non-prescri ption drugs ever made. Americans swallow more tha n SO millio n aspirin tablets a day, mostly for t he pain-relieving (analgesic) effects of the drug. As pirin also wards off heart disease and thrombosis (blood clots), and it has even been suggested as a possible treatment for certain cancers and for senile dementia. Hi ppocrates (460-370 Be), t he ancient Greek physician, recommended an infusion of willow hark to ease the pain of childbirth. It was not until the 19th centuryt hat an Italian chemist, Raffa ele Pi rta. isolated salicylic acid, t he active compo und in t he bar k. Soon there-
Photo of crystals of aspirin ta ken using a microscope and polarized light.
was also a useful drug and had fewer side effects. This is t he com pound we now call "aspirin." Acetylsalicylic acid slowly reverts to salicylic acid, C, H,(CO,H) (O H) , and acetic acid in t he presence of moisture; t herefore, if you smeltthe characteristic odor of acetic acid in an old bottle of aspirin tablets, t hey are too old and should be discarded. Aspirin is a com ponent of various over-the-counter medicines, such as Anacin, Ecotrin, Excedrin, and Alka -Seltzer. The latter is a com bination of aspirin with citric acid and sodium bicarbonate. Sodium bicarbonate is a base and reacts with the acid to produce the sodium salt of acetylsalicylic acid, a form of aspirin t hat is water-
after, it was found that t he acid could be ext racted from a wi ld soluble and quicker acting. flower, Spiraea ulmoria. It is from the name of this plant that t he Questions: name "aspirin" (a + spiraea) is derived. Hippocrates's wi llow bark extract, salicylic acid, is an analgesic, 1. Aspirin has a pK, of 3.49 , and t hat of acetic acid is 4.74 . Which is t he stronger acid? but it is atso very irritating to the stomach lining. It was therefore 2. Identifythe acidic Hatom in aspirin. an important advance when chem ists at Bayer Chemicals in Germany 3. Write an equation for the ionization of aspirin. found, in 1897, t hat a derivative of salicylic acid, acetylsalicylic acid, Answers to these questions are in Appendix Q. 760
Chapter Goals
Chapter Outline
Set' Cha/lln Goals Revisited (/H1gf' 799) lor Stud)' Questions kp)'f'd to these bJ'(}al\ ruul a.nigllahlf' in OH'L
17.1
Adds and Bases: A Review
17.2
The Brensted-Lowry Concept of Acids and Bases Extended
17 .3
Wat er and the pH Scale
17.4
Equilibrium Constants for Acids and Bases
Use the Bransted-Lowry and Lewis theories of acids and bases. Apply the principles of chemical equilibrium to acids and bases in aque-
ous solution. Predict the outcome of reactions of acidsand bases. Understa nd the influence of st ruct ure and bonding on acid-ba se properties.
17 .5
Predicting the Direction of Add-Base Reactions
17 .6
Types of Acid-Ba se Rea ctions
17.7
Ca lculations with Eq uilibri um Constant s
17 .8
Polyprotic Adds and Bases
17 .9
The Lewis Concept of Add s and Bases
17 .10 Mo lecular Structure, Bonding, and Acid-Base Behavior
cid s and bases a re amo ng th e 1I10 st co m mo n substa nces in na tu re . Amino aci d s are th e bui ld ing blo cks o f p ro te ins . T he pH o f th e lakes, rive rs, and oceans is affec ted by d issol ved acids and bases, and yo u r bo d ily fu nctio ns de pend o n aci ds and bases. You we re int rod uced to th e d efi ni tions of aci ds and bases an d to so me of th eir che m istry in Cha pter 3, but th is chapter and t he next ta ke up th e de tailed chemist ry of th is imp o rtant clas s o f su bsta nces.
A
Chemistry-9-NOw'· Throughout the text this icon int roduces an op portuni ty for self-study or to explore interactive tuto rials by signing in at www.thornscnedu .com /Iogtn .
17.1 Adds and Bases: A Review In Chapter 3, yo u wer e int roduced to two d e finitions of acids an d bases: th e Arr heni us d efi nitio n and th e Brensted-Lo wry d efini uon . According to th e Arrheniu s d efin ition , an aci d is an y su bst ance th at , whe n d issolved in water, in creases th e co nce n tra tion o f hydrogen io ns, H + (... page 132) . An Arrheni us base is an y su bstan ce that increase s the co ncen t ratio n of hyd ro xid e io ns, O l l ". when d isso lved in water, Based o n th e Arrhen ius de finition, h yd ro chl ori c acid was th erefore classifie d as an aci d , and sodi u m hydro xid e was cla ssifie d as a base .
HCl(aq) --> W(aq )
+ Cl-(aq)
NaO H(aq) --> Na+(aq) + OW(aq) Usin g thi s de fin ition, reactio ns bet ween acids an d bases involve th e co m bina tio n of 1-1+ and O lI - ions to for m water (and a sa lt) .
NaOH(aq ) + HCl(aq) --> H,O(C) + NaCl(aq) T he Branstcd- Low rv defin ition of acids and bases is more general and views acid-base be hav ior in te rm s o f pro ton transfer from o ne su bstance to ano ther. A Bre nstc d- Lowry ac id is a pro to n (1-1+) d onor, and a Bran stcd-Lowry base is a pro ton acce ptor. T he reactio n be tween an acid and a base is viewed as involvin g an equ ilibrium in wh ich a ne w acid an d bas e (the co nj ugate base of the acid an d t he co njugate ac id of th e base ) are fo rmed . T his defi nitio n ex te nds th e list of bases and th e scope of acid- base reac tions. In th e fo llowing reactio n , lICI acts as a 17 .1
I
Acids and Base s: A Review
761
Bro nstcd-Lowry acid , and wate r acts as a Brenstcd-Lowry base becau se J ICI transfers a H + ion to H 20 to fo rm th e hyd ro ni u m ion , H 30 +.
+ H,O(f )
HCl(aq)
+ Cl-(aq)
H,O+(aq)
T his eq uilibriu m strong ly favors formatio n of HJO +(aq ) and Cl Iaq) : that is, HCl is a stro nger aci d than JI JO +. \f e will begin this chapter by loo kin g at Bronstcd- Lowry acid-base che mi stry in more de ta il. r
17.2 The Bronsted-Lowry Concept of Acids
and Bases Extended A wide var iety of Brans tcd-Lowrv aci ds is know n . T hese include so me mol ecul ar compo unds suc h as n itric acid , HNO,(aq) + H,O(C) ;::::': NO, - (aq) + H,O+(aq) Acid
catio ns suc h as NJ I{" . NH/(aq)
+ H, O(f )
=
NH,(aq)
+ H,O' (aq)
Acid +
and an ions. H,PO, - (aq)
+ H,O(f)
Similarly, man y d ifferen t type s of species ca n ac t as Brenstcd bases in thei r reaction s with water. These incl ude some molecu lar co mpo u nds, NH , (aq)
+ H,O(f )
Base
=
NH, " (eq)
+ OW(aq )
+
anio ns, CO,' -(aq ) + H,O(f )
762
Chapter 17
I
The Chemistry of Acids and Bases
HCO, - (aq) + OW(aq)
and h ydra ted me tal cations. T he se ca tio ns can act as aci ds an d bases. [Fe(H2 0).1' +(aq) + H20(C)
[Fe(H20).( OH)]2+(aq) + H,O+ (aq)
[Al(H 2 0)s(OH) ]2+(aq) + H20 (f )
[Al(H 20). ]3+(aq) + OW(aq)
Acids suc h as IIF, II C), II NO " a nd CII "C0 211 (ace tic acid) are all capable o f d onating o ne pro ton a nd so arc ca lled monoprotic acids. Other ac ids, ca lled polyproti c acids (Table 17.1) , are capable o f do nating two or mo re p roton s. A fam ilia r examp le o f a pol yproti c ac id is su lfuri c acid .
-
+ H20(f)
HSO, " {aq}
+ H20 (t )
=
SO/ - (aq) + H30 ' (aq) Tartaric acid , " 2(4",.06 , is a naturally occurring di proti c add. Tartaric acid and its potassium salt are found in many fruits. The acidic protons are the H atoms of the - C0 2 H or carboxylic acid groups.
2-
+
.Just as th e re are acids th at ca n don at e more th an o ne pro ton , so th ere arc polyprotic bases that can accep t m o re th an o ne proto n . T he full y d eproto nated anio ns of pol yp ro tic aci ds are po lyprotic bases; examp les in clude S0 42- , PO l - , CO :l 2- , and C 20 / - . T he carbo na te ion , for exam ple , can accep t two protons. C0 32-(a q) + H20( t)
~---'-
HC0 3- (aq) + OW (aq)
Base
HC 0 3 - (aq) + H20 (t)
H2C0 3(aq ) + OW(a q)
Base
Some m olecul es (suc h as water) a nd io ns ca n beh ave e ither as Bre ns rcd acid s or bases and are refe rred to as being amphiprotic (... page 136 ). An e xa mple of an
TABLE 17.1
Polyprotic Acids and Bases
Acid Form
Amphiprotic Form
Base Form
H2S (hydrosulfuric acid or hydrogen
HS - (hydrogen sulfide ion)
S2 (sulfide ion)
H2P04- (dihydrogen phosphate ion)
P043- (phosphate ion)
sulfide) {
HPO/ - (hydrogen phosphate ion) HC0 3- (hydrogen carbonate ion or bicarbonate ion)
CO/ - (carbonate ion)
HC20 4 - (hydro gen oxatate ion)
17 . 2
I
The Bransted- Lowry Concept of Acids an d Bases Exten de d
763
am p hip ro tic an io n th at is particul arl y important in bioch emical systems is th e di hydrogen phosph at e ani o n (Ta ble 17.1) . H, PO, -(aq) + H,O(C ) ,
H]O+(aq) + HPO; - (aq)
Acid
H, PO; -(aq) + H,O(C)
H]PO,(aq) + OW (aq)
Base EXERCISE 17.1 Breasted Acids and Bases
(a) Write a balanced equation for t he reacti on that occurs when H3 P04 • phosph oric acid, donate s a proton to wat er t o form t he dihydrogen phosp hate ion. Is t he dihydrogen phosphate ion an acid, a base, or amph iprotic? (b) Write a balanced equation for t he reactio n that occurs when the cya nide ion , CN - , a ccepts a proton fro m water to for m HCN . Is CN- a Brensted acid or base ?
Conjugate Acid-Base Pairs A reaction important in th e co ntro l o f aci d ity in biological syste ms in vol ves th e hydrogen ca rbonate ion , which ca n ac t as a Brensted base o r ac id in wate r. The equati on for H e 0 3 - fun cti oning as an base ex e mplifies a featu re o f all reac tio ns in volvin g Breu sted acid s and oases. conjugate pair 1
I
I
conjugate pair 2
r
HCO]-(aq ) + H,O(C) Base
Acid
+
-
~
1
OW(aq) + H,CO](aq) Base
~
V
Acid
+
In the forward direction , H C0 3 - is th e Bro ns tcd base becau se it ca p tures H + from th e Brousted aci d, H 20 . T he reverse rea ction , ho wever, is also an acid -base reactio n. In th is case, the 1I 2 C0 3 is th e aci d , a nd O lI - is the base . A conjugate acid-base pair co ns ists of two species th at differ fro m each o ther by th e presence of one h ydrogen ion . T h us, H 2C0 3 and HCO s - co m p rise a co njugat e acid- base pair. In thi s pai r, lIe O s- is th e co njugate base of the aci d 1I 2 CO:i , and H 2 CO J is th e conj uga te ac id of th e ba se H CO J - . T here is a sec o nd co nj uga te acid -base pair in thi s re action : H 20 a nd H 30 +. In fact , eoery reaction between a Brenstrd arid and a Brensted base involves tmo tOnjugale acid-base flairs (Table 17.2 ) . Chemislry.ci.Now'"
•
Sign in at www.thomsonedu .com/login and go to Chapter 17 Contents to see Screen 17.2 for an exercise and tutorial on adds, bases, and their conjugates.
EXERCISE 17.2 Conjugate Acids and Bases
In th e followi ng reaction, identify the acid on t he left and its conj ugate base on the right. Similarly, identify t he base o n the left and it s conj ugat e acid on the right .
HNO](aq)
764
C h a p t e r 17
I
The Chemist ry of Acids an d Base s
+ NH](aq) ,
' NH, " (aq)
+ NO ]- (aq)
Acid-Base Reactions and Conjugate Acid-Base Pairs'
TABLE 17.2 Name
Acid 1
Hydrochloric acid
HCl
Nitric acid
+ + + + + + + + +
HN03
Carbonic acid
HzCO J
Acetic acid
CH,C02H
Hydrocyanic acid
HCN
Hydrogen sulfide
H,S
Ammonia
H2O
Carbonate ion
H2O
Water
H2O
Base 2
Base 1
H2O
Cl
H2O
N03-
H2O
HCO, -
H2O
CH3C02 -
H,O
CN -
H2O
HS -
NH,
OW
,
COl -
OW
H2O
OW
Acid 2
+ + + + + + + + +
H,OH 30 ~
H,O' H 30 ~
H3O H3O ·
NH 4 + HC03 -
H,O-
"A ri d I and base I are a co nj uga te pai r, a s are hast- 2 and aci d 2.
17.3
Water and the pH Scale
Be cau se
\\T
ge ne ra lly u se a q ueous so lu tio n s o f ac id s a n d ba se s, an d be cau se th e
acid- hase reacti ons in your body occur in your aq ueo us interior, we want to conside r the beh avio r of wate r in te rm s o f c he m ica l equ ilibria.
Water Autoionization and the Water Ionization Constant, Kw An ac id suc h as H Cl do cs not n eed to be pre sent fo r th e h ydro n ium io n to ex ist
in water. In fact, two water mol ecu les can int e ract with eac h o ther 10 produce a hyd ron iu m io n a n d a h yd roxid e io n by pro to n transfe r from o ne water mol ecu le to th e o ther.
H-O :
I
+ H-O :
H- O- W
I
H
I
H
H
+
+ :0 :
I
H
+.'-"1
T his autoion iza tio n reactio n of water was dem o nstrate d man y years ag o by Friedrich Ko hl ra usch ( 1840- 19 10 ) . J Ie fo un d th a t, even afte r wate r is p ainstaki ngly p u rified , it still co nd ucts electric ity to a vcry small exte nt because au to io niza t.io n produce s very low concen tratio n s of II jO + a n d 0 11- ions. \ Va te r au to ionizatlon is the corn e rstone of o ur conce pts o f aq ueous aci d-base be h avio r. wh en wa te r auroio n izes, th e eq ui lib riu m lies far to th e left sid e. In fa ct , in pure water a t 2:') "C, o n ly ab o u t 1.\ \'0 o u t of a billi o n ( IO!J) water m o lecu les are ionized a t a ny in stan t. To ex p ress this id ea mo re quan titatively, we ca n write th e eq u ilib riu m co nstant ex p ressio n for aut oi on ization . K.
=
[H,O+][OW]
= 1.0
X
10- 14
at 25
O(
(17.1)
• Kw and Temperature The equation Kw = [H30 +)(OW] is valid for pure water and for any aqueous solution. Kw is temperature dependent. Because t he autoionization reaction is endothermic, K" increases with temperature. T (0C)
K.
10
0. 29 X 10- 14
15
0.4 5
20
0.68 X 10- 14
25
1.0 1
30
1.47 X 10- 14
50
5.4 8 X 10 - 14
17 .3
I
x x
10- 14
10 - 14
Water and t he pH Scale
765
Th ere a re seve ra l impor tant aspects o f th is eq ua tion. • Based o n the ru les we have g iven for writin g eq u ilibriu m consta n ts, we wou ld n ot include th e co nce n tratio n of water. • The equili bri u m co nstant is given a spe cia l sym bo l, K w , a n d is kn own as th e autoionization co nstant for wat er. • Because th e a u to io n iza tio n of wa ter is th e o n ly so u rce of hydronium a n d h ydro xid e io ns in pu re water, we kn o w that [JT 30 +] must eq ual [OJT - ] in pure wa te r. Electric a l cond uct ivity measu rem ents o f pure water sh ow th at [11 3 0 +] = [O ll - J = 1.0 X 10- 7 M at 25 °C, so K, has a valu e of 1.0 X 10- 14 at 2S -c . In pure wate r, th e hyd ron ium io n a n d hydro xide io n co n ce n tr ati o ns a re equ al, a n d th e wate r is said to be n eut ral. If so me acid o r base is ad d ed to pu re wate r, h owever, th e eq ui lib riu m
H,O+(aq) + OW (aq)
2 H,O(f)
is disturbed . Add in g ac id raises th e co nc en tratio n of th e H 30 + ions, so th e so lu tio n is aci di c. To o p pose this in cr ease , Le Ch ate lie r's pri n cip le (.. Sectio n IG.G) pred icts th a t a sma ll fractio n of th e JT 3 0 + io n s will react wit h 0 11 - io ns fr om wate r a u toionizati on to form water. T h is lowers [O H -] un til th e produc t o f [ H 30 +} an d [0 11-] is ag ain eq ua l to 1.0 X 10 - 14 at 25 ' C. Sim ilarly, ad di ng a b ase to p ll re wate r g ives a basic so lu tio n beca use th e 011 - ion co nc e n tra tio n h as in c reased. Le Ch a tclicr 's prin ciple p redic ts th at so m e o f th e ad de d O H - io ns will react with H :\O + io n s present in th e so lutio n fro m wate r autoicmi zation , there by lo we rin g [JT 3 0 +] u ntil th e value o f th e product [11 ,,0+] a n d [0 11 -] eq uals 1.0 X 10- 14 at 25 ' C. T h u s, fo r aq u eo us so lu tio n s at 25 "C, we ca n say th at
In a neutral solution, [H,O+] = [OW ]. Both are equal to 1.0 X 10- 7 M. In an acidic solution, [H ,O+] > [OW ]. [H,O+] > 1.0 X 10- 7 M and [OW] < 1.0 X 10 - 7 M. In a basic solution, [H,O+] < [OW ]. [H ,O+] < 1.0 X 10 - 7 M and [OW] > 1.0 X 10 - 7 M.
ChemiSlrv.c).NOw~
Sign in at www.t homsonedu.comj login and go to Chapter 17 Contents to see Screen 17.3 for a simulation of the effect of temperature on K....
Ion Concentrations in a Solution of a Strong Base ProbLem Whatare the hydroxide and hydro nium ion concentrations in a 0 .00 12 M solution of NaOH at 25 °C? Strategy NaOH , a strong base, is 100% dissociated into ions in water, so we assume that the OH - ion concentration is t he same as t he Na OHconcentration. The H3 0+ ion concentration can t hen be calculated using Equation 17.1. SoLution The initial concentration of OH- is 0.0012 M. 0.0012 mol NaO H per liter -. 0 .00 12 M Na{aq)
+
0.0012 M OW(aq)
Substit uting t he OH - concentration into Equation 17.1, we have K.
766
Chapter 17
I
The Chemistry of Acids and Bases
~
1.0 x 10- "
~
[H,O" ][OW]
~
[H, O" ](0.0012)
and so [H0+] = l
1 0 XI 0- 14 . 0.0012
= 8.3 X 10- 12 M
Comment Why didn't we take into account t he ions produced by water autoionization when we calculated the concentration of hydroxide ions? It should add OW and H]O + ions to t he solution. If x is equal t o t he concentration of OH - ions generated by t he autoionization of water. the n, when equilibrium is achieved, [OH -I
= (0.0012 M + OH- from water autoionization)
[OWl
~
(0.0012 M+ x)
In pure water, t he concent ration of OH - ion generated is 1.0 X 10- 1 M. le Ch atelier's principle (.... Section 16.6) suggests t hat t he concentration should be even smaller when OH- ions are already present in solution from NaOH; t hat is, x should be « 1.0 X 10- 1 M. This means x in the term (0.0012 + x) is insignificant compared with 0.0012. (Followi ng the rules for significant figures, the sum of 0.0012 and a number even smaUer tha n 1.0 X 10- 7 is 0.0012.) Thus, t he equilibrium concent ration of OH - is equivalent to the concentration of NaOH in the solution. l ast ly, what about t he Na" ion? As described later (page 774), alkali metal ions have no effect on the acidity or basicityof a solution. EXERCISE 17 .3
Hydronium Ion Concentration in a Solution of a Strong Acid
= 4.0 X 10- 3 M. Whatare t he concentrations of HlO+ and OW in t his solution at 25 °C? (Recall t hat because HCl is a strong acid, it is 100% ionized in water.) A solution of the strong acid HCl has (HCll
The pH Scale Th e pH of a sol ution is defi ned as th e negative of th e ba se- l 0 loga ri th m (lo g) of t he hydron iu m ion co n ce n tratio n (.... Sec tion 4.6, page 179 ) .
pH = - log[H,O+ ]
(4 .3 and 17.2)
In a simi lar way, we ca n define th e pO H of a so lu tio n as the n egati ve of th e ba sc-l O loga r ith m of the hydroxid e io n co nc e n tratio n .
pOH
~
- Log[OW]
(17.3)
In pu re water, t he hyd ro n iu m and hyd ro xid e Ion co ncentra tio n s a re bot h 1.0 X 10- i M. T herefore, for pure water at 25 °C
pH
=-
Log (1.0 X 10- ')
= 7.00
In th e same way, you ca n sh ow th at th e pOl l of p u re wate r is also 7.00 at 25 "C. If we take t he negat ive log a r ith m s of bo th sid es of t he expression K..,. = [ H 30 +)[O H -J , we obtain anot h er usefu l equation.
s; ~ - log x; pKw pKw
1.0 X 10- 14
~
[H,O+][OW]
= - log (1. 0 X 10- 14 ) = - log([H,O+][OW]) = 14.00 = - log ([H,O+]) + (- Log [OW]) = 14.00 = pH + pO H
(17.4)
• The pX ScaLe In general, - log X = pX, so - log K = pK - log[H,O· ] ~ pH - log[OW ] ~ pOH
T he su m of th e pH and pO H of a so lu tio n must be equa l to 14.00 at 25 "C. As illu strated in Figures 4.1 1 a n d 17.1, so lu tio ns with pl I less th an 7.00 (at 25 °C ) a re aci dic , whe re as solution s with p H g re ate r than 7.00 are bas ic. So lut io ns with p H = 7.00 at 25 °C are neutra l. 17.3
I
Water and t he pH Scale
767
CaLcuLating pH
pH
Basic
The ca lcu lat io n o f p H fro m the hydronium io n cou cc ru ra tio n , or th e co nce n tra tio n of hydronium io n co n ce n tratio n fr om pH , was introduced in Chap te r 4 (... pa ge 179 ) . Exercise 17.4 reviews th o se ca lcu latio ns.
14
Neutral
7
Acidic
14
Active Figure 17.1 pH and pOH. This figure illustrates the relationship of hydroni um ion and hydroxide ion concentrations and of pHand pOH.
Sign in at www.thomscnedu.com/ Ioqin and go to Chapter 17 Contentsto see Screen 17.4 for a simulation and tutorial on using pH and pOH .
EXERCISE 17 .4
Reviewing pH CalcuLations
Chemistry..c\Now"· Sign in at www.
(a) What is t he pH of a 0.0012 M Na OH solution at 25 °C?
th omscn ed u.cornyloqtn and go to t he
(b) The pH of a diet soda is 4.32 at 25 "C. What are t he hydronium and hydroxide ion concentratio ns in t he soda?
•
Chapter Cont ents menu to explore an inter active version of t his figure accompanied by an exercise.
(c) If t he pH of a solution containing th e st rong base Sr(OHh is 10.46 at 25 °C, what is the concent ration of Sr(OHh ?
17.4
Equilibrium Constants for Acids and Bases
In Ch a p te r 3, it was stated that aci d s and bases ca n be di vided roughly into th o se th at a re stro ng electrol ytes (su c h as H CI, H N0 3 , a n d NaO H ) a n d th o se th a t are wea k e lectrolyte s (suc h as CI I"C0 211a n d NI l,,) (Figu re 17.2) ( ~ Ta b l e 3.2, Co m m o n Acid s a nd Bases, page 132) . Hydrochloric acid is a stro n g ac id, so 100% of th e acid ionizes to pro du ce h ydron ium a n d c h lo ride io n s. In co n trast, ac e tic ac id a n d am monia are weak elect ro lytes. The y io n ize to o n ly a ve ry small exte n t in wate r, For exa m p le , for ace tic acid, th e acid, its ani on, a nd th e hydronium ion a re a ll present at equ ilib ri u m in so lu tio n, but th e ions are present in vcry low co nc e n tra tio n relative to th e acid co n ceu tra tiou . O ne \va)' to d efi n e th e rel ative stre n gths o f a se ries of ac ids wo u ld be to measure th e pH o f so lu tio ns o f acids of equ al co n ce n tra tio n : The lo wer th e pH , th e grea te r th e co n ce n tratio n o f hyd ro n ium io n , th e stro nge r th e acid . Sim ila rly, fo r a serie s
Strong Add
Weak Add
(a) HCl completely ionizes in aqueous solution.
(b) Acetic acid, CH 3C02H, ionizes onlyslightly in water.
Weak Base
(c) The weak base ammonia reacts to a small extent with waterto give a weakly basic solution.
CHl02H NH3 HCl FIGURE 17 .2 Strong and weak ad ds and bases. (a) Hydrochloric acid, a strong acid, is sold for household use as "muriatic acid." The acid completely ionizes in water. (b) Vinegar is a solution of acetic add , a weak add that ionizes onlyto a small extent in water. (c) Ammonia is a weak base, ionizing to a small extent in water. 768
C h a p t e r 17
I
The Chemist ry of Acids and Bases
of wea k bases, [O H -] will increase , and the pl I will in cre ase as the bases become stro nger. • For a stro ng monoprot ic aci d , [ H :~O +] in so lu tio n is eq ua l to th e origina l ac id concen tratio n . Sim ilarly, fo r a strong mono protic base , [O H -] will be eq ua l to the o rigin al base concen tra tio n. • For a wea k ac id , [H 30 +] will be much less th an th e origin al ac id co ncen tratio n . T hat is, [H 30 +] will he sma ller than if the acid were a strong aci d o f the same concen tra tio n. Similarly, a wea k base will g ive a smaller [O H - l th an if the base were a stro ng base o f th e same co nce n tra tio n . • Fo r a se ries of wea k mono pro tic aci ds (of th e type HA) of th e same concentrat ion , [ H:~ O + ] will in cr ease (a nd the pH will d eclin e ) as the acids become stro nger. Sim ilarly, for a series of wea k bases, [OH - ] will increase (a nd th e pl I will increase ) as the bases become stro nge r. The re lative strength of an acid or base ca n also be expressed qu an ritatively with an equ ilibriu m co nstant , o fte n ca lled an ionization constant. For th e ge ne ra l acid HA, we ca n wr ite HA(aq)
+ H20 (C) , K
•
~
H,O'(aq)
+ A-(aq)
[H,O']W] [HA]
(17 .5)
where th e equi libr ium co ns ta n t, K, has a SU bscrip t "a " to in dicate th at it is an equilib rium constant fo r an aci d in water. Fo r weak acids, th e value o f K" is less than I because the p roduct [11;10+] [A- ] is less th an the equilib riu m co nc en tratio n of the wea k ac id [HA] . Fo r a series of acids, the acid streng th in cr eases as th e value of K" increases. Similarly, we ca n write the eq uilibrium expressio n for a weak base B in wate r. Here , we label Kwith a su bscri p t "b." Its value is less th an 1 for wea k bases. B(aq)
+ H20(£) _ _ BH'(aq) + OH'(aq) K _ [BH'][OW] b [B]
(17 .6)
Som e aci ds and bases arc listed in Ta ble 17.3, each with its value o f K" o r Kh . Th e fo llowing are importan t id eas co ncerning th is table. • Acids are listed in Tabl e 17.3 at the left , an d th e ir co nj ugate bases are o n the right. • A large value of K indicat es th at ionizati on produ cts arc stro ng ly favored , whereas a small value of K in d ica te s that reac tants are favore d . • The strongest acids are at th e u p pe r left. Th ey have th e largest K" values . K; valu es beco me smaller on d escending th e cha rt as th e acid stre ngth declines. • The stro ngest bases are a t the lower rig ht. They have th e largest Kh values. Kh value s become larger Oil d escending the cha r t as base stre ng th increases. • The weaker the arid, the stronw:r its conjugate base. T ha t is, th e sma lle r th e va lue of Ka , th e larger th e value o f Kb • • So me acids or bases are listed as having K, or ~ values th at are large o r very sma ll. Acids th at are stro nger th an H 30 + arc co mp letely io n ized (HN0 3, fo r e xa mp le) , so th ei r K" values are "large." Th eir conj uga te bases (suc h as NO :~ -) do no t prod uce mea ningful co nce n trations o f 011 - io ns, so th eir Kb values arc "very small." Sim ilar argu m en ts fo llow fo r strong bases and their co njuga te acids. 17 .4
I
Equ ilibrium Constants for Acids and Bases
769
TABLE 17.3
Ionization Constants for Some Acids and Their Conjugate Bases at 25 · C
Acid Name
Acid
K.
Base
Kb
Perchloric ad d
HCLO,
large
CLO,
very small
perchlorate ion
Sulfuric acid
H2SO4
large
HSO,-
very small
hydrogen sulfate ion
Hydrochloric acid
HCl
large
Cl-
very small
chloride ion
Nitric add
HNO,
large
NO,-
very small
nitrate ion
Hydronium ion
H3O +
1.0
H,O
1.0 X 10- 14
water
Sulfurous acid
H2S0 3
1.2 X 10- 2
HS0 3-
8.3 X 10- 13
hydrogen sulfite ion
Hydrogen sulfate ion
HSO, -
1.2 x
10- 2
504
10 - 3
~
~
c
.~
~
~ u
oS
sulfate ion
1.6
7.2 X 10- 4
F-
1.4 X 10 - 11
fluo ride ion
4.5 x 10 - 4
N02-
2.2 X 10- 11
nitrite ion
Hexaaquairon(III) ion
6.3
Hydrofluoric acid
HF
Nitrous acid
HNO, HCO, H
1.8
Benzoic acid
C6 HsC0 2 H
6.3
Acetic acid
CH3C02H
1.8
Propanoic acid
CH,C H2C02H
1.3
x 10- 3
x x x x
x
12
10 - 12
dihydrogen phosphate ion pentaaquahydroxoiron(III) ion
10 - 4
HCO,-
5.6 X 10-
10 - s
C6 HsC02-
1.6 X 10- 10
benzoate ion
10 - s
CH,C02-
5.6 X 10- 10
acetate ion
10 - s
CH ,CH 2C02 -
7.7 X 10- 10
propanoate ion
10 - 6
11
10 - 9
formate ion
pentaaquahydroxoalumi num ion
[Al( H,O),P +
7.9 X
[AL(H ,O),OH]'"
1.3 X
Carbonic acid
H2CO ,
4.2 X 10 - 7
HCO,-
2.4 X 10- 8
hydrogen carbonate ion
Hexaaquacopper(II) ion
[Cu (H,O),]2+
1.6 X 10- 7
[Cu(H,O),OH]"
6.3 X 10- 8
pentaaquahydroxocopper(II) ion
~
Hydrogen sulfide
H,S
1 X 10- 7
HS-
1 X 10 - 7
hydrogen sulfide ion
~ ~
Dihydrogen phosphate ion
H2P04 -
6.2 X 10- 8
HPO/ -
1.6
10- 7
hydrogen phosphate ion
.~
Hydrogen sulfite ion
HS0 3-
6.2
50/ -
1.6
10- 7
sulfite ion
Hypoch lorous acid
HCLO
3. 5 X 10 - 8
CLO -
2.9 X 10 - 7
~
'c -c
x
1.3 X 10-
7.5 X
[Fe(H,O),]'-
c;
'"
8. 3
10 - 13
H2 P0 4 -
H3 P04
-5 Hexaaquaaluminum ion
2-
[Fe(H,O};OH]' +
Phosphoric acid
Formic acid
~
Base Name
x
10 - 8
x x
hypochlorite ion
[Pb(H,O) ,]' +
1.5 X
[Pb(H,O),OH] +
6.7 X 10-
[Co( H,O),]'+
1.3
[Co(H,O),OH] "
7.7
x
6
Boric acid
B(OH),(H,O)
7.3
x x
10- 9 10 - to
B(OH), -
1.4 X 10- s
tet rahydroxoborate ion
Ammonium ion
NH 4 -
5.6 X 10 - 10
NH,
1.8 X 10- s
ammonia
HCN
4.0 X 10-
CN -
2.5 X
lO- s 10 - s
Hydrocyanic acid
10
10 - 10
10-
pentaaquahydroxolead(II) ion pentaaquahydroxocobalt(II) ion
cyanide ion
Hexaaquairon(II) ion
[Fe(H,O),]' -
3.2 X
[Fe(H,O};OH],
3.1 X
Hydrogen carbonate ion
HCO, -
4.8 X 10- 11
CO/ -
2.1 X 10- 4
carbonate ion
Hexaaquanickel(II) ion
[Ni(H,O),]' -
2.5 X 10- 11
[Ni(H,O),OH )-
4.0 X 10- 4
pentaaquahydroxonickel(II) ion
Hydrogen phosphate ion
HPO/ -
3.6 X
10 - 13 10 - 14
P0 4 ' -
2.8 X
10- 2
pentaaquahydroxoiron(II) ion
phosphate ion
Water
H,O
1.0 X
OW
1.0
hydroxide ion
Hydrogen sulfide ion*
HS -
1 X 10- 19
S' -
1 X lOS
sulfide ion
Ethanol
C2HsOH
very small
C2 HsO -
large
ethoxide ion
Ammonia
NH,
very small
NH 2 -
large
amide ion
Hydrogen
H,
very smaU
W
large
hydride ion
"T he value s of K, for li S a nd Kb fo r S'.I ar e esti m ates .
770
Chapter 17
I
The Chemi stry of Acids and Bases
~ ~
c;
~
7
Hexaaquacobalt(II) ion
~
c;
~
10- 8
Hexaaqualead(II) ion
-5
~
u
oS
Problem Solving Tip 17.1
Strong or Weak?
How can you tell whether an acid or a base is weak? The easiest way is to remember those few that are strong. All others are probably weak. Strong acids are: HydrohaLic acids: Hel, HBr, and HI
Group 2A hydroxides: Sr(O H), and Ba(OH),
Nitric acid: HN03
Sulfuric acid: H2S0 4 (for loss of first H+ only)
[Mg(O Hh and Ca(O Hh are not considered
Perchloric acid: HCl0 4
strong bases because they do not dissolve
Some common strong bases include the following:
appreciably in water.]
All Group lA hydroxides: liOH, NaO H, KOH, Rb OH, ( sOH
(but not HF)
To illu strate so me of th e se id e a s, let u s co m pa re so me co m m o n ac ids and base s. For exa m p le , HF is a stro nge r ac id th an H eIG , wh ic h is in turn st ro nge r th an I1CO, -,
Decreasing acid strength HF Ka = 7.2 X 10 - 4
HCO,
HClO
Ka
= 3.5 X 10- 8
«, =
4.8 X 10- 11
and th eir co nj ugate bases bec ome stro nge r from F- to ClO- to CO :\2-.
Increasing base stre ngth F
CIO t; = 2.9 X 10- 7
Kb = 1.4 X 10- 11
co,' Kb
= 2.1 X 10- 4
Na tu re abou nds in acids and bases (Figu re 17.3). Ma ny natu rall y occu rrin g acids are based on th e ca rboxy l gro u p (-C0 2 I1) , and a few are illu strated he re . Notice tha t th e organ ic portion o f the molecule has all effect o n its rel ative stre ng th (as d escribed fur ther in Sec tio n 17.10 ).
==============:: :I• •••~
Propanoic acid, CH 3CH 2C0 2H Kb of
conjugate base increases
Ka = 1.3 X 10- 5
Acetic acid, CH 3C0 2H Ka = 1.8 X 10- 5
Ka increases; acid strength increases
Formic acid, HC0 2H Ka = 1.8 X 10 - 4
7) because P04] - is the conjugate base of t he weak acid HPO/ -. The K+ ion, Like the Na + ion, does not affect the soLution pH.
(c) FeCh: An aqueous soLution of FeCt2 shouLd be weakly acidic (pH < 7). The Fe2 + ion in water, [Fe(H 20)6F +, is a Bransted acid. In contrast, ct- is the very weak conjugate base of the strong acid Hct, so it does not contribute excess OH - ions to the soLution. (d) NaHC0 3: Someadditional information is needed concerningsaLts of amphiprotic anions such as HC0 3- . Because they have an ionizable hydrogen, they can act as acids. HCO ,-(aq) + H,O(C) CO,'-(aq) + H,O-(aq) They are also the conjugate bases of weak acids.
Ka
= 4.8
HCO ,-(aq) + H,O(C)
Kb
=
H,CO ,(aq) + OH-(aq)
X 10- 11
2.4 X 10- 8
Whether the soLution is acidic or basic will depend on the reLative magnitude of Ka and Kb• In the case of the hydrogen carbonate anion, Kv is larger t han Ka- so [OH -] is larger than [H 30 +], and an aqueous solution of NaHCO] will be slightLy basic. (e) NH 4F: Whathappens if you have a saLt based on an acidiccation and a basic anion? One exampLe is ammonium fluoride. Here, the ammonium ion would decrease the pH, and the fLuoride ion wouLd increase the pH. NH, ·(aq) + H,O(C) F-(aq)
H,O'(aq)
+ NH ,(aq)
+ H,O(C) Kb (F-), the ammonium ion is a stronger acid t han the fluoride ion is a base. The resuLting soLution should be slightlyacidic. CommentThere are two important points to notice here: Anions that are conjugate bases of strong acids-such as tr and NO l - -have no effect on solution pH. Fora saLt that has an acidiccation and a basicanion, the pH of the soLution will be determined by the ion that is the stronger acid or base of the two. 774
Chapter
17
I
TheChemist ry of Acids and Bases
Problem Solving Tip 17.2
Aqueous Solutions of Salts
Because aqueous solutions of saLts are found in our bodies and throughout our economy and environment, it is important
to know how to predict thei r acid and base properties. Information on t he pH of an aqueous solution of a salt is summarized
in Table 17.4. Co nsider also the following examples:
Cation
Anion
pH of the Solution
From strong base (Na")
From stro ng acid (Ct")
From strong base (K+)
From weak acid (C H3C02 -
= 7 (neutral)
> 7 (basic)
)
From weak base (NH4 +)
From strong acid (Cl -)
< 7 (acidic)
From any weak base (BW)
From any weak acid (A-)
Depends on relative strengths of BW and A-
EXERCISE 17 .6
Add-Base Properties of Salts in Aqueous SoLution
For each of the following saLts in water, predict whether t he pH will be greater than, less t han, or equal to 7. (a) KBr
(c) AICl,
(b) NH,NO,
(d) Na ,HPO,
A Logarithmic Scale of Relative Add Strength, pKa Man y che m ists an d bioch emists use a logarith mic sca le to report an d co m pa re rel ative ac id streng ths . px,
=-
log K.
(17.7)
Th e pKa of an ac id is th e negative log of the Ka value (ju st as p l I is the negative log o f th e hyd ron iu m ion co nce n tra tio n) . For ex am ple, ac etic acid has a pKa value of 4.74. pK. ~ - log (1.8 X 10 - 5 ) ~ 4.74 Th e pKa value becomes smaller as th e acid stre ng th in creases. - - Acid strength increases
~
Propanoic add
Acetic add
Formic add
CH 3CH zC0 2H
CH3CO zH
HCOzH
Ka = 1.3 X 10- 5
Ka = 1.8 X 10- 5
K. = 1. 8 X 10 - 4
pKo = 4.89
pKa = 4.74
pK. = 3.74
f--
EXERCISE 17.7
pKo increases - -
A Logarithmic Scale for Add Strength, pK.
(a) What is the pKa value for benzoicacid, C6H sC0 2H? (b) Is chloroacetic acid (CLCH 2C0 2H). pKa = 2.87. a stronger or weakeracid than benzoicacid? (c) What is the pKa for the conjugateacid of ammonia? Is this acid stronger or weaker than aceticacid?
Relating the Ionization Constants for an Add and Its Conjugate Base Let us look aga in at Tabl e 17.3. Fro m the top of the tab le to th e bottom , th e streng ths o f th e ac ids d ecl in e ( K1. becomes smalle r), and th e stre ngths o f their conj ugate bases increase (the values of Kh increase ) . Exa m in ing a few cases shows 17 .4
I
Equilibrium Constants for Acids and Bases 775
• A Relation Among pK Values A useful relationship for an ad d-conj ugate base pair can be derived from Equation 17.8. pK" = pKa + pKb
th at th e product of K, fo r an aci d and Kh for its co njugate base is eq ua l to a co nsta n t, spec ifica lly K,·. (17 .8)
Co nsid e r th e spec ific case o f th e ion izati on of a weak acid, say H CN, and th e interactio n o f its co nj ugate base , CN - , with H 20 . HCN(aq) + H,O(e) CN- (aq) + H,O(f )
Weak acid: Conjugate base:
2 H,O(e) ,
~
==' H]O+(aq) + CN- (aq)
Ka = 4.0 X 10 - 10
= 2.5 X 10- 5 x; = 1.0 X 10- 14
, HC N(aq) + OW(aq)
Kb
H]O +(aq) + OW (aq)
Ad di ng th e equatio ns gives th e che mical eq uatio n fo r th e au to ionizatio n o f wa ter, and th e numeri cal va lue is indeed 1.0 X 10- 14 . T hat is,
= ( [H]O+][W
K X K
[HeN]
.b
= [HO+][OWj = K ] w
l)( [HeIJ] [OWl) ~l
Equati on 17.8 is useful becau se K h ca n be ca lcu late d fro m K a• T he value of Kb fo r the cyan ide ion , fo r exa m ple , is Kb for CN
EXERCISE 17.8
-
~
Kw - = -
1.0 X 10- 14
K. for HCN
4.0 X 10
~
2.5 X 10- 5
10
Using th e Equation K, X Kb = K.
K, for lactic acid, CH 3 CHOHC0 2H , is 1.4 X 10- 4 • Wh at is Kv for the conjugate base of th is acid, CH 3CHOHC0 2 - ? Where does t his base fit in Ta ble 17.3?
17.5
Predicting the Direction of Add-Base Reactions
Acco rdi ng to th e Bren stcd-Lowrv th eo ry, all ac id-base re actions ca n be writte n as equ ilib ria involvin g the acid and bas e and thei r co nj ugates . Acid + Base
• K and Product- and Reactant-Favored Reactions Reactions with an equilibrium constant greater t han 1 are said to be product-favored at equilibrium . Those with K < 1 are reactant-favored at equilibrium.
==' Conjugate base of t he acid + Conj ugate acid of the base
In Sec tio n 17.4, we used equ ilib riu m co ns tan ts to provid e qu antitative information about the re lat ive streng ths of aci d s and bases. Now we want to show ho w the co nsta n ts ca n be used to d ec id e whe ther a particul ar ac id - ba se reaction is produ ct- o r reac ta n t- favo re d at equilibrium . Hydrochloric ac id is a stro ng aci d . I L"i eq u ilib riu m constant fo r reaction with wate r is very large , with th e eq uilib riu m e ffec tively lyin g co m plet ely to th e rig h t. HCl(aq) + H,O(e) , H]O' (aq) + (l -(aq) St rong acid (= 100% ionized), K » 1 [H30+] = initial concentration of the acid Of the two aci d s here , H Cl is stro nge r th an 11 30 +. Of the two bases, 11 20 an d CI- , wate r is the stro nger ba se and wins out in th e co m pe titio n for th e proton . T h us , th e eq uilibriu m lies to the side of th e ch em ical eq uatio n havin g th e wea ke r acid and base . conju gate pair 1
I
HCl(aq) Stronger acid thanH3O+ 77 6
Chapter 17
I
The Chemist ry of Acids and Bases
conjugate pair 2
+
I
I
H,O(C) Stronger base thanCL-
K>l
H3O +(aq) Weaker acid than HCL
+
I
cq aq) Weaker base than HzO
In co n tras t to H Cl and other stro ng aci ds , ace tic acid , a weak acid , io ni zes to only a very sm all ex ten t Cra ble 17.3). CH lCO, H(aq) + H,O(f )
, HlO+(aq) + CH lCO, -(aq)
100% ionized), K ~ 1.8 X 10- 5
Weak acid «
[H3 0 +] < < initial concent ration of th e acid
Wh en equili bri um is ac hieved in a 0 .1 ~I aq ueous so lu tio n o f CH 3C0 2 H, the co nce n trat ion s of II , O +(aq ) and CII ,CO, -(aq) are each o n ly abo u t 0.00 1 M. Ap p rox imately 99 % of th e ace tic ac id is no t io nized. conjugate pair 1 conjugate pair 2
r
CHlCO,H(aq)
Weaker acid than H,O+
+
r
H,O(C)
Weaker base than CH3C0 2 -
--
K 1, the p Ka inc reases bv abo ut 5 for each sucresxive loss o f a p roto ll.
794
Chapter 17
I
The Che mist ry of Acids and Bases
o
H'-... /I" O-N
..
HN0 3 • st rong acid, pKa = - 1.4
\
q:
HN0 2• weakacid. pKa = + 3.35
A Closer Look
Acid Strengths and Molecular Structure
Although predictions about acid strength in aqueous solution are fairly simple to make, explanations are in fact quite complicated. Acid strength is sometimes correlated with
H· (g)
the strength and the polarity of the H-X
bond, readily identifiable characteristics
J.bondW( HX)
derived from the structure of the acid, the
(endothermic)
+
Slep 3 IE for H(endothermic) X·(g) EA forX(exothermic) Slep 4
W(g)
+
Slep 2
reactant in the ionization process. We need to
Slep 5
Slep 6
.1.solvationHo(H+)
.1.solvationHo(X-)
(exothermic)
(exothermic)
point out, however, that when one is assess-
ing any chemical reaction, it is necessary to consider bot h reactants and products. Looking only at the reactant when dealing with acid dissociation only ta kes you halfway. When evaluating the strength of an acid HX(aq). we are looking at the following reaction HX(aq)
HX(g)
- J.solvationHo(HX)
(endothermic)
HX(aq)
+ H,O(f) ==" H,O+(aq) + X-(aq)
To fully explain the extent of ionization, we must consider characteristics of bot h the acid and t he anion. The ability of the anion to spread out the negative charge across t he ion, for example, and the solvation ofthe anion by the solvent are among the issues t hat must have some relevance in an explanation of the st rengt h of the acid. How ent halpy changes contribute to acid strength can be assess ed using a the rmochemical cycle (such as t he one used to evaluate a lattice enthalpy, page 599). Conside r the relative acid strengths of the hydrogen halides. The entha lpy change for the ionization of an acid in water can be related to ot her enthalpy changes as shown in the diagram. The solvation of H+ (Step 5) and the ionization energy of H(g) (Step 3) are common to all of t he hydrogen halides and do not contribute to the differences among hydrogen halides, but t he four remaining terms are dif-
Slep 1
+ H20(e)
An energ y diagram for the ionization of an ad d HX in aqueous solution .
ferent. For the hydrogen halides, the bond dissociation entha lpy (Step 2) of HF is much larger than the dissociation ent halpies of the other hydrogen halides. However, it is compensated for significantly by the ent halpy of solvation of the anion (Step 6), which for the fluoride ion is much more exothermic than the solvation energies of the ot her halide ions. The electron affinity is also a contributor to the differences in overall ent halpy cha nges. Electron affinity values vary among the halogens, but to a smaller extent t han the variation in bond energy and the enthalpy of solvation ofthe halide ion. Differences in solvation energy for the molecular species (Step 1) are minimal. A complete analysis of strengths of acids in aqueous solution will include consideration of both enthalpies and entropies. Ent ropy has
Le t's apply the bo nd enthalpy/e lec tro n affinity analysis to Il N0 - - Increasing acid strength
yet to be discussed in t his text in detail (.... Chapter 19) but we have noted earlier ("4 page 621) that entropy plays an important role in solution chemistry, specifically in determining solubilities. It is not surprising that entropy has a role in determining acid strength, too . Indeed. differences in entropy changes are significant in accounting for the differences in acid strength of the hydrogen halides. Although aUof these terms contribute to acid strength, acid strength can often be correlated with a subset of this information, as can be seen by the examples presented in this section. We point out, however, that correlations, while highly useful to a chemist because they can be used to make important predictions, are at best only partial explanations.
3
a nd Il N0
I
Molecular Structure, Bonding, and Acid-Base Behavio r 795
2.
~
HNO z
pK,
+3.35
- 1. 4
H-O bond strength (kJjmol)
328
423
Electron affinity of X(kJjmol)
- 219
- 377
Sum (kJ/mal)
10 9
46
17.10
Nit rous acid, HN0 2
Nit ric acid. HN 0 3
FIGURE 17.11 Electrostatic potential surfaces for the nitrogen cxcactds. Bot h surfaces showthe O-H bond is quite polar. More importantly, calculations show that the Hatom becomes mo re positive as more 0 atoms are added to N, and the O-H bond becomes even more polar.
Partial Charges
o atom Molecule
H atom
ofOH
N atom
HN0 1
+ 0.23
- 0.28
+ 0.34
+ 0.27
- 0.37
+0 .66
HN0 3
As in th e case of the Grou p 7A ac id s, we agai n see that acid streng th correlates with th e su m of th e bon d bre akin g energy a nd X g ro u p (N O z or NO j ) el e ct ron afflnl ty, Our analysis of the Grou p 7A acids showed that, in this series, th e II - X bo nd strengt h was th e more impo rt ant fac tor; as th e H- X bond becam e stro nge r, th e aci d became weaker. I Io wever, a g lance at th e data fo r I IN0 3 and I-I N0 2 shows this is not true he re. The 0 - 11 bo nd is stro nger in the stro nge r ac id II NO:\. T he electr on affin ity term is th e more important term in this correlatio n. T hese same e ffec ts are o bserved fo r other ox yacids suc h as th e chlorine-ba sed o xo acids 1I0 Cl < II O CIO < II O CIO , < IIOCIO , and the S-based oxoacids (Table 17.7) . lIo w is this to be interp ret ed ? In II N0 3 , there are two othe r oxygen ato ms bo nd ed to the ce ntra l nitroge n atom , whereas in II NO z o n ly o ne other ox yge n is bon d ed to the ni troge n ato m. ato ms to nitro gen , we are in cre asin g th e By attac h ing m o re el ectron egative electron affinity o f the group attac hed to h ydrog en , an d an ythi ng that increases the electron affin ity of the X grou p sho uld also m ak e II X a stronger ac id an d X- a wea ker co nj ugate base . T his is ano the r way of saying th at if X - has a way to accom modate an d sta bilize a negative charge, it will be a wea ker conjuga te base . In th e case o f oxoacid s, ad d itional o xyge n ato ms have th e effect o f stabilizing t he an io n becau se the negati ve cha rge o n th e an io n ca n be di spe rse d over m ore atoms. In the n it rate ion , for e xample , the negative charge is shared eq ually over t he th re e ox yge n ato ms . Th is is rep resen ted sym bolically in th e th re e resonan ce str uc tu res fo r this io n.
°
[ .)~. ] -.. -.- . : .0.
Correlation of Atom Formal Charge and pK,
TABLE 17 .8
Central atom formal charge
pK.
+2
- 3
:O-H
I
: 0- 5- 0 :
I
:O-H
: 0- 5- 0 :
+1
1.92
: O- Cl- O- H
+1
2
: Cl-O-H
0
7.46
796 Chapter 17
I
r. [ .)~. ].0.
0: ..
: .0.
0 ..
In mtntc ion , o n ly two a to ms share th e negat ive cha rge. T here fore, N O z- is a stro nger conj ug ate base than N0 3 - . In ge neral, grea ter stabiliza tion of the pro duct s fo rm ed by io nizing the ac id con tributes to increased acid ity. A g lance at Ta ble 17.7 and Figure 17.11 show s th at another e m piric al co rre la tio n can be mad e be tween th e structu re o f an acid and its acid ity: In a series o f re la te d aci ds, the larger the fo rm al charge on the ce ntral atom , the stro nge r th e aci d (Ta ble 17.8) . Fo r exam ple , th e N atom formal cha rge in th e wea k acid H NO z is 0, whe reas it is + I in the strong acid II NO :h and these a re refl ected by th e res ults of theore tical ca lcu lations ci ted in Figu re 17.11 . In su m mary, molecules suc h as th e oxo acid s ca n beh ave as stro nger Bre nste d aci ds whe n the anio n created by lo ss o f 11+ is stable and able to accom modate the negative charge. Th ese con d itio ns are p rom o ted by • the presence o f elec tronegative ato ms at tac he d to th e ce ntral atom . • th e po ssib ility of reson an ce struc tu res for th e an io n, wh ich lead to dc localizatio n o f th e negative charge over the anio n and thus to a sta ble io n.
H : O- H
I I
0: ..
Why Are CarboxyLic Acids Brensted Acids? Th ere is a larg e class o f o rg an ic aci d s, typ ifie d by ace tic aci d (C II 3CO zII) (Fig u re 17.1 2 ) , ca lle d carboxylic ac id s because a ll have the carbox ylic acid gro up , - CO zH. T he arg u m e n ts u sed to ex p la in th e acidity o f oxoacids can also be ap-
The Chemist ry of Acids and Bases
pli ed to ca rboxy lic ac ids . The 0 -11 bond in th ese co m po u n ds is pol ar, a prerequisit e for ioni zation. H : 0: ( - H bon~s
H : 0:
I II
~
> - C- C- O- H
I
not broke/ in water
I II + I ..
H-C -C -O :
.. 1
H30 +
H
H
FIGURE 17.12 Electrostatic potential surface and partial charges for acetic add. The Hatoms of the molecule are
Polar O- Hbond broken by interaction of positively charged H atom with hydrogen-bonded H20
In ad d ition, ca rboxylate an io ns are sta bilize d by delocali zin g th e negative charge over th e two o xygen at o ms. H : 0:
[
H-
I II .. ( - ( - O: I ·· H
]
[ ..]H :0 :
~
I I
I .. ··
H- ( - ( =O
H
aU positively charged. but the H atom of the OH is much more highlycharged. As expected. both of the electronegative o atoms have a partial negative charge. The table below givesthe computercalculated partial charges on the acid. Atom or Group
The sim ple ca rboxylic acids, RC0 2I I in which R is a hydrocarbon grou p (.. Section 10.4) , do not differ markedly in acid streng th (co mpare aceti c acid, pKa = 4.74, and p ropanoic acid, pK" = 4.89 , Tabl e 17.3). The acidi ty of carboxylic acids is en hanced, however, if electron egative su bstitue n ts repl ace th e hydrogens in the alkyl grou p . Co m pa re, for exam ple, th e pKa values o f a series o f aceti c acids in which hydrogen is repl aced seque n tially by the more electro negative elemen t chlo rine.
Calc'd Partial Charge
H of OH
+ 0.24
o of OH
- 0.32
H of CH3
+0.12
Add
CH3C0 2H ClC H2C0 2H Cl 2CHC0 2H Cl3CC0 2H
Acetic acid Chloroacetic acid Dichloroacetic acid Trich loroacetic acid
4.74 2.85 1.49
increasing acid strength
0.7
As in the nitrogen o xo acids, in creasin gly elec tr o nega tive su bstit ue nts lead to an in crease in acid stre ng th. Indeed , recent researc h has found th at a key ro le o f e lectro negative su bstituen ts is to sta bilize th e negati ve charge of th e an io n . T hat is, co m pared with CI I3 C0 2 "", th e CI3CC 0 2 - an io n is stabilized by the presen ce of th e th ree CI ato ms, and CI3CC 0 2 - is a wea ke r base th a n CI I3 C0 2 ". Finally, why a re th e C- H hydrogen s of ca rbo xylic acids not di ssoci at ed as H + in stead of (or in add itio n to ) th e 0 -11 hydrogen ato m? The ca lcu late d parti al positive charg es listed in Figure 17.12 show th at the H ato ms of th e e lI 3 gro u p hav e a mu ch smalle r po sitive partial ch arge than th e O -H hydrogen ato m . Fu rth ermore , in ca rboxylic acids , th e C ato m o f th e CII 3 g ro u p is not sufficie n tly e lec tr o nega tive to acc ommodate th e negati ve charge left if th e bond break s as C- H - ? C:- + H +, and the product ani on is not well sta bilized.
Why Are Hydrated Metal Cations Brensted Acids? wh en a co o rd ina te covale n t bon d is form ed between a met al catio n (a Lewis aci d) and a water molecul e (a Lewis base ), th e po sitive charge of th e m etal ion and its sm all size means that th e electrons of th e H 20 -Xl ll + bond are very stro ng ly at-
17.10
I
Molecular Structure, Bonding, and Acid-Base Behavior
797
• Polarization of o-H Bonds Water molecules attached to a metal cation have strongly polarized O-H bonds.
HH
I
tracted to the metal. As a resul t, the O - II bonds of the boun d water mo lecules are polarized , just as in ox oa cids and carboxylic acids. T he net effect is tha t a II ato m of a co o rd inated wate r mole cule is removed as H + mo re re adi ly than in an un co o rdi nated water mo lecule. Thus, a hydrated metal catio n fu nction s as a Bro nstc d acid or p ro ton do no r (Figu re 17.8) .
: O- Hs +
(H20hM "+ -
[Cu (H20),]'+ + H20 (f ) ==" [Cu( H20 ), (O HW(aq) + H,O +(aq)
h-
T he effe ct of th e met al ion increase s with increasing charge . Co nsu lting Ta ble 17.3, you see that the Bransrcd acidity of + 3 ion s (fo r example I\IH and Fe H ) is g rea ter than for + 2 cations (Cu H , Pb 2+, C 0 2+, Fe 2+, NiH ) . Ions with a sing le positive charge suc h as Na " and K+ are not acidic. (T his is sim ilar to th e effect of ce n tra l atom formal cha rge in a series o f re lated acids. See Tab le 17.8.)
Why Are Anions Brensted Bases? TABLE 17.9
Basic Oxoanions
Anion
pK,
P0 43-
1.55
HPO/ -
6.80
H2 P0 4 -
11.89
C032-
3. 68
HCO,-
7. 62
50l -
6.80
HS0 3 -
12.08
An ion s, particul arl y oxoanions such as P0 43 - , are Bronsted bases. T he negative ly charge d an ion in te rac ts with th e positively charged H ato m o f a polar water m olecu le , and an II + ion is transferred to the anion .
:0: 3-
H
I
:
I
I
: O-P -O :" 'H- O: ,
.. I .. \ :a:
0: 2-
: O-P -O -H
..
.. I ..
+
OW
: a:
Hydrogen bond with water. H+ ion moves to phosphate ion.
T he d at a in Table 17.9 show that, in a series of related an io ns, the basicity of an anio nic base increases as the negative charge o f the an io n inc reases.
Why Are Ammonia and Its Derivatives Brensted and Lewis Bases? Ammo nia is th e paren t co mpound of an enormous n u mber of co mpounds that be have as Bransrcd a nd Lewis bases (Figu re 17.13) . T hese mo lec u les all have an elec tro nega tive N ato m with a part ial neg ative charge su rr ou nd ed b y three bo nd s and a lon e pair o f electron s. Owing to thi s negatively charg ed N atom , they can ex tract a proton fro m water.
I
w
H
H
\
I
I
R-N - H + OW
R- N:- "'H- O: ,
~
I
..
R
Hydrogen bond with water. H+ ion moves to Natom.
H2
FIGURE 17.13 Nitrogen-based
Lewis and Bransted bases. All have an Natom surrounded by three bonds and a lone pair of electrons.
/ N. CH; \ ""CH, CH3
H C HC/ ' ""' CH
II
Trimethylamine 798 Chapter 17
I
I
HC....... 'l"'CH N
The Chemist ry of Acids and Bases
Pyridine
C H2C--- \ I CH2
~
HC/ ' "'" C...- }
II
I
H
HC....... 'l"'CH N Nicotine
- (b) HSO .,- + H, O --> (c) Il ,O ' + F- --->
4• • \ Vhat are th e p roducts o f each o f th e foll owing aci dbase re actio ns? Indi cat e th e acid and its co nj uga te base an d the base an d its conjug ate aci d. (a) HCIO , + H, O --> (b) N H 4 + + Ih O ----+ (c) HCO, - + O H - --> 5. Wri te bala nced eq u ation s showing how th e hyd ro g e n o xal ate io n , H C.~ 0 4 - , ca n he bo th a Bre n sted ac id an d a Bro nstcd base , 6. Wri te ba lan ce d eq u ation s showing how the HPO / - io n of sodium h yd rog e n ph o sph at e , Na2H P0 4, ca n he a Bron stcd acid o r a Bren sted ba se . 7. In each o f th e fo llow in g acid-base re ac tion s, id en tify th e Bronsted aci d an d base o n th e le ft an d thei r co nj uga te pa rtne rs on th e rig ht. (a) Il CO,II (aq ) + 11, 0 (£) , II CO , - (aq ) + II , O ' (aq ) (b) NH,(aq) + H,S (aq )
pro ca ine
[IH : " II ,,,N,o, ]' Cl - (aq ) novocaine
T he pK" for novoca in e is S.H5. Wh at is th e pH of a 0.0015 M solu tion of novocain e?
I
805
88 . Pyridin e is a weak o rganic base a nd re adily form s a salt with h ydroc hl o ri c acid. C, H ,N( aq) + HCI(aq ) - > C; H ; N H+(aq) + Cltaq ) pyridin e pyridinium ion Wh at is the pH of a 0.025 M so lu tion of pvridini um h yd ro c hloride , [C "H"N H+ ]Cl-? 89 . • T h e base c thylam inc (C H jC HtN H t) has a Kb of 4.3 X 10- 4 . A clo sel y rela ted base , e th an o lam ine (H OC H t C H 2N H t), h as a K h of 3.2 X 10- 5 . (a) Which of th e two bases is stronger? (b) Calcu late th e pH of a 0.10 M so lu tion of th e stro nge r base.
90. Chl oroacetic aci d, CICll tC O t ll , is a m ode rat el y wea k acid (K;, = l AO X 10- 3 ) . If yo u di sso lve 94.5 mg of the aci d in water to give 125 m L of so lu tio n, wh at is the pH of the so lu tio n? 91. • Sac cha rin (HC 7 H 4N O:1S) is a wea k aci d with pK" = 2.32 at 25 "C. It is used in the fo rm of so d ium sa ccharide , NaC 7 H tN O :1S. Wh at is the pH of a 0.10 M so lution of sodiu m sacc haride a t 25 CC ?
Kb l is 7.0 X 10- 7 and K b2 is 1. 1 X 10- 10 • Ca lcu la te th e ap prox imate pl Lof a 0.020 M so lu tio n.
95. • O xali c acid is a relatively we ak dip rotic ac id . Calcu late the equilibriu m co nsta n t for th e reactio n shown below from K» and K t. (Sec Appendix H for th e require d K. va lues.) fl ,C ,O,(aq ) + 2 fl ,O(t) ,
) C,O / - (aq ) + 2 fl ,o +(aq )
96. .A. The e q u ilibriu m co ns tan t fo r the react ion of hydroc h lori c ac id a nd a mmonia is 1.8 X 109 (page 779 ). Co nfir m thi s value . 97. .... T h e eq ui libriu m cons tan t for th e reac tion of fo r.. m ic acid an d so d iu m h yd ro xid e is 1.8 X lO lO (page 779). Co nfirm th is value. 98. . .... Ca lcu late the pH o f the so lu tio n th a t res ults fro m m ixi ng 25 .0 mL of 0.14 M for m ic ac id an d 50.0 mL of 0.070 ~1 so d iu m hyd ro xid e . 99. . .... To what vo lu me shou ld 1.00 X lOt m l. of a ny weak ac id , Hi\, with a co nce n tra tion 0.2 0 M be d iluted to double th e percen ta ge ionizat ion ? 100. .A. T h e hyd ro ge n p htha lat e io n , CRH"Oj- , is a we ak acid with K" = 3.9 1 X 10- 6 •
saccharin
92. • Given the fo llm...·ing so lu tions: (a) o.! M N H, (c ) o.! M N H ,CI (h) o.! M Na,CO , (I) o.! M NaC fl ,C O, (e ) o.! M NaCI (g) o.! M N H,C H,CO , ( Il ,,\ (~),
8
acidic
106. • A hydrogen ato m in the organic base pyridin e, C:; lI 5~ ' ca n be su bstituted by various ato ms o r g ro u ps to gin' XC"H 1l'\', where X is a n a to m suc h as CI o r a g ro u p suc h as C H j . The follow ing table givl's K, values for th e co nj ugate ac ids of a varie ty of su bsti tut ed pvr idincs.
(a) Describe ea ch product as a Lewis aci d or a Lewis base . (b) If yo u p lan' 1.00 g of the co m p lex in a 565-m L flask a t 25 °C. what is tlu- to tal pressure in till' flask? Wh a t a n' the pa rti al pressures of the Lewis aci d . the Lewis base , a nd tlu- co m p lex?
109. A Su lfan ilic ac id , whi ch is used in making dve s, is made by re act ing an iline with sulfu ric ac id .
w I
~ [H,o+][CH,CO,- ]
[CH,CO,Hj
(x)(0.600 + x) 0.700 - x
As explained on page 814, the value of x will be very small with respect t o 0.700 or 0.600, so we can use the "approximate expression" to find x, t he hydronium ion concentratio n. K ~ 1 8 X 10- > ~ [H,O+][CH,CO,-] ~ (x)(0.600)
,
.
[CH,CO,H]
0.700
x = 2.1XIO- s M
pH
~
- log (2.1 X 10- »
~
4.68
Comment The pH of t he buffer has a value between t he pH of 0.700 Macetic acid (2.45) and 0.600 M sodium acetat e (9.26) .
EXERCISE 18.3 pH of a Buffer Solution What is the pH of a buffer solution composed of 0.50 Mformic acid (HC0 2H ) and 0.70 Msodium format e (NaHC0 2 )?
General Expressions for Buffer Solutions In Example 18. 2, we fo u nd th e hyd ro nium ion co nce n tratio n of t he ace tic acid / acetate ion buffer so lu tio n by so lving for x in th e eq ua tio n K = 1 8 X 10- 5 = [H,O+][CH,CO,- ] = (x)(0 .600) , . [CH,CO,Hj 0.700
If thi s equation is rearran ged , we have a very useful eq uatio n th at can hel p yo u better un d erstand how a buffe r works. [H0+] = [C H,CO,H] X K a [CH,CO, ] e
• Buffer Solutions You will find it generally useful to consider all buffersolutions as composed of a weak ad d and its conjugate base. Suppose, for example, a bufferis composed of the weak base ammonia and its conjugate add ammonium ion. The hydroniu m ion concentration can be found from Equation 18.1 by assuming t he buffer is com posed of the weak acid NH4 + and its conjugate base, NHl • 816
Chapter
18
I
T ha t is, the hydroge n io n co ncen tra tio n in th e ace tic acid /ace tat e ion bu ffer is give n by the ratio of the acid and co nj ugatc base co nc en tratio ns timcs the ac id io nization constant. Ind eed , this is true for all so lutio ns of a weak ad d and its conjugate base. [H,O+] =
[acid] XK [conjugate base] a
(18.1)
It is often co nven ien t to use Equat ion 18.1 in a different fo rm . If we take the nega tive logari th m of each sid e of the eq uation, we have - log[H,O+]
= {-
Princi ples of Reactivity: Other Aspects of Aq ueous Equilibria
lOg
. [acid] } [conjugate base]
+ (- logK.)
You know that - log [H 3 0 +] is d efi ned as pl l. and - log Ka is d efi ned as pKa (.... Se ctio ns 17.3 and 17.4 ) . Fu rthermo re , becau se - 1og
[acid] [conjugate base]
~
,---"j-,ug",a-,-te=-b=-a-,-se"l +1og-.[ c,-on [acid]
th e precedi ng eq uatio n can be rewr itten as p
H= K
px,
+ 10 [conjugate base] 9
(18.2)
[acid]
This equatio n is known as the H enderson- H asselbalch eq ua tion. Bot h Eq uatio ns 18.1 and 18.2 show th at th e pi I of a buffer so lu tion is con tro lled by two fac to rs: the streng th of the ac id (as expressed by K", or p Ka ) a nd the rel ati ve amo u nts of aci d and co nj ugate base . Th e so lutio n pH is es tablishe d p ri marily by th e value of K, or pKa , and the pH is fin e-tu ned by adj usting the aci d-to-co nj ugate base ratio. \Vhe n th e co nce n tra tio ns o f co nj ugate base and acid are th e same in a so lu tio n, th e ra tio [co nj ugate base ]/ [acid ] is I. T he lo g o f I is ze ro, so pl l = pKa u nd er these circu ms ta nces. If there is more of the co nj ug ate ba se in the so lu tio n than aci d, fo r exam ple, th e n pH > pK". Co nve rse ly, if th ere is more aci d th an co nj ugate base in so lu tion, th en p l I < pKa .
• The Henderson-Hasselbalch Equation Many handbooks of chemistry list acid ionization constants in terms of pK. values, so the approximate pH values of possible buffer solutions are readily apparent.
ChemistrY.{J.Now' Sign in at www.t homsonedu.comj login and go to Chapter 18 Contents to see Scree n 18.4 for a simulation and tutorial on the Henderson-Hasselbalch equation.
Using the Henderson-Hasselbalch Equation Problem Ben zoic acid (C6H sC02H, 2.00 g) and sodium benzoate (Na C6H sC0 2 , 2.00 g) are dissolved in enough water to make 1.00 Lof solution. Calculatethe pH of the solution using the HendersonHasselbalch equation. Strategy The Henderson-HasseLbalch equation requires the pK. of the acid, and this is obtained from the K. for the acid (see Table 17.3 or Appendix H).You willalso need the acid and conjugate base concentrations. Solution K. for benzoic acid as 6.3 x 10- s. Therefore, pK,
~
- 109(6.3 x 10-')
~
4.20
Next,we need the concentrations of t he acid (benzoic acid) and its conjugate base (benzoate ion).
, actid ( -1 mal- ) 2.00 9 benzoic 122.1 9
. benzoate ( -1 mal ) 2.00 9 sodium 144. 1 9
= 0.0164 rnol benzcic enzctc acid act . benzoate = 0.0139 mol sodium
Becausethe solution volume is 1.00 L, the concentrations are {benzoic acid] benzoate] = 0.0139 M. Therefore, using Equation 18.2, we have pH
~
4. 20
+ log -0.0 -139 0.0164
~
4.20
+ log(0.84 8)
~
= 0.0164 Mand [sodium
4.13
Comment Notice that the pH is less than the pK. because the concentration of acid is greater than the concentration of the conjugate base (the ratio of conjugate base to acid concentration is less than 1).
18.2
I
Controlling pH : Buffer Solutions
817
EXERCISE 18.4
Using the Henderson-Hasselbalch Equation
Use the Henderson-Hasselba lch equati on to calculate t he pH of 1.00 L of a buffer solution con-
taining 15 .0 9 of NaHCO ] and 18.0 9 of Na2C03" (Con sider this buffer as a solution of the weak acid HCOJ - wit h ( ° 32- as its conj ugate base.)
Preparing Buffer Solutions To b e u sefu l, a b uffer so lu tio n must h ave two c harac te ristics:
• /)11 Control: It sh o u ld con trol th e pl I a t th e d esired value. T h e H e nde rsonHassclbalch eq ua tion sh ows u s how thi s ca n be don e . H = K + Lo [conjugate base] P P a g [acid ] First, a n aci d is c h o se n whose pK<J (or Ka ) is n ea r the inte n ded value o f pf I (or [ 11, 0 +]) . Secon d , the exact value o f I'll (o r [ 11, 0 +]) is th en ac h ieved by adj us ting th e ac id-to -co nj uga te base rati o . (Exa m p le 18.4 illu strat e s thi s a p p roac h .) • Buffer w!mcity: The buffe r sh o u ld h ave th e a b ility to keep the I'll a p p rox im ately co ns ta n t after th e add itio n o f reaso na b le a m o u n ts of ac id a n d base . Fo r example . th e conce n tration o f acetic aci d in an ace tic acid /ace tate ion b uffer must be su fficient to co ns u me all the hyd ro xid e io n that m ay be ad de d and still co n tro l the pH (see Exa m p le 18.4 ) . Bu ffe rs are u su a lly p repared as 0.10 (\.1 to 1.0 M so lutio n s o f reagents. H o weve r, an y bu ffer will lo se its ca pacity if too m u ch strong ac id or base is ad ded .
Chemislry.o.Now~
•
Sign in at www.thomsonedu.comj login and go to Chapter 18 Conte nts to see Screen 18.5 for a simulation and t utoria ls on preparing buffer solutions.
, . Preparing a Buffer SoLution Problem You wish to prepare 1.0 Lof a buffer solution with a pH of 4.30. A list of possibLe acids (and t heir conj ugate bases) folLows:
Add
Conjugate Base
HS0 4-
5042-
1.2 X 10 - 2
1.9 2
CH i0 2H HC0 3 -
CH 3C0 2-
1.8 X 10 - 5
4.7 5
CO/ -
4.8
x 10 - 11
10. 32
K.
pK.
Which combination shouLd be selected, and what shouLd the ratio of acid to conjugate base be? Strategy Use either the general equation for a buffer (Equati on 18.1) or the Henderson-Hasselbatch equation (Equation 18 ,2 ). Eq uation 18.1 informs you that [H 30 +] shouLd be close to t he acid Ka value, and Equation 18 .2 tells you that pH shouLd be close to t he acid pKa value. Thi s wiLL esta blish which acid you wi LLuse. Havi ng decided which acid to use, convert pH to [H 30 +] t o use Eq uation 18.1. If you use Equation 18.2 , use the pKa value in the tabLe. FinaLLy, caLculate the ratio of acid to conjugate base. 818
Chapter
18
I
Pri nciples of Reactivity: Other Aspects of Aqueou s Eq uilibria
SoLution The hydronium ion concentration for the buffer is found from the targeted pH . pH = 4.30, so [HlO+j = to - pH = 10- 4.30 = 5.0 X 10- s M Ofthe acids given, only acetic acid (CHl C02 H) has a Ka vaLue close to that of the desired [HlO+) (or a pKa close to a pH of 4.30). Now you need only to adjust the ratio [C HlC0 2 Hl/[CHl C02-] to achieve the desired hydronium ion concentration.
[H,D' I ~ 5.0 x 10- ' M ~ [CH,CO,H] (1.8 x 10- ' ) [CH,CO, 1
[CH ,CO,H] [CH,CO,- I
.s: [H,O' I
5.0 X 10- s 1.8 X lO s
2.8 moVL
1.0 rno l/ L
Therefore, if you add 0.28 moLof acetic acid and 0.10 mol of sodium acetate (or any othe r pair of molar quantities in the ratio 2.8/1) to enough water to make 1.0 l of solution. the buffer solution wi LL have a pH of 4.30. Comment If you prefer to use the Henderson-Hesselbalch equation, you wouLd have
pH ~ 4.30 ~ 4.74 + Log [CH,CO,-j [CH ,CO,H] [CHCO - ] Log l 2 = 4.30 - 4.74 = - 0.44 [CH,CO,Hl [CH lC0 2 - ] [CH,CO,Hl
= 10- 0.44 = 0.36
The ratio of conjugate base t o acid, [CH lC02- l/[CHlC02 H], is 0.36. The reciprocal of this ratio { = [CHlC02 Hl/[CHlC0 2- ] = 1/0.36)} is 2.8/1. This is the same result obtained previously using Equation 18.1. EXERCISE 18.5
Preparing a Buffer Solution
Using an acetic acid/sod ium acetate buffer solution, what ratio of acid to conjugate base wittyou need to maintain the pH at 5.00? Describe how you would make up such a solution.
Examp le 18.4 illustra tes severa l important po ints co nce rn ing buffer so lu tio ns. Th e hydron ium ion co nce n tra tio n depe nds no t o nly o n th e K.I value o f th e acid b u t also o n the rat io of ac id an d co nj ugate base co nce n tra tio ns . H oweve r, eve n th ough we wr ite th ese rat ios in terms of re age n t co nce n tra tio ns , it is the relative number oj moles oj acid and conjugate base that is important in determining the jJ! I oj a buffer solution. Becau se both reage n ts are d isso lved in th e sa me so lu tion , their co nce n tratio ns de pe nd o n th e same so lu tio n volume . In Exa m ple 18.4, th e rat io 2.8/ 1 fo r ace tic aci d and sod iu m ace ta te implies that 2.8 tim es as man y mo les of acid were d isso lved per liter as mo les of sod iu m ace ta te. [CH,CO,H] [CH,CO, 1
2.8 moLCH,CO,H/
2.8 mol
1.0 moLCH,CO,
L /L
1.0 mol
FIGURE 18.3 Acommercial buffer solution. The solidacid and conjugate base in the packet are mixed with water to givea solution with the indicated pH. The quantity of water used does not matter because the ratio [acid]/[conjugate base] does not depend on the solution votume. (However, if too much wateris added, the acid and conjugate base concentrations will be too low, and the buffer capacity could be exceeded. Again, buffer solutions usually have solute concentrations around 0.1 Mto 1.0 M.)
CH,CO,H CH,CO,
No tice that on di vid ing o ne co nc e n tra tio n by the other, the volu mes "ca nce l." T his mean s th at we on ly need to e nsu re th at th e rat io o f mol es o f acid to mo les o f co nj ugate ba se is 2.8 to I in thi s ex am ple. Th e aci d and its co nj ugate base co uld have been d issolved in any reaso nable am ou n t of wate r. Th is also m eans tha t dilu ting a ln ~/Ter solution will not change its pH. Co m me rcia lly available buffe r so lu tio ns are often so ld as prem ixed, d ry ing re die n ts. To us e th em , you on ly need to mi x the ing red ients in so me volu me of p ure water (Fig u re 18 .3) . 18.2
I
Controlling pH : Buffer Solutions
819
Problem Solving Tip 18.1
Buffer Solutions
The following is a summary of important aspects of buffer solutions.
A buffer resists changes in pH on adding small quantities of strong acid or base. A buffer contains a weak acid and its con-
[H 0'] ~ 3
[acid]
[conjugate base]
X K,
reactions affect t he relative quantities
of weak acid and its conjugate base. Because this ratio of acid to its conjugate base has only a secondary effect on t he pH . th e pH can be maintained relatively
H = K + lo [conjugate base] p
ju gate base.
The hydroni um ion concentration of a buffer solution can be calculated from Equat ion 18.1,
gate base consumes added acid. Such
or t he pH can be calculated from t he Henderson-Hasselbalch equation (Eq uation 18.2). p
a
9
[acid]
consta nt . The buffer must have sufficient capacity to react with reasona ble quantities of added acid or base.
The pH depe nds primarily on the Ka of t he weak acid and seco ndarily on the relative amou nts of acid and conjugate base . The funct ion of t he weak acid of a buffer is to consume add ed bas e, and t he conj u-
How Does a Buffer Maintain pH? No w le t u s ex p lo re quanti ta tively ho w a g iven b u ffe r so lu tio n ca n m a in tain th e p H of a so lu tio n o n a d d in g a sm a ll am o u n t of stro ng ac id . Chemislry~\.Now~ Sign in at www.thomsonedu .com/Ioqtn and go to Chapter 18 Contents to see Screen 18.6 for a tutorial on adding reagents to a buffer solution.
How Does a Buffer Maintain a Constant pH? ProbLem What is the change in pH when 1.00 mL of 1.00 MHCl is added to (1) 1.000 Lof pure water and to (2) 1.000 Lof acetic acid/sodi um acetate buffer with [CH 3C02H] = 0.700 Mand [CH 3C02 -] = 0.600 M? (See Example 18.2, where the pH of this acetic acid/acetate ion bufferwas found to be 4.68.) Strategy HCl is a strong acid and therefore ionizes completely to supply H30 + ions. Part 1 involves two steps: (a) Find the H)O + concentration when diluting 1.00 mLof acid to 1.00 L. (b) Co nvert the value of [H,O+] for t he diLute soLution t o pH. Part 2 involves three steps : (a) Astoichiometry calculation to find howthe concentrations of acid and conjugate base change on adding H30 +. (b) An equilibrium calcuLation to find [H30 +] for a buffer solution where t he concentrations of CH )C 02H and CH )C0 2 - are slightly alte red owing to the reaction of CH)C02 - with added H)O+. (c) Conversion of [H30 -] to pH. SoLution Part 1: Adding Acid to Pure Water 1.00 mL of 1.00 MHCl represents 0.00100 mol of acid. If this is added to 1.000 Lof pure water, the H)O+ concentration of the water changes from 10- 7 to 10- 3 ,
't X Vt = ' 2 X V
2
(1.00MHo.oOl0o L)
~
c, X (1.001 L)
'2= [H)O +] in diluted solution = 1.00 x 10- ) M and so t he pHfaLls from 7.00 to 3.00. Part 2: Adding Acid to an Acetic Acid/Acetate Buffer Solution HCl is a strong acid that is 100% ionized in water and supplies H30 +, which reacts completely with the base (acetate ion) in the buffer solution according to the following equation: H,O-(aq) + CH ,CO, -(aq) 820
Chapter 18
I
PrincipLes of Reactivity: Other Aspects of Aqueous Equilibria
H,O(f) + CH, CO , H(aq)
[H30+] from Added H(l
[(H3(02- ] from Buffer
[(H3(02H] from Buffer
0.00100
0.600
0.700
- 0.00100
+ 0.00100
Initial amount of acid or base (mol = c X V) Change (mol)
- 0.00100
Equilibriu m (mol)
0
0.599
0.700
Concentrations after reaction (c = mol/V)
0
0.598
0.700
Because the added HCl reacts compLetely with acetate ion to produce acetic acid, the solution after this reaction (with V = 1.001 L) is once again a buffer containing only the weak acid and its salt. Now we only need to use Equation 18 .1 (or the Henderson-Hasselbalch equation) to find [H,O +] and the pH in the buffer solution as in Examples 18.2 and 18.3. (H3C0 2H + H,O
Equilibrium Initial (M)
0.700
=
-x
Change (M)
0
+x
0.700 - x
Equilibrium (M)
H3O +
x
+
(H3(0 20.598
+x 0.598 + x
As usual, we make the approximation that x, the amount of H30 -'- formed by ionizing acetic acid in the presence of acetate ion, is very small compared with 0.700 Mor 0.598 M. Using Equation 18 .1, we calculate a pH of 4.68.
[H,O- ]
~
x
~
[CH,CO,H] x K, [CH,CO, ]
~
(0.700 mOI)(1.8 x 10- 5) 0.598 mol
~
2.1 x 10- 5 M
pH ~ 4.68
Comment Within the number of significantfigures allowed, the pH of the buffer solution does not change after adding Hu. Thebuffer solution contains the conjugate base of the weakacid, and the base consumed the added HCl. In contrast pH changed by 4 units when 1 mL of 1.0 MHCL was added to 1.0 Lof pure water.
EXERCISE 18.6 Buffer Solutions Calculate the pH of 0.500 l of a buffer solution composed of 0.50 Mformic acid (HC0 2H) and 0.70 M sodium formate (NaH C02 ) before and after adding 10.0 mL of 1.0 MHCl.
18.3
Add-Base Titrations
A titrati on is o ne of th e most impo rtan t ways o f determ in ing acc ura te ly th e quantity o f an ac id, a base, o r so me o ther su bstance in a mi xtu re or of asce rtain ing th e pu rit y of a su bsta nce. You learn ed how to perform th e sto ich io me try calcula tio ns in volved in tiu-atio ns in Chap ter 4 (.. Sec tio n 4.7). In C hapter 17, we described th e foll owin g impo rtant points regard in g acid-base reactio ns (.. Sectio n 17.6 ): • The pH at the eq uivalence poin t of a stro ng ac id -strong base titration is 7. T he so lu tion at th e equivalence point is truly "neu tra l" only whe n a strong acid is titrated with a stro ng base and vice versa . • If th e su bsta nce bein g titrat ed is a wea k ac id or base , th en the pH at th e equival en ce point is not 7 (see Table 17.5) . (a) A wea k acid titrate d with stro ng base lead s to pH > 7 at th e eq u ivale nce point due to th e co njugate base of th e wea k ac id. (b) A wea k base tit rat ed with strong acid lead s to pH < 7 at th e eq u ivalence point du e to th e co nj ugate aci d of th e wea k base. A knowled ge o f buffer so lutio ns and how they work will no w allow u s to more full y u nd erstand ho w th e pH changes in th e co u rse of an aci d-base reac tio n .
• Equivalence Point The equivalence point for a reaction is the point at which one reactant has been completely consumed by addition of another reactant. See page 185.
• Weak Add-Weak Base Titrations Titrations combining a weak acid and weak base are generally not done because the equiva lence point often cannot be accurately judged. 18.3
I
Acid-Base Titrations
821
Take a Deep Breath!
Case Study Maintenance of pHis vital to the cells of all living organisms because enzyme activity is influenced by pH . The primary protection against harmful pHchanges in cells is provided by buffer systems. The intracelluLar pHof most cells is maintained in a range between 6.9 and 7.4. Two important biological buffer systems control pH in t his range: the bicarbona te/ca rbonic acid system
(HC O,- j H,CO,) and t he phosphate system (HPO; - j H,PO,- ). The bicarbonate/car bonic acid buffer is important in blood plasma. Three equilibria are important here. CO,(g) Na CH,C01(aq ) + H10(f)
Let us focus on th ree important points on thi s cu rve:
• 171£ pH be/ are titration begins. T he pH before any base is ad d ed ca n be ca lcu late d fr om the weak acid 1\\ value and the acid concen tration (... Exam ple 175 ) . • The /JII at the equivalence/JOin t. At th e equ ivalen ce point, th e so lu tio n con tain s only so d iu m ac e ta te, th e C H 3 C0 2 H a nd NaO H havin g be en co ns u m ed . Th e pH is con tro lle d b y the ace ta te ion , th e co nj ug ate base o f ace tic aci d (... Table 17.5 , page 728 ) . • The pH at the halfway point (half-equivalence point) of the titrat ion. Here , the pH is equal to the pl\\ of th e weak acid, a co nclusio n that is d iscussed in mo re d et ail belo w. As NaO II is ad ded to ace tic aci d th e base is co nsu med an d so d iu m ace ta te is pro d uced . Thus, at eyery point between the beginning of th e tit ration (whe n on ly acetic acid is p res ent) and th e equ ivale nce po in t (when o nly so d iu m ace tate is p resen t) , th e so lu tio n co n ta ins both ace tic acid and its salt, sod iu m aceta te. These 824
Chapter 18
I
Principles of Reactivity: Ot her Aspects of Aqueous Equilibria
Active Figure 18.5 The change in pH during the titration of a weak add with a strong base. Here, 100.0 ml of 0.100 M acetic add is tit rated with 0.100 M Na OH. Note espedallythe following: (a) Acetic add is a weak ad d, so t he pH of t he original solution is 2.87. (b) The pH at the point at which half the acid has reacted with base is equal to the pK. for the ad d (pH = pK. = 4.74). (c) Atthe equivalence point, the solution contains the acetate ion, a weak base. The refore, the solution is basic, with a pHof 8.72.
Halfwa y point mol CH 3C0 2H = mol CH 3C0 2 [H30+) = Ka and pH = pKa
pH 6
__---:::~_;~;:, c;o~-~present. CH CO Hand CH. 3 2 . OO 60th 3 2 l Bufferreg • _
Chemistry.o.Now~ Sign in at www.thoms;nedu.com/login and go to the Chapter Contents menu to explore an interactive version of this figure accompanied by an exercise.
2
Volume NaOH added (ml )
are th e components of a b1~[rer solu tion, and th e hydronium ion co nc en tratio n can be ca lcula te d from Equatio n 18.3 o r 18.4. [weak acid remaining]
[conjugate base produced]
(18.3)
or
H= K p pa
+ 10 g
[conj ugate base prod uced] [weak acid remaining]
(18.4)
Th e fact th at a buffer is present at an y poin t bet ween the begin ning of th e titr ation and th e eq uivale nce point is th e reason that th e pH of th e so lu tio n only rises slowly afte r a few millilite rs of titra n t has been ad d ed. Wh at happens whe n exa ctly half of tile aci d has been co nsum ed by base? H alf o f th e acid (C H 3C 0 2 H ) has been co nve rted to the co nj uga te base (C H 3C 0 2 - ) , and half rem ain s. Therefore , th e co ncentra tio n o f weak aci d re mai ning is eq ual to th e co nce ntration of co nj ugate base produced ([CIl 3C O,Il ] = [C Il"CO,- ]) . Using Eq ua tio ns 18.3 o r 18.4, we see th at
[H ,O+]
= (1) X K,
or
pH = pK, + log (l)
Becau se log (1 ) = 0, we co me to th e followin g ge nera l co nclusio n :
At the halfway point in t he titration of a weak acid wit h a strong base [H ,O+] = K, and pH
= pK,
(18.5)
In th e parti cul ar case of th e titration o f ace tic acid with a stro ng base , [H 30 +] = 1.8 X 10- '' 1\1 at the h alf way point, and so th e pf l is 4.74 . This is eq ual to th e P~l of acetic acid . Chemistry.o.Now~
•
Sign in at www.thomsonedu .com/Ioqtn and go to Cha pter 18 Contents to see Screen 18.7 for a simulation and tuto rial on titration curves.
18.3
I
Acid-Ba se Tit ratio ns
825
, • Titration of Acetic Acid with Sodium Hydroxide Problem Consider the titratio n of 100.0 mL of 0.100 Mace tic acid with 0. 100 M Na OH (see Figure 18. 5). (a) What is t he pH of the so lution whe n 90.0 mL of 0.100 M Na OH has bee n added to 100 .0 mL of 0.100
M acetic acid?
(b) Wh at is the pH at the equiva lence point? (c) Wh at is the pH after 110.0 mL of NaOH has been added?
Strategy The problem generally invo lves two majo r steps: (1) Astoichiometry calculation to find the quantity of acid remai ning, if any, and quantity of conjugate base formed after adding NaOH. (2) Before the equivaLence point, in the "buffer regio n," you wi ll do an equiLib rium caLculation to find [H 30+] for a buffer solution where t he quantitiesof CH,C02 Hand CH,C0 2- are known from the first part of the calculation. At t he equivalence point, onlythe conjugate base wi ll remain, so the calcu lation resembles Example 17.7. After the equivalence point, t he solution contains both t he conjugate base of t he weak acidand excess NaOH, but the latter controls the pH . SoLution
Part a: pHbefore the equivalence point. Let us first calculate t he amounts of reactants before reaction ( = concentration x volume) and then use t he principles of stoichiometry t o calculate the amounts of reactants and products after reaction. The limiting reactant is NaOH, so some CH 3C02H remains along with t he product, CH3C 0 2- . Equation
+
(H 3(0 2H
Initial (mol)
0.0100
Change (mol)
- 0.00900
After reaction (mol)
OW 0.00900
=
- 0.00 900
H,O
0 + 0.00900
0
0.0010
+
(H 3(0 2-
0.00900
The ratio of amounts (moles) of acid to conjugate base is the same as the ratio of their concentrations. Therefore, we can use the amounts of weak acid remai ning and conjugate base formed to find the pH from Equation 18 .3 (where we use amounts and not concentrations). [H 30+) -- molCHl02H X Ka -_ ( 0.0010 mOl )( 1.8 X 10- ' ) -- 2.0 X 10-, M mol CH 3C0 2 0.0090 mol pH ~ - Log(2.0 X 10-')
~
5.70
The pH is 5.70, in agreement with Figure 18 . 5 . Notice that this pHis appropriate for a point after t he halfway point (4.74) but before t he equivalence point (8.72).
Part b: pHat the equivalence point. To reach the equivalence point, 0.0100 mol of NaOH has been added to 0.0100 mol of CH 3 C0 2H and 0.0 100 mol of CH 3C0 2- has been formed. Equation Initial (mol) Change (mol) After reaction (mol)
(H 3(0 2H
+
OW
0.0100
0.0100
- 0.0100
- 0.0 100
0
=
(H 3(0 2-
+
H,O
0 + 0 .0100
0
0.0 100
Because two solutions, each with a volum e of 100.0 mL, have been combined, the concentration of CH 3C0 2 - at the equivalence point is (0.0 100 moVO.200 L) = 0.0050 M. Next, we set up an ICEtable for the hydro lysis of this weak base, Equation Initial (M) Change (M)
After reaction (M)
(H 3(0 2-
+
H 20
~
0.00500
-x 0.00500 - x
(H3(02H
o
o +x
x
x
(x)(x) 0.0050 0 - x Chapter
18
I
Pri nciples of Reactivity: Ot her Aspects of Aqueous Equilibria
OH -
+x
and calculate t he concentration of OW ion using Kb for t he weak base.
826
+
Making t he usual assumption thatx is small wit h respect to 0.00500 M,
x ~ [OW l pH
~
~
1.7 x 10- 4 M (pOH ~ 5.7 8)
14.00 - 5.78
~
8.2 2
Port c: pH after the equivalence point. Now the limiting reactant is t he CH3C0 2H, and the solution contains excess OH - ion from t he unused NaOHas well as from the hydrolysis of CH 3C0 2 "". Equation
+
(H 3(0 2H
Initial (mol) Change (mol)
OW
(H 3(0 2-
:o===:::t
0.0110 - 0.0100
+ 0.0100
0.0010
0.0100
o
H20
o
0.0100 - 0.0100
After reaction (mol)
+
The amount of OH - produced by CH 3C0 2- hydro lysis is very smaLL (see part b), so t he pH of the solution after the equivalence point is determined by t he excess Na OH (in 210 mL of solution).
[OW]
~
0.0010 moVO .210 L ~ 4.8 x 10-' M ( pOH ~ 2.32)
pH ~ 14. 00 - 2.32
~
11.68
Titration of a Weak Acid with a Strong Base
EXERCISE 18.8
The titration of 0.100 acetic acid with 0.100 MNaOH is described in the text. Wh at is the pH of the solution when 35.0 ml of the base has been added t o 100.0 ml of 0.100 Macetic acid?
Titration of Weak Polyprotic Acids T he titra tio ns illustrated thus far have been fo r the reaction of a monoprotic acid (HA) with a ba se such as NaO H . It is po ssib le to exte nd the d iscussion of titratio ns to pol yprotic acids su ch as o xa lic ac id, I 12C20 4 . H, ( ,O, (aq)
+
H( , O, " (aq)
H,O(C)
+
~-'-
H,O(C)
+
H,O+(aq)
Ka1
= 5.9
X 10 - 2
+
H,O'(aq)
Ka2
= 6.4
X 10 - 5
H( , O, - (aq)
~='
( ,0, " (aq)
Figure 18.6 illu strate s the curve for the titratio n of 100 mL of 0. 100 M oxalic acid wit h 0. 100 1\'1 NaOH . Th e fir st sig n ifican t rise in pl I is ex per ie nc ed after 100 m L of ba se has bee n ad d ed , ind ica tin g th at th e fir st proton of th e acid has been titra ted. H,( ,O, (aq)
+
OW(aq)
~='
H( , O, " (aq)
+
H, O(aq)
Whe n t he se cond p ro to n of oxalic ac id is titra te d , the p H again rises significan tly. H( , O, - (aq)
+
OW (aq )
( ,O/ -(aq)
+
H,O(C)
The pl l at thi s sec o nd eq uivalenc e point is controlled by the o xalate io n , C;20 42 -
Kb = Kw/ Ka2
=
.
1.6 X 10 - 10
Calculation of th e p l I at the equivalen ce point ind icates that it shou ld be abo u t 8.4, as observed. ChemistrY-9-Now~ Sign in at www.t homsonedu.comj login and go to Chapter 18 Contents to see Screen 18.8 for a simulation on titration of a weak polyprotic add .
18.3
I
Acid- Ba se Titrations
827
FIGURE 18.6 Titration curvefor a diprotic add . The curve for the titration of 100.0 mL of 0.100 Moxatic add (H 2C 204 , a weakdiproticadd) with 0.100 M NaOH. The first equivalence point (at 100 mL) occurs when the first hydrogen ion of H2C204 is titrated, and the second (at 200 mL) occurs at the comptetion of the reaction. The curvefor pH versus volume of Na OH added shows an initial rise at the first equivalence point and then another rise at the second equivalence point.
Substa nce being tit rated 14
12 10
8
Second equivalence point;
pH
pH ~ 8.36
6
4
First equivalence point 2
o
100
200
300
Volume of Na OHadded (mL)
Titration of a Weak Base with a Strong Acid Fin all y, it is useful to co ns ide r the titration of a weak base with a stro ng acid. Figure 18.7 illustrates th e pH cu rve for th e titration o f 100.0 mL o f 0.1 00 Xl N H 3 with 0.100 ;vI n cr.
The initial pH for a 0.100 Xl N H 3 so lu tio n is 11.12. As th e titration progr esses, th e important species in solu tio n are th e weak acid N H 4+ and irs co nj ug ate base, NH 3 •
«, = 5. 6 X 10- 10 FIGURE 18.7 Titration ofa weak
base with a strong add. The change in pHduring the titration of a weak base (100.0 mL of 0.100 MNH l ) with a strong acid (0.100 MHel). The pH at t he halfneutratization point is equal to the pK. for the conjugate acid (NH 4 +) of the weak base (NH,) (pH ~ pK,
~
14
12 10
9.26). At the
equivalence point, the solution contains the NH , + ion, a weak acid, so the pHis about 5.
0.100 MNH 3 pH = 11. 13
pH
Halfway point; [N H3l = [NH/] pH = 9.26
Buffer region
8
6
Eq uivalence point pH = 5.28 4
2
o
40
80
Titrant volume (ml)
828
Chapter 18
I
Principles of Reactivity: Other Aspects of Aqueous Equilibria
120
160
Calculating the pH at Various Stages of an Acid-Base Reaction
Problem Solving Tip 18.2
late (a) t he amount of acid added, (b) the amount of base consumed. and (c) the amount of conjugate base (BH+ ) formed . Step 2. Calculate the concentrations of base, [BJ, and conjugate acid, [BH +J. Recogn ize that the volume of the solution at any point is the sum of the originaLvolume of base solution plus the volume of acid solution added. Step 3. Calculate the pHbe/ore the equiva lence point. At any point before the equivalence point, the solution is a buffer solution because both the base and its conjugate acid are present . Calculate [H30 +] using the con-
Finding the pHat or beforethe equivalence point for an acid-base reaction always involves several calculation steps. There are no shortcuts. Co nsider the titration of a weak base, B, with a strong add as in Examp le 18.7. (The same principles applyto other acid-base reactions.)
H,O' (aq) + B(aq)
-~
BH'(aq) + H,O(f)
Step 1. Solve the stoichiometry problem . Up to the equivalence point, acid is consumed complete ly to leave a solution containing some base (B) and its conj ugate acid (BH+). Use the principles of stoichiometry to calcu-
centrations of Step 2 and the value of Ka for the conjugate acid of the weak base. Step 4. Calculate the pH at the equivalence point. Calculate the concentration of the conjugate acid using the procedure of Steps 1 and 2. Use the value of K, for the conj ugate acid of the weak base and the procedure outlined in Examp le 18.7 (or in Exam ple 18.6 for a weak acid). (For a titra tion of a weak acid with a st rong base. use the value of Kb for the conjugate base of the acid and follow t he procedure outlined in Example 18.6.)
At the halfway poin t, the co nce n tratio ns of NII 4+ and NII 3 are th e same, so [H 0+]
,
[H,O' ]
~
[NH 4 "l __ _ X K [NH,J
~
5,6 X 10- 10
a
= K.
pH = pK. = - 109(5,6
X
10 - 10 ) = 9,25
As the ad d itio n o f l lCl to NII 3 co n tin ues, th e p l I d ecli nes slowly because of th e bufferi ng ac tio n o f th e NII ,/N II 4 + co m bina tio n . Near the eq uivale nce po int, ho wever, the pH drops rap id ly. At th e equ ivale nce poi nt , th e so lu tio n co n ta ins o nly am mo niu m ch lo ride, a wea k Bren sterl aci d, and the so lu tion is weakly acid ic.
ChemistrY.{J"Now'" Sign in at www.t homsonedu.com/ Ioqtn and go to Chapter 18 Contents to see Screen 18.9 for a simulation on titration of a weak base wit h a strong add.
Titration of Ammonia with HCI ProbLem What is the pH of the solution at the equivalence point in the titration of 100.0 mL of 0.100 M ammonia with 0.100 M HCl (see Figure 18.7)? Strategy This probLem has two steps: (a) Astoichiometry calculation to find the concentration of NH 4 + at the equivalence point. (b) An equilib rium calculation to find [H30- j for a solution of the weak acid NH 4+ . SoLution Part 1: Stoichiometry Problem
Here, we are titrating 0.0100 mol of NH ] ( = c x V). so 0.0100 mo l of HCl is required. Thus, 100.0 mL of 0.100 MHCl ( = 0.0 100 mo l HCl) must be used in the titration.
+
Equation
NH 3
Initial (mo l = c X V)
0.0100
0.0100
- 0.0100
- 0.0100
Change on reaction (m ol)
H]O+
After reaction (mol)
0
0
Co ncentration (M)
0
0
NH 4 +
~
+
H20
0
+ 0.0100
0.0100 0.0100 mol (in 0.200L) =
0.0500 M 18.3
I
Acid-Base Titrations
829
Part 2: Equilibrium Problem When the equivalence point is reached, the solution consists of 0.0500 M NH 4 + . The pH is determined by th e hydroLysis of th e acid.
+
NH4 +
Equation Initial (M)
=
H,O
0.0500
-x
Change (M )
0.0500 - x
Equilibri um (M)
+
NH,
H]O+
0
0
+x
+x
x
x
Using Ka for t he weak acid NH 4 " . we have K a
= 5.6
X 10- 10 = [NH ]][H]O+} _ x [NH:] 0.05 00 - x
Simplifying. x ~ [H,o"] ~ .) (5.6
x
10- " )(0 .0500) ~ 5.3
x 10- '
M
pH ~ 5.28 The pH at the equivalence point in this weak base- st rong acid tit ration is indeed slight ly acidic, as expected. EXERCISE 18.9
Titration of a Weak Base with a Strong
Add
CaLcu late th e pH after 75.0 ml of 0.100 M HCl has bee n added to 100.0 ml of 0.100 M NH 3 • See Fig ure 18.7.
pH Indicators Man y organi c co m pou nds , both natural and syn thetic, have a co lo r that cha ngcs with pl I (Figu re 18.8) . No t on ly d oes th is ad d be auty and vari e ty to ou r wo rl d , but it is also a useful property in che m ist ry. FIGURE 18.8 Phenolphthalein, a common acid-base indicator. Phenolphthalein, a weak acid,is colorless. As the pH increases, the pink conjugatebase form predominates, and the cotor of the solution changes. The changein color is most noticeable around pH 9. The dyeis commonly used for strong acid + strong base or weak acid + strong base titrations because the pH changesfrom 3-4 to 10-12 in these cases. For other suitable indicatordyes, see Figure 18.10.
830
Chapter 18
I
OH
6 01 I "":: :
0-
Phenolpht halein. Brensted acid. colorless
~((aq) + ~,
HOY ' )
2 H, O(C)
0
Principles of Reactivity: Ot her Aspects of Aqueous Equilibria
zzzz:
2 H,O" (aq)
o
+
Conj ugate base of phenolphthalein, Bre nsted base, pink
CO, - (aq)
'rou have likely carried ou t an acid-Lase titrat io n in th e laboratory, and, before star ting th e titration , you would have add cd an indicator. The ac id - base indicat or is usu all y an organic co m poun d th at is itself a weak acid or wea k base (sim ilar to th e co mpo u nds th at give th e co lo r to flo wers) . In aq ueous so lut io n , th e aci d fo rm is in eq uilibriu m with its co nj uga te base . Ab brevia ting th e indicator's ac id form ula as IIInd and th e formul a o f its co njugate base as Ind " , we ca n write th e eq u ilibri u m eq ua tion HInd(aq) + H,O(C)
H,O' (aq) + Ind- (aq)
(a)
The importan t charac teristic o f ac id - base indicat ors is th at the acid fo r m of th e co mpound (H ind) ha s on e co lo r and th e co njug ate base (Ind " ) has another. To see how suc h co mpou nds ca n be used as eq uivale nce poin t indicators, let us write th e usu al eq uilib rium co nstan t ex p ress io ns fo r th e d ependen ce of h ydronium ion concentrati on or pH on th e indicat o r 's ionizati on co ns ta n t ( K a ) a nd on th e rel ati ve quan tities of th e aci d and conj ugate base. or
pH = pK
,
[Ind- ] [HIn d]
+ [og-
(18.6)
These eq uations info rm us th at • when th e hydron ium ion co nc en tra tion is eq uivalen t to th e value of (o r when pH = pK, ) , th en [H Ind] = [Ind " ] • whe n [113 0 +] > K, (o r pH < pK, ) , th en [H Ind] > [Ind "] • when [11,0 +] < K, (o r I'll > pK, ) , then [l Und] < [Ind "]
J 18.4
I
Solubilityof Salts
8 37
• Comparing Solubilities When two salts have different cation-anion ratios, you will have to do a calculation to find their solubilities. Simply comparing their K,p values will not necessarily give their relative solubilities.
Similarly, you co uld co m pare 1:2 salts suc h as th e lead halides. PbI, (K" = 9.8 X 10- ') < PbS" (K" = 6.6 X 10- ') < PbCl, (K" = 1.7 X 10- 5 ) - - - - increasing Ksp and increasing solubility ------>
but yo u ca n no t directly co mpare a I: I salt (AgCI) with a 2: I salt (Ag,CrO ,) . EXERCISE 18.14
Comparing SoLubilities
Using Ksp values, predict which salt in each pair is more soluble in water.
(a) AgCI or AgCN • Complex Ions The solubility of a salt will increase if addition of the common ion leads to formation of a complex ion. (See Section 18.6, page 846)
(b) Mg(OH), or Ca(OH), (e) Ca (OH ), or CaSO,
SoLubility and the Common Ion Effect The te st tube o n the left in Figure l R.13 co n tains a p re cipitate o f silver acetate, AgC H 3C 0 2 , in wate r. The so lut io n is saturate d , and th e silver ions and acetate ions in the so lu tio n are in equilibrium with so lid silver acet at e. AgCH ,CO, (s)
==' Ag+(aq) + CH ,CO, - (aq)
But wha t would happen if th e silver ion co ncen tratio n is increased , say by ad d ing silver nitrate? Le Chate licr's principle (.... Section 16.6) sugges ts- and we observe -that more silver ace ta te precipi tate sho uld fo rm be cau se a product ion has been added, cau sin g th e eq uilibriu m to shift to form more silver acet at e. The ionization o f weak acids and bases is affected by the presen ce of an ion comm o n to th e eq uilib rium process (Sec tio n IR.I) , an d the effec t of ad d ing silver ions to a satu ra te d silver acetate so lu tio n is ano ther exa mp le of th e co m mo n ion effe ct. Adding a co m mo n ion to a satu rated so lu tio n o f a salt will lower the salt solubili ty. Chemistry.o..Now~
•
Sign in at www.thomsonedu.comjlogin and go to Chapter 18 Contentsto see Screen 18.15 for a simulation and tutorial on the common ion effect.
FIGURE 18.13 The common ion effect. The tube at the left contains a sat urated solution of silver acetate, AgC HlC0 2 • When 1.0 MAgN0 3 is added to the tube (right), more solid silver acetate forms. 838
Chapter 18
I
Principles of Reactivity: Other Aspects of Aqueous Equilibria
The (ammon Ion Effect and Salt Solubility Problem If solid AgCl is placed in 1.00 L of 0.55 M NaCl, what mass of Agel will dissolve?
Strategy The presence of an ion common to t he equilibrium suppresses the solubility of a salt. To deter-
mine the solubility of the salt under these circumstances, caLculate the concentration of the ion (Ag + ion in t his case) other than the common ion (here Cl"). SoLution I n pure water, the solubility of Agel is equal to either [Ag+] or [Cl"].
AgCI(s)
Ag"(aq)
+
Cl ~(aq)
= ji Ca1+(aq) + 3 H,O(f ) + CO,(9)
«; = 3.4
CO,' - (aq) + H,O+(aq) HCO, "(aq) + H,O+(aq)
HCO, "(aq) + H,O(f) H,CO,(aq) + H,O(f)
Overall: CaCO,(5) + 2 H,O+(aq)
~="
+ H,O(f ) = HS -(aq) + OW(aq) K. ~ IHS -)[OWI/IS' -j
Net reaction:
=
MS(,) + H20( f ) HS- (aq) + MH(aq) + OW(aq) K,p, ~ (MH )[ HS- )[ OW) ~ K,p x K.
Values for K,pa for several metal sulfides are included in Appendix J.
Y ou ca n think of thi s as th e result of a ser ies o f react ions.
Ca' +(aq) + CO,' -(aq)
+ SZ-(aq) K,p ~ IM H )[S' -j
S'-(aq)
In soluble salts co n tain ing suc h anio ns as ace ta te, ca rbonate, hydroxid e , pho sphat e , and sulfide d issolve in stro ng aci d s. Fo r exam ple, you know that if a stro ng aci d is ad ded to a wat er-inso lu ble metal ca rbo nate suc h as CaC0 3 , th e salt d issolves (
Ksp- t he system is not at equilibrium; precipitation will occur.
The co nce n tratio ns of Ag" and CI- in so lu tio n are too high, an d AgCI will p recipita te until Q = K ,p' ChemistryJl"Now~
•
Sign in at www.thomsonedu.comjlogin and go to Chapter 18 Contents to see Screen 18.17 for a simulation and tutorial on when a precipitation reaction can occur.
Solubility and the Reaction Quotient Problem Solid Agel has been placed in a beaker of water. After some time, the concentrations of Ag+ and el- are each 1. 2 x 10 - 5 molfL. Has the system reached equilibrium? If not. will more Agel dissoLve?
r
St rategy Use the experimenta l ion concentrations to caLculate the reaction quotient to decide if the system is at equilibrium (if Q = Ksp) .
a. Compare a and
K~p
18.5
I
Precipitation Reactions
843
SoLution For this Agel case, Q ~ [Ag+][Cl- ]
~
(1.2
x 10- ')(1.2 x 10- ')
~
1.4
x
10- w
Here, Q is less than Ksp (1.8 x 10- 10 ) , Thesolution is not yet saturated, and Agel will continue to dissolve until Q = Ksp, at which point [Ag -] = [Cl"] = 1.3 X 10- 5 M. That is, an additional 0.1 X 10- 5 mol of Agel (about 1.9 mg) will dissolve per titer. EXERCISE 18.17
Solubility and the Reaction Quotient
Solid PbI2 (Ksp = 9.8 X 10- 9 ) is placed in a beaker of water. After a period of time, the lead(II) concentration is measured and found to be 1.1 X 10- 3 M. Hasthe system yet reached equilibrium? That is, is the solution saturated? If not, will more PbI2 dissolve?
Ksp, the Reaction Quotient, and Precipitation Reactions With so m e knowl edge of th e reaction quotient, we ca n decid e ( 1) if a precipitate will form wh en the ion co nc e n tr atio ns arc kn own or (2) what co nc c n tra tio ns of ions are req u ired to begin the p recipitation of a n insoluble salt. Suppo se th e c o n ce n tr a tio n of aqu cous magncsium ion in a so lu tio n is 1.5 X 10- 6 1\1. If e n o ug h NaO H is add ed to make the so lu tio n 1.0 X 10 - 4 ~I in h ydroxide ion , 011 - , will precipitation of Mg (01l) 2 o ccu r ( K ,p = 5 .6 X 1O - 12 )? If not, will it occu r if th e concentration o f' O'l-l" is in creased to 1.0 X 10- 2 !\1 ? Our str atcgy will be as in Example l R1 3. That is, usc th e ion co nc c n tra tio ns to calculate th e value of Q an d th en compa re Q with K sp to decide if th e syste m is at equilibrium. Let us bcgin with the eq u atio n for the di ssolution of in soluble Mg(OH ),. Mg (OH h( s)
-~
Mg 2+(aq)
+ 2 0W(aq)
When th e concentrations ofmagncsium and hydro xide ions arc those sta re d above, we find th at Q is less th an K sp • Q ~ [Mg 2+ ][ O W
]2 ~
(1. 5 X 10- ' ) (1.0 X 10- 4) 2
Q (1.5 X 10- 14)
~
1.5 X 10 - 14
< K,p (5.6 X 10- 12)
This mean s the so lu tio n is not yet saturated , an d p recipitation does not o ccur. Wh en [OH -] is increased to 1.0 X 10- 2 !\'I, the re action qu otient 1.5 X 10- 10 ,
IS
Q = (1. 5 X 10 - 6)(1.0 X 10- 2) 2 Q = 1.5 X 10 - 10
> K,p (5.6 X 10- 12)
The reaction quoti ent is now lmgerthan Ksp . Precipitation of I\..lg (OH h occurs and will co n tin ue until the ~l g2 + a n d 011 - ion co nc e n tra tio ns h ave decli ned to the point wh ere their product is eq u al to Ksp • EXERCISE 18.18
Deciding Whether a Precipitate Will Form
Wi ll SrS04 precipitate from a solution containing 2.5 X 10- 4 Mstrontium ion, Sr2 - , if enough of the soluble salt Na 2S0 4 is added to ma ke the solution 2.5 X 10- 4 Min SO/ -? Ksp for SrS 04 is 3.4 X 10- 7 •
Now th at we know how to decide if a precipitate will form whe n th e co nc e n tration of each ion is known , let us turn to th e problem o f deciding how much o f the precipitatin g age n t is required to begin th e precipitation o f a n ion a t a given co ncentration level. 844
Chapter
18
I
Pri nciples of Reactivity: Other Aspects of Aqueous Equilibria
· Ion Concentrations Required to Begin Precipitation Problem The concentration of barium ion, Ba2 + , in a solution is 0.010 M.
(a) What concentration of sulfate ion, 50/-, is required once BaS04 has started to precipitate? (b) When the concentration of sulfate ion in the solution reaches 0.015 M, what concentration of barium ion will remain in solution? Strategy There are three variables in the Ksp expression: Ksp and the anion and cation concentrations. Here, we know Ksp (1.1 X 10 -10) and one of the ion concentrations. We can then calculate the other ion concentration. SoLution Let us begin by writing the balanced equation for the equilibrium that will exist when BaSQ4 has been precipitated. Ba504 ( 5 )
_ _
Ba2 +(aq)
+ SO/
-(aq)
Ksp
=
[Ba2 +][50/ -]
= 1.1
X 10- 10
(a) When the product of the ion concentrations exceedsthe K,p ( = 1.1 X 10 - IO)- t hat is, when Q > K,p-precipitation will occur. The Ba 2-'- ion concentration is known (0.010 M), so the 50}- ion concentration necessary for precipitation can be calculated. [SO 2- ]
4
K
1 1 X 10- to
[BaH]
0.010
= -"- = .
=
1.1 X 10 - 8 M
The result tells us that if the sulfate ion is just slightly greater than 1.1 x 10- 8 M, Ba504 will begin to precipitate; Q = [Ba2+][50} -] would then be greater than Ksp-
(b) If the sulfate ion concentration is increased to 0.015 M, the maximum concentration of Ba H ion that can exist in solution (in equilibrium with BaS04)is [Ba H]
= - K-'_P - = 1 .1 X [50/- J
10 - t O
0.015
= 7.3 X 10-9 M
Comment Thefact that the barium ion concentration is so small under these circumstances means that the Ba H ion has been essentially completely removed from solution. (It began at 0.010 Mand has dropped by a factor of about 10 million.) The Ba 2-'- ion precipitation is, for all practical purposes, complete.
K,p and Precipitations Problem Suppose you mix 100.0 mL of 0.0200 M Bart, with 50.0 mL of 0.0300 MNa2504' WiLL Ba504 (K~p = 1.1 X 10 - tO) precipitate? Strategy Here, we mix two solutions, one containing Ba 2+ ions and the other 50} - ions. Fi rst find the concentration of each of these ions after mixing. Then , knowing the ion concentrationsin the diLuted solution, calculate Q and compareit with t he K,pvalue for BaS04.
Solution First use the equation clVt = C2V2 ( ... Section 4.5) to calculate Cb the concentration ofthe Ba H or 50}- ions after mixing, to give a newsolution with a volume of 150.0 mL (= V2 ) . [Ba 2+] after mixing
=
(0.0200 moVl) (O .I OOO l) ~ 0.0133 M 0.1500 l
[50/ -] after mixing
=
(0.0300 moVl)(0.0 500 l ) ~ 0.0 100 M 0.1500 l
Now the reaction quotient can be calculated. Q ~ [Ba' +1150!-J
~
(0.0133)(0.0100)
~
1.33
x
10-'
Qis much larger than Ksp, so BaS04 precipitates. EXERCISE 18.19
Ion Concentrations Required to Begin Precipitation
What is the minimum concentration of 1- that can cause precipitation of PbI2 from a 0.050 Msolution of Pb(N0 3h? K~p for PbI2 is 8.7 X 10 - 9 • Whatconcentration of Pb 2+ ions remains in solution when the concentration of 1- is 0.0015 M?
18.5
I
Precipitation Reactions
845
Dim ethylglyoximate complex of NiH ion
EXERCISE 18.20 K,p
and Predpitation
You have 100.0 mLof 0.00 10 M silver nit rate. Will AgCl precipitate if you add 5.0 mLof 0.025 M HCl?
18.6
FIGURE 18.17 Complexions. The green solution contains soluble Ni(HzO)/ + ions in which water motecules are bound to NiH ions byion-dipole forces. TI1i s complex ion gives the solution its green color. Th e NF +-ammonia complex ion is purple. The red, insoluble solid is the dimethylglyoximate com plex of t he NiH ion [Ni{ C4H70 2N2hl (model at top). Formation of this beautiful red insoluble compound is t he classical test for the presence of t he aqueous Nj2 + ion.
• Complex Ions Complex ions are prevalent in chemistry and are the basis of such biologica lly important substances as hemoglobin and vitamin 812, They are described in more detail in Chapter 22.
Equilibria Involving Complex Ions
Metal io ns exi st in aq ueous solutio n as co mplex ions (.... Section 17.10) (Figu re 18.17) . Co mp lex io ns co nsist o f the metal ion and other mo lecules or io ns bou nd in to a sing le en tity. In wate r, met al ion s are always su rr o u nded by water mol ecul es, with th e negat ive en d of the po lar wate r mo lecu le, the ox ygen atom , attracted to the po sitive me tal ion. III the case of NiH , th e io n exists as [Ni( II zO) 6P + in wate r. On add ing am mo nia, water molecul es are d isplaced successively, and in th e p resence o f a high eno ug h co nce n tration of am mo nia, the co mplex ion [ Ni ( N II ~hP + exists. Man y organ ic molecul es also form co m plex ions with m etal ions, o ne example being th e co mp lex with th e dim ethylglyoxim at e io n in Figure 18.17. The mo lecules or io ns th a t bind to m et al io ns are called ligands (.... Chap ter 22) . In aqueo us so lu tio n, m e tal ions and ligands exi st in eq uilib rium , and the eq uilibrium co nsta n ts fo r th ese reactions are referred to as fo r mation constants, K r (Ap pe nd ix K) . Fo r ex am ple , Cu ' +(aq)
+
NH,(aq)
Kf1 = 2.0 X 10 4
[Cu(NH, )]1+(aq)
[Cu (NH, )]1+ (aq) + NH ,(a q)
~='
[C u(NH,),] 1+ (aq)
Kf1
= 4.7
= 1. 1 X 10'
X 10'
[C u(NH , ),] 1+(aq) + NH,(aq) ,
[C u(NH , h ]1+(aq)
K"
[Cu(N H, h] 1+(aq) + NH,(aq) ,
[Cu(N H,) , ]' +(aq)
Kf4 = 2.0 X 10 2
(In th ese reactio ns, Cu z+ begi ns as [Cu (II zO )4P +' Am mo nia succe ssively displ aces the water molec u les.) O verall , the form ation of the tetraam m ine co p per( lI) complex ion has an eq uilibriu m co ns tan t o f 2. 1 X 10 13 (= Kn X K '2 X K" X K14) •
• Complex Ion Equilibria Problem What is the concentration of Cd z+ ions in a solution prepared by adding 0.00 100 mol of Cd(NO ]h to 1.00 Lof 1. 50 M NH]? Kffor complex ion [Cd(N H])d 2 is 1. 3 X 10 7 • +
Strategy The formation constant for the complex ion is very large, so we can assume nearly all of the Cd z+ ions are in that form. That is, t he initial concentration ofthe complex ion, [Cd(NH ])4] 2+, is 0. 00100 M. This t hen dissociates t o produce Cd 2+ ions in solution .
SoLution le t us set up an ICE table for the dissociation of [Cd( NH ])4Y'" Equation
[Cd(NH,) ,j H(aq)
Initial (M)
0.00100
Change Equilibrium (M)
CH,OH(C) If a SO(un ivcrsc ) is po sitive , th e co nversio n o f I m o l o f CO (g) and 2 mol of H 2 (g) I mol of CII 30 II (t') will be spo n taneo us u nd er sta nd ard co nd itio ns.
10
• Using .lSO(universe) For a process that is sponta neous under standard conditions: ilSO(universe) > 0 For a process at equilibrium under standard conditions: .l SO{universe) = 0 For a process that is not spontaneous under standard conditions: ilSO(universe) < 0 Note tha t these conclusions refer to the complete conversion of reactants to products.
Calculating 8S 0(systern ) To ca lcula te a SO(system ), we sta rt by defini ng th e syste m to includ e th e reacta n ts and p roducts. This mea ns th at a SO(system ) corresponds to th e en tro py cha ng e for th e rea ction , a rs o. Calcu latio n o f th is e n tro py change foll ows th e p ro cedure g iven in Example 19.2.
d SO(syst em) = d ,So = :>: :' SO( products) - :>:dS O(reactants)
= (1 mol CH,OH (f) j mol-rxn)[ SO[C H,OH(C)]}
[(1 mol CO (g)jm ol-rxn)[SO[CO(g )]}
+ (2 mol H,(g)j mol-rxn)[ SO[H,(g)])] = (1 mol CH,OH(f) jmol-rxn)(127.2 JjK' mo l) [(1 mol CO(g)j mol-rxn)(197.7 Jj K' mol) + (2 mo l H, (g)j mo l- rxn)(130.7 JjK' mol)] = - 331.9 Jj K' rn ol-rxn
A d ecr ease in en tro py for th e syste m is ex pected be cause three mo les of gaseo us re actan ts are co nver ted to o nc mo lc o f a liquid produ ct. Calculating 8 S 0(surroundings ) \\rc now need to ca lcu late th e e n tropy change for th e su rrou nd in gs. Recall from Eq uation 19.1 that fo r a reversible change, 60 S is eq ual to f/rcvl T Unde r constan t p ressu re co nd itio ns and assu m ing a reversible p ro cess, th e e nt ro py change in the surro u nd ings results fro m th e fact that th e en thalpy cha nge for th e react io n (f/p = all) affec ts th e su rrou nd ing s. Th at is, the energy asso cia ted with an exo the rm ic chem ica l reactio n is d ispersed in to th e su rro u ndi ngs. Recognizing th at aH O(surrollndings) = - arHO (syste m ) , th e en tro py chang e fo r th e surro u nd ing s ca n be calcu lated by th e equatio n :'SO(surroundings)
872
Chapter
19
I
Principles of Reactivity: Ent ropy and Free Energy
~
d W (surroundings)j T ~ - d ,W (syste m)jT
For the synthesis of m e thano l by the reaction given , the en thalpy change is - 127.9 ~ J / mo l -r x n, calculated from enthalpy of form ation data. ti W (system) = 1 tif W (products) - 1 tif W (reactants) ~ (1 mol CH 30H(C )j mol-rxn){ ti W[CH 30H (C)]) f [(1 mol CO(g)j mol-rxn){ tif W [CO (g)]) + (2 mol H,(g)j mol-rxn){tifW [H ,(g)])]
= (1 mol CH 30H(C)j mol-rxn)(-238.4 kJj mol)
[(1 mol CO (g)jm ol-rxn)( -110.5 kJj mol) + (2 mol H,(g)j mol-rxn)(O kJj mol) ]
=-
127.9 kJj mol-rxn
Assum ing that the pro cess is reversible and o ccurs at a co nstant te mperature and pressure , the e ntro py chang e for the surrou nd ings in the met hano l synthesis is + 429 .2 J/ K . m ol- rxn , calc ulated as fo llows. tiSO(surroundings) = - ti ,W (system)j T ~
- [(- 127.9 kJjmo l-rxn)j2 98 K)](1000 Jj kJ )
=
+ 429.2 JjK . mol-rxn
Calculating 6.S 0(universe ), the Entrop y Change for the System and Surro undings The pieces are no w in place to calculate the e ntro py change in the unive rse . For the fo rmatio n of CII , O II (C) from CO (g) a n d 1I, (g ) , ti S O(uni \'e rse ) , is tiSO(universe)
~
ti SO(syst em) + tiSO(surroundings)
=-
331.9 JjK' mol-rxn + 429.2 Jj K' mol-rxn
= + 97.3 JjK ' mol-rxn
The positive value ind icates an increase in the e ntro py of the universe . It fo llows from the seco nd law of the rmodyna mics that this reactio n is spo ntaneous.
Chemislry/ ,.Now'"
•
Sign in at www.t homsonedu.com/ Ioqtn and go to Cha pter 19 Contents to see: Screen 19.5 for a simulation and tutorial on calculating ASfor a reaction • Screen 19.6 for a simulation and tutorial on the second law of thermodynamics
Determining Whether a Process Is Spontaneous ProbLem Show that
~5 ° ( u n i v e r s e)
is positive for the process of dissolving NaCl in water at 298 K.
Strategy The process occurri ng is NaCl(s) --7 NaCl(aq) . The entropy change for the system, ~ 5 ° ( syst e m ) , can be calculated from values of S° for the two species using Equation 19.3. -l5 °(surroundings) is determined by dividing - .;l,Ho for the process by the Kelvin temperature. The sum of these two entropy changes is -lSO(universe). Va lues of :lJHo and S° for NaCl(s) and Na Cl(aq) are obtained from Appe ndix L. SoLution Calculate j,5°(system) -l5 °(system) ~
=
'l 5 °(products) -
~5 ° ( rea cta nts )
(1 mol NaCI(aq)/mol-rxn ){S O[NaCI(aq) I) - (1 mol NaCI(s)/mol-rxn){SO[NaCl(s)lJ
~
(1 mol NaC I(aq)/mol-rxn)( 115.5 JjK ' mol) - (1 mol Na Cl(s)jmol-rxn)(72.11 J/K' mol)
=
+ 43.4 J/K ' mcl-rxn
19 .5
I
Entropy Changes and Spontaneity 873
• as O(universe), Spont aneity, and Standard Conditions It is important to reiterate that when we calculate J.W or J.S Ofor a reaction, t his is the value for t he complete conversion of reactants to products under standard conditions. If J.SO(universe) is > 0, the reaction as written is spontaneous under standard conditions. However, one can calculate values for ~ S(u n ive rs e ) (without the superscript zero ) for nonstandard conditions. If ~S(unive rs e ) > 0, t he reaction is spontaneous under t hose conditions. However. it is possible that this same reaction under standard conditions is not spontaneous [ ~S O( universe) < 0]. We will return to this point in Sections 19.6 and 19.7 .
Calculate j,SO(surroundings) cl,W (system)
~
:UfW(p roducts ) -
~
(1 mol NaCI(aq)/ mol-rxn)( clfW[NaCI(aq)J) - (1 mo l NaCl(s)/mol-rxn)(clfW[NaCI(s)J)
~ clfW( rea ctan ts )
~
(1 mol NaCI(aq)/ mol-rxn)(-407.27 kJ/ mot) - (1 mol NaCI(s)/ mol-rxn)( -411.1 2 kJ/mol)
= + 3.85 kJjmol-rxn The entropy change of t he surroundings is determi ned by dividing - j" W (system) by the Kelvi n temperature. clSO(surroundings)
~
- cl,W (syst em)/ T
~
[(- 3.85 kJ/ mol-rxn/ 298 K)](1000 J/1 kJ)
=
- 12.9 Jj K. mo l-rxn
Calculate ~ S O (un iverse) The overall ent ropy change-the change of entropy in the universe-is t he sum of the values for t he syst em and the surroundings. j,SO(universe)
= j,SO(system) + ;:lS O(surroundings) = +43.4 Jj K' mol-rxn - 12. 9 Jj K' mol-rxn = + 30.5 Jj K. mol-rxn
Comment The sum of t he two ent ropy quantities is positive. indicating t hat th e ent ropy in the universe increases; thus, t he process is spontaneous under sta ndard conditions. Notice, however, t hat the spontaneity of the process results from j,SO(system) and not from j,SO(surroundings).
In Summary: Spontaneous or Not? In th e p recedin g ex am ples, pred icti on s abo u t the spo n ta ne ity o f a process u nd e r sta nda rd co n d itio ns were mad e usin g values o f Ll SO(systcm ) and LlHO(systcm) ca lcu la ted from tables of th ermodynamic d ata. It will be useful to loo k at all po ssibilities th at re sult fro m th e inte rp lay of th ese two q uanti ties. The re are four possiblc outcom es whe n these two qu anti tics arc pai red Cra ble 19.2 ). In two, LlHO(system ) a nd Ll SO(systcm ) wo rk in co nc e rt (Types 1 and 4 in Table 19.2) . In th e o ther two , th e two quan tities are o p posed (Type s 2 a nd 3) . Pro cesses in wh ich both th e sta ndard en tha lpy and c n tro py cha nge s favor c nergy dispersal (Type 1) arc alwa ys spo n tane ous under sta nda rd co nd itio ns . Pro cesses disfavo red by both their standard en tha lpy and entro py changes in th e system (Type 4 ) ca n never be spo n taneo us und er stand ard cond itio ns. Let us co nsid er exam ples th at illustrat e ea ch situatio n. Co m bus tio n reactio ns are always exo therm ic and o fte n p ro d uce a large r num be r of product mo lccu les fro m a few reactant mol ecules. T hey arc Typ e 1 rea ctio ns. T he eq uatio n fo r th e co mbus tio n of butan e is an exa m ple . 2 C, H,O(g)
+ 13 O, (g) -> 8 co,(g) + 10 H,O(g)
TABLE 19.2 Pred icting Whether a Reaction Will Be Spontaneous Under Standard Conditions clW(system)
clS·(system)
Sponta neous Process (Standard Condition s)
1
Exotherm ic, < 0
Positive, > 0
Spontan eous at all te mperat ures. llSO(universe) > O.
2
Exotherm ic, < 0
Ne gat ive, < 0
Reacti on Type
Depends on relative magnit udes of ;i.Ho and Spontaneous at lower temperatures.
~S O.
8 74
C h a p t e r 19
I
3
Endot hermic, > 0
Positive, > 0
Depends on relative magnit udes of .1. HO and .l s o. Spontaneous at higher temperat ures.
4
End ot hermic, > 0
Ne gat ive, < 0
Not spontaneous at any temperat ure. ;:lS O(universe) < O.
Principles of Reactivity: Entropy and Free Energy
For this reaction , D.Jr = - 53 15.1 kJ, a n d D.rSo = 3 12.4 J / K. Both co n trib u te to thi s re action 's being spo n ta n co us und e r sta n d a rd condition s. l Iyd ra zin e , N ;! H -t, is u sed as a hi gh-en e rgy ro cket fuel. Syn thesis of N ;!II -t fro m gaseous N 2 a n d 11 2 wo u ld be a ttrac tive because th ese reactants a re in e xp en sive.
N,(g)
+ 2 H,(g) --> N,H, (C)
However, this reaction fits into th e Type 4 ca tegory. The react ion is en doth e rmi c (D. rH o = + 50 .fi3 kJ/m ol-rxn ), a n d the cntropy c ha nge is ne gati ve (D.1',\'O= - 33 1.4 .11K· m ol -rxn ) (l m ol of liquid is produced fro m 3 m ol o f gases), so th e reaction is n o t sp o n ta n eo us u nder sta n dard co n d itio n s, an d com p le te co nversio n o f react ants to products will not occu r witho ut o u tsid e inte rventio n . In th e two other po ssibl e outcomes, cntropy and enthalpy ch a nges o pp ose eac h oth er. A pro ce ss co u ld be favore d by th e e n tha lpy c ha nge but disfavo red by th e e n tro py c ha ngc (Type 2), o r vice ve rsa (Type 3) . In e ith e r in stan ce , whether a process is spon ta neo us d epends o n wh ic h fa ctor is more important. Temperatu re a lso influences the valu e o f D.SO(uni ve rse ) . Because the en th alpy cha n ge for th e su rro u n d in gs is di vided by th e tempcratu re to ob ta in D. SO(surroundin gs), th e num erical value of D.SO(surroundings) will be smalle r (either less po siti ve or less ne gati ve ) a t hi gh e r temperatures . In co n tr as t, D.SO(system ) and D.HO(systcm ) do n ot vary mu ch with te m pe ra tu re . T h us , th e e ffec t o f D. SO(surroundings) relative to D.,\'O(system ) is diminish eel at higher tempe rature. Stat ed a n o th er way, a t hi gh e r tempe ratu re , th e e n tha lpy c ha nge becomes a less important factor in d etermining the overa ll cntropy chan ge . Co ns ide r the two ca ses whe re D.f[O (system ) and D. .\'O(systcm ) a re in oppo sition (Ta b le ID.2) : • Type 2: Exo thermic processes with D. S O(syste m ) < O. Su ch p ro cesses beco m e less favo rab le with a n increase in temperaturc. • Type 3: En doth e rm ic processes with D. .\'O(system ) > O. These pro cesses become mo re favorabl e as th e temp e ratu re increases.
FIGURE 19.1 0 Thermal decomposition of NH 4Cl(s). White, solid ammonium chloride, NHll(s) , is heated in a spoon. At high temperatures. decomposition to form NH 3 (g) and HCl(g) is spontaneous. At lowertemperatures, the reverse reaction, forming NH4Cl(s), is spontaneous. As gaseous HCl(g) and NH ](g) cool, they recombine to form solid NH4Cl , the white "smoke" seen in this photo.
The effect o f tempe rature is illustrat ed by two ex a m p les. T he first is th e reac tio n of N 2 a n d H 2 to form N H :~ . The reaction is ex othermic, a n d thu s it is favored by e nerg y di sp e rsal to th e su rro u ndi ngs. The e n tr o py c ha n ge for th e syste m is unfavo ra ble , howe ve r, because th e reacti on, N 2( g ) + 3 11 2 (g ) --7 2 N J-I ~(g ) , con verts fo u r moles o f gaseo u s re act ants to two m ol es ofgas COLIS p roduct s. T h e favo ra ble enthalpy effe c t [D. rSO(surroundings) = - D. J JO(syste m )/ " l becomes less important a t h ighe r te m pe ra tures . Furthe rmore, it is re ason abl e to expec t th at th e reacti on will not be sp o n ta neo us if th e temperature is hi gh e n o ug h . The seco nd exam ple co nsiders th e thermal decompositio n ofN I I-tCI (Figure 19.10 ) . At room temperature , NI-J _ICI is a stable, wh ite, crystalline salt. Wh en heat ed stro ngly, it decomposes to NI-I 3( g ) an d HCI(g ). T he reaction is e n doth e rm ic (e n thalpydisfavo red ) but e n tro py-favored because of th e form at ion o f tw« m oles of gall from o ne m ole of a so lid reactant. The reaction is in creasin gly favored a t h igh er tempe ratures. EXERCISE 19.3
Is a Reaction Spontaneous?
Classify t he following reactions as one of t he four types of reactions summarized in Table 19.2 . Reaction
Jo ,H" (at 298 K) (kJ/mol-rxn)
(at 298 K) (J/K 'mol-rxn)
;",5 0
(a) CH .(g) + 2 O,(g) --> 2 H,O(I ) + (O, (g)
- 8 90.6
- 242.8
(b) 2 fe,O,(s) + 3 C( graphite) --> 4 fe(s) + 3 (O, (g)
+ 467.9
+ 560.7
(c) C(graphite) + O,( g) --> (O, (g)
~ 3 93 . 5
+ 3.1
(d) N, (g) + 3 f, (g) --> 2 Nf, (g)
- 264.2
- 277.8
19 .5
I
Ent ropy Cha nges and Spontan eity
875
EXERC ISE 19.4
Is a Reaction Spontaneous?
Is the reaction of hydrogen and chlorine to give hydrogen chloride gas predicted to be spontaneous under sta ndard conditions?
H,(g) + (I,(g) --> 2 HCl(g) Predictthe spontaneity of t he reaction based on .:lrHo and .:l ,S ", and then calculate .i.SO(universe) to verify your prediction.
EXERCISE 19.5
Effect of Temperature on Spontaneity
Iron is produced in a blast furnace by reducing iron oxide using carbon. For this reaction, 2
F" O,(, ) + 3 ((graphite) --> 4 F, (,) + 3 (O,(g),
the following parameters are determined: .;l,Ho = + 467,9 kJjmol-rxn, and ..l,So = + 560.7 Jj K' mol-rxn. Show that it is necessary that t his reaction be carried out at a high temperature.
Module 24 • J. Willard Gibbs (1839-1903) Gibbs
received a Ph.D. from Yale University in 1863. His was the first Ph.D. in science
awarded from an American university.
19.6
Gibbs Free Energy
T he method used so far to d et ermin e whe ther a process is spo n ta neo us require d evalua tion of two quantities. .6. S0(system ) and .6. S0(slllToundings). vVollldn't it he co nven ien t to have a single th ermod ynamic funct ion th at serves the same pu rpose? A functio n assoc iated with the syste m o n ly- o ne th at d oes no t require assessmen t o f th e su rr o u nd ing s- wou ld he even better. Such a fu nct io n exists. It is called th e Gibbs free en ergy, with the name ho noring J. Willard Gibbs (18 39-1903). Gib bs fre e ene rgy, G, o fte n referred to sim ply as "fre e e nergy," is d efined math em atically as G=H - TS whe re II is enthalpy, T is the Kelvin tem perature, and S is e n tro py. In th is eq ua tion , G, II, and S ail re fe r to the system . Because e n thalpy and e n tro py are state fun cti on s (... Section 5.4 ) , free en erbJY is a lso a sta te fun ction . Every su bsta nce possesses free e ne rgy, but th e actual quan tity is seld o m known . Instead , just as with en thalp y (If) and intern al e nergy (lJ) , we are co ncerned with r!ul1lgeJ in free en erh'l', 6.G, th at oc cu r in che m ica l and p hysica l pro cesses. Let us first see how to use free energy as a way to det ermi ne whe the r a reactio n is spo ntaneous. \Ve ca n th en ask furth er q ue stio ns abo ut th e meanin g of th e term "fre e ene rgy" and its use in dec id ing whether a reaction is p rodu ct- o r reactant- favored .
The Change in the Gibbs Free Energy, t>.G Recall th e eq uatio n d efin ing th e en tro py change fo r the universe : tiS(universe) = tiS(surroundings)
+ tiS(system)
T he e n tro py cha nge o f th e su rro u nd ings equals th e negati ve of th e change en thalpy of the syste m di vid ed by T T h us,
III
tiS(universe) = - ti H(system)jT + tiS(system) Mul tipl ying th rough th is eq ua tio n by - 7; gives th e eq uatio n - Tti5(universe)
~
ti H(system) - TtiS( system)
Gib bs d efi ned th e free e nergy fun cti on so th at 6.G(system ) = - T6.S(u n ive rse ) . Co m bin ing term s and sim plifying give the general exp ression re lating changes in free e nergy to th e ent halpy and e n tr o py changes in th e syste m. ti G = ti H - TtiS 876
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Under sta ndard co nditio ns, we can rewrit e th e Gibbs free energy eq uatio n as (19.5)
Gibbs Free Energy, Spontaneity, and Chemical Equilibrium Becau se a rGo is related d irec tly to a SO(un ive rsc ) , th e Gib bs frcc c ncrgy ca n be used as a cr iterion o f spo n ta ne ity for p hysica l an d che mical changes. As shown earl ier, the signs of a rGo an d 6.S0(un iverse ) will be o pposite s [a rG o (system ) - 'fa So(u ruvcrse ) ]. T herefore, we fin d th e fo llowing rela tio ns hips: j, rGO < 0
The process is spontaneous in the d i re ct io n writte n un de r stan da rd condi tions.
Il,Go ~ 0
The process is at equilibrium un der stan dard conditions.
j, rGo
>
0
The process is not spontaneous i n th e direct ion wri tten under standard con ditions.
To bette r understa nd the Gib bs funct io n , let us exam ine th e diag ram s in Figure 19.11. T he free enc rgy of p u re reactants is plo tte d o n th e left, and th e fre e e ne rgy of the pure p ro duct s on the righ t. T he extent of react io n , plo tted on the x-axis, goes from zero (pure reac ta n ts) to o ne (p ure p rod uc ts) : In both case s in Figure 19.11 , th e free e ne rgy initially decl ines as react ants beg in to for m prod ucts; it reac hes a minimum at equ ilibrium and th en increases aga in as we move fro m th e equ ilibri um position to pure p ro d ucts. 77w [ree energy at equilibrium. when' there is a mixture ofreactan ts a nd products, is always lower than thefree energy of th e pure reactants a nd of the [ntre [miducts , A reaction fmJrlwls .\j)(mtrw eolls1y toward the minimum in free rrwrgy, whi ch corresponds to equilibrium.
a rG o is th e cha nge in free e ne rgy accompa nying the chem ica l react ion in whic h the reactants a re converte d co mp le te ly to th e p ro d ucts u nder sta nd a rd
Reaction is product-favored at equilibrium ~ rG c is negat ive, K> 1
Reaction is reacta nt-favored at equilibrium ~ TG o is positive, K < 1
--------------------f---------
-----f-----------------------------------
1
.i.rGo < 0
____1
~ rG o
>0
I i,;, ---------------------
_
~> K
~ ~
O ~K
Slope = 0 ~G ~ 0 Rea ctants only
/
: : :
Equilibrium mixtur e
~
'"
Q> K Slope > 0 ~G > 0
.s
~
c
.~
~
v
Products only
Stope < 0 .l G < 0
Reactan ts only
Extent of reaction
00
0:;;::'
Stope > 0 .l G > 0
: ~O ~ K ,
Slope = 0 ~G ~ 0
Equilibrium mixture
Products only
Exten t of reaction
00
FIGU R E 19 .11 free energy change s in the course of a reaction. The difference in free energy between t he pure reactants in t heir standard states and t he pure products in t heir standard states is .i.,Go. Here, Q is t he reaction quotient. and Kis t he equilibrium consta nt. 19 .6
I
Gibb s Free Energy
877
co nd itio ns . Ma th em a ticall y, it is th e d iffere nce ill free e nergy be tween th e product s a n d th e re act ants u n der sta n d a rd co n d it io ns . If th c free e nergy of th e p roduc ts is less tha n tha t o f th e re act ants, th e n 8.,. G o < 0 , a n d the reaction is spo nta ne ous u nd e r sta nd ard co n d itio ns (Fig u re 19.11 a ). Co nve rse ly, if th e fre e e nc rgy o f the product s is grca te r th an th at o f th e reactants, th en 8.1' G o is positi ve (8.,. G o > 0 ) , an d th e react io n is n ot spo nta neo us u nd e r sta n da rd co ndi tio ns (Fig ure 19 .1lb ). Fina lly, n otice th at in Fig u rc 19.11 a , th e eq u ilibri u m po sition occu rs clo ser to the p ro duct side th an to th e reac ta n t side . This is a p ro d uc t-favo re d reaction a t eq u ilib riu m . In Figure 19 .110, we find th e o p posite . The reac tio n is reacta n t-favored at equilibrium . It is n o acc ide n t th at th c re action with a n egati ve 8.,. (; 0 is pro d u ctfavo red, whereas th e o ne with th e p ositive 8.1G O is reac ta nt-favored a t eq u ilibriu m . It tu rns out th a t thi s is a lways true as th e di scu ssio n bel ow will show. Now let u s now co n sider wha t h appen s to th e in stantaneous slo pe o f th e cu rve in Figure 19.11 as th e reac tion proceeds. Initially, thi s slope is nega tive , co rrespo ndin g to a n e gati ve 8.G in m oving fro m p oint to poi nt. Even tually, ho wever, th e fr ee enc rbT)' reach es a min imum . At this po int, th e in sta ntan eous slo pe of th e g ra p h is zero (8.(; = 0) a nd th e react io n h as reac he d eq uili b rium . If we move p ast th e eq u ilib riu m p oin t, the in stan ta ne ous slo pe is po sitive (8. G > 0) ; proceeding fu rth er to p roduct s is n ot sp o n ta n eo us. In fac t, th e rev e rse reacti on will oc cu r spon ta ncously; th e reacti o n will o nce again proceed toward equ ilibri um . The re latio ns h ip of 8.,. G o (t he value o f 8.1' (; und er sta nd ard cond itio ns) a n d th e value of 8.,. G u n de r n onstand ard co nd itio ns is given by Equation 19 .6 ~ ,G = ~ ,G o
(19.6)
+ RT lnO
whe re R is the un iversal gas co ns ta n t, T is th e tcmpcrature in kel vin s, and reac tio n qu o tient (... Sec tio n 16 .2 ) .
a=
[CY[O]' for aA [A]'[B]'
+ bB -->
cC
Q is th e
+ dO
Eq ua tio n 19 .6 in forms u s tha t, at a give n te m pe ra ture, 8.,. G is d et e rm in ed b y the values of 8.r G o a n d Q. Furthe r, as lon g as the reacti o n is "desce nd ing" fr om the fr ee cnc rbT)' o f th e pu rc reac ta n ts to th c eq u ilibri u m po sition, 8.r G is ne gat ive. a n d th e reac tio n is spo n ta neo us in th e forwa rd d irect ion (a n d Q < k) . Wh en th e system reac hes equ ilib riu m , no furth er n et c ha nge in co ncentra tio n of react ants a n d p ro du ct s will occ u r, a n d 8.,.G = 0 a n d Q = K Substituting th ese values in to Equat ion 19.6 gives
o = ~ ,G o + RTlnK (a t equili brium) Rcarran g in g th is e qua tio n leads to a u seful rel a tionship between th e sta n da rd free energy change fo r a reaction and the eq ui lib riu m co n sta nt, K, Eq ua tion 19 .7: ~ ,G O
=
(19.7)
- RT InK
Fro m th is e q uatio n , we lea rn th at , wh e n 8.1 G is n egative. K is greate r th an I , an d we say the reac tio n is jnndurt-jrw ornl. The m ore n e gati ve th e value o f 8.J ; o, the lar ge r the equilibrium co n stan t. T h is m a kes sense beca u se , as d esc ribed in Cha p ter 16 , large eq u ilibri u m cons ta n ts a re assoc ia te d with product-favored reac tio ns. The co nverse is also tr ue: For rmaaru-faoored reacti o ns , 8.,. G o is posit ive , and K is less th an I . Fin ally, if K = I (a special se t of cond itio n s) , th e n 8.J ; = O. 0
0
878
Chapter 19
I
Princi ples of Rea ctivity: Entropy and Free Energy
A Summary: Gibbs Free Energy (Ll,G and Ll ,GO), the Reaction Quotient (Q) and Equilibrium Constant (10, and Reaction Favorability Let us now su m marize th e re latio ns h ips among 6.r G o• ArG, Q. an d K • In Figu re 19.1 I, you see th a t free e nergy decre ases to a minim u m as a syste m approac he s equil ibrium . The free c ne rbl)' o f th e mi xture of reac ta n ts and product s at equ ilibrium is always 100\'cr than th e free e ne rgy of the pure reac tants or o f the pure product s. • Wh e n A rC < 0, the reacti o n is p ro ceeding spon ta neous ly toward eq ui lib riu m and Q < K. • \Vhen ArC > 0, the reaction is be yo nd th e eq u ilibrium poin t a n d is not spo ntan eous in th e forwa rd dire ction . It will be spon ta n eous in th e reve rse dire ction; Q > K. • \Vhe n A rC = 0 , the reactio n is a t eq ui libri um ; Q = K • \Vhen a rGo < 0, th e reac tio n is stJOulrl1l('ous under standard conditions. The syste m will proceed to an eq u ilib riu m po sition a t whic h po int th e product s will domin ate in the reac tio n mix ture because K > I . That is, the reactio n is product-favored at eq ui librium . • when 6.r Go > 0 , th e reacti on is not sj)()ntrmeous under standard cond itions. T h e syste m will p roceed to th e eq ui lib ri um po sition at whic h th e reactants will d ominat e in th e eq u ilibriu m mi xture becau se K < 1. T ha t is, th e reaction is imrtant-faumnl. • Fo r th e special co n d itio n where 6.J ; 0 = 0, th e reaction is at eq u ilibriu m at sta ndard co n di tio ns , wit h K = 1.
A reactant-favored process. If a sample of yellowlead iodide is placed in pure water, a smaU amount of the compou nd will dissolve spontaneously ( ~ ,G < 0 and Q < K) until equilibrium is reached. Because PbI" is Quite insoluble (KIP = 9.8 x 10- ~), however, the process of dissolving t he compound is reactant-favored. We may conclude, therefore, that the value of j" G o is positive.
What Is "Free" Energy? Th e term "free e nergy" was not at hi tr arilv chosen . In a ny given p ro cess, th e free e neq.,'y rep resents the m aximum ene rgy ava ilable to do useful work (ma the matica lly, 6.G = wma '.;) . In th is co n text, th e wo rd "free" mean s "availa b le ." To illus tra te the reaso nin g behin d thi s rel ati on sh ip, co nside r a reaction carrie d out under sta ndard co nd itio ns a n d in whic h e nergy is evo lved as heat (6.r H o < 0) and e n tro py d ecreases (6.r S o < 0).
+ O, (g) --7 2 H,O(g) il ,W = - 483.6 kJj mol-rxn and il ,5° = - 88.8 Jj K' mol-rxn il ,Go = - 483.6 kJjmol-rxn - (298 K)( - O.0888 kJjmo l-rxn) = 2 H, (g)
457.2 kJj mol-rxn
At fir st g lance, it mi ght seem reaso n abl e th at all th e e n e rh'T rel eased as heat ( -483.6 ~ J / mol - rx n) wou ld be availabl e . T his e nergy cou ld be tr an sferred to the su rr ou ndin gs an d wo u ld thus be ava ilable to d o work . This is n ot th e case, h oweve r. A nega tive e n tr o py c ha nge in th is reacti on means th at e nergy is less di sp ersed in th e pro d uc ts than in th e reac ta n ts. A portion o f the e nergy rel eased as heat fro m th e reactio n must be used to reverse energy dis persal in th e syste m ; th at is, to co nce ntrate e n er6')' in th e product. The e nerh'Y left over is "free ," o r availab le to pe r form work. H ere , the fre e e ne rgy c ha nge am oun ts to - 457.2 ~ J / mo l-rx n .
19.7
Calculating and Using Free Energy
Standard Free Energy of Formation Th e sta n dard free e nergy of forma tio n of a co m p o und, 6./ G o , is th e free ene rb' Y cha nge th at occu rs to form o ne mol e of th e co m po u nd fro m th e co m p o n e n t clements, with p roduct s a nd reac ta n ts in thei r stan dard sta tes. By d efin ing 6./ (; 0 in th is way, the free purrg)' off ormation of an element in its standard state is zero. 19 . 7
I
Calculating and Using Free Energy
879
Just as the sta ndard en thalpy or entropy change for a reactio n can be calc u lated usin g value s of lirlf' (Equatio n 5.6) or .\., (Equatio n 19.3) , the stand ard free en e rgy change for a react io n can be calc u la ted from valu es o f 8 rG o : (19.8)
Calculating il,Go, the Free Energy Change for a Reaction Under Standard Conditions T he free e nergy change for a reacti o n und e r standard co ndi tions can be ca lcu late d from thermodyna mic data in two ways, either fro m sta nd ard e nt halp y and en tropy changes using values o f 8 .flr a nd S o (as we did for the fo rmation of H 20 just ab ove) o r d irec tly fro m values o f Ii .fGo foun d in tables. T hese calc ulations are illustrat ed in the foll owing two example s. ChemislrY-9-Now~ Sign in at www.thomsonedu.com/Ioqtn and go to Chapter 19 Contents to see Screen 19.7 for an exercise and tutorials on Gibbs free energy.
• . Calculating il,G o from il ,H" and il,S o ProbLem Calculate t he sta ndard free energy change, .l ,Go. for t he form ation of methane from carbon and hydrogen at 298 K. using tabulated values of .lJW and ~ 5 ° . Is t he reaction sponta neous under standard conditions? Is it product-favored or reactant-favored at equilibrium? C(graphite) + 2 H,(g) -? CH ,( g) Strategy Th e values of .lJHo and S° needed in this problem are found in Appendix L. These are first combined to find .1. ,Ho and .1.,50. With these values known, .1.,Go can be calculated using Equation 19.5. When doing t he calculation, rememb er that S° values are given in units of J/ K' mot. whereas .1.f Ho values are given in units of kJjmol. SoLution (graphite) Joj W (kJ/mol)
5" (J/ K'mol)
H,(g)
0
0
+ 5.6
+ 130.7
CH ,(g) - 74.9 + 186.3
From t hese values, we can find bot h .1. ,W and ~ ,S o for t he reaction: Jo,W
~ ~ JojW (p ro d u cts)
-
~.1j W ( ,ea ctan ts)
~
(1 mol CH, (g)/ mo!-rxn)pjW[CH, (g)]} - [(I mol C(graphite)/mo l-,xn) p jW[C(graphite)]} + (2 mo l H,(g)/ mo!-rxn){.1j W[H,(g)])]
~
(1 mol CH,(g)/mo!-rxn)( - 74.9 kJ/ mol) [(I mo l C(graphite)/mol-rxn)(Okf/mol)
=-
+ (2 mol H,(g)/ mol-,xn)(OkJ/mo l)]
74.9 kJ/ moL-rxn
.1,5" ~ };.15"(products) - };.15"(,eactants)
880
Chapt er
19
I
~
(I mol CH, (g)/ mo!-rxn){5"[CH,( g)]) [(I mol C(graphite)/ mol-,xn){5"[C(graphite)]) + (2 mo l H,(g)/ mol-,xn){5"[H,( g)])]
~
(I mol CH ,(gl! mo!-rxn)(186.J J/K ' mol) [I mo l C(gra phite)/ mol-,xn](5,6 J/K ' mo l) + (2 mol H,(g)/ mol-rxn)(IJ O.7 J/ K' mo l)]
=
- 80.7 JjK ' mcl-rxn
Principles of Rea ctivity: Entropy and Free Energy
Combining the vaLues of ~ ,H o and ~ ,S o using Equation 19.5 gives ~ ,G o .
..l,G = o
..l,H o -
~
=
• Enthalpy- and Entropy-Driven Reactions In Example 19. 4, t he product TJ.,5 ° is negative ( -24.0 Jjmol-rxn) and disfavors the reaction. However, the entropy change is relatively small. and il ,W = - 74.9 kJjmol-rxn) is the domina nt term. Chemists caU t his an "enthalpy-driven reaction."
T~.s °
- ) 4.9 kJ/mol-rxn - [(298 K)(-80.7 J/K' mo l-rxn)](1 kJ/I OOOJ) -
50.9 kJ/mol-rxn
~ ,G o
is negative at 298 K, so the reaction is predicted to be sponta neous under standard conditions at this temperature. It is also predicted to be product-favored at equilibrium.
, • Calculating d,G' Using Free Energies of Formation ProbLem Calculate the sta ndard free energy change for t he combustion of one mole of methane using values for standard free energies of formation of the products and reactants. Is t he reaction spontaneous under standard conditions? Is it product-favored or reactant-favored at equilibrium? Strategy Write a balanced equation for the reaction. Then, use Equation 19.8 with values of j"f Gc obtained from Ap pendix L. SoLutio n The balanced equation and values of ~f G o for each reactant and product are: CH,(g) + 2 O,(g) -. 2 H,O(g) + CO,(g)
°
- 50.8
-228.6
- 394.4
Because ~t G o values are given for 1 mol of each substance (the units are kJjmo l), each value of ..lf Go must be mu ltiplied by the number of moles defined by the stoichiometric coefficient in the balanced chemical equation. !l,GO(system) ~
~
~ ~ !l/ G O ( p ro d u cts )
-
~!l/G O ( rea cta nts )
+ (I mol CO,(g)/mo l-rxn)[!l/GO[ CO,(g)])J - [(I mol CH,(g )/mol-rxn)[ !l/GO [CH,(g)J) + (2 mol O,(g)/ mol-rxn)[!l/GO[O,(g)]] ] [(2 mol H,0(g)/mo l-rxn)( -228.6 kl/rnol) + (I molCO,(g)/mo l-rxn)( -394.4 kJ/mol)) [(2 mol H,O(g)/ mol-rxn)[!l/ GO[H ,O(g)]]
- [(I mol CH,(g )/mol-rxn)( - 50.8 kJ/ mol) - (2 mol O,(g )/ mol-rxn)(O kJ/ mot)] ~
-800.8 kl/mcl-rxn
The large negative value of j",Go indicates t hat the reaction is spontaneous under standard conditions and that it is product-favored at equilibrium. Comment Commo n errors made by students in this calculation are (1) overlooki ng the stoichiometric coefficients in the equation and (2) confusing t he signs for the terms when using Equation 19.8.
EXERCISE 19.6
Calcu lati ng d,G' from d ,H' and drS '
Using values of ..lf H c and S° t o find j" ,Ho and j" So, calculate the free energy change, j" Go, for the formatio n of 2 mot of NH 3(g) from the eLeme nts at standard conditions and 25 0 (, N2(g ) + 3 H2 (g) -. 2 NH ,(g).
EXERCISE 19.7
Calculating d,G' from d IG'
Calculate the standard free energy change for the oxidation of 1.00 mol of 50 2(g) to form 50 3(g) using values of j,fGo.
Free Energy and Temperature
• Ent ropy- or EnthaLpy-Favored Ta ble 19.2 describes the balance of ilW and ilSo and the effect of temperature on reaction spontaneity.
Th e d efin ition fo r fre e energy, G = Il - 'IS, informs us th at fre e ene rbl)' is a func tio n o f te m perature, so d l' G o will cha nge as th e tem perature cha nges (Figu re 19.12) . A conseq ue nce o f thi s d e pe nd ence o n tempe ratu re is th at, in certain in stances, reaction s call be product-favored at eq uilib riu m a t o ne tempera tu re
19 .7
I
Ca lculating and Using Free Energy 881
Dehydration of CuS0 4 • 5 H20 and other hydrates is favo rableonly at highertemperatures.
Reaction of potassium with water is favorable at all temperatures.
.l GO > 0
.l GO > 0
j, GO > 0
Reactant-favored reactions .l Go = j, W - r.lS o > 0
1o
1o
I
.l W < 0 ..15 positive 0
.lGO < 0
o
I
Product-favored reactions
j, W > 0 .i So positive
j, GO= j, W - TJ. SO < 0
j, GO < 0
j, GO < 0
Increasing Temperature
Increasing Temperature
Increasing Temperature
Blue line: j, W < 0 and .l So < o. Favored at low r. Red line: .l W > 0 and .l So > o. Favored at high T.
< 0 and .l So > o. Product-favored at all temperatures. .l W
> 0 and ..i.So < Reactant-favored at all temperatures. .l W
o.
Thevariation in .6.,Go with temperature. lU
Chemistry.c>.Now Sign in at www.thomsonedu.comjlogin and go to the Chapter Contents menu to explore an i~teractive version of this figure accompanied by an exercise.
a n d reac ta n t-favo red a t an o th e r. Th ose in stan ces arise when th e D.J / 0 an d Tl:i. rS o term s work in o p p osite di re cti ons: • Pr ocesses th at a re e n tro py-favore d (D.rSo > 0 ) a nd e n th a lpy-d isfavo red (!l, H ' > 0) • Processes th at a re e n tha lpy-favo re d (I:i.J r < 0) a n d e n tro py-d isfavored (!l, S' < 0) Let u s e xplore th e relati onsh ip of I:i. Go a n d T fu rthe r an d illu strat e how it ca n be u sed to adva n tage. Ca lciu m carbona te is th e p rim ary co m p o ne n t o f lim eston e , m a rbl e , a n d sea shel ls. Heatin g CaC 0 3 produces lim e , CaO , a n important ch e m ica l, a lo ng wit h gaseo us CO :!. Thc d at a belo w fr o m Ap pe nd ix L are at 2!J8 K (25 °C ) . (.(0,(5)
--7
(.0(5)
+
(0 ,(9)
~J G o
(kJjmol)
- 1129.16
- 603.42
- 394.36
~J H'
(kJjmol)
- 1207.6
- 635.09
- 393.51
So (JjK ' mol)
91.7
38.2
213.74
For the co nve rsio n o f I m ol of CaC0 3 (s) to I mol o f CaO (s) un d er sta n dard co ndi tio ns. !l ,G' = + 131. 38 k], !l,l r = + 179 .0 kj . a n d !l ,S' = + 160 .2 .11K. Alt h o ug h th e react ion is e n tro py-favo re d. th e large po sitive a n d un favorable e n th alpy c h an ge d omin at es a t 298 K. Thus, th e stan d a rd fr ee e ne rgy c h a nge is po sitive a t 29 8 K a n d I bar, ind ica tin g th at the rea c tio n is reactan t-favored a t eq u ilibriu m u nd er the give n co n d itio n s. 882
Chapter 19
I
Princip les of Reactivity: Entropy and Free Energy
T he te m pe rature de pendence of a rGo provides a m ea ns to turn the CaCO:\ d ecompo sition into a product-favo re d reacti on . No tice th at th e entro py c ha nge in the reac tio n is po sitive as a res u lt of the fo rmatio n of CO 2 ga s in th e reac tio n. Thus, raisin g the te m pe ra ture res u lts in the value o f 'I'dr.)·o b ecomin g in cr easin gly la rge. At a high eno ug h temperatu rc , 'I'drS o will o u tweig h th e e ntha lpy effec t, a n d th e process will become prod uc t-favored at eq u ilib riu m . lI ow h igh m ust th e te m pe rature b e fo r thi s react ion to beco me p rodu ct-favore d ? An estima te of th e temperatu re ca n be o bta ined usin g Eq ua tio n 19 .5 , by ca lcu latin g the tem peratu re at whic h 8 r Go = O. Above that te mperatu re , 8 r (; 0 will h ave a negative value . D. rGo= D. rHo - T6.rS °
o ~ (179.0 kJ/mo l-rxn)(1000 J/kJ) T = 1117 K(or 844 "C)
- T( 160 .2 J/K ' mol-rxn)
lIow accu ra te is thi s res u lt? As n ot ed ear lie r, we can o btain on ly an a pp rox im at e answe r from th is ca lcu lat io n. One so u rce o f e rro r is th e assu m p tio n th at aJr an d asodo no t vary with te mperature , whic h is n o t strictly tr u e . T here is always a sm all varia tio n in th ese value s wh en th e te mp e rature c hanges-not large e nou g h to b e impo rt a nt if t he tem peratu re ran ge is na rrow, but p otentially a pro b lem ove r wider tem pe ra tu re ra ng es su c h as seen in th is exa m p le . As an estima te , h o weve r, a te mpe ratu re in t he ra nge o f 85 0 °C for this reac tion is re a son abl e .
• CaC0 1 Decomposition Expe riments show t hat the pressure of CO 2 in an equilibrium system [Ca C0 3 (s) =:t CaO (s) + COz(g); .1Go = 0] is 1 bar at about 900 "C, dose to our estimated temperature.
Chemistry-9- Now'· Sign in at www.thomsonedu .com/Ioqtn and go to Chapter 19 Contents to see Screen 19.8 for a simulation and tutorial on the relationship of aw , 45 °, and T.
. . Effect of Temperature on t.,G o Problem The decomposition of liquid Ni( CO)4 to produce nickel metal and carbon mo noxide has a .l,Go value of 40 kf/m ol- rxn at 25 °C.
Ni(CO ), «( ) --. Ni(s) + 4 CO(9) o Use values of tJ.f H and So for t he reactant and products to estimate t he temperature at which the reaction becomes product-favored at equilibrium.
Strategy The reactio n is react ant-favored at equilibrium at 298 K. However, if t he ent ropy change is positive for t he reaction and t he reaction is endot hermic (wit h a positive value of tJ. ,HO). t hen a higher temperatu re may alLow th e reaction t o become product-favored at equilibrium. Therefore. we first find tJ.,H o and -l ,50 to see if t heir values meet our criteria for spontaneity at a higher te mperature , and then we calculate the te mperat ure at which t he following condition is met: o = -lrH0 - TtJ. ,50. SoLution Va lues for tJ.f Ho and 50 are obtained from t he chemical literat ure for t he substances involved. Ni(CO} ,( I} ~J W ( kJj m o l)
- 632.0
S"( Jj K · mol)
320.1
--.
Ni(s} 0
+
4 CO (g) - 110.525
29.87
197.67
For a process in which 1 moL of liquid Ni(CO)4is converted to 1 mol of Ni (s) and 4 mol of CO( g). we find .1,HO= + 189.9 kf/m cl-r xn .1,5 °
=
+ 500.5 J/K mcl- rxn
19.7
I
Calculating and Using Free Energy 883
At298 K, the reaction is reactant-favored at equilibrium largely because it is quite endothermic. However, a positive entropy change should allow the reaction to be product-favored at equilibrium at a higher temperature. Therefore, we use the valuesof ~ rH o and ~ ,S o to find the temp erature at which .l ,Go = O. -
~ fG O = ~,H o
o
~
T~,s °
(189.9 kJ/mo l-rxn)(lOOO J/ kJ ) - T( 500.5 J/K ' mo l-rxn)
T ~ 379.4 K (or 106.2 "C)
Case Study
Thermodynamics and Living Things
The laws of thermo dyna mics applyto all chemical reactions. It should come as no surprise, therefore, that issues of spontaneity and ~ G arise in studies of biochemi cal reactions. For biochemical processes, however, a differe nt standard state is often used. Most of the usual definition is retained: 1 bar pressure for gases and 1 m concentration for aqueous solutes with the exception of one very important solute. Rather than using a standard state of 1 molal for hydronium ions (corresponding to a pH of about D), biochemists use a hydronium concentration of 1 X 10- 7 M, corresponding to a pHof 7. Thi s pHis much more useful for bioche mical reactions. When biochem ists use this as the standard state, they write the symbol next to the thermodynamic function. For example, they wou ld write .l Go (pronounced delta Gzero prime ). living things require energy to perform their many functio ns. One of the main reactions involved in providing this energy is the reaction of adenosi ne triphosphate (ATP) with water, a reaction for which ~ ,G o , = - 30.5 kJ/mol- rxn (... The Chemistry of Life: Biochemistry) .
a reaction that is reactant-favored at equilibrium. How then do our bodies get this reaction to occur? The answer is to couple the production of ATP with anothe r reaction that is even more product-favored than AlP production is reactant-favored. For example, organisms carry out the oxidation of carbohydrates in a multistep process, producing energy. One of the compounds produced in t he process called glycolysis is phosphoenolpyruvate (PEP).
0-
1+
0-
0-
~ 0
- O- :I.- O- :I.- O- :I.- O- (H, 0-
0-
0-
H
H
1
HO Al P, aden osi ne t riphosphate
--7
~ ,G o , ~
884
Chapter
19
I
ATP + H,O + 30.5 kJ/mo l
Its reaction with water is product-favo red at equilibrium PEP + H20
Pyruvate + HPj ~ ,G o , ~ - 61.9 kJ/mol
This reaction and the ATP formation are linked t hrough the HPj that is produced in the PEP reaction. If both reactions are carried out, we obtain the followi ng PEP + H20
~
Pyruvate + HPj - 61.9 kJ/mol
ADP + HPj + H+ ~ AlP + H20 ~ ,G o ' ~
OH
+ 30.5 kJ/ mol
PEP + AOP + H+ ~ Pyruvate + ATP ~ ,G o , ~
- 31. 4 kJ/ mol
The overalL reaction has a negative vaLue for ~ ,G o ' and thus is product-favored at equilibrium. ATP is formed in this process. The coupling of reactions to produce a system that is product-favored is used in a multitude of reactions that occur in our bodies.
Princip Les of Reactivity: Entropy and Free Energy
Questions: 1. Consider the hydroLysis reactions of creatine phosphate andadenosine-5'-monophosphate. Creatine Phospha te + H20
~
~ ,G o , ~
One of the key functions of the process of respiration is to produce mo lecules of ATP for our bodies to use. ATP is produced in the reaction of adenosine diphosphate (A DP) with hydrogen phosphate (HPi ~ HPO; - ), ADP + HPi + W
PEP, phosphoenolpyruvat e
H
H
Living t hings use Al P to produce energy.
0-
l
0-
II 2 II
- O- P- O- C- C- OH
I
NJ:--~ ~N)LN
CH 0
-7 ~ ,G 0 ' ~
Crea tine + HP,
- 43.3 kJjmol
Adenosine-5'-Monophosphate + H20 -7 adenosine + HPj + H+ ~ ,G 0 ' ~
- 9.2 kJjmol
In wh ich direction does a reaction that ;5 product·favored at eouiiibrium result: for creatine phosphate to transfer phosphate to adenosine orfor cdenosine-S'smonophosphate to transfer phosphate to creatine? 2. Assume the reaction A(aq) + B(aq) ~ C(aq) + H30 +(aq) produces one hydro· nium ion. What is the mathematical retotionship between j,Go' and ~ G o at 25 ° C? (Hint: Use the equation ~ G = .l Go + RT LnG and substitute .l Go' for ~ G. ) Answers to these questions are in Appendi x Q.
EXERCISE 19 .8
Effect of Temperature on d,G o
Oxygen was first prepared by Joseph Priestley (1733-1804) by heating HgO. Use data in Appendix L to estimate t he temperature required to decompose HgO( s) into Hg(f) and O2(9).
Using the Relationship Between Ll,G o and K Eq ua tio n 19 .7 provides a dire ct rou te to d e te r m in e th e stan da rd fre e e n e rgy c hange fro m ex pe rim e n ta lly d etermi ned e q u ilib ri u m co n sta n ts. Alte r n a tively, it a llow s ca lcu lation of an equ ilib ri um co nsta nt fro m thermochemical d at a co n tained ill tabl es o r o bta in e d fro m a n ex perim ent.
Chemislryj'J"Now
N
•
Sign in at www.thomsonedu .com/Ioqtn and go to Chapter 19 Contents to see Screen 19.9 for a simulation
and tutorial on aG o and K.
Calculating Kp from d,G o Problem Determine the standardfree energy change, .1,Go, for the formatio n of 1.00 mo l of NH3(g) from nitrogen and hydrogen, and use t his value to calculatet he equilibrium constant for this reaction at 25 °C. Strategy The free energy of formation of ammonia represents the free energy change t o form 1.00 mol of NH3(g) from the elements. The equilibrium constant for this reaction is calculated from ~ ,G o using Equation 19.7. Because the reactants and products are gases, t he calculated value will be Kp• Solution Begin by specifyi ng a balanced equation for the chemical reaction under investigation. '/, N,(g) + 'h H,(g) ;::::± NH,(g) The free energy change for this reaction is - 16.37 kJ/mol-rxn (j. ,Go = .lfGo for NH3 (g); Appendix L). In a calculation of Kp using Equation 19.7, we will need consistent units. The gas constant, R, is 8.3145 J/ K' mol, so the value of .l ,Go must be in J/mo l-rxn (and not kf/mc l-rxn) . Thetemperature is 298 K. ~ ,G o ~
- RTI nK
- 16.370 JjK ' mol-rx n ~ ( -8.3145 JjK' mol-rx n)(2 98.15 K) InK, LnKp
= 6.604
x, = 7.38 X 102 Comment This example illust rates how to calculate equilibrium constants from thermodynamic data. In fact, many equilibrium constants you find in the chemical literature are not experimentally determined but are instead calculated from thermodynamic data in this way.
. , Calculating d,G o from K,p for an Insoluble Solid
°.
Problem The value of K,pfor AgCl(s) at 25 °Cis 1.8 X 10- 1 Use this value in Equation 19.7 to determine ~ .G o for the process Ag ;(aq) + (l - (aq) ;::::± Agel(s) at 25 "C. Strategy The chemical equation given is the opposite of the equation used to define K,P; therefore, the equilibrium constant for this reaction is l/ Ksp- This value is used to calculate .l ,Go. Solution For Ag' (aq)
+ (l- (aq);::::± Ag(l(s),
K = l / Ksp = 1/1.8 X 10- 10 = 5.6 X 109 ~ ,G o ~
- RTlnK ~ - (8.3145 JjK' mol-rxn)(298.15 K) In(5.6 X 10' )
= - 56 kJ/mol-rxn Comment The negative value of .l ,Go indicates that the precipitation of AgCl from Ag +(aq) and (l - (aq) is product-favored at equilibrium.
19.7
I
Calculating and Using Free Energy 885
EXERCISE 19 .9
CalcuLating
Kp
from 11.,6·
Det ermine the value of .1. ,G" for t he reaction C(s) + ( 02(g) _ _ 2 CO(g) from data in Appendix L.
Use this result to calculate the equilibrium constant.
Chapter Goals Rev isited
Chemislrv.c\'Now'·
Sign in at www.thoms;nedu ocomjlogin to : Assess your understanding with Study Questions in OWL keyed to each goat in t he Goals and Homework menufor this chapter For quick review, download Go Chemistry mini-lecture flashcard modules (or purchase them at www.ichapters.com) Check your readiness for an exa m
by taking the Pre-Test and exploring t he modules recom mended in your Personatized Study plan.
~ Access How Do I Solve It? tutorials on how to approach problem solving using concepts in this chapter.
Now that yO Il have studied this chapter; JO Il shou ld ask whether J'Ol I have met the chapter goals. In panicular; you should beable to:
Understand t he concept of entropy and its relationship to reaction spontaneity a.
Un d e rstand that entropy is a measure o f e ne rgy d isp ersal (Sectio n 19 .2 ) . Study Question(s) assignabLe in OWL: 2.
b.
c.
Reco gniz e th at a n e n tro py cha nge ca n be de te rm in ed ex perime n tally as th e e ncrbT)' tr an sferred as h eat for a reversib le proccss di vided by th e Kelvin ternperature . (A Closer Look. Sect io n 19 .3 .) Study Question(s) assignabLe in OWL: 40. Id e n tify common processes th at are e n tro py favored (Sectio n 19.4 ) .
Calculate t he change in ent ropyfor system, surroundings, and the universe to determine whet her a process is spontaneous a.
b. c.
Calc u late e n tr o py ch a nges from tabl es of standard e n tropy values for co m p ou nds (Sec tion 19.4 ). Study auestion(s) assignable in OWL: 4. 6. 11. Use sta n dard e ntro py a n d e n thalp y ch anges to predict wh e th e r a reacti on will be spo n tan eo us und er stand ard co n d itio ns (Sectio n 19 .5 and Ta ble 19.2 ) . Study Question(s) assignabLe in OWL: 10, 12 , 33, 35, 47,55,67,72. Recognize ho w tempcrature influen ces whet he r a reaction is spo n tan eous (Sectio n 19.5). Study Question(s) assignabLein OWL: 14, 25, 26.
Understand and use the Gibbs free energy a. b.
Un d e rsta n d th e co n ne ctio n be tween e n tha lpy a n d e n trop y cha nges and th e Gibbs fre e e ne rbT)' ch a nge fo r a process (Sectio n 19 .6) . Un d ers ta n d th e rela tio nsh ip of dr G, dr G o, Q, K, reaction sp o n tan eity, a n d p rodu ct- or rc ac ta n t-favorabiliry (Sec tio n 19 .6) . Q
.lG
Spontaneous?
Q< K
4lG< 0
Spontaneous to the rig ht as the equation is written
Q= K
.l G = 0
Reaction is at equilibrium
Q> K
4lG> 0
Not spontaneous to the right; spontaneous to the left
K
Reactant-Favored or Product-Favored at Equilibrium?
Spontaneous Under Standard Conditions?
1
4lGO
c.
Describe a n d use th e rel at ion ship bet ween th e free e n e rgy ch a n ge under stan dard co n d itio ns a n d eq u ilib riu m co nsta n ts, a n d calcu late K from d rG ° (Sec tio ns I U.G and 1Y.? ) . Study Question(s) assignabLe in OWL: 28,30,44,48,50,61;
d.
Calcu la te th e change in free e n e rgy at sta n d a rd co nd itio ns for a rea ction from the e n th a lpy a nd cntropy changes unde r sta n da rd co nd itio ns or from th e sta n da rd free e ne rgy of fo rm a tio n of reactan ts a n d p roduct" (d I G 0) (Sectio n 19 .7) . Study Ouestion(s) assignable in OWL: 16 . ia. 20, 22. 46 . 56. 75 . 79. Kno w h ow fre e ene rgy c h a nges with tempe ratu re (Sec tio n 19.7 ) . Study Question(s) assignabLe in OWL: 24, 57,59,60,63,65,70.
Go Chemistry ModuLe 24.
e.
886 Chapter 19
I
Principles of Reactivity: Entropy and Free Energy
KEY EQUATIONS Equation 19.1 (page 863): Calcu la te the entropy change from th e e ne rg y tr an sfe rred as h eal for a reversib le p roce ss and the tempe ratu re a t whic h it o ccu rs. il S
=
q rev
T Equation 19 .2 (page 866) The Bol tzm a nn eq uation : The en tro py o f a syste m , S, is propo rt io n a l to th e numbe r o f ac cessib le m icro sta tes, H: belongi ng to a given e nergy of a syste m o r su bs ta nce. 5 = k lnW Equation 19.3 (page 870) : Calcu la te th e sta n dard e n tro py change u n de r standard co n d itio n s fo r a proce ss fr om th e ta bu lated e n tro pies o f the prod ucts a n d reacrants.
:',5 °=
~5 °( p roducts)
- H O( reactant s)
Equation 19.4 (page 872): Ca lcula te th e to ta l entropy c ha n ge fo r a system an d its su rro u nd ings , to d etermin e wh e the r a proce ss is spo n ta ne o u s under standard co n d itio n s.
:'5 °(universe) = :'5 °(system)
+ :'5°(surroundings)
Equation 19.5 (page 877): Calc u late th e free e ne rgy c h an ge for a p ro cess fro m en tha lpy an d entropy changes.
Equation 19.6 (page 878): Rela tes th e free e nergy cha nge unde r n on standard co ndition s (d iG) to th e sta n dard free e n e rgy cha nge (Ll,.C 0) and the re acti on quotient
Q :' ,G = :' ,Go + RT lnO Equation 19 .7 (page 878) : Rel at es the stan d a rd fre e e ne rgy c h an ge fo r a reac tio n and i t "; equ ilibriu m co nsta nt.
:' ,GO = - RTlnK Equation 19.8 (page 880): Ca lcu late th e stan dard free e ne rgy c h a nge for a reac tio n u sin g tabulated va lues of Llf ( ; " .
:' ,Go ~
~ :'fG O(prod ucts)
-
~ :'fG O( rea ctan ts)
STUDY QUESTIONS Online homework for this chapter may be assigned in OW l.
... denotes chalLenging questions. •
denotes questions assignable in OWL.
Blue-numbered questions have answers in Appendix 0 and fully-worked solutions in t he Student Solutions Manual.
Practidng Skills Entropy (SI'I' Ex a m/l Ip.\' J 9. J and
J 9.2 (HI d
CJU'11/i.~l ryN()w St:H'('1l.\
/9. 4
and
/ 9.5.) 1. Wh ich substance ha s the h igh e r e n tro py? (a) d ry ice (so lid CO, ) at - 7H "C o r CO , (g ) a t 0 "C (b) liquid wate r a t 25 °C or liquid water a t 50 °C (c) pure a lu m ina . A1',/03(s), or ru by (Ruby is AI',/O j in whic h some AP + io ns in the cr ystall ine la tti ce are re placed with Cr3+ ion s.) (d) on e mo le of r'\2(g) a t 1 ha l' pressure o r one mole of N 2(g) a t 10 ba r pressure (bo t h at 298 K)
887
2. • Whi ch su bs ta nc e h as th e h igh e r e n tropy? (a) a sam ple o f pure silic o n (10 he used in a com p u ter ch ip) or a piece of silico n co n ta in ing a tr ace of a nothe r e leme n t suc h as boron or ph osp horu s (h) O , (g ) a t 0 "C or O , (g ) at - 50 °C (c ) 12(s ) o r 12(g ) , bo th a t room le mpe raturl' (d) o ne mole of 0 2(g ) at 1 bar pressure o r on e mo le o f 0 2(g ) a t OJ)} hal' p re ssllf(' (bo th a t 29H K )
3. Use S o values to calc ulate th e standard e n tropy change, D.,-So, for each of the foll owin g proce sse s an d co mmen t on th e sign of th e change . (a) KOH (s) -> KO H (aq ) (b) Na(g) ---7 Na(s) (e ) Br , (f )-> Br, (g ) (r1) HCI (g ) -> HCI (aq ) 4. • Use So val ues to calculate th e stan da rd e n tropy change, D.,-So, fo r eac h o f th e fo llowi ng cha nges, a nd co m men t on the sign of the chan ge . (a ) N II ,CI(s ) -> N l t,CI (aq ) (h) C II 3 0 Il(f) -> C II , OI l(g) (e) CCt,(g) -> CCI, (£) (r1 ) Na CI (s) -> Na CI(g )
II. Calc ul ate tiS"(un iverse ) fo r th e d ec o mpositi o n o f 1 mol of liq uid wa te r to form gas eo us h yd rog e n an d o xyg en . Is thi s reactio n spon ta neo us und e r th e se co ndi ti o n s at 25 °C? Ex plain yo ur answer briefly.
12. • Ca lcu la te D.SO(un iverse ) fo r th e form ation of 1 mo l H CI(g ) li·OI11 gaseo us hyd roge n a nd chlori ne . Is thi s re ac tion spo n taneous under th ese conditio ns a t 25 °C? Expla in yo ur a nswer briefl y,
13. Classify eac h of th e reactio n s acco rdin g to o ne of the four reac tion types su mmarized in Table 19.2 . (a) Fe,O, (s) + 2 AI(s) -> 2 Fe (s) + AI,03(s)
D.,.lf" (h) N , (g)
=+
85 1.5 kJ / m o l-rx n d rS "
~J / mol-rxn
ti TH" = 66.2
d rS o = - 12 1.6 .1/ K . mol-rx n
14. • Classif}' eac h of th e re actio n s acc o rd ing to o ne of th e four reactio n types summa rized in Table 19. 2. (a) C,;l I"O, (s) + 6 O , (g ) -> 6 co, (g ) + 6 11,0(£)
D. ,Ho= - 673
~ l /m o l-rx n
D. ,S "
tio n of 1.0 mol o f th e fo llowin g co m po un ds fro m t he e lem ents a t 25 "c. (h) Ca( O Il), (s ) (a) IICI (g)
7. • Ca lcu la te t he sta nd ard entropy ch an g e fo r eac h of the foll owing re action s a t 25 "c. Co m me n t o n th e sign of (a ) 2 AI(s) + " CI, (g ) -> 2 AICI, (s) (h) 2 CH,OH (f ) + " O , (g) -> 2 CO, (g ) + 4 H,O (g )
s.s-,
8. Calculate the sta nd a rd e n trop y change for e ac h of the followin g reactio ns a t 25 "C. Co mme n t o n the sign of (a) 2 Na(s) + 2 II , O (£) -> 2 Na O Il (a q) + 1I, (g ) (h) Na ,C O ,(s) + 2 HCI(aq ) -> 2 NaC I(aq) + H, O (f ) + CO ,(g )
s.s-,
(h) :Vl gO (s)
888
I
= 60 .4 .11 K
C(grap h ile ) -> Mg (' )
+
· mol-rx n
CO (g )
~J /nlO l ~rxn
ti ,-S o = 197 .9 J/ K . m o l-rxu
Gibbs Free Energy (See Example / 9.4;
J(>(>
CJmni.\lryNow
SO'fPIl
/9.7.)
15. Using values of ti tJr a nd S o, calculate d rG o fo r ea ch of the fo llowing re actio ns a t 25 "c. (a) 2 Ph (s) + O , (g ) -> 2 I'hO (s) (h) N H, (g) + HNO, (aq ) -> N H.,NO ,( aq)
Wh ich of the se re act io ns is (a rc ) p red icted to be p roduct-favo red at equilibrium ? Are the re acti on s c n thalpy- o r e n tro py-d rive n?
16. • Usi ng valu es o f titll" an d S" , ca lcu la te D. rG " for each of th e foll m..:in g reactions a t 25 °C. (a ) 2 Na (s) + 2 H, O (f ) -> 2 Na O Il (a q) + 1I, (g ) (h) 6 C(grap hi le) + " lI,(g) -> C"II ,(t) Wh ich of these reaction s is (are ) p redi cted to be produ ct-favo red at equi libriu m ? Arc the re acti on s en thalpy. or e n tro py-d riven?
(S()(> Example / 9.3 and Cht'1JIi.\tI)NmlJ Screen 19.6.)
10. • Is the reacti o n Si (s) + 2 112(g ) ---7 Sill..(g ) spon tan eous u nd e r sta ndard con d itio ns at 291'\ K? Answe r th is que stio n h y calcula ting ca lc ula ting D. S" (syste m ), D. SO(su rr oundin g s) , a nd D.S" (universe ) . (Defi ne re acta nt s a nd p ro ducts as th e svstc m .)
+
D.J/" = 490 .7
ti rS°(universe) and Spontaneity 9. Is th e reacti o n Si fs) + 2 CI2(g ) ---7 SiCI..(g) spon tancous u nd e r sta ndard co nd itio ns at 29H K? An swe r thi s q uestion hy ca lcula ting ti .\''' (system ) , D.SO(su rr oundin g s) , a nd D.S" (universe ) . (Defi ne rcacrants a nd p ro ducts as th e syste m .)
- 375 .2 .11 K · mol-rxn
2 O , (g ) -> 2 NO ,(g)
5. Ca lcu la te the sta nda rd en tro py cha nge fo r the fo r ma-
6. • Ca lc ula te t he sta ndard cu tropv change fo r the fo rma tio n of 1.0 m o l o f the foll owing co m pou nds from th e ele me n ts a t 25 "C . (a) H, S (g ) (h) MgCO, (s)
=
17. Using values of ti f H o an d S o, ca lcul a te the stan dard m o lar fr ee e neq.•~ · of form a tio n , tit G ", for e ach of the fo llo win g co m pou nds : (a) CS,(g) (h) Na O Il(s) (e) ICI (g ) Co m pa re yo u r calc u lat ed values of D.f G" wit h th o se listed in Appe ndi x L. Wh ich o f the se fo rm a tio n reactio ns arc p redict ed to be spon tan eous under sta n dard co n di tio ns at 25 "C ? •
more chaLLenging
•
in OWL
Blue-numbered questions answered in Appendix 0
18. • Using values o f dl I" an d So, calc ula te th e sta nd ard mola r free e n ergy of form ati on , dl ;', for each (If th e fo llowing: (a ) Ca (O Il),(s) (b) C l(g) (e) Na,CO ,,(s) Co m pare your calc ula ted values of d JG" with th o se liste d in App endix L. \·Vh ich of th e se form ation rea ctions arc p red icted to b e spon taneous u nde r stan da rd co nditions at 25 OCr Free Energy of Formati on (.)1'1'
Exam/)Ip /9.5;
SPI'
Chl'mL\l1}Nm'll Screen /9.7.)
+
CO 2(g ) , d ,G" = + 219 .7 kJ. Using thi s val ue an d oth e r d at a ava ilable in Ap pe n d ix L, calcu la te th e value o f d j G " for RaCOj (s) .
22 • • For th e reacti on T iC12(s) + C12(g) -? T iC1.j (f ) , d rG o = - 272.8 k]. Us ing thi s value a nd o th e r d at a available in Ap pen dix L, calc ula te tlu - value of d j G for Ti Cl A s) .
23. Dete rmine ' . . hc th c r th e rea cti on s listed bel ow are e n tro py-Favored or di sfavored u nde r stan dard co nditio n s. Predi ct h ov..' all in crease ill tem pera ture wil l atfee t th e va lue of d ,- CO . (a ) N, (g ) + 2 O ,(g ) ---> 2 NO , (g ) (b) 2 C(.s) + O , (g ) ---> 2 CO(g) (e) Ca O(s) + CO,(g ) ---> CaCO ,,(s) (II ) 2 NaCI(s) ---> 2 Na (s ) + Cl, (g ) 24 . • Determ in e whe ther th e reac tions listed bel ow arc e ntropy-favored or disf avored unde r stan dard co ndition s. Pred ict how an increase in te m peratu re will affect the va lue of d ,- Go • (a) I, (g ) ---> 2 I(g) (h ) 2 so, (g ) + O , (g ) ---> 2 S0 3(g ) (e) SiCL, (g ) + 2 II ,O (f) ---> SiO, (s) + 4 II Cl (g ) (II ) P, (s, white} + (; H, (g ) ---> 4 PH 3 (g )
morechallenging
• in OWL
CO , (g)
(a) Calcu la te d ,-H" and ts. S" for th e reac tion . (h) Is th e re a ct io n spon ta neous un de r sta nda rd co ndi-
tion s at 298 K? (c) Is the rea ct ion p re d icted to he spou tam-o us a t hi gh e r temperatures? 26 . • Ca lc ula te tx.H" a nd d ,- S o for th e react ion of tin ( IV) oxi de with carbon.
+ C (s ) ---> Sn (s) + CO ,(g )
(a) Is th e reactio n spo n tan eo us un de r stan d a rd co ndition s a t 298 K? (h ) Is th e reacti o n pred icte d to be sponL.'1Ill'OUS a t h igh er temp eratu res?
(Sf''' Exam/)I" /9.7; use d G 0 Screen 19. 9.)
Blue-numbered questions answered in Appendix 0
= - R T InK; S" I'
r:hnlliJtryVollJ
27. T he sta n d a rd free e ne rgy ch a nge , d ,- G o , for th e form ation o f NO(g ) fro m its elem en ts is + 86.58 kJ /mol at 25 °C . Calcu la te K1> a t t h is temperatu re fo r th e eq u ilibr iu m
Y, N,(g) + Y, O , (g )
NO(g)
o
Co m men t o n th e sig n o f dG a nd th e magnitud e of KI" 28. • Th e stan dard fr ee en ergy cha nge , d , G o , fo r th e form at io n of O :l(g) from 0 2(g) is + 16:-l.2 kJ / lllo l at 25 "C. Calcu la te K p a t thi s te mperatu re for the eq u ilib riu m
0
Effect of Temperatu re on d G (Sf'1' Exam/)I" / 9.6 and ChnnixtryN()w Saven / 9.8.)
•
+
Free Energy and Equilibri um Constants
+ ti H,O (g )
20. • Using val ues of d jG' , calcu late d ,G' fo r eac h of th e foll oc.. . ing re acti on s a t 25 °C. Whi ch a re produ ct-favo red at equi lib ri um ? (a) II gS (s) + O ,(g ) ---> II g (C) + SO , (g ) (h) 2 II ,S (g ) + :> O , (g) ---> 2 1I,0 (g ) + 2 SO ,(g) (e) SiCI, (g ) + 2 Mg (s) ---> 2 MgCl, (s) + Sits) 21. Fo r th e react ion BaC0 3 (s) -? BaO (s)
MgC0 3 (s) ---> Mg O (s)
Sn O , (s)
19. Using values of d I G' , calc ula te d ,e;, for e ach of th e fol lowing reactions at 25 °C . Whi ch are product-favored a t equil ibrium ? (a) 2 K(s) + CI,(g) ---> 2 KCl (s) (b) 2 Cu O( s) ---> 2 Cur s) + O , (g ) (e) 4 N H 3(g) + 7 O , (g ) ---> 4 ;-': O,(g)
25. • He ating so m e m et al ca r bonates . amo ng th em m agnesium carbo na te . lead s to th e ir d e compo sit ion .
:> O , (g )
2 0 3(g )
Co mme n t o n th e sig n o f d G o a nd th e m agnitude of K p .
29. Calcu la te 6"J ; o a t 25 °C fo r th e for ma tio n of on e mol o f C 211 j ,(g) fro m C 211 4 (g) a n d 11 2(g ) . Use th is value to ca lcu la te KI' for th e eq u ilib riu m . C, II ., (g )
+ II , (g )
C, II ,;{g) o
Co m me n t o n the sig n o f d ,.G an d the magn itude of K p .
30. • Ca lcul ate d , G o at 25 °C fo r th e form ati on o f I mol o f C ~ II :IO II (g ) fro m C 2 I1 4 (g) a n d H 20 (g) . Use thi s val ue to calc ula te KI' fo r th e equi lib rium . C, H., (g)
+ H,O (g )
C, H, O H(g ) o Co m me n t o n th e sig n of d , G an d the m agn itude o f Kp .
I
889
General Questions TIU'sp questions ore not designated (is to l)'j)(' or location ill the dwlJfer. The')' l1W)' combine seneml concepts.
37 . Wh en vapors fro m hydrochlo ric ac id a nd aq ueo us am m on ia com e ill co n tact, th ey reac t. p ro ducin g a wh ite "clo ud" of so lid NII 'ICI (Figure 19.10 ) . II CI (g )
31. Co m pare t he fo r m u las in eac h sc i o f co m p o u nds, am i deci de whic h is expecte d ( 0 h ave tlu- h igh e r cn tropv, Assum e a ll are at th e sa me temperatu re . C heck yo ur a nswe rs usin g" data in Ap pe nd ix L. (a) HF (g ) , HCI (g ) , or H Br (g ) (b) N H ,CI(s ) or N H.,CI( aq ) (c) C ~HI ( g ) or N'2(g ) ( IWO su bs tances wi th th e sa m e mola r mass) (a) NaCI (s) or NaCl(g)
32. Using stan dard entropy vallies, calc ula te 1:1,.\-.0 for th e form a tion of 1.0 m ol of N H 3 (g) fro m Nz(g ) and 1I2 (g ) at
~5
0(;.
33 . • Abou t 5 bill ion kilog ra ms or be nzene, Cjil-l j ), a rc m ad e each vca r. Ben zen e is u sed as a star tin g m at eri al fo r m any o ther com poun ds and as a so lve n t (alt h o u g h it is a lso a ca rci nogen, and its u sc is re st rict ed ) . On e co m pound th at ca n he made fro m benzen e is cyclo hcx anc , c'IjH 12. C,;l ldO
D.Jr
=-
=-
34. l lvd ro gcu at ion , th e addi tio n of h yd rogen to a n organic co m poun d , is an in dust riall y importa nt react ion. Calc ula te l:1Jr , I:1 r.\' o, an d 1:1,.{; 0 for the h ydroge n ati o n of oc t cnc , CXH !li, to g ive octan e , CHH u" a t 25 CC. Is th e re actio n prod uct- or re actant-favored a t equi libriu m?
Alo ng with dat a in Appe ndix L. th e following info rm atio n is need ed for thi s ca lcula tio n . Compound
D.JH' (kJ/mol)
Octene
- 82.93
Octane
- 208 .45
so (J/K . mol) 462.8 463.639
35• • Is th e com bustio n of eth a ne , C 2 11h , produ ct-favored a t eq ui libri um a t 25 °C?
C, H,;(g)
38. Ca lculate d SO(syslem ), I:1 S0(su rroundings), I:1S0(un ive rsc) for e ach of th e follo win g- p ro ce sse s a t 298 K, a nd co mme n t on ho w th ese systems diffe r. (a) H NO, (g ) ---> H N0 3 (aq ) (h ) Na O Il(s) ---> Na O Il (aq) 39 . Meth anol is now widely used as a fuel in race cars. Co nsi de r the followin g re act ion as a possibl e syn thetic ro ute to met hano l. C(gra phi lC)
40 . • Th e en th al py of vap o rization of liquid di cthvl et her, (C2 I1:,h O , is 2G.O kJ/m ol a t th e boi ling point of 35.0 "C. Ca lcu late d So for a vapo r-to-liquid tra nsformatio n a t 35.0 -c . 41. Ca lc u late th e e n tro py cha nge , 1:1 SO, fo r the va porization o f e th a no l, C 2H",O H, a t its no rm al boi ling po int, 7H.0 " C. T he e n thalpy o f vap oriza tion of et han o l is 39 .3 kJ/m oL
42 . Using therm ody na m ic dat a , es tim a te th e no rm al boiling point of ethanol. (Reca ll th at liq uid an d vapor a re in e q u ilibriu m a t 1.0 ann p re ssu re at the no rma l boi ling poiru. ) Th e a ct ual n orm al ho iling point is 78 °C, Ho w well docs yo u r calc ula te d res ult agree with th e actu al valu e ? 43. Th e followin g reac tio n is reactan t-favo red a t cqu ilibrium a t ro om temperatu re . COCI,(g ) ---> CO(g)
+ Y, O , (g ) ---> 2C O ,(g) + 3 H,O (g )
Answer th e que stion by ca lcu lating th e value of I:1So (u n iverse) a t 298 K, using values of I:1 j l r an d So in Ap pen d ix L. Do c s the answer agree with yo u r pr econce ived idea of this reaction?
I
+ CI, (g)
\Vill raisin g o r lowe rin g th e tempe rature ma ke it product-favored? 44. • When calciu m carbo nate is he ated strongly. CO 2 gas is evo lved. Th e eq u ilibrium pre ssure of co, is 1.00 bar a t H97 "C, a nd I:1 ,H o at 29H K is 179.0 ~ J.
36. Write a bala nced equa tio n th at depic ts the fo rm ation o f 1 mol of Fe20 3(S) from its c le m ents. What is t he stan dard fre e ('lH'rgy of fo rm ati o n of 1.00 mol of Fe 20 :\(S) ? \-Vh at is th e value of 1:1 (; 0 when 454 g (l Ib ) of FC20 3(s) is form ed from th e clem en ts? 890
+ Y,O ,(g) + 2 H, (g ) =" C H,O H (f)
Calculat e K p fo r th e for ma tio n o f me th an ol at 298 K using th is react ion . Wo ul d a d iffe re nt te m pera ture be bett e r sui ted to thi s re actio n?
36 1.5 11 K
Is thi s reac tio n predi cted to be pro d uc t-favo red a t eq uilib rium a t 25 °C? Is th e re ac tio n e n tha lpy- o r e n tropydriven ?
I\ l l ,CI(s)
Definin g the reactants a nd p roduc ts as th e system un d er stu d,": (a) Pr ed'ict whet her 1:1 ,\'O(system ) , 1:1 ,\'O(surro u ndi ngs). I:1 S0(unive rse ), I:1 .-Ho, an d d .- (; O(a t 29H K ) arc g rea te r th an zero , eq ua l to ze ro , o r less tha n zero ; an d exp lain yo ur pred ictio n . Verity yo ur p red ict ion s by ca lcu lat ing valu es fo r each of th ese qu antiti es. (h ) Calcula te the value of K p fo r thi s reacti on at 298 K.
+ 3 Il ,(g ) ---> C6 1l ,,(C)
206.7 q ; D.,SO
+ 1\1I 3(g )
CaC0 3 (s) ---> Ca O( s)
+ CO ,(g)
Esti m a te th e value o f !i .- S o a t H97 °C for th e reac tion .
•
more rhallenqinq
• in OWL Blue-numbered questionsanswered in Appendix 0
45 . So d iu m reacts vio len tly with wate r accordi ng' to th e eq ua tio n :-Ia(s)
+
H,O (t) -> :-IaO H (a q )
+
" H,( g)
Without d o in g' ca lcula tion s, p red ict th e signs of d rHo an d dr ,\'O for the re ac tio n. Verify yo ur p red icti on with a ca lcula tio n .
46 . • \
B (s)
+
3 HC I(g )
Ca lcu la te d ,Ho, d r,"o, an d d , Go a t 25 °C fo r th is reactio n . Is th e react io n predict ed to he pro duct-favo red a t equil ibrium at 25 °C ? If so, is it e n th alpy- o r e n tropyd riven? [SO fo r R(s) is ,'l.8 G.11K · mol.]
4H. • .... Estimate th e vap or pressu re of et hanol a t :n °C using th erm o d yn a mi c da ta. Ex press th e result in m ill of mercury. 49. T he eq ui libri u m co ns ta nt, K Il , for N 20 ..(g ) ~ 2 NO ';,!(g ) is O. I 4 a t 25 °C. Calcula te d ,(;o for the co nvers ion of N 2 n 1(g ) to NO ';,! (g ) from thi s co ns ta n t, ami com pa re this value with th at d ete r min ed from th e d J(; O values ill Ap pen dix L
50. • ... Estima te th e boi ling po int of wat er ill De nve r, Co lo rado (whe re th e altitude is I ,GO km an d th e atm osp heri c p re ssu re is 6:{() mill 0.8 40 bar) . 5 1. T he equi libri u m consta n t fo r the bu ta ne +=t- iso b utan c eq ui libriu m a t 25 °C is 2.50, Calcula te d rGo at th is te mpe ramre in uni t.. of ~J /m ol.
buta ne
CH,cH,CH,CHJ
isobut ane
-
CHJ
I
52. • A crucia l reac tio n for th e produ ct ion of syn th e tic fu -
els is th e co nversio n o f coal to chemical rea ct ion is C( s)
+
H~
with ste am. T he
+
Il , O (g ) -> CO (g )
Jl 2 (g )
o
(a ) Calcu late d rG for th is reac tio n a t 25 °C , assu m ing' C(s ) is gra p h ite . (b ) Calculate AI' fo r th e reac tion at 25 °C, (c) Is th e re actio n pre d ict ed to he produ ct-favo red a t equilibrium at 25 °C? If no t, a t wha t tempera ture will it bec ome so ?
53 . Calc ula te d ,-Go fo r th e d e co mpo sition of sul fur trio xid e to su lfu r diox ide a nd ox yge n. 2 SO , (g )
2
so, (g) +
O , (g )
(a) Is th e re ac tio n prod uc t-favored a t e q u ilibri u m at 25 °C ? (h ) If th e reactio n is n ot product-favored a t 25 °C , is there a temperatu re a t wh ic h it will he come so? Estim a te th is tem perature , (c) Wh at is th e eq u ilibri um consta n t for th e reaction at 1500 ' C?
54. Me th an ol ca n he m ade by par tial oxi da tio n o f m eth ane by O , (g ) . C H.,(g )
+
y, O , (g)
C H , O H (C)
(a ) Det erm in e d S O(syste m ) , d .\' O(su rround in gs) , a nd d .)' o (universe ) fo r th is p ro ce ss, (h ) Is thi s reac tion p roduct-favo red a t e q ui libriu m at 25 °C?
55. • A cave in Mexico was recen tly d isco ve red to have so me interestin g chem istry. Hyd ro ge n sulfi de, H';,!S , reac ts with oxyge n in th e cave to g ive su lfu ric ac id , wh ich d rips fro m th e ce ilin g in droplets with a pH less th an I. Th e reactio n o ccu rri ng is Jl ,S (g )
+
2 O ,(g ) -> Il ,SO, (C)
Calc ula te s.u-, d rS o, a nd d, Go. Is th e reaction p ro du ct-favored a t equ ilib riu m at 25 °C? Is it e n thalpyor cutropv-drivc n?
56. • Wet limesto ne is used to scru b SO ';,! gas fro m th e exhaust gases o f powe r pl ant s. OIH' po ssible reaction gives h yd rat ed calcium su lfite : CaCO,,(s)
+
SO , (g)
+
" H,O( f}
=
CaS O , . " Il ,O (s)
CH,cHCHJ
+
SO , ( ~)
+ Y,
II ,O (t)
+
" O,( g )
Ca SO ., . Y, H , O (s)
Kc
•
morechallenging
~
• in OWL
[but ane]
~
CO, (g )
An o ther rea ction g'ives h yd ra ted calci u m sulfa te: CaCO, (s)
[isobuta ne]
+
1.50 at 198 K
Blue-numbered questions answered in Appendix 0
+
CO , (g)
(a ) \ Vhich is th e more product-favored reaction ? Use th e da ta in th e tabl e o n the ne xt page an d an y o ther informa tion n eed ed in Ap pen dix L to calc ulate d ,Go fo r eac h reactio n at 25 °C.
I
891
~fW
(kJ/ mol)
- 1311. 7
S ° (J/K' mol)
121.3
- 1574.65 134.8
61. • Calc ula te df Go fo r HI (g ) at 35 0 °C, g ive n the foll owin g e quilib ri um pa rtial p re ssu re s: P(H 2) = 0.132 ha l'. P(l 2) = 0.295 ha l', a nd P(lII ) = 1.6 1 bar. At 350 °C an d I ha l', 1:/ is a gas.
\!., H , (g ) + \!., I,(g )
(h ) Calcula te fl. ,G o fo r th e reactio n Ca SO :\ · ~ I h O (s) +
12 0 2(g )
E
CaSo., · !-2 11 20 (5)
Is thi s reaction p ro d u ct- or react an t-favo re d at eq uilibriu m? 57 . • Su lfur un de rgoes a p hase t ran sit io n betwee n 80 an d 100 °C . S,-; (r hom h ic )
~
d ,- H e = 3.2 1:1 kJ /lllOl.rxn
62 . • Calculate t he eq ui librium co ns ta n t for the form at io n of NiO at 1627 °C . Can the reac tion p ro ceed in th e fo rward d irection if th e ini tia l p re ssure o f O 2 is below 1.0 0 mm ITg ? It-fG o [NiO (s) I ~ - 72. 1 k.J/mo! a t 1627 °C j N i (s)
S,'\ (nlOllocl in ic )
t1 r S c = 8.7 .1 / K . mol -rx n
(a ) Estimate aJ;o fo r th e tr ansi tion a t 80 .0 °C an d 110,0 "C. Wh at do these resu lts te-l l you abo ut th e sta bility o f the two fo r m s o f sul fu r a t each of these tem perat u res? (b ) Calcu la te th e tem pera ture a t wh ich !1,G " = O. Wh at is th e sig n ific a nce of th is te mperature ?
TiO, (s)
Free Energies of Formation at 727 °C, kJjmol
TiOAs)
- 757.8
Ti C(s)
- 162.6
CO (g)
- 20 0.2
(a ) Calcu late dJ; o and K at 72 7 0(: (b) Is th e reacti on p ro duct-favo red a t e q ui libri u m at this te mpe ratu re ? (c) How can the reactan t or product co nce ntratio ns be adj us te-d for th e reactio n to he produc t-favo red a t 72 7 °C ?
59 . • Some meta l ox ides ca n he d e composed to th e me tal an d o xyge n unde r reasonab le co nd itio ns. Is th e decompositio n of silver( I) o xide prod uct-favored a t 25 °C?
64. Ci splatin [ci.rd ia m m ine rlic h lo ro pla tin u m (II) ] is a po te nt treatme nt fo r certai n types of cance rs, but th e trans iso m er is n ot effe ctive. (T h ey are ca lled isomers be cause th e 1\\"0 co m po u nds h ave th e sa me fo rm u la hut a d iffe ren t a rr a nge m e n t o f a to m s.) Wh a t is th e equi librium co ns ta n t a t 29 8 K fo r the tra n sfo rm a tion o f t he cis to th e trans iso m e r? \Vh ich is more thermodynam ica lly sta ble , t h e cis or th e tra ns iso m e r?
If no t, ca n it become so if th e temperature is raised ? At wha t te m pe ra tu re docs th e reaction be come p ro ductfavored ? 60 . • Copper( I1) o xid e , CuO. can be re duce d to co p pe r me ta l with h yd ro gen a t h igh e r te mperatures. CuO (s)
0 2(g) +====! N iO( s)
+ 3 C (s) -> 2 CO (g ) + TiC (s)
Compound
In the Laboratory
+ O , (g )
+ 12
63. • T ita niu m (l V) oxide is converted to tita n ium carbide with ca r bon a t a h igh temperatu re.
58. Ca lculate t he e n tro py ch an ge for di sso lving H e l gas in wa ter a t 25 °C. Is the sig n of t1S o wha t yo u expected? \Vhy or why HO I?
2 Ag,O (s ) -> 4 Ag (s)
HI (g)
+ H,(g ) -> Cu rs) + H, O (g )
Is thi s reaction p rod uct- or reactant-favored a t equilibrium a t 298 K?
~lI o
Compound
(kJ/mo!, 298 K)
~r; o
(kJ/mo!, 298 K)
Cis-Pt( NH l ) 2ClZ
- 467.4
- 228.7
Trans-Pt( NHlh Cll
- 480. 3
- 222.8
H3N'I". Al' ·(aq) (Oxidation number of Al increases.) Step 3. Balance each half-reactionfor mass. Both half-reactions are already balanced for mass. Step 4. Balance each half-reaction for charge. To balance t he equations for charge, add electrons to t he more positive side of each half-reaction. Reduction : 2 eEach CU2 ~ ion requires two electrons.
+ Cu 2 +(aq) ---+ Cu(s)
Oxidation: Al(s) ---+ Each At atom releases three electrons.
Al 3 ~ (a q )
+ 3 e"
Step 5. Multiply each half-reaction by an appropriate factor. The reducing agent must donate as many electrons as the oxidi zing agent acquires. Three Cu 2 + ions are required to take on t he six electrons produced by two Al atoms. Thus, we multiply t he Cu2 +ICu halfreaction by 3 and the AVAl3 '" half-reaction by 2.
900
Chapter 20
I
+ CU"(aq) --> Cu(s)]
Reduction:
3[2 e
Oxidation:
2[AI(s) --> Al' -(aq) + 3 e- ]
Principles of Reactivity: Elect ron Transfer Reactions
Step 6. Add the half·reactions to produce the overall balanced equation. Reduction: 6 e- + 3 Cu 2+(aq) ~ 3 Cu(s) 2 AL(s) -7 2 AL' - (aq) + 6. 3 ( u2+(aq) + 2 At(s) -) 3 Cu(s)
Oxidation: Net ionic equation:
+ 2 Al3+ (aq)
Step 7. Simplify by eliminating reactants and products that appearon both sides. This ste p is not required here. Comment You should always check t he overall equation to ensure there is mass and charge balance. In t his case, three Cu atoms and two At atoms appear on each side. The net electric charge on each side is + 6. The equation is balanced. EXERC ISE 20.1
Balandng Oxidation-Reduction Equations
• Balandng Equations in Add Solution To simplify equations, we shall use W instead of H10" when balandng equations in add sotution.
Aluminum reacts wit h nonoxidizing acids to give All +(aq) and H2(g). The (unbalanced) equation is AL(s) + W( aq) -7 AL'-(aq ) + H,(g ) Write balanced equations for the half-reactions and t he balanced net ionic equation. Identify t he oxidizing agent, the reducing agent, the substa nce oxidized, and the substance reduced.
Balandng Equations in Add Solution
Whe n balan cin g eq uatio ns for redox reac tio ns in afJ ucous solu tio n , it is so me times n ecessary to ad d water m olecul es (H 20) an d eithe r l l t Iaq ) in aci d ic solution o r O H - (aq ) in ba sic so lution to th e eq uatio n. Eq ua tio ns th at incl ude oxoanion s such as 50.j2- , N0 3 - , CIO-, CrO.j2- , and I\.·I nO.j- and o rga n ic co m pounds fall into thi s cat egory. T he p ro cess is ou tlined in Examp le 20.2 for the red uc tion of an oxocation in ac id so lu tio n and in Example 20.3 for a reactio n in basic so lu tio n . Balandng Equations for Oxidation-Reduction Reactions in Add Solution Problem Balance t he net ionic equation for t he reaction of the dioxovanadium(V) ion, V0 2" . with zinc in acid solution to form V02+ (see Figure 20.3). VO, " {aq) + In (s) -7 VO' -(aq) + I n" (aq) Strategy Follow the strat egy outlined in the t ext and Examp le 20.1. Note that water and H+ ions will appear as product and reactant, respectively, in t he half-reaction for t he reduction of V02• ion. (But H20 and H- witt never appear on the same side of the balanced hatf-reaction.) SoLution
Step 1. Recognize the reaction as an oxidation-reduction reaction. The oxidation number of Vchanges from + 5 in V0 2+ to +4 in V0 2+. The oxidation number of Zn changes from 0 in t he metal t o + 2 in ZnH .
Step 2. Separote the process into halj.reoctions. Oxidation:
Zn(s) ~ ZnH (aq)
In(s) is oxidized and is the reducing agent. Reduction:
V02 - (aq)
~
V02+(aq)
VOl- (aq) is reduced and is the oxidizing agent. Step 3. Balance the half-reactions for mass. Begin by balancing aUatoms except Hand O. (Th ese atoms are always the last to be balanced because t hey often appear in more than one reactant or prcduct.)
Zinc half-reaction:
Zn(s) ~ ZnH (aq)
This half-reaction is already balanced for mass.
Vanadium half-reaction:
V02 -(a q)
~
V02 +(aq) 2 0 .1
I
Oxidation- Reduction Reactions
901
The VO z+ ion is yellow in acid solution.
With time, t he blue VO H ion is further reduced to green V3+ ion.
In added. With time, the yellow VOz+ ion is reduced to blue VO z+ ion.
Finally, green V3+ ion is reduced to violet V2+ ion.
VO/ FIGURE 20.3 Reduction of vanadium(V) with zinc. See Example 20.2 for the balanced equation for the first stage of the reaction.
Th e Vatoms in this half-reaction are already balanced. An oxygen-containing species must be added to t he right side of the equation to achieve an a atom balance, however.
va," (eq} -> VO' +(aq) + (need 1 °atom)
In acid solution, add H20 to the side requiring 0 atoms, one H20 molecu le for each 0 atom required.
va,"{aq} -> VO ' +(aq) + H,O(f) There are now two unbalanced Hatoms on the right. Because the reaction occurs in an acidic solution, W ions are present. Therefore, a mass balance for H can be achieved by adding W t o the side of the equation deficie nt in Hatoms. Here, two H+ ions are added to the left side ofthe equation. 2 H' {aq) + VO,+(aq) -> VO' -(aq) + H,O(f)
Step 4, Balance the half-reactions for charge by adding electrons to the more positive side to make the charges equal on bothsides. Two electrons are added to the right side of the zinc half-reaction to bring its charge down to the same value as is present on the left side (in this case, zero).
linc half-reaction:
In(s)
~
ln 2+(aq) + 2 e"
The mass-balanced V0 2- equation has a net charge of 3 + on the left side and 2 + on the right. Therefore, 1 e" is added to the more positive left side.
Vanadium half-reaction:
e"
+ 2 H' {aq) + VO/(aq)
~
V02-(a q)
+ H20 (t')
As a check on your work, notice that the vanadium atom changes in oxidation number from + 5 to + 4 and so needs to acquire one electron.
Step 5. Multiply the half-reactions byappropn"ate factors so that the reducing agentdonates as many etectrans as the oxidizing agentconsumes. Here, the oxidation half-reactio n supplies 2 mol of electrons per mol of In, and the reduction half-reaction consumes 1 mol of etectrons per mol of V02" . Therefore. the reduction half-reaction must be multiplied by 2. Now 2 mol of the oxidizing agent (V0 2" ) consumes the 2 mot of electrons provided per mo le of t he reducing agent (In).
+ 2 e+ 2 H'{aq) + VO, +(aq) -> VOH(aq) + H,O(f)]
Zn(s) -> Zn' +(aq) 2[e-
902
Chapter 20
I
Princip les of Reactivity: Elect ron Transfer Reactions
Step 6. Add the half·reactions to give the balanced, overall equation. Oxidation half·reaction:
Zn (s) -) Zn 2 +(aq) + 2 e"
Reduction half·reaction:
2 e- + 4 W( aq) + 2 V0 2 +(aq) -) 2 V 0 2 ~(aq ) + 2 H20(f)
Net ionic equation:
In (s) + 4 H' {aq) + 2 V0 2 +(aq) --? Zn 2 -(aq) + 2 V02+(aq) + 2 H20(l')
Step 7. Simplify by eliminating reactants and products that appear on bothsides. This step is not required here. Comment Check the overall equation to ensure that there is a mass and charge balance.
zn. 2 V, 4 H, and 4 0
Mass balance:
1
Charge balance:
Ea ch side has a net charge of 6+ .
EXERCISE 2 0 . 2
..
Balancing Oxidation-Reduction Equations
The yellowdioxova nadium(V) ion, VO/(aq). is reduced by zinc metal in t hree steps. The first step reduces it to blue V02+(aq) (Example 20.2). This ion is further reduced to green V3+(aq) in the second step. and V3 + can be reduced to violet V2+(aq) in a third step. In each step. zinc is oxidized to ln 2+(aq) . Write balanced net ionic equations for Steps 2 and 3. (This reduction sequence is shown in Figure 20.3.)
EXERCISE 20.3
FIGURE 20.4 The reaction of purple permanganate ion (MnO~ - ) wit h iron(II) ions in ad d solution. The prod-
ucts are the nearly colorless Mn 2+ and Fe 3 + ions.
Balancing Equations for Oxidation-Reduction Reactions
in Acid Solution A common laboratory analysis for iron is to titrate aqueous iron(II) ion with a solution of potassium permanganate of precisely known concentration. Use t he half-reaction method to write the balanced net ionic equation for the reaction in acid solution. Mn O, - (aq) + Fe" (aq) -> Mn"(aq) + FeH (aq) Identify the oxidi zing agent. the reducing agent. the substance oxidized, and the substa nce reduced. See Figure 20.4.
Balancing Equations in Basic Solution
Exa mple 20 .2 and Exercises 20.2 and 20 .3 illustra te th e tech nique of ba lancing equations fo r redox reactio ns invol ving o xocatio ns and o xoanions that occ ur in acid so lu tio n. Un der th ese co nd itio ns, H + io n or the H +/ H 2 0 pa ir can be used to achi eve a ba lanced equ atio n if re qu ired . Co nvers ely, in basic so hu to u. only O J l ion o r the O II - / II 20 pair ca n be used.
Problem Solving Tip 20.1 Hydrogen balance can be achieved only wi th W / H,O (in acid) or OW/ H,O (in base). Never add H or H2 to balance hydrogen. Use H20 or OH- as appropriate to balance oxygen. Never add 0 ato ms, 0 2- ions, or O2 for 0 balance.
Balancing Oxidation-Reduction Equations: A Summary Never include H+ (aq) and OH - (aq) in the sa me equation. A solution can be either acidic or basic, never both . The number of elect rons in a half-reaction reflects t he change in oxidatio n number of the element being oxidized or reduced. Electrons are always a component of halfreactions but should never appear in the overa ll equation.
Include charges in the formulas for ions. Omitti ng the charge, or writing t he charge incorrectly, is one of the most common errors seen on st udent papers. The best way to become competent in balancing redox equations is to practice, practice, practice.
2 0 .1
I
Oxidation- Reduction Rea ctions
903
Balancing Equations for Oxidation-Reduction Reactions in Basic Solution ProbLem Aluminum metal is oxidized in aqueous base. with water serving as t he oxidizi ng agent. The products of t he reaction are [A l(O H)4t (aq) and H2(g). Write a balanced net ionicequation for this reaction.
Strategy First. identifythe oxidation and reduction half-reactions, and then balance them for massand charge. Finally, add the balanced half-reactions to obtain the balanced net ionic equation for the reaction.
SoLution
Step 1. Recognize the reaction as an oxidation-reduction reaction. The unbalanced equation is Al(s) + H,O(£) -7 [AI(OH),naq) + H, (g) Here, aluminum is oxidized, with its oxidation number changing from 0 to + 3. Hydrogen is reduced, with its oxidation number decreasing from + 1 t o zero.
Step Z. Separate the process into half-reactions. Oxidation half-reaction:
Al(s)
-7
[AI(OH),] - (aq)
(Al oxidation number increased from 0 to +3.) Reduction half-reaction:
H20(f) -; H2(g)
(H oxidation number decreased from + 1 to 0.)
Step 3. Balance the half-reactions far mass. Addition of OH - and/or H20 is required for mass balance in both half-reactions. In the case of the aluminum half-reaction, we simply add OH- ions to the left side.
Oxidation half-reaction:
Al(s)
+ 4 OW(aq) -7 [AI(OH), n aq)
To balance the half-reaction for water reduction, notice that an oxygen-containing species must appear on the right side of t he equation. Because H20 is a reacta nt, we use OH", which is present in this basic solution, as the other product.
Reduction half-reaction:
2 H, O(t)
-7
H,(g) + 20H -(a q)
Step 4. Balance the half-reactionsfar charge. Electrons are added to balance charge. Oxidation half- reaction:
Al(s) + 4 OW (aq)
Reduction hatf-reaction:
2 H, O(f) + 2 e-
-7
-7
[AI(OH ),] - (aq)
+ 3 e-
H, (g) + 2 0H- (aq)
Step 5. Multiply the half-reactions by appropriate factors so that the reducing agent donates as manyelectrons as the oxidizing agent consumes. Here, electron balance is achieved by using 2 mot of Alto provide 6 mol of e-, which are the n acquired by 6 mol of H20.
+ 4 OW (aq) -7 [AI(OH),naq) + 3 e- ] + 2 e- -7 H, (g) + 20W(aq) ]
Oxidation half-reaction :
2[AI(s)
Reductio n half-reaction:
3[2 H, O(f)
Step 6. Add the half-reactions. 2 Al(s) + 8 OW(aq) 6 H,O( f) + 6 e: Net equation:
2 AI(s)
-7
-7
2 [AI(OH ),) - (aq) + 6 e:
3 H, (g) + 6 0W(a q))
+ 8 OW (aq) + 6 H,O(f)
-7
2 [AI(OH) , ]-(aq)
+ 3 H,(g) + 6 0W(aq)
Step 7. Simplify by eliminating reactants and products that appearon both sides. Six OH - ions can be canceled from t he two sides of t he equation: 2 Al(s)
+ 2 OW (aq) + 6 H,O(f ) -> 2 [AI(OH ), n aq) + 3 H,(g)
Comment The final equation is balanced for mass and charge.
904
Chapter 20
I
Mass balance:
2 At, 14 H, and 8 0
Ch arge balance:
There is a net - 2 charge on each side.
Principles of Reactivity: Elect ron Transfer Reactions
Problem Solving Tip 20.2
An Alternative Method of Balancing Equations in Basic Solution
Balancing redox equations in basic solution. which may require you to use OH- and H2 0, can sometimes be more challenging than doing so in acidic solution. One of the ways to balance such equations for reactions in basic solution is to first balance i t as if it were in acidic solution and then add enough OH - ions to bot h sides of the equation so that t he H+ ions are converted to water. Taking the half-reaction for the reduction of eLO - ion to ([2' we have: (a) Balance in acid 4 H' {aq} + 2 ClO - (aq)
+ 2 e- --> CI,(g) + 2 H,O(f)
EXERCISE 20.4 Balancinq
(b) Add four OW ions to bot h sides 4 OW (aq)
+ 4 H'' {aq) + 2 CLO "(aq) + 2 e" --> CI,(g) + 1 H,O(C) + 40 H-(aq)
(c) Combine OH - and H+ to form water where appropriate 4 H, O(f)
+
1 C10-(aq)
+ 2 e- --> CI,(g) + 2 H,O(f ) + 40 W(aq)
(d) Simplify
2 H, O(f)
+ 2 CIO -(aq) + 2 e" --> Cl,(g) + 4 0 W(aq)
Equations for Oxidation-Reduction Reactions
in Basic Solution Voltaic cells based on the reduction of sulfur are under development. One such cell involves the reaction of sulfur with aluminum under basic conditions.
AI(s) + S(s) --> AI(OH),(s) + HS-(aq) (a) Balance this equation, showing each balanced half-reaction. (b) Identify the oxidizingand redudng agents, the substance oxidized, and the substance reduced.
20.2
Simple Voltaic Cells
Let us use the reactio n of co p pe r metal and silver ions (Figure 20. 1) as the basis of a volta ic cel l. To d o so, we p lace the co mpo nc nt'i o f th e 1\"0 half-reactio ns in scparatc co mpa rt m e n ts (Figu re 20.5 ). T his prevcnts t he co p per met al fro m tr ansferrin g e lec tro ns d irec tly to silver io ns. Iu stead , e lec tro ns will be transferred tilro ug h an extern al circ uit, and useful work ca n poten tially be do ne . T he co p per hal f-cell (o n th e left in Figu rc 205 ) holds co p pe r metal th at se rves as o ne el ec trode and a so lut io n con ta ining co ppe r( lI) io ns. The hal f-cell O il the rig h t uses a silver el ec trode a nd a so lu tio n co n ta in ing silvcrf l) io ns. Im po rtan t featu res o f thi s sim ple ce ll are as follows: •
The two half-adls are connected with a salt bridge that allows rations and anions to move benoecn the two half-cells. The e lec tr o lyte chosen fo r the salt b ridge sho u ld co n ta in ion s that will not rcact with che m ical reage n ts in both half-cells. In th e exam pic in Figu re 20.5, N aNO :~ is used . • In all rlearorhemical cells. the anode is the electrode at tohich. oxida tion occurs. 17w electrode at which reduction occurs is always the cathode. (In Fig u re 20.5, th e u >pper elec trod e is th e anode, and th e silver e lec tro d e is th e ca rhodc.) • A negatioe sign can be assigned to the anode in a voltaic cell, and the cathode is marked with a positiue sign. T he che mical oxidatio n occurring at the an ode , which p roduces electron s, g ivcs it a negative cha rge. Electri c cu rre n t in the ext ern al circ uit of a volta ic ce ll co nsists of el ectrons m oving from the nega tive to the positive el ectrod e . • In all electrochemical cells. electrons jlow in the external circuit [nnn the anode to the cathode.
• Salt Bridges Asimple salt bridge can be made by adding gelatin to a solution of an electrolyte. Gelatin makes the contents semi-rigid so that the salt bridge is easier to handle. Porous glass disks and permeable membranes are alternatives to a salt bridge. These devices allow ions to traverse from one half-cell to the other while keeping the two solutions from mixing.
20 .2
I
Simple Voltaic Cells
905
FIGURE 20.5 Avoltaic cell using
(U(5) I (u" (aq) and Ag(5) I Ag ' (aq) half cells. Electrons flow through the external circuit from the anode (the copper electrode) to the cathode (silver electrode). In the salt bridge. which contains aqueous NaNO], negative NOl-(aq) ions migratetoward the copper half-cell. and positive Na '{aq } ions mig rate toward the silver half-cell. Using 1.0 MCu2+ (aq) and 1.0 MAg +(aq) solutions, this cell will generate 0.46 volts.
(+)
Cu anode
_
e-~
Salt bridgecontains Na NO) Na +------;. NO -
Ag cathode
+
' ~~:;lI- Po rou s
plug
Net reaction: Cu(s)
+ 2 Ag+(aq)
------;. Cu 2+(aq) + 2 Ag(s)
T he che m istry occ u rr ing in the cell pictu red in Figu re 20.!') is su m marized by th e foll owing half-reactio ns and ne t io nic eq ua tio n:
+ 2 e- ->
Cathade (reductian):
2 Ag '(aq)
2 Ag(s)
Anade (axidatian) :
Cu rs) -> Cu" (aq)
Net ionic equation:
Cu rs) + 2 Ag'(aq) -> Cu" (aq) + 2 Ag(s)
+ 2 e"
The salt b ridge is requ ired in a voltaic ce ll for the react io n to proceed. In th e
Cui Ag + vo lta ic ce ll, an ions mo ve in th e salt b rid ge toward the co p per ha lf-ce ll, an d
• Electron and Ion Flow It is helpful to notice that in an electrochemical cell the negative electrons and negatively charged anions make a "circle." That is, electrons move from anode to cathode in the external circuit. and negative anions move from the cathode compartment. through the salt bridge, to the anode compartment.
cations move toward th e silver hal f-cell (Figu re 2(5). As Cu 2+ (aq) ions a re fo rm ed in the coppe r hal f-cell by ox ida tio n of co p per met al , negative ions enter that ce ll fro m the salt b rid ge (and positive io ns leave th e cell), so th at th e n u m bers of positive and negati ve charges in the half-cel l co m pa rt men t remain in balance . Likewise , in the silver half-cell , neg ative ion s move o ut o f th e hal f-cell into th e salt b ri dge , an d positive ion s m ove in to the ce ll as Ag ' (aq) ion s are red uced to silver metal. A co m plete circ uit. is required fix cu rre n t to flow. If the salt b ridge is removed, re actio ns at th e elec tr o des will cease . In Fig ure 20 .!i, th e electrodes are co n nec te d by wires to a voltme ter. In a n al tern ative se t-u p, t he co n nec tio ns mi g ht be to a ligh t bul b o r o t he r d evice that use s e lec trici ty. Elec tr o ns are pro d uced by o xidation o f co p per, and Cu 2 +(a q) ion s e n ter t he so lu tion . T he elect rons traverse th e ex terna l circ u it to th e silver el ectrode , whe re th ey re d uc e Ag +(aq) io ns to silve r me tal. To ba la nce the ex te n t of o xidat ion and re d uc tio n , two Ag +(a q) io n s are red uce d fo r every Cu 2 +(a q) ion for m ed . T he mai n fea tures of th is and o f a ll othe r vo ltaic ce lls are su mma rize d in Figu re 20.6.
Chemislry/ :>.Now'· • Sign in at www.thomsonedu.com/Ioqtn and go to Ch apter 20 Contents to see Screen 20.4 for an animation of a cell based on zinc and copper.
906 Chapter 20
I
Principles of Reactivity: Elect ron Transfer Reactions
Voltmeter
tt )
( -)
Electrode (- )
Direction of electron flow
Electrode (t )
Satt bridge
OXid~zed
Electrolyte: ions in ~ p solution ~
)
FIGURE 2 0 . 6 Summary of terms used in a voltaic cell. Electrons move from the anode. the site of oxidation. through the external circuit to the cathode. the site of reduction. Charge balance in each half-cell is achieved by migration of ions through the salt bridge. Negative ions move from the reduction half-cellto the oxidation half-cell and positive ions move in the opposite direction.
SP'"" ) Reduced species ANODE compartment OXIDATION occurs
CATHODE compartment REDUCTION occurs
• . Electrochemical Cells Problem Describe howt o set up a voltaic cell t o generate an electric current using the reaction Fe(s) + CuH (aq) --> Cu(s) + FeH (aq) Which electrode is the anode. and which is the cathode? In which direction do electrons flow in the external circuit? In which direction do the positive and negative ions flow in the salt bridge? Write equations for the half-reactions that occur at each elect rode. Strategy First. identify the two different half-cells that make up the cell. Next. decide in which half-cell oxidation occurs and in which reduction occurs. SoLution This voltaic cell is similar to t he one diagrammed in Figure 20.5. One half-cell contains an iron electrode and a solution of an iron(II) salt such as Fe(N03h. The other half-cell contains a copper electrode and a soluble copper(II) salt such as CU( NO,h. The two half-cells are linked with a salt bridge containing an electrolyte such as KN 0 3 • Iron is oxidized. so t he iron electrode is t he anode:
Oxidation. anode:
Fe(s)
---7'
Fe2+ (aq)
+ 2 e"
Because copper(II) ions are reduced. the copper electrode is t he cathode. The cathodic half-reaction is
Reduction. cathode:
Cu2+ (aq)
+ 2 e"
---7'
Cu(s)
F, (-)
Anode
Cathode
In the external circuit, electrons flow from the iron electrode (anode) to the copper electrode (cat hode). In the salt bridge, negative ions flow toward t he Fe I Fe2+ (aq) half-cell, and positive ions flow in t he opposite directio n.
20 .2
I
Simple Volt aic Cells
907
EXERCISE 20.5
Electrochemical CeUs
Describe how to set up a voltaic cell using t he following half-reactions: Reduction half-reaction:
Ag +(aq) + e" -) Ag(s)
Oxidation half-reaction:
Ni(s) -) Nj2+(aq) + 2 e-
Which is the anode, and which is the cathode? Whatis the overall ceLL reaction? Whatis the direction of electron flow in an external wire connecting the two electrodes? Describe the ion flow in a salt bridge (with NaNOll connecting the ceLL compartments.
Voltaic Cells with Inert Electrodes In th e hal f-cell s d escr ibed so far, th e metal used as an el ectrode is a lso a react ant or a p roduct in th e redox re ac tio n. Not a ll hal f-reactio ns invo lve a me ta l as a reac ta n t o r produ ct , how ever. With th e exception o f ca rbo n in th e fo rm of gra p hite , m os t no n me tals are u nsuitab le as electrod e ma teria ls becau se t hey d o no t co n d uct e lec trici ty. I I is n o t possib le to m ak e an e lec tr ode fro m a gas, a liq ui d (exce p t m e rcu ry) , o r a so lut io n . Ionic so lids d o no t m ak e sa tisfac to ry e lec trodes becau se the io ns are lo ck ed tightly in a crysta l latt ice , an d th ese materials d o no t co n d uct e lectrici ty. In situa tio ns whe re reac ta nts and prod ucts ca n no t se rve as the e lec tro de m aterial, an inert o r chem ica lly u n reacti ve electrode must be used . Such elect rodes are made of mat e rials that cond uc t an electric curre n t but th at a re neither o xid ized nor re d uced in th e ce ll. Co ns ide r co nstructing a volta ic ce ll to accommo d ate th e fo llowing p ro d uctfavored reaction : 2 FeH (aq )
FIGURE 20.7 Hydrogen electrode.
Hydrogen gas is bubbled over a platinum electrodein a solution containing H~ ions. Such electrodes function best if they havea large surface area. Often, platinum wires are woven into a gauze, or the metalsurfa ce is roughened either by abrasion or by chem ical treatment to increase the surface area. 908
Chapt er
20
I
+
H,( g) -> 2 FeH (aq)
+
2 H'' {aq)
+ e: -> FeH(a q)
Reduction half reaction:
FeH(a q)
Oxidation half reaction:
H, (g) -> 2 W(aq)
+ 2 e:
Ne ithe r the reac ta n ts n o r th e product s can be used as an e lec tro de mat eri al. T he refore , the two hal f-cell s are se t up so tha t th e reactants and products co me in co n tac t wit h an el ectro d e suc h as g ra p h ite wh ere they can ac ce p t o r g ive u p electro ns. Gra phit e is a co m mo n ly used e lec tro de mat eri al: It is a con d uc to r o f e lec tricity, an d it is in ex pe nsive (esse n tia l in co m mercial ce lls) and no t read ily oxidi zed u nd er the co nd itio ns e n co u n te re d in most ce lls. Mercu ry is used in certai n typ es of ce lls. Platinum and go ld are also commo n ly u sed becau se bot h are chemically ine r t un d e r most circumstan ces, bu t they are g ene ra lly too co stly fo r com m ercia l ce lls. Th e hylrogen electrode is p arti cul arl y importan t in th e fie ld of e lectroc he m istry becau se it is used as a refe ren ce in ass ign ing ce ll voltages (see Sect ion 20.4) (Fig u re 20.7) . T he electrode ma teri a l is platin u m , chose n becau se hyd rogen adsorbs on t he metal's su r face. In th is h alf-cell's ope ra tion , h yd rog en is b ubb le d over the e lectrod e , an d a large su r face area maxim izes the co n tac t of the gas an d the el e ctro d e . T he aq ueo us so lu tio n co n tai ns H 't Iaq}. T he hal f-reactio ns in vo lving I P (aq ) and II , (g) 2 W( aq )
+ 2 e-
-> H, (g)
or
H, (g) -> 2 W( aq)
+
2 e-
take place at the elec trode su rface , and th e e lectro ns invol ved in the re action are co nd uc ted to or from the reac tio n site by th e met al e lec tro de . A half-cell using the reduction of Fe 3 + (aq) to FeH (aq) ca n also be set u p with a platinu m elec tro de. In thi s case , the so lu tio n su rrou nding th e elec tr o de co n tai ns
Principles of Reactivity: Elect ron Transfer Rea ctions
Anode (- )
H, (g)
(I bar)
Salt bridge
+ - Anions
Cat hode
Cation s.--.
FIGURE 2 0 .8 Avoltaic cell with a hydrogen electrode. This cell has Fe 2 +(aq, 1.0 M) and Fe3 +(aq, 1.0 M) in the cathode compartment and Hz(g) and H' {aq, 1.0 M) in t he anode compartment. At 25 0(, the cell generates 0.77 V.
(+)
Fe" (aq) Chemically inert Pt
1HI'ii-
~
elect rode
1M
Chemically inert Pt elect rode
H+(aq) (I ~
H, (aq) ---+ 2 H+(aq)
+ 2 e"
iron io n s in two different o xidation sta tes . Tra ns fe r o f el e ctrons to or fro m th e reacta n t occ u rs at th e e lectrode surface. A volta ic ce ll invo lving th e reducti on of Fe H (aq, 1.0 M) to Fe 2 + (a q , 1.0 ~I ) with H 2 gas is illu strated in Figure 20.8. In thi s cell , th e hydrogen electrode is the an ode (lI 2 is ox id ized to JI +) , and th e iro n-containing compa rtm e nt is th e cath o de (Fe 3 + is reduced to Fe 2 +) . Th e ce ll p rodu ces 0.77 V.
Electrochemical Cell Notations C he m ists often us e a sh o r t ha n d n o ta tio n to sim p lify ce ll de scription s. For e xam p le, th e ce ll in vo lving th e re d uc tio n o f silver ion with co p pe r metal is wri tte n as Cu(s)ICu ' +(aq, 1.0 M)I IAg+(aq, 1.0 M)I Ag(s)
Th e ce ll usin g II 2 gas to reduce Fe 3 + io ns is wr itte n as PtlH , (P = 1 bar)I W( aq, 1.0 M) II Fe"(aq, 1.0 M) , Fe' +(aq, 1.0 M) IPt An ode information
By con ve n tio n ,
Cathode information
the left we write th e a n od e a nd inform ation wi th resp ect to th e so lu tio n with wh ich it is in co ntact. A sin gle ver tical line (I) indicat es a phase bounda ry, a n d double vertica l lin es (II ) indicate a salt bridge .
20.3
O il
Commercial Voltaic Cells
Th e ce lls d escribed so far a re un likely to h ave practical lise. The y are ne ither co mpact nor robust , high prioritie s fo r mo st a p p lica tio ns. In mo st situa tio ns, it is a lso impo rt an t th at th e cell produce a co ns ta n t vo ltage , but a problem wit h the ce lls d escribed so far is tha t th e vo ltage produced va ries as th e co n ce n tratio ns of ion s in so lu tio n ch a nge (sec Section 20.5 ) . A lso , th e cu r re n t produc tio n is low. 2 0.3
I
Commercial Voltaic Cells
909
Historical Perspectives
Frogs and Voltaic Piles
Voltaic cellsare also called galvanic cells after an Italian physician Luigi Galvani (1737- 1798). who carried out early studies of what he called "animal electricity," studies that brought new words into our language-among them "galvanic"and "galvanize." Around 1780, Ga lvani observed that the electric current from a static elect ricity generator caused t he contraction of the muscles in a frog's leg. Investigating furthe r. he found he could induce contractions when the muscle was in contact with two different metals. Because no external source of electricity was applied to the muscles, Galvani concluded the frog's muscles were themselves generating electricity. This was evidence, he believed, of a kind of "vital energy" or "animal electricity," which was related to but different from "nat ural electricity" generated by machines or lightning.
I
.::.-
.."1'1 ..
1
5
~
•
~
j
" .....--===-Volta's "volt aic pile." These drawings done by volta show the arrangement of silver and zinc disks used to generate an electric curre nt.
Alessandro Volta, 1745-1827.
• Cell Potentials and Reactant and Product Concentrations Concent rations of species in a ceU affect t he pote ntial, as discussed in Section 20.5. 910
Chapter
20
I
Alessandro Vo lta repeated Ga tvani's experiments, wit h the same results, but he came to different conclusions. Volta proposed that an electric current was generated by the contact between two different metals-an explanation we now know to be correct-and that t he frog muscle was simply detecting t he small current generated . To prove his hypothesis, Volta built the first "electric pile" in 1800. This was a series
of metal disks of two kinds (silver and zinc). separated by paper disks soaked in acid or salt solutions. Soon after Volta announced his discovery, Carlisle and Nicholson in En gland used the electricity from a "pile" to decompose water into hydrogen and oxygen. Within a few years, the great Eng lish chemist Humphry Davy used a more powerful voltaic pile to isolate potassium and sodium metals by electrolysis.
Atte m p ting to d raw a large cu rre n t results in a d rop in voltage because the cu rre n t depends o n how fast ion s in so lu tio n rnlgrarc 10 the e lectrod e. Ion co ncen tratio ns near the electrode beco me de ple te d if curre nt is drawn rap id ly, resu ltin g in a decli ne in voltage. The elect rica l work thai can be drawn fro m a voltaic ce ll de pends o n the quan tity of reage nts consu me d. A voltaic ce ll must have a large mass of reactants to p ro d uce curre n t over a pr olonged period . In ad d itio n, a voltaic cell th at ca n be rech arged is
Principles of Rea ctivity: Electron Transfer Reactions
FIGURE 20.9 Some commercial voltaic cells. Commercia l voltaic cells provi de energy for a wide range of devices, come in a myriad of sizes and shapes, and produce different voltages. Some are rechargeable; others are discarded after use. One might think that t here is nothing further to learn about batteries, but t his is not true. Research on these devices is actively pursued in t he chemicalcommunity.
attractive. Rech argin g a ce ll means rc tu rn ing th e reagents to th eir o rigina l sites in th e ce ll. In th e ce lls d escribed so far, th e movemen t of ions in the ce ll m ixes th e reage n ts, and the y ca nnot be "u n m ixed " after the ce ll has been running. Batteri es ca n be classified as p rim ary and seconda ry. Primary batteries ca n no t he re turned to their original state by rec hargi ng , so whe n the reactants are co nsu me d , th e batte ry is "dead" and m ust be d iscarded. Se condary batteries are often ca lled storage batteries o r rechargeable batteries. T he react ion s in these batteries can be reverse d; th us , the ba tteries can he rec harged. Years of d evel opme nt have led to man y d ifferent co mmercial voltaic ce lls to meet specific need s (Figure 20 .9), and several co mmo n o nes are desc ribe d bel ow. All ad he re to the p ri nciples that have bee n deve loped in earlier discussions.
Primary Batteries: Dry Cells and Alkaline Batteries If yo u b uy an in expe nsive flashli gh t ba ttery or dry ce ll battery, it mig h t be a mo dern versio n of a volta ic ce ll inven ted by Geo rge Lcf.l an ch e in 1866 (Fig u re 20. 10). Zinc serves as the a nod e, an d th e cat ho de is a grap hite rod placed down the cente r of the d evice . T hese ce lls are o fte n ca lled "d ry ce lls" because th ere is no visib le liqu id p hase. However; wate r is p re se n t, so th e cel l co nta ins a mo ist paste ofN H.tCI, Znel:!, an d :\,l n02' The mo isture is necessary because the io ns p resen t m ust be in a medi um in whic h they can mig rat e fro m o ne electrode to th e o ther. T he ce ll genc ra tes a potential o f 1.5 V usin g th e foll owing hal f-reaction s:
Cathode, reduction: Anode, oxidation:
Anode I _
Cathode
I
Insulating w sher
+
Wax seal Sand cushion Carbon rod (cathode) NH 4 CL.
2 NH, " (aq) + 2 e: -; 2 NH ,(g) + H,(g)
ZnCl 2 • Mn02 paste
Zn(s) -; Zn ' +(aq) + 2 e:
Porous separator
T he two gases fo rmed at th e ca thode will build up p ressure and co u ld ca use th e ce ll to rup tu re . Thi s p robl em is avoided , howeve r, by two other reactio ns that ta ke place in the cel l. Ammon ia mol ecules bind to Zn H ions, and hyd rogen gas is ox idized by 1\l n0 2 to water.
Zn1+(aq) + 2 NH,(g) + 2 Cl- (aq) -; Zn(NH ,),CI,(s) 2 MnO,(s) + H,(g) -; Mn,O,(s) + H,O(C)
Steet cover
Zinc can (anode) Wrapper
FIGURE 20.10 The common Let tanche dry cell battery.
20 .3
I
Com mercial Voltaic (e lts
91 1
• Batteries The word batte ry has become part of our common language, designatin g any self-contained device t hat gene rate s an elect ric current . The term batte ry has a more precise scientific meaning, however. It refers to a coltection of two or more voltaic cells. For example, t he t z-volt battery used in automobiles is made up of six voltai c cells. Each voltaic cell develops a voltage of 2 volts. Six cells connected in series produce 12 volts.
LeClan che ce lls were wide ly used becau se of th eir lo w cos t, but th ey have seve ra l disad vantag es. If cu rre n t is drawn from th e battery rapidl y, th e gaseous product s ca n no t be co nsu med rapidl y e no ug h, so the ce ll res istance rises, and th e voltage d rops. In ad d itio n, th e zinc electrode and am mo niu m io ns are in co n tac t in th e cell , an d these che m icals react slowly Recall th at zinc reacts with aci d to form hydrogen . The am mo niu m ion , N I I"+ (aq) , is a weak Bron srcd acid and reac ts slowly with zinc. Becau se of thi s re act ion , th ese vo lta ic ce lls ca n no t be sto red indefin ite ly, a fact y O li may have learn ed fro m ex pe rienc e . When the zin c outer shell deteriorat es, th e battery can leak aci d and pe rh aps d amage the ap plia nce in wh ich it is co n ta ined. At th e present tim e , yo u are more likel y to use alkaline batteries in yo u r ca me ra or flashli gh t. They ge ne ra te cu rre n t up to SO% lon ger th an a d ry cell o f th e sa me size . Th e che mis try of alka line cell s is quite sim ilar to th at in a LeClan ch e ce ll, except th at th e mat eri al in side th e cell is basic (alkaline) . Alkaline cells lise th e ox ida tio n of zinc and th e re d uct io n of T\.Jn02 to ge nera te a curren t, but NaOII o r KOII is used in th e ce ll inste ad of th e acid ic salt NI LtCI.
Cathode, reduction:
2 Mn0 2(S)
Anode, oxidation:
Zn (s)
+ H20(f) + 2 e- --> Mn203(S) + 2 0W(a q)
+ 2 OW (aq) --> Zn O(s) + H20(f) + 2 e-
Alka line ce lls, whi ch prod uce 1.54 V (a p prox ima tely th e sa me vo ltage as th e LeClanche cell ), have the furth e r advantage th at th e cell potential do cs not decline unde r high cu rr e n t loads be cau se no gases are fo rmed. Prior to 2000, m ercury-containing batt eri es were wid e ly used in ca lculators, ca me ras, watches, heart pacem akers, and oth er d evices. Howe ver, th ese small batteri es were bann ed in the Uni te d Stat es in th e I990s becau se of en viro n me n ta l problem s. They have been re p laced by several other types of batteries, su ch as silver o xid e batteri es and zinc-o xyge n batteries. Both o pera te u nder alkaline condition s, and both have zinc an o des. In th e silver ox id e batt e ry, wh ich produce s a voltage of abo ut 1.5 V, th e ce ll reactions are
Cathode, reduction:
Ag 20(s)
Anode, oxidation:
Zn(s)
+ H20(f) + 2 e- --> 2 Ag(s) + 2 0W(a q)
+ 2 OW (aq) --> Zn O(s) + H20(f) + 2 e:
Th e zinc-oxyge n battery. which produces a bo u t 1.15-1.35 V, is unique in th at a tmo spheric oxygen and not a metal oxide is the o xidiz ing agent.
Cathode, reduction:
02(g) + 2 H20(f ) + 4 e- --> 4 OW( aq)
Anode, oxidation:
Zn(s)
+ 2 OW (aq) --> ZnO (s) + H20(f ) + 2 e:
These batte rie s ha ve found usc in hearin g aids , pa gers, and medical d evice s.
Secondary or Rechargeable Batteries When a Lcf.Ian ch e cell or an alk aline ce ll ceases to produce a usable el ectric cu rre nt, it is discard ed . III co n trast, so me types o f ce lls ca ll be rec harged , often hundreds o f times. Rcchargin g requires ap plying an electri c current from an external sou rce to re store the ce ll to its original sta te. An au to mo bile battery-the lead storage battery-is probabl y th e best-known re ch argeable battery (Figu re 20.11) . Th e 12-V versio n o f thi s battery contain s six vo ltaic cells, each ge nerating about 2 V. The lead sto rage batte ry ca n produce a large initial cu rre n t, an esse n tial fea ture when startin g an au to m o bile e ng ine .
912
C h a pt e r 2 0
I
Princip les of Reacti vity: Elect ron Tra nsfer Reacti on s
FI G URE 20 .11 Lead storage battery, a secondary or rechargeable battery.
The negativeplates (anode) are lead grids filled wit h spongy lead. The positive plates (cathode) are lead grids filled with lead(IV) oxide, Pb02 • Each cellof the battery generates 2 V.
Positive plates: lead grids filled
with PbO,
Negative plates: lead grid s filled with spo ngy lead
The anode of a lead storage battery is m et allic lead . The ca tho d e is a lso mad e o f lead , but it is co vere d with a layer o f co m p ressed , insoluble lead (IV) o xide, PbO z. The electrodes, arran ged alternat ely in a stac k a nd se para ted by thin fibergla ss shee ts, are immersed in aqueo us sulfu ric acid . \Vhen th e ce ll su p p lies electr ical ene rgy, th e lead a node is oxidized to lead (lI ) sulfate , an inso lubl e su bstance th at adheres to th e el ectrode su r face . T he two el ectro ns p roduced per lead ato m move th rough th e ex te rna l cir cuit to th e cathode, whe re PbO z is reduced to Pb 2 + ion s th at , in th e presen ce o f H 2S04, also form lead (lI ) sulfate .
Cathode, reduction:
PbO,(s) + 4 W (a q) + SO,' -(aq) + 2 e: --> PbSO,(s) + 2 H,O(f)
Anode, oxidation:
Pb(s)
Net ionic equation:
Pb (s) + PbO,(s ) + 2 H,SO, (aq) --> 2 PbSO, (s) + 2 H,O(C)
+ SO.'-(aq) -->
PbSO, (s)
+ 2 e-
Wh en curre n t is gene rated, su lfu ric acid is co nsu m ed a nd wate r is fo rmed . Because water is less d en se th an su lfu r ic ac id , th e d en sity o f th e so lu tio n d ecreases du rin g th is p ro cess. Th erefore , on e way to d ete rmi ne whether a lead sto rage battery n eed s to be recha rged is to measure t he d e nsity of th e so lu tio n. A lead sto rage batte ry is recharged by su p plying electrical energy. T he PbSO" coating th e su r faces of th e el ectr o d es is co nverted back to m etall ic lead and PbO z, and su lfu ric ac id is re gen erated . Rech argin g thi s battery is po ssibl e becau se th e reac tants and products rcmain attach ed to the el ectrode su rfac e. T he lifetime of a lead sto rage battery is lim ited , how ever, becau se , with tim e , the coati ng s of PlrO , and PbS0 4 flake o ff o f th e surface and fall to th e bottom o f th e battery case . Scientists a nd e ngi nee rs would like to find an alte rna tive to lead sto rage batteries, es pecially fo r use in cars. Lead sto rage batteries have tile di sadvantag e of bei ng large an d he avy. In ad d itio n , lead and its co m po u nds are to xic and th eir di sp o sal adds a further complicati on . Neve rt h eless, at thi s tim e , th e ad vantage s o f lead sto rage batteri es outweig h th ei r d isad vantages. Nickel-cadmium ("Ni-cad") batteries, used in a var iety of co rd less ap p liances suc h as telephones, video ca mc o rd ers, and co rd less power tool s, are lightweight and rechargeabl e . The che m istry of the ce ll u tilizes t he o xid ati on o f cad m iu m an d
20 .3
I
Commercial Voltaic Cells
913
the red ucti on o f n ickel (II l) ox ide un d er basic conditio ns. As with th e lead storage battery, t he reacta n ts an d products fo rm ed whe n pro du ci ng a curren t a re so lids th at adhe re 10 the elec tro des.
Cathode, reduction:
NiO(OH)(s) + H,O(C) + e-
Anode, oxidation:
Cd(s) + 2 OW (aq)
-7
-7
Ni(OH),(s) + OW( aq)
Cd(OH ), (s)
+ 2 e-
Ni-cacl bat teries prod uce a nearly co ns ta nt vollage. H oweve r, their cost is re latively high , an d th e re are res tric tions on their disposal because cadm iu m compo unds are toxic and present an en viron me n ta l hazard .
Fuel Cells and Hybrid Cars
• Energy for Automobiles Energy available from systems that can be used to power an automobile. W ' h/kg" (1 W • h = 3600 J)
Chemical System Lead-acid battery
18-56
Nickel-cadmiu m battery
33-7 0
Sodium-sulfur batte ry
80-14 0
Lithium polymer battery
150
Ga soline-air combustion engine
An adva ntage o f voltaic ce lls is th at th ey are small and po rtable , b u t their size is also a lim itatio n . T he amo u nt o f electric cu rre nt produ ced is limited by the q uan tity o f reagen ts co n ta ined in th e ce ll. Wh en o ne of th e reac ta n ts is co mp letely co nsu med , th e cell will no lo nge r genera te a curre nt. Fuel ce lls avoid this limit ati on because the reactan ts (fue l and ox idant) can be sup plied co nt in uo usly to th e ce ll fro m an ex terna l reservo ir. Altho ug h the first fue l ce lls were constr ucted m o re than 150 yea rs ago, litt le was d on e 10 d evelop thi s tec h no logy u n til th e space p rogram re kind led in te rest in th ese d evices. H ydrogen-oxygen fuel ce lls have been used in NASA's Gemi n i, Apo llo , an d Space Sh u tt le programs. No t o nly are th ey lightwe ight and efficien t, but th ey also have th e ad d ed ben efit t ha t th ey ge nerate drin kin g water for the ship 's crew. Th e fue l ce lls o n bo ard th e Space Sh u ttle d eli ver th e same power as batteri es we igh ing 10 times as mu ch . In a hydrogen -o xygen fue l ce ll (Figure 20. 12) , h ydrogen is p um ped onto th e ano de o f th e cell , and O 2 (o r air) is d irect ed to th e ca t ho de where th e followin g reac tions occur:
12.200
* watt-hour/kilogram
Cathode, reduction:
O,(g)
+ 2 H,O(f) + 4 e-
Anode, oxidation:
H, (g)
-7
2 W (aq)
FIGURE 20.12 Fuel celldesign. Hydrogen gas is oxidized to H'{aq ) at the anode surface. On the other side of the proton exchange membrane (PEM), oxygen gas is reduced to DW(aq). The H' {aq) ions travel through the PEM and combine with DW(aq), forming water.
-7 4
OW (aq)
+ 2 e-
Electrical energy output
:-! J .... Hydrogen fuel
PROTON EXCHANGE
MEMBRANE
( 91 4
C h a pt e r
20
I
) ( O, (g)
-----'-----~
Principles of Reactivity: Electron Transfer Reactions
CATHODE
+ 2 H, O( £) + 4 e-
---->
4 OW (aq))
Th e two halves of th e cel l are separat ed by a specia l mate rial ca lled a p roton excha nge membrane (PE1\'1). Protons, 1-1 + (aq), form ed at the ano de traverse th e PliM and react with the h ydro xid e ion s p roduced at the catho de, formin g water. Th e ne t reacti on in th e cell is thus the form ation of water from 11 2 and 0 2' Ce lls cu rrently in use run at tempe ratures o f 70- 140 °C and produce abou t 0.9 V. Hydrogen-oxygen fu el cells operate at 40-60 % efficie nc y and meet mo st o f th e requi rem en t" for use in au to mo biles: they ope rat e at ro o m temperature or sligh tly above , start ra pid ly, and d evelop a high cu rren t d en sity. Cos t is a serio us problem , however, and it ap pears th at a su bstan tial sh ift away from the internal co m bus tio n eng ine re mains a lo ng way off. For thi s reason , severa l m ajor ca r manufacturers have d esigned electric ca rs th at use vario us types o f batteri es to provid e th e power to d rive th e car. Th e most co m mo nly em ploye d type fo r auto motive use is th e lead sto rage battery, but th ese d evices are p robl em ati c, owing to thei r mass. To p roduce on e mol e o f el ectron s requi res 32 1 g o f rea ct ants in lead storage batteries. As a result , th ese batt eries rank ve ry low amo ng various o p tions in power per kilo gr am of battery weigh t. In fact , the powe r ava ilable from any type o f ba ttery is mu ch less th an th at available from an equi valent mass of gasoline. Hybrid cars ap pe ar to offer an in te rim so lu tio n. These veh icle s co m bine a small gasoline-fueled engine with an elec tric m ot o r and batteri es fo r storage of elec tric en c rgy. Cu rren tly, hybrid ca rs use re ch argeable ni ckel-metal hydride batteries. Elec tro ns are ge nerated whe n H ato ms int eract with 01-1 - ion s at th e metal allo y anode. Alloy(H) +
Hybrid car. This car combines a gasolinefueled engine with an electric motor and recharqable batteries. Its fuel efficiency is about double that of t he current generation of cars using only gasoline engines.
ow --> Alloy + H20 + e:
Th e re action at th e catho de is the same as ill Ni-cad batteries. Ni OOH + H20
+ e- -->
Ni( OH),
+ OW
ChemislrY.(J.Now'" Sign in at www.thomsonedu.com/Ioqtn and go to Chapter 20 Contents to see Screen 20.5 to view animations of various types of batteries.
20.4
Standard Electrochemical PotentiaLs
Different electroch emi cal ce lls produce different vo ltages: 1.5 V for th e LeClancb e and alka line cells, abou t 1.25 V for a Ni-Cd battery, and abou t 2.0 V for th e individ ua l cells in a lead sto rage batt e ry. In thi s sec tio n , we wan t to ide n tify th e vario us factors affecting ce ll vo ltages and d evelop procedures to calculat e th e volt age of a ce ll ba sed o n th e che m istry in th e ce ll and th e co nd itio ns used.
ELectromotive Force Elec tro ns ge nera ted at th e anode of a n e lec tr oc he m ica l ce ll move th rough th e ex te rna l circ ui t toward th e ca tho de, and th e force need ed to mov e th e electron s ari ses from a difference in th e potential e nergy o f el ectrons at th e two electrodes. This difference in po te n tial e ne rgy per e lec trica l charge is called th e electromotive force or emf, fo r which th e literal m eaning is "forc e cau sin g el ectrons to move." Emf has units of volts (V) ; 1 vo lt is th e potential difference ne eded to impart one jou le o f e nergy to an electric charge of one co ulo mb ( I J = I V X I C). One coulomb is the quantity of chmge that jmsses a point in an electric circuit when a current of J ampere [ unos [or 1 secon d ( 1 C = 1 A X 1 .I). 2 0 .4
I
• Electrochemical Units The coulomb (abbreviated C) is the standard (51) unit of electrical charge (Appendix (3 ). 1 joule = 1 volt X 1 coulomb. 1 coulomb = 1 ampere X 1 second. Standard Electrochemical Potentials
915
Measuring Standard Potentials Im agin e you plan ned to stu dy ce ll voltages in a lab o ratory with two obje ct ives: ( I) to unde rsta n d th e fac to rs that affect th ese value s and (2) to be able to p redict th e pote nti al o f a voltaic cel l. You m ight co nstruc t a n um ber of di ffe rent half-cells, link th em together in various combinatio ns to form voltai c ce lls (as in Figure 20.13) , and measure th e cell po ten tials. After a few ex perim e n ts, it would become ap parent that ce ll voltage s de pend on a num be r of facto rs: the hal f-cells used (i.c., th e react ion in eac h half-cell a nd th e overall or net reaction in the cell) , the co nce ntratio ns of rea ctants and p ro d ucts in sol u tion , the pressu re o f gaseous reactants, a nd the te m pe ra ture . So th at we ca n lat e r co mpa re th e po te n tia l of o n e h al f-cell with a no the r, let u s measure a ll cell vo lta ges und e r standard conditions: • React an ts a nd p ro d uc ts a rc p resen t in their standard sta tes. • So lu tes in aq u eo us so lu tio n have a concentration of 1.0 I\L • Ga seous react a n ts or prod u c ts have a p re ssure of 1.0 ba r. A ce ll po ten tia l m easured u n d er th ese cond itio n s is ca lled th e standard potential a nd is d e no ted by E \ ell' U n less o therwise sp ecifi ed. all values o f E On ' lI refer to measurements a t 29 8 K (25 °C) . Suppose yo u se t up a n u m ber of sta n d a rd h alf-cell s a n d conn ec t eac h in turn to a standard hydrogen ele ctro de (SH E) . Your apparatus wou ld look like the vo ltaic ce ll in Figure 20 . 13. For n o w, we will co n ce n tr a te o n th ree asp ec ts of th is ce ll: 1. The reaction that ou un . Th e reac tio n oc curri ng in the ce ll pictured in Figure 20 .13 could be either t he redu c tio n o f Zn 2 + ions with H 2 ga s Zn2+(aq)
+
H, (g) --> Zn(s)
+ 2 W(aq )
Zn 2+(a q) is the oxidizi ng agent, and H2 is the reducing agent.
Standard hydrogen electrode would be the anode (negative electrode).
ort he redu ction of l l t Iaq) io n s by Zn(s) . Zn(s) + 2 W (aq) --> Zn2+(aq) + H,(g) Zn is the reducing agent, and H''{aq} is the oxidizing agent . Standard hydrogen electrode would be the cathode (positive electrode). Voltmeter ...- = = =...
A ctive Fi gur e 20.13 Avolt aic cell using ZnIZn2+(aq, 1.0 M} and H2 1H''{aq, 1.0 M) half cells. (a) Zinc metal reacts readily with aqueous HCl. (b) When zinc and acid are combined in an elect rochemical cell, the cell generates a potential of 0.76 Vunder standard conditions. The electrode in the H2 1H' {aq. 1.0 M) half-cell is the cathode, and the Zn electrode is the anode. Electrons flow in the external circuit to the hydrogen half-cell from the zinc half-cell. The positive sign of the measured voltage indicates that the hydrogen electrode is the cathode or positive electrode.
,--
+
Zn anode
, -- > Cathode (+ )
Chemically inert Pt electrode
li" i\"- -..J,I!:"
Chemistry.¢.NOW'" Sign in at www. thom sonedu. comjlogin and go to the Chapt er Contents menu to explore an interactive version of this figure accompanied by an exerdse . 916
C h a pt e r 2 0
I
Zn(s) ----.Zn1+(aq) + 2 e-
2 H+(aq) + 2 e- ----. H1(g)
Net reaction: Zn(s) + 2 H'{aq) ----. H1(g) + Znl+(aq) {a]
Principles of Reactivity: Electron Transfer Reactions
(b)
All th e su bstances nam ed in th ese eq ua tio ns are presen t in th e ce ll. The reaction th at actually o ccurs is th e o ne th at is product-favored. T hat is, th e re ac tio n o ccurr ing is the on e in whic h th e reactan ts are th e str onger reducing and oxi d izing agents. 2. Direction C!! electron./low in the externa l circu it. In a volt aic cel l, e lectrons a lways flo w from th e ano de (nega tive e lectrode) 10 th e ca tho de (positive el ectrode ) . That is, electrons move from the electrode of higher potential energy to the one of lower potential en"rgy'. \ Ve ca n tell th e di re ction of e lec tro n movement by placin g a voltmet e r in the circui t. A positive pote ntial is observed if the voltme te r termi nal with a plus sig n ( +) is con nec ted to th e positi ve el ectrode [a nd th e termina l wit h the minus sign ( -) is connect ed to th e negative electro de ] . Co n nec te d in the o p posite way (pins to minus and minus to plus) , the voltmete r will show a negative value on th e d igit al reado ut. 3 . Cell potential. In Figure 20. 13, th e vo ltmete r is ho ok ed up wit h its po sitive te rmin al co n nected to th e h ydro ge n half-cell , and a read ing o f + 0.76 V is o bserved . Th e hydrogen e lect rode is thus the positive el ectrode o r cathode, and th e reacti on s occurring in thi s ce ll mu st be
Reduction, cathode:
2 W( aq) + 2 e: -> H,(g)
Oxidation, anode:
Zn (s) -> Zn 2+ (aq)
Net cell reaction:
Zn(s) + 2 W(aq) -> Zn2+ (aq) + H, (g)
+ 2 e:
Th is resu lt co n firms th at , o f th e two o xidi zing ag en ts present in th e cell , JI + (aq) is better th an Zn 2 + (aq ) and th at Zn met al is a bette r re d uci ng age n t th an 1-1 2 gas. A potential o f + 0.76 V is measu re d fo r th e oxidation of zinc with hydrogen ion . This value reflec ts th e difference in potential energy of an ele ctron at each el ectrode . Fro m th e di rection o f flow o f electro ns in the ex ternal circuit (Zn elec tro de ~ 11 2 e lectro de) , we co nclude that th e po ten tial energy of an electr o n at th e zinc electrode is higher th an the pot ential enerhT)"o f the electron at th e hydrogen electr ode. Hund reds of e lec troc he mica l ce lls lik e that show n in Figure 20. 13 ca n be se t up , a llowing us to de te rm ine the re lati ve ox id izing o r red uc ing ab ility of vario us chem ica l species an d to d et ermine th e el e ct rical potential ge nera te d by th e re actio n un de r sta ndard con di tio ns . A few results are give n in Figure 20. 14, whe re half-re act ion s are listed as reductions. Th at is, th e che m ica l species on th e left arc oxidi zing age n ts and are listed in d escending oxidi zin g ability.
Standard Reduction Potentials By d oing ex perimen ts suc h as th at illustra ted by Figu re 20. 13, we not only have a not ion of th e rel ati ve ox id izing and re d uci ng abilities of vario us che mical species, but we ca n also rank them quantitat ive ly. If R Oed l is a measu re o f the standa rd potential for th e cell, th en E Oe;lthod", and E Oatlod ", ca n he tak en as measures of electr od e pot ential. Becau se E Ocell reflec t" the differen ce in electro de pote nti als, E On "1I must be the differen ce be tween E OcarllOde and E Oallou t". E Ocell
= E Ocathode - EOanode
(20.1 )
H ere , E Ocathod e and E Oanode are th e sta nd ard reduction potentials for th e half-cel l rea ctio ns th at occu r at the ca tho de a nd ano de, res pectively, Eq uatio n 20. 1 is im portant for three reaso ns :
• Equation 20.1 Equation 20.1 is an other example of calculating a change from XIi "" t - X initiat. Electrons move to the cat hode (t he "final" state) from t he anode (t he "initial" state). Thus, Equation 20.1 resembles equations you have seen previously in this book (such as Equations 5.6 and 6.5).
• If we have value s for EOc a thode and EOanodes we can calc ulate th e sta ndard potential , E Or l'l1. fix a voltaic cel l. 2 0 .4
I
Standard Elect roche mica l Potentials
917
A Closer Look
EMF, Cell Potential, and Voltage
Emf a nd cell potential (Ecell) are often used synonymously, but the two are subtly different. Ecell is a measured quantity, so its value is affected by how the measurement is made. To understand this point, consider as a n analogy wate r in a pipe under pressure. Water pressure can be viewed as a nalogous to emf; it repre-
to move. If we open a faucet, water will flow. Opening the faucet will, however, decrease the pressure in the system .
Emf is the potential difference when no current flows. To measure Ecell, a voltmeter is placed in t he exte rnal circuit. Alt hough voltmeters have high internal resistance to minimize current flow, a small current flows none-
sents a force th at will cause water in t he pipe
theless. As a result. the va lue of Ecell will be slightly different than the emf. Finally, there is a difference bet ween a potential and a voltage. The voltage of a cell has a magnitude but no sign. In contrast, the potential of a half-reaction or a cell has a sign (+ or - ) and a magnitude.
ll71en the calculated value of EOcell is posltioe. the reaction as unium is [nedicted to IN:' jm)(luct{avornl at equilibrium. Co n verse ly, if the calcu la ted value of EOn .1I is negative, th e reacti on is p redicted to be reactan t-favored at equi lib rium . Suc h a rea ct ion will be p roduct-favo red at eq ui lib riu m in a direction o pposite to the way it is writte n . • If we m easure EOcdl a nd know ei t he r E" cathode o r EOanode s we can calcu late th e ot her value . This value wou ld te ll us how one half-cell reac tio n co m pa res with o thers in ter ms of re lative oxid izing o r redu cin g ability.
•
But he re is a dil em ma. On e ca n no t measu re indi vid ual hal f-cell po ten tials. Just as va lues fo r 8 JH o and 8 JGo we re es tablished by choosing a re fere nce point (the ele me nts in t heir sta nda rd sta tes), scie n tists have selec te d a refe rence poin t for hal f-reacti ons. \Ve assig n a po ten tial of exac tly 0 V to th e half-reacti on th a t o ccu rs at a sta nda rd hydro gen electro d e (SHE) . 2 W (aq, 1 M)
+
2 e : -> H,(g, 1 ba r) EO = 0.00 V
With th is sta ndard, we can now d et ermine EO values for hal f-cells by measu rin g R Oeel l in ex pe rime n ts suc h as those d escr ibed in Figures 20.8 and 2 0 . 1~~, whe re one of the e lec trodes is th e sta ndard hydrogen electrode. \Ve ca n the n quan tify the in fo rm atio n wit h reduction po te nt ial tables suc h as Figu re 20 .14 and use these values 10 make p redictions abo u t E \ ell for new vo ltaic ce lls.
Tables of Standard Reduction Potentials T he experime n ta l ap p ro ac h j ust d escribed lead s to lists of EOvalues suc h as see n in Fig u re 20. 14, Table 20.1, an d Ap pe n d ix ~ 1. Let us list so me im portan t po in ts co nc erning these tables and the n illustrate them in the di scussio n and e xa m ples tha t fo llo w. 1. Reaction s are wr itte n as "ox idize d fo rm + electro ns ---7 reduced fo rm ." T he species on th e left side o f th e reaction arrow is an ox id izing ag en t, and th e spe cies on th e rig h t sid e of the reactio n arr ow is a re d uci ng agent. T he refor e , all poten tials are for reduction reactions. 2. T he mo re po sitive th e value of t:..-o fo r the reactio ns in Figure 20. 14, Table 20. 1, and sim ilar tabl es, the be tter the o xid izin g ability o f the io n o r co mpou nd 0 11 the left side of the reactio n . T his m eans F2 (g) is the best oxidizing agent in the table. Lit hium ion at th e lower-le ft co rner of Table 20. ] is th e poo rest ox id izing age n t because its EOvalue is th e mo st negative .
918
Chapt er
20
I
Principles of Reactivity: Electron Transfer Reactions
Reduction Half- Reacti on
Reduction Pote ntial. V
+2.87
F,( g) + 2 e-
f-
"• . C' .".,:;:; .::J:
~
2 F- (g)
~
c
Il
+ 1.36
f-
•
~
~
~.
c o
+0.799
f-
+ 0.535
f-
+0.337
+-
£ :E
~
2 Cl- (aq)
~
Ag(s)
c
12(s) + 2 e-
~
2qa q)
0
Cu2+(aq ) + 2 e - _
•
•
ec
~
-c 'ii -c
..
c" ·G , c
"
Ag+(aq) + e"
1J
~
~
c •
Q;'5
• •
~
Cl,(g) + 2 e-
~
~
~
0
e
g :E
• ,
~
c ·G
Cuts)
~
~
.~
0.00
f-
- 0.44
r-
- 0.763
f-
•~ v .s
~
2 H+(aq ) + 2 e - _
-• •. c C' _
~
g> "
~c
c
,
:~
.
~
•
+ 2 e"
Fe2+(aq)
------+
Fe(s)
Zn2+(aq)
+
2 e--
Zn(s)
• ~ v .s
"•g> ~
.
c" ·G c es ~
'5
'¥
.~
"
~
+-
.~
c
0
- 1.66
~
c
H, (g )
:v •
~
1J
0 0
Al3+(a q)
~
+ 3 e"
-
Al(s)
~
FIGURE 2 0 .14 A potentia l ladder fo r red uction half-reactions. The relative positio n of a half-reaction on th is potential ladder reflects t he relative ability of t he species at the left to act as an oxidizing agent. The highe r the compound or ion is in list, th e better it is as an oxidizing agent. Converse ly, the atoms or ions on t he right are reducing agents . The lower t hey are in t he list, the bet ter t hey a re as a reducing age nt. The pote ntial for each half-reaction is given with its reduction potential, P cathode. (For more information see J . R. Runo and D. G. Peters, Journal of Chemical Education, Volume 70, page 708,1993.)
3. The more negative th e value of the reduction pot en tial, EO, th e less likely th e half-reactio n will occur as a red uction , an d th e mo re likel y th e reve rse half-reaction will oc cu r (as a n oxid ation ) . Thus, Li(s) is th e stro ngest reducin g age n t in the tab le , and F- is the weakes t red uci ng ag e nt. The reducing age n ts in th e tabl e (the io ns, e le me nts , a nd co m po u nd s a t th e right) increase in stre ngth fro m th e top to th e bottom . 4. \Vhen a reaction is reversed (to g ive "red uc ed fo rm -7 ox id ized fo rm + electr on s") , th e sig n o f EOis reversed , but th e m agnitude of E" is un affected. Fe3+ (aq, 1 M)
+ e- -> Fe2+(aq, 1 M)
Fe2+ (aq, 1 M) -> Fe"(aq, 1 M)
+ e:
EO
• EOValu es An extensive listin g of EOvalues is found in Appendix M, a nd still larger tables of data can be found in che mistry reference books . A common convention, used in Appendix M, lists standard reduction potentials in two groups, one for acid and neut ral solutions and the ot her for basic solutions.
+ 0.771 V
- 0.771 V
5. Th e reactio n betw een a ny substan ce o n th e left in th is tabl e (a n ox idi zing age n t) wit h a n y su bsta nce lower th an it on the ri ght (a re d ucing agent) is p roduct-favored at eq u ilib ri u m. This has be en called th e northwest-sou theast
2 0 .4
I
Sta ndard Electroche mica l Poten tia ls
919
TABLE 20 .1 Standard Reduction Potentials in Aqueous Solution at 25 · C· Reduction Half-Reaction
"•"
F,(g) + 2 e-
-> 2 F-(aq)
H,O,(aq) + 2 W(aq) + 2 e:
-> 2 H,O(I )
+ 1. 77
PbO,(s) + SO.'-(aq) + 4 W(aq) + 2 e-
-> PbSO,( s) + 2 H,O(I )
+ 1.685
MnO, -(aq) + 8 H'' {aq) + 5 eAu 3-(aq) + 3 e-
-> Mn' +(aq) + 4 H,O(f)
+ 1. 51
-> AU( s)
+ 1. 50
CI, (g) + 2 e-
-> 2 Ct-{aq) -> 2 Cr 3-(aq) + 7 H,O(f)
+ 1.3 6
Cr,O,'-(aq) + 14 H' {aq} + 6 e: O,(g) + 4 H'' {aq) + 4 .-
-> 2 H,O(I )
+ 1. 229
Br,(f) + 2 e-
-> 2 Br-(aq)
+ 1.08
N0 3-(aq) + 4 H' {aq) + 3 eOCl-(aq) + H,O(I) + 2 e-
-> NO(g) + 2 H,O(f)
+ 0.96
-> Cl -(aq) + 2 OW (aq)
+ 0.89
Hg"(aq) + 2 e-
-> Hg(f)
+ 0.855
Ag +(aq) + e"
-> Ag( s)
+
•
Hg/ +(aq)
"
Fe 3 +(aq) + e-
~ ~
'N
eo x 0
P (V)
-> 2 Hg(1)
2 e:
~
Fe 2 +(aq)
-> 2 l ' {aq)
I,(s) + 2 e-
~
0
'S
O,(g) + 2 H,O(f) + 4 e-
-> 4 OW(aq)
" ~ '0 " •
Cu' +(aq) + 2 e-
-> Cu(s)
Sn' +(aq) + 2 e:
-> Sn' +(aq)
~
~
~
oS
2 H'' {aq) + 2 eSn2+(aq )
+ 2 e+ 2 e-
~
• '0 ",
+ 0.7 99 + 0.789
~
~
"
'0
~
b,
" ~
~
-> H,(g)
" •
-> Sn(s)
.s
+ 0.771 + 0.535 + 0.40
+ 0.337 + 0.15
~
0.00 - 0.14
-e Ni(s)
- 0.25
-> V' +(aq)
- 0.255
PbSO, (s) + 2 e-
-> Pb(s) + SO,' -(aq)
- 0.356
Cd"(aq) + 2 e-
-> Cd(s)
- 0.40
Fe' +(aq) + 2 e:
-> Fe(s)
- 0.44
+ 2 e-
-> Zn(s)
- 0.7 63
2 H,O(I) + 2 e-
-> H,(g) + 2 OW(aq)
- 0.8277
Al3+(aq) + 3 e-
-> AI( s)
- 1. 66
Mg"(aq) + 2 e-
-> Mg(s)
- 2.37
Na ' {aq) + e-
-> Na(s)
- 2.7 14
K'{aq] + e-
-> K(s)
- 2.925
W(aq) + e-
-> li (s)
- 3.045
* In \'olls (V) VcfS US
I
11•
V3-(aq) + e:
Zn 2+(aq)
Chapter 20
+ 1.3 3
'0
Nj2+(aq)
920
+ 2.87
th e sta nd ard h ydrogc lI ck-c rrodc.
Prin ciples of Reactivity: Elect ron Tran sfer Rea ctions
rule: Product-favored reactions will always invol ve a reducing age n t th at is "so u t heast" of th e propo sed ox id izing age n t.
Reduction Half- Reaction 1, (s) + Cu2+ (aq) + 2 W (aq) + Fe' +(aq) + Zn2+ (aq) +
2 e2 e2 e2 e2 e-
-
-
• Northwest-Sout heast Rule This guideline is a reflection of the idea of moving down a poten tial energy "ladder" in a product-favored reaction.
2 I- (aq) The northwest-southeast Cu (s) rule: The reducing agent H2 (g) always lies to the sout heast Fe(s) of the oxidizing agent in a Zn(s) ~duct-favo red reaction.
Fo r exa mp le, Zn can red uce Fe2+ , 11+, Cu 2 +, and 12 , but, o f th e species on thi s list, Cu ca n red uce o n ly 6. T he algebraic sign o f th e half-reaction reduction potential is the sign of the e lec tro de when it is attac he d to the 11 2 /11 + sta nd ard ce ll (see Figu res 20.8 an d 20.13) . 7. Elec troc he m ical pote nti als d epend on th e na ture of th e rea ctan ts and p ro duCLIi an d t he ir concen tra tio ns , no t on th e quan titie s o f mat e rial used . The refo re , chang ing the stoi ch iometric co efficien ts for a half-react ion does no t cha nge the value of EO. For exam ple , the reduction of Fe 3 + has a n EO o f + 0. 77 1 V, whe the r the reaction is writt en as
'2-
• Chan ging Stoichi ometric Coeffidents The volt is defined as "energy/ charge" (V = J/ C) . Multiplying a reaction by some number causes both t he energy and t he charge to be multiplied by t hat number. Thus, the ratio "energy/charge = volt" does not change.
E' = + 0 .77 1 V
or as
2 Fel+(aq, 1 M) + 2 e: -> 2 Fe2+ (aq, 1 M)
E' = + 0 .771 V
Using Tables of Standard Reduction Potentials Tab les or "ladde rs" of sta ndard reduction pot en tials are immen sely useful, They allow yo u to p re di ct the po te n tia l of a new volt a ic ce ll, provide info rmatio n that ca n be used to ba la nc e re d o x eq ua tio ns, and help predict wh ich redox re act ion s are product-favo red.
Calculating Standard Cell Potentials, f' coli Th e stan da rd re d uc tio n pote n tia ls for half-reaction s were obtain ed by measu ring ce ll potentials. It m akes sen se , therefore , t hat these values ca n be co m bined 10 g ive the potential o f so me new cel l. T he net reactio n o ccu rrin g in a volta ic ce ll using silver and co p per hal f-cells is
2 Ag ' (aq) + Cu(s) -> 2 Ag(s) + Cu2+(aq) Th e silve r elec tr o de is th e ca th ode, and th e co p per elec tro de is th e an ode . \Ve kno w this be cause silver io n is red uced (to silver met al) and co p pe r metal is o xid ized ( to Cu 2 + io ns) . (Recall that oxidations always occu r at the anode and reduction s at the catho de .) Also not ice tha t th e Cu H ICu hal f-reactio n is "so u theas t" of the Ag ' lAg half-reactio n in th e po tentia l ladder (Fig u re 20. \4 and Table 20. \) .
EOcat hode = + 0.799 V "Distance" from EOcathode to EOanode is 0.7 99 V - 0.3 37 V =
j
Ag +(aq)
+ e"
-
Ag(s) Cu is "sout heast" of Ag+
0. 46 2 V.
EOanode = + 0.337 V
Cu2+ (aq)
+ 2e-
-
Cu (s) 2 0 .4
I
Standard Elect roche mica l Potentials
921
Th e potential fo r th e vo lta ic ce ll is the differen ce between th e sta n dard reduction potential s for the two half-reactio ns. EOcell
= EOcat hode -
EOanode
EO" " ~ ( +0.79 9 V) - ( +0 .33 7 V) £0cell = + 0.462 V
No tice tha t th e value of R Oed l is re late d 10 the "d ista nce" bet ween th e ca th ode and anode reac tio ns on th e potential lad d e r. Th e product s have a 100\'e r pot e nti al ene rgy th an th e re act ants, and the ce ll potential, E Ocd b has a positive value . A po sitive potential calculated fo r the Ag+IAg and Cu 2+ICu ce ll (ROeell = + 0.462 V) co nfirms th at th e reduction of silve r ions ill water with co p pe r metal is productfavo re d at equilibrium (Figu re 20.1) . We mi ght ask, however, abo u t th e valu e o f R\ oell if a reactan t-favored equat io n had been selec ted. For example , what is f7°{O ell fo r the reduct ion of co p per io ns with silver metal?
Cathode, reduction:
Cu 2+ (aq) + 2 e- -> Curs)
Anode, oxidation:
2 Ag(s) -> 2 Ag +(aq)
Net ionicequation:
2 Ag (5) + Cu 1+(aq) -> 2 Ag+(aq) + Cu(s)
+ 2 e"
Cell Vo lta ge Calculation E \ at hode
EOcell
= + 0.337 V a nd £0a node = + 0.799 V
= £Ocat hode -
E Oa node
= ( + 0 .337
V) - (0 .799 V)
EOcell = - 0.462 V
Th e negati ve sign for R Oedl ind icates tha t th e reaction as wri tte n is react ant-favored at eq uilib riu m . T he products o f th e reactio n (Ag " and. Cu) have a hig her pote nti al energy th an th e reacta nts (Ag and Cu2+). For th e in d icat ed reactio n to o ccur, a potentia l of at least 0.462 Vwo uld ha ve to be imposed 011 the system by an ex te r na l so u rce o f electri cit y (see Sectio n 20.7) .
ChemistrY.{J.Now'· Sign in at www.t homsonedu.com/ Ioqtn and go to Ch apter 20 Contents to see: • Screen 20.5 for demonstration of the potentials of various cells • Screen 20.6 for a simulation and tutorial on sta ndard potentials
EXERCISE 20.6
CaLculating Sta ndard Cell PotentiaLs
The net reaction t hat occurs in a voLtaic celt is In(s)
+ 2 Ag "(aq) --> In ' -(aq) + 2 Ag (s)
Identify the haLf-reactions that occur at the anode and the cat hode, and caLculate a potentiaL for the cell assuming standard conditions.
922
Chapter 20
I
Principles of Reactivity: Elect ron Transfer Rea ctions
Relative Strengths of Oxidizing and Reducing Agents Five h al f-re action s, sel e cte d from Ap pe n d ix XI, are arrange d [rom th e h al f-re acti on wit h th e h igh est (m os t positive ) / 2 Cl-(aq) Ag +(aq) + e " - > Ag(s) 2 W (aq) + 2 e: - > H, (g) Ni 2+(aq) + 2 e: - > Ni (s) Zn 2+ (aq) + 2 e" - > Zn (s)
• T he list 011 th e left is h e ade d by C12 , a n d ement th a t is a stro ng o xid izin g agent and th u s is easily re du ce d . At t he botto m of t he list is Zn 2 +( aq), an io n n ot easily reduced and th u s a poor o xid izin g age n t. • O n the righ t, the list is h e ad e d by CI-(aq) , a n ion th at can be o xid ize d to CI2 on ly wit h d iffic ul ty. It is a vc ry po or red ucing agent. At th e bo tt om of the list is zinc metal, which is qu ite easy to o xidi ze and a goo d reducin g age n t.
By arra nging these ha lf-react ions based on f:o va lues, we h ave a lso a rran gerlth e ch e m ica l species on th e tw o sid es of th e eq u ation in o rd e r of th ei r stre n gt hs as oxid izin g o r re d ucing age nts. In t h is list , from stro ng es t to weakest, t he o rd er is Oxidizing agents: ( [, > Ag + > W > Nil + > Zn' + l weak strong - - - - - - -->
Reducing agents: Zn > Ni > H, > Ag > ( l strong - - - - - - - - » weak
Fin all y, notic e that th e value of f; °ct'll is grea te r th e fa rth e r a part the oxid izin g a n d re ducin g ag e n ts a re o n th e potentia l ladde r. Fo r exam p le ,
Zn(s) + Cl,(g) ---> Zn 2+ (aq) + 2 Cl - (aq)
EO ~ + 2.12 V
is m o re stro ngly produc t-favored th an th e red ucti on of hyd rogen io ns with ni c kel meta l
Ni(s) + 2 W (aq) ---> Ni 2+(aq) + H, (g)
EO = + 0.25 V
Ranking Oxidizing and Reducing Agents Problem Usethe table of standard reduction potentials (Table 20.1) to do the following:
(a) Ra nk the halogens in order of their strength as oxidizing agents. (b) Decide whether hydrogen peroxide (H202) in acid solution is a stronger oxidizing agent than (l2'
(c) Decide which of the halogens is capable of oxidizing gold metal to Au 3 +(aq). Strategy Theability of a species on t he left side of a reduction potential table to function as an oxidizing agent declines on descending the list (see points 2 and 3, pages 918·9 19). Solution (a) Ranking halogens according to oxidizing ability. The halogens (Flo ([ 2. Br2• and 12) appear in t he upper-left portion of t he table, with F2 being highest. followed in order by the other t hree species. Their strengths as oxidizing agents are F2 > el2 > Br2 > 12, (The ability of bromineto oxidize iodide ions to molecular iodine is illust rated in Figure 20.15.) 20 .4
I
Standard Electrochemical Potentials
923
FIGURE 20.15 The reaction of
bromine and iodide ion. This experiment proves that Br2is a better oxidizing agent t han 12 , The presence of12 in the bottom layer in t he photo on t he right indicates that t he added Br2was able to oxidize t he iodideions originally present to molecula r iodine (12) ,
The test tube contains an aqueous solution of KI (top layer) and immisci ble Cel4 (botto m layer).
After adding a fewdrops of Brz in water, the Ja produced collects in the bottom CC14 layer and gives it a purple color. (The top layercontains excess Brz in water.)
Add Br z to solution of KI, and shake.
( b) Comparing hydrogen peroxide and chlorine. H20 2 lies just below F2 but well above Cl2 in t he potential ladder (Tab le 20.1) . Thus, H20 2 is a weaker oxidizing agent t han F2 but a st ronger one than Cl2• (Note that the E O value for H202 refers to an acidic solution and standard condtttcns.) (c) Which halogen will oxidize gold metal to gold(TII) ions? TheAu H IAu half-reaction is listed below the FI t F- half-reaction and just above the el21el- half-reaction. This tells us that, among the halogens, only F2 is capable of oxidizing Au to Au H under standard conditions. That is, for the reaction of Au and Fl , Au(s) --> Au' -(aq)
Oxidation, anode:
+ 2 e: --> 2 F-(aq) + 2 Au(s) --> 6 F (aq) + 2 Au ' "(aq) fO aoDde = + 2.87 V - ( + 1.50 V) = + 1.3 7 V
Reduction, cathode:
F,(g)
Netionic equation:
3 F, (g)
fO cell
= fO eathDde -
+ 3 e-
FI is a stronger oxidizing agent than Au H , so the reaction proceeds from left to right as written. (Thi s is confirmed by a positive value of £0cell' ) For the reaction of elz and Au, Table 20.1 shows us that Cl, is a weaker oxidizing agent than Au3 + , so the reaction would be expectedto proceed in the opposite direction. Oxidation, anode:
Au(s) --> Au'"(aq)
Reduction, cathode:
([ ,(aq )
Net ionic equation:
3 ([ ,(aq )
E" cell
= EO c~thode
-
E Daoode
=
+ 3 e"
+ 2 e- --> 2 ([ -(aq ) + 2 AU( s) --> 6 ([ -(aq) + 2 Au' "(aq )
+ 1.36 V - ( +1.50 V)
=
- 0.14 V
This is confirmed by the negative value for E"cell , Comment In part (e), we calculated EOcell for two reactions. To achieve a balanced net ionic equation, we added t he half-reactions, but only after multiplying the gold half-reaction by 2 and the halogen, halfreaction by 3, (This means 6 mol of electrons was transferred from 2 mol Au t o 3 mol el2. ) Notice that this multiplication does not change t he value of E O for the half- reactions because cell potentials do not depend on t he quantity of material. EXERCISE 20.7
Relative Oxidizing and Reducing Ability
Which metal in the following list is easiest to oxidize: Fe, Ag, Zn. Mg, Au? Which metal is the most difficult t o oxidize?
EXERCISE 20.8 Using a TabLe of Standard Reduction PotentiaLs to Predict Chemical Reactions
Determine whether the following redox equations are product-favored at equilibrium. (a) NF "(aq ) + H,(g) --> Ni(s) + 2 H'{aq)
+ 2 I - (aq) --> 2 Fe' "(aq) + Iats} (e) Br, (I) + 2 ([-(aq ) --> 2 BrTaq} + ([ ,(g) (d) er,O,' -(aq ) + 6 Fe' "(aq) + 14 H'{aq } --> 2 Cr' - (aq) + 6 Fe' "(aq) + 7 H,O(I)
(b) 2 FeH(aq)
924
Chapter 20
I
Principles of Reactivity: Electron Tran sfer Rea ctions
Chemical Perspectives
An Electrochemical Toothache!
It was recently reported that a ee-year-ctd woman had intense pain t hat was traced to her denta l work (New England Journal of
ing on one tooth slightly closer to a gold alloy crown on an adjacent tooth . Ea ting acidic foods caused her intense pain. Whe n dental amalgams of dissimilar metals come in contact with saliva, a voltaic cell is formed
Medicine, Vo l. 342, p. 2000, 2003) . A root canal job had moved a mercury amalgam fill-
20.5
that generates potentials up to several hundred millivolts-and you feel it! You can do it, too, if you chew an aluminum foil gum wrapper with teeth that have been filled with a dental amalgam. Ouch!
Electrochemical Cells Under Nonstandard Conditions
Ele ctroc h e m ica l cells seld o m o pe rat e un d er sta n dard co nditio ns in t he real wo rld . Even if the ce ll is co ns tructed wit h all d isso lved species a t I M, reactant co ncentratio ns d ecrease a n d p rod u ct con ce n tra tio n s increa se in th e co u rse o f th e reactio n. Chang ing co nce n tra tio n s of reacta n ts an d p ro d u cts will affect th e ce ll vo ltage. T h u s, we n eed to ask what h ap pen s to ce ll po te ntials u n d e r n o n stan d a rd co n d itio ns.
The Nernst Equation Based o n bo th th eory an d ex pe rime n ta l res u lts, it h as be en d eterm in ed tha t ce ll po te n tials are rel a ted to co nc e n tra tio ns o f reactan ts a n d p roducts a n d to tem peratu re , as foll o ws: (20 .2)
E = EO- (RT/nf) In Q
In th is equat ion , wh ich is known as th e Nernst equati on, R is th e gas co ns ta n t (R3 I 447 2 .11K . m o l) ; T is th e te m pe ratu re (K) ; a nd n is th e num be r o f mo les o f e lec tr o n s tran sfe rred be tween ox id izing an d re d uci ng age n ts (as d eterm in ed by the ba la n ced equation for t he rea c tio n) . T he sym bol F represents th e Faraday constant (9 .6485338 X 10-1 C / mo l). One Famdas is the fJlum titJ nleln;trir dwrf!:e ramer! by one mole (1 electrons. T h e te rm Q is the reac tio n q uo tie n t. a n ex pressio n re lating the con ce n tra tio ns of th e p rod ucts a n d reactants raised to a n a p p ro p riate powe r as defi ned by th e sto ich io me tric co efficien ts in the balan ced, n e t eq u a tio n [see Eq ua ti o n 16.2, Sec tio n 16.2 ) . Substituti ng values for the co ns ta n ts in Equat ion 20 .2, a n d u sin g 298 K as the te mperat u re , g ives
E = £0 - 0.0257 1nQ
at 25
O(
• Walther Nernst (1864- 1941) Ne rnst was a German physicist and chemist known for his work relating to the third law of thermodynamics.
(20 .3)
n
or, in a co m mon ly used fo rm u sin g base-l 0 loga ri thm s,
E = EO_ 0.0591 1ogQ
n
In esse n ce , th e term ( RTlnF) lnQ "co rrec ts" the sta n dard p o ten tial EOfo r n on stan d a rd co n d itio ns or co nc e n tra t ions.
• Units of Rand F The gas constant R has units of Jj K. mol. and F has units of coulombs per mol (Cjmol). Because 1 J = 1 C . V, the factor RTjnF has units of volts.
Sign in at www.thomsonedu.com/Ioqtn and go to Chapter 20 Contentsto see Screen 20.8 for a tutorial on the Nernst equat ion.
20 .5
I
Elect rochemical Cells Under Nonstandard Conditions
925
• . Using the Nernst Equation ProbLem Avoltaic cell is set up at 25 O( with the half-cells: Al3- (O.0010 M) IAland NF-(O.50 M) I Ni. Write an equation for the reaction t hat occurs when the cell generates an electric current. and determine the cell potential. Strategy The first step is to determine which substance is oxidized (Al or Ni) by looking at the appropri-
ate half-reactions in Tab le 20.1 and decidi ng which is the better reducing agent (Example 20.5). Next, add the half-reactio ns to det ermi ne the net ionic equatio n, and calculate fO. Finally, use the Nernst
equation to calculate E, the nonstandard potential. SoLution Alumi num metal is a stronger reducing agent than Ni metal. (Conversely, NF · is a better oxidizing age nt tha n Al3+.) Therefore, At is oxidized, and the Al3-!AL compartment is the anode. Cathode. reduction:
Ni2+ (aq) + 2 e :
Anode. oxidation:
Al(s) ~ Al3+(aq)
Net ionic equation:
2 Al(s)
EOcell = EOcathode
-
~
Ni(s)
+ 3 e-
+ 3 NF+(aq) ~ 2 All· (aq) +
3 Ni(s)
EOanode
(-0 .25 V) - (-1.66 V) ~ 1.41 V The expression for Q is writte n based on the cetl reaction. In the net reaction. AP+(aq) has a coeffiEO"II
~
cient of 2, so t his concentration is squared. Similarly, [NF·(aq)] is cubed. Solids are not included in t he expression for Q(... Section 16.1).
The net equation requires tra nsfer of six electrons from two Al atoms to three NF· ions. so n Substituting for P , nand Q in t he Nernst equation gives E
= EO _
cell
cell
= 6.
H
0.0257 ln IAl ]2 n [Ni2+p
~ + 1. 41 V _ 0.0257 In [0 .001 0J' 6 [0.50 ]' ~
+ 1.41 V - 0 .00428 In (8 .0 X 10 - ')
~
+ 1. 41 V - 0.004 28 (-11.7)
~
1.4 6 V
Comment Notice t hat Ecell is larger tha n EO cell because the product concentratio n, [AlHo l , is much smalLer than 1.0 M. Generally. when product concentrations are smaller initially than the reacta nt concentrations in a product-favored reaction, t he cell potential is more positive t han E"' . EXERCISE 20.9
Variation of E,," with Concentration
A voLtaic cell is set up with an aluminum eLectrode in a 0.025 MAt(N01h(aq) solution and an iron electrode in a 0.50 M Fe(N03h( aq) solution. Det ermine the cell pote ntial. Ecell • at 298 K.
Example 20 .6 d emonstrat es th e calcu lati on of a ce ll potentia l if co ncentration s are kn o wn. It is also usefu l to appl y the Nc rns t eq uation in the o p posite se nse, using a m easured cell potential to determ ine an unknown co nc en tra tio n . A device tha t docs j ust this is t he p H meter (Figu re 20 .16) . In an electroc hemical ce ll in which H + (aq ) is a reactan t or p rod uct, the cell voltage will vary predi ctably wit h th e h yd rogen ion co nce n tratio n . T he cell voltage is measu red and th e value used to calcu la te pH . Exa mple 20.7 illustrates how En .1I varies with the hyd roge n ion concentration in a sim ple cel l.
926
Chapter 20
I
Principles of Reactivity: Elect ron Transfer Reactions
Variation of E"" with Concentration Problem AvoLtaic cell is set up with copper and hydrogen half-cells. Standard conditions are employed in t he copper half-cell. Cu2 +(aq, 1.00 M) ICu(s). The hydrogen gas pressure is 1.00 bar, and [W(a q)) in the hydrogen half-cell is the unknown. Avalue of 0.490 Vis recorded for E(eu at 298 K. Determine the pH of t he solution.
Strateg y We first decide which is t he better oxidizing and reduci ng agent so as to decide what net reactio n is occurring in the cell. With this known, P celL can be calculated. The only unknown quantity in the Ne rnst equation is the concentration of hydrogen ion, from which we can calcu late the solution pH. SoLution Hydrogen is a better reducing agent than copper metal. so Cu(s)I Cu2+(aq, 1.00 M) is the cathode, and Hig, 1.00 bar) I H'' {aq. ? M) is the anode.
Cathode, reduction:
Cu2+(aq)
+ 2 e-
Anode, oxidation:
H2(g)
~
2 W( aq)
Net ionic equation:
H2 (g)
+ Cu ' ' {aq) ~ Cu(s) + 2 H''{aq)
[0" " ~ (+0.337 V) - (0.00 V)
~
~
Cu(s)
+ 2 e-
+0.33 7 V
The reaction quotient, Q, is derived from the balanced net ionic equation.
The net equation requires t he t ransfer of two electrons, so n = 2. Th e value of [Cu!" ] is 1.00 M, but [H+] is unknown. Substitute this information into the Nernstequation (and don't overlook the fact that [H+] is squared in the expression for Q) . [ ~ [ " _ 0.0257 Ln~ n [(U2-+ ]PHI 0.490 V ~ 0.3 37 V _ 0.0257 Ln [H'l' 2 (1.00)(1.00) - 11.9
[W]
~
~
2.6
Ln [Wl' X
10- ' M
pH ~ 2.59
EXERCISE
20 .10 Using the Nernst Equation
The half-cells Fe>+ (aq. 0.024 M) Ife(s) and H'{aq, 0.056 M) I H,( 1.0 bar) are Linked by a saLt bridge to create a voltaic cell. Determi ne the cell pote ntial, f oell' at 298 K.
In th e real wo rld , using a h ydrogen elec trode in a p l I meter is no t practic al. T he a p pa ratus is cl um sy; it is an ythi ng but robust; and plat inum (fo r th e el ectrode ) is cos tly. Common pH met ers use a glass electrode , so ca lled because it co ntains a thi n glass membran e se pa ra ting th e ce ll from th e so lution whose plI is to be measured (Figu re 20 .16 ) . In sid e the g lass electrode is a silver wire coa ted with Agel and a so lutio n of H CI; o u tside is th e so lu tio n of unk now n pH to be evaluated. A Ag/ Ag el or ca lo mel e lec tro de-s-a co m m o n refe ren ce elec tro de using a me rcury(l )me rcu ry redox co u ple (I IgzClzII-Ig )- serves as tile sec o nd e lect rod e of tile ce ll. T he potential across the glass membrane d epe nd s on [H +]. Co m mo n p H met ers give a di re ct read o u t of pl l.
20.5
I
Elect rochemical Cells Under Nonstandard Conditions
927
Non-conductive glass or plastic electrode body
Reference electrode
Porous ceramic diaphragm Internal solution
....,....-
lnternal electrode
'C~liI''---- AgCl precipitat e
[r}
(b)
(a}
FIGURE 2 0 . 16 Measuring pH. (a) A pH portable meter that can be used in the field. (b) The tip of a glass electrode for measuring pH. (c) Aschematic diagram of a glass electrode. (See Chemi st ryN ow Screen 17.4 for an animation of the operation of a glass electrode for pH measurement.)
20.6
Electrochemistry and Thermodynamics
Work and Free Energy The firs t law of th erm o d ynam ics [Sect ion 5.4J states th at t he in te rn al en ergy ch a n ge in a syste m (dlJ) is rel ated to two quan titi e s. h e at ( q) a nd wo rk (w): Do U = q + w. T his e qua tio n also ap plies to chemica l changes tha t oc cur in a vo ltai c ce ll. As curren t flo ws, e ne rgy is tran sferred from the syste m (the vo ltai c ce ll) to th e su rrou nd ings. In a voltaic ce ll, the d ecrease in internal energy in the system will m an ifest itsel f id eall y as el e ctrical work do ne o n the su r ro u n d ing s by the syste m . In p ractice , ho wever, so mc c ne rgy is u suall y evo lved as heat by th e vo lta ic cel l. T he maximum work do ne b y an elec tro che mical system (id eally. assumi ng no h eat is generated) is pro po rtio n al to th e po tenti al differen ce (vo lts) and t he q uanti ty o f charge (co ulo m bs): Wmax
= nFE
(20 .4)
In th is equatio n . E is th e ce ll vo ltage . an d nF is th e qu antity o f elect ric chargc tr an sferr ed fro m a no d e to catho de. The free e nergy change fo r a p rocess is, by d efin ition , t he maximum amount o f work th at ca n be o btained {Sec tio n 19.6] . Becau se th e maximum wo rk and th e cel l po te n tia l are relate d , F O and d J Jc can be rel ated mat he matically (taking ca re to assig n sig ns correctly). T he max im u m work d o ne o n the su rro u nd ings whe n electri city is produced by a vo lta ic ce ll is + n Fl~~ with th e po sitive sign d en oting an in crease in energy in th e su rrou ndi ngs . The energy co nten t of the ce ll d ecreases by th is am o u n t, Th us, 6.r G for t he volta ic ce ll has the o p posite sign . ::',6 928
Chapter 20
I
Principles of Reactivity: Electron Transfer Reactions
~
- nFE
(20.5)
Under sta ndard co nd itio ns, th e ap pr o p ria te eq ua tio n is (20 .6) O
This ex p ressio n sho ws that , the mo re po sitive the value of R the large r and m ore ne ga tive th e va lue o f a J ;o fo r t he reacti o n . Also , because of th e rel ati onshi p between a rGoand K, th e fa rt her apart the hal f-reactio ns o n the potent ial lad d e r, t he more strong ly product-favored the reac tio n is a t eq u ilib r iu m . ,
• Units in Equation 20.6 n has units of mol e", and F has units of (Cjmo l e'). Therefore, nF has units of coulombs (C). Because 1 J = 1 C. V, the product nFE will have units of energy (J).
• , The ReLation Between fO and Il ,G' Problem The standard cell potential. E" c~ ll for t he reduction of silverions wit h copper metal ( Figure 20.5) is + 0.462 Vat 25 -t . CaLculate j" Go for this reaction. Strategy We use Equation 20.6, where Fis a constant and E \~ ll is given. The only problem here is to determine t he value of n, the number of moles of electrons transferred between copper metaL and silver ions in the balanced equation. Solution In this cell, copper is the anode, and silveris t he cathode. The overall cell reaction is Cu(s) + 2 Ag+(aq) --; Cu ' +(aq) + 2 Ag(s) which means that each mole of copper transfers two moles of electrons to two moles of Ag+ ions. That is,
n = 2. Now use Equation 20.6. ~ ,G " ~
- nFE"
~
- (2 mol e - )(96.485 C(mol e -)(0. 462 V)
~
- 89.200 C. V
Because 1 C' V = 1 J, we have ~ ,G " ~
- 89.200 Jar - 89.2 kJ
Comment This examp le demonstrates a very effective method of obtaining the rmodynamic values from reLatively simple electrochemica l experiments.
EXERCISE 20.11 The Relationship Between E' and 1l,G" The following reaction has an Eo value of - 0.76 V: H,(g) Calculate
j" Go
+ l nH(aq) --; In(s ) + 2 H''{aq)
for this reaction. Is the reaction product- or reactant-favored at equilibrium?
EO and the Equilibrium Constant Wh en a vo ltaic cell produces an electr ic cu rren t, th e re actant co nce nt rations decrea se, and th e pr oduct co nc en tratio ns inc rease. T he cell vo ltage also chang es; as reac tants are conve rted to p ro d ucts, th e value o f f~e ll decreases. Even tually, the ce ll po te n tial reaches zero; no furthe r net reaction occu rs; and eq uilibriu m is ac hieved . T h is situatio n can be ana lyzed using th e Ncrnst eq uatio n. Wh en l\ 'ell = 0, th e re ac ta nts and prod ucts are at eq u ilibriu m, and th e reactio n quo tien t, Q is equal to th e eq ui lib rium co nstan t, K. Su bstitu ting th e appro priate sym bo ls and values in to the Ner nst equatio n. - E' - 0.Ol5 7 ln K -E -- 0-
n
an d co llec ting terms gives an eq ua tio n th at relate s the cc ll pot e nt ial and eq u ilib riu m co nstant:
nE'
lnK = - 0.0257
at l5 ' (
(20.7)
• K and fO The farther apart half-
reactions for a product-favored reaction are on the potential ladder, the larger the value of K.
Eq ua tio n 20. 7 can be used to de te rmi ne val ues fo r equ ilib riu m co nstan ts, as illu stra te d in Exam ple 20.9 and Exercise 20. 12. 2 0 .6
I
Elect rochemist ry and Thermodynamics
929
• . P and Equilibrium Constants ProbLem Calculate the equilibrium constant for the reaction Fe(s) + Cd' +(aq)
FeH(a q) + Cd(s)
Strategy First determine P',ell from EOvalues for the two half-reactions (see Example 20.5) and from those the value of n. t he other parameter required in Equation 20.7.
SoLution The half-reactions and [0 values are Cathode, reduction: Cd2+ (aq) + 2 e: Anode, oxidatio n:
Fe(s)
Net ionic equation:
Fe(s) + Cd 2 -(a q)
['" ll ~
( -0.40 V) - ( -0.44 V)
~
---?
~
Feh (aq)
Cd(s)
+ 2 eFe2+(aq)
+ Cd(s)
+ 0.04 V
Now substitute n = 2 and EOcell into Equation 20.7.
InK ~ ~ ~ (2)(0.04 V) ~ 3.1 0.025 7 0.0257 K ~ 20
Comment The relatively small positive voltage (0.04 V) for the cell indicates that the cell reaction is only mildly product-favored. Avalue of 20 for the equilibrium constant is in accord with this observation. EXERCISE
20 .12 P and Equilibrium Constants
Calcu late the equilibrium constant at 25 O( for the reaction 2 Ag - (aq) + Hg«()
2 Ag(s) + Hg ' ·(aq)
T he relatio nshi ps betwee n J~' o , K, and d rG' , wh ich is su m ma rized in Ta ble 20.2, ca n be used to obtain eq ui librium co nstan ts fo r ma ny d ifferent ch e mical syste ms. For exam p le, le t us construct a n electrode in wh ich an inso luble io n ic compound (suc h as Agel) is a com po ne n t of the hal f-cel l. For th is purpose , a silver electrode wit h a su rfac e laye r of Age l can be pr epared. T he react io n occu rri ng at this elect ro de is th en AgCl(s) + e" -> Ag(s) + Cqaq ) The stan d ard re d uction potential fo r thi s h alf-cell (Ap pe n d ix 1'' '1) is + 0.222 V. Wh e n thi s h alf-reactio n is paired wit h a sta n da rd silver elec tro d e in a n ele ctro chem ical ce ll, the cel l reac tio n s are
Ag(s) -> Ag+(aq)
Net ionic equation:
AgCl(s) -> Ag +(aq) + Cl- (aq)
K K»
1
K«
I
1
= ( +0.22 2 V)
+ e" - (+0 .799 V)
20.2 Summary of the Relationship of K, a ,Go, and f
=-
0.577 V
O
d ,Go
EO
Reactant-Favored or ProductFavored at Equilibrium?
Spontaneous under Standard Conditions?
.:l,Go < 0
£0 > 0
Product-favored
Spontaneo us under stan dard conditions
=0
EO = 0
[C] ' [O]d ~ [A]'[B] ' at equilibrium
At equilibrium under standa rd conditions
.:l,Go > 0
Eo < 0
Reactant-favored
Not spontaneous under standa rd conditions.
j" Go
K ~l
20
+ Cl - (aq)
Anode, oxidation :
TABLE
Chapter
-> Ag (s)
AgCl(s)
P eell = EOeathode - Panode
930
+ e:
Ca thode, reduction:
Principles of Reactivity: Elect ron Transfer Reactions
Th e equation for t he ne t reaction rep res en ts th e eq uilib r iu m o f so lid Agel an d its io ns. T he ce ll po te ntia l is nega tive. indi catin g a r eac ta n t-favo red pro ce ss, as wou ld be e xpe cted based o n th e lo w so lu bilit y of Age l. Usin g Eq ua tion 20.7, the va lue of th e eq ui lib ri u m co ns tan t [ Ks p , Se ctio n 18 .4 ] can be o b ta ine d fr o m E On.lI • lnK = ~ 0.0257
EXERCISE 20.13
= (1)(-0.577 V) = -
22.5
0.0257
Determining an Equilibrium Constant
In Appendix M, the followi ng standard reductio n potentia l is reported :
[In (CN ).J ' - (aq )
+ 2 e" -.. In (s) + 4 CN -(aq)
[0 ~
- 1.26 V
2
Use th is informatio n, a long with the data on th e Zn "'(aq)lZn half-cell , to calc ulate t he equilibrium consta nt for th e react ion
ln H (aq )
+ 4 CW (aq) -.. [I n(CN )d ' - (aq)
The value ca lculated is t he formation co nstant for t his complex ion at 25°C.
20.7
Electrolysis: Chemical Change Using Electrical Energy
T h us fa r, we have de scrib ed electrochem ica l cel ls th at usc product-favor ed red o x re actions to gcn eratc an el ectric current. Eq ua lly im po rtan t, however, is the o pposite p rocess, e lectro lysis, the use o f e lec trical en ergy to b rin g about che mical change. Elec t ro lysis o f water is a classic che m istry expe rime n t. a nd th e el ectro plating o f met als is ano t her e xa mp le o f electro lysis (Figu re 20. 17) . In electroplating , an electric cu rr e n t is passed th ro ugh a so lu tio n co n ta in ing a salt o f thc met al to be p lat ed . Th e o b jec t to he plated is th e catho de. \Vhen metal ion s in so lu tio n are red uced , the me ta l d epo sits on the o bject's su r fac e . Electrolysis is an important proced u re be cause it is widely used in the re fining of metals suc h as alu mi nu m a nd in th e productio n o f chemica ls suc h as chlo rine. FIGURE 20.17 Ele ctrolysis. (a) Electrolysis of water produces hydrogen and oxygen gas. (b) Electroplating adds a layer of metal to the surface of an object , either to protect t he obj ect from corrosion or to improve its physical appearance. The procedure uses an electrolysis cell, set up with the object to be plated as t he cat hode and a solution containing a salt of t he metal to be plated.
(a)
(b) 20 .7
I
Electrolysis: Che mica l Chan ge Usin g Elect rical Energy
931
Manganese in the Oceans
Case Study Manganese is a key component of some oxidatio n-red uction cycles in the oceans.
Acco rding to an article in the journal Science, it "can perform this role because it exists
in muLtip le oxidation states and is recycled ra pidly between thes e states by bacterial processes." figure A shows how this cycle was thought
to wo rk. Ma nganese(II) ions in subsurface water are oxidized to form manganese(IV) oxide, Mn02' Particles of t his insoluble solid sink toward t he ocean floor. However, some encounter hydrogen sulfide, which is produced in t he ocean dept hs, rising toward the surface. Another redox reaction occurs, producing sulfur and manganese(II) ions. The newly formed Mn 2 + ions diffuse upwa rd. where t hey are again oxidized. The manga nese cycle had been thought to involve only the + 2 and + 4 oxidation states of manganese. and ana lyses of water samples assumed the disso lved manganese existed only as Mn 2+ ions. One reason for this is that t he i ntermediate oxidation state, Mn H , is not predicted to be stab le in water. It should disproportionate to the + 2 and + 4 states. 2 MnH(aq) + 2 H,O(f ) --. MnH(aq) + MnO"s) + 4 H' {aq) It is know n, however, t hat MnH can exist
when complexed with species such as pyrophosphate ions, PZ0 14 - .
Concentration -----+ Figure A Manganese chemistry in the oceans. Relative concentrations of important species as a function of depth in the oceans. See K. S. Johnson. Science, Vol. 313. p. 1896. 2006 and R. E. Trouwborst, et at.. Science. Vol. 313, pp. 1955- 1957, 2006.
SeveraLyears ago, geochemists suggested that MnH ions could exist in natural water. They could be produced by bacterial action and stabilized by phosphate from ATP or ADP. They speculated that the Mn H ion could play an important part of the nat ural manganese cycle. Now, other researchers have indeed discovered t hat, in oxygen-poor waters, t he manganese(III) ion, Mn 3 + , can persist. These ions were found in anoxic zones (zones wit hout dissolved oxygen) below 100 m in the Black Sea and below about 15 m in t he Chesa peake Bay. It is now clear that Mn 3 + ions. which had previously been know n only in the laboratory, can exist in natu ra l waters under the right circumstances a nd tha t t he manganese cycle may have to be revised.
Questions: 1. Given the following reduction potentials, show that Mn 3 + should disproportionate to Mn 2+ and Mn02 at standard conditions. 4 W (aq) + MnO,(s) + e- --. Mn H (aq) + 2 H,O(f) EO = 0.95 V
Mn 3 +(aq) + e" -) Mn2+(aq) Eo ~ 1.50 V
2. Balance the following equations in acid solution. (a) Reduction of Mn02 with HS- to Mn 2+ and 5 (b) Oxidation of Mn 2+ with O2 to Mn02 3. Calculate EOfor the oxidation of Mn 2+ with O2 to Mn02' Answers to these questions are in Appendix a.
Electrolysis of Molten Salts All elec tr o lysis ex pe rime n ts are se t up in a sim ilar mann e r. T he materi al to be c k-ctro lyzcd , either a mo lte n salt or a so lu tio n, is co n ta ined in an electro lysis cel l. As was th e case with voltaic ce lls, io ns must be pr esen t in th e liquid o r solu tio n fo r a curren t to flow. Th e movem en t of io ns con stitu tes the elec tric cu rren t within the cel l. T he ce ll has two e lec tro des th a t are conn ec te d to a so u rce o f DC (d irectcu rre n t) voltag e. If a high e no ug h voltage is a p plied , ch em ical reactions oc cu r at the two el ectrodes. Reduction o ccu rs at the negatively charge d ca thode , with electrons bein g tran sfe rr ed from th at e lectrode to a che m ical species in th e cel l. Oxidatio n oc cu rs at the po sitive anode, with electro ns from a che m ical spec ies bein g transferred to that elec tr o de. Let us first focu s o ur atte n tio n o n th e chem ica l re actio ns th at oc cu r at eac h e lectrode in th e el ect ro lysis o f a molten salt. Sodi um ch lo rid e mel ts at abo u t 800 °C, and in th e m ol ten sta te so d iu m io ns (Na" ) and chloride io ns (CI-) are freed fro m th e ir rig id a rra ng eme n t in t he crys ta lline lattice . The refo re , if a po ton-
932
Chapter
20
I
Princip les of Reactivity: Elect ron Transfer Reactions
Cathode
Anode (+ )
( -)
Sodium ion mig rates to cathode.
FIGURE 2 0 .18 The preparation of sodium and chLori ne by the eLectroLysis of moLten HaCI. In the molten state, sodium ions mig rate to the negative cathode, where they are reduced to sodium metal. Ch loride ions mig rate to the positive anode. where they are oxidized to elemental chlorine.
Reduced to sodium metal
Chloride migrates to anode.
Oxidized to chlorine
tia l is ap plied to the e lec trodes, so d iu m ion s are attrac te d to th e nega tive elec trode, and chlorid e ions are att ra cted to th e po sitive el ectrode (Figure 20.18). If the poten tial is high c no ug h , chemical re acti ons occu r at each elect rode . At the ne gati ve ca tho de, Na " io ns acce p t electro ns an d are reduced to sod iu m m et al (a liquid at thi s temperatu re ) . Simu ltan eously, at th e po sitive anode , chloride ion s g ive u p e lec tro ns a nd form ele me n tal ch lo rine .
Cathode ( -), reduction:
2 Na + + 2 e - -> 2 Na(f)
Anade ( + ) , oxidation:
2
Net ionic equation:
2 Na + + 2 n- -> 2 Na(f)
n- -> ( l,(g) + 2 e+ ( l,(g)
Electron s move through th e external cir cui t under th e force ex erted by th e applied potential , and the moveme n t o f po sitive and negative ions in th e molte n salt co nstitu tes t he cu rre n t with in the cel l. Finally, it is important to rec og niz e th at th e re ac tio n is not spo ntaneous. The e nergy required for thi s reaction to o ccur has be en provided by th e elec tric current.
Electrolysis of Aqueous Solutions Sod iu m ions (Na " ) and ch lo ride ions (C I-) arc th e p rimary spec ies p resent in molten NaC I. Only ch lo ride ions ca n be o xid ized , and o nly sodiu m ion s ca n be re d uce d . Electr o lyses of aq ue o us so lu tio ns are mo re co mp lica ted th an e lec tro lyses o f molten sa lts, howeve r, becau se wate r is now present. Wat er is an elearoactioe su bsta nce; th at is, it ca n be oxi d ized or reduced in a n electro chemical pro cess. Co ns ider th e electro lysis of aq ueo us sodi um iod ide (Figu re 20.19) . In thi s experime nt, th e electr ol ysis cel l co n ta ins Na t fa q) , I- (aq ) , an d H 2 0 molecul es. Possibl e reduction reactions at the n egative cathode includ e Na +(aq)
+ e" ->
Na(s)
2 H,O(f ) + 2 e: -> H, (g) + 20 W(aq) 2 0 .7
I
Electrolysis: Chem ical Change Using Electrical Energy
933
Problem Solving Tip 20.3
Electrochemical Conventions: Voltaic Cells and Electrolysis Cells
Whether you are describing a voltaic cell or an electrolysis cell. the terms "anode" and "cathode" always refer to the electrodes at which oxidation and reduction occur, respectively. The polarity of the elec-
Type of Cell
Electrode
function
Vo ltaic
Anode
Oxidation
Cathode
Reduction
Electrolysis
Anode
Oxidation
Cathode
Reduction
trades is reversed , however.
Polarity
+ +
Possible oxidation reactions a t tile posuioe an ode are 2 I-(aq) -> 12(aq) 2 H20(C) -> 02(g)
+ 2 e:
+ 4 W(aq) + 4 e-
In t he electro lysis of aq ueous Na I, experi me n t shows th at I12 (g ) a n d O II - (aq ) a re formed. by wate r redu cti on at th e cathode, a n d iodine is form ed a t th e a node (Figu re 20. 19). Thus, the overall ce ll p rocess can be su m marized by th e followin g equations:
Anode ( + ) , oxidation:
+ 2 e" -> H2(g) + 20 W(aq) 2 I-(aq) -> 12(aq) + 2 e:
Net ionic equation:
2 H20(C)
Cathode (-), reduction:
where
E Ord 1
2 H20(f)
+ 2 I-(aq)
-> H2(g) + 2 OW (aq) + 12(aq)
h as a negativ e value .
E(Jcell = E(J Sn(s)
Anode ( +), oxidation:
2 H, 0 (t) --> O,(g )
EO" "
= (-0.14 V)
+ 4 W (aq) + 4 e"
- ( +1.23 V)
=-
1.37 V
a sm alle r a p p lied po ten tial wo u ld see mingly be req ui red . To ex p lain th e fo rm ati on of c h lo ri ne instea d o f ox yge n , we m us t tak e in to ac co u n t rat es of react io n . T h is prob lem occu rs in the co m m e rc ially impo rt ant electro lysis of aq ue o us NaC I, wh ere a vo ltage h igh e n o ug h to ox id ize bo th CI- an d H 2 0 is used . Howeve r, because ch lo r ide ion is o xidi zed mu ch faste r th an II 20 , th e result is th at CI2 is the majo r p rod u ct in th is el ectro lysis. An oth e r in sta nce in whic h rat es a re im porta n t co n ce rns e lectrod e materia ls. Gra p h ite , co m mo n ly used to mak e in e rt elec tro des, ca ll be o xid ized . Fo r th e h al freac tio n CO, (g) + 4 H +(aq ) + 4 P- ---; C (s) + 2 H ,O (C) , EOis + 0 .20 V, indica tin g th at carb o n is slig h tly easier to ox id ize th a n co p pe r (/ Sn(s)
II fo llows that if the num ber of moles of elec tro ns flowing th rough the elec tro lysis ce ll co u ld be m easured, th e num ber of m ol es o f silver or tin pro duced co u ld be ca lcu la te d. Co nve rsel y, if th e a mou nt of silve r o r tin produced is known , th en th e nu mber of moles o f elec tr o ns moving throug h th e circ uit co uld be calcu lated. T h e n um be r of m ol e s o f e lectrons co ns u med o r produce d in a n el e ct ro n transfer react io n is o btai ned by measuring the cu rre n t flowing in th e exte rna l electric circuit in a given time . Th e current flO\\Ting in an elec trical circuit is th e amou n t of cha rge (in u n its o f cou lo mbs, C) per uni t time , and the usual un it for curren t is th e ampe re (A). (O ne a mp ere equa ls the passage of o ne cou lomb of charge per seco nd .) Cu rrent (amperes, A) =
electric charge (coulombs, C) time, t (seconds, s)
(20 ,8)
The curre n t passing thro ugh an electro che mical cell and th e tim e for which th e cu rre n t flows are easily measu red qua ntiti es. T he refore, the cha rge (in co ulom bs) th at passes th ro ugh a cell ca n be o btained by multiplying the cu rre n t (in ampe res) by the lime (in seconds) . Kno wing th e charge and using the Faraday co nsta n t as a co nversio n facto r, we can calcu late th e num ber o f mo les o f electrons that pa ssed th rou gh an electroc he mical ce ll. In turn , we ra n use this qu ant ity to calculate th e quant ities of re ac tants and pro d ucts. T he fo llowing examp le illustr ates this type o f calculation . Chemislry.o.Now'"
•
Sign in at www.thomsonedu.comjlogin and go to Chapter 20 Contents to see Screen 20.12 for a tutorial on quantitative aspects of electrochemistry.
Historical Perspectives
Electrochemistry and Michael Faraday
Th e terms anion, cation, electrode, and electrolyte originated wit h Michael Faraday (1791- 1867), one of the most influential people in the history of chemistry. Faraday was apprenticed to a bookbinder in Lo ndon when he was 13. This situation suited him perfectly, as he enjoyed reading the books sent to the shop for binding. By chance, one of these volumes was a small book on chemistry, which whetted his appetite for science, and he began performing experiments on electricity. In 1812, a patron of the shop invited Faraday to accompany him to the Roya l Instit ute to attend a Lecture by one of t he most famous chemists of the day, Sir Humphry Davy. Faraday was so intrigued by Davy's Lecture that he wrote to ask Davy for a
position as an assistant . He was accepted and began wo rk in 1813. Faraday was so ta Lented that his work proved extraordinarily fruitfuL, and he was made the director of the La boratory of the Royal Institute about 12 years Later. It has been said that Faraday's contributions were so enormous that, had there been Nobel Prizes when he was alive, he wou ld have received at least six. These prizes could have been awarded for discoveries such as the following: Electromag netic induction, which led to the first transformer and electric motor The laws of eLectrolysis (the effect of electric current on chem icals) The magnetic properties of ma tter
Benzene and other 'E .~ organic chemica Ls i-~ (which led to impor- ~ ~ tant chemica l _; industries) "'.= The "Faraday effect" ~ ~ (the rotation of the ~ plane of polarized light by a magneticfield) Michael Faraday (1791 -1867) The introduction of the concept of eLectric and magnetic fields In addition to making discoveries that had profo und effects on science, Faraday was an educator. He wrote and spokeabout his work in memorable ways, especially in lectures to t he general publicthat helped to popularize science.
f
5
20 .8
I
Counting Electrons
937
Using the Faraday Constant
r
ProbLem Acurrent of 2.40 Ais passed through a solution containing Cu 2+(aq) for 30.0 minutes,with copper metal being deposited at the cathode. What mass of copper, in grams, is deposited?
Strategy Aroadmap for this calculation is as follows:
amount reactant (or product)
Faradays (Mol e-)
current (A) x time (s)
-
mass (g) reactant (or product)
SoLution 1. Calculate the charge (number of coulombs) passing through the cell in 30.0 min.
Ch arge (C) = current (A) X time (5) ~
(2.40 A)(30.0 min)(60 .0 s/min)
= 4 .3 2 2.
Calculate the number of motes of electro ns (i.e .. t he number of Faradays of electricity) . (4.3 2 X 10 3
• Faraday Constant The Faraday constant is the charge carried by 1 mol of electrons:
X 10 3 C
q(
1 mol e- ) 96, 485 C
= 4.48
X 10- 2 mot e-
3. Calculate the amount of copper and, from this. the mass of copper. mass of copper
9.6485338 x 104 Cjmol e-
= (4.48 x 10- 2 mol e) (
l mOI CU) ( 63.55 9 CU) =
2 mol e
1 mol Cu
1.42 9
Comment The key relation in this calculation is current = charge/time. Most situations witt involve knowing two of these three quantities from experiment and calculating the third. EXERCISE
20,1 5 Using the Faraday Constant
Calculate the mass of O2 produced in the electrolysis of water. using a current of 0.445 Afor a period of 45 minutes.
EXERCISE
20,16 Using the Faraday Constant
In the commercia l production of sodium by electrolysis, the cell operates at 7.0 Vand a current of 25 x 103 A. Wh at mass of sodium can be produced in one hour?
938
Chapter
20
I
Princip les of Reactivity: Elect ron Transfer Reactions
C hapter Goals Revisited m
Now that you have studied this chapter; you should ask uihether you have met the chnpter
Chemislry.o.Now Sign in at
goals. In porticulai; )'OU should be able to:
www.t homs~nedu .com/ login to:
Ba lance equation s for oxid ation- red uctio n reactio ns in acidic or basic solut ions using the ha lf- reaction app roach Study Question(s) assignable in OWL: 2, 4-6, 16; Go Chemistry Module 25.
Understand the principles underlying voltaic cells a. In a vo ltaic ce ll, id entify th e hal f-reacti on s occu rring a t th e a n ode a n d the ca th o de , the pol arity of th e e lec trodes, th e dire cti on o f el e ctron flo w in th e ex terna l circu it, a n d th e directi on of io n flo w in th e sa lt bri d ge (Sec tio n 20.2) . Study Question (5) assignabLe in OWL: 8. 10. 16 , 51.
b.
c.
Ap p recia te th e che m istry and a d van tages and di sad vantage s of dry ce lls, a lkalin e b att e ries, lead storage batte ries, an d N i-cad bat te ries (Sect ion 20 .3) . Un ders ta n d h ow fu el cel ls work, a n d recogn ize th e diffe ren ce between batteries a n d fuel ce lls (Sec t io n 20.3) .
Assess your understanding with Study Questions in OWL keyed to each goal in the Goals and Homework menu for this chapter For quick review, download Go Chemistry mini-lecture flashcard modules (or purchase them at www.ichapters.com) Check your readiness for an exam by taking the Pre-Test and exploring the modules recommended in your Personalized Study plan.
'6 Access How Do I Solve It? tut orials on how to approach problem solving using concepts in this chapter.
Underst a nd how to use ele ctrochemical pote ntia ls a. Un derstan d the p ro cess by whic h sta n d ard re d uc tio n pot entials a re d et e rmined, and identify sta n da rd co n d itio ns as a p p lied to elec troc he m ist ry (Section 20 .4) . b. Describ e th e sta n dard h ydrog en e lectrode (EO = 0 .00 V) , an d ex p la in h ow it is used as th e sta nda rd to d etermi n e th e sta n d ard potentia ls of half-reactions (Sec tio n 20.4). c. Know h ow to lise standard red uc tio n po ten tia ls to d e te r min e ce ll vo ltages for ce lls und er stan dard co n di tio ns (E q ua tio n 20. 1). Study auestion(s) assignabLe in OWL: 14, 16 , 18, S6, 57. d. Kn ow how to lise a ta bl e o f standard red uctio n pot en ti als (Ta b le 20. 1 a n d Ap pe n dix M] to ra n k th e stre ngths o f oxidizing a n d redu cing ag ents, to predi ct wh ich su bs ta nc es ca n re d uce or oxidize a not he r species , and to predic t whet he r re d ox reactions will he produc t-fa vo red or re acta n t-favored (Sec tio n s 20 .4 a nd 20 .5) . Study Question(s) assignable in OWL: 20, 22, 24, 55, 56, 82. e. Use th e Nc rnst eq u a tio n (Equatio ns 20 .2 a nd 20.3) to ca lcu late th e ce ll poten· rial unde r no ns tan d ard con di tions (Sec tio n 20.5) . Study auestion(s) assignabLe in OWL: 26, 28, 30, 36. f . Exp lain how ce ll voltage re lates to ion co nce n tra tio n , and ex p lain how th is all ows the d et e rm in a tio n o f p l l (Sec tion 20.5) a n d other io n conce n trations. g. Use th e rel at ionships betwee n ce ll voltage (l~' on'lI ) a n d free energy (.6.J;") (E q ua tio ns 20.5 a n d 20.6) a n d be tween E Ord l and a n eq u ilib r iu m co ns ta n t for the ce ll reac tio n (Eq ua tio n 20.7) (Sec tion 20.6 a n d Tabl e 20.2). Study Question(s) assignable in OWL: 32, 34, 60, 61, 63, 78, 86 , 88.
Explore electrolysis , t he use of electrical energy to produce chem ical change a. Describe th e c h e m ica l processes occurrin g in a n elec tro lysis. Reco gnize th e factors th at d et ermin e whic h su bs ta nces are oxidized a nd reduced a t th e elec tr odes (Sec tion 20 .7) . Study auestion{s) assignabLe in OWL: 40 , 41. b. Relate th e a m o u n t of a su bs ta nce ox idiz ed or reduced to th e amount of cu rrent a nd the time the cu rrent flows (Sec tio n 20 .8) . Study Question{s) assignabLe in OWL: 44, 46 , 48, 64-66 . 73, 81.
Chapte r Goals Revi sited [
939
KEY EQUATIONS Equation 20.1 (page 917) : d ard half-cell poten tials.
Ca lcu lating a sta nda rd ce ll potential , EOcell
Equation 20.3 (page 925):
= EOcathode -
E OC/:lh
fr o m sta n-
EOanode
Ne rns t eq uation (at 298 K) . E
= EO_
0.0257
n
lnQ
E is th e ce ll pot ential und e r non stan d ard co ndi tio ns; n is th e numbe r o f e lec tro ns tr an sferred from th e reducing ag e n t to th e o xidizin g ag e n t (acco rd ing to th e balan ced e q uatio n) ; and Q is the reacti on quotien t.
Equation 20.6 (page 929) : Rel ati on shi p be twee n sta nda rd free ene rgy cha ng e a n d th e sta ndard ce ll po ten tial.
F is th e Faraday consta nt, 96 ,485 e / mo l
«,
Equation 20.7 (page 929): Rel ationship bet ween the eq uilibr iu m constan t a nd th e sta nda rd. cel l potential fo r a reacti on (a t 298 K) . nEO 0.0257
lnK = Equation 20.8 (page 937):
Relationship be tween cu rren t, electri c charge, an d tim e.
Current (amperes, A)
STUDY QUESTIONS Online homework for this chapter may be assigned in OWL. ... denotes challenging questions.
•
denotes questions assignable in OWL.
Blue-numbered questions have answers in Appendix a and fully-worked solutions in the Student Solutions Manual.
Practidng Skills Balandng Equations for Oxidation-Reduction Reactions (Ser Exnmples 20.1 -20.3 fllld O W1!l i.\'tl) Now StH'fl/ 20.3.) a 7/t'fI balanci ng the foll owing redox equations, it ma)' be nece.Hm)' add t r (a'!) or t r (((q) plu s / f 2 0 P'" reactions in add, a nd OH - (ll '!) or OH - (rlq) plus HJ) jor reactions in haw', (0
1. Write balan ced eq ua tions fo r th e fo llowing hal f-re acti o ns. Spec ify whether eac h is an ox idation or re d uc tio n . (a) Cr(s ) -> Cr ' +(aq ) (in aci d ) (h ) AsIl 3(g ) -> As (s ) (in aci d ) (ill acid ) (c) V0 3 - (a'1) -> V2+(aq ) (ill base ) (d) Ag (s) -> Ag, O(s ) 940
=
electric charge (coulombs, C) time, t (seconds, s)
2. • WriLe ba la nced eq uations for th e fo llow ing half.. reac tion s. S pe cify whet her eac h is an oxi dation or re duction. (a ) 1I,0, (a q ) -> O , (g ) (in aci d ) (h ) II , C, O , (aq ) -> CO ,(g) ( in aci d) (in aci d ) (c) NC}, - (a'1) -> NO(g ) (in base ) (d) MnO.,- (aq ) -> Mn O , (s)
3. Bala nce the foll o w ing redox eq ua tio ns . All occur in acid so lu tio n . (a ) Ag (s) + NO, - (aq ) -> NO, (g) + Ag +(aq ) (b ) ~lnO, - ( aq ) + HSC}, - (aq )-> M n' +(a q ) + SO ,' - (aq ) (c) Zn (s ) + NO,,-(aq) -> Zn H(aq) + N,O (g) (d) Cr(s ) + NO,,- (aq ) -> CrH (aq) + N () (g) 4. • Balance th e followi ng red ox eq uati on s. All occur ill acid so lu tion . (a ) Sn (s ) + H t faq) -> Sn 2+(aq ) + H 2(g ) (h) Cr20 7' - (aq) + FcH (aq) -> Cr3+(aq ) + Fe3+(a q) (c) ~l nO , ( s ) + CI- (aq ) -> Mn H (aq) + C l,(g) (d) CII 20(a'1) + Ag +(a'1) -> II CO,Il (a'l ) + Ag(s ) 5. • Balance th e fo llowi ng redox eq u ati on s. All occur in basic sol utio n . (a ) A1 (s) + 11,0 (1.) -> I\ I(O Il ), -(aq ) + ll , (g ) (b) Cr O .,' - (aq ) + SO ,'- (aq ) -> Cr (O ll h (s) + SO ,' - (aq ) (c) Zn(s) + Cu (O Il),( s) -> [Zn (O Il) ,F-( a'1) + Cu( s) (d) HS- (aq ) + CI0 3 -(aq) -> S (s) + Cl- (aq )
6. • Balan ce th e foll owin g re dox eq ua tio n s. All o ccur in basic so lu tio n . (a) Fe(Oll ), (s) + Cr (s) ---> Cr (O ll ), (s) + Fe(OH ), (s) (1)) NiO,(s) + Zn( s) ---> Ni( O ll),(s) + Zn (O llh(.s) (c) Fe( O H ), (s) + CrO .'- (aq ) ---> Fe (O ll ), (s) + [Cr (O H) ..J - (aq) (d) N, H ,(a q ) + Ag,O (s) ---> N, (g ) + Ag (s) Constructing Voltaic Cells (See Ex a mple 2 0.4 a nd Chn niJ/l)N oul Screen. 20. 4.) 7. A vo lta ic ce ll is co ns truc te d using th e reac tio n of ch ro m iu m metal and irn n (lI ) ion . 2 C r (s)
+
~
FeH (aq ) ---> 2 Cr 3+(aq )
+ S Fe (s)
Co m ple te th e fo llowing se n te n ces : Electrons in th e exelect rode to th e tern al ci rc uit flo w fro m the ele ctrode. Nega tive io ns move in th e salt brid ge from th e half-cel l to th e half -cell . Th e h alf -re act ion a t the a node is _ , a nd th at a t the ca tho de is _ . 8. • A volta ic ce ll is constructed usin g th e re act ion Mg (s)
+ 2 II ' (a'l ) ---> Mg" (a'l ) + II , (g )
(a) Write equ at ion s for th e o xid ation a nd reducti on hal f-reacti ons. (h) Wh ich h alf-reacti on o cc urs in th e a node co m pa rt llle ll t, and which o ccurs in th e cath ode co m pa rtme nt? (c) Co m plete the followin g sen ten ces: Ele ctron s in e lec tro de to th e ext ern al ci rcui t flow from th e th e _ ele ct rode . xcgarivc io ns m ove ill th e salt b rid ge fro m the _ half-ce ll to th e _ half-cell. Th e h alf-reaction a t th e a no d e is _ ' a nd tha t at th e catho de is 9. T he half-c el ls Fe 2+(aq) I Fe( s) and O A g-) I H 20 (i n aci d so lu tio n) are linked to cr ea te a vol taic cell. (a) Writ e eq uatio ns for t he o xid ation a nd reduction half-react ion s a n d f(JI ' th e overall (cel l) reactio n . (h ) Whi ch half-reacti o n o ccu rs in th e a nod e co m part men t, and which o ccu rs in th e ca th ode com partmen t? (c) Co m p let e the followin g se n ten ces: Ele ctro ns in electrod e to th e e x te rn al circui t flow from th e th e _ e lectro de . Xe ga tive io ns mo ve in th e salt half-ce ll to th e _ Ilalf·cell . h ridg e from the _
10. • Th e hal f-cell s Sn' (aq ) 1Sn (s) and CI, (g ) 1 CI- (aq ) are linked to create a voltaic ce ll. (a) Write equ at io n s for th e o xid a tion an d red uc tion ha lf-reacti ons ami for th e overall (ce ll) re ac tio n . (h) Whi ch half-reacti o n o ccu rs in th e an ode co m pa rt me n t, and wh ich o ccurs in th e cathode co m pa rt me n t?
•
more challenging
• i n OWL
Blue-numbered questions answered in Appendix0
(c) Com ple te the following se n te nces: Ele ct ron s in e lec trode to th e ex te rna l circu it flo w from th e th e _ ele ct ro de . N eg a tive ion s move in th e sail half-cel l to th e _ half-cell . bridge fro m th e _ Commercial Cell s II. Wh at a rc the sim ilar ities and d ifference s between d ry ce lls, alkali ne hau crics , an d Ni-cad batt e ri es? 12. Wh at reac tio ns occur whe n a lead sto rage batte ry is rc cha rge d ? Sta ndard Electrochemical Pote nti als (See Exam/llp 2(J.j an d Ch pm i~1r)'No w S(;W('1/ S 2 0. 6 awl 20. 7.) 13. Calcu late th e value of EOfor ea ch o f th e followin g re actio n s. De cide whe ther each is produ ct-favored in the di rection writte n. (a) 2 I- (aq ) + Zn " (aq ) ---> I, (s) + Zn (s) (h) Zn" (aq ) + Ni(s) ---> Zn (s) + Ni" (aq) (c) 2 C I- (aq) + Cu" (aq ) ---> Cu (s) + C I, (g ) (d) Fe" (aq) + Ag ' (aq ) ---> F" H (a q ) + Ag (s) 14• • Calcula te th e value of f.'" fo r e ach of th e followin g re acti on s. De cide whethe r e ach is produ ct-favored in th e d irec tion writ te n . [Re act io n (d) occurs ill basic so lu tio n.] (a) I\r, (t) + Mg (s) ---> Mg" (aq ) + 2 1\r " {aq ) (h) Zn" (aq ) + Mg (s) ---> Zn (s) + Mg" (aq ) (c) SIl2 +(aq ) + 2 Ag +(a q ) ~ Sn 4 +(a q ) + 2 Ag (s) (d) 2 Zn (s) + o ,( g ) + 2 H,O (f) + 4 OH - (aq ) ---> 2 [Zn( O Il ). F - (aq ) 15. Balan ce each of th e foll owing u nb ala nced eq ua tions; then calcu la te the sta nd ard potent ial , EO, a nd deci de wh e the r each is produ ct-favored as writt e n . (All re actio ns occu r in acid so lu tio n.) (a) Sn' " (aq } + Ag (s) ---> Sn (s) + Ag ' (aq ) (h) AI(s) + SnH (aq ) ---> Sn" (aq ) + AIH (aq ) (c) C IO ,-(aq) + Ce'H (aq) ---> CI,(g ) + Ce' +(a q ) (r1 ) C n(s) + NO , - (a q ) ---> Cu'" (aq) + N O (g ) 16. • Balan ce eac h of the follo wing unbalan ced equ ation s; th en ca lcu la te the sta nd ard potent ial, EO, a nd de cid e wh et her each is product-favored as wruu-n. (All rcactiou s oc cu r in acid so lu tio n .) (a) I, (s) + I\r - (a 'l) ---> I- (a'l ) + I\r, (f) (h) FeH (aq ) + Cu " (aq ) ---> Cu (s) + FeH (a q ) (c ) Fe" (aq ) + Cr , 0 7' - (a q ) ---> Fe " (aq ) + Cr' +(aq) (d) MnO,- (aq ) + H NO , (aq ) ---> Mn H (aq ) + NO , - (aq )
I
941
17. Co nsi der the foll owing h alf-re actions:
21. Whi ch of th e fo llowing io ns is m os t eas ily red u ce d? (a ) Cu ' +(a q) (d ) Ag +(a q ) (b ) Zn H (aq ) (e ) AI3+(aq) (e) Fe' +(aq )
E"(V)
Half-Reaction Cul+(aq)
+ 2 e:
----t Cu (s)
Sn' t'{aq)
+ 2 e:
----t Sn(s)
- 0.14
+ 2 e" -e Fe(s) Zn2+(aq) + 2 e- -l- Zn(s)
- 0.44
Al' ·(aq) + 3 e- -> Al(s)
- 1.66
Fe2+(aq)
+ 0.34
22 . • Fro m th e following list , iden tify the io ns that are m ore e asily redu ce d than H +(a q ) . (d ) Ag +(aq ) (a) CuH (aq) (b ) Zn H(aq ) (e ) AI3+(aq) (e) Fe' +(aq )
- 0.76
(a) Based on EOvalues, which m et al is t he m ost eas ily o xid ized ? (b) Whi ch m e ta ls o n th is list arc ca pa b le of reducin g F(·2+(a q ) to Fe(s )? (c ) Writ e a ba la n ce d che m ical eq uatio n for the reaction of Fe2+ (aq ) with Sn tx). Is th is re acti o n p ro d uct-favo re-d or reac tant-favored? (d) Wr ite a balan ced che m ica l eq uatio n for th e re actio n of Zn 2+(a q ) with 511 (s) . Is th is rea ctio n pro duct-favo red or re actant-favored ? 18. • Co ns ide r the fo llo wing ha lf-re act io ns: E"(V)
Half-Reaction
+ 8 W(aq) + 5 e" ----t Mnl+(aq ) + 4 HzO(t)
23. (a) Wh ich h alogen is most easi ly red uced: F2 , C12, Rr2' or 12 in acidic so lution? (h ) Iden tify the halo ge n s th at are lx-tu-r o xidizi ng age n ts than '\-ln02(s) in ac id ic so lut io n . 24. • (a) Whic h ion is m ost easily o xidiz ed to the e le menta l ha lo ge n : F-, CI-, Br' . or 1- in ac idic so lu tio n ? (b) Ide n tify th e ha lide io ns tha t arc m ore easily o xidi zed th an 11 20 (0 in ac idi c so lut io n . Electrochemical Cells Under Nonst and ard Conditions (See Exam/des 20.6 and 20.7 and Ch fm iJfr)Now Screen 20.8.)
25. Calc u la te t he po ten tial del ivered hy a vo lta ic ce ll using
+ 1.47
th e fo llowing reacti o n if all d isso lved species are 2.5 X \0 - 2 M a nd the p re ssure of H 2 is 1.0 bar.
CrzO/ -(aq ) + 14 H'{aq ) + 6 e" -l- 2 CrH(aq ) + 7 HzO(l}
+ 1.33
Zn (s)
NO ,-(aq ) + 4 H'{aq ) + 3 e - -> NO(g) + 2 H,O(t)
+ 0.96
50.z-(aq ) + 4 H+(aq) + 2 e - ----t 50 z(9) + 2 HzO(f)
+ 0.20
Mn 0 4 -(aq)
BrO]-(aq ) + 6 H'{aq ) + 6 e" ----t Br-(aq)
+ 3 HzO(f}
+ 1.51
+
2 H,O (f)
+
2 OH - (aq ) -> [Z n (O Il) ,j' - (aq )
+ II ,
(g )
26. • Calcu late th e po ten tia l devel oped by a vo lta ic ce ll u s-
in g the fo llowin g re ac tion if a ll d isso lved spec ies are (a ) C ho osi ng from am ong the reactan ts in th ese halfreacti ons, identity the stron ge st an d wea kes t o xid izin g age n ts. (b ) Whi ch of th e ox id izing age n ts list ed is (are ) capahie of oxi d izing Br - (aq ) to BrO:\- (aq ) (in aci d sol u tion)? (c) Write a bala nced che m ical equat ion for th e reaction of Cr20 72- (aq ) with 5 0 2(g ) in acid so lu tio n . Is th is rea c tio n p roduct-favo red or reac tan t-favored? (d ) Write a bala nced chem ica l eq uat io n for th e reactio n of Cr20 l - (aq ) with ;"1 n 2+(aq ). Is this reaction p ro duct-favo re d or reacta n t-favo red ? Ranking Oxidi zing and Redudng Agents (SI'I' Es.. .amjJII' 20.5 and Chnuist1),Now Screen 20.7.) Use a table ~r standard reductio n potentials (Table 20.1 or AjJfJnulix A1) to an.mlf'r Study QueJt;orJs /9-24 .
19. Whi ch o f the followin g e le m ents is the be st red ucin g age n t u nde r stan d ard co ndi tio ns? (a ) Cu (d) Ag (h ) Zn (e) Cr (e) Fe 20. • Fro m the foll owing list, ide n tity th ose ele m en ts th at arc easier to o xid ize than 11 2(g ) . (a ) Cu (eI) Ag (h ) Zn (e) Cr (e) Fe 942
I
0.015 M .
2 Fe H (aq )
+ II ,O, (aq ) + 2 11+(aq) -> 2 Fe 3 +(aq )
+
2 I1, O(f)
27. One hal f-cell ill a vo lta ic ce ll is co nstruc te d fro m a silver ,v-irc dipped into a 0.25 ~'1 so lu tion o f AgNO:\. T he othe r half-ce ll consists ofa zinc e lect ro d e in a 0.01 0 M so lutio n of Zn(N O jh. Calcu la te th e ce ll po te nti al.
28. • On e half-ce ll in a vol ta ic ce ll is constr ucted from a co p per wire dipped into a 4.8 X 10- 3 ,\1 so lu tio n of C u (NO .~ h T h e o ther half-cell co nsists of a zinc ek-e trod e in a 0.4 0 M so lu tion of Zn (N O j h. Calcula te the cell pot ential.
29. O n e hal f-cel l in a vo lta ic cell is co nstruc ted fro m a silver wire di p pe d in to a AgN O j so lu tio n of unk n own concen tratio n. T h e oth e r ha lf-ce ll consists of a zinc electrode ill a 1.0 M so lu tio n o f Zn (N 0 3 h A pot ent ial of 1.48 V is measured for th is cell. Usc th is informatio n 10 ca lcul ate the conce ntrat io n of' Ag +(a q ) , 30. • On e ha lf-c e ll in a vo ltaic ce ll is co ns truc te d from a n iro n wire d ipped in to an Fe(N 0 3 h so lution of un kn own concen t ra t ion. T he ot he r hal f-cel l is a sta n da rd hyd ro ge n ele ct rod e . A poten tial of 0.4 9 V is m easu red fo r th is ce ll. Use th is info rm at io n to calc ul at e th e co ncen tr a tion of Fe2+(aq ).
•
more chaLLenging
•
in OWL Blue-numbered questions answered in Appendix 0
ELectrochemistry, Th ermodynamics, and Equilibrium (See Examples 20. 8 and 20. 9 an d CllPmistryNow Sorm 20. 9.)
31. Cal cula te dI G' an d th e equil ibrium con sta n t for th e fo llowing reac tions . (a) 2 Fe"' (a'l) + 2 \- (a'l ) 2 Fe" (a'l ) + I, (a'l ) (b) I, (a'l ) + 2 Br r f aq} 2 \- (a 'l ) + Ilr, (a'l ) 32. • Ca lcu la te d , G' an d th e e q u ilibriu m co nstan t for th e followin g re action s. (a) Zn 2+(aq ) + Ni fs) :o:===:t Zn(s) + Ni2+(aq ) (h ) C u (s) + 2 Ag ' (a'l ) C u" (a 'l ) + 2 Ag (s)
3:l. Use stan d ard re d ucti on pot en tials (Ap pe nd ix M) fo r the h alf-reactio n s Ag Br( s) + c " ---7 Ag (s) + Br - (aq } an d Ag +(a q ) + c " ---7 Ag (s) to calculat e th e value o f K ,l' for AgTk
34. • Use the sta n d a rd reduction potential s (Ap pe nd ix .\1) for th e half-re action s I1g 2CI2 (s) + 2 c " ---7 2 I1g (f) + 2 C1- (aq ) and Hg) l+ (aq ) + 2 e" ---4 2 Hg (f) to ca lcula te the val ue of K' 1l for II g';!CI2. 35. Use the sta ndard red uc tio n pot e n tials (Ap pen d ix M) for th e half-rea cti on s [AuCtd - (aq ) + ;l c " ---7 Au (s) + 4 Cl- (aq ) an d All 3 + (aq} + :l e - ---7 Au i s) to calc u late the valu e o f K fo n n a lio ll for the co m plex ion [AuCI4]-( aq ) . 36 . • Usc the sta nd a rd reduc tion potentials (A p pe n d ix M) for th e h alf-reactio ns (Zn( O F1hP-( aq ) + 2 c " ---7 Zn (s) + 4 OH - (a'l ) an d Zn " (a'l ) + 2 c : -> Zn (s) to calc u la te th e value of K fo n n dl io rl for th e co m plex ion [Z n( O H ), J' - .
ELectroLysis (See Seciiou 20.7, Exa mple 20. 10, an d CJwmiJll)'Nm.ll Screen 20,10.)
37 . Dia g-ram the ap pa ra tus u sed to e le ctrolyze m olten NaCI. Iden tify th e anod e an d th e ca tho d e. Trace the move m e n t of electron s th ro ugh t he ex te rnal circuit a nd the movemen t of ion s in th e eh-c tro lvsis cel l. 38 . Diag ra m th e ap pa ratus used to el e ctrol yze aq u eo us CuC I2. Id e n tity the reacti o n products, th e a no de , am i rlu - ca th o de. Tr ace th e m O\'eITH'TH of electrons th roug h the ex te rnal c ircu it a nd the movement of io ns in the elec tr o lysis cel l.
39 . Whi ch p roduct, 0 , o r F" is mo re likely to form a t th e a no de in t he ele ctrolysis o f an aqueous sol u tio n of KF? Ex plain yo u r reasonin g , 40• • \Vhich product . Ca or H t • is m ore likely to fo rm a t the ca tho de in th e electro lysis of CaC12? Explain yo u r
reason ing.
41 . • An aq ueo us so lu tion ofKRr is pl aced in a be ak e r with two ine rt plat inum el ectrode s. When the cell is a ttach ed to a n ex tern al so u rc e o f elec trica l e nergy. el e ctr o lysis occ urs. (a) Hydrogen gas and hydroxide io n fo rm at the ca tho d e. Write a n e q ua tio n fo r the half-rea ct io n tha t oc curs a t th is el ec tro d e. (h) Brom in e is rlu- p rim a ry product a t rl n- a no d e, Wri te a n e q uatio n fo r its form ation . 42 . An aqueous so lu tio n o f Na2S is pla ce d in a beak er with two inert pl a tinum electrodes. Whe n the cel l is a tta ched to an e x te r nal battery. electro lysis o ccurs. (a) Hydrogen gas an d h yd roxid e ion fo r m a t the cat hode . Writ e a n e q ua tion fo r the hal f-re a ct io n th at oc cu rs a t th is electrode. (h) Su lfu r is th e p rim ary product a t th e anode . Writ e a n e q uatio n for its for mati on .
Counting ELectrons (See Exa mjll" 20. 11 and ChnnislryNoU! Screen 20.12.) 43 . In III I..' elec trol ysis o ra so lu tion co ntain ing Ni 2 +(aq ) , metalli c Ni (s) d e posits o n th e ca thode . Usin g a cu rre n t of 0.150 A fo r 12,2 min , wha t m ass of n ick el will fo rm ? 44• • 111 th e e lec tr olysis of a so lu tio n co n ta in ing Ag +(aq }, m etallic Ag (s) d ep osits o n the ca thode . Usin g a curre n t o f 1. 12 A for 2.40 h , wha t m ass of silver fo rm s?
45 . Electro lysis of a so lu tio n of Cu S0 4(aq ) to give copper m etal is car rie d o u t u sin g a current o f 0.66 A. How lo ng sh ould el ect rol ysis co n tin ue 10 produce 0.50 g of co p per? 46 . • Electrolysis o f a so lu tio n of Zn (N 0 3 )z(aq) to g ive zinc me tal is ca rri ed out using a cu rre nt of 2.12 A. How lo n g- sh o u ld e lec tro lysis co n tin ue in orde r to prepare 2.5 g o f zinc? 47 . A vo ltaic ce ll ca n be built using th e reac tion betwe en Al metal and O 2 from th e ai r. If t he Al a no d e o f th is ce ll co nsists of 8 4 g of alum inum , h ow m a llY h o u rs can th e ce ll produce 1.0 A of ele c trici ty, assu mi ng an unlimit ed su p ply of 0 2?
48. • Assu me th e spe ci fica tions o f a Ni-Cd vo lta ic ce ll in elude del ive ry o f 0.25 A of cu rr e n t for 1.00 h . What is th e mini mu m m ass of the ca d m iu m th a t mu st be used to ma ke th e an od e in t hi s ce ll?
General Questions Th ese questions are 1101 designated as to 1)1)1' or location in the chapter: Till)' ma)' combine seoeml concepts. 49 . Vh ite balan ced e q ua tio ns for the foll o win g h alfrea c tio n s. (acid so lu tio n ) (a) UO , ' (a'l ) -> U O - (a 'l ) (basic so lu tio n) (e) N, H ,(a'l ) -> N, (g) (d) C10 - (a'l ) -> C1 - (a'l ) (ba sic so lutio n )
•
more challenging
• in OWL
Blue-numbered questions answered in Appendix 0
I
943
50. Bal an ce th e foll owin g equ ati o n s. (a) Zn (,) + YO " (a q) ---> Zn 2+(aq) + VJ+(a q ) (h) Zn(s) + YO , - (aq ) ---> V2+ (aq ) + Zn H (a q )
55. • In th e tab le o f sta n da rd red u ction p ot enti al s, locat e th e h alf-reacti on s fo r th e red uctions of th e foll owin g m et al ion s to th e m et al: SI1 2+(a q ) , Au t Ia q}, Zn 2+(aq ) , Co 2 +(aq ) , Ag +(a '1) , Cu 2 +(aq ) . Amon g th e met al io n s and metal s th at make up th e se h alf-react io n s: (a) Wh ich m et al ion is th e wea ke st o xidizing ag e ll t? (b) Wh ich m e tal ion is th e stro ng es t o xid izin g ag e n t? (c) Whi ch me tal is th e stro ng es t re d uc ing age nt? (d) \Vhich metal is th e weak e st redu ci ng agen t? (e) W ill Sn(s) red uce Cu2+(a q ) to e u(s)? (0 Will Ag( s) red uc e C02 +( Zn( O Hh(s) + ClT aq ) (bas ic so lu tio n) ( c' \-(aq) + c'r O,/2-(aq) (basic so lu tio n) 51. • Mag ne siu m me tal is o xidi zed, a n d silve r io ns are re-
d uced in a volta ic cell using Mg 2+ (aq. I M) I Mg a n d I M) I A ~ half-cel ls
A~ + (aq.
56. • ... In th e ta ble of sta n d a rd re duction potentials, loca te th e h al f-re action s for th e re d u c tion s of th e followin g n onme ta ls: F2, C12, Br2' 12 (red u ction to h al id e ions ) , a n d O 2• S, Sc (red uc tio n to 11 2X in aq ue o us aci d ) , Amo n g th e cl e me nt s, io ns, an d co m p o u n ds th at make u p th ese h a lf-reactio n s: (a ) W hi c h cl em ent is th e we a kes t ox idiz in g agen t? (h) Wh ich el em ent is th e wea kes t red u c ing ag e llt? (c) Wh ich of th e e le m e n ts liste d is (a rc) ca pa b le o f ox idi zin g H 20 to 0 2? (d) Which of th ese e lemen t!'> listed is (a rc) ca pa b le of ox id izing H 2S to S? (e) Is O 2 ca pa ble of ox id izin g 1- to 12, in acid so lu tio n? (f) I!'> S ca p a b le of ox id izing 1- to I2? (g) Is th e reaction H 2S (aq ) + Se(s) ---7 H 2Se( aq ) + S (s) p ro duct -favo re d ? (h) Is th e reaction I12S (aq ) + I2(s) ---7 2 l l t fa q ) + 2 I- (aq ) + S(s) p roduct -favo red?
(a) Label each pan of tlu- cell. (h) Write equat io ns fo r th e half-reacti o n s occurrin g at th e a nod e and the ca thod e . an d write a n e qua tion fill' th e n et re actio n in th e ce ll. (c) Tr ace th e m ovem en t of elec t ro ns in th e ex te rnal circu it. Assum in g the sa lt b ridg e con ta ins NaN O J, trace the m ove me n t o f th e Na" an d N O j - ion s in
the salt bridge th at occu rs when a voltaic cell pro· d uces cu rre n t. Wh y is a sa lt b rid g e re q uired in a ce ll?
52. You wan t
57. • Four vo lta ic ce lls a re ...ct u p. In each , on e hal f-cel l co n tains a sta n dard h yd ro ge n elec tr o d e , T h e sec o n d h alf-cell is one of th e fo llowin g: Cr3+(aq , 1.0 M ) I Cr(s), Fe ' +(aq . 1.0 M) I Fe( , ) . Cu' +(aq , 1.0 M) I Cu ts) , o r Mg' +(aq , 1.0 M) I Mg ("), (a) In whic h o f th e volta ic ce lls does the h yd ro gen e lec tro d e se rve as the ca tho de? (b) Wh ich vo lta ic ce ll p rodu ce s the h igh est vo lta ge? Which p ro d u ce s the lowest vo ltage?
sCI u p a se rie s of vo ltaic cell s with speci fic ce ll vol tages. A Zn 2+(a q. 1.0 M ) I Zn (s) half-ce ll is in o ne com p artme nt. Id entify seve ra l h alf-cell s that yo u cou ld u se so th at th e ce ll pote nti al will be clo se to (a ) 1. 1 V and (b ) 0.50 V. Cons id er ce lls in wh ich zinc ca n b e ei th er th e ca thode o r th e a node . [0
53. You
wan t to se t up a se ries o f vo ltai c cel ls with sp eci fic (('11 p oten tial s. T h e Ag + (aq, 1.0 M ) I Ag (s) h alf-ce ll is on e o f t h e co m pa n me n ts. Id e nt ify se ve ral h al f-cell s th a t yo u could u se so th at th e ce ll p o tenti al will h e clo se to (a) 1.7 V a nd (b ) 0.50 V. Cons ider cel ls in whi ch silver can be e ither th e ca thod e or th e a nod e.
58. T he followin g h alf-Cl' lIs are availab le: Ag + (aq , I.O 1\1) I A~ ( s ) , Zn ' +(aq , 1.0 M) I Zn(s) , Cu2+ (aq , 1.0 M) I Cu ts) , an d Co 2 +(aq , 1.01\.1) I Co(s) . Linking an y two h alf-cell s makes a vo ltaic ce ll. Given fo u r different hal f-ce lls, six vol taic cells ar e possible. T h ese are label ed , for sim p licity, Ag-Zn , Ag-C u , Ag-Co , Zn-C u . Zn-Co . an d Cu -Co. (a) In whic h o f tln- volta ic ce lls does the co p pe r e lectrode se rve a!'> th e ca th o de? In wh ich o f th e vo lta ic ce lls doe s the co b a lt electro d e se rve as th e anode ? (b) \Vhich com biu a tio u of h a lf-ce lls ge ne ra tes th e h igh . est poten tia l? Wh ich com b inatio n g e ne ra tes the lo west p otent ial ?
54. Whi ch o f th e foll owin g re acti on s is (a rc) p roduct favo red? (a) Zn(s) + I, (s) ---> Zn2+ (aq ) + 2 (- (aq ) (h) 2 CI- (aq ) + (, (s) ---> Cl, (g) + 2 1-(aq) (e) 2 Na t taq ) + 2 CI-(aq) ---> 2 Na(s) + C l , ( ~ ) ( 2 K+(aq ) + H, (g ) + 2 0 W(aq)
944
I
•
more chaLLenging •
in OWL
Blue-numbered questions answered in Appendix 0
59. T h e reactio n occu rring in the ce ll in whic h AI ~03 a nd alu m in um salts are e lec tro lyzed is Ap -t (aq) + :) e " ---7 Al (s) . If th e ele ct ro lysis ce ll ope ra te s at 5.0 V and 1.0 X 1O.~ A, wha t mass of alu min u m m etal ca n be pro d uced in a 24-h day? 60. . ... A ce ll is co nstruc te d using the fo llo wing halfrea c tio n s: Ag +(aq )
+ c " -> Ag(s)
Ag~ S O I (s)
+2
c " ---7 2 Ag (s)
+ SO / - (a q )
E"
= 0.65 3 V
(a) Wh at reac tions sh o uld he observed at th e a node a nd ca thode? (h) Calcula te th e solubi lity p ro d uct COnSGlI lt, Ksp , fIJ I' Ag,S O .,. 61. . ... A pot e n tia l of 0.142 V is recorded (un der standard co nditions) fo r a voltaic ce ll construc ted usin g th e following h alf reactions:
Ca th o de:
Pb'H (aq) + 2 e " ---7 Pb (s)
Ano d e:
Pb CIA s)
Ne t:
Pb ' +(aq )
+ 2 e - ---7 Ph (s) + 2 C1-(a'1 ) + 2 Cr (aq ) --> PbCI, (s)
(a) Wha t is th e stan dard re d uc tio n po te nt ial for the a node reacti on ? (b) Estimate th e sol ubil ity p roduct. Ksp , for PbCl';!. 62 . Wh a t is the value of /-;0 for the fo llo wing h alf-re act ion ? Ag,CrO .,(s)
+ 2 e:
--> 2 Ag (s)
+ CrO / -(aq )
63. • T he sta ndard vo ltage, EO, for th e re ac tio n of Zn(s) a nd C I ~(g) is + 2. 12 V. wh at is th e sta nda rd fre e e neq.,'.Y cha nge, ~ , (;''' , fo r the re action ? 64 . . ... An e lec tro lysis cel l for alumi n u m pro duc tion o pcrates a t 5.0 V an d a curre-n t of 1.0 X 10" A. Calcu la te th e number of kilowatt-hou rs of e nerg y re q u ired to produc t' 1 m et ric to ll ( 1.0 X 10 3 kg ) o f alumin um. (I kWh = 3.6 X 10'.1 a nd 1.1 = 1 C . V.) 65 . . ... Electro lysis of mo lte n Na C I is do ne in ce lls o pera ting a t 7.0 V an d 4.0 X 10 1 A. 'V"h at m ass of Na (s) an d CI';!(g") ca n be produced in o ne day in suc h a cell? Wh at is the energy consump tion in kilowa u-h o urs? (I kWh = :1.6 X 10'.1 an d 1.1 = I C . V.) 66 . . ... A cu rren t of O.D I OO A is passed th roug h a so lu tio n of rh o d iu m sulfa te, ca usi ng red uc tio n of th e me ta l ion to th e me tal. After 3.00 h , 0.038 g o f Rh has been deposite d . wh at is th e cha rg e o n the rhodi u m io n , Rh "+? What is the form u la for rhodi um su lfate ? 67 . ... A cu rr ent of 0,44 A is passe d throug h a so lu tion of ru th e niu m nitrate causing reduct io n of th e m e tal ion to th e m et al. Afte r 25 .0 m in , 03 4:; g of Ru has be en deposited. W hat is the ch arg e o n th e ruth c ili um ion , Ru v"? Wh at is the for m ula fo r ruth enium n itr a te?
•
morechallenging
• in OWL
Blue-numbered questions answered in Appendix 0
68. T he total charge th a t ca n he deli vered bv a large d ry ce ll ha u crv befo re its vo ltage d rops too low is usually aho ut 35 am p-hou rs. (O ne am p-h o ur is the cha rge tha t passe s th ro ug h a ci rc ui t when I A flows for I h .) Wha t mass of Zn is con su me d wh e-n 35 a mp- hours a re d rawn fro m the ce ll?
69. Chl orin e gas is o btai n ed comm e rcially hy elec tro lysis of bri ne (a conce n trated aqu eo us so lu tio n o f Na C I). If th e e lectrol ysis ce lls ope ra te a t 4.6 V and 3.0 X 10" A, \v·ha t m ass o f chlorine can he p ro d uced in a 24- h day?
70. ... Write ba la nced eq uatio ns fo r the foll owing reductio n h alf-reacti on s involving organic co m po un ds. (acid so lutio n ) (a) H C() ~ H ---7 C H ~O (b) C h ll"C0 2 11 ---7 C 6 11[,C II :\ (aci d so lu tio n) (ac id so lutio n) (c) C H 3C H ~C H O ---7 C H 3C H 2C H ~O H (d ) C 113 0 1I ---7 C I 11 (acid so lutio n)
7 1. ... Bala nce the foll o wing eq ua tio ns involvin g orga n ic co m pou nd s. (a) A~ + (aq ) + C"II "CII O(aq) --> Ag( s) + Cl;H"CO ';! H (a q) (acid so lu tio n ) (h) C II 3C II,O H + Cr, O ,'- (a q) --> C H:\CO ';! H (aq ) + Cr3+(a q ) (acid so lu tio n)
72. A vo ltaic ce ll is constru cted in whic h o ne half-cell co nsists of a silve r wire in an aqu eo us sol u tion of Ag N0 3. T he ot her h alf-cell co ns ists o f a n inert plat in um wire in a n aque o us so lu tio n co n taini ng Fe';! +(aq ) an d Fe 3+ (aq }. (a) Calculat e th e cell po te nt ial. assuming standard con d itions. (h) Wri te the net io n ic e q ua tio n for the reacti o n occ urri ng in the ce ll. (c) In this vo ltaic ce ll, whic h el ectrode is th e a no d e , a nd wh ich is t he ca t hode? ( C,, (s)
+
C ,,2+(a q)
(a) Wh at two h a lf-reactio n s make up t he d isp roportion.
a rion re ac tion ? (h) Use va lu es of th e sta n d ard re d u ct io n po tenti als for the two half-reaction s in part (a ) to d etermin e wh eth er th is di sproportiou ati on reactio n is spon taueous. (c) \Vh at is th e equilibrium cons ta n t for thi s re action? If you h ave a so lu tio n th a t initially co n ta in s n.1 n mol of ell + in I.n I. of wat er; what are th e conc entration s o f ClI + a n d Cu 2 + at eq u ilibriu m ? 946
I
In the Laboratory 79. Con sider a n e lec troc hem ica l ce ll b ased o n th e halfre action s N iH(aq) + 2 c " ~ Ni fs) an d Cd 2+(aq ) + 2 e - --> Cd(s) . (a ) Dia gram th e cell , and labe l e ac h of the co m ponents (i n clu d ing the a n o de, ca thode , a n d salt hri dgc} . (b ) Use th e equ ation s for th e h al f-re act io n s to wr ite a bal an ced, net io n ic e q ua tion fo r th e o verall cel l reaction , (c) What is the p o larity of e ach el ectrode? (d) What is th e va lue o f E O ~..lI? (I..') In wh ic h d ire c tion do ele ct ro ns flo w in the exte rnal circuit? (f) Assu me th at a sa lt bridge co n taini ng NaN O j conn ec ts the two h al f-cell s. In wh ich dire ction do th e :":a+( aq) ion s move? In wh ich directi on do th e N0 3 - (aq} ion s move ? (g) Calcu la te th e e q ui lib rium co ns ta n t for th e rea ctio n . (II) If the co n ce n tr a tio n of Cd 2+ is re d uced 10 0.010 1\.'1 a nd [N iH ] = 1.0 M , wh at is the va lu e o f I:~_ ..II? Is th e net re actio n still the re action g ive n in p a rt (h )? (i) If O.O'iO A is drawn from th e bat tery, h ow lo n g ca ll it last if yo u be gin with 1.0 I. of eac h of the so lu tio n s and each was initi ally 1.0 1\1 in di sso lved sp ecies? Eac h electrode weig h s 50.0 g in th e beginnin g . 80. An o ld meth od o f measuri n g th e cu rre n t flowin g in a circuit was to usc a "silver co ulom ct cr," T h e cu rre n t passed fir st th rough a solu tio n o f Ag +(aq ) an d th en in to an o th e r so lu tio n con taining all clc cu'oa cuve species. The a mo u n t o f silver metal d eposited a t th e cat h ode was 'we ig h e d . From th e m ass of silve r; the num her o f ato ms of silve r was calcu lated . Since th e red ucti on of a silve r ion req uires o n e e lec tro n. thi s valu e equalle d th e numher of e lec tro ns p assin g th rough th e circ u it. If th e time was noted , th e ave rage cu rren t co u ld be calc ul at ed. lf in su ch an ex p erim en t, 0 .052 g of Ag is d eposited duri ng 450 s, wha t was th e current flowin g in th e circu it? 8 1. • A "silve r co ul o m et c r" (Stu dy Que stion 80) was u sed in th e pa st to me asure th e cu rre n t flowin g in a n electro ch em ica l ce ll. Suppose you found th a t th e current flowin g th ro ugh a n electrol ysis cel l depo sited O.OR9 g of Ag metal at th e cathode afte r ex actl y 10 min . If th is sam e cu rren t th e n p assed th ro ug h a ce ll co n taini ng go ld ( lI l ) io n in the form of [AuC l tJ - , h ow m ilch gold was d eposited a t th e ca th o d e in th at el ec trol ysis ce ll? 82. . ... Fou r m et al s, A, R, C, and D, ex h ibit th e foll o win g properties: (a ) O nl y A an d C re act with 1.0 M h yd ro chl o ric acid to g ive H 2 (g ) . (h ) Wh e n C is ad d ed to so lu tions of th e ion s of th e o t h e r meta ls, me tallic B, D , an d A are formed . (c) Metal n red u ces R"+ to g ive m e ta llic Rand D '" . Rased on th is in fo rma tio n , arran g e the fou r metal s in order of increasing a b ility to ac t as red u c ing agent". •
more chaLLenging
•
in OWL
Blue-numbered questions answered in Appendi x 0
83 . ... A so lu tio n of KI is ad de d dropwisc to a p ale bl u e solu tion of C u(N O j h . Th e so lution changes to a b ro wn col or, an d a p recipitate o f Cu i fo rm s. In contras t. n o ch a n ge is observed if solutio ns of KCI and KBr are ad ded to aque ous Cu (N 0 3 h . Consu lt th e ta ble o f standard reducti on p o tenti al s to ex p lai n the d issimilar resu lts se e n with th e d iffere nt h al id e s. Writ e an eq ua tio n fo r the reac tio n that oc curs \v-lu-u so lutio ns of KI a nd C ll (N O :~ h arc mi xe d .
84 . ... T he a m o u nt of oxygl'n , O 2, di ssol ved in a wa ter sample a t 25 °C ca n he d e te r min ed by titra tio n. Th e firs t ste p is to ad d so lu tio n s o f ~l nS0 4 and Na O H to th e wau-r to conve rt th e d issol ved ox yge n to t\1n0 2' A so lutio n of H 2S0 4 an d KI is then ad de d to convert th e ~1 n 0 2 to M n-" , an d th e iod id e io n is conve rte d to 12, T he 12 is th e n tit rated wit h stan d a rd ized Na2520 j . (a) Bal ance th e e q ua tion for th e reaction of Mn 2+ io n s with O 2 in b asic so lution . (h) Bal ance th e e q ua tion for the reaction o f Mn 0 2 with 1- in aci d sohu.ion, (c) Bal a n ce th e e qua tion for th e reaction of S20 .~ 2 wit h 12 , (d ) Calcu la te th e am o u nt 01'0 2 in 25.0 m L of wa ter if the tit rati on requ ires 2.45 mL of 0 .0112 M :\'a2520 3 so lu tio n .
Summary and Conceptual Questions The fo llowing questions fila)' 11.\(' rfHu:f'jJls )1WJi this ruul invuious dmpters.
Consi der th e h yd rogen-o xygen fu el ce ll, where th e n et reactio n is
H, (g)
+
t
O , (g ) ---> H,O (f )
(a) Calcu lat e th e effi ci en cy o f th e fuel dard co n dit ions. (b) Calcu la te th e effic iency of th e fuel u ct is water vapor ins tea d of liquid (c) Do es th e efficiency d epe n d on th e tio n p rod u ct? Wh y o r wh y n o t?
ce ll under sta nce ll if th e prod -
wate r, sta te of th e re ac-
87 . A hydro gen -oxygen fu el (('11 o pera tes o n th e sim ple reaction
H, (g)
+
t
O ,( g ) ---> H,O (f)
If the ce ll is d e sig n ed to p ro duce 1.5 A o f curre n t and if th e hyd ro gen is co n tain ed in a 1.0-1. tan k at 200 atm p re ssu re a t 25 °C , h ow long can th e fu e l ce ll o perate b efo re th e h yd ro g CF, SO,F + 'I
u,
is carried o u t elec troc he mically ill liq uid I1F as th e solvent. (a) If yo u elec tro lyze 150 g o f ClI j S0 2F, wha t m ass of H F is req u ired . a nd wha t mass o f each product ca n h e isolated? (b) Is H 2 produced a t the anod e or t h e ca thode of th e elec trolysis cell? (c) A typi ca l ele ctro lysis cel l o pera tes a t H.O V a nd 250 A. H ow ma llY kilowa tt-hou rs o f e neq.,'")' d o es o ne suc h ce ll CO !\ Sl Ul W in 24 II? 86 . . ... T he free en ergy ch a nge fo r a reaction , d rG' , is th e maximum e nergy th at can b e ex trac te d fro m th e p ro cess, whereas d rH o is the total chemica l p otential e nergy cha nge . T he e ffic ie ncy of a fuel ce ll is th e ra tio o f th ese two quan tit ie s.
•
morechallenging
• in OWL
Blue-numbered questions answered in Appendix 0
Electro ns in this red ox p ro cess arc transfe rred from g lucose to oxyge n in a series o f a t le ast 25 ste ps. It is ins tructive to calcu late th e total d a ily current flow in a typ ical organism an d th e rate o f energy ex pen dit u re (p ower), (Sl'C T. P, C hirpic h : journal of Chemical Education. Vol. 52 , p. 99 , 19 75 .) (a) Th e mol a r e n t hal py of com bustion of gl ucose is - 2800 kJ/ m ol-r xn . If you are o n a typi cal da ily di et o f 2400 Cal (kilocalories ) , what am o u n t of g lucose (i n mo les) must be co ns u m ed in a day if g lucose is th e on ly so urce o f energy? Wh at amount of O 2 must he consumed in th e oxi da tion process? (h) How m a n y moles of elec tr o ns m ust h e su p pli ed to re d uce th e a mo u n t o f O 2 ca lculate d in p a rt (a )? (c) Based on th e a ns wer in pa rt (h) , calcula te th e curre n t flowin g , pe r seco n d, in yo u r body from the com b u stion o f glucose. (d ) If the average stan dard pot enti al in th e elec tron transp ort chain is 1.0 V, what is t he rat e o f e nergy ex pe n d iture in wa tts?
I
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P
la net Ea rt h ca n be d iscussed from lllall Ypoints of view: g eologica l, po litic al, histo r ical , am i e n vironm en tal . T he last of th ese h as lI O W become a m a in topi c of co ncern. If we a rc to understand wh y th is is so , th e n we n e e d to u nd e rstan d th e che m ica l co m posi tion o f ou r e nviro n me nt a n d the \\'a)' this is c ha ngi ng. Land , sea, air-co nti nen ts, ocea n s, at mo sp here-c. so lid , liquid , gas. T hese a re the th re e m a in co mpone nts of th e c u vironme n t, a nd e a c h is g ove r n e d h y t h e laws of che mis try th at \ VC associate with th e se ph ysical sta te s. Lo o k a li ttl e cl o ser at the cont ine nts, a n d we ca ll see m o rt' a nd mo r t' space heing d evoted to h um a n occ upatio n an d fan ni ng. \ Vc lila)' th ink the wate r 0 11 the pl a ne t is e ith e r seawa ter or fre sh ' »ater, hut ho w d o we g el fro m o ne to th e o ther? \Ve kn ow th e a tmosphere is a mi x tu re of gases a nd \'apo rs a nd is re st less ly in moti on , so m e tim es vio le n tly so , h u t wha t will happen if its co m posi tio n changes? Un d e rsta nd ing the e nvirotn nen t is crucia l, h ut witho u t a ch e m ica l perspective we are flo u nderin g in th e d ark. In this inte rch ap ter; we will lo ok part icul arl y at th o se parts of the e nvironm ent that a re most susc e ptible to change: the at mosphere and the aq ueous sp here (wate r ). \\'h at h um a ns are d oi ng to th e at mosp here appears to he leadi n g to climate cha ng e a nd in a d irectio n that co uld d est roy muc h of what we have cr ea ted. T he water o f th e pl ane t is also of concern : we often poll ut e it, a nd we n eed to purify it. Nature ca n su p ply wa ter in ab u nda nc e as rain , rivers, an d lakes b ut 110t necessa rily whe re it is m o st needed. \\'h ile we in th e d evel oped wo rld have a see mi ng ly a bund a m sup ply o f pure , fr es h water, others obtain wat er on ly with diffi culty- a nd eve n the n it may no t be til to dri n k. As the world 's pop ul ation increases, m any m o r e pe o pl e
• Earth as seen from space.
will fin d the mse lves having to rel y 0 11 seawa ter fo r thei r su pply o f fre sh water. Seawa ter, o f course, is not d rin kabl e , h ut chemistry can m ak e it so . If we are to u nderstand atm os ph eric poll ut ion and its effec ts on cl imate, an d to un de rsta n d water sup p lies and th e need to m ake th em dri nkable , th en we nee d to u n derstand th e chem istry that is involved .
The Atmosphere T he an nosp b crc lIlay he th e least weigh ty part o f th e e llvironmen t, b u t this doc s no t mea n it h as the least effect. In fac t, it ha s a di sp ro po rt io natel y lar g e infl ue nce O il o u r sta nd ar d of li\'ing. \Ye kn ow it is chang ing because of wha t we are ad d ing to it, a nd we fea r th e co nsequelK es o f tha t cha nge. Ea rt h has cha nge d drama tica lly ove r the billions of ye ars th e pla ne t has existed, an d so h as the a u uosphere . T h e first at mosp he re is thought to have heen m ainl y amm o ni a, water \'apor, an d carbon dioxide . Today, it is main ly n itroge n , oxyge n , and argo n . At so me stage , m o st of th e water co nde nsed o nto the su rface of th e plane t, an d seas were for me d. It was iII th e seas th at life probabl y e merged , and life h ad to ada pt to co ndi tions as they change d . Sometimes th e chang es we re so great th at man y species cou ld not ada pt and were wipe d out. Bu t, d e spite these m ass ex tinctio ns , th e re were always a few su rvivors. Ta ble I give s th e co mposit io n of th e a tmosphe re in terms o f the gases it co n ta ins . T he values in th e table are fo r d ry air at sea level. No r mally, air also co n tains wat er vapor, an d th is can vary co nsi derably from place to place an d fro m d ay to d ay and can be as high as 40 ,000 pp m (pa rts pCI' milli on ) , hut is generally abo III half th is.
I 949
950
I
The Chemistry of the Environment
TABLE 1.
The gases of the atmosphere>
Gases
Proportion (ppm)
Nitrogen (N ,)
780,840
Oxygen (0,)
209,460
Argon (Ar) Carbon dioxide (CO 2) Neon (Ne)
9,340 385 18
Helium (He)
5.2
Methane (C H,)
1.7
Krypton (Kr)
1.1
Hydrogen (H,)
0.5
Dinit rogen oxide (nitrous oxide) (N 20 )
0.55
Ozone (0,)
0.4
Ca rbon monoxide (CO)
0.25
Xenon (Xe)
0.086
Radon (Rn)
traces
" Da ta on ga s con cc ru ra rions refe r to n-Iative numbe rs of particles (a n d he n ce art' related to mole fraction s) "
In fact , th ere are m an y more co m po ne nts o f the a u nosp here than th ose listed in Ta ble L Th ere are trace amou n ts o f che m ica ls e m itte d by plants, ani mals, a nd mic robes, as well as those rel eased hy hun tan activity. O ur ad d itio ns to th e atmosphere com e fro m hurnin g fuels to ge nerate electricity, from vehic le ex haust" from industry, an d fro m ou r d ay-to-day activities. T he at mosp here is changing, an d we n eed to understand what thi s m ean s for th e futu re . We mi ght im ag in e th at a ir is a h omoge n eous m ixture o f gases an d particle s because it is co ns tan tly be ing stirred hy th e moti on of th e p la net a n d the h eat o f the su n. In fac t, th is mi xi n g is far from co m p le te , Th ere is rel ati vel y littl e exc hange of air b et ween th e north and south reg ions o f the pl an et, an d th ere are di stin ct layers in the a tmosp here (.... p age 535 ). As far as hum an s an; co ncerned. th e m o st importan t layer is th e lower layer ill wh ic h we liveth e tropo sphere , wh ich goes tip to around 7 km at the p ol es a n d 17 kill at th e eq u a to r. Above th e tro pos ph ere is th e stra tos p here ; thi s ext ends to a bo ut 30 km , ge tting th in n er an d thinn e r as we r ise to hi gh e r alti tudes. T here are eve n hi gh er region s, hut th ere th e a ir is ver y th in , although n ot so thin it ca n n o t interact with in comin g space d eb ris o r veh icles like th e Sp ace Sh u ttle and ca use th em to g low red h ot. In th is interch apter, , »c will co n ce r n oursel ves on ly wit h th e tropo sph er e a n d th e stra tos p here. Th e to ta l m ean m ass o f th e atmosp here is es timated to b e a ro un d 21. 15 X lOIS kg (5. 15 X 10 1:) m etri c ton s, or 5 .15 X lOti gigato ns ). (Giga tc n s are often used in en viro nm en tal stud ies because th e m asse s we are di scu ssin g are so large . A metric ton is 1000 kg , but a gtga ro n is a bi llion me t-
ric to n s, that is, If)!' m etri c tons.) Th re e quarters o f the mass of th e atmosp here is in th e troposphere , an d th e re its average temperat u re is H "C (h u t th at is an m"erage of a wide ra nge of temperatures) . It ca n he as low as - R9 "C, th e lowest tem p eratu re eve r recorded, in An tarctica in 1 9 R ~1 , or as hi gh as 218 °C, th e hi gh est temperature ever recorded, in Nort h Africa in 1922. Afte r temperature, th e n ext m ost o h.. vious property o f th e at m os p here is the pressu re it exerts, ,»hi ch a t sea level is arou n d 101 kPa ( I atmosp here ). So me o f th e gases of th e atm osp he re arc u nreactive an d n ever cha nge chem ica lly. T hese are th e n obl e gases: helium , n eon , argon , kryp ton , an d xe non. Two of th ese , n am ely h elium an d argon, are b eing added co ntin uo usly to th e atmosp here b ecause the y are products o f rad ioactive d ecayh elium co mes from alpha p art icles e mitted by cle ments suc h as uranium , an d argo n is produced b y d ecay o f -10K, a longlived rad ioactive isot ope of p otassium . Hel ium is suc h a light gas th at it is co n tin ually bein g lost into space. Th e o th e r gases o f th e atmosp he re , n amely ox yge n, n itrogen , an d carbon dio xid e , are che m ica lly re acti ve and ca n b e cha nged n aturall y in to o ther m ol ecul es hy reactions dri ve n hv lightnin g , ultraviol e t rays fro m th e sun , o r the influ en ce of livin g thin g s, an d es pec ially the ac tio ns of humans .
Helium. Ultraprecise elect rical measurements require extremely stable temperature, humidity, and vibration control. In this photo, NIST physicist Rand Elmq uist fills a cryogenic chamber with liquid helium in preparation for measuring t he international standard for electrical resistance. U.S . laboratories and manufacturers use NIST electrical standards and calibrations for all kinds of measu rements from home electricity usage to electrocardiograms to gene-mapping.
Nitrogen and Nitrogen Oxides T he n it rogen o f th e at mos ph e re arose fro m th e Ollt-gassing o f Earth when it was simply a m olten m ass. Even to day, so me nitrogen escapes whe n volca noes erup t. Ni trogen gas is rela tively un re acti ve , hut every livin g th in g o n th e pl an et n eeds to in co rp orat e nitrog en int o its ce lls as , a mo ng oth er
The Atmosphere
co m pou n ds, ami n o ac ids, th e bu ildin g blo cks o f proteins. The process by whic h nitroge n gas is co n ve rt ed to use ful nit rogen -containin g co m p o u n ds, suc h as a nn uo ni a (N I 13 ) an d ni tr at e ion s (N 0 3 - ) is kn own as nitrogen fixation (Figu re 1) . Bacteria in th e so il a n d algae in th e o cean s are th e p rimary "fixe rs" of nitrogen . On a glo hal sca le, th e co nvers io n of nitro gen to various co m poun ds is acc o mpli sh ed to the ex te n t o f around 50 % hy h io logical proceases, 10% by fertili zer m a nufacture , 10% hy lightning , an d 30% from th e burnin g o f foss il fue ls, Once n it ro gen is "fixed " it is in corporat ed in to livin g th in gs, passes through lIlan y o the r life fo rm s, an d e n ds up in th e so il, which co n rai n s sign ifica n t amou n ts o f n itrogen-con ta ini ng chemi ca ls suc h as ammonium salts, n itrates, an d ni trit es. T hc two an io ns are co nve r te d b ack to n itroge n gas or to N tO an d re turned to th e a tmosp h ere . Dinitrog en mon o xid e (n itrous oxide) , N tO , is o n ly one o f several nitr ogen ox id es importan t in o u r e nviron me n t. Oth ers are n it roge n m o n o xide (n itr ic ox ide) , NO , and nitrogen dio xide, NO t. Co llec tivel y, th ey a re re fe rred to as NO x co m p o u n ds. T h ere are a num b e r of n a tu ral so u rces of NO x co m po u n ds.
• N O produced from th e co m b inatio n of N, a n d O 2 whe n a ir is heated o r spa rked by a b olt o f lightin g • NO an d NO t p ro d uced b y th e photo chemical ox idat ion o f N t O in th e a tmosp here • O xidation of N H 3 in the atm o sph ere • Product ion o f NO and 0:2 0 in so il b y m icrohi al processes In ad d itio n , fo ssil fuel burnin g an d b iomass co m b us tio n by hum a n s p roduce abo u t 65% o f th e NO x in th e a tmosp h e re .
I
951
Th e m o st a b u n da n t nitro ge n oxi d e , b ecause it is th e least reac tive, is d lnl u'o gen m ono xid e . .Mo st of th is ox ide co mes fr om mi cr obes in th e so il. Ten million ton s o f NtO are rel eased each yea r, so me fro m h um an activity. T he other two nitrogen ox ides m ake up onl y O.OOOO!j ppm of th e atm os p here. \Vh ile 100 million to ns of NO an d NO:! co m e e ac h year fro m th e ex haus ts of "c hi cles , they are qui ckly was hed out of th e atmosphere hy rain . Wh en th ey do lin ge r, a n d in p arti cul ar ove r ci ties ill su u llY elimate s, th ey ca n react with hydrocarbon s in th e a tm os p he re, such as trace s of unburn t fuel , to produ ce irritat in g p hoto ch emi cal smog. Am o ng th e m o st polluting co m p o n e nts o f sm og a re PANs, p eroxvacvl nitrates, repre sented h ere by pe ro xyace tyl n itrate .
One route to its formati on is as foll ows: a) O rg an ic co m p o u nds called aldehyd es (s uc h a s C II :~C H O) are p roduced b y co m p lex re actio n s in sm og. Th e se mole cule s re a ct with h ydroxyl rad icals (0 11·) , wh ich a re a lso th e re sult of photochem ical reactions in th e atmosp h e re . Th ei r interaction produce s a no the r radical sp ecies C H :~C O· .
CH,CHO + OH' --> CH ,CO' + H,O h ) T he n ewly produced radi cal reacts with o xyge n to produce th e peroxyacetyl radical.
CH ,CO'
+ 0, --> CH,cOOO'
c) Th e p eroxyaceryl r adic al r e act s wit h N 0 a ir to giv e p cro xya cctyl nitr ate.
CH,COOO' + NO,
-->
2
in polluted
CH, COOO NO,
PANs are to xic an d irritating . At low co nce n tra tions, th ey irritat e the eyes, hut a t high er co n ce n tra tions they ca n ca use m ore se r ious d am age to bo th an imals an d n :g· e ta uon . Th ey are rela tively stable , di sso ciatin g o n ly slow ly into ra d icals a n d NOt. T h us, it is po ssibl e th at th ey ca n trave l SO lllC distan ce fro m where tlw ), arc fo rm ed b efore d isso ciatin g in to h arm ful radi cals.
Oxygen
Figure 1 The nitrogen cycle. The nitrogen cycle involves nitrogen-fixation by soil bacteria or, in aquatic environments. by cyanobacte ria. The NH 4 + ions produced are converted to nitrate ions, the main form of nitrogen absorbed by plants. Nitrogen is returned to the atmosphere by denitrifying bacteria, which convert nitrate ions to N2 >
Plant life h ad a d ram ati c effect on the early at mosp he re, ch ang ing it fro m a reducin g to an o xid izing o ne , in o the r words, fro m there be ing n o oxygen present to o ne wh e re th is reactive gas is the sec o nd most abundant species presen t. Th is change m eant life had to cha nge. Th e percent o f oxy· gen in the air is n ow m idway between two ex tre m es that wou ld m ak e life 011 Ea rth impo ssible for hum an s: h elow 17%
952
I
The Chemistry of the Environment
we wo uld su ffoca te, and above 25 % all o rga nic materi al burns ,'ery easi ly, although the re is evidence th at oxygen in Eart h's atmo sphere did at o ne Lim e exceed this level. Th e total III ass of oxygen in th e at m os p here is a mill ion g igato ns. Even tho ug h th e h u rn ing of 7 g iga to lls of fossil fuel ca rho n a yea r uses up I S g iga to lls o f ox ygen , thi s m akes alm ost n o di scern ahl e d iffe re nce to the amou n t of oxyge n in th e a tm os p he re . O xygen is also moderately soluble ill wa ter C'" pag e ( 26) , wh ich m ak es life in rivers, lakes a nd oceans po ssihl e . Oxyge n is a by-pr oduct o f plant photosynthesis. Ca rbo n di o xid e is th e so u rce o f th e ca rbon plants need and wh ich th ey cap tu re fro m th e air a nd turn into ca rbohyd ra tes suc h as glucose (C(i H l:l0d by photo synth e sis. T he o verall c he mical reactio n IS:
6 CO,
+ 6 H,O---. ( 6H 1106 + 6 0,
am i th e net result is th e release of an oxygen mol ecule back in to th e a tmosp he re fo r eac h CO 2 abso rbe d. O xygen mol ecules rele ase d by ph otosynth esis r ema in in th e aunosp h ere , o n ave r age, fo r aro u nd 3000 years before bein g re abso rbed. Blue-gr een algae or cyanobac teri a first beg an p roducin g o xygen as lon g ago as 3..1"> hilli on years. But, m ysterio usly, hund reds o f milli on s of years elapsed before th ere ' . . as a sign ifican t amo u n t of oxyge n in the at m os p here . Astrohio log ists are not at all ce rta in why thi s occurred, h u t it d oc s see m plausib le th at th e oxyge n fir st produ ced did n ot remain ill th e a tmosp here because it re acted with m et als, es pec ially cOllve r ting iron (II) to iron (I II) . Th is process co n tin ue d for about a h illion yea rs, but th e n the level in th e a tm os p here bega n to rise 2 billi on yea rs ago, and abo u t 500 millio n years ago it ro se relativel y rapidly to a ro u nd 20 % whe n th e fir st land pla nts start ed to a p pea r. Th ere are th ree natu rall y occ u rri ng iso to pes of o xygen : oxyge n- 16 acc o u n ts fo r 99 .76% of 0 ato ms, oxyge ll- l 7 fo r a m ere 0.04 %, a nd o xyge n-I S fo r 0.2%. Beca use o xyge n-I S is 12% heavie r than o xygen-16 , it ca n infl uence th e behavior of wate r in th e e nviro n me nt. T he ra tio o f o xygen-I S to o xyge n-I 6 in th e world 's oceans has var ied sligh tly ove r geological tim e , and thi s h as left all imprint on parts of th e e nvir o n me n t, p ro vidin g evid e nce o f past climat es. wh en th e wo rld is in a coole r period, wate r m ol ecule s with o xygen-If eya porate m ore easily from th e ocea ns th an th ei r heavier o xygen-I S counterpart. T h us, snow th at falls is ve ry sligh tly ri ch er in oxyge n-l 6, and the wate r th at rem ain s in the o cean s is very sligh tly ri ch er in oxygen -I S. Mar-in e creatu r es th erefo re lay down she lls th at h ave more oxyge n-IS th an e xpected, and th e se are preserved in se diments. An a lyzing th e ra tio o f th e two isotopes in suc h d epo sits reveal s th e cycle o f g lohal cooling a nd warm ing th at has cha rac te r ized th e past half milli on yea rs with its five ice ages.
Ozone
Oz on c (0 3 ) h as a key rol e to play for life o n thi s planet, but it is also a threat. In th e tr opo spher e, it is a pollutant, wh ile in th e stra tos p he re it ac ts as a sh iel d , protecting th e planet from d amag in g ult ra violet rays from the su n . This sh ield is kn own as th e ozone layer. Th ere is a natu ral low level o f ozon e in th e air we breath e , about 0.02 pplll (pa rts pe r milli on ), hut ill su mm er th is can inc rease to 0.1 pplll o r m o r e as a re sult of su n ligh t acting o n the nitrogen dioxid e e m itte d hy ve h icles. In th e prescnce of su n ligh t, N O ~ di ssociat es to NO and 0 a to ms , and th e a ato ms ca n r eact with O ~ (in th e presen ce of a third m ol ecul e ac ting as an energy "sink"] to p roduce ozo ne. NO,
+
energy ---. NO + 0 (A < 240 nm)
Because ozone d am ages th e lun gs, th e legal limit for expo sure to ozone in the wor k place is 0.1 ppm . Some growin g plants are also susce pt ib le to th e gas, and eve n though th ey d o not sh ow visibl e signs of stress, th ei r gr owth is redu ced in propo rtion to th e level of ozo ne in th e a ir. In co n tras t with ozo ne in the tropo sphere , 0 3 in th e stra tosp here is vita l to th e pl an et because th e mol ecul es absorb ult ra viol e t radiati on before it re ach es Earth 's su rface . Radi ation with wavele ngths sho rte r th an 240 mn in teracts with O :! m ol ecules and cleaves the m in to two 0 ato ms. Eac h 0 a to m co m bines with anot he r O ~ m ol e cul e to produce an o zon e m ol e cul e. 0,
+ energy ---. 0 + 0
o + 0, ---. 0, O zone in turn abso rbs ultraviol et radi ati on with wavelen gths o f le ss th an 320 1lI1l , a nd th e 0 3 is decompo sed to 0 1 a nd a ato ms. 0,
+ energy ---. 0, +
0 (A
HClO(aq) + HCI(aq) which dissociates to yield the hypochlorite ion (ClO-), HCLO(aq) + H,O(C)
H,O' (aq) + ClO -(aq)
Ka = 3.5 X 10- 8
Two of the chemical species formed by chlorine in water-hypoch lorous acid and the hypochlorite ion- are known to environmental engineers as free available chlorine, mean ing they are available for disinfection. Because chlorine gas is quite toxicand can easily escape from damaged containers, most water treatment facilities are now switching to solid hypochlorite salts such as sodium or calcium hypochlorite [NaCiO and Ca(C LO)" respectiveLy]. Hypochlorous acid and hypochlorite ion are effective in killing bacteria. However, they will also oxidize reduced inorganic ions such as iron(II), manganese(II), and nitrite ion, as welt as organic impurities. These com poundsare collectively known as the chlorine demand. To be available to kill bacteria, sufficient chlorine must be added to exceed the chlorinedemand. The advantages of chlorination include its relatively low cost and simple application as well as its ability to mai ntain residual disinfection throughout the distribution system . However, there are also disadvantages. Ch lorine can react with dissolved, naturally
HClO (aq) + NH,(aq) -. NH ,CI(aq) + H,O(C) Chloramines are called combined available chlorine. li ke free available chlorine, com bined available chlorineis retained as a disinfectant throughout the water distribution system, but it is a weakerdisinfectant overa ll. Combined available chlorine, however, has the advantage that it is less likely to lead to chlorinated organic compoun ds. Recently, however, chemists in Washington, D.C., which uses t he ammonia/chlorine method of water treatment, found increasing lead levels in the drinking water in area homes. [Several hundred homes had lead levels of 300 parts per billion (ppb), whereas the allowed levelis 15 ppb.] Water treated with only chlorine is highly oxidizi ng, and lead in the water is oxidized to lead(IV) oxide. Th is insoluble oxide is trapped in the mineral scale that inevitably forms inside water pipes, and
Mexico , the An tofagas ta n~ g i on of no rt her n Ch ile, the Codoba n regio n of Argen tina , and espec ially in the \Vest Bengal reg io n of Ind ia and the neighbo ri ng co un try of Bangladesh . T hi s had go ne unn o ticed for man y yea rs, but th e enviro n me ntal effec ts of arse nic in d rinkin g wate r can be very se rious. In Taiwan , a n e p ide miological study in the 1960s sh owed th a t the prevale nce of skin ca n ce r was rel ated to the a mo un t of arse n ic in wate r. In on e reg io n whe re th e level of a rse nic was 500 microgra ms pe r liter (500 p pb ), 10% of peo ple age d 60 or over h ad cance rs , ma in ly 0 11 th eir skin . Than kfull y, these types o f ca ncers a re »o r u sually fatal and ca n be rem o ved . In te rn al cance rs, however, are no t as easy to d iag nose o r treat , a nd ill o ne part of Ta iwan where th e level of arsenic was ROO ppb, th ere was a hi gh in cid e nce of bladder ca nce r. Unfortu nately, eve n when
6~ '--_ _.........._
..........
Chlorine tanks in the water t reatment plant of a small city.
lead is thus removed from the water system. However, switching to the less-oxidizing ammonia/chlorine method apparently means lead ions are not removed by oxidation, so lead(II) ions remain in the system. The odd-electron molecu le chlorine dioxide (Cl0 2) is also an effective disinfectant. The primary advantage of Cl0 2 is that it completely oxidizes organic com pounds (forming CO2 and H20 ) and therefore does not produce trihalomethanes. However, Cl0 2 provides no residual disinfection in t he distribution system. Also, C10 2 must be generated on site because it is explosive when exposed to air.
measu res are take n to purify th e water, th e incide nce o f ca ncer ca n co n tin ue to be h ig h for many yea rs bec au se of th e lon g late ncy per io d befo re th e di se ase appea rs. T h e large st mass poison ing by arse n ic involved more than 30 m illion people (Figu re 7) . Pe o ple ill West Be ngal , In d ia , and in Ban glad e sh go t th ei r dri nking wa te r fro m tu be we lls th a t were drilled in the 19 70s. Th e wells were installed hy th e Un ite d Na tio n 's C hil dre n 's Fu n d (UN ICE F) to provide safe drin king wate r for a populatio n th at h ad trad itionally take n its water from co n ta mi na te d streams, rivers, an d po nds an d wh ich had su ffe re d acco rdi ng ly fro m wate r-bor ne d iseases suc h as gas tr oe n te r itis, typ ho id , and chole ra. Even tually, 5 m illio n suc h we lls were d ri lled, and they in deed served the purpose of hrin gin g cho lera under co n trol. In its place , h oweve r, they broug h t low-level arsen ic poi son in g to large areas of Ban glad esh .
Green Chemistry
Tube wells a re 5 em in diame te r an d tap th e abun dan t g ro u n d water abo ut 200 m eters belo w th e sur face. A few h ad be en in sta lle d III Ban gladesh in th e 19 30s wh e n th e reg ion was part of th e British Empi re. Th ose villages that h ad installe d them sh owed a d ecr ease in illn ess, es pec ially a mong ch ildren , UN ICEF co n tin ued to su p po rt th e progr am after indepen d en ce , an d hy 1997 it co u ld repo rt that it h ad alread y su rp asse d it" go al o f p rovid -
I s
':: ~
I
959
sig ne d by ch em ists for d o ing this, such as pa ssing th e water through layers of sa n d, pieces o f iron , an d ch arcoal. The ino rgan ic arsen ic, wh ich is present as arsen ite io ns, As0 33 - , stick s to th e pieces of iro n wh ile th e o rga n ic form s o f arse nic are absorbed in to th e ch arcoal. FLuoride in Drinking Water
It is g e ne r ally ack nowledge d tha t a little flu ori d e in d rin ki n g wa te r ~ ~ str e n g th e ns toot h e n ame l a n d preve n ts d ecay, an d to thi s e n d man y ing 80 % of the po p u latio n in th o se Figure 7 A Bangladeshi villager shows his hands affected p u b lic wa te r su p p lies a re flu o rireg ions wi th "sa fe" wa te r, Su ch h ad by arsenic contamination in the water suppLy. bee n th e suc cess o f th e p ro g ra m d a ted to a le vel o f 1 pp m . th at m an y village rs in stalle d pr ivate wells. Then d oc to rs heNeverthe les s, the re arc dan gers in d r in ki ng wa te r tha t gan to notice a worrying in cr ease in skin ca ncer an d sush as mu ch hi gh e r level s, a n d in so m e lo calit ies th e level pected tha t arse n ic was th e ca use . An alysis o f th e wate r from o f fluo r id e is n at u r ally h igh . H um a n s, as we ll as an im als, th e wells sh owed th at m o re than 20 m illion p eo pl e were can b e affec te d an d su ffe r ivcnw flu orosis (Fig u re 8) , wh ich dri n kin g ' . . ater co n tai ni ng arse n ic at th e :')0 p p b level, an d is a h ard enin g of th e hon es lead in g 10 a d efo rm ed skelas lIla ny as :') m illio n were d ri n kin g water co ntain ing e to n. III cert ain p arts of In di a, su ch as th e Pu nj ab . the co n d itio n is e n de mi c , es pe cia lly wh ere village rs d r in k wa300 p pb arse nic. It d oes a p pear, th ough , th at 80 % of the Ban glad esh i pop ulati on h as access 10 safe dri n kin g wa ter. ter fro m we lls with hi gh level s of fluo ri de , up 10 15 p pm . In 199 7, th e Ba ngladesh i goye nJ me n t in stituted a Rapi d Abou t 2:) m illion In di a n s suffer a m ild for m of flu o ro sis, Actio n P ro g ra m th at foc u sed o n a gro u p o f villag e s whe re with m any th ousa n d s sh owing ske le tal d efor mities. con ta mi nation was seve re. Almost two thi rds 01'30 ,000 tube wells ill th ese "illages we re dcb ven ng wat er with arse n ic in Green Chemistry excess o f 100 ppb . A te a m fro m th e Dh ak a Co m m u n ity Ou r e nviro n men t is a co mplex che m ical syste m but not o n e Ho spital visit ed 18 a ffe cted ar eas a n d ex a m ined 2000 th at is so co m p lex th at we ca n never u nderstan d it. It is also ad u lts an d ch ild re n ; th ey co n cl u d ed that m ore th an h alf a robust system th at Illay h c capable o f h e in g h ealed wh e n h ad ski n lesio n s d ue 10 ar sen ic. it is da maged . (O ne ex am p le is th e slow reversal of th e So wh at ca n th e peo pl e of thi s regi on d o ? O n e ansv....e r An tarctic ozo ne h ole wh en CFC p rod ucti on was sro p ped.) is to dig wells that go d e epe r than 200 m eters o r shallow e r O ur stewa rd sh ip of Eart h d e mands a go od knowledge of wells th at re ach d own o n ly 20 m et ers. In th is way, th e co nchemistry an d se nsible b eh avior. No thing less will su ffice. ta m ina te d laye r of grou n d wa ter ca n he avo id ed . Once O n e way c h e m ists h ave re spo nd ed to e nviro n m ental people h ave acce ss to un co nta mi n ated wa te r, th e a rse n ic p roble ms is th e so-calle d g re e n che m isin th ei r bod ies is qu ickl y lo st. tr y m ovem e n t. In 199 6, the U.S. Rem edi ati o n steps have to be taken Environmen ta l Pr ote ct ion Ag e n cy (EPA) when d r inking water is co n tam in ated, initia led its G reen Che m istry Pro gra m , and ye t it is a pro blem tha t can he cos tly whi ch includ es re search , ed u ca tio n , a n d 10 sol ve. Wh en half a milli on p eopl e in o u treach effo rt s as we ll as th e Pr esid ential New Mexico were ex pose d to hi gh arscG ree n Che m istry C h alle nge Awa rd s, a n n ic levels in th ei r drin king water, action an n ual p ro g r am recog n izin g in n ovation s was take n hut at a cost o f $ 100 milli o n . in "cl ea n er, ch ea per, sm arte r che m istry." Tak ing sim ilar actio n in Banglad esh Figure 8 FLuorosis. Here, tile teeth have T he Amer ica n C h e mi cal Society activel y wou ld h e beyond th e abilityofthat cou nwhitish patches, indicating mild fluorosis. p rom o tes gree n ch emistry (Fig u re 9 ) , U) ' to pay. Even repl acin g a sh allow well Teeth and bones are hydroxyapatite, and th e Ro yal So ciety o f C he m istry in with o ne that draws water {i-OI Il Ca 5( OHj(PO.)]. If one drinks water with a Eng la nd p u bli sh e s the rese a rc h j o u rn al th e lower, safe r, wate r tabl e costs $ 1000. hiqh F- ion concentration, the OW ion in Green Chemistr)'. SO IlI C unive r sitie s n ow (Sh allow tube we lls cos t o n ly S IOO to apatite can be replaced by the F- ion. This o ffe r d egrees in g re e n che m istry. drill , wh ich is why they h ave proved so can have the beneficial effect of hardening If we assu me th e wo rd "gre e n" mean s p opula r.) T h e alt ern ative is to re move t he surface of teet h, but in excessive cases it en viro n me ntally so u n d o r fr ie n d ly, how the arsen ic fro m th e water. Various simcan lead to mottling. And in very hiqh conca n ch c m istry be gre e n? Peo ple h ave centrations, skeletal deformities can result. ple a nd ch eap device s h ave be e n de-
960
I
Th e Ch em i stry o f th e Envi ronment
Chemical Perspectives
Particulates and Air Pollution
The words fog, haze, mist, and smoke all describe atmospheric particulates that can range in size from about 0.1 millimeterto less than t urn (f um = 1 micrometer). Large particulates, more than t u m in diameter, generallyoriginate from the disintegration or dispersio n of even larger particles from volcanoes, wind blown dust. or sea spray. Many of these originate as soil or rock, and their composition is similar to the Earth's crust; that is, they have high concentrations of Al, (a, Si, and O. Fine particulates are mainly formed by chemical reactions and by the coalescing of small molecules. These particulates often contain carbon that originates in power plants burning fossil fuels, inci nerators, home furnaces, fireplaces, combustio n engines, and forest fires. Black soot particles, main ly crysta llites of carbon, are present in the exhaust from diesel trucks and coal-fired power plants. Most other fine particulates in the atmosphere consist of inorganic sulfur and nitrogen compounds. Sulfur compo unds originate as sulfur dioxide, SOb a gas produced naturally by volcanoes as well as by coal-fired power plants, metal smelters, and, in some parts of the world, by cars. Sulfur dioxide is oxidized in air to yield gaseous sulfur trioxide, 503, in a process catalyzed by particulates. The trioxide then reacts wit h water to form sulfu ric acid, H2S04, which t ravels in the air as an aerosol of fine droplets. Nitric acid can be formed by the oxidation of nitrogen-containing compounds such as NO (page 991) and ammonia. Automobile exhaust gases provide a major source of NO, which forms in a reaction between N2 and O2 at the high temperatures reached in an internal combustion engine. Sulfuric and nitric acids react with ammon ia in the atmosphere to produce particles of ammonium sulfate or ammonium nitrate, respectively. Acommon measurement of atmospheric particulates is the particulate matt er or PM index, which is defined as the mass of particulate matt er in a given volumeof air, usually micrograms per cubic meter (pq/m ' ). Sma ller
particulates have a greate r det rimentaleffect on human health t han do larger ones. Therefore, onlythose particulates having a diameter smatter than a given value are measured. This cut-off diameter is reported as a subscript to the PMsym bol. In the past. the U.S. has monitored particulates having diameters less t han 10 u rn (PM IO) . Monitori ng of even smaller particulates, such as those with PM Z•5, has begun more recently. Particulates can have several effects, the most obvious of which is a decrease in visibility. Particles wi th a diameter near the wavelength of visible light, 0.3 to 0.8 u m (300-800 nm),
can interfere with the transmission of light through air or scatter the light, which reduces the amount of light reaching the ground. Particulates also provide active surfaces for chemical reactions, suchas t he oxidation of 502, Finally, particulates provide a surface for the condensation of water vapor,thereby exerting a significa nt influence on weather. Particulate air pollution has caused health problems for centuries. Ma nystudies have linked breathing particulate matter to such health problems as aggravated asthma,
increases in respiratory symptoms such as coughing, difficu lt or painful breathing, chronic bronchitis, and decreased lung function . Particles smaller than 10 urn in diameter are considered inhalable particulates because they may enter our respiratory system when we breathe. However, particles larger than z.s um in diameter are efficiently filtered by the cilia and generally do not enter the lungs. These larger particles can still pose a problem, however, because 50 2, produced along with particulates from burning coal, paralyzes the cilia in the respiratory tract and thus increases the damage caused by particulates. Respirable particulates (those sma ller than 2.5 11 m in diameter, PM 2.5 ) seem to be responsible for most health problems. Indeed, PM Z•5 correlates most strongly with increases in the rate of disease or morta lityin most regions. Particulate air pollution levels were recently related to morta lity across 151 metropolitan areas in the U.S. for the period 1982-1985. This study of 500,000 people found mortality rates (deaths/yr/l,OOO,OOO people) to be cor-
related to both sulfate particulate concentrations and to PM 2.5•
Haze. A high particulate concentration produces haze. This viewof the Taj Mahalin India is partly obscured by haze from natural and human sources.
Study Qu e stion s
I
961
3 , \ Vhat mass o f NaCI co u ld he ob tai ned byevaporatin g 1.0 I. of seaw ater? (No te : th e a m o u n t of Na + io ns ill seawater lim its th e a m o u n t o f :\aC I th at ca n be ohrained ; Sl'e..' Q uesti o n I.) 4. In a n y so lu tio n , the-re m ust he a bal a nce between the positive a nd nega tive cha rge s of th e io ns tha t are p re ~ cn t. Usc the d a ta in Table 2 10 determine ho w close to a balance of cha rge is ac hieved for the ions in seawa ter; (T h e table lisls only the major ions in seawater and the bal ance between positive a nd negative charge will only be approxim ate .]
j
j
j "0
i Figure 9 Green chemist ry. The American Che mical Society has bee n actively encouraging programs in green chemistry, See wwwacs.orq.
many differe nt opinions concerni ng ch emist ry and che micals, Some use th e word "chem ica l" as a synon ym for a toxic substa nce. O th e rs believe that a ll our proble ms ca ll he so lved th rough resea rc h in c he mist ry a nd o ther sciences, C hemists realize th a t a ll matter. incl ud in g o ur foo d , wa ter, am i bodies as well as to x ic com poun ds, is com posed of c he m ica ls, T he stu dy a nd pra ct ice o f ch emis try. li ke m o st asp ec ts of life , h a ve crea te d p rob lem s at t imes , bu t che mistry h as a lso so lved m a ny, m a llY 1II0 re .
SUGGESTED READINGS 1. 2.
3,
4. 5. 6.
.J.
Emslcy, Natu re's Building motk~. En g la nd: Ox fo rd Universit y Pre ss, 200 I. S, Ma na han , ed ito r: Enmnnunenta! Chf'llli.\(1Y, Srh ed . Bo ca Ra ton: CRe Press, Flo rida, 20(H . Sa fe Drinkin g Wal e r Act, in clud in g a list o f cu m-r u ma xim u m co n tami nant level s: vlww.qla .gov /safl'W 2 H, (g) + O,(g) Any sam pic of nat ural wate r a lways co n tai ns a tin y co nce n tratio n o f [)20. Wh en e lectrolyzed , 1120 is elec tro lyzed mo re rapidly th a n DiO . Thus, as the elec trol ysis p ro ceeds, th e liquid remaining is e nric hed ill D 2 0 . Re pe atin g this process man y tim es will even tua lly give purc 0.;.10, often ca lled "heavy wate r." Large am o u n ts o f D 2 0 are now produced because th is compou nd is u sed as a mode rato r in some nu clear reactors th at are use d fo r pm ve r genc ra tion . Hydrogen co mbi ne s chem ica lly with virtu ally cvcry o ther c lem en t, cxcc p t th e noble gases. There are th ree di fferent types o f bin ary hydrogen-con tainin g co mpoun d s. Ion ic metal tiydvides arc form ed in th c reaction of H 2 wit h a Gro u p l A o r 2A me tal. 2 Na (s)
+ H,(g) --> 2 Na H(s)
FIGURE 21.4 ElectroLysis of water.
Electrolysis of water (containing dilute H2S04 as an elect rolyte) gives O2 (left) and H2 (right).
Ca(s) + H,(g) --> CaH,(s) Th ese co mpo u nd s co n tain the h ydrid e io n, 11- , in whic h hydrogen has a - I o xida tion Humber. Molecular compoun ds (suc h as H 20 , H I', and N HC!) are ge ne ra lly form cd by d irect com b inatio n of hydrogen with no nm etallic e leme nts (Figu re 2 1.Pi ). The oxi d a tio n n u m ber o f th e hydrogen ato m in th ese co mpou nds is + I, b u t cova le n t bonds to hydrogen are th e rul e .
N,(g )
+ 3 H,(g) --> 2 NH ,(g)
F,(g)
+ H,(g)
--> 2 HF(g)
Hydrogen is ab sorbed by ma ny metal s to form in terstitial hyd rides, th e third general class of hydroge n co mpoun ds. Th is name refe rs to the str uctu res o f th ese species, in which th e hydrogen atoms resid e in th e spaces bet ween the me tal at oms (called interstices) in th e crystal latti ce. Palladium metal, for exam pic , ca n soa k up 1000 tim es its vo lu me o f hydroge n (at STP ) . Most interstitial hydride s are non- sto ich iom etr ic; that is, th e ra tio of met al an d hydrogen is no t a who le Humber. \Vhen interstitial metal hydrides arc heated, H 2 is driven o u t. This phenomen on allows th ese rnatcrials to store 11 2 ,just as a sponge ca n store water. It suggests o ne way to store hydro gen for usc as a fuel ill auto mo biles (... page 264 ) .
Preparation of Hydrogen Abo u t 300 billio n liters (ST P) of hydrogcn gas are produced an n ua lly wo rldwide, an d virt ually all is used immedi ately in the manufactu re of am mo nia (... Sect ion 2 1.8) , me th anol (... Sec tio n 10.3) , or o ther chem icals. Some hydro gen is m ad e from coal and steam, a reac tio n th at has been used fo r mo re than 100 years.
C(s) + H,O(g) -----> H, (g)
+ CO (g)
'-----------v--
water gas or synthesis gas
:,.,w = + 131 kJ/ mol-rxn FIGURE 21.5 The reaction of
Th e reactio n is carried o u t by injecting water in to a bed o f re d-ho t coke. Th e mi xtu re o f gases produced . ca lled umler gas o r synthesis gas, was use d until abo u t 1950 as a fuel for co o king, heatin g. and lighting. However, it has ser io us drawbacks. It
H2 and Br2' Hydrogen gas burns in an atmosphere of bromine vapor to give hydrogen bromide. 21.3
I
Hydrogen
969
p ro duces on ly abo u t half as mu ch heat as all eq ual am o u nt of m eth an e does, an d th e flam e is nearl y invisibl e . Mo reove r, becau se it co n ta ins ca r bo n mo no xid e , wate r gas is to xic. The largest quant ity of hydrogen is no w produced by th e cataly tic steam ref ormalion of hsdvocarbons su ch as m etha ne in natu ral ga s (F igure 21.6). Meth an e react s with steam at high tempera ture to give 112 an d CO . CH,(g)
+ H,O(g) ->
3 H, (g)
+ CO(g)
tl ,W
= + 206 kJj mol-rxn
Th e reactio n is rapid at 90 0- 1000 "C a nd go es nearly to co mple tio n . Mo re h ydr oge n call be obtain ed in a seco nd ste p in whi ch the CO form ed in th e first ste p reacts with mo re wate r. T h is so-ca lled water gas sh ~jt reaction is run at 400-500 °C and is sligh tly exothermic. H,O(g)
F IGURE 21.6 Production of water gas. Water gas, also called synthesis gas, is a mixture of CO and H2• It is produced by treating coal, coke. or a hydrocarbon like methane wit h steam at high temperatures in plants such as that pictured here. Methane has t he advantage t hat it gives more total Hz per gram than other hydrocarbons, and the ratio of t he by-product CO 2 to H2 is lower.
+ CO(g) -> H, (g) + CO,(g)
tl ,W
=-
41 kJjm ol-rxn
Th e CO 2 fo rmed in th e process is rem oved by re act ion with CaO (to give solid CaC O j ) , leaving fairl y pure hydrogen . Pe rh ap s the clea ne st way to mak e hydrogen on a rel ati vely large scale is th e e lectrolysis o f water (Figure 2 1.4) . Th is ap p ro ac h provides no t o nly hydroge n gas but also h igh-purit y O 2 , Beca use el ect ricity is qui te ex pe nsive , howeve r, thi s m ethod is not ge ne ra lly used co m me rc ially. Tabl e 2 1.4 and Figu re 2 1.7 give ex amples of react ions u sed to p roduce 1-12 gas in th e lab ora tory. The mo st often used method is the reactio n of a met al with an acid . Alternatively, th e reaction of alu m inu m with aq ue ous NaO H (Fig ure 2 1.7b) also ge nerates hyd roge n. During \ Vorl d \ Var II, this reaction was used to o b tain hydrogen to inflat e small balloons for wea the r observa tion an d to ra ise radio an ten nas. Met allic aluminum was plentiful at th e tim e becau se it ca me from dam aged airc ra ft. The co mb inatio n of a met al h ydride and water (Figure 2 1.7c) is an efficien t but expensive way to syn thesize 11 2 in th e labo rato ry. The reaction is more commo n ly used in laboratories to d ry o rgan ic so lven ts becau se th e metal hydride reacts with traces o f water presen t in the so lvent. EXERCISE 21.5 Hydrogen Ch emi st ry
Use bond enthalptes (page 389) to estimate t he enthalpy change for th e reactio n of methane and water to give hydrogen and carbon monoxide (with all compounds in t he gas phase).
FIGURE 21.7 Produdng hydrogen gas.
(a) The reaction of magnesium and add. The products are hydrogen gas and a magnesium salt. 9 70
Chapter 21
I
(b) The reaction of aluminum and NaOH . The productsof this reaction are hydrogen gas and a solution of Na[A I(OH), ].
The Chemist ry of t he Main Group Elements
(c) The reaction of CaHz and water. The products are hydroge n gas and Ca(OH )z.
TABLE 21.4 Methods for Preparing H, in the Laboratory 1. Metal + Acid
~
metal salt + Hz
Mg (s) + 2 HCI(aq) --+ MgCl,(aq) + H,(g) 2. Metal + H20
~
metal hydroxide or oxide + Hz
2 Na(s) + 2 H,O« () --+ 2 NaO H(aq) + H, (g) 2 Fe(s) + 3 H,O(t ) --+ Fe, O,(s) + 3 H, (g) 2 AI(s) + 2 KO H(aq) + 6 H,O«( ) --+ 2 K[Al(O H) ,](aq) + 3 H, (g)
3. Metal hydride + H20
~
meta l hydroxide + Hz
Ca H, (s) + 2 H,O(!') --+ Ca(OH),( s) + 2 H,(g)
21.4
The Alkali Metals, Group lA
Sodium a n d potassium a rc, respectively, th e six th a n d eigh th most a bu n d a n t elements in the ea rth 's cr ust by mass. In co n trast, lithium is rel at ivel y ra re , as a rc r ub id iu m a n d ce siu m . Only traces of radioacti ve fran cium o ccur in n ature . Its lo ngest-lived isotope (22:iFr) has a ha lf-life o f only 22 minutes. The Grou p IA ele ments are metals, a n d all are highl y reactive wit h o xygen , wate r, an d o ther o xidi zin g age n ts (see Figure 7.] 3 , p age 32 9) . In all cases, compounds o f th e Gro u p l A met als co n ta in th e element as a 1 + ion . T h e free metal is n eve r found in n ature. Mo st sod iu m a n d potassium co m p o u n ds a re wate r-sol u b le (.... so lu b ility g u idelin es, Fig u re 3 .10 ) , so it is n ot su rp ris ing th at so d iu m a n d potassium co m pou nds are fou nd eith er in th e o cean s or in un d e rgr ound d eposits th at a re the resid u e of a nc ie n t seas . To a mu ch smalle r exte n t, th ese e leme n ts are also found in mi n erals, suc h as C h ilea n saltpe te r (NaNO J) . Despit e th e fact th at so d iu m is on ly sligh tly mo re a bu n dan t th a n potassium o n t he earth , sea water co n ta ins sign ifican tly mo re so d iu m th an potassium (2 .8 % NaC I versu s 0.8 % KCI) . \Vhy the grea t di ffe rence? Most co m po u n ds of both clements are wate r-so lu b le , so wh y d idn 't rai n diss o lve Na - a n d K-containin g m in e rals ove r t he ce n tu ri es a n d carry th em down to the sea, so th at the y a p pea r in t he same p roportions in the ocean s as o n land? T he a nswe r lies in th e fac t th at potassium is a n impo rt ant facto r in plan t gro wth . Most plants co n ta in four to six times as m uc h com b ine d p ot assium as sodiu m . T hus, m ost of th e potassiu m ions in g ro un dwa rer from di sso lved mine ral s a re taken u p p referentially by p lants, wherea s sod iu m salts continue o n to th e oc ean s. (Bec a use plan ts re q uire p o tassiu m, co m mercial fertilizers usu ally co ntai n a sig n ific a n t a mo un t of potassium salts.) Some NaCI is essential in th e di e t o f h u ma ns a n d ot her a n ima ls because man y b io lo gical fu nc tions a re co n tro lled by th e concen tra tio ns o f Na " a n d CI- ion s (Fig u re 2 1.8) . The fact th a t sa lt h as lo n g bee n recog n ized as impo r tan t is evident in su rp rising lvays. Fo r ex am p le , we a re paid a "sa la ry" for wo rk d one . T h is wo rd is d e rived from the La tin salarium, whic h mean t "sa lt money" because Rom a n sold ie rs were p aid in salt.
FIGURE 21.8 The imp ort ance of salt. All animals, including humans, need a certain amount of salt in t heir diet. Sodium ions are important in maintaining electrolyte balanceand in regulating osmotic pressure. For an interesting account of the importance of salt in sodety, culture, history, and economy, see Salt, A World History, by M. Kurlansky, NewYork, Penguin Books, 2002.
Group lA Alkali metals Lithium
3
Li 20 ppm Sodium 11
Na 23,600 ppm Potassium 19
K 21,000 ppm Rubidium
37
Rb 90 ppm Cesium
55
Cs 0.0003 ppm Francium 87
Fr trace
Preparation of Sodium and Potassium
Element abundances are in parts per mi llion in the earth's crust.
Sodi um is p rod uced by red ucing so d ium ions in sod iu m salts. Howeve r, com m o n che mi ca l re d uci ng age n ts are no t powerfu l e nough to co uvert so d iu m io ns to sod iu m me tal. Bec ause of this, th e metals a re usu a lly p repared by el ectrolysis. 2 1.4
I
The Alkali Metals, Group lA
971
-
FIGURE 21.9. A Downs cell for pre-
paring sodium. Acirculariron cathode is separated from the graphite anode by an iron screen. At the temperature of the electrolysis. about 600 0(, sodium is a liquid. It floats to the top and is drawn off periodically. Chlorine gas is produced at the anode and collected inside the inverted cone in t he center of the cell.
Cl 2 output
liquid Na metal
-
Na outlet
Iron screen
Cathode ( -)
The Engli sh che m ist Sir IIu mph ry Davy first isolat ed sod iu m in 1807 by th e e lec tro lysis of m ol ten so d iu m carbonate . H owever, the e lem en t re maine d a laborato ry cu riosity until 1824, whe n it was fo u nd so d iu m co uld be used to reduce alumi nu m ch lo ride to alu m in u m met al. At th at time , me tall ic alu m in u m was rare and very valua ble, so thi s d iscove ry inspir ed consid e ra ble in terest ill m an ufactu ri n g sod ium. By 1886, a p ra ctica l metho d of so d iu m prod uctio n had bee n devis ed (the red uc tion of NaO I I with carbo n) . U n fo rt u nate ly for so d iu m p roducers, in this same yea r Cha rles l lall and Pa ul l le roult inven ted th e electrol ytic me tho d fo r alu m in u m prod uc tio n (... page 980 ), th ere by el iminatin g th is ma rket for sod iu m . Sodiu m is cu rre n tly p roduced by the elec trolysis of mol ten NaC I (... Section 20.7) . T he Downs ce ll fo r the electro lysis of molt en NaC I operates at 7 to 8 V with cu rre n ts o f 25,000 to 40,000 amps (Figu re 21.9) . T he cell is filled with a mixture of d ry 1\aCI, CaC I~, and BaCI2 • Addi ng o ther salts to NaCl lowers the meltin g poi n t fro m that of pu re NaC I (800.7 "C) to abo u t 600 "C. [Recall that so lu tio ns have lower melting poin ts th an pure solven ts (Cha pte r 14) .] Sod iu m is produced at a cop pe r or iro n ca thode that su rr ou nd s a circu lar gra p hit e anod e. Directly ove r the ca thod e is an inverted tro ug h in which the low-density, mo lten sod ium (me lting point, 97.8 "C) co llects. Chlo rine , a valuabl e by-pro duct , co llec ts at th e anode. Po ta ssiu m can also be made by el ect ro lysis. Mo lte n po tassium is so luble in mo lten KCI, however, makin g sepa ra tio n of the metal d ifficu lt. Th e p referred m eth od for preparat io n o f po tassiu m uses th e reactio n o f sod iu m \"apo r with molten Ke l, with potassium be ing co n tin ually removed fro m the equi lib riu m m ixture . • Alkali Metals React with WaterThermodynamics and Kinetics The alkali metals all react vigorously with water, and it is easily observed that the violence of t he reaction increases with atomic number. This seems counter to the argument presented in A Closer Look that lithium is the best reducing agent. However, the reducing ability of a metal is a t hermodynamic property, whereas the violence of the reaction is mainly a consequence of reaction rate. 972
Chapter
21
I
Na(g)
+
KCl( 1:),
' K(g)
+ NaCl(f )
Properties of Sodium and Potassium So d iu m and potassiu m arc silve ry met als th at arc so ft and easily cu t with a knife (see Figu re 2.6). Th ey are just a bit less d e nse th a n wate r. Th ei r meltin g po int s are qui te low, 9 7.8 "C for sodiu m and 63.7 "C for potassium . All of th e a lkali me tals arc highl y reac tive. when expo sed to mo ist air, th e met al surface qu ickly becomes coa ted with a film of ox id e or hyd ro xide . Co nseque n tly, thc me tals must be sto re d in a way th at avoids co n tac t with a ir, typicall y by pl acing th cm in ke ro sen e o r mine ral o il.
The Chemist ry of t he Main Group Elements
A Closer Look
The Reducing Ability of the Alkali Metals
The uses of the Group lA metals depend on their reducing ability. The values of EO reveal that Li is the best reducing agent in the group, whereas Na is the poorest; the rema inder of these metals have roughly compa rable reducing ability.
Li'
+ e: -e li + e: -; Na
-----=="'-----., W (aq) + e-
- 2,7 14
- 2.925 - 2.925
-7
(s
mic. From Hess's law (page 233), we know that the overall energy change should be
- 3.045
Rb+ + e: -; Rb
+ e-
F"
While this ana Lysis of the problem gives us a reasonable explanation for the great reducing ability of lithium, recall that E" is directly related to :1 ,Go and not to :1 W . However, .:lrGo is largely determined by :1rW , so it is possible to relate variations in EOto variations in !irHo.
Eo (V)
K+ +e --;K
(5 +
i:~
M(s)
Reduction Potential Element
Na'
IE (ionization energy)
Do rHoet = Do subH + IE
- 2.92
Analysis of EOis a thermodynamic problem. and to understand i t better we can break the process of metal oxidation, M(s) -7 M+(aq) + e", into a series of steps. Here, we imagine that the metal sublimes to vapor, an electron is removed to form the gaseous cation, and the cation is hydrated. The first two steps require energy, but the last is exother-
+
Do hydH
The element that is t he best reducing agent should have the most negative (or least positive) value of :1 H. That is, the best reducing agent should be the metal that has t he most exothermic value for its hyd ration energy because this can offset the energy of the endothermic steps (!isubH and IE). Forthe alkali metals, enthalpies of hydration range from - 506 kJjmol for Li" to - 180 kJjmot for Cs" . The fact that !ilrydH is so much greater for Li" than for Cs" largely accounts for the differencein reducing ability.
Potassium is a very good redudng agent and reacts vigorously with water.
The high reacti vity o f Group IA metals is ex e m plified by th cir re action with wate r, wh ich ge ne ra tes a n aq ueo us so lu tio n of th e me ta l hydroxide a n d hydrogen gas (F igu re 7.13, page 329) , 2 Na(s) + 2 H,O(f ) -> 2 Na +(aq) + 2 OW(aq) + H,(g) a n d th ei r reactio n with an y o f th e h alo ge ns to yiel d a metal h al ide (Figu re 1.4 ) , 2 Na(s) + Cl,(g) -> 2 NaCI(s) 2 K(s)
+ Br,(f ) -> 2 KB r(s)
C h e m istry ofte n p rodu ce s su r pr ises. Group l A m et al ox ides, ~1 20 , a rc kn own , but th ey a rc not th e princip al product s o f reacti on s bet ween th e Group 11\ clements a n d oxygen . Instead, th e primary product o f th e reaction of so d iu m a n d o xygen is so d iu m peroxide, Na 202, whe reas th c principal prod u ct from th e reacti on of potassiu m and oxygen is K0 2 , potassium supemxide. 2 Na(s) K(s)
+ O,(g) -> Na,O,(5) + O,( g) -> KO,( s)
Both Na20 2 a n d K0 2 a re ionic co m pou n ds. The Group 11\ catio n is pa ired with either th e p e roxid e io n (° 22 - ) o r th e supe roxid e io n (0 2 - ) , Th ese co m po un ds are not merely laboratory cu riosities . The y a re used in oxyge n ge neratio n d evices in places whe re people are co n fin cd, suc h as su b mari nes, airc ra ft, and spacec ra ft, o r when a n eme rge ncy su p p ly is needed. Wh en a perso n b reath es, 0.82 L o f CO 2 is ex ha led for every I L o f O 2 inh aled . T h us, a requirement o f an O 2 ge ne ratio n syste m is th at it shou ld produce a la rger volu me o f O 2 than the vo lu me o f CO 2
FIGURE 21.10. Ac1osed -drcuit breathing apparatus that generates its own oxygen. One source of oxygen is potassium superoxide (K0 2) . Both carbon dioxide and moisture exhaled bythe wea rer into the breathing tube react with the KO l to generate oxygen. Because the rate of the chemical reaction is determined bythe quantity of moisture and carbon dioxide exhaled, the production of oxygen is regulated automatically. With each exhalation, more oxygen is produced byvolume than is required bythe user.
21.4
I
The Alkali Metals, Group l A 973
taken in . This requirem ent is met with supero xides (Figu re 21.10) . With K0 2 th e re action is
4 KO, (5 ) + 2 (O,(g) --> 2 K,CO,( 5) + 3 O,(g)
Important Lithium, Sodium, and Potassium Compounds Elec tro lysis o f aq ueo us so d iu m ch lo rid e (b rine) is th e oasis of one o f the largest chemical industries in th e Un ited Sta tes.
2 Na Cl(aq) + 2 H,O( C) --> Cl,(g) + 2 NaOH(aq) + H,(g) Two of the product s fro m thi s process-chl orine and so d iu m hvdro xid c-c-givc th e ind ustr y its name : the chlor-alknli industry. Mo re th an 10 billion kilogr am s of CI2 and Na O J I is produced annually ill th e Un ited States. Sodiu m carbo nate , Na2CO J, is ano ther co mmercially im portant co m pou nd of sodiu m. It is also know n by two co mmo n nam es, soda ash. a nd washing soda. In th e pa st, it was largely man ufactured by co m bining NaCl, am mo nia, and CO 2 in th e Solvay jnoress (which rem ain s th c method o f choice in many cou n tr ies) . In th e Un ited Stat es, however, sod ium carbo nate is ob taine d fro m naturally OCCUlTing deposits of th e m ineral trona, Na, CO , · Nail CO " . 2 11 ,0 (Figu re 21.1 1). Owin g to th e en vironmental problem s assoc iate d with the chlo r-alka li pro cess, co nsid era ble in te rest has arise n in the po ssibili ty of manu facturin g sodium h ydro xide by o ther methods. This has led to a revival of th e o ld "soda- lime process," which p roduces Na O H from inex pens ive lime (CaO ) an d soda ash (Na2C0 3)'
Na ,CO,(a q) + ( aO(5) + H,O( C) --> 2 NaOH(aq) + (a (O,(5) T he inso lubl e ca lciu m ca rbonate by-produ ct is filtered off, thcn heated (ca lcining) to co nvert it to lim e , whic h is rec ycle d in to th e reaction syste m.
(a ( O,(5) --> ( aO(5) + ( O,(g) So di u m bicarbon at e , NaJICO J, also kn own as baking soda, is ano ther co m mo n com po un d of so d iu m . No t on ly is Na J-IC O :~ used in coo king , but it is also ad de d in sm all amou n ts to tabl e salt. NaC I is oft en co nt ami nated with sm all am ounts o f (\1gCI2 • The magnesium salt is h ygroscopic; th at is, it p icks water up from th e air
FIGURE 2 1.11 Producing soda ash. Trona mined in Wyo ming and Catifornia is processed into soda ash (NazCO]) and other sodium-based chemicals . Soda ash is the ninth most widely used chemical in the United States. Domestica lly, about half of all soda ash production is used in making glass. The remainder goes to make chemicals such as sodium silicate, sodium phosphate. and sodium cyanide. Someis also used to make detergents, in the pulp and paper industry, and in water treatment.
(a) (Above) Aminein California. The mi neral trona is ta ken from a mine 1600 feet deep.
(b) (Right) Blocks of trona are cut from the face of the mine. 9 74
Chapt er
21
I
The Chemist ry of t he Main Group Elements
and , in d o ing so , ca uses the NaCI to clump . Ad d ing Na IlCO:! co nve rts magnesium car bo nate , a non hyg ro sco pic salt.
~ tgC l 2
to
MgCl, (s) + 2 Na HCO,(s) --> MgCO, (s) + 2 NaCl(s) + H,O(f) + CO,(g) Large depo sits o f sodi u m n itrate , NaN O :l , are fo u nd in Chi le, whi ch ex p lains its com mo n nam e of "Chi le saltpe ter." T hese d epo sits are tho ng h t to have fo rm ed by bacte rial ac tio n on o rgan isms in sha llow seas. The in itial prod uct was ammon ia , which was su bseq ue n tly oxidized to nit rat e ion ; co m bina tio n with sea salt led to sodi um ni trate . Becau se nitrat es in ge neral, and alka li metal n itrat es in part icu lar, are h igh ly wate r-solu ble , d epo sits o f NaN0 3 are found o n ly in areas with very litt le rainfall. Sodi um nit r at e is im portant becau se it ca n be co nverted to potassium nit rat e by an exc ha nge reac tio n .
NaNO,(aq) + KCl(aq)
KNO,(a q) + NaCl(s)
Eq uilibriu m favors th e products here becau se , o f the fo u r salts invol ved in th is reac tio n , NaC I is least so lu ble in hot watc r. Sodium ch loride precipitates. and the KN0 3 that remains in so lu tio n ca n be recovered by evaporating th e water. Po tassiu m nitrate has be en used for ce n tu ries as th e o xid izing age n t in gu n powder. A m ixtu re o f KJ'\JO:\o cha rcoa l, and sulfu r will spo nt aneo usly react whe n ignited . 2 KNO, (s)
+ 4 C(s) --> K,CO,(s) + 3 CO(g) + N,(g)
2 KN O,(s)
+ 2 5(s) --> K,50,(s) + 50,(g) + N,(g)
No tice tha t both reactio ns (wh ich arc dou b tless more co mp lex th an those written here ) produce ga ses. These gases p ropel the b ull e t from a gu n o r ca use a firecrac ker to ex plo de. Lithium ca rbonate, Li2C0 3 , has been used fo r mo re than 40 yea rs as a tre atme nt fo r bip olar di so rder, an illness th a t in volves alte rnating periods o f d epressio n or overexcit em ent th at ca n ex te nd over a few weeks to a year or mo re . Altho ug h the alka li met al salt is efficien t in co n tro lling th e sym p to ms of bipo lar di sorder, its mech an ism of actio n is not understood.
Group 2A
Alkaline earths
,
BeryLLi um
Be 2.6 ppm Magnesium
EXERCISE 2 1.6
Brine Electrolysis
12
Mg
What current must be used in a Downs celL operating at 7.0 Vto produce 1.00 metric ton (exactly 1000 kg) of sodium per day? Assume 100% efficiency.
23,300 ppm Calcium 20
Ca
21 .5
41,000 ppm
The Alkaline Earth Elements, Group 2A
The "earth " part of the name alka line ea rth dat es back to the d ays of med ieval alc he my. To alchemists, any so lid th at did no t melt and was not cha ng ed by fire into anot her su bsta nce was called an "eart h ." Co m po u nds of the Grou p 2A eleme n ts, such Ca O , were alka line acc ording to ex perime n ta l tests co nd uc ted by the alche mi sts: th ey had a bitter taste a nd neu tral ized ac id s. \Vith vcry hig h melting po in ts, th ese co m pou nds were unaffected by fire . Ca lciu m an d m agn esium ran k fifth and eigh th, res pec tively, in abu nda nce on the ea rth. Both c le men ts form m an y co m me rc ia lly important co m po u nds, an d we sha ll fo cu s our anen tio n on th ese species. Like the Gr o up lA el emen ts, the Gro up 2A e leme nts are very reac tive, so th ey are found in na tu re as co mpo u nds. Un like mo st of the com poun ds o f th e Gr o u p l A me ta ls, ho weve r, man y com pou nds of th e Gro u p 2A elemen ts have lo w water so lu bility, which ex p lains th ei r occurren ce in vario us mine rals (Figure 2 1. 12) . 21.5
I
Strontium 38
Sr 370 ppm Barium 56
Ba 500 ppm Radium 88
Ra
6 x 10- 7 ppm
Element abundances are in parts per mi llion in the earth's crust.
The Alkaline Eart h Elements, Gro up 2A
975
limestone CaC0 3 ,.
Common minerals of Group 2A elements.
The walls of the Grand Canyon in Arizona are Icelandicspar. This minera l. one of a number of crystalline forms of CaC0 3 , displays birefrinlargely limestone or dolomite. gence, a propertyin which a doubleimageis formed when light passes throughthe crystal. FIGURE 21.12 Various minerals containing calcium and magnesium.
Calciu m m ine ra ls include lim estone (CaCO j ), bTypsu m (Ca SO I' 2 H 20 ) , and fluorite (CaF,) . Mag nesite (Mg CO ,), tal c or so ap sto ne (3 MgO . 4 5iO, . Il,O) , and asbesto s (3 ~t gO . 4 SiO z . 2 H 20 ) a re co m mo n m agnesium-containi ng mi ne ra ls. Th e min eral d ol o m ite, I\tgCa (CO j h , contain s both magnesium and ca lciu m . Lim eston e , a sedi me ntary rock, is found wide ly o n th e earth 's su rface . Man v of these d e posits co ntain tile fossilized remains of marin e life . Other fo rm s of calciu m ca rb on at e include marbl e an d Icel andi c spar, th e latte r o cc u rr ing as large , clear cr ystals with the in terestin g o p tica l property of birefringen ce .
Properties of Calcium and Magnesium Calciu m and. magn esium are fairly high-melting, silvery metals. The che m ica l propert ies of th ese e lem en ts present few surp rises. Th ey are ox id ized by a wid e ran ge of ox id izing ag en ts to fo rm ionic co m poun ds that co n ta in th e ~1 2 + ion . Fo r example, these elements co m bine with halogen s to fo r m ]\iIX2 , with oxygen or su lfur to form M'O o r ~:IS , and with wate r to form hydrogen and th e metal h ydro xid e , M (O il ), (Figu re 2 1.13) . With acids , hydrogen is evo lved (see Figu re 2 1.6) , and a salt o f th e meta l ca tio n and th e an io n of th e aci d results.
Metallurgy of Magnesium
FIGURE 21.13 The reaction of calcium and warm water. Hydrogen bubbles are seen rising from t he metal surface. The other reaction product is Ca(OHh. The inset is a model of hexagonal closepacked calcium metal (see page 595).
9 76
C h a p t e r 21
I
Several h undred thousand tons of magn esium are produced ann ually, largely fo r lise in ligh tweight alloys. (Mag nesiu m ha s a very low d en sity, 1.74 g /cm J . ) Mo st alu min u m used today co n ta ins abo ut 5 % magn esium to imp rove its mechan ical p roperties an d to mak e it more re sistan t to co rrosio n . Othe r alloys having mo re magn esium th an alu m in u m arc used whe n a high stre ng th-to-we igh t ratio is needed and whe n corrosio n resistan ce is import an t, suc h as in aircra ft and auto motive parts and in lightweight tools. Interestin gly, magn esium- contain in g minerals are not th e sou rce of thi s element. Most magn esium is obtain ed from sea water, in whic h (\[g2+ io n is prese nt in a co nce n tra tio n of abo u t 0.05 1\.1. To o b tain magn esium metal , magn esium ions in sea wate r are fir st precipitat ed (Fig u re 21.14) as th e rel ati vely insoluble hydrox ide [ K,p fo r Mg (OIl J, = 5.6 X 10- 12] . Calciu m h ydr o xid e, the source o f 011 - in this reactio n, is p re pared in a seq ue nce o f reac tio ns beginn ing with CaC O:1> wh ich may be in th e for m of seashe lls. H eatin g CaCO J gives CO 2 a nd Ca O , and ad d itio n o f
The Chemist ry of t he Main Group Elements
Oyster shells CaC0 3
Producing Magnesium from Sea Water and Seashells
FIGURE 21.14 The process used to produce magnesium metal from the magnesium in sea water.
Ocean water intake
j Lime kilns CaC0 3 CaO + COz
j
Precipitate Mg(OH h
Staker CaO + HzO _ Ca(OH )z
-
MgCl z + Ca(OH h -
Mg(OH)z + CaClz - - -
Strainers
I Settling tank I Fitter
!
Co nvert Mg(OH h to MgCl z MgC lz + 2HzO
Mg(OH)z + 2HCl _
Evaporators
j
j
Convert MgCl z to Mg by Electrolysis MgClz _ Mg + Clz
Dryers
HCl
Hydrochloric acid plant
~
J Mg
water 10 Ca O gives ca lciu m hydroxid e . Wh en Ca( O H h is added to sea wa te r, Mg (O I I), p recipitates: Mg' +(aq )
+ Ca(OH),(s)
==' Mg(OH ),(s) + Ca1+(aq)
Magnesium hyd roxide is iso lated by filt ration and th e n co nve rted to m agn esium ch loride by reaction with hydro ch lo ri c ac id. Mg (OH ),(s)
+ 2 HCl(aq) --> MgCl, (aq ) + 2 H,O(C)
Chemical Perspectives Plants and animals derive energy from the oxidation of a sugar, glucose, with oxygen. Plants are unique, howeve r, in being able to synthesize glucose from COz and H20 by using sunlight as an energy source. This process is initiated by chlorophyll, a very large, magnesium -based molecu le. In your body, the metal ions Na ". K+. Mg2+, and Ca 2 + serve regulatory functions. Although the two alka line earth metal ions are required by living systems, the other Group 2A elements are toxic. Beryllium compounds are carcinogenic, and soluble barium salts are poisons. You may be concerned if your physician asks you to drink a "barium cocktail" to check the condition of your digestive tract. Don't be afraid. because the "cocktail" contains very insoluble BaS04 (Ksp = 1.1 x 10- 10) , so it passes through your digestive tract without a signifi cant amount being absorbed. Barium sulfate is opaque to x-rays, so its path through your organs appears on the develo ped x-ray.
Alkaline Earth Metals and Biology
A molecule of chlorophyll. Magnesium is its central element.
The calcium -containing compound hydroxyapatite is the main component of tooth enamel. Cavities in your teeth form when acids (such as soft drinks) decompose the weakly basic hydroxyapatite coating. Ca,( OH)(PO,h(s) + 4 H,O' (aq) --> 5 Ca1+(aq) + J HPO/ - (aq) + 5 H20 (C )
21.5
X-ray of a gastrointestinal t ract using BaS04to make the organs visible.
This reaction can be prevented by converting hydroxya patite to the much more acidresistant coating of fluoroapatite. Ca, (OH)(PO,j,(s) + f -(a q) --> Ca,f(PO,j,(s) + OW(aq) The source ofthe fluoride ioncan be sodium fluoride or sodium monofluorop hosphate (Na 2FP0 3, common lyknown as MFP) in your toothpaste.
I
The Alkali ne Earth Elements. Group 2A 977
Afte r eva pora ting th e water, an hydro us magnesium chlori de rem ains. So lid ~ t gC l2 melts at 714 "C, and th e molten salt is e lectr o lyzed to give th e metal and chlorine. MgCl,(f) --> Mg(s) + Cl,(g)
CaLcium MineraLs and Their AppLications The mo st co m mo n calcium mi nerals are th e fluo rid e, ph o sph ate, and ca rbo nate salts of th e elemen t. Fluorite , Ca F2 , and Iluoroapatitc . Ca;,F( PO -th, arc important as co mmercial sou rces o f fluo rine . Almost half of the Ca P;! mi ne d is used in th e steel ind ustry, where it is adde d to the mi xture of materials th at is m elted to make cr ude iron. The CaF 2 acts to rem ove some im pu ri ties and imp roves th e separation o f molt en metal from silica tes an d o ther by-products resu lting from th e reduc tio n of iron ore to th e me tal (Chapter 22). A second major use o f fluorite is in the manu factu re of hydrofluori c ac id by a reaction of th e min eral with co ncen tra te d su lfu ric acid. (o F,(s) + H,50, (f ) --> 2 HF(g) + (050,(s) Hyd rofluori c ac id is used to mak e cryolite, Na;v\IF1i , a ma te ria l need ed in alu m in u m p roduct ion (... Section 2 1.6) a nd in the man ufac ture of fluorocarbon s suc h as tet ra flu oroethylene, th e p recu rsor to Teflo n (Table 10.12) . Apa tites have th e ge ne ra l formula Ca"X (l'O ,), (X = F, C l, OH ) . Mo re than 100 milli on to ns of apatite is m in ed an n ually, with Flo rida alo ne accou nting for abou t o ne th ird o f th e worl d 's out p ut. Most of thi s mat eri al is co nve rte d to phosphoric ac id by reac tio n with su lfu ric acid . Ph osphoric ac id is ne ed ed in th e ma n ufacture of a multitude o f p ro d ucts, includin g fert ilize rs and d et ergen ts, bak in g po wder, and vario us food p rodu cts (... Sect ion 2 1.8.) .
Chemical Perspectives
Of Romans, Limestone, and Champagne
The stones of the Appian Way in Italy, a road conceived by the Roman senate in about 310 B.C., are cemented with mortar made from limestone. The Appian Way was intended
to serve as a military road linking Rome to seaports from which soldiers could embark to Greece and other Mediterranean ports. The road stretches 560 kilometers (350 mites) from Rome to Brindisi on t he Adriatic Sea (at the heel of the Italian "boot"). It took almost 200 years to construct. The road had a standard width of 14 Roman feet, approximately 20 feet. large enough to allow two chariots to pass, and featured two sidewalks of 4 feet each. Every 10 miles or so, there were horsechanging stations with taverns, shops, and lotrinoe, the famous Roman restrooms. AUover the Roman Empire, buildings, temples, and aqueducts were constructed of blocks of limestone and marble. and the mortar to cement the blockswas made by heating the chips from stone cutting (to give CaD). In central France, the Romans dug chalk (also CaCD3 ) from the ground for cementing sandstone blocks. This activity created huge caves that remain to t his day and are used for aging and storing champagne.
The Appian Wayin Italy.
1 ~
j
S
~ =o~
-s. liiliOl:Ii.......;......;i;;;;iliiiiii Champagne in a limestone cave in France.
978
Chapter
21
I
The Chemistry of the Main Group Elements
Ca lcium carbonate and calciu m ox ide ( limt') are of special in terest . T Ile thermal d ecompo sitio n of CaC0 3 to give lim e (and CO 2 ) is o ne of the o ldest chemical reactio ns known . Lime is one of the top 10 ind ustrial che m ica ls produced today, with about 20 billio n kilograms prod uced an nually. Limeston e, which co nsists mostly of calciu m carbonate, has been used in agriculture for centuries. It is spread on fields to neutr alize acidic compounds in th e soil and to supply Ca 2 +,an essenti al nutrient. Becau se magnesiu m carbonate is often present in limestone, "liming" a field also supplies .Mg 2+, another important n utr ien t for plants . For seve ra l thousand years, lime has been used in mortar (a lime , sand, and water pa ste ) to secure sto nes to one another in b ui ldi ng houses, walls, and ro ad s. T he Chinese used it to set sto nes in th c Great Wall . T he Romans perfected its usc, and the fact that many of th eir co nstructio ns still stan d today is testamen t both to their skill and to the usefu lness of lime . T he famous Appian \Vay used lime mortar be tween several layers of its sto nes. T he u tility of mortar de pends on som e sim p le che m istry. Mo rtar consists of one pan lim e to th ree parts sand , with WaLeI' added to make a thick pa ste . T he first reac tio n , referred to as slaking, occurs after t he so lid s arc mi xed wit h water. T his produces a slurry co ntaining calcium hyd ro xide , whic h is known as slaked lime. CaO (s)
Apatit e. The mi neral has t he generat formula of Ca sX(P04h (X = F, Cl, OH). (The apatite is t he elongated crystal in the center of a matrix of other rock.)
• Dissolving limestone Figure 3.6 illust rates the equili brium involving (aC0 3, CO 2, H20, CaH , and HC0 3 - .
+ H,O(C) ~_ Ca(OH),(s)
Wh en the wet m orta r mix is placed bet ween bricks or sto ne blocks, it slowly reacts with CO 2 from the air, and the slaked lime is converted to calcium carbo nate . Ca(OH ),( s)
+ CO,(g)
CaCO,( s)
+ H,O(C)
The sand grains are bo und tog ether by the particles o f calciu m ca rbon at e . EXERCISE 2 1.7
Beryllium Chemistry
Beryllium, t he lightest element in Gro up 2A, has some important industrial applications. but exposure (by breathing) t o some of its compounds can cause berylliosis. Search t he Wo rld Wide Web for the uses of the element and t he causes and symptoms of berylliosis.
Group 3A
21.6 Boron, Aluminum, and the Group 3A Elements \Vit h Group 3A, we see th e first evide nce o f a chan ge fro m metallic be havior of the elem e nt s at th e left sid e of th e pe rio d ic table to no nmet al beh avior on th e rig ht side of th e tab le. Bo ron is a metall o id , whereas all th e other e lements o f Group 3A are metals. T he c leme nts of Gro up 3A vary widely in the ir rela tive abu ndances o n earth . Aluminum is the third most ab undan t element in the eart h 's cr ust (82 ,000 pp m ) , whereas th e othe r elements o f the grou p are re latively rare , and, ex ce p t for bo ro n , their co m po u nd s have limited co mme rc ia l uses.
Boron S
B 10 ppm Aluminum 13
Al 82.000 ppm Gallium 31
Ga 18 ppm
Chemistry of the Group 3A Elements It is ge nerally recogni zed that a chemical sim ilarity exists be tween so me eleme nts d iago nally situated in the periodic tabl e . This diagonal relatio nsh ip means that lithi u m and magnesiu m share so me chem ica l p ro perties, as d o Be a nd AI, and B and Si. Fo r ex ample:
• Boric o xid e , B~O :l> and bo ric aci d , B(O II h , are weakly acidic, as are Si0 2 an d its acid , onhosi lic aci d (I I.t Si0 4 ) . Boron-oxygen co m po u nds, borates, are often che mica lly simi lar to silico n- o xyge n compounds, silicates. • Both Be (OII )2 and AI(Oll h are am photeric, dis so lving in a str o ng base such as aq ueous Na O Il ( 5i(C) + 2 CO(g)
2.2 ppm
l ead 82
Pb 14 ppm
Element abundances are in parts per million in the earth's crust. 986
Chapt er
21
I
The mo lten silicon is drawn off th e b o tto m of th e fu rnace a nd allowed to cool to a sh in y b lue-gray so lid . Because extreme ly high-puri ty silicon is n eeded for the e lec tro n ics industry, pu r ifying raw silico n requires severa l steps. First, the silico n in the impure sample is allowed 10 reac t with ch lo ri ne to convert the silicon to liq uid silico n tetrac h loride,
The Chemistry of t he Main Group Elements
Si(s)
+ 2 Cl,(g) --> SiCl, (C)
Silicon tetrac hloride (boiling poin t o f 57 .6 °C) is ca refu lly p urifi ed by di stillati on and th en re d uced to silico n using m agnesiu m .
5iCl,(g) + 2 Mg(s) --> 2 MgCl,(s) + 5i(s) Th e magn esium chloride is washed o ut with water, and th e silico n is re mel ted an d cas t into ba rs. A fina l p urifica tio n is carried out by zo ne refi ning, a pro cess in whic h a specia l hea tin g device is used to melt a narrow seg me nt of the silico n rod . T he heat er is m oved slowly down the ro d . Impu rities co n ta ined in th e silicon ten d to re ma in in the liqui d p hase because the melting po in t of a m ixtu re is lower tha n that o f the pu re elemen t (Chap ter 14 ) . T ile silico n that crysta llizes a bove the heat ed zo ne is therefo re of a h igh er p uri ty (Figure 21 .19) .
Silicon Dioxide The sim plest oxide of silico n is Si0 2, com mon ly called silica, a co nstituent of man y rocks such as gra nite and sandsto ne. Q uartz is a pu re cr ystalline form o f silica , but im purities in quartz prod uce gemsto nes such as ame thyst (Figure 2 1.20). Silica and CO 2 are oxides of two elements in th e same che mical g rou p, so similari ties b e tvv'eeu them migh t be expec ted . In fact , Si0 2 is a high-mel ting so lid (q uartz melts at 1610 °C) , whereas CO 2 is a gas at ro om tcmpc rat u rc and 1 bar. This g reat disparity arises fro m the d iffe re n t str uctures of the two oxides. Carbon d io xid e is a molecul ar co m po u nd , wit h the carbon atom link ed to each ox yge n ato m by a dou ble bo nd . In contras t, Si0 2 is a netwo rk so lid , whic h is th e prefe rre d str uct ure becau se the bon d energy of two Si=O double bo nds is much less tha n the bo nd energy of four Si-O sing le bo nd s. Th e contrast be tween Si0 2 and CO 2 exe mpl ifies a more ge nera! ph en omeno n . Mult iple bonds, ofte n e ncoun tered between secon d-period eleme nts , are rare amo ng e leme nts in the thi rd and h igh e r pe riods. T he re are man y compounds with multipl e bo nd s to carbon b u t very few compo un ds fea tu r ing m u ltip le bo nd s to silico n . Q uartz crystals are used to control the fre q uen cy of radio and te levision transmi ssio ns. Because th ese and re la ted ap plicatio ns lise so m uc h quartz, the re is not enoug h na tu ra l qu artz to ful fill de ma nd , and quartz is th e re fo re synt hesized . No ncrystalli ne , or vitreo us, qu artz , made by melting pure silica san d , is place d in a steel "bom b ," and d ilute aq ueous NaO IJ is ad d ed. A "seed" crysta l is placed in the mixture, just as you mi ght use a seed cr ysta l in a hot sugar so lu tio n to grow
FIGURE 21.19 Pure silicon. The manufacture of very pure silicon begins with producing the volatile liquid sitanes SiCl or SiHCl3 • After carefully purifying these by distillation, t hey are reduced to elemental silicon with extremely pure Mg or In . The resulting spongy silicon is purified by zone refining. The end result is a cylindrical rod of ultrapure siliconsuch as those seen in this photograph. Thin wafers of silico n are cut from the bars and are the basis for the semiconducting chips in computers and other devices.
FIGURE 21.20 Various forms of
quartz .
Quartz (a) Pure quartz is colorless, but t he presence of small amounts of impurities adds color. Purple amethyst and brown citrine crystals are quartz with iron impurities.
(b) Qua rtz is a network solid in which each Si atom is bound tet rahedrally to four 0 atoms, each o atom linked to another Si atom. The basic st ructure consists of a lattice of Si and 0 atoms.
2 1.7
I
Silicon and t he Group 4A Elements
987
rock candy. \Vhen th e mixture is heated above th e critical tem perature of water (above 400 °C and 1700 atm) over a pe riod of days , pure q uartz cr ysta llizes . Silicon dioxide is re sistant to attack by all acids exce pt IIF, with which it reacts to give SiFt and H 2 0 . 5iO,(5) + 4 HF(f) --> 5iF,(g) + 2 H,O(C) Silicon d ioxid e also dis so lves slowly in hot, molten NaO I lor Na 2COJ to give Na -tSiO-t, sodium silica te. 5iO,(5) + 2 Na,(O,( C) --> Na,5iO,( 5) + 2 (O,(g)
Synthetic quartz. Th ese crystals were grown from silica in sodium hydroxide. The colors come from added Co2+ ions (blue) or Fe2+ ions (brown).
After the m ol te n mixture has coo led , h o t water under pressure is added . T his partia lly d isso lves the material to give a so lu tio n of sodi u m silicate . After filtering off in so lub le sa nd or g lass, th e so lvent is evaporated to leave so d iu m silica te, ca lle d umter glass. The biggest sing le use of this material is in househ ol d an d industr ial d etergents, in whic h it is in cl ud ed because a sodi u m silica te so lu tio n ma in tai ns p H by its b uffe r ing abi lity. Additio na lly, so d iu m silicate is used in va rio us ad hesives and hin d e rs, especia lly fo r g luing corrugated cardboard boxes. If sod iu m silica te is treate d with ac id , a gela tino us p recipita te o f Si0 2 ca lled silica gf'l is obtained . Washed and dried , silica ge l is a h ighl y porous material with d ozens of uses. It is a drying age n t, read ily absorbing u p to 40% o f its own weigh t of wate r. Small packet s of silica gel are often p laced in packing bo xes o f m erch andise d uring storage. T he m ater ial is frequen tly stai ne d with (N H -t hCoCI-t, a h um idity d et ector that is pink whe n hyd ra ted and bl ue when d ry.
Silicate Minerals with Chain and Ribbon Structures
'"11l~1~" I'lOT!"'''' "'Q
HVIWQIf! SILICA Gel
il-'11l7 .. t:....4 tL.. 'tA.
00 NOT ,,-,U
HUM/DR/ SIUCA GEL ~.,,,'J
t:..... 6 tL "
P, (g )
+ 6 ( a5iO, (5) + 10 CO (g )
Whi te ph o sphorus is th e mo st sta b le allo trope of phosphorus. Rath er th a n occurring as a di at omic mol ecul e with a tripl e bond , like its sec o nd-pe rio d rel ati ve nitrogen (N 2 ) , phospho ru s is made up of tetrah edral P-t mol ecul es in wh ich eac h P ato m is joined to th ree o the rs via single bo nds. Red pho sphorus is a p ol ymer o f Pol units.
White phosphorus, P.Ii
Polymeric red phosphorus
Nitrogen Compounds The red and white allotropes of phosphorus.
992
Chapt er
21
I
A n ot abl e feature o f th e chemistry o f nitrogen is the wide di versity o f its compounds. Co m po u nds wit h nitrogen in all o xida tion numbe rs bet ween - :1 a n d + 5 a re kn own (Fig u re 2 1.23) .
The Chemistry of the Main Group Elements
I
Hydrogen Compounds of Nitrogen: Ammonia and Hydrazine
Ammo nia is a gas at ro om tem perature a nd pre ssure . It h as a vc ry pe ne tratin g o dor ~
a n d co n de nses to a liquid at - 3:1 °C u nder I bar o f pressu re . So lu tio ns in water, ~ ofte n r efe rred to a s a m m o n iu m h ydrox id e , a rc basic d u e to th e re action of a m- ,~ m onia with wa te r ("4 Se c tio n 17 5 ~n d Fig u re 3. 13).
i
~
~
Am mo nia is a maj o r ind u st r ial c h e m ica l a nd is p re pared by th e H abe r pro ce ss
(... page 749 ) , la rgel y for lise as a ferti lize r. Hydrazin c , N 2H 4, is a co lo rless, fu m in g liquid with an am mo n ia-like odo r (m p, 2.0 "C; b p. 1135 "C ) . Almost I millio n kilog ram s of hydrazin c is produ ced a n n ually by th e Rasch ig pro cess- th e ox ida tio n of am mo nia wit h alka line sod iu m h ypochlorite in th e presence o f gel atin (wh ich is ad d ed to su p p ress me tal-ca ta lyzed sid e reac tio ns th at lower th e yield of h yd razin e ) .
2 NH,(aq) + Na ClO(aq)
NO,(g) + 3 H,O(C)
E'
+
V
0.80 V
Co nce n trate d nitric ac id attacks and ox id izes most meta ls. (Alu m in um is an exce ptio n; see page 985 .) In th is pr ocess, the nitrate ion is reduced to o ne of the ni tr oge n o xid es. Whi ch o xid e is fo rm ed depends on th e metal and on reactio n co nd ition s. In the case of co p per, fo r exam ple , either :.'\0 or N0 2 is produced , d epending on th e co ncen tra tion of th e aci d (Figure 2 1.24b ) . In dilute aci d : 3 Cu(s)
+ 8 H,O+(aq) + 2 NO, -(aq) --> 3 Cu' +(aq) + 12 H,O(f) + 2 NO(g)
In co nce n tra te d ad d : Cu(s)
+ 4 H,O+(aq) + 2 NO, - (aq) --> Cu' +(aq) + 6 H,O(C) + 2 NO,( g)
Four met al s (Au , PI, Rh , and II') that are not atta cked b y nitri c acid a re ofte n d escribed as th e "noble met als." The a lchem ists o f the 14th ce n tu ry, ho wever, kn ew th at if they mixed H NO J wit h H CI in a ra tio o f a bou t 1 : 3, th is aq ua regia, or "king ly water," wou ld attack eve n go ld , the noblest of me tals. 10 Au(s)
+ 6 NO, - (aq) + 40 Clfaq) + 36 H,O+(aq)--> 10 [AuCl,naq) + 3 N,(g) + 54 H,O(C)
FIGURE 21.24 The preparation and properties of nit ric add. (a) Nitric acid is prepared by the reaction of sulfuricacid and sodium nitrate. Pure HNOl is colorless, but samples of the acid are often brown because of NO l formed by decomposition of t he acid. This gas fills the apparatus and colorsthe liquid in the distillation flask. (b) When concent rated nitric acid reacts with copper, the metal is oxidized to copper(lI) ions. and N02 gas is a reaction product.
(a) Preparation of nitric acid. 996
Chapter 21
I
The Chemi st ry of t he Main Group Elements
(b) Reaction of HN0 3 with copper.
A Closer Look
Making Phosphorus
He stoked his smallfurnace with more charcoal and pumped the bellows until his retort glowed red hot. Suddenly something strange began to happen. Glowing fumes filled the vessel and from the end of the retortdripped a shining liquid that burst into flames.
from t his source. Humans consume much more phosphorus, in the form of phosphate, t han they require, and t he excess phosphorus (about 1.4 g per day) is excreted in the urine. It is nonet heless extraordinary that Brandt was able to isolate the element. According to a n 'l Sth-century chemistry book, about 30 g of phosphorus could be obtained from 60 gallons of urine. And the process was not simple. An other 18th-century recipe states t hat "50 or 60 pails full" of urine was to be used. "Let it lie stee ping . . . till it putrefy and breed worms ." The chemist was the n to reduce the whole to a paste and finally to heat the paste very strongly in a retort. Afte r some days, phosphorus distilled from the mixture and was collected in wat er. (We know now that carbon from the organic compounds in the urine would have reduced t he phosphate to phospborus.} Phosphorus was made in this manner for more than 100 years.
J. Emsley: The 13th Element, p. 5. New York, Joh n Wiley, 2000. Jo hn Emsley begins his sto ry of phosphorus, its discovery, and its uses, by imagining what t he German alchemist Hennig Brandt must have seen in his laboratory that day in 1669. (See page 338 for an artist's conception of t he discovery of phosphorus by Brandt.) He was in search of the philosopher's stone, the magic elixir t hat would turn the crudest substa nce into gold. Brandt was experimenting with urine, which had served as the source of useful chemicals since Roman times. It is not surprising that phosphorus could be extracted
EXERCISE 21.10
The glow of phosphorus burning in air.
Nitrogen Oxide Chemistry
Dinitrogen monoxide can be made by the decomposition of NH 4N 0 3 • (a) A l ewi s electron dot structure of N20 is given in Table 21.5. Is it the only possible structure? If other structures are possible, is the one in Table 21.5 the most important? (b) Is the decompositio n of NH4 N0 3(s) to give N20(g) and H20 (g) endothermic or exothermic?
Hydrogen Compounds of Phosphorus and Other Group SA Elements Th e phospho ru s a na log o f am mo n ia, phosph ine (PH j ) , is a po isonous, highly rea ctive gas with a fa int ga rlic-like od o r. Ind ust rially, it is m ad e by th e reactio n o f white ph osphorus and aq ueo us Na O H . P, (s) + 3 KO H(aq) + 3 H,O(f ) --> PH,(g) + 3 KH, PO,(aq) T he other hydrides of th e heavier Gro up 5A e le men ts are a lso toxic and beco me mo re un stabl e as the ato mic n um ber o f the e lem en t in creases. No ne the less, arsine (AsI I3 ) is used in the se m ico nd uc to r industry as a sta rt ing mat e rial ill th e p re paratio n of ga lliu m arseni d e (GaAs) sem ico nd ucto rs.
Phosphorus Oxides and Sulfides Th e mo st im po rta n t co m po u nds of pho spho ru s are those with o xygen , and th ere are at least six sim p le binary co mpou nds co n ta in ing j ust ph osphorus and o xygen. All of th em can be th ought o f as be ing d erived stru ctu ra lly fro m t he P4 te tra hed ro n of white p ho sph orus. Fo r exa mp le, if Pol is ca re fu lly o xid ized , P.,O ti is formed ; an o a to m has been placed in to each P-P bond in the tetr ah ed ron (Fig u re 2 1.25 ) . 21.8
I
Nitrogen, Phosphorus, and t he Group 5A Elements
997
.-.._~ 165.6
pm
143 pm
H3P03 H3P04 Phosphorous acid Phosphoric acid FIGURE 21.25 Phosphorus oxides. Ot her binary Pc-O compounds have formulas between P40 6 and P40 10 • They are formed by starting with P4 0 6 and adding atoms successively to t he P atom vertices.
°
The most co mmo n and important pho spho ru s o xid e is P"O w, a fine wh ite powd er co m mo n ly called "p hos p ho rus pcntao xid c " because its e m pirica l fo rmula is P20 ". In P4 0 IO , each phosphorus ato m is su rrou nded tetrah edrally by at oms. Phospho rus also forms a series of compounds wit h sulfur. Of th ese, th e most important is P4S3'
°
4 P(5,red allotrope) + + 58(5) ---t~
209 pm
233 pm
P, 53
Matches. The head of a "strike anywhere" match contains P4 S3 and the oxidizing agent KClO]. (Other components are ground glass. FeZ03' ZnO. and qlue.) Safety matches havesulfur (3- 5%) and Keto 3 (45-55%) in the match head and red phosphorus in the striking strip.
998
Chapter 21
I
In thi s phosph o ru s su lfid e, S atoms are pl aced into o nly three of th e P-P bonds. Th e principal use of P"S:l is in "strike an ywhe re" mat ch es, th e kind th at ligh t whe n yo u rub th e head aga inst a rou gh obj ect. The act ive ing red ients are P.,S3 and th e powe rful ox id izing age nt potassium chlora te , KCI0 3• The "safety mat ch " is now more co mmo n th an the "strike anyw here " match . In safe ty m at ch es, th e head is p redomin antly KClO :h and th e mat eri al o n th e mat ch bo ok is re d ph o sphorus (abo ut 50% ), 5b ,5" Fe203, and glu e.
The Chemi st ry of t he Main Group Elements
Phosphorus Oxoacids and Their Salts A few o f the ma ny kn own pho spho rus oxoacids a rc illustr at e d in 'Table 2 1.6. In de e d , there arc so man y acids a n d their sa lts in th is category tha t str u c tu ra l principles h ave bee n deve lo ped to o rga nize a nd u ndersta nd th em . (a) All P a toms in th e oxoacid s and t he ir a n io n s (co nj uga te bases) arc fourcoo rd inate a n d tetrah e dral. (b) All the P atoms in th e acids h ave at least one P- Olf gro up (a n d this occurs o fte n in the an io ns as we ll). In every ca se , th e H a tom is io n izabl e as 1-1 +. (c) Some oxoacids h ave o ne or mo re P- H bonds. Th is H a tom is not ioniza bl e as 1-1+, (d) Polymerizati on can occu r by P-O- P bond for ma tio n to g ive bot h lin ea r and cycl ic spe c ie s. Two P ato ms arc neve rj o in ed by m ore th an o ne P- O -P b rid ge . (e) Whe n a I' a tom is su rrou nded o nl y by 0 at o m s (as in H ~P O 'I ) , its oxidatio n nu mber is + 5. For eac h P- OH th a t is replaced by P- H , the ox idatio n n um ber drops by 2 (b e ca u se P is co n sid e re d mo re e lectronega tive th a n I I) . Fo r example, th e ox idation n u m be r of P in H :IPO Z is + I .
FIGURE 21.26 Reaction of P4010 and wat er. The white solid oxide reacts vigorously with water to give orthop hosphorie acid, H3PO•. (The heat gene rate d vaporizes t he water, so stea m is visible.)
O rt hop hosp horic aci d, H:\PO ,h a n d its sa lts a re far m ore im portant co mm ercia lly than ot her P- O acids. Millions of to ns o f phosphoric acid a re made a nnually, so m e using white phospho ru s a s the starti ng m a te r ial. T h e e leme n t is burned in oxygen to give 1'.,1 0 10 , a n d th e oxi de reacts with wate r to pro duce th e aci d (Fig u re 2 1.26).
P,OlO(S) + 6 H20( C) --> 4 H, PO, (aq)
TABLE 21 .6
Formula
H, PO,
Phosphorus Oxoacids Name
Orthophosphoric acid
Pyrophosphoric add (diphosphoric acid)
Structure
o I
HO II~"· P,
/ OH
OH
o I
0
HO ""'·' P,
/
OH
I 0
....... p ··"''' OH '\ OH
O,p . . . . . O.. . . . p.. . . . O (HPO,h
Metaphosphoric add
Phosphorous acid (phosphonic acid)
I'
OW/ I
0,
p
OH
/ 0
a................. OH o I H"1"" P, / OH OH
o Hypophosphorous acid (phosphinic acid)
I
H"""· P,
/
H
OH
21.8
I
Nitro ge n, Phos phor us, and t he Group SA Ele ments
999
Phosphoric acid (impure)
Phosphate rock
Fertilizers
80%
Food phosphates Industrial phosphates
Elemental phosphorus
Detergent phosphates
Phosphorus sulfides fl:tiosptiorus cHtorides Organic pnospnorus comRounds (a) Mining phosphate rock. Phosphate rock is primarily Ca3(P04h, and most mined in the United States comes from Florida.
Used as such in metaltreatment, etc.
(b) Usesof phosphorus and phosphoric add.
FIGURE 21.27 Uses of phosphate rock, phosphorus, and phosphoric add.
T his ap proach g ives a purc pro du ct, so it is em ployed to mak e ph o sph o ric ac id fo r use in foo d p ro duc ts in particular. T he aci d is no n to xic, and it gives the tart o r so u r taste to carbona ted "so ft drinks," such as vari o us co las (abo u t 0.05 % H 3P0 -t) o r ro ot beer (abo u t 0.0 1% H jP0 -t) . A major use for p ho sp horic acid is to impart corrosio n resista nce to met al objec ts such as nu ts and bolt s, tools, and ca r-engi ne parts by plung ing th e o bjec t into a hot acid bath. Car bod ies arc simi larly treated with phospho ric acid co n ta in ing metal ions such as Zn 2+ , and alum inum tri m is "po lished" by treating it with the acid. T he reaction o f H 3PO t with strong bases produce s salts suc h as Na H2P0 -t, Na2H P0 4, an d NajP0-t. In indu str y, the mo nosod iu m and d isodium salts arc produced using Na2C0 3as th e base , b u t an ex cess o f th e stro nger (and more expens ive) base NaO H is requi red to re move th e thi rd proton to give Na d ' Oj . Sodi u m pho sphate (Na3P0 -t ) is used in sco u ring powders a nd pain t str ip pers because th e an io n PO l - is a relative ly stro ng base in wate r ( K b = 2.8 X 10- 2) . Sodiu m rno nohydro gcn phosphate , Na2H P0.J, whic h has a less basic anio n th an PO l - , is wide ly used in food pr odu cts. Kraft ha s patented a process using th e salt in the manufacture of pasteurized cheese, fo r ex am ple. T ho usands o f tons of Na2IJPO -t are still used for thi s purpose , even tho ug h th e fu nc tio n of th e salt in thi s pnKcss is not co mplet ely understood. In addi tio n, a small amo u n t of I\'a2H P04 in pu d d ing mixes enables th e m ix to gel ill co ld water, and the basic anio n raises the pI I of ce rea ls to provide "q uick-coo king" b reak fast cerea l. (T he O H - io n fro m HP 0 4 2- hydrolysis acc ele ra tes the br eakdown o f the ce llu lose mate rial in th e cereal.) Ca lciu m pho sph at es are used in a broad spec t ru m of p rod ucts. For example, th e weak ac id Ca( H 2PO .1l2 . H 20 is used as th e acid leavening age n t in ba king powd er. A typica l ba king po wder co n tai ns (a lo ng with ine rt ingredien ts) 28% Na ! ICO ", 10.7% Ca (l1 2PO , ),· H 2 0 , an d 2 1.4% NaA I(SO .,), (a lso a weak ac id ) . The weak ac id s reac t with so d iu m bicarbon at e to prod uce CO 2 gas. For exa mp le,
Ca( H,PO,),' H,O(s) + 2 NaHCO](aq) --> 2 CO,(g) + 3 H,O(C) + Na, HPO,(aq) + Ca HPO, (aq) Fina lly, calc iu m mo no hydrogcn pho sphat e , Ca H P0-t, is used as all abrasive an d po lishi ng age nt in tooth pa ste. 1000
Chapter 21
I
The Chemistry of t he Main Group Ele ments
21.9 Oxygen, Sulfur, and the Group 6A Elements
Group 6A Oxygen 8
Ox ygen is by far th e most a bu n d a n t c le m e n t in the eart h's crust , re p re sen tin g slig h tly le ss than :>0% of it by we ig h t. It is present a s el emental oxygen in the atmosphere a n d is co m b ined with othe r e le me n ts in wat er a n d in m an y mineral s. Scientists be lieve that elemental o xyge n did not appear 011 this p lane t u nti l about 3.5 billion ye a rs ago, wh e n it was formed on the planet by p lants throu gh the process of p h o to syn th esis. Su lfu r, seventeenth in abundance in the ea r th's crust , is also found in its elemental form in nature, but only in certain co nce n tra ted deposits. Sulfur-containing compounds occur in n atu ral gas and oil. In min e rals, su lfu r occurs as th e su lfide ion (fo r ex am p le, in cin na ba r, HgS, a nd gale n a, PbS ), as the di su lfide io n (in iron pyrite, FeS~b or "fo o l's gold") , and as sulfate ion (e.g., in gyp su m, CaSO..j, · 2 H 20 ) . Sul fur oxides (S02 and S03) a lso oc cur in nature, primarily as produc ts of vo lca n ic ac tivity (Fig ure 21.28) . In the Un ited States, most su lfu r- a bo u t 10 mi llio n ton s per yea r- is obtained from d eposits of th e e le m e n t found a lo ng th e Gulf of Me xico . These d eposit s occu r typi cally at a depth of 150 to 750 rn below th e su r face in laye rs about 30 m thick . They are th ought to have been fo rmed by a naerobic (twith o u t e le m e n tal ox ygen" ) bac te ria acting on sed ime n ta ry su lfa te d eposits suc h as gyps u m.
0 474,000 ppm Sulfur 16
5 250 ppm Selenium 34
5e 0.5 ppm Tellurium 52
Te 0.005 ppm Polonium 84
Po trace
Element abundances are in parts per millionin the earth's crust.
Preparation and Properties of the Elements Pu re oxygen is obtai ned by th e fraction al di sti llation of air a n d is among the top five indu stri al ch e m icals produced in the Un ited States. Oxygen ca n be made in th e la bo ra to ry by e lec tro lysis of water (Figu re 2 1.4) a nd by the ca ta lyzed de compositio n o f m eta l c h lo ra tes suc h as KC103 . 2 KClO, (s)
~~IY' ) 2 KCl(s) + 3 0,(g)
At room temperature a n d pressure , ox ygen is a co lorless gas, hut it is p ale bl u e whe n co n d e ns ed to th e liquid a t - 183° C (... Figure 9. 15) . As d escribed in Section 9. 3, diatomic ox ygen is para mag n etic be cause it has two unpaire d el ectron s. An allo tro pe of o xygen , o zone (0 3), is a blue , diam agn etic gas with a n odor so stro ng th at it ca n be detected in co ncentrations as low as 0 .05 ppm . O zone is synthe sized by passin g O 2 through an e lec tr ic dis ch arge or by irradiatin g O 2 with u lrr aviol et light. It is often in th e ne ws because of th e realiz ati on th at th e ea rt h 's p rotecti ve layer of ozone in the stratos p here is being disrupted by c h lo ro fluorocarbons and ot her c he m icals (... page 953) . Su lfur h as n um erous allo tr o pes. The most co m mo n a n d most sta b le allo tr o pe is th e yellow, orthorhombic for m , wh ich consists of SR mo lecu les wit h the sulfur atoms arra nged in a crown-shaped rin g (F igure 2 1.29a). Less sta b le allotropes a rc known th at h ave rings of f to 20 su lfur a to ms. Another form of su lfur, ca lled pl astic su lfu r, h as a mol ecu lar struc tu re with c ha ins of su lfu r atoms (Figure 2 1.2gb) . Su lfur is obtain ed from und erground d eposits by a p rocess developed by Herman Frasch ( 18:; 1- 19 14) a bo u t 1900 . Superheated wate r (a t 16:; ' C) a n d then air are fo rced into the deposit. The su lfur melts (m p , 113 °C) a nd is for ced to the su rface as a frot hy, yellow strea m, from whi ch it so lid ifies . Se leni um a nd tellurium a re co m p ara tively rare on eart h, h aving a b u n dan ces about the sa me as th o se of silve r a n d go ld, re sp ecti vely. Because their c h e m istry is sim ila r to that of su lfu r, they are often fou nd in mine ra ls asso ciat ed wit h the su lfides of copper, silver, iro n, and arsen ic, and th ey a re reco vered as by-product s of th e industries devot ed to those meta ls. 21.9
I
i" ~
Ie
0:...
"""'-'''-'
FIGURE 21.28 Sulfur spewing from a volcano in Indonesia.
Oxygen, Sulfur, and the Group 6A Ele ments
1001
FIGURE 2 1.29 Sulfur allotropes.
(a) At room temperature, sulfur exists as a bright yellow solid composed of 58 rings. (b) When heated, the rings break open, and eventually form chains of S atoms in a material described as "plastic sulfur."
(a)
(b)
Seleni u m has a ra ngc of uses, including in g lass makin g. A cadmiu m su lfid e/ selenide mi xtu re is added to g lass to give it a b r illiant red co lor (Figure 2 1.30£1) . Th e most fa miliar use of seleniu m is ill xe rograp h y, a word m eani ng "d ry pri nting" and a proccss at t he heart o f th e moder n co py m achine . Most photoco py machin es use an alu mi n u m plat e or ro ller co ated with sele niu m. Ligh t co mi ng fro m the im agi ng lens selectivel y d ischarges a sta tic elec tric chargc on the selen iu m su r face , and th e black tone r sticks o n ly on th e areas th at rem ain charged . A co py is mad e whe n th e toner is transferred to a shee t of plain paper. The heaviest ele me nt o f Group 6A, pol onium . is rad ioactive an d found o n ly in tr ace amo u n ts on ea rt h . It was d iscovered in Pari s, Fra nce, in 1898 b), Marie Sklodowska Curie ( 1867- 1934 ) and her husband Pierre Curie ( 1859-1906). The Curies pains taking ly separate d this e lement from a large qu anti ty of pi tchblend e , a uranium-containin g ore.
Chemislry.o.Now'"
•
Sign in at www.thomsonedu.com/login and go to Chapter 21 Contentsto see Screen 21.8 for an exercise on sulfur chemistry.
FIGURE 21.30 Uses of selenium.
(a) Glass takes on a britliant red color when a mixture of cadmium sulfide/ setenide (CdS, Cd Se) is addedto it. (b) These sample bottles hold suspensions of quantum dots, nanometer-sized crystals of CdSe dispersed in a polymer matrix. The crystals emit light in the visible range when excited by ultraviolet light. Light emission at different wavelengths is achieved by changing t he particle size. Crystals of PbS and PbSe can be made that emit light in t he infrared range.
2 PbO(s)
• Bad Breath Halitosis or "bad breath" is due to th ree sulfur-ccntaininq compounds: H2S , CH 3SH (methyl mercaptan). and (CH 3hS (dimethyl sulfide). AUt hree can be detected in very tiny concentrations. For example, your nose knows if as little as 0.2 microgram of CH ,SH is present per liter of air. The compounds result from bacteria's attack on the sulfurcontaining amino acids cysteine and methionine in food particles in the mouth. A genera! rule: if something smells bad, it probably contains sulfur! (See Case Study, page 524.)
+ 2 S02(g)
Here, lead su lfide is converted to lead (II ) oxide, and thi s is fu rther reduced to lead usin g carb o n o r ca rbo n mo n oxi de in a blast fu rn ace. PbO (s) + CO(g) ---> Pb(C) + CO,(g)
Alternati vely, th e o xid e ca n be reduced to e lem e n ta l lead by co m b in ing it with fresh lead su lfide. 2 PbO(s)
+ PbS(s) ---> 3 Pb (s) + S02(g)
so,
H2S
so,
Models of some common sutfur-containinq molecules: H2S, S02' S03' and H2S0 4 ,
Sulfur d io xide (S02), a co lo rless, to xic gas with a sharp odo r, is produced on an enor mo us sca le by the combustion of su lfu r and by ro asting su lfide orcs in air. The co m bus tio n of su lfu r in su lfur-co n ta in ing coal and fu el oil creat es particularly large envir o n me n ta l problems. It h as been es tima te d th at a bo u t 2.0 X 10K ton s of su lfur oxides (p rima rily S02) a re rel eased into th e atm o sph ere eac h year by hum an activities; thi s is more than h alf of th e total e m itte d by all other natural sou rces o f su lfu r in the environ ment. Sulfur dioxide readily dissolve s in wat er. The most important re acti on of thi s gas is its ox idatio n to S03' SO, (g)
+ 'h 02(g) ---> SO, (g)
t. ,W
~
- 98.9 kJjm ol- rxn
Sulfur trioxide is almost n ever isol at ed but is co nve rte d directl y to su lfuric acid by re action with wate r in th e "co n tac t p ro cess." T he largest lise of su lfu r is th e producti on of sulfuric acid, H 2SO.h th e co m pou n d p rodu ced in lar gest quantity by th e c he m ical industry (.... p ag e 135) . In th e Un ite d Stat es , roughl y 70% of th e acid is used to manufacture su p e rp hos p hate fertilizer fr o m ph osphate rock. Plants n eed a so lu ble form o f pho sphorus for g rowth, but ca lciu m ph osphat e and apat ite [Ca"X (p O , h X = F, Oil , Cl] are in soluble . Tr eatin g phosph ate- containin g mi nerals with su lfu ric ac id p roduces a mi xture of so lu b le 21.9
I
Common household products containing sulfur or sulfur-based compounds.
Oxygen, Sulfur, and t he Gro up 6A Elements
1003
A Closer Look
Snot-tites and Sulfur Chemistry
Sulfur chemistry can be important in cave formation, as a spectacular exam ple in t he j ungles of southe rn Mexico amply demonstrates. Toxic hydrogen sulfide gas spews from t he Cueva de Villa LUI along with water that is milky white with suspended sulfur particles. The cave can be followed dow nwa rd to a large underground stream and a maze of actively enlarging cave passages. Water rises into the cave from underlyi ng sulfur-bearing st rata. releasing hydroge n sulfide at concentrations up to 150 ppm. Yellow sulfur crystallizes on the cave walls around t he inlets. The sulfur and sulfuric acid are produced by the following reactions:
th rive on sulfur compounds in acidic environments. The chemical energy released in t heir metabolism is used to obtain carbon for th eir bodies from calcium carbonate and carbon dioxide, bot h of which a re abu nda nt in the cave. One result is t hat bacterial filaments hang from the walls and ceilings in bundles. Beca use t he filaments look like something coming from a runny nose, cave explorers refer to them as "snot-tites." Ot her microbes feed on th e bacteria, and so on up the food chain-w hich includes spiders, gnats, and
+ O,(g) --> 2 5(s) + 2 H,O(f) 2 5(s) + 2 H,O(C) + 3 O,(g) --> 2 H,50,(aq) 2 H,5(g)
The cave atmosphere is poisonous to humans. 50 gas masks are esse ntial for wouldbe explorers. But surprisingly, t he cave is teeming with life. Several species of bacteria
pygmy snails-a ll the way to sa rdine-like fish t hat swim in the cave stream. This entire ecosystem is supporte d by reactions involving sulfur within t he cave.
Snot-tites. Filamentsof sulfu r-oxidizing bacteria (dubbed "snot-tites") ha ng fromthe ceiling of a Mexican cave containing anatmosphere rich in hydrogen sulfide. The bacteria thrive onthe energy released by oxidation of the hydrogen sulfide, forming the base ofa complex food chain. Droplets ofsulfuric add onthefi laments have an average pHof 1.4, with some as lowas zero! Drops that landed on explorersin the cave burnedthei r skin and disintegratedtheir dothing.
p ho sp hat es. Th e balan ced eq uatio n fo r the reaction o f excess su lfu ric acid and calc iu m ph o sphat e, for exa mp le, is
but it does no t tell th e whole sto ry. Co nc en tra ted su pe rp ho sp ha te fertil izer is act ually mo stly CaH P0 4 o r Ca( H 2 P0 4h plus so me H:l P 0 4 an d CaS04' (Notice that th e chemical p rin cipl e beh ind th is reactio n is t ha t su lfu ric acid is a str o nge r acid th an II "PO , (Table 17.3), so th e PO ,3- io n is proto nat ed by su lfu ric acid .) Smaller amou n ts o f sulfu ric acid are used in th e co nversio n o f ilm en ite , a titaniumbeari ng ore, to T i0 2, which is then used as a white pigme n t in pain t, plastics, an d pap e r. T he acid is a lso used to manufact ure iron and steel as well as pe troleum prod ucts, syn thetic pol ymers, and pape r. ChemistrY.{J.Now· Sign in at www.thomsonedu .com/Ioqtn and go to Chapter 21 Contents to see Screen 21.9 for an exercise on the st ructural chemistry of sulfur compounds.
EXERCISE 21.11
SuLfur Chemistry
Meta l sulfides roasted in air produce metal oxides. 2 Zn 5(s) + 3 O,(g) --> 2 ZnO(s) + 250,(g)
Use thermodyna mics to decide if the reaction is product- or reactant-favored at equilibrium at 298 K. Will the reaction be more or less product-favored at a high temperature?
1004
Chapter 21
I
The Chemistry of t he Main Group Elements
21.10 The Halogens, Group 7A
Group 7A
Halogens Flu ori n e a n d ch lo rin e a rc th e m o st a bu n da n t hal ogen s in th e earth 's cr us t, with flu o rin e so m ew h a t m ore abunda nt than c h lo rin e . If their a bu n d a n ce in sea water is m easured, howeve r, th e situa tio n is q uite d iffe re nt. Ch lo rine has an abu ndance in sea wa te r of 18,000 ppm , whereas t he a bu n dan ce o f flu o r in e in th e sa m e so u rce is o n ly 1.3 p pm . T h is variatio n is a re sult of th e d iffe re n ce s in th e so lu b ility of thei r sa lts a n d p lays a ro le in th e m et hods used to reco ve r th e el eme n ts th e m sel ve s.
Fluorine 9
F 950 ppm Chlorine 17
Cl 130 ppm Bromine
Preparation of the Elements
35
Br
Fluorine
0.37 ppm
T he wa te r-insolu b le m ine ral fluo rs pa r (calc iu m fluo ride , Ca F2 ) is o ne o f th e m an y so u rces o f fluo r ine . Because the mine ral was o r iginally used as a flu x in metalworkin g, its n ame co mes fro m t he La tin wo rd m eani n g "to flo w." In th e 17th ce n tu ry, it was d isco vered that so lid CaF 2 wo u ld e mi t light wh e n h e a te d , an d the p henome non was ca lle d fluorescence. In the e a rl y 1800s, whe n it was re co gn ize d th a t a n ew c le m e n t wa s co n ta in e d in flu orsp ar, A. M. Am pere (1 775- 1836) sugges te d th at th e c le m e n t be cal le d flu o ri n e. Alt h o ug h fluorin e was recog ni ze d as a n e le m e n t by 181 2, it was not u nti l 1886 that it was iso lated by th e Fre n ch c he mis t H en ri Moi sson ( 1852- 1907) in c lc mc ntal fo rm as a vcry pal e ye llow gas by th e e le c trolysis of KF d isso lved in a n hy d rous IIF. Ind e e d , be ca u se F;! is suc h a powe rfu l o xidi zing age n t, c h e m ical o xid atio n o f F- to 1'2 is no t fea sib le , a n d e lec tro lysis is th e only pra ctical way to obtain gaseous F, (Figu re 2 1.31 ). T he prepa ra tion of F2 is d ifficul t be cause F2 is so re ac tive . It o xidi ze s (co r rod e s) t he e q u ip me n t a n d re a cts vio le n tly with tra ce s o f g rcase or ot h er co n ta m in a n ts . Furt h e rm o re , t he pro du ct s of el e ct ro lysis, F 2 an d H 2 , ca n re co m b ine ex p los ive ly, so the y m ust n ot be al lowed to co m e in to con tac t with each o ther. (C o m pa re with t he reac tio n of H 2 a n d Br 2 in Figu re 2 1.5.) Cu rre n t U.S. prod u ctio n of flu o ri n e is ap p roxima tel y ~) O O O metric to n s per yea r.
Iodine 53
I 0.14 ppm Astatine 85
At trace
Element abundances are in parts per mi llion in the earth's crust.
Chlorine Ch lorine is a stro ng o xidizin g agen t, a nd to p repa re thi s e le men t from ch lo ride ion by a c hemical reac tio n requi re s a stro nge r o xid izin g age n t. Pe r m an gan a te o r di ch rom at e ion in ac id so lu tio n will serve thi s p urposc (Figure 2 1.32 ) . Ele me n ta l ch lori n e was first mad e by tbe Swedis h cbe mist Karl Wilh el m Schee le ( 1742- 1786) in 1774, wh o co m b in e d so d iu m ch lo rid e with a n o xid izing age n t in an ac id ic so lu tio n . Indu str iall y, c h lorin e is m ade by e le ctro lysis o f bri ne (co nc e n tra te d aq uc o us NaC,I) . Th e o the r p rod u ct o f the e lec tr o lysis, NaO Il , is a lso a va luab le industri a l che m ical. Abo u t 8 0 % o f th e c h lorin e p roduce d is m ad e usin g a n e lec troc he m ical cel l sim ila r to th e o n e d e pi c ted in Fig u re 2 1.33. O xid ation o f ch lo rid e ion to CI2 gas o ccurs at the a nod e a n d reducti o n of wat e r occu rs a t th e ca th ode.
+
An ode reaction (oxidation)
2 W(aq) --> Cl, (g )
Cat hode reaction (reduction)
2 H,O(f ) + 2 e: --> H2 (g) + 2 0W(aq)
2 e-
Activate d titanium is used fo r th e a n o de , a nd stain le ss stee l o r n ic ke l is p re ferred for th e cat h o de in the e lec tro lytic cel l. T h e a node a nd cathode co m par tme n ts a re se p a ra te d by a m em b ran e th at is n ot pe rm e a bl e to wa ter b u t a llow s Na + io n s to pass to m ai ntain the c ha rge ba lance . T hus, the membra n e fun cti o n s as a "sa lt " b rid ge betwe e n th e a n o de a n d cath ode co m part m e n ts. The ene rg y consu m p tio n of the se cel ls is in th e ra nge o f 2000-2500 kw h pe r ton of Na O H p rodu ce d.
Anode
Cathode
Skirt Cooling tube
FIGURE 21.31 Schematic of an
electroLysis ceLL for produdng fluorine. 21.10
I
The Halogens , Gro up 7A 1005
Bromine The sta n d a rd reduction potentials of the halogens indicate that their stre ng th a s oxidizing age n ts decreases go ing from F2 10 12, Reduction Potential (EO V)
Half-Reaction
I
F, (g)
+ 2 e- -e
( l,(g)
+ 2 e- ->
2 (l -( aq)
1.36
~
2 Br- (aq)
1.08
BrAt) + 2 e"
I, (s)
+ 2 e- ->
2 F' {aq)
2.8 7
2 ]- (aq)
0.535
T his means that CI2 will oxidize Rr- ions to Rrz in aqueous so lutio n , for example. Cl2(aq)
EOnet
=
P
+
2 Br-( aq) -> 2 W(aq)
cat hode -
EOa node
=
+ Br2(aq)
1.3 6 V- (1.08 V)
=
+ 0.28 V
In fact , this is the commercial method of p reparing brom ine when NaBr is obtaine d from natu ral brine wells in Arkansas a nd Michigan .
g-- - - -
Iodine
FIGURE 21 .32 Chlorine prepara-
tion. Chlorine is prepared by oxidation of chloride ion using a strong oxidizing agent. Here, oxidation of NaCl is accomplished using K2( r20 7 in H2S 04 , (The el2 gas is bubbled into water in a receiving flask.)
Iod ine is a lustrous, purple-black soli d , easily su blime d at room temperature an d atmospheric p re ssure (Figu re 13.17). T he cl eme n t was first iso lated in 18 11 fro m sea weed and ke lp , ex trac ts of which had lo ng been used for tre atmen t of go iter, the e n largement of th e thyroid g land . It is now known tha t the th yroid g land p rodu ce s a grow th-reg u lati ng ho r mo ne (th yrox ine) that co n tains iodin e. Co nseq uen tly, mo st table sa lt in the Un ited States has 0.01 % Na l ad d ed to p ro vide the nece ssary iodin c in th e d iet.
Anode ( +)
Cat hode (-) Ion-permeable membrane
Depleted brine
-
~ Cl- -
)
2
-
Wate r
'\
Na +
OW
Bri ne
H2O
H2O
~
NaOH(aq) ~
Active Figure 21.33 Amembrane cell for the production of NaOH and CI 2 gas from a sat urated, aqueous soluti on of NaCI (brine) . Here, the anode and cathode compartments are separated by a water-impermeable but ion-conducting mem brane. Awidely used membrane is made of Nation. a fluorine-containing polymerthat is a relative of polytetrafluoroethylene (Teflo n). Brine is fed into the anode compartment and dilute sodium hydroxide or water into t he cathode com partment. Overflow pipes carry the evolved gases and NaO Haway from the chambers of the electrolysis cell.
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1006
Chapter 21
I
The Chemist ry of t he Main Group Elements
FIGURE 21.34 The preparation of iodine. A mixture of sodium iodide and manganese(IV) oxide was placed in tile flask (left). On adding concentrated sulfuric acid (riqht), brown iodine vapor is evolved.
2 NaI(s)
+ 2 H,S04(aq) + MnOz(s) ~ Na zS04(aq)
+ MnS0 4(aq) + 2 H,O( f ) + I, (g)
A labo ratory met hod for preparing I;! is illustr at ed in mercial preparat ion depe nds o n th e sou rce o f 1- and metho d is int e restin g because it involves so me che m istry bo ok . Iodid e io ns are first precipitated with silver io ns to I - (aq )
+ Ag+(aq) -->
Figure 2 1.34. T he co mits concen trat io n . O ne descri bed ea rlie r in thi s give inso lu ble AgI.
AgI(s)
This is re d uced by clean scrap iron to give iron (II ) iod ide and metallic silve r. 2 AgI(s)
+
Fe(s) --> Fel,(aq)
+
2 Ag(s)
T h e silve r is re cycled by ox idizing it with n itr ic acid (form ing silver ni tra te ) which is th en reu sed . Fin ally, iodid e io n fro m water-solu ble Fel 2 is ox id ized to io d in e with ch lo rine [with iro n( lI l) ch lo ride as a by-pr od uct] . 2 Fel ,(aq)
+
3 Cl, (aq ) --> 2 I, (s)
+
2 FeCl, (aq)
Fluorine Compounds Fluori ne is the m ost react ive of all of the e leme nts, formi ng compou nds with eve ry e leme n t exce pt He and Ne . In mos t cases, th e el em e nt s co mbine directly, and so me reactio ns can be so vigo ro us as to be exp losive. T his react ivity can be exp lained by at least two featu re s o f flu o rine chemistry: the re latively weak F-F bo nd compared with chlo r ine and b ro mi ne , and , in particular, the relatively strong bonds fo rm ed by fluo rine to o ther cleme nts . T his is illus trated by the tab le o f bo nd disso ciation ent halpies in the ma rgin . In additio n to its oxi diz ing abi lity, anot her notable characte ristic of fluorine is its sma ll size. T hese p ro perties lead to the fo rma tion of compounds where a n umber of F ato ms can be bon d ed to a central eleme nt in a h igh oxidation state . Exa mples in clud e PIFl), UFo, IFi t and Xe F4 • Hydrogen flu orid e is an impo rtan t ind u strial che m ical. Mo re th an I million IOns of hyd ro gen flu o rid e is prod uced ann ually wor ldwide, almo st all by the action o f co nce ntrated su lfuric acid o n flu orspar. CaF,( s)
+
H,50, (f) --> Ca50,(s)
+ 2 HF(g)
Bond Dissodation EnthaLpies of Some HaLogen Compounds (kJjmoL)
x
X- X
H- X
F
155
565
Cl
242
4 32
339
Br
193
366
285
151
299
213
T he U.S. capacity fo r H F p rodu ctio n is ap p ro ximatel y 2 10,000 metric to ns, but curre n tly d em a nd is exceedi ng su p ply for thi s che m ical. An hydro us HF is used in 2 1. 10
I
The Haloge ns. Group 7A 1007
a broad ra nge of in d ustries: in t he p roductio n of refrigeran ts, herbicid es, pharm aceu ticals, high-o ctan e gaso lin e, alu m in u m, plastics, electr ical co m po nen ts, an d fluo rescen t ligh tbul bs. The flu o rspa r used to p roduce H F m ust be very p ure an d free of SiO z because HF reacts re adi ly with silico n di oxid e. 5iO,(s) + 4 HF(aq) --> 5iF, (g) + 2 H, O( C) 5iF,(g)
+ 2 HF(a q) -->
H,5iF, (aq)
This series o f reactio ns exp lai ns why H F ca n be used to e tc h or fros t glass (such as the ins ide of flu orescent lig h t bulb s) . It a lso ex plai ns why H F is no t sh ipped in glass co n taine rs (un like II Cl , fo r exa m ple) . T he alu m in u m industry co nsu mes abo u t 10-40 kg of cr yolite , Na j A1F6, per metric ton of alu mi n u m produced. T he reason is that cryo lite is ad de d to alu m in u m oxid e to p roduce a lower-melting mi xture that ca n be el ec tro lyzed. Cryo lite is fo und in only small qu an tities in nat ure, so it is ma de in vario us ways, amo ng them the fo llowing re actio n : 6 HF(aq) + AI(OH),(s) + 3 NaOH(a q) --> Na ,AIF,( s) + 6 H,O(C)
Abo u t 3 % of the h ydrofl uo ri c acid p roduced is used in u ran iu m fu el p roductio n . To sep arate uran ium isotopes in a gas ce n trifuge (... page 523) , the u ran iu m must be in the fo rm o f a vo latile co m po u nd . Na tu ra lly o ccu rr ing u ran iu m is proce ssed to give U0 2 • T his ox id e is tre ate d with hydroge n fluo ride to give U F.-t, whic h is the n reac ted with F2 to p rod uce th e vo latile so lid U Fc,. UO,(s) + 4 HF(aq) --> UF,(s) + 2 H,O(C) UF, (s)
+ F,(g ) --> UF, (s)
T his last ste p consu mes 70-80% o f fluo r ine produced an n ually.
Chlorine Compounds Hydrogen Chloride Hyd ro chl o ric acid, an aqueous so lu tio n o f hydro gen chlo r id e , is a valua ble indu str ial che mical. l lydrogen ch lo ride gas ca n be p repared by the reaction of hyd roge n and ch lo ri ne, but the ra pid , exo the rm ic reac tio n is d ifficu lt to con tro l. T he class ica l met hod of mak ing H e l in th e labo rat o ry uses the reacti o n of NaC I and su lfuric acid, a p ro cedure th a t takes adva n tage o f th e facts th at I ICI is a gas an d th at I12SO .J will no t oxi d ize th e chlo ride io n . 2 Na CI(s) + H,50 ,(C) --> Na,50,(S) + 2 HCl(g)
Hyd rog en ch loride ga s has a sharp, ir ri tatin g o d o r. Both gaseous and aq ue o us II CI react with met als and m e ta l ox ides to g ive metal chlo rides and , d epe ndin g o n th e re ac ta n t, h ydrogen o r water. Mg(s) + 2 HCI(aq) --> MgCl , (aq) + H,(g ) ZnO(s) + 2 HCI(aq) --> ZnCl,( aq) + H, O(C)
Oxoarids of Chlorine O xo acid s o f ch lo ri ne ran ge from H CIO , in which chlo rine has an oxid at ion n u mber o f + I, to II CI0 4 , in which the o xid ation n umbe r is equal to th e gro u p nu mber, + 7. All are stro ng o xid izing age nts.
1008
Chapter 21
I
The Chemi stry of the Main Group Elements
Oxoadds of Ch lorine Add
Name
Anion
Name
acto
Hypochlorous
ClO-
Hypochlorite
HClO z
Chloro us
ClOz-
Ch lorite
HC lO ,
Chloric
Cl03-
Ch lorate
HCl04
Perchloric
Cl0 4 -
Perchlorate
Hyp ochlo rous acid, H ClO , fo rm s whe n ch lo rine disso lves in wa ter, In th is rea ctio n, h alf o f th e c h lorine is oxidized to hyp ochlo rite io n a nd h alf is red uced to ch lo r ide ion in a disproportionation rea ction. Cl,(g) + 2 H,O(i') (rl ) C(s) + O , (g ) -.
Practidng Skills
14. • Co mple te and b al an ce t he e q u a tio ns fo r th e fo llowin g reac tio ns: (a) K (s) + I, (g ) -. (h) Ba (s) + O , (g ) --> (c) A I(s) + Ss(s) --> (rl ) Si (s) + CI,(g) -.
Properties of the Elements
Hydrogen
1. Give e xa m p les o f two ba sic o xides. Write eq uation s illu str a tin g th e fo r ma tio n of each o xide fro m its co m p o· uc nt elements. \Vrite ano ther chem ical eq u a tio n th at illust ra te s th e b asic cha racte r of each o xid e . 2.
(~iv l' e xa m ples o f two acidi c o xid es. Wri te e q ua tions illu stratin g th e form ati o n o f ea ch o xide from its comporu-n t dements. \ \trite a no ther chemical eq ua tio n that it· lu stra tc s th e ac id ic cha ra cter o f eac h oxid e .
15. Write hal an ced ch em ical equati on s for th e reactio n of h yd rog en g-as wit h oxyp;e n , ch lorin e , an d nitro g e n . 16. • Write a n e qua tion for th e reac tio n of p ot assiu m an d h yd rog en . Na me th e prod uc t. Is it ionic o r cova len t? P redict o ne ph ysical p rope rt y a nd o ne c he m ica l p ro perty of th is co m pou n d.
3 . Give the n a me a n d sym bol o f e ach el em e nt h avin g th e valence co n fig u ra tion [n o b le gas] n.s2I1J/ .
17. Write a b ala n ce d che m ical equa tion for th e p repa ra· tio n o f H 2 (a n d CO) hy th e reactio n of C I 1.1 a nd wat e r. Us ing data in Ap pen di x L. calcu la te !3. J fO , !3. r G" , an d !3. r S o for th is react ion .
4. Give sym bo ls and n a m es for four mon a to m ic ion s t hat h ave t h e sam e electron con fig ura tio n as argon.
18. • Usin g- d a ta in Appen d ix L, ca lculate !3. r H o , !3. r Go , a n d !3. rS" fo r th e re ac tio n of ca rb o n and wa te r to g ive
5 . Sele ct o ne o f th e alkali m et al s, a n d wri te a b alan ce d ch e m ica l equa tio n fo r its reaction with chlorin e . Is th e re act io n like ly to b e exo the r mi c o r e n do the rm ic? Is th e p ro d u ct ioni c or mole cular? 6. Selec t o ne o f th e alk aline e a rt h metal s and wri te a ba la n ce d ch e m ica l equ atio n fo r its reactio n with o xyg en . Is th e reactio n lik ely to be exo th e r m ic o r en d ot h erm ic? Is th e p rodu ct ion ic o r mole cular? 7. Fo r th e p ro d uct of th e re action YO Il se le ct ed in Stu d y Q u e stio n 5 , p red ict the followin g p h ysica l p ro p e rti es: co lo r, slate of ma tter (s, f. o r g ), so lub ility in wa ter. 8. Fo r th e p ro duc t o f the reac tio n yo u sel e ct ed in Study Q uestio n 6 , pred ict the foll owin g p hysical p ro p e r ties: co lor, state of m at ter (s , C, o r g ) , so lu b ility in wa te r. 9 . w ould yo u ex pe c t to fin d ca lciu m oc cu rri ng n a turall y in th e eart h 's crust as a fre e c le m e n t? Wlry or why no t?
10. Wh ich of th e fi rst 10 c le men ts in th e pe ri od ic table a re fou nd as tree de me n ts ill th e eart h 's crust? Whi ch dem e nts in thi s g ro u p o ccur in th e ea rt h 's cru st o n ly as p a rt o f a chemi ca l com pound ? 11. Place th e followin g ox ides in ord er of in cr e asing b asicity: CO 2, Si0 2, Sn 0 2'
12. • Place the fo llowin g o x id es in orde r of in cre asin g b asici ty: Na20 . AI20 :1, Si0 2, SO j .
CO a nd H 2 • 19 . A m e th od re ce n tly suggr-stc d for th e p re pa ratio n of h yd ro ge n (and ox ygen ) fr om wa ter p ro cee d s as follows: (a) Su lfu ri c ac id a n d h yd rog en io d id e are fo r med from su lfu r d ioxide , wa ter, an d iodine . (b) Th e su lfu ric ac id fro m the fir st ste p is d e co mpo sed by h eat to water, su lfur dioxide , and o xygen . (c) Th e h yd ro g-e n iodi n e fro m th e fir st ste p is de co mposed with h e a l to hydro ge n a n d iod ine . Write a b ala n ce d eq uat ion for each of th e se ste ps, an d show th at th e ir sli m is th e d eco m p o sitio n of wa ter 1.0 fo rm h yd ro gen an d o xygen .
20. Co m pare th e m ass of 112 expected from th e reac tion of steam ( H 20 ) p e r mo le of meth an e , p etrol eu m , and co al . (Assu m e co m ple te reaction in each case. Use C Il 2 a n d C H as re p re se ntative formu las for p e tro leu m an d co al, res p ecti vcl y.)
Alkali Metals 2 1. Write eq ua tio ns fo r til e rea cti o n o f so di u m with eac h o f til e h al o g en s. P red ict a t le ast two physical properties th at arc co m m o n 1.0 a ll of the alkali m e tal h alid es.
22. • Write bala n ced eq u ati o n s for til l' reactio n o f lith iu m , so d iu m, and p ot assium with 0 2' Spec ify wh ich me tal form s a n ox ide , wh ich for ms a p ero xid e , an d wh ich fo rm s a su pc rox id c.
1011
23. Th e e lec trolysis o f aq ueo us NaC I gi ves Na O II , C12, a nd 112 , (a ) Wri te a hal a nc ed equa tio n fo r th e p ro cess. (h ) III th e U ni te d Sla tes, 1.19 X 1010 kg o f Na G l 1 a n d 1. 14 X 10 10 kg o f CI2 we re produced in a re ce nt yea r. Do es th e ratio of m asses of N aO lI an d Cl 2 p rodu ced ag ree with th e ra tio o f m asses ex pe cte d from th e ba lanced equa tio n ? If not, what d o c s th is tell yo u ab o u t th e way in wh ic h Na O H an d O 2 arc ac tu al ly p rod u ced ? Is the e lectro lysis o f aq ueo us Nae l th e o nly so urn' of th ese ch e m icals? 24 . • (a ) Wr ite eq u a tion s for th e half-rea cti on s th at o ccur a t th e ca th o d e a n d th e anod e when a n aq u eous so lu tio n o f KCI is e le ct ro lyzed . Whi ch ch e m ical spe cies is o x id ize d , and whi c h ch em ica l sp ecies is re d uced in th is reactio n ? (b) Pred ict th e p roducts fo r med when an aq ueou s so lutio n of Cs i is e le ctro lyzed.
Alkaline Earth Elements 25. When m ag ne sium bu rn s in a ir, it form s b oth a n o xide a nd a n itrid e . Write ba la nc e d e quations for the fo r m ation of bo th com pou n d s.
26 . • Ca lciu m rea ct s with hyd ro gen gas a t 300- 400 °C to fo r m a h yd ride . T h is co m pou n d re acts readi ly wit h wat e r, so it is an ex ce lle n t d ryi ng ag e n t fo r organ ic so lve nts. (a) Write a balan ced equa tion sh owing th e fo r m a tio n o f ca lc iu m h yd rid e from Ca a n d 1I2. (1)) Write a balanced equa tio n fo r till' re action o f calci um h yd ride with wa te r (Fig u re 2 1.7) .
27. Na m e th re e u ses o f lim e ston e . Write a bal an ced equation fo r th e re ac tion of limestone with CO 2 in wa te r.
32 . • Th e b o ron tri halidcs (e xce p t BF3 ) h yd rolyze co mp let e ly to bo ric acid and th e aci d Il X. (a) Wri te a b alan ced eq u atio n for th e re actio n of RCh with water, (b) Calc u la te d rH o for th e hydrolysis of BCh u sing d at a in Ap pe nd ix L a n d the foll owin g in formati o n: 6.JW [BCl,,(g )] ~ - 40'1 kJ/mol; 6.JW [B(OH ),(s)] ~ - 1094 ~ J / 1Tlo1. 33 . Wlu -u b o ro n h yd rides hurn in air, th e reactio n is vc ry exot hermic . (a) ,,,rrite a b a la n ced eq ua tio n for th e co m b us tio n o f B., H 9 (g) in ai r to g ive B2 0 :~ ( S ) a nd H 20 (t') . (b ) Calcu la te th e e n th alpy o f co m b u stio n fo r B,,1I 9 (g ) (!J.f H o = 73.2 kJ / mo l), a nd co m p a re it wit h the enth alp y o f com bus tio n of B2Hl i o n page 984 . (T h e e nth alp y o f formatio n o f B..i ) 3(S) is - 127 1,9 ~ I / mo l. ) (c) Co m pare th e e n thal py o f co mbust ion of Cd ~ (,( g) with th at o f R21 Ifj (g ) . Wh ich tran sfe rs m o rt' e n e rgy as h eat per g ra m? 34 . • Dibo ran e can b e p repared by th e reactio n o f N a BH j a n d 12. Wh ich substance is o xid ized, a n d whi ch is re du ced?
35 . • Write bala nced equa tio n s for th e re acti on s of aluminum with H CI (aq ) , C12, an d 0 2' 36 . (a) Writ e a bala n ced eq ua tio n for the re actio n of Al an d I120 (C) to p ro duce I12 a n d A1 20 3 . (h) Usin g th e rm o d yn a m ic d a ta in Ap pendix L. calculat e tJ.,1 r , !J. rS o• an d 6.,(;0 for th is re act ion . Do thes e d at a indicate th at till' re action sh o u ld favo r th e p roduc ts? (c) "\ih y is a lu m in um rrn-ta l un affe ct ed by wa te r?
37. Alum in u m di ssol ves read ily in hot aq ue o u s Na O H to give the alu m in a te io n , [AI (O H )11- , a nd H 2. Write a h al anced equation for thi s re ac tio n . If yo u beg in with 13 .2 g o f AI, wh a t vo lu m e (i n mill ilit e rs) of H 2 g as is p roduced when th e gas is m e asu re d a t 22.5 °C and a pressu re o f 735 mm H g?
28. Ex p la in what is meant by "h ard wa te r." Wh at ca u ses h a rd wa te r, an d wh at p ro bl em s are assoc iate d with it? 29 . • Ca lci u m o xid e , Ca O, is u sed to rem ove S0 2 fr om power plant ex h au st. Th ese two co m p o u nds react to gi ve so lid Ca SO :\. Wh at mass of S0 2 can b e re moved usin g 1.2 X lO 3 kg of CaO ?
30. Ca( O Hh ha s a «.; of 5.5 X 10- 5 , wh e reas K,I' fo r 7v1 g (Oll h is 5.li X 10- 12. Calc ula te th e eq uilibriu m co nsta n t for th e reacti on Ca (O H ), (s)
+
~l g2+ ( aq )
Ca2+ (a q )
+
Mg (OH ),( s)
Exp lai n wh y th is re act io n can h e use d in th e co m m ercial isol ation o f m agnesiu m fro m sea wa te r.
Boron and Aluminum 3 1. Dr aw a p o ssib le stru ct u re for the cycli c a n ion in th e salt K:;R;10tj and th e ani o n ill Ca 2R20 ",.
1012
I
38. Alum in a, A120 3 , is am phote ric. Am o n g exa m ple s o f its a m p h o te ric ch aracter a re th e rea cti on s th at o ccur wh e n AI20 :'l is h eated stro ngly or "fu sed " with ac idic oxid es a n d b asic o xides. (a) Wr-ite a hala u ccd eq u ation fo r th e re action o f alumin a with silica . an ac id ic ox ide , to g ive alu m in u m met asilicatc. A12 (Si0 3h (b) Write a bal an ced eq u ation fo r th e re actio n of alum in a with til l' b asic o xid e Ca O to gi ve ca lci u m alumin at e , Ca (AI0 2) 2' 39. Alu mi n u m su lfa te is th e m o st co m me rc iall y impo rt ant al u m in u m co m p o u n d , afte r aluminum o x ide a nd alumin um hydroxid e . It is p roduced fro m the rea ction o f alu m in u m o xid e a n d su lfu ric aci d. Wh at m ass (in kilogra ms) o f alu m in u m ox ide an d su lfu ric acid m ust be us e d to m a nufacture 1,00 kg of alum in u m su lfa te ?
•
more challenging • in OWL
Blue-numbered questionsanswered in Appendix0
40. "Ae ra ted " concrete bricks are wid ely used build ing m ate rials. T hey a re ob tai ned by mixin g gas-fo rmi ng ad ditive s with a m o ist m ix ture of lime , ce me n t, a nd possibly sa n d . Ind ustrially, th e fo llo wing reactio n is im po rtan t: 2 Al (s) + ~ Ca (O H ),(s ) + () H,o (f) ---> ~ CaO . Al,03 . 6 H, O (s) + ~ H, (g ) Assum e th at tlu- m ix ture of reacta n ts co n tains 0.56 g of AI (as we ll as exce ss calci um hyd ro xide an d wate r) fo r eac h brick. wh at vo lu m e of h yd ro ge n gas do you ex pe ct a t 26 °C an d a pressur e of 74!1 mm JIg ) ?
Silicon 41. De scri be t he stru ctu re of pyro xe nes (sec p age 988 ) . wh a t is the ra tio of silico n to o xyge n in thi s type of silica te ? 42 . De scr ibe how u ltrapu re silico n can be produced fro m san d . 43 . • Silicat e stru ct u res: Draw a struc tur e , a nd give th e charge for a silicate a ni o n with the fo rmu la [Si6 0 18 ]n- . 44. Silicate s often h ave cha in , rihbon , or shee t st ruc tures . O ne of th e sim ple r ribbo n stru ctu res is [Si20 / - ] Il' Draw a struc tu re fo r th is a n ion ic ma teri al.
Nitrogen and Phosphorus 45. Co nsult th e d ata in Ap pen di x L. Are an y of the n itro ge n o xides listed th e re stable with respe ct to de co mpositio n to N 2 an d 0 2? 46. US(' da ta in Ap pen d ix L to calcu late th e e n tha lpy a nd free e neq..,,;· c ha nge for th e re ac tio n 2 NO , (g ) ---> N, O , (g ) Is thi s re ac tion ex o therm ic or e n d o therm ic? Is th e reaction p roduct- or react a nt-favo re d? 47 . Usc da ta in Ap pe nd ix L to ca lculate th e e n tha lpy an d free e ne rgy ch a ng e fo r th e reac tio n 2 NO (g )
+ O , (g ) ---> 2 NO ,(g)
Is th is reaction exo th e rmic o r e n d o th er m ic? Is th e reacti o n product- o r rea ctant-favo re d? 48 , • T he o verall reac tio n invo lved in the ind us trial synth esis of n itri c aci d is
Ca lcula te ~ rr. ° fo r th is re actio n and its e q u ilibriu m co nsta n t a t 25 °C. 49 . A m ajor use of hyd razinc , N 2H 4 , is in stea m bo ile rs in powe r pla n ts. (a) T he reac tion of lryd raziuc with O 2 d isso lved in wate r gives N 2 a nd wa ter. Wri te a balan ced e q ua tio n for thi s reaction .
•
morechallenging
• in OWL
Blue-numbered questions answered in Appendix 0
(h) O 2 diss o lves in water to the exte nt of 0.00 44 g in Ion. ml , of water a t 20 °C. To co n sume all of the d isso lved O 2 in 3.00 X 10-' L of wa te r (e n o ugh to fill a sm all swim m ing pool) , what m ass of N:! IJ 4 is needed ? 50 . • Be fo re hyd razine ca me in to usc to re m ove d isso lved ox yge n in th e wate r in ste am bo ilers, l':a :!SO j was co mmo n ly used fo r thi s purpose : 2 Na , S0 3(a q ) + O , (aq ) ---> 2 Na,S(}, (aq) Wh at rn ass of Na2S0 3 is req u ire d to re m ove O 2 fro m :1 00 X 10-1 L of wate r as o u tline d in Study Q uestion 49? 51. Review th e structu re o f ph ospho ro us ac id in Table 2 1.6. (a) Wh at is th e ox ida tio n n umber of th e p ho sp h orus a to m in thi s ac id ? (h) Draw the struc tu re of di phosp horous aci d, 114P20:;. \ Vhat is th e m a xim u m n u m ber of pro ton s th is ac id can d issocia te in water? 5 2, U n like ca rb o n , whic h ca n fo rm ex tended ch ain s of a to m s, nit ro ge n ca n fo rm chai ns of \'el]' limi te d lengt h . Draw the Lewis electro n dot str uct u re of th e az ide ion , 1'\3- ' Is th e io n lin e ar or bent?
Oxygen and Sulfur 53 . In th e "co n tac t pro cess" for ma king su lfur ic acid , su lfu r is first burned to SO:!. Envir o nm en tal re stric tio ns allow no m o re th a n 03 0% of thi s S02 to be vented to the a tm osph e re. (a) If e n o ug h su lfu r is burn e d in a pl a nt to prod uce 1.80 X 10 1; kg of pu re , a nh yd ro us IJ 2SCl, per day, wha t is the ma xi mu m amou nt of S02 th at is allowed to be ex hauste d to the a tmosp he re ? (b) O ne way to preve nt an y S0 2 from re ach ing th e a tm os ph e re is to "scr u b " th e ex ha us t gases wit h slake d lim e , Ca (O Ilh: Ca (O Il ), (s)
+ SO , (g ) ---> CaSO ds ) +
2 CaS03(s)
1I,o (t )
+ O , (g ) ---> 2 Ca SO , (s)
What ma ss of Ca (O H h (i n kilog ra ms) is needed to rem ove th e S0 2 ca lcu lated in pa rt (a)? 54. • A sulfu ric acid pl an t pro duces a n e n ormous am ou n t of hea t. To keep costs as low as po ssible , mu ch of thi s h ea t is used to m ake ste am to generate el ec trici ty. So me of th e e lect rici ty is u sed to ru n th e plan t, a nd th e ex cess is so ld to th e local elec trical ut ility. T h ree rc actio us are impo rt an t in su lfu ric ac id pro du ctio n : ( I) bu rnin g S to S0 2; (2) o xida tio n of SO :! to SO:;; an d (3) re ac tio n of 50:\ with H 20 : So, (g)
+
H,o (in 98 % H,SO .,) ---> H,SO , (t)
T he e ntha lpy chan ge of th e th ird reaction is - 130 ~ J/m ol. Estima te the ent h al py ch a nge whe n 1.00 m o l of S is use d to p ro d u ce I .OO mo l o f HzSO j • H ow much ('n e rgy is p rod uced per m et ric LO ll of Il zS0 4?
I
1013
(c) the sta te o f th e el e m ent (s, f , or g) unde r sta nd ard co nd itio ns
55. Sulfur fo rm s an io n ic ch ains of S a to ms call ed polvsultid e s. Dra w a Lewi s e lec tron d o t struc ture for the S/ion . T h e S/ - ion is th e di sulfid e ion . an analogue of th e pe ro xid e ion . IL occurs in iron pyri te s, FeS2' 56. Su lfur form s a rang e of co m po u nds wit h fluorine . Dra v.. Le wis electron do t struc tu res for S2F2 (co n nec tivity is FSSF) , SF2, SF 4, SF(j, and S2FIO' Wh at is the ox idation number of su lfu r in ea c h o f the se co mpound s?
Fluorine and Chlorine
57. Th e ha log en o xid es and o xoanions a re go o d o xidizin g agents. For example , th e reduction of brom ate ion has an EO value o f 1.44 V in acid so lu tio n: 2 IIrO , - (aq )
+
12 1I' (aq )
+
10 e ---->
IIr, (aq ) + Ii H,O (f) Is it possib le to ox idi ze aq ueous 1.0 M Mn 2+ to aq ueous ~1n04 - with 1.0 M b rom a te io n? 58 • • T he h ypoh alite io ns , XO - , are tilt' an ion s of weak acid s. Calcula te the p IT of a 0. 10 M so lu tion o f NaCIO. What is the co nc eu tra tio u of IICIO in thi s solut ion ? 59 . Bromine is ob ta in ed from brine we lls. T he p ro cess in vo lves trea tin g wa ter containin g bromi d e io n with C I2 and ex tracting the Hr2 fro m the so lu tio n us in g a n org a nic so lve n t. Write a balan ced equa tion for th e reaction of C12 an d Br -" . What are th e oxi di zin g a n d re d ucin g ag e nt s in thi s re ac tion ? Using the tab le of sta ndard reduction p otentials (Ap p e nd ix M) , ve rify that thi s is a p ro d uct-favo red reacti on . 60. To prepare ch lorine from chl oride ion a st ro ng o xid izing agent is req ui red. Th e d ic hrom a te io n , Cr2072- , is o ne example (see Figure ~ 1.32 ) . C o ns u lt the tabl e of standa rd redu ction po tentials (Ap pe nd ix M) , a n d id en t ify seve ra l other oxidizin g agents th at may he su itable . Writ e balanced equation s fo r th e reactions of th ese su bst an ces wit h chloride ion . 6 1. • If an elect rolytic cell for p ro d u cin g F2 (Figu re 21 .31 ) ope rates at 5.00 X 10:1 A (a t 10.0 V) , wha t m a ss of F2 can be p rodu ced per 2 4-hour d ay? Assume th e COI1\'e rsio n of F- to F2 is 100 % . 62 . Halo ge n s com bine with on e a no ther to p roduce interlutlOgl'lH suc h as HrF3 . Ske tch a possible m olecular struc tu re fo r thi s mole cule , and d ecide if the F- Br-F bond a ngles will he less th an or grea te r th a n id eal.
For help in th is ques tio n , co ns u lt Figure ~.4 o r use th e periodic tabl e "to o l" in Che m istryN ow. Th e latter provides a pic ture of eac h element a nd a listing of its properties. 64. Consider th e che m istries of C , Si, G e , and Sn . (a) Writ e a balan ce d che m ical equat io n to depict th e reac tion of eac h el e ment with e le me n tal ch lo rine. (h) Des cribe the hond in g in e ach o f th e p roducts of the reactions with chlorine as ion ic o r covalent. (c ) Co m pare the re ac tio ns. if an y, of some C roup 4A ch lorid es - CC14 • SiC14 , and SnCI 4 - with wat er. 65. Con sider the ch emistries of th e clem en ts potassium , ca lc iu m, g alliu m , ge r ma n iu m , a nd arse n ic. (a) \Vrite a balanced chemical equatio n d epicting th e reac tio n of e ac h cle m e n t wit h e le m e n ta l ch lo rine . (h) Des cribe the bon di ng in e ach o f the prod ucts of the reactio ns with ch lo rin e as io n ic or cova lent. (c) Draw Le wis elec tron d ot struct u res fo r th e products o f th e reaction s o f g allium and arse n ic with ch lorim'. wh at are the ir elec tron-pai r and mol ecula r geometrie s? 66. When HCI:1 g as is p assed throug h an electric di scharge , sma ll amounts of the reactive m o le cul e B2CI.1 a rc produced. (T h e molecu le has a B-B covalent bond.) (a) Draw a Lewi s elec tron do t st ructu re for B ~CI 4 ' (b) De scr ibe th e h ybridization or the H atoms in th e m o le cule an d the geo m etry a ro u n d ea ch R a to m.
67 . Co m p le te a nd ba lan ce th e foll owin g equation s. (a) KCIO" + h e a l ---> ( h) Il , S (g ) + O , (g ) ---> (c) Nafs) + O , (g ) ---> (d ) PI (S) + KOII (aq ) + II ,O (t) --->
(e) N I I4I'\O J(s) + heat -)o (I) In (s) + IIr, (f ) ---> (g) SnCI , (f)
+ H, O (f)
--->
68. (a) H ea tin g barium o xide in p ure ox ygen give s bariu m peroxide . \Vrit e a bala nce d equati on for thi s re action . (b) Barium peroxid e is an ex ce llen t ox id izing age n t. \ Vrite a balanced equation for t he reaction of iro n wit h ba rium peroxid e to give iron (Ill ) oxi de an d barium o xide .
77u' questions a re not rlf'S igllfftnl as to l)j)(' or [oration in the dlfljr fer: Tlw)' ma)' combine severa l concepts.
69. Worldwid e production o f silico n car bide , SiC, is seve ral h u n d re d th o usan d ton s annually. If yo u wa n t to produce 1.0 X 10" m e tric ton s of SiC, wha t m ass (m e tri c to ns) of silico n sa nd (SiD 2) will yo u use if 70% of th e sa nd is converted to SiC?
63 . For eac h o f the third-period elements (Na t hroug h Ar ) , identify th e follo win g: (a) whe ther the el ement is a metal , nonmetal, o r meta llo id (b) the col o r a nd appearance of the elemen t
70 . To store 2.88 kg of gasoline with an enl''fg}' e q u ivale nc e of 1.43 X I O ~ .J req ui re s a vo lu m e of 4. 1 L. In compa riso n . 1.0 kg- o f I-1 :1 has the sa me energy equivalence. Wh at volum e is required if th is quan tity of [12 is to be sto red a t ~5 "C a n d 1.0 an n of pressu re ?
General Questions
1014
I
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more challenging • in OWL
Blue-numbered questionsanswered in Appendix0
71. Using data in Ap pen dix L, calcu la te ~ l G o va lu es for til e decompos itio n o f ~1C0 3 to MO and CO 2 where M = Mg , Ca. Ba. \ Vha t is the rel ative tend en cy o f these
(b) Th e same p ropulsio n syste m 'was use d by th e Lunar Lande r on moon m issio ns in th e 1970s. If th e Lander used 4 100 kg of 11 2 NN (CH :1h . wha t m ass (in kilogra ms ) o f N t 0 4 was required to rea ct wit h it? Wh at mass (in kilogra ms ) of each o f the re actio n p rodu cts was ge nerated?
carbo nates to decompose?
72. Am moni u m perchlora te is use d as the oxidizer in th e so lid-fu e l booste r rocke ts of the Space Shu ttle . Assume th at o ne la u nc h re q ui res 700 to ns (6.35 X 10" kg ) of th e sa lt, an d the sa lt deco m poses accord ing to the eq ual ion on page 1009 . (a) Wh a t mass of wa ter is produced ? What mass of O 2 is p ro d uced ? (b) If th e O 2 p ro du ced is assu med to react with th e pow de red alum inum present in th e rocke t engine , what m ass of a lu minum is req u ired to usc u p all of the O t ? (c) Wh at mass of A120 :1 is p roduced ? 73 . .... Metals reac t with h yd rog e n h alide s (suc h as Hf.l ) to give th e m etal h alid e a nd h yd ro gen :
M(s) + " IIX (g ) ---> MX,,(s) +
Y,n 1I, (g )
The free l' lH'rgy cha nge for th e reaction is
~J; o
t./ ;O(MX,,) - " t.j G" [HX( g)] (a)
~f G o fo r HCl (g ) is - 95 . 1 kJ /m o l. wh at must be the value fo r ~J G o fo r MX " for th e reaction to be
product-favored? (b) Which of th e fo llowi ng metals is (are ) p redicte d to haw product-favo red reaction s wit h Il CI(g ): Ba, Pb , IIg , Ti ? 74 . Ha lo ge n s fo rm polyhali dc ions. Sketc h Lewis e lec tro n dot structu res and m o le cu lar stru ctu res for th e fo llowing io ns: (a) 13 (b) BrC l, (c) C lF, + (d ) An iodide ion an d two iod ine m o le cu les form th e [5- ion . He re , th e io n has five [ atoms in a row , hut the ion is not line ar. Draw th e Lewis dot structure for the ion, a nd propo se a stru ctu re for th e ion .
78 . .... Liq ui d HCN" is dangerously unstabl e with respect to trimer fo rmation-that is, fo rmat io n of ( HCNh wit h a cyclic struc ture . (a) Propo se a stru ct ure for thi s cycl ic trimer. (b) Estimate th e e nergy of th e tri m e rizat io n reactio n us ing bond di ssociation cntha lpics (Table 8. 10) . 79 . Use ~JHo da ta in A p pendix L to ca lcula te th e en t halpy change o f th e reaction
2 N, (g) + 5 O, (g) + 2 Il , O (f) ---> 4 II N0 3 (aq ) Spe cu late on whe th er suc h a reaction could be used to "fix" uiuogt-n . Would re searc h to fi n d ways to acc om pl ish th is reactio n be a u seful e nd eavor? 80 . .... Ph o sph o r us form s a n ex tensive seri es of o xoanions . (a) D raw a struc tu re, and give the cha rge fo r an oxop hos p hate anion with the form ul a [P,jO I:d n-. How man y ionizable H a toms sho u ld th e com plet ely p roton au-d ac id h ave ? (b) Dr aw a stru ctu re , an d give the cha rge fo r an oxop hosp hate an io n with the formul a [P/) It ]" -. How m an y ionizable H atoms sho u ld t he co m ple tely p roton atcd aci d h ave ? 81 . Boron and h yd ro ge n form an exte nsive fa mi ly of compounds, an d the di agram be low shows how they are related by re act io n .
S,10 0' 10
75 . • T he sta nd a rd enthal py of form a tio n of O Ft ga s is + 24.5 ~J /mo l. Calcula te th e average O -F bond entha lpy. 76 . Calc iu m flu o rid e can be used in the fluoridation of mu n ici pal wa ter supplies. If yOll wan t to ac h ieve a tluo ride io n con cen tration of 2.0 X 10 - 5 M, wha t mass of CaFt m ust yOll use fo r 1.0 X 10(' L o f wa ter? 77 . T he steerin g ro ckets in the Span' Shu ule lise N 20 4 and a de rivative of h vdrazinc , 1, l-d ime thylhvdrazi nc ( pa ge 249 ). This m ix ture is called a 'l)'/m KOlir/ll d bec ause it igni te s when th e reac tan ts come in to co ntact:
H, NN (CH 3 ), (C)
+ 2 N,O, (f)---> 3 N, (g ) + 4 1I,o (g) + 2 CO , (g )
(a) Id e n tity th e o xidizing age nt and th e reducing agent in th is reacti on .
•
morechallenging
• in OWL
Blue-numbered questions answered in Appendix 0
C/
C10 100'1 ' C
/
/ c
A~
15' C
1:0:
11 0 'C ~
/
o
E 95° C
T he fo llowi ng table give s th e we igh t pe rce nt of boron in each o f the com poun d s. De rive th e em pirical an d m o le cul ar form ul as o f com p ou nds A-D. Substance
State (at SiP)
Mass Percent B Molar mass (gjmo l)
A
Ga s
78.3
17.7
B
Gas
81.2
53.3
C
Liquid
83 .1
65. 1
Liquid
85. 7
63.1
Solid
88.5
122. 2
D
I
10 15
82 . A In 177 4 , C. Sc h e ele o b tai n ed a g as by re acting pyrolu sitc (Mn O\!) with sulfu ric aci d. T h e gas. wh ic h h ad b een o b ta in ed th a t sa me year by J o se ph P riestley b y a d ifferent m e thod, was an ele men t. A. (a) What is th e elem ent iso lated by Sc hee le and P riestley? (b) Elem ent A co m bi n es with a lm os t all other e lement s. FOf ex ample , with ces iu m it g ives a co rnpound in whi ch th e mass percen t o f A is 19.39 % . T h e cle me n t co m b ines 'with h yd ro gen to give a co m p o u nd with a mass percent of cl e m en t A o f 94 . 12 % . Determine the fo rmulas o f the cesiu m a n d h ydrogen co m po u n ds. (c) T h e co m p o u n ds of ce siu m a n d hyd rog e n with clem ent A react wit h o n e another, Write a b al an ced e q uatio n fo r th e re actio n .
Ad dition o f aq ueo us ll CI to th e so lu tio n of B a nd eva pora tion o f th e f esull in g so lut io n to dryness yield 1.0 55 g o f a whi te so lid D. Wlu-u D is pl aced in a Bu nse n b u rner Fla me, it colors the flam e grccn. Fin all y. if th e aq u eous so lu tio n of B is t re ated wit h su lfu r ic ac id, a wh ite p re cipitate, E, fo rm s. Id en tify the le ttered com pou n ds in th e reactio n sche m e.
In the Laboratory 83 . On e m a terial n eeded to make sil ico nes is di ch lorodimcth vlsila n c , (C H 3hS iCI\!. It is ma d e by tre ating silico n po wd er at a bou t :100 °C with C H :~C I in th e p resence of a coppe r-eontaining ca ta lyst. (a) Wri te a b alan ced equation fo r th e reac tio n . (b) Assum e yo u ca r ry out the reaction o n a small scale ",. ith 2.65 g of silico n. To mea su re th e C1I 3CI gas, yo u fill a 5.60-L flask at 24.5 -c. What p re ssu re of C H 3Cl gas mu st yo u h ave in th e flask to h ave th e sto ich io met rica lly co rrect am o u n t o f the comp ound? (c) What m ass of (C H 3h SiC I2 ca ll he produ ced from 2.65 g of Si a n d exce ss C H 3C I?
84 . Sodium boro hvdri d c, NaB H " reduce s m an y m e ta l io n s to th e metal. (a) Wri te a b al an ced e quation fo r th e rea ction o f Na B II~ ,v-ith AgN0 3 in wa ter to give silver metal, 112 gas, bo ric a cid , and so d iu m n itrate (page 984 ) . (h) What m ass of silve r ca ll b e produced from 575 m l, of 0 .0 11 M AgN () :~ an d 13.0 g of Na BH ,t?
The salt s (aCl2, SrCl2, and BaCl2 were suspended in meth anol. When the methanolis set ablaze. the heat of combustion causes the salts to emit light of characteristic wavelengths: calcium salts are yellow; strontium salts are red; and barium salts are green-yellow.
87. A In 19 :17, R. Sch wa rtz an d ~1. Schmi csscr prep ar ed a yellow-orange bromine o xide ( B r O~) h y tr ea tin g Br2 with oz o ne in a fluoro carbon so lve n t. Many ye a rs la te r, .J. Pascal fo und th at, on h e a ting , thi s o xide d e comp o sed to two oth er ox ides, a le ss vo la tile go ld e n ye llow ox id e (A) a n d a m ore vo lat ile de ep-brown oxi de (B ). O x ide B was late r id entified as 1\1"20. To d etermine th e formul a for oxide A, a sa m p le was tr eated with iod ide ion . T he rea cti o n liberat ed io din e . which was titra te d to an e q uiv ale nc e poin t with 17.7 mL of O.(Hi5 M sodi u m th iosulfat e.
Co m po un d A was also trea ted with AgN 0 3 , an d 14.4 mL of 0. 020 M Ag!'\()3 was requ ired to com pl e tely p recip ita te th e bromin e from th e sa m p le. (a) Wh at is th e fo rmula of th e unkn own bro mine o xid e A? (h ) Dr aw Lewis structures for A aud Rr20. Sp ecu la te on th e ir mole cular geom etry.
85 . A co m m o n a n alytical m et hod for hyd razin c invo lves its o xida tio n with iod a te io n. 10 3 - , in aci d solu tio n . In the p ro cess, hvd razin c ac ts as a fou r..el e ctro n re d uc ing ag en t.
N,(g)
+ 5 11,,0 ' (aq) + 4 e N, H, '(aq)
-->
+', H,O (f)
Write th e ba lan ced equation fo r th e re action of hyd razinc in acid so lu tio n (l'\2H r'+) wit h IO j - (aq ) to give:":2 and 12, Ca lcu la te EO for thi s re act io n. 86 . • \ Vh e n 1.00 g o f a wh ite so lid A is stro ngly h ea te d , yOll ob tai n anot h er whi le soli d. B. and a gas. AI1 e x pl'r~ int e n t is carrie d o u t on th e gas, showi ng tha t it exerts a p fe ssu re of 209 mm II g in a 4 50~lTlL flask a t 25 "C. Bubbling th e g as in to a so lu t io n of Ca (O H h g ives a not h er wh ite so lid, C. If th e whi te solid B is ad ded to waLe I', th e re su lting so lu tio n turn s red litm us p aper b lu e . 1016
I
88. • A mixtu re o f PC1" ( 12.4 1 g) an d excess N II ~ (:I was h eated at 145 °C fo r six ho u rs. Th e two reacted in cquirn c lar amo u n ts a n d evo lved 5. 14 L o f l l Cl (a t STP ) . T h re e su bs ta n ces (A, B, and C ) we re iso lat ed fro m th e reac tion m ix tu re . Th e th re e su bs ta nc es h ad th e sa me e le men ta l co m posi t io n hut differed in thei r m olar mass. Substan ce A, h ad a mol ar mass o f 347.7 g / mo l a n d B h ad a mo lar m ass o f 46:1.5 g/ m o l. G ive th e e m pi rica l a nd molecul a r fo rmul as for A and B an d d raw a reaso na ble Lewis structu re for A.
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more challenging • in OWL
Blue-numbered questionsanswered in Appendix0
Summary and Conceptual Questions T he j(JllowiliK qurs tions mil)' W oP Wll("~f)tJ lmlll this a nd I npI1;ou,\ d W / J{PH .
89 . Din itro gcn trio xid e , :"\203, has th e MrHCIUH' show n IW f (, .
96. Th e stru ct ure o f n itr ic acid is ill us tra ted on page 99 3, (a) Wh y a re the N- O bo nds till' sanu- le ng th , an d why are bo th sho rte r than th e ~ - O ll bond It'ngth? (b ) Rati on ali ze th e bon d ang les in th e m ol ecu le , (c) "11at is th e hyb ridi zat io n of the ce n tra l X a tom? Whi ch o rbi tals overla p to fo rm t he :\- 0 -rr bond ?
97. Assu m e an elec trolysis cel l tha t p rodu ce s chlo ri ne
Th e oxide is unstable deco m posi ng to
~O
an d :\'0 2 in
ti ll' gas ph ase at 25 "C. ~ ,O j ( g )
--> NO (g ) +
~O , ( g )
(a) Exp lai n why one X-O bo nd d ista nce in :"\203 is 114.2 pm , wh ereas th e o ther two bon ds a n." lo ng er (1 2 1 p m ) a nd nea rly eq ual to each 01 1)("1'.
(h) Fo r th e cit-com pos ition reac tion . ~ rll o = + -to.5 kJ / 1110 1 a nd Il r G o = - 1.59 kJ /m o l. Calculate d .\'O a nd K for the reactio n at 298 K. (e) Calc ula te al fO fo r N 20 3 (g) .
from aq1lt'ous so di u m chlo ride opera tes a t 4.6 V (with a cu rren t o f 3. 0 X 10" A) , Cal cu la te the n u m ber o f kilowatt-hours of t'lw q..,T)· req uired to p roduc e 1.00 kg of c hlorine ( 1 kWh = I kilowa tt-hour = 3,6 X 101; .I). 98. Sodi um metal is p roduced hy elec trolysis of molte n sod ium c hloride , T he cell o perates at 7.0 V wit h a curre nt of 2::' X 10:1 A, (a) wh ar mass of so di u m ca n he prod u ced in I h ? (b) How ma n y kilowa tt-hou rs of electric ity a re used to p roduce 1.00 kg of sodi um meta l ( I k\\11 =
3.t> X IO".I)? 99. Th e reduct io n pot entia ls fo r the Group :lA meta ls. EO,
3
90. ... Th e d en sity o f le ad is 1 1.350 g/cm , and th e.' m et a l crys ta llizes in a face-cente red cu bi c un it (l-II. Estimate th e radiu s o f th e le ad ato m.
a re gin- n be low, Wh a t tren d o r tre nd s d o yo u o bserve in th e se d a ta? Wh at ca ll yo u lea rn a bou t the c hem istry of th e G ro u p 3.\ clenu-uts from these d at a? Red ucti on Potential Half- Reaction
9 1. \ h tl have a 1.0-1. fl ask that co n ta ins a m ix tu re o fargo n and h yclro gt'l l. The p ressu re inside the flask is 745 mm II g-, a nd t he temperature is 22 °C , Describe H,(g) + 2 H,O(t ) 2 H,O(t) + 2 e: -----> H,(g) + 2 0W(aq) O,(g) + 2 H,O(t) + 4 c: -----> 4 OW(aq) The rate of iron corrosion is controlled by the rate of the cathodic process. Of the three possib le cathodic reactions, the one that is fastest is determined by acidity and the amount of oxyge n present. If little or no oxygen is present-as wh en a piece of iron is buried in soit such as moist clay-hydroge n ion or water is reduced. and H2 (g) and hydroxide ions are
the products. lron(II) hyd roxide is relativelyinsoluble and will precipitate on the metal surface, inhibiting the further formation of Fe2+.
Anode
Fe(s) -----> Fe' +(aq) + 2 e"
Cathode
2 H,O(f) + 2 e- -----> H, (g) + 2 0W(aq)
Precipitation
Fe Fe(O H),(s)
Net reaction
Fe(s) + 2 H,O(f) -----> HAg) + Fe(OH ),(s)
If both water and O2 are present, the chemistry of iron corrosion is somew hat different, and the corrosion reaction is about 100 times faster than wit hout oxyg en.
Anode
2 Fe(s) -----> 2 Fe" (aq) + 4 e"
Cathode
0,(9) + 2 H,O(t) + 4 e" -----> 4 OW (aq)
Precipitation
2 Fe 2 Fe( OH),(s)
Netreaction
2 Fe(s) + 2 H,O(f) + O,( g) -----> 2 Fe(OH ),(s)
If oxygen is present but not in excess. further oxida tion of the iron(II) hydroxide leads to the formatio n of magnetic iron oxide Fe304' (which can be t hought of as a mixed oxid e of Fe 20 3 and FeD).
6 Fe (O HM s) + O, (g) -----> 2 Fe,O, . H,O(s) + 4 H, O(t) green hydrated magnetite
Fe,O, . H,O(s) -----> H, O( t) + Fe, O, (s) black magnetite
It is the black magnetite t hat you find coating an iron object that has
corroded by resting in moist soil. If the iron object has free access to oxygen and water, as in the open or in flowing water, red-brown iron(III) oxide will form. 4 Fe(OHM s) + O, (g) -----> 2 Fe,O,' H, O(s) + 2 H, O(t ) red-brown
This is the familiar rust you see on cars and buildings. and t he substance that colors the water red in some mou ntain streams or in your home. Anode and cathode reactions in iron corrosion. Two iron nails were placed in an agar gel that contains phenolphthalein and K][Fe(CN)"l. Iron(I1) ion. formed at the tip and where the nailis bent. reactswith [Fe(CN )"p - to form blue-green Fe4[Fe(CN6)] l' 14 H20 (Prussian blue). Hydrogen and OW (aq) are 2! " formed at the other parts of the surface of the nail. the latter being detected by the redcolorof the ~ acid-dase indicator. In this electrochemical cell, regions of stress-the ends and the bent region of the nail-act as anodes. and the remainder of the surface serves as t he cathode.
! j
The corrosion or rusting of iron results in major economic loss.
22 .1
I
--------- - - - .1
Properties of the Transition Elements
1023
so u rces o f mol ybd enu m and tu ng ste n are th e o res mol ybd enite ( ~ 1oS ~) an d wolfra m itc (\VO:l ) . Th is ge nera l tren d is ca rr icd ove r in t he .f block. T hc lanth anid es fo rm p r im arily 3 + io ns. In co ntrast, acti ni de ele me nts usuall y have h igh e r oxi da tio n numbers in th eir co mpou nds; + 4 and eve n + 6 are typi cal. Fo r examp le, UO j is a co m mo n o xide o f u ra niu m, a nd UF fi is a co m po u nd important in pro ce ssin g u ranium fue l fo r nuclear reac tors [... Section 23.6] .
ChemiSlrY.(J.Now'" Sign in at www.thomsonedu.com/login and go to Chapter 22 Contents to see Screen 22.2 for an exercise on transition metal compound s.
Periodic Trends in the d-BLock: Size, Density, MeLting Point Th e periodic tabl e is the mo st u seful sing le re fere nce so urce for a che mi st. No t o n ly d oes it provid e d at a th at have everyday use , but it also o rga nizes th e ele me n ts with rc spect to th e ir che m ical and p hysical properties. Let us look a t three ph ysical p roperti es o f th e tran sition ele me n ts th a t vary pe riodi call y: ato mic ra dii, d en sity and m elting poin t.
Metal Atom Radii Th e var iatio n in atom ic radi i fo r th e tra nsitio n ele me nt s in the fou rth , fifth , an d sixth periods is illu strat ed in Figu re 7.9 . T he rad ii of th e tran sition elem ents vary ove r a fairly narro w ra nge, with a small decrease to a m inimum be ing o bserved aro u nd the middl e of this g ro u p of eleme nts. This sim ilari ty of rad ii ca ll be u nd ersto od base d on electr on co n figu ra tio ns. Ato m size is d et ermined by the elect rons in the o u termost orbital, whic h fo r these ele ments is th e ns o rbital ( n = 4,5, or 6) . Progressing fro m le ft to rig ht in th e pe rio d ic tabl e , th e size decline ex pected from increasing th e number of proton s in th e nucleus is mostly ca nc eled o ut by an opposi ng effect , re p ulsion fro m additi onal e lectrons in the ( 11 - I ) d orbitals. T he radi i of the d-block ele me n ts ill the fifth and sixt h peri od s ill ea ch gro u p are almos t identical. Th e reason is th at th e lanth anide elem en ts imm edi atel y precede the th ird series o f d-block elements. The filling o f 4Jorhitals is acco mpa nie d by a steady con tractio n in size, co nsis te n t with th e gene ra l tre nd of d ecreasin g size fro m left to right in th e period ic table. At th e point whe re th e 5d o rb itals begin to fill again , th e radii have d ecreased to a size simi lar to th at of e lements in the p revio us pe rio d . T he d ecrease in size th at re su lts from the filling of th e 4J orbitals is give n a specific nam e , th e lanthanide contraction. The simi lar sizes of the second- and thi rd-p e riod d-hlock ele me nts have significant co nseq ue nces for thei r che mistr y. Fo r exam ple , th e "platin u m gro up m et als" (Ru , Os , Rh , II', I'd , and Pr] form simi lar co m po u nd s. T hus, it is no t su rp rising th at min e rals co n tai ni ng these metals are fo u nd in the sa me geological zo nes o n ea r th . No r is it su r prising that it is di fficult to se parate th ese e leme nts fro m one ano th er.
Density Th e varia tio n in met al rad ii ca uses th e d en sities of th e tr an sition cleme n ts to first increase and th en d ecrease across a per iod (Figure 22.5a ) . Althoug h the overall change in radi i a mo ng these eleme n ts is sma ll, th e effec t is m agnified because the volu me is ac tua lly chang ing with the cu be of th e radius [ V = (4/:i ) 7T' r:I ] . Th e lan th an id e contrac tion e xp lai ns wh y e le me nts in th e sixt h perio d have th e hi gh est d e ns ity. Th e rel a tivel y sm a ll radii of sixt h-perio d tran sitio n m et al s, com bine d wit h th e fac t th at th eir ato mic m asses are co ns ide rably large r th an 1024
Ch apt er 22
I
The Chemist ry of t he Transition Element s
o Fourth period
[J
4000
Fifth period 0 Sixth period
f- -r====-, --====--: :-===,. . ,I·_, I D
fourth period
[J
fifth period
D
Sixth period
3500
20 3000
g
2500
C
'~ 2000
'•"
~
c
'~ 10
~
o
>:
1500 1000
500
38
48
58
68
3B
48
58
6B
w
(~ FIGURE 2 2 . 5 Periodic properties in the transition series. Density (a) and melting point (b) of the d-block elements.
the ir co u n te rparts in the fifth period , ca uses six th-perio d m et al de ns ities to be ve ry large . MeLting Point
T he melting po int of any su bsta nce reflect s th e forces of attrac tio n between th e atoms , molecul es, or io ns th at com pose th e so lid. With tra ns itio n cleme n ts, th e melting poi n ts rise to a maxim um arou nd the m idd le o f th e seri es (Fig ure 22.5b), th en d escend. Aga in, th ese cleme n ts' electro n co n figu ra tions p rov ide us with an ex planatio n. T he variatio n in mel ting point indicates that t he stro ngest metall ic bo nds occur when the d su bshel l is abo u t hal f filled. This is also the poi nt at which th e largest nu m ber of electro ns occu py th e bonding mol ecular o rbita ls in th e met al. (See th e di scu ssion of bondin g in me ta ls o n page 657.) ChemislrY~J.NowN Sign in at www.thomsonedu.com/Ioqtn and go to Chapter 22 Contentsto see Screen 22.3 for more on tran sition metal chemistry.
22.2
Metallurgy
A few me tals occ u r in natu re as the free e leme nts . T his grou p incl ud es copper (Figu re 22.6) , silver, and go ld . Mos t m et als, however, are fou nd as ox ides, su lfid es, hali d es, ca rbo nates, or o ther ionic co m pou nds (Figu re 22.7) . Some metal-eon taining min eral deposits have little econom ic value, eithe r because the concentrat ion o f the met al is too low or becau se th e met al is difficult to separa te fro m impu rities . T he re lative ly few minerals fro m whic h e lements ca n be obtai ned profitabl y are called ores (Figu re 22.7) . Metallurgy is th e genera l name given to th e pro cess of o btain ing met als fro m th eir o rcs. Ve ry few ores are chemically pu re substa nces. Instead , th e d esired mineral is usually mixed with large quant ities o f impuri ties suc h as sand an d clay, ca lled gangue (p ro no u nced "gang "). Ge ne ra lly, th e first ste p in a me tallurgical pro cess is to se p-
FIGURE 22.6 Naturally occurring copper. Copper occurs as t he meta l (native copper) and as minerals such as blue azurite [2 CuC03 • Cu(OHh l and malachite [CuCO, . Cu (OH h l . 22 .2
I
Metallurgy
1025
FIGURE 22. 7 Major sources of the element s. Afew transition metals, such as copper and gold. occur naturally as the metal. Most other elements are found naturallyas oxides. sulfides, or other salts.
H
He
li
Be
B
C
N
0
F
Ne
Na
Mg
Al
5;
P
5
Cl
Ar
K
Ca
Sc
n
V
Rb
Sr
y
Zr
Nb Mo
Cs
Ba
La
HI
Ta
Cr
W
Mn
Re
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
Ru
Rh
Pd
Ag
Cd
In
Sn Sb
Te
I
Xe
Os
Ir
Pt
Au
Hg
Tl
Pb
B;
Rn
'-----
Lathanides Key
D D Sulfides
Oxides
D Can occur uncombined
D D Halide salts
Phosphates
D
D
D
Silicates
Cfrom coal, Bfrom borax
Carbonates
ara te the mine ral from t he ga ng ue. T hen the ore is converted to t he m etal , a redu ction p ro cess. Pyro mc ta llurgy and hydrom crallu rgv are two methods o f recovering met als fro m thei r ores . As th e names im ply, pyrometallurgy involves h igh temperatu res and hydrometallurgy uses aq ueous so lu tio ns (a nd thus is limited to th e relatively low tempe ratu res at which wate r is a liqu id ) . Iron and co p per me ta llurgy illustra te these two met hods of met al prod uction .
Pyrometallurgy: Iron Production T he product ion o f iron from its ores is carried o ut in a blast furn ace (Fig u re 22.8) . T he fu rnace is charg ed wit h a m ixture of ore (usually hem at ite , Fe203) , co ke (which is prim aril y carbo n), and limesto ne (CaC 0 3 ) . A blast of hot air forced in a t the bo tto m o f the fu rnace causes the co ke to burn wit h such an int en se hea t th at th e temperatu re at th e bottom is almost 1500 "C. T he quant ity o f air in put is co n tro lled so that carbo n mono xide is the p rim ary p rod uct. Both ca rbon and carbo n m o no xide par tici pate in th e red uction of ir on (HI) ox ide lo g ive imp u re me tal. Fe,O,( s) + 3 C(s) -----> 2 Fe(f) + 3 CO(g) Fe, O, (s) + 3 CO(g) -----> 2 Fe(C) + 3 CO, (g) • Coke: A Redudng Agent Coke is made by heating coat in a tall. narrow oven that is seated to keep out oxygen . Heating drives off volatile chemicals, including benzene and ammonia. What remains is nearly pure carbon.
Much of the ca rbo n dio xid e for med in the reduction pro cess (a nd fro m heating the limestone ) is red uced o n contact with unburn ed co ke and produ ces more red uci ng age n t. CO,(g) + C(s) -----> 2 CO (g) T he m ol ten iro n flows down throug h the fu rnace and co llects at the bo t to m , whe re it is tapped off throu gh an o pen ing in the sid e . T his im pure iro n is ca lled cast iron or jJig iron. Usu ally, the im p u re me tal is ei th er brittle or so h (u nd esira ble p ro perties fo r most uses) due to th e p resen ce o f imp uri ties suc h as el e me nt al carbo n, p ho sp ho ru s, and su lfu r. Ir on o res gen erally co n tai n silica te m ineral s a nd silico n d io xid e . Lime (CaO), for med whe n lim esto ne is hea ted , reac ts wit h th e se m at e rials to give ca lciu m silicate . 5iO, (s) + ( .O( s) -----> C.5iO,(C) T hi s is an acid- base reaction because CaO is a basic o xide a nd Si0 2 is an acid ic o xid e. T he calc iu m silica te , molt en at the te m peratu re of the blast furn ace and less
1026
Ch a pt er 22
I
The Chemist ry of t he Transition Elements
Charge of a re, coke, and limesto ne
Act ive Fi gure 2 2 .8 A blast furnace. Ihe largest modern furnaces have hea rths 14 mete rs in diamete r. They can produce as much as 10,000 tons of iron per day.
Flue gas
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at www. th omso ned u:comj logi n and go to the Cha pter Contents men u t o explore an inte ractive versio n of th is fi gure accom panie d by an exercise .
Hot gases used to preheat air
- --iH-945 -c
Reducing zone
)
/7~:;~~~~~~= ~.~-- Heated air - --51'9 Molte n iron
d en se than mo lten iro n , floats 011 th e iron . Other no nmetal o xides d issolve in thi s laye r and th e mix ture, called slag, is easily re moved. Pig iron from th e blast furn ace may co ntain as mu ch as 4.5 % ca rbon, 0.3% phosp horus, 0.04% su lfu r, 1.5% silico n, and so me other e lements as well. The impu re ir on must be pu rified to remove th ese no nmet al co m po nen ts. Several processes are available to accom plish th is tas k, but th e mos t im porta nt uses th e basic oxygen furnace (Figu re 22.9) . The p ro cess in th e fur nace rem oves mu ch o f the carbon and a ll o f the ph ospho ru s, su lfur, a nd silico n. Pure oxygen is bl own in to the molten pig iron and o xidi zes phosph orus to P-tOIO, su lfu r to S02, and carbo n to CO 2 , These no n metal ox id es ei t he r cscape as gases o r react with basic oxi des such as Ca O th at are ad de d or are used to lin e th e furnace . For exam ple ,
Th e re su lt is o rdin ary carbon steel: Almos t any d egree offlexibility, hardn ess, stre ng th, and ma lleability can be ac hieve d in ca rbo n stee l by reheat in g and cooling ill a process ca lled tn njJering. T he resulting material ca n then be used in a wide var iety of ap plicat ions. The major d isadvan tages of ca rbo n stee l are th at it co rrodes easi ly and th a t it lo ses its prope rties whe n heat ed stro ng ly. Oth er tr an siti on m etals, suc h as chro m iu m, m an gan e se , and ni ck el , ca n be added d uri ng the stee l-maki ng p ro ce ss, giYing alloys (so lid so lu tio ns o f two o r more me ta ls; see T he Chemis try of Alar/ern Ma terials. page 656) th at have
FIGURE 22.9 Molt en iron bei ng poured from a basic oxygen fu rnace.
22 .2
I
Metallurgy
1027
specific p hys ica l, che mical, a nd m ech an ic al p ro p e rtie s. O ne we ll-known alloy is stai n less stee l, wh ic h co n tai ns 18% to 20% Cr and 8 % to 12% Ni. Stain less stee l is muc h more res ista nt to co rros ion tha n ca rbo n steel. Ano ther allo y o f iro n is Aln ico V. Used in loud sp e ak e r m agn ets because of its perm an en t m agn ctism , it co n tain s five e leme n ts : AI (8 %) , N i (1 4 % ), Co ( 24% ) , Cu (3 %) , a n d Fe (5 1%) ,
Hydrometallurgy: Copper Production
FIGURE 22.10 Enriching cop per ore by the flotation process. The less dense particles of Cu2S are t rapped in t he soap bubbles and float. The denser gangue settl es to t he bottom.
In co n trast to iron o res , whic h are mostly ox ides, most co p pe r m ine ral s are su lfides. Co p per-bearing min eral s includ e cha lco pyrite (Cu FeS2)' cha lcoc ite (C U2S), an d cove llite (CuS) . Beca use ores co ntaining th ese mi nera ls ge nerally have a very lo w percen tage o f co p per, e nrich me n t is necessary. This ste p is carrie d o u t by a pro cess kn own as flotation. First, th e ore is finely powdered. Nex t, o il is ad de d and th e mi xtu re is agi ta te d with soa py water in a large ta nk (Figure 22. 10). At th e same tim e , co m pressed air is fo rced t h ro ug h the mi xture , so that th e lightweight, oilcove red co p pe r su lfide particles are ca rried to th e top as a frothy mi xture . The heavier ga ng ue se ttles to th e bottom of th e ta nk, and the co pper-lade n fro th is skim me d o ff. Hydromcrallu rgv can be used to o bta in co p per from an enriched o re . In o ne me thod, e nric hed chalco pyrite o re is tr eate d with a so lutio n of co p per (II) ch lo ride . A reaction ensue s tha t leaves co p pe r in th e fo rm o f so lid , in so luble CuC I, which is easily se pa ra ted from th e iron th at remain s in so lut io n as aqueou s FeCI2.
Cu FeS,(s) + 3 CuCl, (aq) ----> 4 CuCl(s) + FeCI,(aq) + 2 S(s) Aq ueous NaC I is then added , an d CuC I d isso lves becau se o f th e fo rm ation of th e co mplex ion [Cu CI2 ] - .
CuCl(s) + Cl- (aq) ----> [CuCl,]-(aq) Co p per (I ) com po un ds in so lu tio n are u n stabl e with res pect to Cu (O) and Cu (II ). Thus, [CuCI2 ] - d isproportion at es to th e me tal an d CuC I2, and the lat ter is used to tr eat fu rth er o re .
2 [CuCI,n aq) ----> Cu(s) + Cu2+ (aq) + 4 ([- (aq) Ap p rox imately 10% of th e co p per produced in t he Un ited Stat es is obtained with th e aid o f bacteria . Acid ified wa te r is sp rayed o n to co pper-mi ning waste s th at co n tain low levels o f co p pe r. As th e wate r tr ickles down through th e crus he d ro ck, th e ba cterium Tbiobacitlusfenr oxidrm s b re ak s down the iron sulfid es in th e ro ck an d co nverts iro n (II) to iro n (III) . Iron (III) ions o xidi ze the sulfide ion o f co p per su lfide to sulfate ions, leaving co p per( II) ion s in so lu tio n . Then the copper (I I) ion is reduc ed to metall ic co p per by reactio n wit h iron .
Cu2+ (aq) + Fe(s) ----> Cu(s) + Fe2+ (aq) The purity o f th e co p pe r obtained via th ese me tallu rg ica l pro cesses is abou t 99 %, hut thi s is not acce ptable because eve n traces of im purities greatly dimin ish th e elec trica l co nd uctivity of the met al. Conseq uently, a fu rth er pu rificatio n ste p is ne eded-one invo lving el ectrolysis (Figu re 22. 11). T h in shee ts o f pure co p pe r me tal and slabs of impu re co p pe r are imm ersed in a so lutio n co ntai ning ellso.} and II 2S0 4. The pure co p per shee ts serve as th e catho d e of an elect ro lysis ce ll, an d th e impure slabs are th e a node. Co p pe r in th e im pu re sam ple is o xid ized to co p per( II) ions at th e ano de , and co p per (II) ion s in so lutio n are reduced to pu re co p per at the cathode.
1028
Ch apt er 22
I
The Chemist ry of the Transition Elements
Anode + - Cathode
FIGURE 22 .11 Electrolyticrefining of copper. (a) Slabs of impure copper, called "blister copper, form the anode, and pure copper is deposited at t he cat hode. (b) The electrolysis cells at a copper refinery. U
22.3
Solution of CuS04 and H2S04
Stabs of impure copper
Coordination Compounds
Wh en m et al salts di ssol ve , wate r molec ules cluste r aro u nd th e ions (page 6 17) . T he nega tive e nd o f eac h pol ar wate r mol ecul e is attrac te d to th e po sitively cha rge d me tal ion , a nd th e po sitive end o f th e wa te r mol ecul e is attracted to the anio n. As noted earl ier (.... Sectio n 12.2 ), th e e ne rg y o f th e io n- solvent interaction (so lvatio n energy) is an im porta n t as pc ct of th e so lu tio n p ro ccss. Bu t there is mu ch mo re to this story.
Complexes and Ligands A gree n so luti o n formed by di ssol ving n ickel( I1 ) ch lo ride in wate r co n ta ins NiH (aq) and CI- (aq ) ion s (Figu re 22. 12). If th e so lven t is re moved , a green crystalline solid is o b taine d. The fo rm u la o f this so lid is o ften wri tten as NiC l2 • f 1120 , and t he co m po u nd is ca lled nick el (II ) chlo ri d e hexah ydrat e. Add itio n o f ammon ia to th e aq uco us n icke l( lI) chlo ride so lu tio n g ives a lilac-col ored so lu tio n from which anot he r co m pou nd , NiC l2 • 6 Nils, ca n be isolated. Th is fo rmula lo ok s \'ery sim ilar to th e fo rmula fo r the h ydrate , with am mo n ia su bst itut ed for water. Wh at are th ese two nickel spec ies? T he formulas identify th e com pos itio ns of the co mpo u nds b ut fail to give in fo rm ation abo u t thei r str uc tu res . Becau se properties of co m po u nds deri ve fro m the ir structu re s, we need to evaluate th e str uctu res in mo re d etail. Typica lly, m et al co m po u nd s are io nic , and so lid ionic co m poun ds have struc tu re s with catio ns and an io ns arranged in a regu lar array. The str uct ure of hyd rated ni ckel chlorid e co nt ain s catio ns with th e formula [N i( 1I 20 ) rJ 2 + an d ch lo ride an io ns. T he structu re of th e ammon ia-co n tai n ing co m pou nd is sim ilar to the h ydr at e ; it is m ad e u p o f [ Ni (N II :~) (iF + cations an d ch lo ride an io n s. Ions such as [N i(1I 20 )r.P + and [N i( N II 3) r.F \ in whic h a m et al io n an d either wate r o r am mo nia molecul es co m po se a sing le struc tu ral unit, are exam ples of coordination complexes, also known as complex ions (Figure 22 . 13). Co m poun ds co n taini ng a co o rd ina tio n co m plex as part of th e st ruc tu re are ca lled coordination compounds, a nd th eir che m istry is kn own as coordination chemistry. Althoug h th e old er "hydrate " fo rmul as are still used, the p re ferred m eth od of writi ng th e formula fo r coo rd inatio n co m po u nds places th e met al atom or io n an d the molec ules or an io ns directl y bonded to it withi n bracket s to show th a t it is a single structu ral enti ty. Thus, the fo rmul a fo r the n ickel fll y- am ruo u ia co m po u nd is bette r writte n as [N i (N ILJ ) ,, ] C12 .
22 .3
I
Coordinati on Com pounds
1029
Add Na OH
Add
ethylenediamine NH 2CH2CH2NH2
I
Insoluble Ni( OH )2 dimethylglyoxime Add (dmg)
Ni(dmgh
-'ol;l --
:-4 UlO~
--- -
J....
_. •
--
..,
FIGURE 22.12 Coordinationcompounds of NiH ion. The transition metalsand their ions form a wide range of compounds. often with beautiful colors and interesting structures. One purpose of this chapter is to explore some commonly observed structures and explain how these compounds can be so colorful.
1030
Chapter 22
I
The Chemist ry of t he Transition Elements
All co o rd inat io n co m plexes contain a metal a to m or io n as the cen tra l part o f th e stru ctu re. Bo nd ed to th e met al are mol ecul es o r io ns ca lled ligands (fro m th e Lat in verb ligm«; mean ing "to bind ") . In th e p re cedi ng e xam ples, water and ammonia a re the ligands. Th e n u mber o f ligand ato m s attac he d to the metal d efin es th e co ordination number o f th e me ta l. T he geometry described by th e attache d ligands is called the coo rdination geometry. In the n ickel co m plex iI1 [N i(N I 13)0] 2+ (Fig u re 22.12) , th e six ligan d s are arra ng ed in a reg ular oct ahedral geom et ry aro und th e cen tral metal ion . Liga nds can be eithe r neutral mol ecul es or an io ns (o r, in rare instances, catio ns) . Th e charac te ristic featu re of a ligan d is that it co n ta ins a lon e pair of el ectr on s. In th e classic d escrip tio n of bo nd ing in a coo rd inatio n complex, th e lone pair o f e lectrons on a ligand is shared with the m e tal io n . T he attac h men t is a coo rd ina te cova le nt bon d (.... Section 8.5) , because tile e lec tro n pair bei ng shared was o riginally on th e ligand . T he nam e "co o rd inatio n co m p lex" de rives fro m th e nam e given to thi s kind of bo nd in g. The n et charge o n a co ord ina tio n co mp lex is th e su m of th e cha rges on the metal an d its attached gr ou ps. Com plexes ca n be ca tio ns (as in th e two nickel comple xes u sed as examp les he re ) , anio ns, or un ch arged . Liga nds su ch as H 2 0 and NH ;h which coo rd ina te to th e m etal via a sing le Lewis base a to m, are term ed mono dentate. The wo rd "de nt ate " comes fro m th e Lat in deutis, mea nin g "too th, " so NH:'1 is a "one-too thed " ligand. Som e ligan d s attach to the me tal with more than o ne d on or ato m . T hese ligand s are called polydentate. Ethylenediam ine (l ,2-d iam ino et ha ne ) , H 2 N CH 2 CH 2 NH ~h o ften abbrevia ted as e n ; oxalat e ion , C20 / - (ox 2- ); a nd p he nam h ro line, C I2H,'l N 2 (p he n) , are exam ples o f the wide varie ty of bid en tat e liga nd s (Figu re 22 . 14) . Structures and exa m p les o f so me co m plex ion s with bidentate ligan d s are shown in Figure 22. 15. Pol yd en tate ligand s are also called chelating ligands, o r j ust chelates (pro nou nced "key-la res"). T he na me derives from th e Greek chelc; meanin g "claw." Because two or more bo nd s are bro ken to separate the ligand fro m th e met al, co m plexes with chc lated ligands have gre ater stab ility than th o se with mon od en tat e liga nd s. Che1at ed co mp lexes are importan t ill everyday life . One way to clean th e rust o ut of watercoo led au to mo bile engi nes and stea m bo iler s is to ad d a so lu tio n of oxalic acid. Iro n oxid e reacts with oxa lic acid to g ive a water-solu ble iron o xalat e co mplex ion : 3 H, O(f)
+
Fe,O, (s)
+
6 H,c ,O, (aq) ----> 2 [Fe(C,O,hp -(aq)
+
6 H, O' (aq)
Et hyle ne diami net etraacetate io n (EDTA 4- ), a he xad e nt ate ligand , is an exce llen t che lating ligan d (Fig u re 22. 16) . It ca n wra p aro u nd a met al ion , enca psu lating it. Salts o f this anio n are oft en ad d ed to co mme rc ial salad d ressing s to rem ove trace s of fre e metal io ns fro m so lu tio n ; otherwise , th ese m e tal io ns can act as ca ta lysts fo r the oxid atio n of the o ils in t he dressing. With o ut EDTA.J -, the dressin g wo u ld
(a) H2NCH 2CH2NH 2, en
Sum of metal ion and ligand charges Coordination complex
Ag (s ) + [Zn (C N )"F - (a'l ) (d ) Mn3()' (S) + AI(s) -----> Mn (s) + AI, ()3(S) 6. Id e ntity till' p ro ducts of e ach re action. an d balan ce th e eq uat io n . (a) Cu SO ,,( a'l ) + Zn(s ) -----> (b) Zn (s) + HC I(a'l ) -----> (c) Fe rs) + CI, (g) -----> (d ) V(s) + O , (g ) ----->
Formulas of Coordination Compounds (Sp(' fxam/llf$ 22. 1 a nd 22. 2.)
7. Which of th e fo llow in g lig an d s is ex pe cte d to be mo nod entate , and wh ic h m igh t be polvd c nta te ? (a) C H 3N H, (d) en (b) C Il 3CN (e) Br" (c) N, (I) ph cn 8. O ne o f th e fo llowing n itrogen co m po u nds or io ns is no t ca pa bl e o f' se rving as a lig and : N II 4 +, N H:1• NIl :;/ - . ld e mify th is spe cies, and ex plai n yo u r a nswer. 9. Give the oxi d a tion n um be r of the m e ta l ion in eac h o f the fol lowin g co m po u nd s. (c) [Co (N H, ) ,CI, ICI (a) [Mn(N H 3 )6]SO , (d ) Cr( en),CI, (1)) K,[ Co (CN )6] 10. Give the o x id a tion nu mbe r o f the met al ion ill each of th e fo llowing co m plexes. (a) [Fe( N I 13),;]" (c) [Co (NIl , ); (NO, ) ] ' (b) [Z n (CN), F (d ) [Cu( e n ), ]" 11. Give the fo rmu la of a co m plex con str ucted from on e NiH ion , one e th yle ned ia m ine lig an d , th re e a m mo n ia mo le cul es. an d o ne wa te r mo le cule. Is th e com plex neutral or is it charged ? If charged . g iw the charge . 12. C;ive the fo r mula of a co m ple x co nstr uc ted from on e Cr'H ion . two e th yle nediamine ligan d s, a nd two am m on ia molecules. Is th e com p lex ne u tra l o r is it ch a rged ? If charge d . g ive the charge .
Naming Coordination Compounds (8('(' h>.:a m/ll(' 22. 3.)
15 . Nam e th e following io ns o r com pound s. (a) [Ni (C,O, ), (II ,O ), )' - (c) [Co (m ),( NII , )CI]" (h) [Co (en ),Br, ] + (d ) pL(NH , ), (C,O..) 16. Nam e t he fo llowing ions or co m poun d s. (c) [!' L(NH, )B r, ] (a) [Co( H , O ).C1, ]' (b) Co ( H, O h F, (d ) [Cofcn ) (N H 3hC I)' + 17. Give the n am e or fo rm ul a fo r eac h ion o r co m po u n d . as ap p rop riat e . (a) pcntaa quahydroxo iro n (I II) io n (1)) K,[ Ni (C N)..J (c) K[ Cr (C,O ,),( II ,O ), ] (d) a m m o n iu m u-trachl oropl ann atefl l ) 18. • Give the name or form ul a for each io n or co rnpou n d , as a p p rop riat e . (a) tetraaq uadi ch loroc hromiu m (II I) ch lorid e (b) [C r (N H 3 h SO ..] CI (c) sodiu m tet ra ch lo ro cobal ta tc t l l ) (d ) [Fe (C, O ,hl" -
Isomerism (S('(' Exa m/lIp 22.4.)
19. D raw all possible g eome tri c isom ers o f th e foll owing . (a) Fe (NH , ), Cl, (h) pL(NH , ), (SCN ) (Br) (SC1\ - is bo nd ed to throug h S) (c) Co (N H 3 h (N 0 2 h (N O:;/ - is bonded to Co 3 + throug h N) (d ) [Co (e ll )C I,] -
1',"
20 . • In wh ich o f the foll owing co m plexes ar e g eo m e tr ic isom e rs po ssible? If isomers a rc possible. draw th e ir structu res and labe l th em as cis o r trans. o r as fa c o r mn.' (a) [Co (I I, () ).CI, ] ' (c) [p L(N II , )Br3] (b) Co (N H 3 ), F, (d ) [Co(ell ), (N H 3)C l j" 2 1. De termine whe ther the fo llowin g co m plex es have a ch iral meta l ce n te r. (a) [Fe( m h l " (h) tran s-[Co (ell ):;/Rr:;/ ] + (c) (ru' [Co( e ll) (H,O) CI3 ] (d ) sq uare-pla na r !' , (NII , ) (11,0 ) (C I) (NO,) 22 . Fou r geom etric iso me rs a re possible fo r [Co(e n) (N H 3) :;/ ( H:;/O )C l j +. D raw th e st ru ctures o f all fou r. (Two o f the isomers a re ch ira l. m eani ng th a t e ac h has a no n-supc ri m po sabl c m irro r im age .)
Magnetic Properties of Complexes 13. Write form ul as fo r th e foll owin g ion s o r co m pou nds . (a) di ch lorohi s(c thylcncdia mi uc j ni ckcl f l l ) (I» po tassium u-trachloroplarin atet l l ) (c) pota ssium d icya n o cuprat c (l ) (d ) tctra ammiru- di aq uai ro n ( II) 14. 'Vrite fo r mulas fo r the foll owin g ions o r co m pou n ds. (a) d ia mmi netriaqua hyd roxoc hromillm ( l l ) nitra te (b) he xaam mi nc iron (III ) n itrat e (c) pentaca rhon yliron (O) (wh ere the liga nd is CO ) (d ) a m m o n iu m te tra d] loro cuprat c ( II) •
morechallenging
• in OWL
Blue-numbered questions answered in Appendix 0
(Se(' Fxam/lle 22.j. )
23. Th e foll owing a re low-spin co m ple xes. Use th e ligan d fiel d model to find the ele ctron co nfigu ra tion of each io n . De term in e wh ich a rc di amag ne tic. Give the n um be r of unpa ired eh-ctrous fo r the pa ram agnet ic co m plex es . (a) [Mll (CN ),J ' (c) [Fe (II ,o )"l" + (1)) [Co (NII , ),,]CI3 (d ) [C r (ell ),JSO.
I
1055
24 . Th e followin g arc h igh -spin com plexes. Usc the ligand field model 10 find the e lec tron co nfig u ra uou o f each io n . a nd deter mi ne th e n u m ber of unpaired el ectro ns in each . (a ) K,[ FeF , ] (c) [C r (H ,O) 6)' + (h) [Mn F is d efi ned as the ene rgy re q uired to se para te th e nucleus o f an ato m in to p roto ns a nd neut ro ns. For exam ple, the nuclear bind ing energy for d euteri um 23 .3
I
Stability of Atomic Nuclei
1069
is th e e ne rhT)' required to co nve rt o n e mole of d e ut erium of proto n s and o n e mo le o f ne u trons.
(iJ I) nuclei into o n e mol e
Eb = 2.15 X 10' kJ/ mol Th e po sitive sign for E h ind ica tes th at e n ergy is required for thi s p roc ess. A deuteriu m nucleus is more stab le tha n an iso lat ed pr oton a n d a n iso late d n eu tron , just as th e 112 mo lecule is more stable th an two isol ated II ato ms . Recall , h owe ver, th at the H -H bond e n e rgy is o n ly 436 ~J/mo l. T he en e rgy hol d in g a proton a n d a ne ut ron tog e th e r in a d eut erium nucleus, 2. IPl X l Oll ~J /mo l , is abou t 500,000 tim es la rge r th a n th e typ ical cova le n t bond e n e rg ies. To furt her un de rstand nucl ear bindi ng e n e rgy, we turn to a n ex pe rime n tal obse rvatio n a n d a th eory. The ex p eri me n tal obse rvati on is th at th e m ass of a nucleus is always less than th e su m of th e masses of ito; co ns titue n t p roton s a n d n eut rons. The t heory is th at th e "m issing mass," ca lled the mass d efect , is eq u ated with ene rgy th at holds th e nucl ear pa rt icle s toge th e r. The mass d efect for d euterium is th e d iffe ren ce between th e mass of a de u te riu m nucl e us a n d th e su m o f the masses of a proto n a n d a n eutron . Ma ss spectro me tric me asu rem e nts ("4 Sect ion 2.3) give the acc u rate masses o f th ese pan icles to a high level o f p recision , providin g th e n u m b e rs n eeded to carry o u t ca lcu latio ns of mass defect s. Masses of ato mic n ucle i are n ot ge nerally listed in re fe re nce tables, hut m asses of ato ms are. Ca lcu lat ion of th e mass d efect ca n be carried out using masses of ato ms in stead of masses o f nucl ei . By usin g ato m ic masses, we a rc includin g in th is ca lcu lat io n th e masses of extran uclear e lec tro ns in th e reac ta n ts a nd th e p roduct s. Because th e same numbe r o f ex tra n u clear elec tro ns a p pears in product s a n d reactants, this does not affect th e res u lt. T hus, fo r o n e mo le o f d euterium nucl ei , th e mass d efe ct is found as fo llows: ~H 2.0 1410 q/ rn ol
Mass defect
=
1H
+
1.007825 q/rnol
!n 1.008665 q/mol
am = mass of products - mass of reactants = [1. 007825 q/rnol + 1.008665 q/rnol] = 0.0 023 9 q/mc l
- 2.01410 q/mol
T h e re lationsh ip bet wee n mass a n d e ne rgy is con ta ined in Albe rt Eins tei n 's 190.ii th eory o f special rel ati vity, wh ich ho lds th at mass a n d e n e rgy a re differe n t man ifesta tio ns o f th e sa me q uantity. Eins te in defin ed th e e ne rgy- mass re lat io nship : energy is equ ivalen t 10 mass tim es th e sq uare o f th e speed o f lig ht; that is, E = me2 • In th e ca se of a to m ic nuclei, it is assu me d that th e m issin g m ass (the m ass d e fect, Llm) is eq u ate d wit h th e bin din g ene rgy h o ld ing th e nucleus toge th e r. (23.1)
If Is m is given in kilog ram s an d th e speed o f light is g h'en in met ers pe r sec o n d, Eh will have un its o f j o ules (because 1.1 = I kg' rn 2 / s2 ) . For th e de composition of on e mo le of deuterium n u cle i to o ne mo le o f pr oto n s a n d one mo le of n eu tr on s, we h ave
Eb = (2.39 X 10 - 0 kg/mol)( 2.998 X 10' m/s )' = 2.15 X 1011 J/mo l of lH nuclei ( = 2.15 X 10' kJ/mo l of ~ H nuclei)
The nuclea r sta bilit ies of di ffe re nt ele ments a re co m pa re d usin g th e b in d ing energy per m ol e o f nu cleons. (Nu cleon is th e gene ra l n am e g iven to nucl ea r particles- th at is, p ro to ns and n c u tro ns.) A d eu terium nucl eus con tains two u n cle1070
Chapter 23
I
Nuclear Chemistry
9.0 X10' I~::::=:::'-~;:=======J ~56 Fe
8.0 X 108
26
-;;;- 7.0 X 10 8
"o
~,
" o'0
6.0 X 10 8
./' Region of greatest stability
5.0 X 10 8
/'
-
E
~
FIGURE 23.4 Relative stability of nuclei. Binding energy per nucleon for the moststable isotope of elements between hydrogen and uranium is plotted as a function of mass number. (Fission and fusion are discussed on pages 1080-1082.)
Fusion
4.0 X 10 8
~
:r 3.0 X 108
-
Fi ssion
•"
g' 2.0 X 10 8 '6 " j£ 1.0 X 10 8
o
50
100
150
200
250
Mass number
ons, so th e b inding ene rgy per mole of nucleons, J~'IJ n, is 2.15 X l Ol'l kJ / m o l di vided by 2 , or LOS X 10' ";j/mol n uele nn . 2.15 X 10' kJ )( I moll H nU clei )
Eb/n
= ( mol ~H
Ebln
~
nuclei
2 mo l nucleons
1.08 X l a' kJ/mo l nucleons
The bindin g ene rb,)' p e r n u cleon ca n be ca lcu late d for an y ato m whose mass is known . T he il, to co m pa re nucl ear sta b ilities, b inding en ergies per n u cle o n a rc p lotte d as a fun ct ion o f mass nu m be r (Figu re 23.4) . Th e grea ter th e b inding energy per nucl eon , th e gr eat er th e sta b ility o f th e nucleus. From th e gra p h in Figu re 23.4, th e point o f max imu m nucl ear sta bility occ urs at a m ass o f 56 (i.e., at iro n in th e period ic tabl e ).
ChemislrY.(J.Now'" Sign in at www.thomsonedu.comjlogin and go to Chapter 23 Conte nts to see: Screen 23.4 for a simulation on isotope stability • Screen 23.5 for a tutorial on calculating binding energy
. Nuclea r Binding Energy Problem CaLculate the binding energy, Eb (in kl/rncl}. and t he binding energy per nucleon, Etin (in kJ/ mol nucleon), for carbon-12. Strategy First, determine t he mass defect. then use Equation 23,1 to determi ne t he binding energy. There are 12 nuclear particles in carbon-az. so dividing t he nuclear binding energy by 12 will give the binding energy per nucleon. SoLution The mass of tHis 1.007825 q/ mol. and the mass of ~n is 1.008665 q/m ol. Ca rbon-12 , 1~ c. is t he standard for the at omic masses in the periodic ta ble, and its mass is defined as exactly 12 q/ mcl
am = [(6 x mass iH) + (6 x mass ~ n ) l - mass ~
l~ C
[(6 X1.007825 gjmol) + (6 X1.008665gjm ol)J - 12.000000 gjmol
= 9.8940
X 10- 2 q/mol nuclei 23 .3
I
Sta bility of Atom ic Nuclei
107 1
The binding energy is calculated using Equation 23.1. Using the mass in kilograms and the speed of light in meters per second gives the binding energy in joules: Eb ~ ("'role' ~
(9.8940 X 10-' k9/mol)(2 .99792 X 10' m/s)'
= 8.89 X 1012 J/mol nuclei (= 8.89 X 109 kJ/mol nuclei)
The binding energy per nucleon, Eo/n, is determined by dividing the binding energy by 12 (the number of nucleons) fb
n
_
8.89 X 109 kJ/m ol nuclei 12 mol nucleons/ mol nuclei
= 7.41 X 108 kJ/mol nucleons
EXERCISE 23 .6
Nuclear Binding Energy
Calculatethe binding energy per nucleon, in kilojoules per mole, for lithium-e. The molar mass of ~ Li is 6.015125 q/mcl .
23.4
Rates of Nuclear Decay
HaLf-life \Vhen a new rad ioac tive iso tope is ide n tified , its half/ife is usu ally measu red. H al flife ( t l / 2 ) is used in n uc lea r chem istry in th e sa me way it is used when d iscu ssing the kin et ics o ffirs t-order che mical reac tio ns (... Sect io n 15.4 ) : It is the time req uired for hal f of a samp le to d ecay to p rod ucts (Figu re 23.5) . Recall th at fo r first-ord er kin eti cs th e half-life is indepe nd ent of the amoun t o f sam ple. H alf-lives fo r ra di oactive isot o pes cover a wid e ran ge o f values. Ura ni u m-23R has one of the longe r hal f-lives, 4.47 X 10'" years, a len g th of time clo se to the agc o f th e earth (es tima ted a t 4.5-4 .6 X IO" years) . Ro ugh ly hal f o f t he urani um 238 prese n t whe n the pl a net was form ed is sti ll a ro u nd. At the o ther en d of t he ra nge o f half-lives are isotopes suc h as e le me n t 112, whose 27 7 isot ope has a hal flife of 240 microseconds (1 p.s = 1 X 10 - 6 s) . H alf-life p ro vides an easy way to es tima te the time required befo re a ra di oac tive e le me nt is no longe r a health hazard . Stron tiu m-90, fo r exam ple, is a {3 e m itter
FIGURE 23.5 Decay of 20.0 mg of oxygen-IS. After each half-life period of 2.0 min, the mass of oxygen-15 decreases byone half.
20
_ 15 ~
oS
•• 0
~
10
•
a;
5
0
r.r-----~ II o
2
First half-life
Second half-life Time (minutes)
1072
Chapter
23
I
Nuclear Chemistry
6
4
Third half-life
etc.
with a half-life of 29. 1 years. Sign ifica n t quantities o f str o n tiu m-90 were di spersed int o the e nviro n me n t in atmosp heric nuclear bomb tests in the 1960s and 1970 s, and from the h alf-life , we know th at a little less th an half is still aro u nd. The health p robl ems assoc iate d with str o n tiu m-Dtl arise becau se calcium and stron tiu m have sim ilar che m ica l p roperties. Stro ntiu m-DO is taken into th e body and depo sited in bone , tak ing th e place of ca lciu m . Rad iation d am age by stro n tiu m-90 in bon e has be en d irect ly link ed to bon e-related cancers.
• Half-life and Temperature Unlike what is observed in chemical kinetics. temperature does not affectthe rate of nuclear decay.
Using Half-Life Problem Radioactive iodine-tat. used to treat hyperthyroidism. has a half-life of 8.04 days. (a) If you have 8.8,..,.g (micrograms) of this isotope. what mass remains after 32.2 days? ( b) How long will it take for a sample of iodine-131 to decayto one eighth of its activity? (c) Estimate the length of time necessary for the sample to decay to 10% of its original activity. Strategy This problem asks you to use half-life to qualitatively assess the rate of decay. After one halflife, half of the sample remains. After anot her half-life, the amount of sample is again decreased by half to one fourth of its original value. (This situation is illustrated in Figure 23.5 .) To answer these questions. assess the number of half-lives that have elapsed and use this information to determine the amount of sampLe remaining. SoLution (a) The time elapsed, 32.2 days, is 4 half-lives (32.2/8.04 = 4). Theamount of iodine-131 has decreased to 1/16 of the originalamount [1/2 X 1/ 2 X 1/2 X 1/2 = (1/2)4 = 1/ 16J. The amount of iodine remaining is 8.8,..,.g X (1/2) 4or 0.55 Jlg . ( b) After 3 half-lives ,the amount of iodine-tat remaining is 1/8 (= 1/ 2}3 of the original amount. The amount remaining is 8.8,..,.g x (1/2}3 = 1.1 Jlg . (c) After 3 half-lives. 1/8 (12.5%) of the sample remains; after 4 half-lives, 1/16 (6.25%) remains . It will take between 3 and 4 half-lives, between 24.15 and 32.2 days, to decrease the amount of sampLe to 10% of its originaL value. Comment You willfind it useful to make approximations as we have done in (c). An exact time can be calculated from the first-order rate law (pages 683 and 1074).
EXERCISE 23 .7 Using Half-Life Tritium (fH), a radioactive isotope of hydrogen has a half-life of 12.3 years. (a) Starting with 1.5 mg of this isotope, how many milligrams remain after 49.2 years? (b) How long will it take for a sample of tritium to decayto one eighth of its activity? (c) Estimate the length of time necessary for the sample to decay to 1% of its original activity.
Kinetics of Nuclear Decay Th e ra te of nuclea r decay is d e termined from measurements of the ac tivity (A) o f a sam ple. Acti vity refers to the number of disintegrati on s observed per un it time , a q uan tity th at ca n be measured readi ly with d evice s suc h as a Ge iger- Mu lle r CO U Il ter (Figu re 23.6). A ctiv it)' is proportion al to the number of mdioactioe atoms !Jres('nt (N) .
Ax N
(23.2)
If the nu m ber o f radioactive n ucl ei N is reduced by ha lf, the activity of the sam pic will be ha lf as large . Do ublin g N will d ou ble the activity. Th is evide nce indicat es that the rat e o f decompositio n is first order with respect to N. Co nseq ue n tly, th e equation s de scribing rates of radioact ive deca y are the same as th o se used to d e-
• Ways of Expressing Activity Common units for activity are dps (disintegrations per second) and dpm (disinteg rations per minute).
23 .4
I
Rates of Nuclear Decay
1073
+
Thin window t hrough which
radiation enters FIGURE 23.6 AGeiger-Muller counter. Acharged particle (an a or (3 particle) enters t he gas-filled tube (diagram at the right) and ionizes the gas. The gaseous ions migrate to electricaUy charged plates and are recorded as a pulse of electric current. Til e current is amplified and used to operate a counter. Asample of carnotite. a mineral containing uranium oxide. is also shown in the photograph.
scr ibe first-order c hemical reac tions; th e c h ange in th e nu m b e r of radioactive a toms N pe r u n it of tim e is propo rt ion al to N : !!.N ~ - kN !!.t
(23.3)
T he in teg rat ed rate eq ua tion ca n be wr itte n in two ways d epending o n th e data us ed: (23.4)
or (23.5)
He re , No and Au are the n u m ber o f a to ms and the activity o f the sample initially, respect ively, an d N a nd A are the n u mber of a to ms and the act ivity of the sam p le after tim e t, re spectively. T hus, NINfl is th e frac tion of atoms re main ing after a glven tim e (I) , a n d A I Au is the frac tio n o f the activity remain ing after the same period. In these equa tio n s, k is the rat e co n sta n t (d ecay con sta n t) fo r th e isotope in qu est ion . T he rela tionsh ip be twee n h alf-life and th e fir st-o rd e r rat e co ns ta nt is the sa m e as seen with c hemical kine tics (Equ a tio n 15 .4, p age 690) : 0.693
t i/ 2 = - k-
(23.6)
Eq uatio ns 23 .3-23.6 are usefu l in seve ral ways: • If the activity (A ) or the n u mber o f radioactive n uclei (N) is measu red in the labo rat ory ove r so me perio d t, th en k ca n be calc u la ted. T he decay co nsta n t k can the n be u sed to d e te rm in e the h alf-life of the sample. • If k is known , th e fract io n of a radi oactive sa mple (N/N o) sti ll p rese n t afte r so m e time t h as ela psed ca n be calcu lated. • If k is known , the time re quired for that isoto pe to d eca y to a fra ction o f the origina l ac tivity ( A I A n) can be ca lculated .
1074
Chapter 23
I
Nuclear Chemistry
· Determination of Half-life Proble m A samp le of radon-222 has an initial a particle activity (A o) of 7.0 X 104 dps (disi ntegrations per second). After 6.6 days, its acti vity (A) is 2.1 X 104 dps. What is the ha lf-life of radon-222?
Strategy Values for A, Ao, and t are given. The problem can be solved using Equatio n 23.5 wit h k as t he unknown. On ce k is found, the half-life can be calculated using Eq uation 23.6.
Solution In (2.1 x 10' dps/7.0 X 10' dps) In (0.30) k
~
- k (6.6 day)
~
- k(6.6 day)
= 0.18 days" !
From k we obtain t I/2: t l/ 2
= 0.693/0. 18 days-t
=
3.8 days
Comments Notice that the activity decreased to between one half and one fourth of its original value. The 6.6 days of elapsed time represe nts one full half-life and part of an other half-life.
Time Required for a Radioactive Sample t o Partially Decay Problem Galliu m citrate, containing the radioactive isotope qalti um-er. is used medically as a tumorseeking agent. It has a half-Life of 78.2 h. How long will it take for a sample of gallium citrate to decay to 10.0% of its original activity? Strate gy Use Eq uation 23.5 to solve t his problem. In this case. the unknown is the time t. The rat e constant k is calculated from the half-life using Eq uation 23.6. Although we do not have specific values of activity, t he value of A/A o is known. Because A is 10.0% of Ao• the value of A/A o is 0.100. SoLuti on First, determine k: k
~
k
= 8.86
0.693/ t'l'
~
0.693/ 78.2 h
X 10- 3 h- 1
Then substit ut e t he given values of A/ Ao and k into Equatio n 23.5: In (A/A,)
~
- kt
In (0.100)
~
- (8.86 X 10- ' h- ')t
t
=
2.60 X 102 h
Comments The time required is between three half-lives (3 X 78.2 h = 235 h) and four half-lives (4 X 78.2 h ~ 313 h). EXERCISE
2 3.8 Determination of Half-Life
Asample of Ca 3(P04h containing phosphorus-32 has an activity of 3.35 X 103 dpm. Two days later, the activity is 3.18 X 103 dpm. Calculate the half-life of phosphorus- 32.
EXERCISE
23.9 Time Required for a Radioactive Sample t o Partially Decay
A highly radioactive sample of nuclear wast e products with a half-life of 200. years is stored in an underground ta nk. How long will it take for the activity t o dim inish from an initial activity of 6.50 X 1012 dpm to a fairly harmless activity of 3.00 X 103 dpm?
Radiocarbon Dating In ce rtain situatio ns . the age o f a mat erial ca n be d e ter mi ned base d on the ra te o f d ecay o f a ra d ioac tive isot ope . T he best-kn own exam ple of this procedu re is the use o f ca rbo n-14 to d a te h isto rical artifac ts. Carbo n is pri mari ly ca rbo n-12 and ca rb o n-I 3 with isoto pic abu ndances of 98 .9% and 1. I %. respecti vel y. III ad d itio n , tr aces ofa third iso to pe , carbo n- I-l, are p rese n t to th e ex ten t of abo u t 1 in 10 12 ato ms in atmosp he ric CO 2 and in living mat eria ls. 23 .4
I
Rate s of Nuclea r Decay
1075
FIGURE 2 3.7 Variation of at mospheric carbon-14 activity. The amountof carbon-14 has varied withvariation in cosmic ray activity. To obtain the data for the pre-1990 part ofthe curveshown in this graph, scientists carried out carbon-14 dating of artifacts for which the age was accurately known (often t hroug h written records). Similar results can be obtained using carbon-14dating of tree rings.
E
.g t:,
o
~ ~
:;; ~ u
Be/AD
- 5 '---
7000
6000
5000
4000
3000
2000
1000
o
_
1000
2000
Year of tree ring growth Source: Hans E. Suess. La J olla Radiocarbo n Labo rat ory
• Willard Li bby ( 1908-1980) li bby received the 1960 Nobel Prize in chemistry for developing carbon-14 dating techniques. Carbon-14 dating is widely used in fields such as anthropology.
Carbo n-14 is a f3 e mi tter with a half-l ife o f 5 730 years. ;\ I-g ram sam ple o f carbo n fro m living mate ria l will show abo ut 14 di sintegr ati ons p CI' mi n u te , not a lo t o f rad ioactivity but nevertheless detecta ble by modern methods. Carbo n-14 is formed in th e u pper atm os p here by nuclear reac tio ns in itiated by ne u tro ns in cos m ic ra d iatio n: l~ N Desper Collection inthe
History 01 Chemistry University ol Cincinnati
FIGURE 23.8 The Ice Man. The world's oldest preserved human remains were discovered in the ice of a glacier high in the Alps. Carbon-14 dating techniques allowed scientists to determine that he lived about 5300 years ago.
1076
Chapter 23
I
+
~n -? l~C
+
~H
On ce fo rm ed , ca rbo n-14 is o xidi zed to Heo 2. This p rod uct en te rs the ca r bo n cycle , circu lating th rough th e atmos p here. ocean s, and bio sphere . The usefuln ess o f carho n-14 fo r d atin g co me s abo u t in the foll owing way. Plan ts abso rb CO 2 and co nvert it to o rganic com po u nd s, thereb y inco rporating car bo n-14 into living tissue . As lo ng as a plan t rem ains alive, thi s p rocess will continue . and the percentage of carbo n th at is ca rbo n-I4 in th e plant will eq ua l the percentage in th e atmosp he re. Wh en the plan t dies, carbo n-14 will no longer be taken up. Radioactive d ecay co n tin ues, however, with the carbo n-l -t activity d ecreasing over time. After 5730 years, th e activity will be 7 d p m/g; after 11,460 yea rs, it will be 3.:'> d p m/g; and so on . By measuring tile activity of a sample, and knowing th e half-life of ca rbo n- I4, it is po ssible to calculate whe n a plant (o r an ani ma l that was eating plan ts) died. As with all ex perime n tal p rocedu res, ca rbo n- I4 dating has lim itati ons. The p roced u re assu mes th at th e amou nt of carbo n- l 4 in the atmo sp here h undred s o r th ousands o f years ago is the sam e as it is no w. We kn ow that thi s isn 't exa ctly tru e; the pe rce n tage has varied by as much as 10 % (Figure 23.7) . Fu r th erm o re , it is no t possib le to use ca rbo n- i -t to d at e a n obj ect th at is less than abo ut 100 years o ld; th e rad iat ion level from carbo n- 14 will not. chang e e no ug h in thi s sho rt time per iod to per m it accu rate d etect ion o f a d iffe ren ce from the in itial value, In mos t in sta nces , th e acc u racy of tile measu rem ent is, in fact , o nl y abo u t ± 100 years. Fina lly, it is not po ssible to determine ages of o bj ec ts mu ch o lde r than about 40,00 n years. By th en , after nearl y seve n half-lives, the radioactivity will have d ecreased virt u ally to ze ro. But fo r the spa n o f tim e between 100 an d 40 ,000 years, th is technique has provid ed important. info rm atio n (Figu re 23.H).
ChemistrY-9- Now'"
Nuclear Ch emistry
Sign in at www.thomsonedu.com/ Ioqtn and go to Ch apter 23 Contents to see: Screen 23.3 for an exercise on t he Geiger counter • Screen 23.6 for a tutorial on half-life and radiochemical dating
, Radiochemical Dating Problem To test the concept of carbon-14 dating, J. R. Arnold and W. F. libby applied this technique t o analyze samples of acacia and cypruswood whose ages were already known . (The acacia wood. which was supplied by the Metropo litan Museum of Art in New York, came from the tomb of Zcser, the first Egyptian pharaoh to be entombed in a pyramid. The cyprus wood was from the tomb of Sneferu.) The average activity based on five determinations was 7.04 dpm per gram of carbon. Assume (as Arnold and Libby did) that the activity of carbon-14, Ao. is 12.6 dpm per gram of carbon. Calculate the approximate age of the sample.
Strategy First. determine the rate constant for the decay of carbon-tv from its half-life (t 1/ 2 for 14( is 5.73 X 10 3 years). Then, use Equation 23.5.
SoLution k ~ 0.693(t' I' ~ 0.693(5730 yr
= 1.21 X Ln (A(A,)
~
10- 4 yr- t
- kt
7.04 dPm(gJ Ln ( d ( ~ ( -1.21 X 10- ' yr- ' )t 12.6 pm 9 t
= 4.8
X
103 yr
The wood is about 4800 years old. Comment This problem uses real data from an early research paper in which the carbon-14 dating method was being tested. The age of the wood was known to be 4750 :!:: 250 years. (See J. R. Arnold and W. F. Libby: Science, Vo l. 110, p. 678, 1949.) EXERCISE 23.10
Radiochemical Dating
Asample of t he inner part of a redwood t ree felled in 1874 was shown to have He activity of 9.32 dpmjg . Ca lculate the approximate age of the tree when it was cut down. Compare this age with that obtained from tree ring data , which estimated that the tree began to grow in 979 ± 52 Be. Use 13.4 dpm jg for the value of Ao.
23.5
Artificial Nuclear Reactions
Ho w man y different isoto pes are found on ea rt h? All of the stable isotopes o ccu r naturally. A few u nstab le (rad io ac tive) isotopes tha t have lo ng hal f-lives arc fou nd in nature; the be st-known ex am ples are u ranium-Z'B, u rani um-238, and thorium232 . Trace qu ant ities of o ther radioact ive isoto pes with short half-lives are p rese nt because th ey are being for med contin uo usly by nu clear re ac tion s. T hey include iso to pes o f rad iu m , po lo niu m , a nd rado n, alon g with o the r eleme nts p ro du ced in var io us radi o active d ecay se ries, and ca rbo n- I4, fo r med in a nuclear react ion ini tiated by co sm ic radiatio n . Natu ra lly o ccurri ng isoto pe s accou n t fo r o n ly a very small frac tio n of th e cu rren tly known rad ioactive iso to pes, however. T he rest- several thousand- have be en syn thesized via artificial nuclea r reactio ns, so metimes referred to as tra nsmutatio n. T he first artificial nuclear reaction was iden tified by Ruth erfor d about 80 years ago . Recall the classic ex perime n t that led to the n uc lear m od el o f the atom (See Milestones in the Deoelopment (!/ Chemistry, pa ge 338) in which gold fo il W'd S bom bard ed with Q' parti cles. In the years following th at experiment, Ru therford and his cowo rkers bombarded many other ele ments with Q' particles. In 1919 , o ne of these ex perim en ts led to an u nexpected result : when nitroge n ato ms were bo mbard ed with a particles, p roto ns were detected amo ng th e produ cts. Ru therford correctly co nclude d th at a n uclear reactio n had occu rred. Nitrogen had undergon e a transmutation to oxygen : ~ He
+
l~ N ---t l~O
+
~H
23 .5
I
Artificial Nuclear Reactions
1077
Durin g the nex t d ecad e, other nuclear reactio n s we re d isco ve red by bombardi ng other elements with a particles. Progress wa s slow, however, be cau se in m ost cases a pa n icles are sim p ly sca tte red by target n uclei. The bo mbardin g particles ca n no t get close e no ug h to the n ucleus to react because o f th e stro ng repulsive fo rce s between th e positively cha rged a particle a n d the positively ch arged a to m ic n u cle us. Two adva nces were m ade in 1932 that g re a tly exte n ded nucl ea r reaction che mistry. The first in volved the use of p article accelera to rs to create h igh -en ergy pa rticles as projectil es. The seco nd was th e lise o f neutrons as th e bombard in g particles. The a p ani cles us ed in th e ea rly stu d ies o n nucle ar react ions ca m e fro m n atu ra lly rad io ac tive m at e ri als suc h as uranium a n d h ad relati vel y low e n ergies , a t least by tod ay's sta n d a rd s. Particles with h igher e n e r6'1' we re n eeded, so .J. D. Coc kcr o ft ( IR97- 1967) a n d E. T. S. Walt on ( 1903- 199,';), working in Ruth e rford 's laborat o ry in Ca m brid ge , Eng la n d , tu rned to proto ns. Proto ns a re fo rm ed wh e n hyd rogen a to m s ioni ze in a ca th od e-ra y tube , a n d it was known th at they could bc ac ce le ra te d to h igh e r e n e rgy by a p p lying a h igh vo ltag e. Cock er-o ft. a n d \ Valto n found that whe n ene rgetic p roton s str uc k a lith iu m ta rget, th e foll owin g reaction o ccurs:
§Li + [p ---t 2 ~ H e • Discovery of Neutrons Neutrons had been predicted to exist for more than a decade before they were identified in 1932 by James Chadwick (1891-1974). Chadwick produced neutrons in a nuclear reaction between a particles and beryllium: ~ + ~Be ~ l~C + ~ n .
• Glenn T. Seaborg (1912 -1999) Seaborg figured out that thorium and the elements that followed it fit under the lanthanides in the periodic table. For t his insight, he and Edwin McMillan shared the 1951 Nobel Prize in chemistry. Over a 21-year period, Seaborg and his colleagues synthesized 10 new t ransuranium elements (Pu through Lr). To honor Seaborg's scientific cont ributions, the name "seaborqium" was assigned to element 106. It marked the first time an element was named for a living person.
T h is was the first example o f a reacti on initiated by a pa rt icle th at h ad been a rt ificially accele ra ted 10 h igh e nergy. Since th is ex pe rim e n t was co n d uc ted , th e technique has been devel o ped mu ch fu r the r, a nd th e use o f p a rticle accel e rat ors in nucl ear ch c m istry is n ow co m mo n p lace . Particle accele ra tors o pera tc o n th e princi p le th at a c h a rged p art icle p laced bet ween c ha rged p la tes will be accele rate d to a h igh speed a n d hi gh e ne rgy. Mod e rn ex ampl es o f th is proce ss a re see n in th e syn th esis of th c transuranium el ements, seve ral of wh ich a re de scr ibed in more d e tail in A Closer Look: Th e Search for Ne u Ele ments. Ex pe rime n ts u sin g ne u tro ns as bombard in g p articl es \ "'CIT first ca rried out in both the Un ited Sta tes an d Great Britain in 19 32. N itroge n , ox yge n , flu orine , a n d n eon we re bombard ed with e n e rge tic n eutrons, a n d a particles we re d et ect ed a mon g th e products. Usin g n eut ro n s m adc se nse : because n e u tro n s h ave n o c h a rg e, it was reasoned th a t the se particle s wo u ld n ot be re pell ed by th e po siti vely c h a rge d n u cle u s pa rticles. Thus, n eut ron s did n ot n eed hi gh e nergies to reac t. In 1934, En r ico Fer mi (190 1- 1954 ) a n d hi s cowo rke rs sho wed th at nucl ea r reactions u sin g ncu tr on s are m ore favorab le if the n e utron s h ave low e ne rgy. A low e nergy n e utron is sim p ly ca p tu red by the n ucleu s, giving a p ro duct in whic h th e m ass n um be r is in c re ased by o ne uni t. Bec au se of th e low e n e rgy of th e bomb ard in g parti cle , the p roduc t nucle u s d oc s not h ave su ffici e nt e ne rg y to fra gment in th e se re ac tions. The new nucle u s is prod uced in a n e xcited sta te, h oweve r ; whe n th e nuc leus re tu rn s to th e gro un d sta te , a y-ray is e m itte d. Reac tion s in wh ic h a n eutron is ca p tu re d a n d a y-ray is e m itted are ca lle d (n, 'Y ) reactions . The ( n, )I) reactio n s a re th e so u rce of man y of th e ra dioiso to p es u sed in m edicin e an d ch e mi stry. An exam p le is radio active pho sphoru s, i~ p, wh ic h is used in ch emical stu d ies suc h as tr acin g the uptake of phosphorus in th e b od y.
UP+ 6n ---t i~ p + • Transuranium Elements in Nat ure Neptu nium, plutonium , and americium were unknown prior to their preparation via these nuclear reactions. Later, these elements were found to be present in trace quantities in uranium ores.
Y
Transuranium elements, ele m e nts with an a to m ic numbe r g rea ter th an 92 , we re first m ad e in a nuclear react ion seq u e n ce be g inning with a n ( n , y) re acti on . Scien tists a t the U n ive rsity of Ca lifo rni a a t Berkel ey bomba rd ed uranium-23R with n e utrons. Amon g th e produc ts ide n tified we re n eptunillm-239 an d p lll ton illm-239 . T he se n cw el ements we re fo rmed whe n 239U d ecayed by {3 ra d ia tio n . 2~~ U
+ An ---t 2~~ U
2~~ U ---t 2~~ N p + - ~f3 2~ ~ N p ---t 2~~ PU + - ~f3 1078
Chapter 23
I
Nuclear Ch emistry
A Closer Look
The Search for New Elements
By 1936, guided first by Mendeleev's predictions and later by atomic theory, chemists had identified all but two of the elements with atomic numbers between 1 and 92. From this point onward, all new elements to be discovered came from artificial nuclear reactions. Two gaps in the periodic table were filled when radioactivetechnetium and promethium, the last two elements with atomic numbers less than 92, were identified in 1937 and 1942, respectively. The first success in the search for elements with atomic numbers higher than 92 came with the 1940 discovery of neptunium and plutonium. Since 1950. laboratories in the United States (Lawrence Berkeley National Laboratory). Russia (Joint Institute for Nuclear Research at Dubna, near Moscow),
and Europe (Institute for Heavy Ion Research at Darmstadt, Germany) have competed to make new elements. Syntheses of new transuranium elements use a standard methodology. An element of fairly high atomic number is bombarded with a beam of high-energy particles. Initially, neutrons were used; later, helium nuclei and then larger nuclei such as 11B and 12( were employed; and, more recently. highly charged ions of elements such as calcium, chromium, cobalt, and zinc have been chosen. The bombarding particle fuses with the nucleus of the target atom, forming a new nucleus that lasts for a short time before decom posing. New elements are detected by their decomposition products, a signature of particles with specific masses and energies.
Fermilab. The tunnel housing the four-mile-long part icle accele rator at Batavia, Illinois
By using bigger particles and higher energies, the list of known elements reached 106 by the end of the 1970s. To further extend the search, Russian scientists employed a new idea, matching precisely the energy of the bombarding particle with the energy required to fuse the nuclei. This technique enabled the synthesis of elements 107, 108, and 109 in Da rmstadt in the early 1980s, and the synthesis of elements 110, 111. and 112 in the following decade. Lifetimes of these elements were in the millisecond range; the 277 isotope of element 112, for example, mass has a half-life of 240 1'5.
Yet another breakthrough was needed to extend the list further. Scientists have long known that isotopes with specific magic num· bers of neutrons and protons are more stable. Elements with 2, 8, 20, 50, and 82 protons are members of this category, as are elements with 126 neutrons. The magic numbers correspond to fitled shells in the nucleus. Their significance is analogous to the significance of filled shells for electronic structure. Theory had predicted that the next magic numbers would be 114 protons and 184 neutrons. Using this information, researchers discovered element 114 in early 1999. The Dubna group reporting this discovery found that the mass 289 isotope had an exceptionally long halflife, about 20 s. At the time this book was written, 117 elements were known . Wilt research yield further new elements? It would be hard to say no, given past successes in this area of research, but the quest becomes ever more difficultas scientists venture to the very limits of nuclear stability.
Four years later, a simi lar reactio n sequence was used to m ake americi u m-24 1. Plutoni um- 239 was fo u nd to add two neutron s to form pluton iu m-241, which decays by f3 em issio n to g ive americiu m-z-l l . , Nuclear Reactions
Problem Write equations for the nuclear reactions described below. (a) Fluorine-19 undergoes an (n. y) reaction to give a radioactive product that decays by -?f3 emission. (Write equations for both nuclear reactions.} (b) Acommon neutron source is a plutonium-beryllium alloy. Plutonium-239 is an U' emitter. When beryUium-9 (the onlystable isotope of beryllium) reacts with a particles emitted by plutonium, neutrons are ejected. (Write equations for both reactions.) 23 .5
I
Artificial Nuclear Reactions
1079
Strategy The equatio ns are writte n so tha t bot h mass and charge are balanced. SoLution
+ bn ~ 2gF + Y
(a)
l~ F
(b)
t8Ne + - ~{3 2Jfpu ~ 2Jiu + ~a 2~ F -e
23.6 • Fission Reactions In the fission of uranium -236, a large number of different fission products (elements) are formed. Barium was the element first identified, and its identification provided the key that led to recognition that fission had occurred.
Nuclear Fission
In 193R, two chem ists, O tto H ah n ( IR79- 19fi8) an d Fr itz Strassma n ( [ 902- [ 9RO), isolated and ide ntified barium in a sam p le o f uraniu m tha t had been bo mbard ed with neutron s. How was barium form ed ? The answe r to th at question expl ain ed o ne of th e m ost sign ifica n t scie nt ific d iscoveries o f th e 20th ce n tu ry. The u ran ium n ucle us had split int o sma lle r pieces in the process we now call nuclear fission. The details of nuclear fission were un raveled th rou gh th e work of a number o f scie n tists. They d et e rm in ed tha t a u ranium-235 nucleus in itiall y ca p tured a ne u tro n to form u rani u m-236. This isotope u nderwen t nuclear fission to p roduce two new nuclei , one with a mass nu mber arou nd 140 and th e o t he r with a mass aro un d 90 , alo ng with several ne utro ns (Figu re 23.9). Th e nucl ear reaction s t hat led to form atio n of bariu m whe n a sam p le o f 235 U was bombard ed with neu tr on s are 2~~ U
+
In ---7 2~~U
2~ ~ U ---7 l~J B a
• li se Meitner (1878-1968)
Meitner's greatest contribution to 201hcentury science was her explanation of the process of nuclear fissio n. She and her AlP-Emilio SejreVisual Arctllves, Herzteld nephew, Otto Frisch, Collection also a physicist, published a paper in 1939 t hat was the first to use the term "nuclear fission." Element number 109 is named meitnerium to honor Meitner's contributions. The leader of the team that discovered this element said that "She should be honored as the most significant woman scientist of [the 20th] century."
• The Atomic Bomb In an atomic bom b, each nuclea r fission step produces 3 neutrons, which leads to about 3 more fissions and 9 more neutrons, which leads to 9 more fission steps and 27 more neutrons, and so on. The rate depends on the number of neutrons, so the nuclear reaction occurs faster and faster as more and more neut rons are formed, leading to an enormous output of energy in a short time span. 1080
Chapter 23
+
j~ Kr
+ 3 6n
An important aspec t of fission reac tio ns is th at th ey p roduce more neutron s th an are u sed to initiat e the p ro cess. Un d er th e righ t cir cu ms tances , th ese neut rons the n serve to contin ue th e reaction . If one o r m ore o f th ese neutrons are cap tu re d by ano ther 2~j U n ucleus, th en a fu rt her re action ca n oc cu r, releasing still mo re neutron s. Th is seq ue nce re peats over and over. Such a m echan ism , in whic h each ste p ge ne ra tes a reac tan t to contin ue th e reaction , is ca lled a chain reaction. A nu clea r fission cha in reactio n has thre e gen eral ste ps: 1. In itiation. T he re action of a sing le a tom is ne eded to start th e cha in. Fissio n of 235U is ini tia le d hy the abso rp tio n o f a neutron . 2. Propagation. Th is part of th e process re peats itself over an d over, with each step yiel d ing more p roduct. T he fission o f 2:lfiU rel eases neutrons th at initia te the fissio n of o ther uranium ato ms . 3. Termination. Even tua lly, th e cha in will e nd . Te rmin ati on co uld o ccu r if th e re actant (2::EJ U) is used up, or if th e neutron s th at co n tin ue th e chai n escape fro m th e sam p le witho u t bein g ca p tu re d by 2:l.'",U.
To har ness the e ne rgy p rod uce d in a nuclear reactio n , it is necessary to control the ra te at which a fission re ac tio n oc curs. This is man aged by balan ci ng the p ropaga tio n and te rmin ation ste ps by limiting th e number o f ne u t ro ns available . In a nuclear reactor, th is balan ce is accomp lishe d by usin g cad mi u m ro ds to absorb neu tro ns. By withdrawing o r inse rti ng the rod s, th e n umber o f neu tro ns available to pr opagat e th e chai n can be cha nge d, and th e rate of th e fission reactio n (an d the ra te o f energy production ) can he in creased or d ecr eased. Ura nillm-235 and plutonillm-239 are the fission abl e iso topes most co mm o n ly used in powe r reacto rs. Na tu ra l uranium co n ta ins o n ly O.72 % o f uran ium-Z'Ih; more than 99 % o f th e natural e lem en t is u raniu m-238. T he pe rce n tag e o f uran ium- 235 in na tural uranium is too small to sustain a chain reactio n , however, so t he u ranium used fo r nuclear fuel mu st be e n ri ched in th is isotope . O ne way to d o so is by
Nuclear Ch emistry
FIGURE 23.9 Nuclear fission. Neutron capture by 2~iu produces 2~~ U . This isotope undergoes fission, which yields several fragments along with several neut rons. These neutrons initiate further nuclear reactions byadding to other 2~i u nuclei. The process is highly exothermic, producing about 2 X 1010 kJjmol.
gaseo us di ffusio n ( Ni' +(aq)
1.77
+
2 H20
PbO"s) + 50,' -(aq) + 4 H'' {aq) + 2 . - ----> Pb50,( s) + 2 H,O Au+(aq)
O(
+ e" ----> Au( s)
1.7
1.685 1.68
2 HClO(aq) + 2 W (aq) + 2 e: ----> Cl,(g ) + 2 H,O
1.63
Cev'{aq} + e-
1.61
------?
Ce3+ (aq)
NaBiO,( s) + 6 H' {aq) + 2 e" ----> Bi 3+(aq) + Na ' {aq) + 3 H20
= 1.6
MnO, " (aq) + 8 H' {aq) + 5 e: ----> Mn' +(aq) + 4 H20
1.51
Au ' - (aq) + 3 . - ----> Au(s)
1.50
( IO,- (aq) + 6 H' {aq) + 5.- ---->
+Cl2(g) + 3 H,O
1.47
BrO,- (aq) + 6 H'' {aq) + 6. - ----> Br- (aq) + 3 H,O
1.44
( I, (g) + 2 . - ----> 2 Cl (aq)
1.36
(r20,'-(aq) + 14 H'{aq}
+ 6 e: ----> 2 (r ' +(aq) + 7 H,O
1.3 3
N2H , +(aq) + 3 H' {aq) + 2.- ----> 2 NH,+(aq)
1.24
Mn0 2(S) + 4 H'{aq) + 2 . - ----> MnH(aq) + 2 H,O
1.23
O, (g) + 4 W(aq) + 4 e- ----> 2 H,O
1.229
Pt' +(aq)
+ 2 e:
IO, -(aq )
+
----> Pt( s)
6 W (aq)
+
5 e- ---->
1.2
+I, (a q) + 3 H,O
1.195
(continued)
A-36
TABLE 21
Standard Reduction Potentials in Aqueous Solution at
25 . (
(cont inued)
Standard Reduction
Potential fO (volts)
Acidic Solution CIO, -(aq) + 2 W(aq) + 2 e- ---> CIO, -(aq) + H,O
1.19
Br,(C) + 2 e: ---> 2 Br-(aq)
1.08
AuCl, "{aq) + 3 e: ---> Au (s) + 4 Cl-(aq)
1.00
Pd' -(aq) + 2 e- ---> Pd(s)
0.987
NO, -(aq) + 4 H' {aq) + 3 e: ---> NO(g) + 2 H, O
0.96
NO , -(aq) + 3 H'' {aq) + 2 e - ---> HNO,(aq) + H, O
0.94
2 Hg'(aq) + 2 e " ---> Hgr(aq)
0.920
Hg' +(aq) + 2 e: ---> Hg(C)
0.855
Ag '(aq) + e" ---> Ag(s)
0.7994
Hg, ' +(aq) + 2 e - ---> 2 Hg(C)
0.789
Fe' +(aq) + e" ---> Fe' +(aq)
0.771
SbCIo -(aq) + 2 e - ---> SbCI, -(aq) + 2 Cl-(aq)
0.75
[PtCl,J' -(aq) + 2 e - ---> Pt(s) + 4 Cl-(aq)
0.73
O,(g) + 2 W(aq) + 2 e : ---> H, O, (aq)
0.682
[PtCl6J' -(aq) + 2 e - ---> [PtC4J' -(aq) + 2 Cl-(aq)
0.68
I,(aq) + 2 e - ---> 2 I ''{aq)
0.621
H,AsO, (aq) + 2 H' {aq) + 2
e ---> H,AsO , (aq) + H, O
0.58
I,(s) + 2 e: ---> 2 I-(aq)
0.535
TeO, (s) + 4 H'{aq) + 4 e " ---> Te(s) + 2 H, O
0.529
Cu+(aq) + e: ---> Cu(s)
0.521
[RhCl6P -(aq) + 3 e " ---> Rh(s) + 6 Cl-(aq)
0.44
Cu ' +(aq) + 2 e- ---> Cu(s)
0.337
Hg, CI, (s) + 2 e: ---> 2 Hg(C) + 2 Cl-(aq)
0.27
AgCI(s) + e " ---> Ag(s) + Cl-(aq)
0.222
SO,'-(aq) + 4 H"{aq) + 2 e - ---> SO,(g) + 2 H,O
0.20
SO,'-(aq) + 4 H'{aq) + 2 e : ---> H,SO,(aq) + H, O
0.17
Cu +(aq) + e" ------? Cu +(aq)
0.153
Sn4 +(aq) + 2 e " ---> Sn' -(aq)
0.15
S(s) + 2 W + 2 e : ---> H,S(aq)
0.14
AgBr(s) + e " ---> Ag(s) + Br-(aq)
0.0713
2
2 H'{aq)
+ 2 e - - 7 H2 (g)( refere nce electrode)
0.0000
N,O(g) + 6 H'{aq) + H, O + 4 e : ---> 2 NH ,OW(aq)
- 0.05
Pb' -(aq) + 2 e- ---> Pb(s)
- 0.126
Sn2+(aq) + 2 e " ---> Sn(s)
- 0.14
AgI(s) + e - ---> Ag(s) + I-(aq)
- 0.15
[5nF6 J' - (aq) + 4 e - ---> 5n(s) + 6 F-(aq)
- 0. 25
Ni2+(aq) + 2 e" ---> Ni(s)
- 0.25
Co' +(aq) + 2 e" ---> Co(s)
- 0.28
(continued) Appendix M
I
Sta nda rd Reductio n Potentials in Aqueos Solutionat 25 O(
A-37
TABLE 21
Standard Reduction Potentials in Aqueo us Solution at 25 O( (continued)
Addic Solution
Standard Reduction Potential EO (volts)
W(aq) + e- - > TI(s)
- 0.34
Pb50, (s) + 1 e: - > Pb(s) + 50.'-(aq)
- 0.356
5e(s)
+ 1 H' {aq} + 1 e " - > H,5e (aq) Cd' +(aq) + 1 e: - > Cd(s)
- 0.40
Cr' +(aq) + e - - > Cr' +(aq)
- 0.41
+ 1 e- - > Fe(s)
- 0.44
Fe"(aq)
1 CO, (g) + 1 H' {aq) + 1 e: - > H,C, O, (aq)
+ 3 e: - > Ga(s)
Ga ' -(aq)
- 0.403
- 0.49 - 0.53
+ 1 H' {aq) + 1 e - - > Hg(t ) + H,5(g) Cr' +(aq) + 3 e - - > Cr(s) In ' +(aq) + 1 e: - > In( s) Cr"(aq) + 1 e - - > Cr(s) Fe5(s) + 1 e: - > Fe(s) + 5' -( aq)
- 0.763
Mn' +(aq) + 2 e" - > Mn(s)
- 1.1 8
+ 1 e - - > V(s) Cd5 (s) + 1 e - - > Cd(s) + 5' -(aq) lnS (s) + 1 e: - > I n(s) + 5' -(aq) l r" (aq) + 4 e" - > l r(s)
- 1.1 8
Hg5(s)
V' +(aq)
- 0.71
- 0.74
- 0.91 - 1.01
- 1.1 1
- 1.44 - 1.53
Al' +(aq) + 3 e" - > Al(s)
- 1.66
Mg ' -(aq)
+ 1 e" - > Mg( s) + e" - > Na(s) Ca"(aq) + 1 e- - > Cars)
- 1.37
Na '(aq)
- 1.714
Sr' +(aq) + 1 e: - > Sr(s)
- 1.89
Ba"(aq)
+ 1 e - - > Ba (s) + e - - > Rb(s) K' {aq] + e" - > K(s) W(aq) + e- - > Li(s)
- 1.90
Rb '(aq)
- 1.915
- 1.87
- 2.915 - 3.045
Basic SoLution ClO -(aq)
+ H,O + 2 e: - > Cl-(aq) + 1 OW(aq)
+ H,O + 1 e- - > 3 OW(aq) 1 NH,OH(aq) + 1 e - - > N, H,(aq) + 1 OW (aq) CIO,-(aq) + 3 H,O + 6 e" - > Cl-(aq) + 6 OW(aq) MnO, -(aq) + 1 H,O + 3 e: - > MnO, (s) + 4 OW(aq) MnO, -(aq) + e- - > MnO.'-(aq) NiO ,( s) + 1 H,O + 1 e- - > Ni(O H),( s) + 1 OW(aq) Ag ,CrO,( s) + 1 e: - > 1 Ag(s) + CrO.'-(aq) O,(g) + 1 H,O + 4 e- - > 4 OW (aq) OOW(aq)
0.89 0.88 0.74 0.61 0.588 0.564 0.49 0.446 0.40
(continued)
A-38 Appendix M
I
Standa rd Reduction Potentia ls in Aq ueos Solution at 25 "C
21 Standard Reduction Potentials in Aqueous Solution at 25 . ( (continued)
TABLE
Standard Reduction Potential EO (volts)
Acidic Solution
+ H20 + 2 e - ----> ClO, -(aq) + 2 OW(aq)
CIO, -(aq)
+ H20 + 2 e: ----> 2 Ag(s) + 2 OW(aq)
Ag20(s)
2 N02-(aq)
0.34
+ 3 H,O + 4 e - ----> N20(g) + 60W(aq)
0.15
+ 2 H20 + 2 e: ----> 2 NH,(aq) + 2 OW(aq)
N, H,(aq)
[Co(NH,) .p·(aq)
0.10
+ e: ----> [Co(NH,) .]'·(aq)
0.10
+ H20 + 2 e - ----> Hg(f ) + 2 OW(aq)
HgO(s) O,(g )
0.36
0.0984
+ H20 + 2 e - ----> OOW(aq) + OW(aq)
NO, -(aq) MnO,(s)
0.076
om
+ H20 + 2 e- ----> N0 2- (aq) + 2 OW(aq)
+ 2 H,O + 2 e - ----> Mn(OH),(s) + 2 OW(aq)
- 0.05
+ 4 H20 + 3 e" ----> Cr(O H),(s) + 50 W(aq) CU(OH),(s) + 2 e: ----> Cuts) + 2 OW(aq) CrO; -(aq)
S(s)
+
- 0.12 - 0.36
2 e - ----> S2-(aq)
Fe (OH ),( s)
+ e: ---->
Fe (OH),(s)
- 0.48
+
- 0.56
OW(aq)
+ 2 e: ----> H2(g) + 2 OW(aq) 2 NO , -(aq) + 2 H20 + 2 e - ----> N,O,(g) + 4 OW(aq)
- 0.8 277
2 H20
Fe (OH ), (s) SO; -(aq) N, (g)
- 0.85
+ 2 e - ----> Fe (s) + 2 OW(aq)
- 0.877
+ H20 + 2 e: ----> SO,z-(aq) + 2 OW(aq)
- 0.93
+ 4 H20 + 4 e" ----> N,H,(aq) + 4 OW (aq)
[In(OH).J 2-(aq) In(O H)2(s)
- 1.1 5
+ 2 e - ----> In(s) + 4 OW(aq)
- 1. 22
+ 2 e : ----> In(s) + 2 OW(aq)
[I n(CN), ]'- (a q)
- 1. 245
+ 2 e - ----> In(s) + 4 CN-(aq)
- 1. 26
Cr(O H),(s)
+ 3 e - ----> Cr(s) + 3 OW(aq)
- 1.3 0
SiO,z-(aq)
+ 3 H, O + 4 e: ----> Sit s) + 60W(aq)
- 1.7 0
Appendix M
I
Sta ndard Reductio n Potentials in Aqueos Solutionat 25 O(
A-39
APPENDIX
Answers to Exercises
N
Chapter 1 1.1
(a) Na = so d iu m ; Cl = chlorine; Cr = chromium (1)) Zin c = Zn ; nickel = Ni; pot assium = K
1.2
(a) Iro n : lust ro us sol id . metallic , good con ductor o f
hea t a nd electricity, mall e able , duct ile , attracted 10 a mag net (h) Wa LeI': co lorless liq uid (a t room te mperature ) ; melting point i s 0 "C , a nd hoiling po in t is IO(l "C, de nsity - 1 g /Cln \! (c ) Tabl e salt: so lid . whi te crystals. so lu ble in wa ter ('1.930 788 1l) ( 1.titil X IW " gi ll) ~ 9.955 X I W " g (c) 6-'Zn has 30 proton s, 30 elect rons, a nd (64 - 30) = 34 neutrons. (d) T he mass o f a .i-'Zn at om is 63.929 / 12.0... , o r 5.3 274 tim es th e mass o f a I ,;!C a tom . (No te th at m ass of 12C is d e fin ed as an ex act val uc .)
2.2
T h e m ass nu m be r o f th e sec o n d silve r iso to p e is 109 (62 + 47 ). Symbol: 1(~)Ag, ab un da nce = 48.16 1%.
2.3
Us c Eq u a tio n 2.2 fo r th e calcu la tio n . Ato mi c mass = (34. 96885) (75. 77/ 100) + (36.96590) (24. 2:V IOO) = 35.45. (Acc uracy is li m ited by th e va lu e o f th e p erce nt a huu dan cc to 4 sig n ifica n t fig ures.)
2.4
The force of attra ct io n be tween ion s is pro p o rtio nal to t he p rodu ct of th e ion ch ar ges (Co u lo m b 's law) . The force of attra ct io n betwee n Mg 2+ a n d 0 2- io n s in MgO is app rox imately fo u r tim es g re ate r th an th e fo rce of a ttra c tio n b et wee n Na " a nd Cl " io ns in NaCl, so a mu ch hi gh e r tem p era tu re is r e q u ir ed to di sr upt th e orderly a r ray o f io n s ill crystalline Mg O .
Th ere a rc eig h t el em ents in th e third p eri od. So di u m (Na), magn e siu m ( I\'fg), an d alumi n um (A1) a rc m eta ls. Silico n (Si) is a metall oid. Ph osphoru s (P) , su lfu r (S) , ch lo ri n e (CI), a n d argon (Ar) a rc non me tals.
2.5
T he m o lecula r for m ul a is c' 3H 7NOtS. \ hu will ofte n se c its form ula wri tten as HSC,Ht CH (N +H 3) C0 2 - to bet ter id entity th e mok-cuh-s struc tu re .
2.6
(a) K+ is fo r med if K loses one electro n . K+ h as th e sa me nu m b er o f el e ct ron s as Ar. (h) Se 2 - is for m ed by ad d in g two el ectro ns to an a tom of Se. It h as th e sa m e n u m ber o f elect ron s as Kr.
trifluo ride 2.11
(a) ( 1.5 11101 Si) (28 .! g /nw l) ~ 4 2 g Si (h ) (45 4 g S) ( 1.00 mol S/ 32 .07 g ) = 14. 2 mol S ( 14.2 m ol S)(6. 02 2 X 10 2:1 a to m s/ m o l) = 8.S:} X 10';1 ·\ a to m s S
2.12
(2. 6 X 10" a tom s) ( 1.000 m o1/6.022 X 10" a tom s) ( 197.0 g Au / I .()OO mo l) = S50 g Au
Volume = (850 g Au ) ( 1.00 cm 3 / 19.32 g) = 44 em" Volume = 44 crrr' = (th ickn ess)( arc a) = (0. 10 cm)(arca) A re a
= 440
em"
Le n g th = wid th = '1,/440 cm '' = 2 1 cm
2.1 3 (a) Citric a cid : 192.1 g / mo l; mag ne siu m ca r bo n a te: 84.3 g / m o l (h ) 454 g citric aci d ( 1.000 moll 192.1 g ) = 2.36 m ol citric acid (e) 0 .12511101 MgCO , (84.3 g / m ol) ~ 105 g MgC O , 2.14 (a) 1.00 mo l (N H ,hC0 3 ( mo la r mass 96 .09 g / mol) ha s 28 .0 g o f N (2!1.2 % ) , 8 .0(J g of H (8.39%) , 12.0 g o l C (1 2.5 % ) , a n d 48. 0 g ofO (5 (J.(J %) (h) 454 g C"H HI ( 1 mol CI{ If 18/ 114.2 g) (8 m ol C/ l m ol CI\H J f()( 12.0 1 g e /I mo l C) = :182 g C 2.1 5 (a) C,I I ,
(h ) C, I I ,O ,
2. 16 (88. 17 g C)( I m o l C/ 12. 0 11 g C) = 7.3 4 1 mo l C
Appendi x N
I
Answers to Exercises
A - 41
(l I. H ~
g 11 )(1 lllo l lI / l.OOH g II )
= 11.74 III 01
(e) NiCI, (a q ) + 2 1'0 11 (a q ) ----+ 1'i (O Il ),(s) + 2 KCI(aq )
II
11.74 mo l li /7 .:HI mo l C = 1.6 mol Hi t m ol C mo l C) = 8 mo l HI S mol C
= (H/ 5) ; (m o l II /I
3. 6
T lu- empirica l fo rm ul a is C 5 11l.l' Th e m o la r mass. 68 .1 J g / mol. closely m at ches this formula. so C[,II ", is a lso the.' moh..cula r fo r mula .
2. 17 (78.!)[) g C) ( I mol CI 12.0 II g C) (1 0.59 g il) ( I m ol 1l / 1.0 0H g II ) (10.5 1 g 0 ) ( 1 mo l 0 / 16.00 g 0 ) 10.51 mol Il j O.fi569 mol 0
~
°
= 16 m ol 11/1 mo l ()
Ph" (aq) + 2 CI- (aq) ,
°
(a ) 1I,o (aq ) an d :-;O, - (aq ) (h) Ba" (aq ) an d 2 O ll - (a q )
3.8
(a) II ,PO, (aq ) + 1I, 0 (f) (
10 SI
(0) Acting as a base:
II , PO , - (aq ) + II , O (f)
on he ating- is 0.235 g- - 0.128 gre mai n
l
Acting as an acid: II ,PO . -< aq ) + 1I, 0 (f) (
(0. 107 g 11, 0 ) ( J mol 11 ,0 11 H.O16 g 11, 0 ) = O.OOil94 mol 11:/0
l
II PO o'- (aq ) + II , O ' (C) Because H 2 P O ~ - (aq ) ca ll reac t as a Bronsted aci d and a s a ba se , it is sa id 10 he am ph ip ro tic. l II CN (aq ) + O l l Iaq) : (e) e N -( aq) + II ,O (C) I cyan ide ion is a Breus tcd base.
(O. 12H g :-liC!, ) ( I mo l :-liCI ,1 129.6 g NiCI, ) = 0 .000988 mol NiCI 2
T he formu la fo r th e h yd ra te is :\fiC h . () IltO .
I
Il , PO ,(aq ) + O ll - (aq )
= n.107 g ; O, I2H g ~i C 1 2
Mo le ra tio = 0.00594 m ol H 20 / O.OOOYHH mol N iCI 2 = H.Ot : T hercforc x = 6
l
1I,0 ' (aq ) + II ,PO,- (aq )
TIH' ra tio of moles K to mo les 0 ato ms is I to 2; the empirical form ula is RO t.
2. 19 Mass of wan-t
r
3.9
~l g ( OIl ), ( s )
+ 2 II CI ("q ) ----+ MgC I, (aq ) + 2 II ,O (C)
l':et io nic eq ua tion : Mg (O ll h (s)
Chapter 3 3. 1
(a) Stoic hiometric coefficie nt s: 2 fo r AI, 3 for fir:!. and 1 fo r I\ I:/Brl ; (h) XOOO a to ms of A I requi res (3 /2)8000 = 12 ,000
molecu les of Hr2
3. 2
I~
(a ) 2 C, II ",(g) +
O , (g ) ----+ 8 co, (g) + III 1I, 0 (f)
(h) 2 Ph (C, II ,.), (f) + 27 O , (g ) ----+ 2 PhO (s) + 16 CO,(g ) + 20 1I,0 (f) 3.3
Epsom salt is a n e lec t rolyte, a n d methan ol is a nonel ectrolvu-.
3.4
(a) LiXO j is so lu ble an d gives Li t Iaq} an d :"0 3 " {aq }
3.5
A~42
s KCI("'I )
) Ph CI, (s)
3.7
The.' e m p iric al fo rm ul a is C lUfl ll,O.
2. 18 (0 .586 g 1') (1 m ol 1'/ 39 .10 g 1') ~ 0.0 150 m ol I' (O.4HO g 0 )(1 m ol 0 /1 6 .00 g 0 ) = O .O ~ OO m o l
Na Cl(aq )
F,·H (aq ) + 3 O ll - (aq ) ----+ Fe(O I I), (s) (e ) Ph (:-IO , ), (aq ) + 2 KCI(aq ) ----+ PhCI, (s) + 2 K:-;O, (aq )
10.51 mol II 0.65 69 m ol
6.569 mol C / 0.6;')69 m ol 0 = 10 m ol C / I m ol 0
+ :\
AI" (aq ) + PO ,' - (aq ) ----+ AIPO , (s) (h) FeCI, (aq ) + 3 KO II (aq ) ----+ Fe(O I I), (s) +
= 6.5 69 mol C ~
(a) AICI, (aq ) + Na,PO, (aq ) ----+ AIPCl , (s)
+ 2 1I '" (.5 g) (4 .184 .J1g · K)(1 8 .9 "C - 165 ' c) ~ 0 C Ill ("lal
5.4
= 0,44 Ji g '
K) ]
~
1> ,11, CO, (g ) + 2
5 .9
Mass of in ' remainin g = 75 g - 57 g = 18 g ~
I
Answers to Exercises
q
1> , ir.;
= 1> ,11,
Fe (s) +
1;, CI, (g ) ------> Fl'CI ,,(s)
+ 1>, If, + 1>, II;
+
= + 116. 8
'v, O , (g )
kj
------>
~
0
5. 10 l:1,.1r = (6 mol/mol-rxn ) l:1;lr [C0 2 (g ) ] + (3 mol! mol-rxn )8f H O[H 20 ( C)] - 1(1 mol/ l mo).rxn )l:1 j H O
[C"Il,;{f) ] + ('% mol ym ol -t-xu) 1>;11" [0 2 ( g )JI ~ (6 mol/mol-rxn ) (-393.5 k:J/ mo l) + (3 mol/ mol-rxn ) ( -285 ,H.I! mol) - ( 1 mol ym ol-rxu ) ( +49 .0 k/lmol) - 0 = - 3 26 7.4 k.l! mol-rxn
Chapter 6 0.4988
J l'l/ = 0.4988 m ol C 2 H lj(l mol-rxn !2 m o l C 2 11b ) ( - 285 73 ~ I /mol -rxn )
- 713 kJ
- 59 3 .6 k.J
C l2H 220 11(s)
Ma ss of ice re m ai n in g = mass of ice in itiall y - mass of icc mel ted
( 15 .0 g C, H ,,) (I mol C, H ,,/'10. 07 g C, H ,;) mo l C 2 H t)
~
------> CS, (g) + 3 O , (g )
12 C(s , g ra p h ite) + J J H,(g )
+ energy as h ea t
57 g' of ice me lts wit h e ner~y as h e at su p p lie d by co o lin g 250 g of le a from 18 .2 °C (29 1.4 K) t o 0 °C (2 73 2 K)
Appendix N
2 ( - 296.8 )
Nl't: C(s ) + 2 S (, ) ------> CS, (g )
x = 57 g
A -44
so, (g)
~
1>,Ir, = + 1103 .9
K
Energy tra n sfe rred as h ea t from te a ex pended to m e lt ice = 0
~
1>,If, = - 39 3 .5 k/
2.39.J / g . K
(250 g ) (4.2 ./1 g . K) (2 73 .2 K - 29 1.4 K) + xg U133 ./1g)
5 .5
C (s ) + O ,(g ) ------> CO, (g ) 2 [S (s) + O , (g ) ------> SO , (g ) ]
(a ) (3800 calories ) (4. 184 .J1 calo rie) ~ 1.6 X 10 '.J (h ) (25 0 calo ries ) ( lOOO ca lo ries /calorie ) (4. 184 J/ calor ic) ( I k/ I J0(0)) ~ 1.0 X 10" k l
5 .2
(a ) Energy evolved a s h eat in reactio n + energy as h e at absorbed by 1120 + e neq.,'. Y as h e at a bsorbed by bo mb = 0 '/"" + ( 1.50 X 10 3 g) (4. 20 J/g . K) (27 .32 "C 25.00 "c) + (837.J1K)(27.32 K - 25.00 K) = 0
().OO ~WO
5 .1
0
- 4 .50 X i o-j
Energy tra nsfe rred as h ea t per mole = 1:1 ,11 = - 4.50 kj/ 0 .0800 mol H CI ~ - ;;6 .3 k/ lmol H CI
0,00199 mol 12 is in excess, and was n o l used in till' re actio n wi th asco r b ic acid .
= (0.05 20 m ol 12 /1
~
+
T hi s represen ts th e elH.-'rgy t ran sfe r re d as h e at in the reactio n o f 0 .0800 mol H Cl.
mol
(0 .00398 mol Na~5 2( 3 ) ( 1 mo l [2/2 mol Na2S:l)3) = 0.00 199 mol It
12 originally ad d ed mol Iz
Mass of fin al so lu tio n = 40 0. g I:1T = 27.78 °C - 25 .10 °C = 2.G8 °C = 2.G8 K
6.1
(a) H igh est freq u e nc y, violet; lowest fr e q u e n cy. red (b ) T he FM ra dio fre q uen cy, 91.7 M l-lz, is lower tha n the frequen cy of a m icrowave oven , 2.45 G Hz . (c) T he wave le n gt h of x-rays is sh o rt e r than the wave length of u lt ravio let lig ht.
6.2
Oran g e lig ht: 6.25 X 10 2 11m = 6.25 X 10 - 7 V
111
6.8
(a)
= (2.998 X 101\ m /s) / li.25 X 10 - 7 m = 4.80 X 1 0 1 ~ s- I
E = (6 .626 X 10- 3 .1 .1
' s/ p ho to l1) (4.80
X 10 1-1 S-I) (6. 022 X 10 2:1 ph o to n s/ mol )
~ 1.9 2 X 10' .1111101
Microwave : E = (6.626 X 10 - 3.1 .1 . s/ p h o LOn) (2.45 X 10" , - ') (6 .0 22 X 10 2:1 ph o to n s/ m o l)
=
0.9 78 .1 / m o l
Orang e (6 25-n m ) light is abo ut 200.000 ti m e s mo re e n e rgetic th an 2.45-G III. microwaves.
6.3
(b)
(a) E (per at om)
= - Rhc/ n2 ~
(-2.17!! X l o- JR) / (3' ) .J/ato l1l = - 2.42 1 X 1O- "' .J/atol1l (b) E (per 11101) ( - 2.42 1 X IO- " .J/at o m ) (6.022 X 10 23 a toms / m o l) ( I kj /IO './ )
Orbita l
n
I
6,
6
0
4p
4
sd
s
2
4f
4
3
A 41) o rbital h as o n e n o d al p la n e ; a 6d orb ita l h as two nodal p la n e s.
Chapte r 7 7.1
7.2
(a) 4s t n + (' = 4) filled b ef o re 41J (11 + ( = 5 ) (b) 6.5 ( 11 + (' = 6 ) fill ed b efo re 5d (1/ + ( = 7) (0 ) 5, ( n + f ~ 5) filled bef o re 4( ( '1 + (' ~ 7) (a) ch lo ri ne (C l) (0)
1 .52 2 .5 2 2 1/; 3 J 2 31) ~1
- 14 ').8 q /mol 3J 6.4
Th e le ast ene rgetic lin e is from t he electron tra n sition fro m 1/ = 2 to 11 = I.
M>
=
- /il,, [ I/ I ' - 1/2' ] - (2.179 X IO- JR .J/at o m ) ( 3/ 4) - 1.634 X 10 - 11\ J / at om
~
M;;"
~
v
e
( - 1.634 X 1O- ,s.J/ at o m ) / (6. 626 X 10 - " ./ ' s) = 2.466 X 10 15 s-I A = C/ V = (2.998 X lOl'< l1I / s- I) / (2.466 X lO L' s- l) = 1.2 16 X to - 7 m (o r 121. 6 nm ) En e rgy per a to m = li E = - Rhr[l/ x 2 = 2. 179 X 10 - 11\ .I/ a to l1l En c rgy p e r m ole 1023 a to ms / m o l)
= (2. 179
-
7.3
4.\'
3d
VU
[Ar] I t It It I
VH
[Ar] l t l t l I
1/ 12 ]
0 45
3d
0
X 1O- 1i'.J/ ato m ) (6. 02 2 X
3d
First , calcula te th e velo cit y o f th e neu tron :
= 2720 m . S- I
USC thi s val ue in th e d e Brogli e eq uation : A = II/mv = (6.626 X 1O- 3~ kg . m 2 S- 2)/
0
All three ion s a re p a ra m agn et ic with th re e , tv.... o , an d fou r lin pa ired e le ctro n s, re sp e ctivel y.
< Si < AI
7.5
In creasin g a to m ic ra d ius: C
7.6
(a) In cre asing a to m ic radi us: C < B < Al (h) In cre asin g io n izati on e n ergy: Al < R < C (c ) Carbo n is p redic ted to h ave th e m o st n egative el e ct ron affi ni ty.
7.7
Trend in io ni c radii: S2- > CI - > K +. T hese ion s are iso cl cc troni c (they a ll h ave the A I' config u ra tio n ) . T h e size decre ase s wit h in creased n u cle a r charge, the h igh er n ucl ear charge resulti n g in a g reate r fo rce o f a tt raction of th e elec trons by th e n ucleus.
( 1.675 X 10-" kg) (27 20 m s-') = 1.45 X 10 - 1; 111
(a) ( = 0 o r 1; (h) ril l' = - 1, 0 , or + 1, IJsu hs hdl; (c ) d su bs h e ll; (d) t = 0 an d 1111' = 0 ; (e ) 3 o rb ita ls in th e IJ su hs hc ll: (I) 7 va lues of tIIf an d 7 0 rhita ls
4J
[Ar] lll l t lt l t l t l
v = [2 1i/ m] , /2 = [2 (6. 2 1 X 10- " kg ' Ill' , -') /( 1.675 X 10 - 27 kg)]' /2
6.7
O b tain th e answers fro m Tabl e 7.3.
7.4
~ 1.3 12 X 10"./I mol ( ~ 1312 kj/ m o l)
6.6
31'
D.IJ ITIIIII
(c ) Calc iu m h a s two valence e lectrons in th e 4s su bshel l. Q u antum nu m bers for these two el ec tro n s are n = 4, = 0, mt = 0, an d 1ItJ = ± 1/ 2
~
6.5
[Ne ]
Appendix N
I
Answers to Exercises
A-45
7.8
Mg CI:1• if it cxi sn-d . wou ld presumabl y con tain one Mg3+ ion (an d th ree CI- io ns) . T he fo rma tio n of ~ l g:H is em-rgc rica llv u nfavorabl e, ,...ith a hug e in p ut o f e llt'rgy hdn g requ ired to re m ove th e th ird e le ct ron (a co n ' e lec tron }.
Chapter 8 8.1
- 0 .·]· .·C- O·. [ . ",,,=
:0:
~=
II
I -X-
].
[ 8.2
.. ..1 ..
[
II
]'-
: O - S- O : ..
1
..
:0:
II 1
I
II - C- O -II
1I- :-:-0-H 1
1
II
11\111", ' la mine -
1Ilt' I I I, t l lt ,1
8 .7
[:~-0)-~ :J +
11[ 8.4
8.5
8.6
] 11
+,2bo nd pai rs au d ~ lon e pa irs.
[: ~---;~"t.-:- ~ :J - CIF
2- ,
~
bo nd pa irs an d :\ Io ne pair s.
8 .8
Te trahe d ral ge o me try a ro u nd ca rb on . Th e CI-C-CI bo nd a ngle will he close to 109.5°.
8 .9
Fo r each species, the el ectron-pair gt'ollll' t ry an d the molecula r sha pe are th e same. BF3 : trigo nal plana r; RF~ " : tetrah edral. Ad ding F- to RF3 ad d s a n el ec tro n pair to the cen tra l atom a nd ch ange s the shape.
8. 10 T he e lectron-pair g eome try a ro u nd I is trigonal hipvra mi d al . T h e molecu la r gt'ollw try of the ion is lin ear.
[:-l .]I"'.:
:CJ :
8.3
.. : 0: I .. 0 -1'- 0 .. I ..
Cl F2
8 . 11 (a) In
rOl -, the n ' is tetrahedral elec tr o n-pa ir gcom c-
try. T he mo lecul ar
:0:
g e o m t' t l~'
is also tet rah ed ral .
.. ]3-
:0:
(a) T he acct ylidc ion , C} - , a nd t he N 2 m o lt-cul t, haw the sam e num ber o f va le n ce el ec tron s (10 ) a nd identi ca l ele c tron ic structu res ; th a t is. th ey a n' isoch-crroui c. (h) O zon e . 0 :30 is isock-c tronic with NO z" : h yd ro xid e ion . 0 11- , is isoc lect ro u ic with H F. (a) CN- : fo rm al c harg e o n C is - 1; forma l cha rge o n N is 0, (h) S0 3:1 - : for m al cha rg e O il S is + 2; for m al cha rge Oil ea c h 0 is - I.
Reso nance stru ct u re s for the H e 0 3 - io n :
[i=C-ii:]- [.o-c=o]-
·.
1
..
1
..
: O - p-O:
[
·.
:0 ;
(h) III S () :~:1 - , th ere is rctral n-dral e lec tro n-pair ge o m etr y. Th e mole cu la r gl'Ollll'l ry is trigo lla l pyram ida l.
:6-s-6 ·. I .. :] ' [ :0 : (c) In IF". there is oc ta hed ra l el ectron-pa ir gt' o me try. Th e m o le cul ar gco m crrv is . L) = (6.8 7 mol) (O.OH205 7 L . a tm / mol . K)
(298.2 K)
P(NI13 ) = 1.35 a tm 11.10
(!l.OO L ) (0 .0 760 m ol ) (0.08 20 6 L . a tm / mo! . K) (298. 2 K)
n aln l hall kJ/lllol (Ap pe nd ix F)
Diagon al di stance = side dim e nsion ( 13) = 4 rf "
= (2.9 10
= 6.HS1t'p
/11~llkc H
+
Ste p l a . Enth al py of formatio n of I(g ) ~ + 106.8 kJ/ lllo l (Appe nd ix L )
Side dime n sion of cu be = 4 ( 1.26 X 10- l-l cm ) / (13 ) ~ 2.9 10 X 10- " em U n it ce ll vo lu me X J o- 23 cm :1
Use th e Bo rn-H ub er cycle eq uatio n shown on pages 600 -602. T he un known in th is p rohlcm is th e e n tha lpy of fo rma tio n of Na l (s).
t. "",;"H
t.fW [Na l (s) ]
~
~
+ e"
- 702 kl/mol (Tabl e 1:1,2)
- 28 7 kJ/mo l
Chapter 14 14.1
(a ) 10.0 g suc rose = 0.0292 mol; ~5 0 g 11 20 = 13.9 mol
x.un".... ~
C,ur row
=
= ( ). 0 2 ~ ) 2
mo l) /( O.0292 mol
+
l ~U)
mo l)
0.00 210
= (0.0 29 2 mol
0.117
sucrose ) / (0.25 0 kg so lvent)
III
\Ve ight % su cro se = ( 100%) = 3.85 %
(lo.n g
Appendix N
I
s llcrose /~60
g so ln )
Answers to Exercise s
A-51
(b ) 1.08 X 1O'! p pm = l. 08 X 10 4 mg NaCI pe r 1000 g so ln ( 1.08 X IO'! m g Na / lOOO g so ln) ( 105 0 g so lo l 1 l )
Chapter 15 15.1
-v, (tJ. [N O CI]l tJ.I) = Y, (tJ. [NO ) / tJ. I) = tJ. [ CI, ]ltJ. 1
15.2
Fo r the fir st two ho u rs:
1. 13 X I (),! m g Na / L
- tJ. [sucros e] 1 tJ. 1 ~ [ (0 .033 - 0.(50 ) m ol/ I.] 1 (2.0 It )
11.3 g Nai l
~
( 11.3 g Na i L) (58.44 g NaC 1/ 23.0 g Na) 28.7 g NaC I /I. 14.2
~
tJ." ,,,/I'
~
For the last two hou rs: tJ.[ -'l l crose] /tJ. I ~
s .n: [NaO II (aq ) ]
~
=
- tJ.;l I ' [Na Oll (s) ] - 469 .2 kJ/mol - (- 425. 9 kJ/ mo l) - 43.3 q /m ol
14.3
Sol u bility of CO 2 = kf-l Pg = 0.0:14 m ol/kg ' bar X 0.:J3 ba r ~ 1.1 X 10- ' M
14.4
T he so lu tio n con ta ins sucrose [ (10.0 g )(l m o1/ 342.3 g) = 0.029 2 mol] in wain [ (22 5 g )( l mo1/ 18.0 2 g ) = 12.5 mol] .
=
=
15.3
0.998
=
14.5
=
= 0 ,998(1 49 .4 m m
H g)
=
Cgl\'lol = (5 25 g )(l mo1/ 62.07 g) / (3. 00 kg ) = 2.82 m
k 15.4
15. 5
Mol a r mass =
= 1.40
~
15.6
=
= O . 81 .~
Ke X 2.80 " 'C
tJ. Tep ~
=-
A- 5 2
Itt
In X;~
Appendix N
I
(- 1.86 ' CI IIl) (0 .8 15 1Il)( 1.85)
Answers t o Exercises
s
~
k[ P, (N II , ),C1, ]
=
0.00 54 m ol/ I. . h
~
(0.27 1t- 1)( 0.020 m ol/ I. )
=
In ( [slic ros C]/ Isllc rose L,)
= 0. 003 5
~
- Ill
- (0 .2 1 1t- 1) (5.0 It )
rnul / L
(a) T he frac tio n re m a in ing is [C I I, N,C1 I,] I [C I I,N, CII , ]".
(b ) Afte r the re action is 99 % co m p lete [C I ilN, C II , ] I [CII , N, CII , ]" ~ 0.010 . In (0.0 10) ~ - (3.6 X 10- 1 , - 1) (1)
g/ 1.00 X 10- " mo l
(25.0 g NaCI) (ImoI/58.44 g ) / (0.525 kg )
1110 12 .
[C II ,N,C H, ]I[CII ,N,CII 3 1o = 0.95
1.0 X 10- 5 mol
1.4 X 10" g / m o l
'N,C1
10 3 1.2/
In ( [CII., N, C II 3 ] I [C II , N, C II , ],,) ~ - (3.6 X 10- 4 s-I)( 150 s)
(Assu m ing the polyme r is co m posed of CH 2 u n its, th e pol ymer is a bou t 10,000 un its lo ng.) 14.8
Ra[l'
X
[sucro se ]
c (mo l/ I.) ~ II / lfr ~ [ ( 1.8 Ii rrnn II g ) (1 a[ml 760 m m II g ) ]I [( 0.082 06 I. . atm l m ol , K) (298 K) ] ~ 1.00 X 10- 1 M (1. 00 X 10- ' mo l/ l. )( 0. 100 I.)
= 7.0
In ( [sllcmse ] /[ IU JIOj)
K fl) X m = ( - 1.86 "C/ m) (2.82 m) = - 5 .24 "C
\ 'i.m will he pro tect e d o n ly to abo u t - 5 "C a nd not to - 25 ' C. 14.7
k[NO ]' [O, ]
0.02 8 m ol/ I. . s = k[0 .020 m ol/ l. ]'[O.OIO moil l. )
' ,I,wl = tJ. Topl K op = 1.0 ' CI (0.512 ' CI Ill) 1.95 11/ = 1.95 mol/kg
~ Tfl' =
~
Usin g th e dat a in experi m e n t I to determin e k:
149 m m H g
m aSSglycol = ( 1.95 mol/kg ) (0.125 kg )( G2.02 g / mo l) = 15 g 14.6
Compare ex perimen ts I a nd 2: Doubl ing rOd cause s tlu- rate 10 d oubl e, so the rat e is first orde r in [0 2] . Co m pa re ex pe ri me n ts 2 a nd 4: Dou bl ing [N O ] ca uses th e rat e to in cre ase by a fac to r of 4, so the rat e is seco nd ord e r in [ NO ] . T h us , th e ra te law is Rate
Xwaler P "wal t'r
- [ (0.0 10 - 0.(1 5 ) m o l/ l. ] / (2.0 It ) 0.00 25 mo l/ L . h
Instant an eo u s rat e a t 4 h = 0.00 45 mo l/I. . h. (Calc ula ted from th e slo pe of a lin e ta nge n t to th e curve a t th e d efine d co ucc n t ratio u .)
X.'"".r = (12.5 mol H 2 0 ) / ( 12.5 mo l + 0.(J29 2 mol ) P wa lt>r
0.00 80 m o l/I. . It
t = 1.3 X 10'· s (220 m in ) 15.7
1/ [111 ] -
1/ [111 10 = kl
1/ [111 ] -
1/ [0.0 10 M)
[111 1 = 0.0022 M
~
('10. Lz'mol . m in ) (I 2 m in )
15.8
(h ) 125 1 d ecays m uc h faster. (c ) III [(11) / (1.6 X 10 " a lo los) J = - (IWII d - ' )
Co ncen t rat io n versu s time
(2 .0 d)
~ ,------------....,
,1/1.6 X 1015 a to m s
= O.9i H; 11 = 1.57 X
Iu" a toms
Since th e answer should han- two significan t figu res. we sho u ld round th is o ff to 1.6 X 10 1:' a toms. Th e a pproxi mately 2% that has decayed is not di sce rn abl e within th e limi ts of acc Hnlc)' o f th e da ta presen ted.
15.10 III (k,/ k,)
= (- 1-.". / /1) ( I/ T,
- I /T,)
III [ (1.00 X 10' ) / (4.5 X IO' ) J = - (1-.".18 .3 15 X 10-' kJ/lllOl . 1')(1 /28:\ I' - 1/ 2H 1') £ ;0 = 57 kJ /mol
Time (m ill)
15 .11 All th ree su-ps a re bimo lecul ar;
For ste p :\ : Ra il· = k[N , O ][ H, J. The re are two inn-rmcd iatcs, :" 20Ag) and :" 20 (g ) . Wh e n ti ll' three equ atio ns a re ad d e d , ~ 20 2 (a produ ct in the first step a n d a reactant in th e second ste p ) an d ~20 (a p ro duct in the secon d ste p an d a re act ant in th e th ir d ste p ) ca n cel . leaving th e net equation : 2 N O (g ) + 2 11 2 (g-) ------7 N, (g) + 2 H, O (g).
6'
7.'
_ I'L--
-
-
-
-
15.12 (a ) 2 NII , (alj) + O CI- (alj ) ---> N, II ,(alj) + CI-(alj) + H, O (f) (h) T h e seco nd ste p is the ra te-dv n-rmi ni ng ste p. (c) Ra w ~ k[i'\ II,CI] [N I I, J (d) N112C I, N2 11,'j +, a nd 011 - a r e in term ed ia tes .
_
n Tim e (m in )
15,13 Overa ll rea ction : 2 N0
2C
l(g)
------7
2 NO , (g) + Cl,( g ) Rate
1.50'- - - - - - - - - - - - - - -:..-- ,
~
k' [N O ,CIj'/ [:'\ O , J (whe re k'
In creasin g [N (
2)
= k,k, / k ,)
ca uses the reac tio n rate to d ec r ease .
Chapter 16 16.1
(a} K = [CO ]'/[CO , J (b) K = [CII' ·][ :'\ II , J' / [CII(:'\II ,) ,"1 (c) K ~ [ H ,O ·][ CH, CO , - I![ CII , CO, HI
16,2
(a ) Ro th react ions arc rea ctant-favored ( K< I ) . (0 ) [:"\H j ] in the second solution is g rea te r. K for this re action is la rg e r; so the rea ct ant, Cd C~~ H 3 ) 12 + , dissoci ates to a g r ea ter ex tent .
16.3
(a)
40 Time (m in)
The plot of In [;\'20 ,;] versus time has the best line ar lit. indicating th at this is a first-order re action. T he ra te co ns ta n t is determined from the slo pe: k = - slo pt· = 0.038 m in - I .
15.9
(a) Fo r :! lI Am . t i l:! = O.693 /k
= ~ 3"
Y
For ' ~' 1.
= O.6!t\/ (O.OOI6 y - I)
"I' = 0 .693/ (0 .0 11 d -' ) ~ 6:1 d
Q = [2,lH]I[O.97] = 2.:t T he sysle m is not at equilibrium: Q < K. To reac h equilibri um.
[isohu tau e ] wi ll in crease a n d [b utan e] will decrease , (h ) Q= [2.1)0]1[0.75] = :t 5. T he sy!'ltem is 110t a t e quilibri u m ; Q > K. To reac h equilibri um , [butan e] will in crease and [ iso b u ta n e ] will
decrease.
Appendix N
I
Answers t o Exercises
A-53
16.4
Q = [l'O ]'/[N, ][O,) = [4.2 X 10- ' ]'/[0.50 ][0.25]
K
~ 1 .4 X l o- '
[butan e ]
(2 < K, so the reac tion is not at equilib ri um . To reach equ ilibrium, [1'\0 ] will increase and [N 2 ] and [0 2 ] will d ecre ase , 16.5
~H lOI 2 (
(a) Equation
Initial (M)
Equilibrium (M) K~
16.6
l ( 6H10
0.050
Ch ange (M)
(h)
+ 0.035
+ 0.035
0.015
0.035
0.035
~ O .O~2
(0. 035) (0 .035) / (0 .015 )
+
I,
H,
Initial (M )
6.00 X 10- 3
6.00 X 10- 3
Change (M)
-x
-x
+ 2x
Eq uilibrium (M)
0.006 00 - x
0.00600 - x
+ 2x
(
2 HI
)
0
(2x)' (0.00600 - x)'
x ~ 0.0045 M. so [ H, ] 16.7
~
~
0.0 21
[CO,)
(a ) K (h ) K"
~ ~
~
+ 2x
0.0057 M a m i 2 x
2x
[CO ]
~
~
lI (g ) (
' Y, II , (g )
lUll I M
17.4
(a) 1'011
Ilr (g ) (
) Y, llr,(g)
Net: H (g )
+
Ilr (g ) (
A -54
butane
c'
X
0.50
After adding 2.0 M more isobutene
0.20
2.0 + 0.50
Change (M)
+x
-x
Eq uilibrium (M)
0.20
I
Answers to Exercises
~
14.00 - 1'011
17.6
(a) (b) (c) (d )
isobutane
0.20
Appendix N
2.92 ; I' ll
An swer t h is que stion by com parin g va lue s of K, and fro m Tahle 17.3. (a ) H 'tS04 is stronge r th a n H 'tSO :I' (b) C 6 H 5COJ-1 is a stro ng e r aci d than C H 3C O't H. (c) T he conjuga te base of boric acid , R(OH ) ,! - . is a stro nge r base th a n the conjuga te base of acetic ac id. C II 3 C0 2 - . (d ) Amm on ia is a stro nge r base than ace ta te ion . (c ) T h e co nj ugat e acid of ace ta te ion , C fl 3 C0 211 , is a stro nge r acid th a n the conjugate acid of am moni a . NIL!" ,
urn
Initial (M)
+x
~
Kw/ [H ,O' ]
17.5
10'"
) H llr (g
1\'1 K2/(j = 2.6
- log [0 .0012 ]
NI 14 +
(h ) [11,0 '] = 4. ~ X IIJ- " 11101/ 1.; [(JII -] = 2.1 X 10 - 10 m o l/ I. (c) 1'011 ~ 14.00 - 1IJ.46 ~ 3.54; [ ( JII - ] ~ 2.9 X 10- -1 mol/L. T he so lu b ility of Sr(O H h is h alf of thi s valu e (b ec a u se I m o l Sr(O H h give s two m o l 0 11- when di s...olved}, or 1.4 X 10 - -· mol/I..
K3 ~ 1 / ( K, ) ' I ' ~ 2 . 1 X 10'
K'WI = 16.10 Equation
= 1.4 X
~
~
3;
~ II.O~
~ ( K,)' I ' ~~. 9 X In'
( K, ) '; '
+ O H -( aq )
[H ,O +] ~ 4.0 X 10- 3 M ; [O H- ] ~ 2.5 X 10 - 12 M
' 1IIlr (g )
K; = I/
) HC N (aq )
17.3
Mani pul a te the equa tions and equilibrium co nsta n ts
Ki
H,O (C) (
NO j - is the cn uj uga u - base of th e aci d IIN0 is th e co nj u g a te ac id o f the b ase NH :t •
as foll ows: Il r, (g ) (
+
17.2
K2 ~ (2.5 X 10 - " ')' ~ 6.3 X 10- " 1/ K2 ~ 1/ (6.3 X 10-"' ) ~ 1.6 X 10,,7
Y, II , (g) + Y,
+ H, PO , - (aq )
CN - is a Bronstcd base ; it is ca p a b le of ac cepting a pro to n .
(2 2 V1+(aq) + Zn H (aq)
20.2
2 YO H (aq )
20.3
Oxidation (Fe 2 + , the reducin g ag en t. is oxidized) : Fe2 + (aq) ~ Fe3+ (aq) + c "
+ (3 mol Zuiol-
198. 1.l/ K · mol-rxn ( = 0. 198 k.I/K · mol-rxu )
+ 6 H t faq)
2 AI" (aq)
6.,S"
- [(I
AI3+ (at1) +:1 c " H,(g )
Overal l reaction: 2 AI(s)
=
= 2 S" (N 1I3 ) - [S " (N,) + 3 S" (lI ,) ] 6.,S" = (2 mol/m o l-rxo )(192.77 J1 m ol , K)
~
Reduction half-rea ctio n: 2 H + (aq) + 2 e "
2 ~lr for N II:\ (g) = (2 mol/m ol-rxn ) (- 45 .90 kJ/m o l) ~ - 9 1.80 kJ/mol-rx n
~ rS o
120.02 ~J /mo l-rxn
120.020.11 mol-rxn ~ - (8.3 145 .J/ mol-rx n . K) (29H K) (In l\.i
= ~S O ( system ) + ~ .~' O ( su rro u n d i n gs ) = 560.7 J/ K · mol- rxn - 1570J /K ' m ol-rxu = - IOI0 .I/ K · mol-rxn
The negative sign indi cates th at the process is no t SpOil· tancous. At high e r temperature, the value of - ~ 1/0 ( sys te lll ) I T wi ll he less negative. At a high enough tem pe rature . ~ .~' O ( su rroundings ) will outweig h ~ S O ( syste m ) and the reaction will he spo ntaneous.
=
~ ,G o =
- (467,900 .l/ mt>l-rxn) / 298 K ~So lln i\'
kJ/mol-rxn
+ Y, SOlO, ) - S"( lIg0 (s) ]
+ [ (0.5 mol/m o1-rxn) (205.07 .I/ mol . K)
+ S"(CI,»)
(2 mol Zm ol-rx u ) (I 8fi.2 J /1l101 . K)
~ .\q s nrro l1n din gs )
A - 58
= [ -6. / r
~ lS o =
[ ( I m ol/ m o l-rx n ) ( 130, 7 Jl mol . K)
19.6
~ I1g ( C) + \r2 0 2(g) ; determin e the n-mpcratu rc a t which ~ J; o = ~J r - T~ rSo = O. Tis
6.,S " = S"[lIg (C) ]
~SO ( sys L('m ) =
19.5
~ J /mo l -r xn
IIgO (s)
ts.H "
2 3 I 2
6.S"(system )
- 70.9 1
=
An increase in entro py is expected becau se there is a increa se ill th e number of moles of gases. 19.3
(-300. 13 kJ/ lTlol-
rxn + 0)
K) ]
+ 96 .09 .1 / K . mo l-rxn
+ 0.5 mol/ mol-rxn )
mo l-rxn)~ G'[S02 (g) ]
(2 mo l/moJ.rxn) (2 1;l.74 .lIm o] . K)
+
~ J / mol- rxn
~ 0 .44 -
Reduct ion ( MnOj " . th e ox idi zing ag en t. is re d u ced ) MnO , - (aq )
+ 8 l l t Iaq) + "
~
------>
e"
Mn 2+(aq )
+
4 Il ,O ( C)
20,11 Cl, r. '
0 .4 4 V - 0 ,026 V
~
O verall re action: MnO, - (aq )
+8
+5
11+(a q )
+5
Mn 2+(aq ) 20.4
~
F,,' +(aq ) ------> F,,3+(aq )
+
+ 3 OH - (aq ) ------> AI(OH ), (s) + 3 ,, -
20.1 2
+
H,O (C)
+ 2,, - ------>
HS- (aq )
+ ()H - (aq )
+ 3 OW (aq) ------> AI (OI I ),(s ) + 3 Il S - (aq )
H,O (C)
2 (h) Aluminum is th e re d u cing ag en t a nd is o xid ized ; su lfur is th e o xidiz in g agent and is reduced. 20 .5
20.6
Co ns truc t two h alt-cel ls, th e first with a silve r cl ectrodc and a so lu tio n co n ta in ing Ag +(a q ), a n d t he sec o n d wit h a ni ck el e lec trode a n d a so lu tio n con tai n ing Nj'H (aq}. Co nn ect th e two h alf-cell s with a sa lt bridge. Wh en th e el ect rodes are con nec ted through a ll ex te rnal ci rcui t, el ectron s will fl ow from th e a n o de (t h e n ickel e le ctro d e) to th e ca thode (t h e silve r e le ct ro de ) . Th e overall ce ll r ea ction is "' i (s) + 2 Ag +(aq ) ------> ",j2+(aq ) + 2 Ag (s), To main tain el e ctrical ne u tr ali ty in the two h alf-ce lls, negative ion s will flow from th e Ag I Ag + half-ce ll to th e Ni I Ni2 + hal f-cell , a n d po sitive ion s will flow in th e opposite d irection . A n o de re actio n : Zn (s) - ) Zn 2+(a'1) Ca th o de re acti on : 2 Ag' (aq) E o,e ll
= 1..' o,alh" de
-
E Oa " " " ,.
+ 2 e-
= 0.80
+2
Use the "n o rt h wes t-so u theas t ru le " o r cal cul a te the ce ll vo lt ag e to d etermine whe ther a reaction is p roduct-favored. Re actions (a) an d (c) are reac tan tfavored ; reacti o n s (h) a nd (d) are product -favored .
20.9
O verall re action : 2 Al(s)
f:", II
= f:0"," -
11
=
+
3 Fe 2 +(a q ) - ) 2 AI3+(aq) + 3 Fl'(s )
0 .50
11
=
n.855 V
(0,0 257 / 2 ) In K
(W1 4 c" - ) E \ al hod e
=
~
- 0.76 V
(0 ,0257 / 2 ) In K
+
2 ,, - ------> 2 Ol l " (aq )
+ Il , (g )
1:."0ca lh" cl,> = - 0.83 V
Anode : 4 Of-l Iaq) ------> O , (g ) r
E Oall" d,'
=
+
2 H,O ( C)
+ 4 c"
0.40 V
Overall: 2 H,O (C) ------> 2 H , (g ) E\ "II = E Ol ;' lh<J(k ~ - 1.23 V
-
+ O ,(g )
E O,"wd.. = - 0 .83 V - 0.40 V
Thi s is th e minimum vo ltag e n eeded to ca use this reac tio n to oc cu r.
20. 15 O 2 is fo r med a t th e anod e , hy the reac tio n 2 Il ,O ( C) ------> 4 11+(a q )
+ O , (g ) +
4 ,, - ,
(0 .4 45 A ) (45 m in ) (60 s/ m in )( 1 C / I A, s) ( 1 m o l c : / 96 ,500 C ) (l mol O 2 / 4 m o l ('- )( 32 g O 2 / 1 mol 0 , ) ~ (!.IO g 0 ,
20. 16 T h e ca th o d e re act io n (e le ctr o lysis o f molte n NaCI) is Na t fm cl t}
+ c"
- ) Na (f ) ,
(2" X 10 A ) (6 0 m in ) (60 s/m in ) (I CI I A , s) ( 1 mol e - / 96 ,500 C ) ( I mol Na / m o l e - ) (2:\ g Na/ l m ol Na )
(O,0257 In) In IIAI3+]'/ {F,,' +J'I
V,
~
20, 14 Ca thode: 2 Il ,O (C)
li)
1.22 V - ( - 0 ,02 3 ) V
= 0,44
= 0.80 V -
3
~
~
2 1,45 0 g Na
~
21 kg
1.24 V
20.10 Overall re actio n : Fc( s) + 2 H 't faq ) - ) F,,2+(aq ) + Il , (g ) ( E o( e ll
f : oa" ode
K = 7.9 X 10 1(;
1.22 - (0 ,025 7/ 6 ) In I[O,02 5 ]'/ [0.5 o J"l ~
-
Zn( s)
V - ( - 0.76 V)
20.8
= F. ° , a l h ,,"~
+ 4 CN - (a '1) - ) [Zn (CN ) _,2- ] + 2 e - E O"" ", le = - 1.26 V Overall: Zn H (a q ) + 4 CN - (a 'l ) ------> [Zn (eN) 12 -] £ \ .. 11 = 0.50 V EO~ (O.lJ25 7 I n) In K
- ) 2 Ag (s)
Mg is easies t to o xidize , an d Au is th e most d ifficult . (S" " Tahle 20 , I.)
150 kJ)
Anode : Zn (s )
c:
20 .7
(~
20.13 Cat hod e : Zn H (aq ) + 2
= 1.56 V
(E\ "II = 1.22 V,
E O( ell
K~
+ 3 S (s) + 3
- nFlo" ~ - (2 mol ,, -) (96,5 00 C/ mol,, - ) ( - 0 ,76 V) (I Jl I C ' V)
146 ,1>80)
- (1.055
O verall re act io n: 2 AI(s)
0 .41 V
= - 0.055 V; n = 2 EO~ (O.lJ25 7 I n) In K
Red ucti on h alf-reactio n : S (s)
~
The m-ga tivc va lue o f 1':0 and the posit ive val u e of 6.(; 0 both indi cate a reacta n t-favo red re actio n .
4 1l,o ( C)
(a ) Ox idation h alf-re act io n : AI(s)
«(J.()25 7/2 ) In I[O,02 4 ] 1.0 / [O, 056]'J
2)
f:,, " ~ f:' ",," - (O,025 7 In) In IIF,,2+]P n / III +]'1
Chapter 21 2 L1
(a) 2 Na (s) + Rr,( C) ------> 2 "'aRr (s) (b ) Ca(s) + Sds ) ------> Ca Sd s) (c ) 2 Pb (s) + O , (g ) ------> 2 PhO (s) Appendix N
I
Answers to Exercises
A-59
Lead (ll ) o x id e . a red co m po u n d co m m on ly called lit h ar g e , is the most wid el y u sed in org ani c le ad co m p o u n d . Maroon -col o red lead ( IV) oxide is the p rodu ct o f lead ox id a tio n in le ad -acid sto rage ba tteries (C h a p ter 21 ) . Oth e r o xide s suc h as Pb 30~ a lso e x ist (rl) 2 AI (s) + 'I C I,(g) ------> 2 AIC I, (s)
2\.2
(a ) H , Te (h) Na,As O , (c) SeCI" (,I) IlBrO,
21,\0 (a) :N = l':-O:
(a) N H 1 + (am m o n iu m io n ) (b) O ~~ - (p erox id e ion ) (c ) N 2 H 4 (hvd raz inc ) (.
E = l (6.626 X I O- :H.I . s/ p h o to n) (3 .00 X 10' m / s) J1 (2.0 X 10- " )
(iI ) Co (N Il , j"CI, (h) ( i) Kj [ C o (N 0 2) ti]: a co mplex o f cobalt f l l l) with a
E = 9 .94 X 1O- II.l/photoll
co o rd ina tio n n umber of 6 ( ii) ~ tn (N H 3) 'ICI 2 : a co m p lex o f mangancscf l l} with a coordination n umber of 6
22 .2
(a) lu-xaaquani ckclfl l ) sulfate
(h) d icvanobis fcthvlc ncd iami ne )ch ro miu m ( III) chlori de
23.3
(e) p ot as sium a m mi nctri ch loroplati na tcf l l )
(d) p o ta ssium di chlorocupratc fl ) 22 .3
22.4
( a) Ge om etri c isomers are possible (with th e ~H 3 Iigan d s in cis a n d trans p osition s) , (h) Only a single structure is po ssibl e . (e) Only a single structure is po ssible. (d) Thi s com pou n d is chiral : the re aft' two o ptica l iso m ers. (c) Only a single struc tu re is possible. (f) Two stru ct u ra l isom ers a re po ssib le based o n coordina tion o f the :\0 2 ~ ligand th ro u g h oXy~t'n or nitrogen .
(a ) IR u ( H 20 )6F +: An o cta hedral co mplex o f r u thc n iumf l l } ( d 6 ) . A lo w-sp in co m plex ha s no un p aired e lec tro ns a n d is diamagneti c. A h i~ h spin co m p lex h as fo ur unpai red el ec tron s and is pa ra m agn e tic.
.i. i
dcry
d,n
«;
h ij.( Ii ·'pi l1 Rtl ~ '
23.4
( a)
.}Jl f3
(h )
23.5
(iI) ?~ Si
-----+
dX L
22.5
(c) ~t Pl1 ------4
«;
;4~l g
~'l
f3 ii' Sc + JII f3
or
;g Ti
+
~l c
-----+
+ ~U -----+ J~ Ca + ~Il f3 if
= 3.090 = 5. 150
X 10 " .J/llIo! ( ~ :1.09 0 X 10" k.J/llIol)
X 10" ~ J /ll10 1 n ucl eo ns
23. 7
(a) 49 .2 yt';u s is exactly 4 half-lives : quan tity re main ing = 1.5 m ~ ( I/ 2 ) ' = 0.09 4 m g (b ) 3 h a lf-lives. 36.9 yea rs (c) I % is between () ha lf-lives, 7:l8 yea rs (1 /64 remain s), a nd 7 h alf-lives. H6.1 yt'ars (1 / 128 re mains)
23 .8
In ([ ,1]1 [,1. ])
~
- kl
In ([ 3 . 18 X 10"]1 [3 .:15 X 10"]) ~ -k(2 .00 ell
(rl~)
k
(a) T Ill' C"H 5 - lig an d is a n a n io n (6 elect ro ns }, C"H 6 is a ueutral Iigand (6 e lectrons) , so ~ I n must 1)(' + 1 (6 val en ce electrons) . There is a tot al o f 18 va len ce electrons. (h) The ligand s in this com p lex are all n e u tral. so tilt' ~' o a to m must h an' n o charge . The (:.,H6 lig an d con tribu tes six electron s. e ach Co conu-i but cs two electron s fo r a tot al o f six. a n d ~I o h as six valence electrons. The tot al is 18 e lec tro n s.
~
0 .0260 el- I
' , 12 ~
23. 9
O.6!J3/ k ~ 0 .1i9:1/ (O.02 60 el- I )
k = 0.693/ 11/2
In ( [A]I [A .])
= 0.693/ 200 r = :t 47 = kt
(a) ~ Rn ------>
'ill' Po + l "
(h ) 2~: Po
2Jt At +
~\
f3
= 26.7
d
X 10- 3 y- l
-r
In ( [:1.00 X 10"]1 [1;.50 X 10 " ]) = - (3 .47 X 10 - ' y- I) I In ( ~. 62 X 10- '" ) ~ -(3.47 X 10 - ' y-I) I t =
6 190 v
23 .10 In ( [,1]/ [,.1 .])
~
- kl
In ([ 9 .32 ]/ [ 13.4 ]) = -( 1.2 1 X
------4
(el)
Ism = O. 0343H g- /m ol dE ~ (3.4 :18 X 10-" kg / llIo l) (2.99 8 X 10' )' 1'4.,
d)z
Chapter 23 23 .1
(c) .", f3
4i Sc
h,w-spill RlI ~ '
H H -H -
+ ia + ~l e ~f:: T h + ~I f3
------4
lJ K ft. p +
(h ) ~Ti ------4
23. 6
dX),
X 1010 .I/mol
(a ) Emission of six a p anicles leads to a decrease of 2-1 in the ma ss number an d a decrease of 12 in the a to m ic numbe r. Em issio n o f fo u r f3 particle s in creases the ato mi c number by -I. but d o e sn 't a ffe ct the m ass. TIlt' fin al p roduct of th is proce ss h as a m ass n u mbe r o f 2:l2 - 24 = 208 a n d a n a to m ic number o f 90 - 12 + -1 = 82 . identifyin g it as 2lJ Ph .
Step 3: ~S Ac
_ i_ .t,
xr" i""
= 5.99
(9 .94 X 10- H J / pho ton ) (6.022 X 102:J p hotons/mol)
Step 2: ~ Ra -----+ t'~f Ac
has two u n paired electron s an d is para magnetic.
dX L
E (p e r m ol t')
(d) t!~ K
(b ) !K i(N H 3)tiF +: An octahe dra l com ple x o r nickclf ll ) «(f ). On ly one electron co n figu ra tion is po ssible : it
d:q
=
(b) Step 1: ~f Th ------4 ~ Ra
.L .i. H
E (per mol e}
I ~ 3 .00 X 10'
1O - ~
y- l ) 1
Y
23 .11 :3000 dp m / x = 1200 dp m / f.() .O m g
x = 150
m~
Appendix N
I
Answers to Exercises
A· 61
APPENDIX
o
Answers to Selected Study Questions
CHAPTER 1 1.1
(a) C, carbon (h) K, potassium
(c) (d ) (e) (0
1.3
1.5
C l, ch lorine P, p hosph oru s Mg, m agn e sium Ni, nickel
(a) Ra, bariu m (1)) T i, titan ium (c ) Cr, ch ro mi u m (d ) Ph , lead (e) As , a rsen ic (f) Zn . zinc (a) Na (c le m e-nt) an d NaC I (co m p o u n d)
(b) Suga r (co m p ou n d) a n d ca r bon (c le me n t) (e ) Gold (el e me n t) a n d gold chlo rid e (com pou n d)
1.7
(a) Ph ysical p roperty
(h) (e ) (d ) (c)
C h e m ica l p rope rt y Che m ica l prope rt y P hysica l p roperty Physica l prope rt y (0 Ph ysica l property 1.9
1. 15 calc iu m , Ca; fluorin e , F The crysta ls a rc cubi c ill sh ape be cau se the a toms a rc arran g ed in cubic struc tures. 1.1 7 TIle m acro scopi c view is the ph olograp h of Na C I, an d th e p articu late view is the drawing of th e ions in a cuhie arra n gem en t. T he str u ctu re of th e co m p o u n d at the p articulate level determi n e s t h e properties th at are ob served a t th e macroscopi c level . 1.19 TIle den sity of tln- p la sti c is less IIIa11 th a t of CC14 , so the plastic will flo at on the liq uid CC t j • Aluminu m is mo re de n se th an CC I.j, so alu m in um will sin k when placed in CC I.j . 1. 2 1 Th e t h re e liq u id s will fo rm three se p arate layers with h exa n e on th e top , wa te r in th e middle , an d p crfluoro hexa n e o n the bottom . The HDPE will float at th e in terface of tile h ex anc an d wa ter laye rs. Th e PVC will float at the in terface of th e wa te r and pcrfluoroh cxan e laye rs. Th e Teflon will sin k to the b ott om o f th e cvlinder. 1.23 HDPE will floa t in e t hyle n e g lyco l, wa te r, ace t ic ac id , an d g lyce rol. 1.25
(a)
(a) Ph ysica l (co lo rle ss liq u id ) and chemical (b u rns in
air) (1)) Ph ysical (sh in)' m et a l, o ran ge liqu id ) a nd ch e m ica l
(reac t" with bromin e) 1. 11 (a) Q ual ita tive: blue-green co lo r, sol id p hysical sta ll' Quanti tati ve: density = 2.65 g / ern" an d ma ss = ~.5 g (b) Densit y, p h ysical sta te, a n d co lo r are inten sive properties, whe reas m ass is a n e xte n sive p rope rt y. (c) Volume = 0 .94 cm ' 1. 13 Obse rvation s c. e. a n d f are ch e m ica l proper ties
A·62
1.2 7 T he sam pl e 's d en sity a n d m elt ing p oint co u ld be comp a red to those of pu re silver. 1.29 If to o m u ch su g a r is ex cre ted , th e d ens ity of t h e urine would b e hi gh er th an no rm al. If too m uc h wa te r IS excre te d , th e d en sity wo u ld h e lower than n o rmal.
(h) Method A: e rror = 0.3 g /c m 3 o r a bou t 10 %
1.31 (a) Solid potassium meta l reacts with liquid water to produce gaseo us hyd ro g e n a nd a ho m og e neo us m ixtu re (so h ui o n) of po ta ssium hyd roxid e in liqu id wa ter. (b) T he reac tio n is a ch em ica l ch an g-c . (c) T he re actan ts a re po tassiu m a nd wa te r. T he products a re h yd ro g en gas a nd a wa te r (aq ueous) so lutio n of pot assium h ydroxid e . Hea t and ligh t arc also evolved. (d) Among the q ualitative obse rvation s arc 0) th e reactio n is vio le n t, a nd ( Il) he at an d ligh t (a p urple tlam e ) arc p roduced. 1.33 (a) The wa ter co u ld be eva po ra te d by hea tin g t he solutio n , le avin g the salt beh ind. (b) Usc a m ag net to a ttract the iron away from lead , whic h is n ot mag-ne tic. (c) Mix ing the so lids with wate r will disso lve only the sugar. Filtrati on wo uld se pa ra te t he so lid sulfur fr om the sugar so lu tio n. Fin ally, th e sugar could be se pa rated fro m the \...-atc r by evapo ra ting th e water. 1.35 Separat e th e iro n fr om a wei g h ed sa m ple o f ce real by passing a m agn et th ro ug h a mixture of th e ce real an d wa ter afte r the fla kes have be co me a go o ey paste . Rem ove th e iron flake s fro m th e m ag ne t am i we igh them to dete rmin e t he mass of ir on in thi s m ass o f ce real.
Method R: error = O.()()I g /cm:l or about 0.04 % (c) Met hod A: stan dard d evia tio n
= 0.2 g /c m 3
Method R (incl ud ing all d at a poin ts) : st. dcv = 1.554 g / CJll 3 Method R (ex cl uding th e 5 .8 1 1 g / c m 3 d at a po in t) : st. d ev, = 0.002 g /c m 3 (d) Method B's average value is both m o re p re cise a nd more acc u ra te so long as th e 5.8 11 g / cm 3 d at a po int is ex clu d e d.
5.4 X 10 - 2 g, two sig n ific an t figures 5.4 62 X IO ~ g, four sig-n ifica n t ligu re s 7.9 2 X 10- -1 g, three sig n ifican t figu re s 1.6 X l(r~ rnl ., two sign ifican t figure s
19
(a) (h) (c) (d)
21
(a) 9.44 X 10- 3 (h) 5694 (c) lUI (rl) 0 .122
23 Popcorn kern els 7.000 6.DOO 5.000
1.37 Ph ysical cha nge
:§
4.000
,
~
:E" 3.000
LET'S REVIEW: THE TOOLS OF
2.000
QUANTITATIVE CHEMISTRY
1.000
y
~
O.1637x + 0.096
O.DOO
298
3
K
0
(a) 289 K (h) 9 7 -o (c) 3 lO K (3. 1 X ]0' K)
5
4 2 .1~) 5
7
5.3 ern"; 5.3 X 10- 4 m 2
5
10
15 20 25 30 Nu mber of kern els
40
35
Slope : 0.16:\7 g / kern l' l T he slo pe repre sents th e average m ass of a popcorn ke rn el.
m ; 26.2 19 miles
Mass of 20 popcorn kern els = 3.37 0 g
9
T here ar c 127 kernels in a sa m ple with a mass o f 20.88 g.
II
2.52 X lO' g
13
555 g
15
C hoice (c), zinc
17
(a) Meth o d A wit h all d ata in cl uded : average = 2.4 g /cm:l Meth o d R with all da ta includ ed : average = 3.48 0 g /cm:1 g /c m 3
For B, th e 5.8 11 d at a po int ca n he excluded because it is more th an twice as la rge as all o the r points for cas e Me t hod B. Using on ly th e fi rst three points, average = 2. 70~~ g /cm 3
y ~ - 4 .00 x + 20. 00 Y = - 4 .00
25
(a) (h)
27
C
~
29
T
= 29 5
31
0.197 n m ; 19 7 pm
33
(a) 7.5 X lO- (' m ; (1)) 7.5 X l03 11 m; (c) 7.5 X 106 pill
35
50. mg- proc ai n e hyd ro ch lorid e
37
The volu me of th e ma rbles is 99 ml , - 6 1 m l . = 38 mL T hi s yie lds a density of 2.5 g /Cl1l ~.
0 .08 2 3
Appendi x 0
I
Answers to Selected St udy Questi ons
A-53
39
41
43
(a) 0. 178 11m3; 1.78 X 10- 22 em .'! (h) 3.HIi X 10 - 22 g (c) u.ss X 10- 23 g
63
Relationship between copper and absorb.mce 0.6
S
0.4
0 0
-e
0.3
~
0.2
~
~
(a) 15 %
-c
(h) 3.63 X 10 3 ke rnel s 45
8.0 X 10 1 kg of so d iu m flu oride pe r yea r
47
245 g sulfuric aci d
49
(a) 272 m l , ice (h ) T he ice ca n n o t he co ntain ed in th e ca n .
51
7.99 g/c m:l
53
(a) 8.7 g/cm 3 (h) T h e meta l is probab ly cad miu m , h ut th e ca lcula te r! density is cl os e to th a t o f co ba lt , n ickel, a n d co p p e r. Fu r ther te stin g sh o u ld h e do n e 0 11 th e m e ta l.
0.1 0
0.09 27 em
57
(a) 1.143 X lO 2 1 a to ms; 54.9% of the lan ic is fille d with a to ms ; 24 % o f th e la tt ice is o pen space. Ato ms arc sp h e res . W h en sp h e re s are p acked toge th er, th ey to uch o nly at certain poi nts, th erefore leaving space s in II H.~ str uctu re . (b) Fo ur a to ms
2.1
Ato m s co n ta in th e fo llo win g fu nda m e ntal p an icles: pro ton s (+ 1 charge) , n e u tron s (zero charge) , an d electron s ( - I charg e) . P ro to n s a n d ne u tro n s are in th e nucle u s o f a n a to m . Electron s arc th e least m assive o f the th r ee p articl es.
2.3
(a) g Mg (b) JtT i (c) %Zn
2.5
Element
Spec trophotome tric analysis of copper 1.200
S ee; -e 0
~
~
-c
12
CHAPTER 2
Electrons
Protons
Neutrons
12
12
12
1l9Sn
50
50
69
m Th
90
90
142
24 Mg
BC
y = 248x + 0.002
1.(){)()
10
6 8 Copper (g IL )
Sl o pe = 0.054 ; j-i n tc rcc p t = 0.00 4; th e a b so r b a n ce fo r 5.00 g i L of co p p e r is 0.27
Al, alu m in u m
61
4
2
0
55
59
s = O.0542x + 0.0039
0.5
Yo ur n o rmal body tempe rature (a bout 98 .fi OF) is :{7 "C. As this is h igh e r than ga lliu m 's m elt ing poi nt, th e m e tal will m el t in yo u r h an d .
63CU
0.800 2°~ B i
6
6
7
29
29
34
83
83
122
0.600 0.400 0.200
2.7
~~Co, ~~Co, ~4C()
2.9
2{)"T l is m o re a b u nda n t than 2H3T I. Th e a to m ic m ass o f th allium is clos er to 205 th an to 203.
o.coo lUXIO
Concentration (g i l )
2.11 ( 1.0 750 )( 6.01 5 12 1)
+ (0.9 250 )( 7.0 11i0 0:l)
2.13 (c ) , Ab o ut 50%. Actu al p e rce llt W7Ag Wh e n a bso rb a n ce = 0.6 35, co n ce n tra tio n 2 .55 X 10- 3 g i L = 2 .55 X 10- 3 m g /mL
2.15
6~ ) G a ,
~
6.!J4
= 5 1.H:l9%
60.l 2%; 71(;_a , 39.8H% Symbol
Atomic No.
Atomic Mass
Group
Titanium
T;
22
47 .867
4B(IUPAC 4)
Thallium
Tl
81
2.17
204. 3833 3A(IUPAC 13)
Period 4
Metal
6
Meta l
2.19 Eig h t c le m e nts: p e ri ods 2 an d 3. 18 cleme n ts: p erio d s 4 and 5 . 32 cl emen ts: period 6.
A -54
Appendix 0
I
Answers to Selected St udy Questions
2.21 (a) ( h) (c ) (d) (e) (f)
No n m e ta ls: C , C I Ma in grou p elemen ts: C , Ca , Cl, Cs Lanth a nides: Ce Tran si tion c leme nts: Cr, Co , Cd , Ce , em , Cu , Ac tin id e s: Cm , Cf Gases : CI
cr
2.2 3 Metal s: Na , Ni , !': p Me ta lloi d s: N one in this list N onmetals: N , !':e 2.25 Mo le cu lar Fo r mula : H 2S0 Structural Formu la :
4,
o I I
O -S- O - H O- H T he struc tu re is not flat. T he 0 a to m s are arrang ed a ro u n d th e su lfur a t the corn e rs o f a tetrah ed ro n . The h yd ro gen atoms arc co n n ec ted to two o f th e oxyge n ato ms.
2.2 7 (a) (h) (c ) (d )
Mg H Zn 2 + Ni 2 + ( ;a3+
(N H,1,CO , Ca l, Cll llr, AIPO , (e) AgC II ,CO,
2.4 3 (a) (il) (e ) (d)
2.4 5 Com pounds with Na +: Na 2CO j (so d iu m carbo nat e ) and N a l (so d iu m io di de ) . Co mpoun d s with BaH : BaC O j (hariu m car b o n a te ) am i Ra l2 (b a riu m iod id e ) . 2.47 T h e force o f att ra ctio n is stro n ge r in NaF than in N a I he ca u se th e di stance between ion ce n te rs is sm a lle r in NaF (235 pm ) th a n in Na l (3 22 pm ) . 2.49 (a) ni trogen trifluoride (h) h yd ro ge n iodi de (c) boron u-iiod id c (d ) ph o sph orus p cntafluo rid c 2.51 (a ) SCI, (h) Np " (c ) SiCL,
(d)
liP ,
2,53 (a) 67 s Al (il) (J.()698 g Fe (e) 0.60 s Ca (d ) 1.32 X 10' g Ne
2.29 (a) 11'1 " (h ) Ti H (e) PO / (d) HCO , (c ) S' (f) CIO,(g ) Co 2+ ( h ) SO ,'-
2.5 5 (a) 1.999 8 m ol Cu (h) 0.001 7 mol Li (c) 2. 1 X 10- 5 m o l Am (d ) 0.250 mol AI
2.31 K lo ses one elec tr on p er a to m to form a K+ ion . It h as the same n um b e r o f e lec tron s as a n AI' ato m .
2.59 (a) 159.7 g / lllo 1 (h) 117.2 g / m o l (c) 176.1 g / lllo 1
2.3 3 Ba2+ a nd Br : ions. T he com p o u n d's for m ula is BaBr2'
2,57 Of th ese el e ments, I Ie h as th e sma lle st mol ar mass , a n d Fe h as th e largest molar m a ss. T herefore, 1.0 g of l l c h as tile la rges l nu m b e r o f a to ms ill these sa m p les, a nd 1.0 g o f Fe h as the sm a lles t number of ato ms.
2.35 (a) Two K+ io n s an d o n e 52- ion (h) O n e Co 2+ ion a n d o ne 5°-1 2- io n (c ) On e K+ io n a n d o ne ~l n o., - ion (d ) T hree N II / ions a nd o ne PO ./ - io n (c O ne Ca 2+ ion a n d two ClO - io n s (I) On e Na + ion an d one C H 3C0 2- ion
2.61 (a ) 290.8 g / m o l (h) 24\1 .7 g / mol
2.37 Co 2+ g ives C oO an d Co3 + g ives C0 20 3
2.6 5 Amoun t o f S03 = 12.5 m o l Nu m be r o f mo le cu les = 7.52 X 1024 m ol ecules N u m b e r of S a tom s = 7.52 X lO2-1 a to ms Nu mber o f 0 atom s = 2.26 X 10 25 a toms
2,39 (a ) AlC l2 sh o u ld be AlC 1:1 (based 0 11 all AJ3+ io n and th ree Cl" io n s) . (h ) KF 2 shou ld be KF (b as e d on a K+ io n a n d an F - io n ) . (c ) (; a20 3 is co rrect. (d ) MgS is co rrect. 2.4 1 (a) (h ) (c ) (d)
po ta ssiu m su lfid e cobalt (lI) su lfa te am m o n iu m ph osphat e calc iu m hyp o chl orit e
2.6 3 (a) l S I (il) 4.60 (e) 4.GO (d ) 1.48
g g g g
2.67 (a) 8G.60 % Ph a nd 1~ .4 0 % S (il ) 8 1.7 1% C a nd 18.29 % II (c ) 79 .96 % C, 9.3 94% II , and 10 .65 % 0 2.69 66 .46 % co p per in C u S. 15.0 g o f CuS is n eed ed to o btain 10.0 g o f Cu.
2,71 C4 11,,0 .,
Appendi x 0
I
Answers to Selecte d St udy Questions
A- 5 5
2.73 (a) C H, 26. 0 g / m o l; C, H, (b) C HO , 116. 1 g / mol; C, H, O , (c) en; 112.2 g /m o l, C!lIl I6
2.75 Em p irica l fo rm ula, e ll; mol ecula r formula. C 2 11 2 2.77 Em p irical fo rm ula, C 3 H 4 ; molecu la r form u la ,
C~J HI 2
2.103 T he m olar mass of aden ine (C"H"N 5 ) is 13!i.l :) g / mo l. 3.0 X 1023 m ol e cu les re p re sen ts 67 g. T h us, :H ) X J OZ3 m o lecule s of adenine has a la rger m ass than 40.0 g of the co m poun d , 2.105 1.7 X 10ZI m o lecule s of wate r
2.81 Xe F,
2.107 245.75 g/ mol. Mass pe rce nt : 25.86 % Cu , 22 .80% N, 5.7 42% II , 13.05% S, and 32 .55 % 0 . In 10.5 g of com pou nd there are 2. 72 g Cu a nd 0.770 g H 2 0 .
2.83 Zn 1z
2.109 Em p irical fo rm ula of mal ic ac id : C 4 H h O "
2.79 Em p irical an d molec u lar for mu la s a re bo th
C~ H ~ O :\
2.8 5 Symbol
s8Ni
"s
"N.
sSMn
Protons
28
16
10
25
Neutrons
30
17
10
30
Electrons
28
16
10
25
nickel
sulfur
neon
manganese
Name
2.111 Fe , (CO ), 2.113 (a) C; H,N O , S H
II
II
H
2.87 S
0
b II 'C/-'~C""""' C\ I
N- II
/ C 5 CO ,(g)
+
3.1
C"III2(f)
6 11/J (g)
3.3
(a ) 4 Cr (s) + :\ O , (g) ---> 2 Cr,o, (s) (b) Cu-S'(s) + O , (g ) ---> 2 Cu (s) + SO , (g ) (c ) C,;II,CIl ,(C ) + 9 O ,(g) ---> 4 1I,0 (f) + 7 CO, (g)
3.5
(a) Fc,o, (s) + :\ Mg (s) ---> :\ MgO (s) + 2 Fe( s) Rea ctan ts = iro n ( III) oxi de , magn esium Pro duct s = magnesium o xide , iron (h) AICI,(s) + :\ NaO H (aq ) ---> AI(O Il ),( s) + :\ NaCI(aq ) Reactan ts = a lum in um ch lorid e, sod ium hyd ro xid e P ro d u ct s = alu m in um hyd ro xid e , so diu m ch lo ride (c) 2 NaNO ,(s ) + II , SO , (C) ---> :-Ia,SO, (s) + 2 H NO, (C) Reactan ts = sodiu m n itra te , sulfuric acid Prod uc ts = so d iu m su lfa te , ni tric acid (d) NiC O ,(s) + 2 II NO , (aq ) ---> N i(N O ,J , (aq) + CO, (g ) + H,O (C) Re actants = nick el (l l ) ca rbona te , n it ric aci d Pro ducts = n ickel (I1) ni tra te , carbo n diox id e , wa te r
3.7 T h e re actio n invo lving H CI is more p roduct-favored a t _ _'_' NiS (s)
N i2+(a q)
(b ) S Mn (NO ,,), (aq ) :\ Mn" (aq ) 3.21 H :-IO,,(aq )
CHAPTER 3
+
+
+
2 Na "I'O ..(aq) ---> Mn 3(Po.lh (s ) + 6 NaI\' 0 3(a q )
2 I'O / - (a'l ) ---> Mn ,,(I'O .,), (s)
+ H, O (f) ---> H"O +(aq ) + :-IO ,,- (aq)
3.23 H ,C,O ,(a q )
+
H,o (f) ---> H 3 0 +(aq )
+
HC, O ,- (aq )
H C, O ,- (a q )
+
H, O (C) ---> H,O +(a q )
+
C,o ,' -(aq )
3.2 5 MgO (s)
+
Il/J ( C) ---> Mg (OIl ), (s)
3 .27 (a ) Acetic aci d reacts with m ag n esium h yd ro xid e to g ive magn e sium ace ta te a n d wa te r.
2 C Il,CO , 1I (a q)
+ Mg (O Il ), (s) ---> Mg (CIl, CO, ), (aq )
+ 2 Il ,o (l')
Bre usted acid : ace tic aci d ; Breus ted base: magnesium hydro x ide (h) Pcrchlo ri c acid reacts with a m monia to gi ve am m on ium pe rchlora te
H CIO, (aq )
+
NH , (aq) ---> :-IH.,CIC), (aq)
Bronstcd acid : p crchl o ric ac id ; Bran stc d ba se : amm onia
3.29 lIa (OH ), (aq )
+ 2 HN O ,,(aq ) ---> lIa (NO, ), (aq )
+
2 Il ,o ( C)
3.31 Strong Bro n stcd aci d exa m p les: h yd ro chl oric aci d , n itric acid Strong Bronstcd b ase exam ple: so di u m h yd ro x id e Appendi x 0
I
Answers to Selected Study Questions
A -57
3.33 (a) (N H,),CO ,( ag) + Cu (N0 3), (ag ) -> Cu CO ,(s) + 2 N H ,NO, (ag ) C O / - ta g ) + Cu ' +(a q ) -> CuCO, (s) (b) I'b (OIl ), (s) + 2 II Cl( ag ) -> I'b CI, (s)
+2
f1,O (C)
I'b (OIl ), (.s) + 2 11 , 0 +(a q } + 2 Cl " (a q ) -> I'bCI, (s) + 4 H,O (C) (e ) BaCO , (s) + 2 H CI(ag ) -> BaCI, (ag ) + H,O (C) + CO , (g )
BaCn , (s) + 2 H,O + (aq ) -> Ba h (aq ) + 3 11, 0( 0 + CO , (g ) (d ) 2 C Il ,CO , II(ag ) + Ni( O Il ),(s)-> Ni (C H ,CO, ), (aq ) + 2 H,o (O 2 C H , C O , H( ag ) + N i(O H ), (s) -> Nih (ag ) + 2 CII 3C O , - (a q ) + 2 11,0 (0
3.35 (a ) Ag NO , (a g ) + KI (ag ) -> Ag I (s) + KNO , (ag )
(b ) Ba (OH ), (ag ) + 2 H N0 3(aq ) -> l\a (NO, ), (aq ) + 2 1l, 0 (f)
(c ) 2
+ Il "O +(a q ) -> 2
i\'a:I P O~( a q )
+
11,0 (0
:l N i(l'\ O:\h (aq) ~
Ni , (I'O , ), (s) + 6 NaNO , (a g ) 2 I'O / - (aq ) + 3 Nih (a g ) -> Ni,,(I'O , ), (s)
3.37 FeC O , (s) + 2 Il NO , (aq ) -> FdN03), (ag ) + CO , (g ) + H,() (C) lro n (II ) ca rbo na te re act s wi th ni tric acid to g ive iron ( II ) n itrate, carbo n dioxide , an d ,v'a tc-r. 3.39 (N H,),S( ag) + 2 HB r (ag ) -> 2 N H , Br (ag ) + H ,S (g ) Am m o n iu m sul fide react s wi th hydrobro m ic aci d to
give amm onium bromid e and Ilytl nlgl' 1l sulfid e. 3 .4 1 (a) HI' = + 5 and 0 = - 2 (b) C = + 3 each and 0 = - 2
2 H NO,,(ag )
+ CoCO ,,(s) -> Co (N0 3), (a g ) + H,O (C) + CO , (g )
(c) Precipitation 2 Na "I' O ,(ag ) + 3 C ll( NO ,,),(ag ) -> Cu 3 (P(}.}z(s) + 6 N aN O j(aq ) 3.49 a) Precipi tation Mn CI, (ag ) + Na -S '(aq ) -> MnS (s) + 2 NaCI (a q) Mn ' " (aq) + S' - (aq ) -> Mn S (s)
K, C03 (ag ) + ZnCI, (ag ) -> ZnC 0 3 (s) + 2 KCI (ag ) C0 32 - (aq)
+ Zn2+(aq )
----7
ZnC0 3 (s)
3.51 (a) CuCI, (aq ) + II ,S (aq ) -> Cu S (s) + 2 II CI (a g ) pre cipitation (b ) H 31'0 , (ag ) + 3 KO H (a q ) -> 3 11,0 (0 + K,I'O, (aq ) acid - base (e ) Ca ts) + 2 HII r (ag ) -> H , (g ) + CaB r, (ag )
o xida tio n-red ucti o n and ga s-fo rmi ng (d ) MgCI, (ag ) + 2 II ,O (C) -> Mg (O H ), (s) + 2 H CI (ag ) precip itation
3.53 (a) C O ,( g) + 2 N II , (g ) -> N II , C O N II ,(s) + 11,0 (0 (b) lJ O ,(s) + 4 H F(ag ) -> lJ F , (s) + 2 H,o ( f )
(e) TiO, (s) + 2 CI, (g ) + 2 C (s) ->
°= ° ~
T iC I, ( 0 + 2 C O (g ) - 2
- 2
3.43 (a) Oxidation -reducti on Zn is oxi dized from 0 to + 2. and N in NO J - is re d uced from + !l to + 4 in NO';! . (b) Ar id - base reacti on (c ) O xid ation -reducti on Calc ium is o x idized fro m 0 to + 2 in Ca (O Hh , and II is red uced from + 1 in [120 to 0 in 11 2 3.45 (a) O:.! is the o xid izing age n t (as it always is) an d so C 2 H, is the re d uci ng agent. In this proce ss, C2H 4 is ox idize d . an d O 2 is red uced .
Appendix 0
l\ a (OIl ), (a'l ) + 2 II CI (a g ) -> BaCI, (a q ) + 2 11, 0 (0
lJF, (s) + F, (g ) -> lJ F,;(s)
(e) F ~ - 1
(d ) Ca ~ + 2 a nd H ~ - I (e) II = + 1, Si = + 4, an d (I) H ~ + 1, S ~ + 6. and
A-58
3 .4 7 (a ) Acid-base
(b ) Precipi tation
Ag +(a g ) + I - (ag ) -> Agl (s)
Oll - (aq )
(1)) Si is o xid ized fro m 0 in Si to + 4 in SiCI4 . C12 is reduccd fro m 0 in CI2 to - 1 in Cl " . Si is th e red ucing age nt , and C1 2 is tile oxidizing agen t.
I
An swers to Selected Study Questions
T iCI, ( f) + 2 Mg (s) -> T i (s) + 2 MgCI, (s) 3.55 (a) N aBr, RBI', or other alkali met al bromi de s; Gro u p 2A bro mid es; o the r metal brom ide s exce pt AgHr, Hg 2Hr2' and PbHr2 (b) AI(0 11)3 an d tran sition m e tal hyd rox ides (c ) Al kalin e ea rth carbo nate s (CaCO :j ) or transition metal ca rbo na tes (N iCO j ) ( 4 Il ,O «() + Mg" (aq ) 3.6 1 (a) CI2 is reduced ( 10 CI- ) and Br -" is o x idi zed ( 10 Rr2 ) . (h ) C I ~ is the o xid izin g agent a nd Br" is the reducing •rgc n t.
3.63 (a)
~ l gCO , (s )
+ 2 1I,0 ' (aq ) -> CO, (!:) + ~ I!:" (aq) + :1 II P ( 0
Ch lorid e io n (CI - ) is the spec ta tor io n . (h) C as-fo rming: reactio n
3.69 (a ) Reactan ts: Na ( +I ) , 1( - 1).11 ( +1 ). S (+6 ) , 0 ( -2 ), Mn ( + 4) Products: Na ( + I ). S( +6). 0 ( - 2).
~ ln ( +2 ). I (O ) .
f1 ( + I )
(h ) TIll" o xidiz in g ag e n t is MnOt ' and Nal is o xidized. Th e reducin g agcm is Na l. and ~ 11l0t is reduced . (c) Based 011 the pi cture, th e reac tio n is p roduc tfavored. (d) So d ium iodid e, sulfu ric acid , a nd m an g:anese(lV) oxid e reac t to form sodium sulfa te . m a n ga nc se (ll ) sulfa n-, a nd wa te r.
3,7 1 Am o ng th e reac tion s tha t co u ld he used a re the foll owin g: wea k Bro nst e d base ( h) H ~O , C H , C0 2 H , C H jC O t (a nd a trace o f 0 11 - )
~ 1 !:CO , ( s ) ,
a nd 1-1,0 "
weak Bronstcd ac id
~ l gS ( s )
+ 2 II C\( aq ) -> MgCI, (aq ) + CO , (!:) + 11, 0 (0
+ 2 HCI(aq ) ->
~ l gS O , ( s )
~ l !:C I , ( a q )
(et) 1120 . 1-1 3 0 +, and Br - (a nd a tra ce of 0 11- )
strong- Bron su-d acid
3.67 (a ) K,CO, (aq ) + 2 II CIO .,(aq ) -> 2 KClO ,(aq ) + CO, (!:) + 11 ,0 (0 Po tassium carbonate an d pcrchl crk aci d reac t to fo rm pot assium pe rchl o rate , carbon diox ide , CO , (g) + :1 1I,o (0 (h) F"CI, (aq ) + (:'\ II,),S( aq ) -> FeS (s) + 2 :'\II,CI(aq ) precipitation Il'On (lI ) ch lorid e and ammon ium su lfid e react to fo rm iron ( II) su lfid e a n d am mo niu m chlo ride
Fe" [aq ) + S' - (aq) -> FeS (s) (e) Fe (:'\O, ),(aq ) + J\"a,CO, (aq ) -> FeCO, (s) + 2 :-;a:-;O, (aq ) prec ip itati on Iro n ( II) ni tr a te a nd so dium carbo ua n- react to fo rm iron ( II) ca rbo na te and so d iu m n itra te
+ SO , (!:) + 11,0 (0
In each caw. the re sult ing so lu t io n co uld he evaporat ed to ohtain th e d esired mag nl'siu lll ch loride.
3.73 Th e Ag- " was red uced ( to silver meta l) , and th e glucose was ox idi zed (to CI;II I I!()7) ' T he Ag " is the o xidi zing ag e n t, and the g l uc o.~ t' is tln- reducin g- age nt. 3.75 We ak el e ct rol yte n-st: Co m pa re th e co nd uc tivity o f a so lu tio n o f la ct ic acid a nd th at o f all e q ua l co n ce n tration of a su-o ng acid . T he co nd uc tivity of th e lacti c aci d so lu tio n sho uld he sig- ni fica n tly less. Reve rsible reaction: T he fa ct that lac tic acid is an el ectrol yte indica tes th at rh e reaction proceed s in the fo rwa rd d irec tio n . To tes t whet her the- ionization is rcversible , on e co u ld P!"l' IM!"l' a so lu t io n contain ing as m u ch lactic aci d as it will hol d a nd then add a stro ng: ac id (to p rovid e H :\O " ) . If th e reac tio n proceed s in th e re verse direc tion , this will cause so me la ctic a cid to p re cip itate .
3.77 (a ) Several p re cipitati o n rea cti o ns a re possibl e : i.
R"CI, ('''I) + II ,SO ,(" 'I ) -> R"SO, (s) + 2 II CI(aq )
ii. BaCl t (a q )
+
i':al''')() I(aq )
----7
lIa50 ,(s) + 2 J\"aCI(aq ) ii i . Ra (OH ),(aq ) +
1I ,.~O , ( " q )
F"" (aq ) + CO ,' - (a q) -> F,'CO , (s) (d) :1 :\'a O Il (aq ) + FeCI, (aq )-> 3 :\'aCI (aq ) + Ft"(OIl ), (s) precipi ta tion
H ,.~ ( g)
+ 2 II CI(aq ) -> ~l gC I , ( a q )
str o ng Bronsu-d ba se
+
-> lIaSO , (s) + 2 H ,O (C)
(h ) G Ra50 , (s) + CO , (g) + H, O (C)
Sodium h ydroxide an d iron ( lI l ) c h lorid e react to fo rm so d ium ch lorid e a nd iron ( III) h yd ro x id e
:1 O ll - (aq ) + Fe" (aq ) -> Fe(O Il ), (s)
App endix 0
I
Answers to Selected Study Question s
A - 69
CHAPTER 4
4 .2 1 (a) 14.3 g Cu (NH 3),SO, (b) HH.3 % yield
:~ 1 0
g A L;/0 3
4.1
4.5 mol 0 2;
4.3
22.7 g fi r 2 ; 25 .3 g 1\1 2R r 6
4 .5
(a) en b carbo n dioxid e , and H 20 , wate r (b) CIl,( g) + 2 O , (g) ---> CO ,(g) + 2 Il ,o (f) (c) 102g0, (d ) 12H g products
4. 7
Equati on
2 Pb5(s)
Change (mol)
- 2.5
Final (mol)
4.25 H4.3% Ca C0
+ 3 O, (g) ->
2.5
Init ia l (mol)
4 .23 9 1.9 % hyd rat e
0
2 PbO (s) + 250,(g)
3
4 .27 1.467 % TI,SO, 4 .29 Em p irical fo rm ula = ell 4 .3 1 Empiri cal fo r mul a = CH 2 ; molecular for m u la
= C"H \()
4 .33 Em pirical for mu la = CH sO; m o le cul a r formu la = C'j! lI jj O 'j!
3.8
0
0
- %(2.5) ~ - 3.8
+%( 2.5) = + 2.5
+ %(2.5) = + 2.5
0
2.5
2.5
4 .35 Ni (CO) , 4 .3 7 [Na,Co,,] 0.254 M
~
0.25 4 M; [Na ' ]
~
050H M; [C0 3'
- ]
4.39 0.49 4 g KMnO, Th e am o unts ta bl e show s th at 2.5 m ol of PhS re q uires ~ ( 2 .5 ) = 3.8 mol of O 2 and p roduce s 2.5 mol of PbO an d 2.5 m ol of 502'
4.9
+3
(l;/ (g ) ~ 2 Cr';!O j (s) (h ) 0.175 g orCr is e q u ivale n t to 0.00337 mol
(a) Balan ced e qua tion : 4 Cr(s)
Equation
4 (,(s)
+ 3 O, (g)
2 Cr203 (s)
->
Initia l (mol)
0.00337
0.00252 mol
0
Change (mol)
- 0.00337
- %(0.00337)
lU D.OO337)
Fi nal (mol)
0
0
= - 0.00 252
= + 0.00168
0.00168
Th e 0.00 168 m o l Cr 20 3 produ ced co rrespo nds to 0.256 g e ftO :;_ (c) O.OHI g O,
4 .11 0.11 m ol of Na2S01 and 0.62 m ol of C ar e mixed . So d iu m su lfa te is th e limiting rea ct ant. T he refore , O.ll mol of Na2S is fo r me d, or 8.2 g.
4.41 5.08 X 1O ~ mL 4 .43 (a) 0.50 M NH ,' an d 0.25 M SO ,' (b ) 0.246 M Na ' and 0. 123 M col (c ) (1.056 M 11+ an d 0.056 M \\'0 34 .45 A mass of l.OG g of Na2CO J is req ui red. Aft er weig h. ing o u t thi s q ua n tity of N at C 0 3, tran sfer it to a .')00 .· 1111. vo lu m e tric flask. Rin se allY so lid from th e n eck of the fl ask wh ile fillin g the fla sk with di still ed water. Disso lve the sol u te in wa te r. Add water until th e bo ttom of the meniscus o f t he wa te r is a t the to p of I he scribed m ark on the neck of th e flask . T horoug hl y mi x th e so lu tio n . 4.47 0.0750 M 4.49 Met hod (a) is co rrec t. Me thod (b) gives an ac id concen [r atio n o f 0.15 ~1 . 4 .5 1 0.00 340 M
=
IO- pll = 4.0 X 10- 4 M ; the so lu tio n is
4 .13 F2 is th e lim it in g reacta n t.
4 .53 [ H sO +] acid ic.
4 .15 (a) C H ,\ is the lim iting reactan t. (b) 375 g II , (c ) Excess H 20 = I :{90 g
4.5.1 II NO J is a stro ng aci d , so [1130 +] = 0 .001 3 M . p ll = 2.H9.
4.17 (a ) 2 C"H ,, (f) + 19 O , (g) ---> 12 co, (g) + 14 H,o (g) (h) O 2 is the limit ing reactan t. Prod uc ts arc 187 g o f CO 2 a nd 89.2 g o f H 20 . (c) 154 g o f h exa ne rem ain s (d ) 2 ( ,H,, (f ) + Ig O, (g) -> 12 (O, (g)
Equation Initial (mol)
2.49
Cha nge (mol)
- 0.70 7
- 6.72
1.78
0
Final (mol)
6.72
+
14 H,O(g)
0
0
+ 4.24
+4.95
4.24
4.9 5
4.57
pH
(a) 1.00 (b) 10.50 (c ) 4.89 (d) 7.64
[H,O'j 0.10 M 3.2 X 10- 11 M
x 10- 5 M 2.3 x 10- 11 M
1.3
4 .59 268 m L 4 .6 1 2 10 g Na O H and 190 g CI, 4 .63 174 mL o f NatSt 0 3 4. 65 1.50 X 10 ' mL of r h i NO, ),
4 .19 (332 g /4 07 g) 100 %
~
H1.6%
4 .67 44.6 m l. 4 .69 1.05 2 M II CI
A-70
Appendix 0
I
Answers to Selected St udy Questions
Ad dic/Basic Acidic Basic Acidic Basic
s.1:1 g
4.71 104 g/ mo l
4. 107
4. 73 12.8% Fe
4. 109 (a) (b) (c) (d)
4.75 Ca libration plot for dye 0.80 0.70
Na2S20 j, !Ui .8%
pH ~ 0.97!) l H 3 0 +] = 0.00 28 M; t he so lu tio n is ac id ic. [113 0 +] = 2. 1 X 10- 10 1\.'1; th e so lution is ba sic. T he ne w so lu tio n 's conce n tr ation is 0.102 M FlC!; th e pll = 0.990
4. 111 Th e conc e n tra tion of h ydro chlo ric aci d is 2.92 t he pH is - 0.465
o.en
~1 ;
0.50
0.30
4. 113 1.56 g- of CaC 0 3 requ ired; 1.00 g CaCO j remain ; 1.7:1 g CaC I2 produce d .
0.20
4. 115 Volume o f water in the pool
u-n
0.10
=
7.6 X 10" L
4. 117 (a) Au , gold , has he en ox idi zed and is the reduci n g agen t.
0.00 oo Co ncent ratio n (M)
(a) slo pe = 1.2 X 10" M (b) 3.0 X 10-" M
1
;
j-in te rc c p t
= 0.18
\1
(b) 26 I. NaCN so lu tio n
4.77 (a) Products ~ CO,(g) an d H,O (g) (b) 2 C"II,,(f) + 15 O , (g) ---> 12 CO,(g) (e) 4!).28 g 0 , (d) 65.:12 g products (= 0 ';.1 m a ss)
0 2' oxygen, h as beeu reduced an d is th e oxidi zin g agen t.
SU Ill
+ 6 11,0(g)
ofCtiH/; mass a nd
4, 119 T Ile conc e nt ra tion of Na 2CO j in th e- first so lut ion p repa red is (J.0275 M, in the sec o nd so lu tio n p re· pa red the con cen tr ati on o f Na 2CO j is 0.00110 M. 4. 121 (a) First reac tio n : o xid izing ag en t = CuH an d red ucing agen t = 1-
4.79 O.2S g ar ginin e , 0.21 g o rn ithine Se co nd reaction: o xid izing agen t = 13 - a nd red uci ng ag e n t = S20 / -
4.81 (a) titanium (IV ) chloride , wa te r; tit anium ( IV) o xide , h ydrogen chloride (1)) 4.60 g H,O (c) 10.2 TiO, . 18.6 g II CI
4. 123 x
4.83 8.33 g NaN ,
4. 125 11.48% 2.4-D
4.85 Mass percen t sacch arin = 75.92%
4. 127 3.3 m ol H 20 / m o l CaCl2
4.87 SiH ,
4. 129 (a) Slo pe = 2.06 X 10'"; 0.024 (1)) 1.20 X 10- ' g/ L (c) 0.41 3 rn g 1'0,' -
4.89 C ,H, O
(b ) ~
67 . ~ %
co p pe r
6; Co (NH j),;C!, .
4.9 1 1.85 kg II ,SO , 4, 131 T he to tal m ass o f the beakers a nd p roducts after re ac tio n is equal to th e tota l m ass befo re t he re act io n ( 16 1.170 g ) beca use no ga ses 'were produ ced in the reacti on and the re is conse rva tio n o f mass in ch e mical re action s.
4.93 The ca lcu la ted m o la r m ass of the m et al is 1.2 X lO 2 g/m oL T he metal is pro bably tin (118.67 g/m o l).
4.95 479 kg CI, 4.97 66 .5 kg CaO 4.99 1.29 g
c.u, (45. 1% ) an d
4. 101 62.2% CU2S an d 26.8 % 4.1 03 (a) MgC0 3 (s)
1.57 g C.IE w (54.9 %)
c.s
+ 2 H,O ' (ag )---> CO, (g ) + Mg H
(a q)
+ 2 11,0 (f)
(h) Oas-fo rmi ng reaction (e) 0.15 g
4.105 15.0 g o f Na l reo, req uire 1190 ml . of 0.15 M ace tic a cid. Th erefore , ace tic a cid is th e lim iting reacta n t, (Conversely, 125 mL o f 0.15 M ace tic acid re q u ires only 1.58 g of Na HCO :~ .) 1.54 g of NaC H:,C0 2 p roduced.
4.133 T h e bal an ced che m ica l e q ua tio n indicat e s that the stoichio me tric rati o of HCl to Zn is 2 mol [ ICI/I m ol Zn. In ea ch reaction , the re is O.IO() mol of HCI prese nt. In re ac tion 1, there is 0. 107 m o l of Zn p re sent. This g ives a 0.93 mole HCl/ m ol Zn ratio , indi cat ing th at lI el is th e lim itin g re actant. h i re acti on 2, there is 0.050 mol o f Zn , g iving a 2.0 mol I re I/mol Zn ra tio. Th is ind icat e s th at th e two re actants are prese n l in exac tly the correct stoi ch io m e tric ra tio. In re actio n 3 , there is 0.020 mol of Zn , g ivin g a 5.0 m o l HC I/ mo l Zn ratio . T h is indi cates th at th e lIel is pre sent in excess a nd that the zinc is th e limiting re acta nt.
Appendix 0
I
Answers to Selected Study Questions
A-71
4.135 If hoth studen ts base their calculations on the amount of Hel sol ution pi pe ted into the flask (20 ml .) . then th e secon d student's resul t will he (e ) , the same as th e first studen t's. However, if th e l ICl conce ntratio n is calculated using the diluted solutio n volu me , student I will usc a volume of 40 m l, and stu den t 2 will use a volume of 80 m l. in the calculation . Th e second student 's result will be (c), half that of th e first student's.
5.43 (a)
s.u- =
- 126 kJ / mol-rx n
(h )
CH 4( g) + \ / 2 O , (g)
+ 3/ 2 O , (g ) ~
CII , O Il (g)
4.137 150 mg / dL. Person is into xica ted. d , ~O =
- S02.4
5.1
106 .1
5.3
5.0
5.5
170 hal is eq u ivalen t to 7 10 k]. considerably greater
0.14 0 J/g· K
5.9
2.44 kJ
5.11 32.S
-c
5. \3 20.7
-c -c
5. \ 5 47.S
5.45 D..,-Ho = + 90.:3 5.47 C (, )
~I
~ J /mo l-rxn
+ 2 H, (g ) + 1/2 O , (g)
d/ r = -
than 2RO kJ . 5.7
+ 676
co, (g ) + 2 H,o (e )
Mec ha nica l e nergy is used to move the leve r, which in turn moves gear s. The device produ ces ele c trical e nergy and radiant e nerh~·' X
D.. , H ~ =
~J
CHAPTER 5
CH,O H (e )
23S.4 k.J/mol
5.49 (a) 2 CI'(' ) + 3/2 O , (g) dJW
-t
~
-t
CI',o, (')
- 1134.7 kJ/m ol
(h) 2.4 g is eq uivale nt to (l.046 mo l of Cr. This will produ ce 26 ~J of cn crgv transfe rred as heal. 5.5 1 (a) d /r = (h) su - ~ (c) 6./1 ° = NaCI(s) (d ) 6.Ho = iron
5.\7 0.40Jl g · K
24 kJ for 1.0 g of phosphorus IS k.J f or 0.2 mol NO 16.9 ~J for the fo rmation of 2.40 g of 1.8 X 10:1 k] fo r the ox idation o f 250 g of
5. \ 9 330 k.J 5.53 (a) d,H ' ~ - 906.2 k] (1)) Th e heat evolved is 133 kJ fo r the oxida tion of 10.0 g o f N il ,
5.2 \ 49.3 kJ 5.23 273 .1 5.25 9 .97 X 10 5.1 5.27 React io n is e xo therm ic because D..,-Ir is negative. The heat evo lved is 2.38 kJ.
5.55 (a) d ,H o = + 161. 6 ~ l / mo l-rxn ; the reaction is e ndot he rmic . (h) 1\a (s) + O , (g )
5.29 3.3 X 10' k.J 5.3 1 iHI
s.u;
= - 56 kJlmol C,O Il
= - 553.5 k.J
5.33 O.52J1g · K
5.3 5 d ,H
=
+ 23 kJ/mnl-rxn
5.37 297 kJ/mo l SO,
1\,,0 (, ) + 1/2 O ,(g )
d rH j = - 634.3 k J
5.39 3.09 X 10" q /m ol C6 Il,CO, 1I
s.n- ~
5.4 1 0.236.J1g · K 1\aO , (s)
5.57 d jH o= + 77.7
A -72
Appendix 0
I
An swersto Selected Study Questions
~J /mo l
for nap htha lene
+ SO.8 k J
5.5 9 (a) Exoth er mi c: a proc ess in wh ich e n e rgy is tran sfe rred as h ea t from a syste m to its su rrou n d ings. (T h e co m bustion o f meth ane is c xo thcrmic.)
End ot hermi c: a p rocess in whic h cn ergy is tra nsfe r rcd as h ea t from th e surro u n di ngs to th e syste m . (Ice nwll ing is en d oth c r mi c.) (h ) System : th e object o r coll ec tio n o f objects b e in g stud ied . (A chemi ca l re ac tion-th e syste m- ta king p lace in sid e a ca lorimeter- t he su r ro u n d ing s.) Surrou ndin gs: everyth ing o u tside th e syste m th at can exc hange m ass o r en e rgy with th e syste m. (The ca lo rime ter a nd everyth ing outside th e ca lorim ctcr co m p rise th e su rrou n d ings. ) (c ) Sp ecifi c h e at ca paci ty: th e quantity o f en e rgy th at m u st he tra n sfe r red as hea t LO raise th e te mpe ra tu re of I gra m of a su bs ta n ce I ke lvin . (T h e specific h e at ca pac ity o f wa te r is 4 .184 .11 g . K) . (d ) Sta te fu n cti o n: a quant ity th at is ch arac terized by changes th at do no t depend on th e pa th chose n to g o from th e initial sta te to th e final sta te. (En thalpy and in ternal e Ill'J'gy a re sta te fun c tio u s.) (e ) Sta nda rd sta le ; th e most sta b le form of a substance in th e ph ysical sta te th at exi sts a t a p re ssu re of I b a r a n d a t a speci fied te mpe rature . (T h e sta ndard sta te o f carbo n at 2.5 °C is g ra p h ite .) (I) Enth a lp y change , /1H: t h e ene rgy tran sferred as h ea t at co ns tan t pressure . (The e n t hal py cha nge fo r m eltin g ice a t 0 °C is 6.00 k.l/ m o1.) (g) Sta n dard e n tha lpy o f for mati on : th e e n tha lpy change fo r th e fo rm atio n of 1 mol o f a co m p o u n d in its sta n da rd sta te dire ctl y fro m th e com po ne n t e le m en ts in th e ir sta n da rd sta tes. (D./ I" for liquid wa ter is - 2 8 .5 . 8 :~ kJ/mol )
5. 63 Sta ndard sta te of ox yg en is gas,
O~ (g ) .
0 2(g") ----+ 2 O (g ) , D.,H o = + 498.34
3/2 0 2(g ) ---> O ,,(g ) , tJ.,H'
~
~I ,
e n dot h e r m ic
+ 142.67 kj
5.65 S n!l r, (s)
+ T iCI, (s ) ---> S"CI, (s) + Tillr, (s )
tJ.,l1" = - 4.2 k]
Sn CI, (s)
+ CI,(g) ---> S"CI, (f)
s.u : = + CI, (g)
Ti CI, (f )
---> TiCI, (s )
Snllr, (s)
+ T iCI,(f) ---> Sn CI, (f) + T iBr2(s)
5.67 C"
~
195 kf
kJ
tJ. ,H'
~
+ 273
tJ.,W
~
+ 74 kf
0.24 1/ g . K
5.69 Mass o f ice me lted
5.7 1 Fin al tempe ratu re
= 7.5.4 g = 278 K (4.8
°C)
5.73 (a) W hen su m m ed , the fo llowin g e q u a tio n s gi\'e the halanced equati on for the fo rmatio n of B2H r.(g ) fr om the el em en ts.
:> H 2(g)
+ V 2 ()2(g ) ~
:> H/) (g)
H20 j (s)
+ :~
H2H ti (g) +
H 20 (g)
---?
~
0 2(g )
~,/l' ~
- 1271.9 kJ
:J.,Jl o =
- 72 .~l.'I
kJ
:J.r/l o = +20:>2.9 kJ :J.J[O= + 3::'.0 kJ
(b) T h e e n thal py of for m ati on of B 2 H 6 (g ) is
+ 311 .6
~l / m o l
(e)
2 R(s)
+ 3 H 2 (g )
+ 3/2 0 2(g) :J. /l o = - 127 U I kJ
+
3/2 0 2( g)
su - =
- 725.4 kJ
J./I 0 = + 20:\2.9 kJ
5.6 1 (a) System : react ion bet ween me th an e a nd ox yg e n Su rroundin gs: th e furn ace an d th e res t o f th e unive rs e . En ergy is tr an sfe r red as h eat from th e sys· t ern to th e su rro u n d ings. (h) Syste m: wa ter d rops
Su rro u ndin gs: skin an d th e res t o f th e un ive rse En crgv is tr an sfe r red as h e at from th e su rrou ndin g s to th e syste m
(c ) Syste m: wa ter Surro u n d ings: freezer a nd th e re st o f th e universe En ergy is trans ferre d as h e a t from the syste m to th e su rroun dings (d ) Syste m: re action of a lu m in u m a nd iron (I II ) o xid e Su rro u ndin gs: flask , lab o ra tory ben c h , a nd res t of th e un ive rse
(d) The fo r m at io n o f R2H d g) is reac tan t-favored.
5. 75 (a) tJ. ,W
~
+ 1:11.3 1 k]
(h ) Rea cta n t-favo red (e) 1.09 :12 X 10' kf
5.77 Assum ing CO 2(g) a nd I-1 20 (t) are th e p ro ducts o f combus tion :
D.Jr for iso octam- is - .546 1.3 q /mol o r - 47.8 1 kJ per g ra m
D. rH o fo r liquid m e th an ol is - 726.77 - 22.6H2 kJ p e r gram
~I / m o l
or
5.7 9 (a) Ad di ng the eq ua tions as th ey ar t' given in the que sti on res u lts in th e d e sire d e q u at io n fo r th e fo rm at io n of SrC0 3 (s) . T he calculate d tJ. ,W ~ - 1220. kj/ mo l.
En ergy is tr an sferred as h e at fro m t h e syste m to th e surrou n d ings . Appendi x 0
I
Answers to Selected Study Questions
A-73
(h)
5.10 1 (a) Sr(s)
+ 1/ 2 O, (g ) + C(grap h itc ) + O ,(g )
tJ. / I" = - 592k J
tJ./ I" = - 39 4 kJ
.1(H (i~-~-b u l('J1('
tJ./, O~ - 1220 k J
CO, (g)
+ fi O ~( g)
AJI
SrO (s)
tJ.,l1" = - 234 kJ
4 CO , lg )
SrCO,,(s) ~
5.8 1 tJ.,W
+ ·1 1l,D (g )
(h ) ci.....2·11111l·11e: b./r = 146.1 kJ /mol
' ran....2·lllllelle: b./fO
- 305.3 kJ
= 142B
kJ /mol
o
l -bu tcnc: b.JH = 155 .3 ~ J /mol
5.83 C"b =0. 121J1g·K
(c)
l-b ute u e
5.85 tJ. ,1l
~
- 69 kJ/ mol AgCI
5.8 7 36.0
~J
evolved per mol of i\ II 4NO :\
5.8 9 T he stan da rd e nthalpy change , IJ. rH ", is - :152.8H kJ. T he q uan tity of m ag-n e sium n ee d ed is 0.43 g-.
riJ-2-h u tc nc
5.91 (a) product-favored (h) reacta n t-favored 5 .93 T he en thalpy c hange for eac h o f th e th re e reac tion s below is known or can he me asu re d by calo rimetry. Th e three eq ua tio ns su m to give the en th al py of form ati o n o f CaSO.•(s) .
+ 1/ 2 O , (g)
Ca (s)
~
CaO (s)
~ , II"
+
CaO (s)
kJ
i:J./ r
= = - :>9S.77 kJ
~ ,II" =
SO ~ ( g )
- 402.7 kJ
s.u : = .J./r =
5.95 Metal
- 1 4 ~ ~ . r.
kJ
24.2
Fe
25 .1
Cu
24.5
Au
25.4
5.103 (a ) - 726 kJ/mo l Mg (1)) 25. 0 -c 5.105 (a ) Meth a ne (h ) Methane (c ) - 279 kJ (rI) C H ,(g) + 2 O , (g ) -> C H"O H (f ) 5.107 (a) Metal H eated = 100.0 g o f AI; Me ta l Cooled = 50.0 g o f Au ; Final Temperature = 26 °C. (h ) Metal H eat e d = 50 .0 g of Zn ; Meta l Cooled = 50.0 g of AI; Final Tem perat u re = 21 °C
Molar Heat Capadty (Jjmol . K)
Al
+ 4 11, (g )
(d ) - 3.3 kJ/m o l-rxn
= ~/ I' = - fi ~ 5 '(19
s.tr
4 C( s)
CHAPTER 6 All the metals h ave a molar h ea t ca pacity of 24.8 .J/mol . K plu s or minus 0.6 .J/mol . K. Th e r e fo re , assu min g th e m o la r h e at ca paci ty of Ag is 24.HJ /m o l . K, it." spe cific h ea l capaci ty is 0.230 J i g ' K. T his is very close to the experimental valu e of 0.2 36 Ji g ' K.
5.97 120 g of CI14 required (ass um ing
II ~P ( g )
Appendix 0
I
Answers t o Selecte d Study Questio ns
(a) microwave s (h) rc rl lig ht
(c ) infrared 6.3
(a) (;l'eCI1 light has a h igh e r fre quen cy th a n amber light (b) 5.04 X 10 " s-'
6. 5
Fre q uency = 6.0 X 10 1-1 S-I; enerh~' per p hoton = 4.0 X lO - l ~l .I; energy pe r mol of ph o ton s = 2.4 X 10"'.1
as pro d u ct)
5.99 1.6 X 10 " kJ released 10 the surroundings. T his is eq uivalent to :3.8 X 101 ton s of dynam ite .
A- 7 4
6.1
6.7 6.9
Frequen cy = 7.5676 X 10 11 ",- J; en erh~' per p h ot on 5.0 144 X 10 - 19 J; 302 ~J /mo l of p hoton s In order o f in creasin g energy: FM sta tio n < ye llo w light < x-rays
m~ = 0 is not po ssible. of :!:I /2 ,
6.37 (a)
< micro-
6.13 (a) T he light of sho r tes t wave le n gt h ha s a wavelengt h of 25:-t652 run . (b) Frequency = 1.18190 X 1 0 1.~ S- I. Energy per p hoto n = 7.83 1:~ 9 X 10- 19 JI p ho ton . (c) T he lines at 404 (vio le t ) and 436 nm (blue) are in til l' visibl e re g ion of the sp ec tr u m. 6.15 The co lo r is violet.
TJinilial
= Ii an d
IItin ,,1
n =5 to u = 1 (c) Longe st wavelength (lowes t e n e rgy) , n =S tou = 4
(h)
to 11 = 2 to n = 1; T he en ergy levels are p ro gressively clo ser a t h igh e r level s, so th e e n ergy differcncc from n = 4 to n = 1 is g reat er than from Jl = 5 to 11 = 2. Jl
= :)
=4
6.21 Wavel engt h = 102.6 nm an d frequ ency = 2,923 X 10"; S-I. Light wit h these propertie s is in the ult raviolet re gi on , 6.23 Wavelengt h = 0.29 n m 6.25 The wavele ng th is 2.2 X 10 - 2.~ nm . (Calcu la ted fro m A = h i m ' v, where 111 is th e hall's mass in kg and lJ is the vclocity.) To have a wavelength of 5. 6 X 10 - 3 nm , th e ba ll wou ld haw to tr avel at 1.2 X 10- 2 1 m / s.
6.27 (a)
/I
= 4. e = 0,
I , 2. 3
(h) When ( ~ 2, 111( ~ - 2, - 1, 0,1 ,2 (c) Fo r a 4J orbital, 11 = 4, ( = 0 , an d (d) For a 4f orbital, u = 4, ( = :~ , and - 1, 0, 1, 2, 3
6.29 Set I : n = 4, Sct 2:
1/ =
4,
Set :): n = 4,
t =
I, a nd
!II(
( ~
1, and
111/
t
1, and
~
Itl/
=~
~
UI/ ca n n o t equal - 3 ill th is case. If (: = I , only be - I , 0, or I .
One po ssib le se t o f quantu m num bers: f 1, rII t = - 1, tn,= - 1/2
m (
= 0
!II(
= - :~ ,
- 2,
1
4,
11//
II ::::
ca n
3,
(c) (' 3 is n ot po ssib le in thi s case, T he m aximum value of (:' is n - 1. On e possible se t of quantum numbe rs: (' = 2, lilt = - 1, fII, = + 1/ 2
6.41 (a) For 2/J: 11 = 2, ( = 1, and (b ) For :)d: u = 3, f = 2, and + 1, or + 2 (c) For 4f: 11 = 4, { = 3. an d + 1, + 2, or +3
II
= :),
+1
6.31 Four su bshc lls. (T he number o f su bs hcl ls in a shell is a lways equal to n.) 11 -
1.
6.35 (a) None . T he quantum number se t is no t po ssible . When (' = 0 , nit ca n on ly equal O. (b ) 3 orbitals (c) 11 orbitals (d) I orbi tal
1//(
=
fJI /
=
tn f
- 1, 0, or + 1 - 2, - 1, 0,
= - :~ ,
- 2. - 1, 0 ,
6.43 4d
6.45 (a) 2.\ h as 0 no d al su rfaces that pa ss th roug h the nucleus (( ~ 0) . (h) SrI has 2 nodal sur face s th at pa ss th ro u gh th e nucle us (f ~ 2) , (c) 5/ has th re e nod al su r faces that pa ss throug h t he nucleus ( f = 3) .
6.47 (a) Correct (h) Inco r re ct. T he inte n sity of a lig ht beam is indcpendent of freque n cy and is related 10 th e number of p ho tons of lig h t wit h a certain energy. (c) Correc t 6.49 Considering on ly an gu lar n odes (n o d a l su r faces th at pa ss th roug h the nucleu s) : s orbital
o no d al su r fa ces
IJ orbitals
1 nodal su r face or plane passin g throug h th e n u cleu s
d orbitals
2 no d a l su r faces or planes pa ssing throug h th e nucleus
f
3 nodal su rfaces o r planes pa ssin g throug h the nucleu s
0
6.33 (a) f must h ave a value no g rea te r th an (b) When (' = 0 , m( ca n on ly equal O. (c) When (' = 0 , ",( can o n ly eq ual O.
=
6.39 2d a n d 3f orbital s cannot exist. T he Jl = 2 sh e ll co nsists o nl y of san d IJ subslu-l!s. T he 11 = 3 shell consist" only of s, p, and d su bs hc lls.
6. 17 (a) 10 line s po ssible (h) H igh e st frequen cy (h ig h est e n ergy) ,
6. 19 (a) n
(h)
2
=
m ay only h ave values
One po ssib le se t o f q ua n tum n umbers: n C= 2, /lit = 0 , m~ = + I I ~
waves
6.1 1 Ligh t with a wavele n g th as lo n g as GOO nm would be su ffic ie n t T h is is in the visible regi on.
nl J
orbitals
6.51 f value
Orbital Type
f
3
o P d
2
Appendi x 0
I
Answers t o Selecte d St udy Questions
A-75
6.53 Co ns ide ri ng o nly a ng u la r node s ( no da l su rfac es that pass throug h th e nu cleu s) : Orbital Type
Numberof Orbitals in a Given Subshell
Numberof Nodal Surfaces
o
s P
3
d
5
2
f
7
3
6.55 (a ) Green ligh t (b) Red light has a wave len g th of 68 0 11m , a nd g reen ligh t has a wave le ng th of .~ () () n m . (c ) G reen ligh t has a h igh e r frequen cy than red ligh t. 6.57 (a) Wavel en g th = 0.35 m (h) En e rgy ~ 0.34 jlm n l (c ) Blu e lig ht (with A = 42 0 n m } has a n e nergy of 2HO kJ/mol of ph o to ns. (d ) Blu e lig ht has an e nergy (pe r mol of p ho ton s) that is B40,OOO tim es greate r th an a mol e of ph oton s from a cell ph one. 6, 59 T he ion izat ion ene rgy for He + is 5248 ~J / m o l. T h is is fou r ti m es t ill' io n iza tio n ell l'l"gy for th e I I a to m. 6.61 1.\ < 2 s = 2 j) < :'h = :ljl = :ld < 4 s In t he H ato m orbital s in t he sa m e sh e ll (e.g., 2s an d 2jJ) have the sam e e n ef!.,. ,)·.
6.73 (a) A = 0.0005 em = 5 I' m (b ) T he left sid e is the high e r e ne rgy sid e , an d the righ t side is tlu- 10 \,,.'(' 1' energy side. (c) T h e in terac tion \v·ith O - H requi re s more en ergy.
6,75 (e) 6.77 An ex perime n t ca ll be d one th at shows th at th e cl ectron ca n be have as a pa rt icle , and an ot her e xperim e nt ca n be do ne to show th at it has wave propert ies. (H oweve r, no sing le ex pe rime n t shows hoth properties of the c lec tro n.) Th e m od ern view of ato mi c structu re is based o n the wave properties of the elec tro n. 6,79 (a) a nd (h) 6.81 Radi at ion wit h a wavele ng th of9:1.B nm is suffi cie n t to raise the electron to the rl = 6 quantum level (see Figure fl. 10 ) . T h ere sh o uld he 15 e m issio n line s in vo lving tran sition s fro m II = 6 to lowe r e n e rgy levels. (T here are five lines fo r tran sition s from 1/ = 6 to lower level s, four lin es for 11 = 5 to lower leve ls, three fo r u = 4 to lowe r levels, two lin es fo r 1/ = 3 to lowe r levels, an d o ne line for n = 2 to n = 1.) Wave lengt hs fo r m a n y of th e lim' s are g in 'll ill Fig u re 6.10. Fo r ex am p le, th ere will he an e m issio n invo lving a n e lectro n movin g fro m n = 6 to n = 2 with a wavele ng th of 410.2 11m . 6.8 3 (a ) G ro u p 71\ (lU PAC Gro u p 7) ; Pe riod 5 (1)) n = 5, ( = 0, nil = 0, In, = + 1/ 2 (c ) A = K 79 X 1O- 1'.! m ; v = :1.4 1 X 10 19 J- 1
= 2.B36 X IO'.!(I S-I a nd wavel en gth 1.057 X 10- 12 m
6.63 Fre q ue n cy
6.65 260
S
(d) (i) H TeO , (aq )
NaO H(aq) -> II , O (C) + NaTc O ,(aq ) (ii) H.5 X lO - :l g Na Tc04 p roduced; 1.8 X 10- 3 g Na O n ne eded (e) 0.28 mg NaTeO,; 0.000 15 M
or 4.3 min
6.67 (a) size a nd el1eq.,.J}·
e
(h ) (c) m ore
(d ) 7 (wlu -u f = 3 th ese a re I o rbit a ls) (c ) on e o rb ita l (f) (le ft to righ t) d, .\', a nd II (g) f ~ 0, I, 2, 3, 4 (h ) 16 o rbit als (1,\', 3jJ, 5ri , an d 7J) ( = 1/2) (i ) parama g ne tic 6.69 (a) Dr awing (a) is a ferrom agn et ic so lid, (b) is a di amagne tic so lid , an d (c) is a para magn e tic so lid . (b) Substan ce (a) wo u ld be most stro ngly a ttracted to a m ag n et , wh e re as (h) wo uld be lea st strong ly a ttra cted.
6.85 Six emi ssio n lin es a rc observed. More th an one lin e is o bserv e d beca use th e fo llo wing ch anges in ene q.,'. Y levels a rc possible: fro m n = 4 to n = :l, n = 2, a nd n = 1 (th ree lin es), fro m n = 3 to 1/ = 2 a nd /I = 1 (2 lines) , and from 11 = 2 to n = 1 (o n e line ).
CHAPTER 7 7.1
(a ) Ph ospho ru s: 1.~ 2 .~2 IP 3 .f2 3j/'
[ill [ill 1111111 til [ill h
6.71 T he pick le g lows bec ause it was mad e by soak ing a cucum ber ill brine, a co nc en trat ed so lut ion of NaCI. T he so d iu m a to ms in th e p ickle a rc exc ited hy t he electric cut-n-u t a nd release ene rgy as yellow light as th ey re tu rn to th e g ro u nd sta te. Excited so d iu m ato ms are the so urce o f th e ye llow lig ht yo u see ill fir ewo rks an d ill ce rtain kin ds of str e et lig htin g.
+
2.>
21'
'J.>
ITIIIIJ 31'
T he c le men t is in th e third period in G ro u p SA. T h e refore , it ha s five elec tro ns in th e th ird shell. (h ) C h lo rine; 1.i'2.i'2IP3.i'3IF
[ill [ill I1II1IIHI [ill !illillIJ 1S 2J
2jJ
3J
3/J
Th e e leme nt is in th e third period and in G ro u p 71\. T he-refore , it has seve n electro ns in th e th ird sh ell. A -76
Appendix 0
I
Answers to Selected St udy Questions
7.3
(a) C hromi um : I s2 2 s2 2 pt; 3 s2 :{ jh-\ (J5 4 .~ 1 (h) Iron: I s2 2s22 pli:Js23 p63 d li4s2
7.5
(a) Ars en ic: I s22 s22pli 3 .~23p6 3d I 0 4s 24 j)j ; [Ar] :~ d l u 4 .~2 4 jJ3 (II) Krypton : I s22 s22 pti3 s23 pti3 d ltJ4s24 j)'i; I A r ] :~ dItJ4s2 4j/i = [Kr]
7.7
7.9
(d)
02 ~
io n
[jj] [jj]
rn:mrm
1.\ 2.\
2/J
7. 19 (a) V (paramagne tic ; th r('l' unpai red el ect ron s}
(a) Tantalum : This is l i lt' third e leme n t in the transitio n se ries in the six th peri od. Therefo re , it has a core e quivalent to Xc plus two 6 s e lec tron s. 14 4/ electrons , and th re e electrons in 5d: [X c H/"5d'6i' ( h) Plat inum : Th is is the eighth c le me nt in the transitio n se ries in the six th pe ri od. Th erefo re , it is pn'dieted to han' a core e quivalen t to Xc plus two 6s el ec tron s. 14 4/ elec tro ns. an d eig ht electrons in 5 d : [X (·Hf 45d I'6s2 • In reality, its ac tua l co n fig ura tio n (Ta ble 7.3) is [Xe]4f4S(f'651 .
[Arll till 1 1
I [jj]
:\ d
4s
(b) V'l+ io n (pa ra m ag n e tic , th n'e unpa ired el ectrons )
[At]
(c)
7.21 (a)
ion . This ion has a ll electron configura tion eq uivale nt to argon . [Ar] . It is d ia magnetic with no unpa ire d elec trons .
\ T;')+
~ l a n g-a n e S("
[Ar I IIIII (b) Mn l +
[Ar]
[jj]
(c) T he 4 + io n is param agm- tic to the ex tent of th r ee u npaired elec tron s. (d) 3
3.\
Q ua n tu m numbers fo r the two elect rons in th e :h orbital : 1/
= :\. ( =
0,
tnt
=
O. and
111,
n
= :\. (
0,
11/,
~
0, a nd
11/,
~l g2 +
+ 1. a nd
III .
=
+\'2 01' - \'2
ion
[jj] [jj]
rn:mrm
Is 25
2p
(h) K+ io n
[jj] [jj] I s 25
rn:mrm [jj] rn:mrm 2p
3s
3j)
(c ) Cl " io n (No u- th a t both CI - a nd K+ h ave th e sa me co n fig uratio n ; hoth arc equivalent to Ar.)
[jj] [jj] Is 2s
rn:mrm [jj] rn:mrm 2 jl
3s
->
Il-F
II-I
+ .5 - [,
+ .5
- .5
BF is more po la r
All han ' two a to ms attac h ed to the cen tra l a tom . As the bond and 10Ill' p airs vary. till' elec tro n-pai r g co nu-. tries rd ry from linear to tetrahedral. a nd th e molecular g eol1ll' tr ie s \'ary from linear to bent.
8.21 (a) Elec tron-pai r g eo m etry a ro u n d CI is trigo nal bipyra midal . Mo le cu lar g eom e try is linea r.
8 .29 (a ) C H and CO bonds art' polar. (b) T he CO bo nd is m o st polar, a nd 0 is th e most negative a to m .
8.3 1 (a ) O H ": Th e formal charge on 0 is - I an d on H it is O. BH ~ " : Even th o ugh the formal ch arge o n B is - 1 an d 0 11 (J is O. " is slig ht ly m ore el ec tron eg a tive th an R. T he fou r I I a to ms an' th erefo re m o re likel y to hea r th e - I cha rge o f the ion . T he BH bonds are pol ar with th e I I a to m 'h e nega tive e ncl. (c) T h e C II an d C O bo nds are all po la r (h ut l i lt' C-C bond is not ) . T he llegat iw ch a rg e in the CO bon ds lit'S on th e 0 a to m s.
(h )
ji---':CI::-ji :] [ : .. .. .. (h) Electro n-pair geome tr y arou nd CI is tri g on al hipyramidal . Mo lecul a r gcom cu-y is T-sha pl'd ,
(c) Electron-pair geollll' lry around CI is o c ta hedral. Mol ec u lar geometry is sq ua re p lan a r.
sm a ll as possible a nd th e n eg a tive c ha rge res ides on th e m o rt' clcc tro ncg mivc a to m .
: F : .. ] .. ..I : F- C1-,- F : [
8.3 3 St r ucture C is most reaso nable. T he ch a rg es are as
- 2
.. I · .: F:
+I
+1
- I
.,
+1
o
u
"
+1
- I
: N - :-I==O:
N=N =O
: ~ ==~- O:
A
1\
C
(d) Electron-pair g eo m e tr y a round CI is o c ta hed ra l. Mol ec u lar geo metry is a sq uare p yramid .
8.35 (a)
.. : F : .. : ...... F C1 I ~ ..F : : ji.... -- ..... ji : 8.23 ( a) Ideal 0-5- 0 angle
~
120
(h) If a n 11+ ion were 10 a ttack :\'02 - , it would a ttach to an 0 a to m bec ause the () a to m s he ar the neg atin' charge in thi s ion.
0
(c)
( h) 12()0 (c ) 120 0 ( CII,NII,' + CI -
CI II
10.65
I
1,3-D id llt>ro propalW
I
.
CI
I
H - C- C- C-H
I
H
I
H
I
H
II CI II 2,:l-B id II orop ropa l i t'
10 1
()
" 1I0CH ,CII, OH
+
n 1I0 - C
j
C-OII ----->
10.71 CH ,
+ n H,o
) - tOC H'CH,o t
I
I
I
I
C Hj
~CII,
+ O- C- (
I
II CI II
()
"
I
H- C- C- C-H
V
¢r''''
CH ,
CII,
10.67 (a ) 2, :1-Dime thylpe ntanc
1,2A-lrilll t'thyI1 1{Olll t'll t'
1 . 2. 3·l rinWl h ~' l ht' 1 1Z t' l l t'
CH ,
10.73 Rep lace the carboxylic acid gro up with an " ato m.
I
I I,. C- CI CI I, ClI, CII .,
10.75
II
\
CII,
/
H
II
p=\
11 , C
(b) :l, :1-Dime thylpc ntanc
..:':..'.'.1...
CII,
I I
H
11-6I- -6I-11 CH , CII,
hruaue (!10 1 chiral )
CH ,C H , ~; II,
CI I,C1I, - C-ClI, CI I ,
.
I ,3,5-lri lilt' [II y l b t 'Il It' l lt'
.
H-
CII,C II ,
- CH 3 r CH "
(c) 3-Ethyl-2-mc thylpc ntanc H
CH ,CII,
I
I
.
CI I, -C-C-CI I.,0 13
I
I
.
CH 3 H
(0) :l-Ethylhc xane CII.,CII,
I .
CII , CH'-r - CH,CH,CH ,
10.77
Ir
11, C-0 - C- R 0
11, C- OII
I HC-O-~-R ~ H(\ -OH + 3 RI Ir H , L oH
0 i \- o ' Na +
H , C-O - C-R g:l)l :t'ryl trilaurare
sodium laura te
H
Appendi x 0
I
Answers to Selected Study Questions
A- 89
10.79
H ad d
I
Il- C- C-Il
II ~
I
I
II
II
II
\
oxi d izt'
\C = CrCI I.,O : Il /
\
C -C
/
CO,H
po lyJnn jn '
i
II H
C.H,o H H
I
I
I
II
Oil
I
H
I
+ 11, 0
o
H
I I
j
/I
H "C-C-O -CH,CH=CH , H
I I
Il, C- C-
I II
Br H
I I
+ II B,
CH , Il
3,3-d imd hyl-l-pelltelle
I
HII
II
10.81 (a)
x=
I
- C-C
I
II
CH,o H }
I
C-C-
I
II
I I 1[ , C ~ C - C - C - Cll , I I I .
II
II
en,
10.91
r>.H
H
II
10.89 2· Pro pa n o l will rea ct with a n ox id izing ag en t suc h as KMno.l (to g ive th e ketone ) , wh ereas methyl ethyl et her (C H 3 0 C;!lI s) will no t rea ct. In ad d itio n , tlu - a lco ho l sh ou ld be more so lu ble in water th an th e ethe-r.
CH,o H
I
Y=
I I
en,
Il
C-
C- Cll,
I
I
I
o
o xid il illg"
I
age ll t
II
1l,C- C-
Cll.• Il
I
C-
I
C-Cl [,
I I err, II
CH " II
3" 3-di me th yl -~-p l"n(anon('
3 .3-d imelh yl -~-pe 1ltan(J1
Il - C- C= C- I l - > H - C- C- C- I l
I
I[
II
II
10.93
H II
II
I I
ll - C- l l
2-bro1l1o pn 'p ,Ult>
nu-rhauc
fo ur single bonds
formaldehyd e
Olll' d o u ble b on d and (W O sing le bon ds
alle-m-
two d ouble b o n d s
;l(t' tykllt'
01 1('
H (h )
Il CH ,. Il
I
[I
I . I
I
I
I
H
II
I ' I
+ 1J10 -- CH H jC'- ('~ - C-,. > H ,C - C- C-
H
o
CIl , Il
H
C-H
I
I
OH
H
H
\
2-1l1("th yl-2-h utan o l
/
H
(c)
H
CH , H
I
I ' I CC-H I
H "C- C=
.
H +11,0
~
H, C-
H
H
-
OH
I C-H I H
10.83 Co m pou nd (h ) , acetal de hyde , and (c), etha nol , du cc ace tic acid whe n o xidi zed.
p ro~
10.85 Cvcloh excue , a cyclic alken e, will add fi r2 read ily (to g ive Ct,HI 2Brt ) . Benzen e , however, ne ed s much mo re stringe n t co nd itio ns to reac t with bromin e ; then fir 2 will su bstitu te fo r H a to ms on be n zen e an d not add to th e ri ng.
10.87 (a ) Th e com po un d is ei the r propano nc , a keton e. or propan al , a n aldeh yde . H
H-
0
H
I II I C- C- C-H I I Il H p r o p al10 1l{, (a ke-tnue ]
H
H-
H
H
propa na l (all aldt'hyck )
(h) Th e keto ne will no t undergo o xid ation, hu t th e aldehyde will be oxi d ized 10 th e acid , CIJ 3 Cll t CO tll. Thus, th e un known is likely p ropan al. (c) Propan oi c acid A -90
Appendix 0
I
II
/
C= C= C
\
II single bon d and omtr ip le bond
10.95 (a ) C ross-lin king ma ke s the mate rial very rig id a nd in flex ible . (h ) T ile OH g ro u ps g ive th e pol ym e r a high affin ity fo r wa te r, (c) Hyd ro gen hond in g a llows th e chains to form co ils an d shee ts with high te nsile stre ng t h.
= - 47.5 1 kJ /g Eth anol h eat of combustion = - 2G.8 2 ~J / g (h) Til l' hea t o bt a ined from the co m b us tio n o f e thanol is less neg ati ve th an fo r e thane , so p arti ally o xidi zin g e than e to form e tha no l d ecrcasc s th e a m o u n t o f en erh~' pe r mol l' avail abl e from th e co m b us tio n o f th e su bs ta nce.
10.97 (a ) Eth an e heat o f co mh ustio n
0
I I II C- C-C-H I I Il
H
ll- C= C- l l
CH , H
I I C- CI I
/ C"
Answers t o Selecte d St udy Questio ns
10.99 (a) Empiri cal formu la , C H O (h ) Molecul a r fo rmula , C1 1110 (c) 0 0
II
1
II
[[O - C- C=C- C-OII
H H
( 8 CO, (g ) + 10 H,O (g)
+
2 SO, (g )
Total p ressu re = 3 7.3 mm Hg . Partial p re ssu re s: C O z = 14.9 mm I-I g , H 20 = 18 .6 mm I-I g , an d SOz = 3.73 m m H g .
11. 27 57 .5 g/mol
11. 29 Molar ma ss = 74.9 g /mol ; R1i1l 1l) 11. 31 0.0 :19 mol H 2 ; O.O!Ui a nn ; 7:1 mm Hg
11.59 4 mol
11. 33 170 g NaN ,
11. 61 N i is th e limi ting re actant; 1.3 1 g
11. 35 1.7 atm 0 ,
11. 63 (a . h ) Sample 4 ( li e ) h as the larges t number of molecules and sa m p le 3 ( Hz at 27 "C and 760 mm Il g ) h as th e fewest number of m o le cul es. (c) Sa m p le 2 (A r )
11. 37 4. 1 a n n 11 2 ; 1.6 ann Ar ; to ta l pre ssure
= 5.7 atm
11. 39 (a) 0 .30 mol h a lo th an e / I mol O 2 (h ) 3.0 X lO2 g h a lo th a n e 11. 41
(a) CO 2 h as the h igh e r kin e tic energy. (h ) The average sp eed of the II ';! molecules is g rea ter than th e average sp e ed o f t he CO';! molec u le s. (c) Th e number o f CO:! m o lecul e s is g rea te r th an the numbe r of 112 mole cules [ 1I (C0 2 ) = 1.8 ,,(11, ) ] . (d) T h e mass of CO 2 is g re ater th a n th e mass of H 2 -
11.43 Average sp eed
or CO';! molecule
= 3 .65 X IO ~ cm / s
~ i( C O) ..
11. 65 8 .54 g Fe (CO )., 11. 67 S2FlO 11.69 (a ) 28 .7 g /mol = 29 g /mol (b) X of o, = 0.17 and X of N z 11.71
= 0 .8:)
Mol ar mass = 86 .4 g / m o l. T h e gas is p ro bab ly C IO,F.
11.7 3 I/(He )
~
0.012H mol
11.75 "Ve ight percellt KCl0 3
Appendix 0
I
=
69 .1%
Answers to Selected Study Questions
A-91
11. 77 (a ) 1\0, < 0, < NO (h ) 1'(0 , ) ~ 75 mm Hg (c) 1'(1'0 , ) = 150 m m Ilg
11.79 P (N l-lj )
= 69
mm lI g and P(F 2 )
= 51 111m
11. 10 1 T he sp ee d of gas m ol e cul es is rel a ted to th e sq uare roo t o f the absolu te lem pe ra ture , so a doub lin g of th e te m p e ra ture will lead to an increase o f abo u t (2) 1/ 2 or 1.4.
JIg
P ressu re afte r reacti o n = 17 m m Il g
11.81 At 20 ' C, there is 7.8 X 10 - 3 g H ,O / L. At 0 ' C. t here is 4.6 X 10- 3 g H 2 0 / L
CHAPTER 12
11.83 T h e m ixture con tain s 0 .22 g CO 2 and 0.77 g CO.
12.1
I'(CO , ) = 0.22 atm ; 1'(0, ) 1.2 2 at m
= 0.12
atm ; I' (CO )
=
11.85 T h e formul a of the iro n com po und is Fe(CO );;.
11.87 (a) 1'(Il,H,) ~ 0.0160 a nn (h ) j J(H 2) = 0.0 :l20 a tm , so
12,3 I~,, (a l
O,04HO atm
11.89 Amount of Na 2 C0 3 = 0.00 424 mol Amount o f N a I IC0 3 = 0.0095 1 mol Amou n t of CO 2 p ro d uced = 0.013 8 mol Volu m e o f CO:! pro du ce d = 0. 34 3 L 11.91
De co mpo sitio n of I m o l of C u (N 0 3 h should gi ve 2 mo l }'\02 a n d ~ m o l of 0 2' To ta l act ual amount = 4.72 X 10 - :1 m ol o rgas . (a ) Average molar mass = 41.3 g / m o l. (b) Mole fracti o ns: X(N0 2 ) = 0 .666 a nd X(O , ) ~ 0.3 34 (c ) Amoun t of eac h gas: 3 .13 X 10 - 3 mol NO :! a n d 1.57 X 10 - 3 mo l O 2 (d ) I f so m e X0 2 m o le cules combine to form N 20~ , the a p p a re n t mole fract io n of N0 2 wo uld he sm alle r th an expec ted ( = 0.8 ) . As this is th e case , it is app are n t tha i so m e N 2 0 1 h as been forme d (as is o bse rve d in th e experiment) .
11.9 3 (a) 10.0 g of 0 2 repre senL I'(CO ) (b) Th e re arc m o re mol e cules in th e C 2 H 2 containe r th an in th e CO cont ainer,
11.97 (a ) No t a g as. A g as would ex pa n d to a n in fini te volum e . (h ) No t a gas. A den sity of 8.2 g /mL is typ ical of a so lid . (c) Ins uffi cient info r m ation (ct) Ga s 11.9 9 (a) Th e re arc mo re mol e cules of H 2 th a n a toms of He. (h ) T he m ass of H e is g rea te r th an th e mass of H 2 •
A -92
Appendix 0
I
An swers to Selected Study Questions
(a ) Dipol e-d ip o le in teraction s (a n n hyd ro ge n bonds) (h) Ind u ced di pole-induced dipole for ces (c) D ipol c- clip ol e inte raction s (a n d h yd ro ge n bonds) (a) In d u ce d di pole-i nduced d ip ol e forces (b) In du ce d di pole-ind uced dipo le force s
(c ) D ipol e-d ip o le force s (d ) D ipol e-d ip o le force s (a n d hydrog e n honding)
12.5
T h e p red icted order of in creasin g stre ng th is Ne < C H ~ < CO < Cel. l . In this case , pred ict io n d o e s n o t qu ite ag re e with reali ty. Th e bo iling p o in ts arc Nt' (- 246 "C) < CO ( - 19 2 "C) < en, ( - 162 ' C) < CC l, (77 ' c) .
12.7
(c) HF ; (ct) ace tic acid ; (I) CH 30 H
12.9
(a ) LiCI. The u : ion is sm alle r than Cs + (Figu re 7. 12) , which make s th e io n- io n for ce s of attraction stro nger in LiCl. (h) Mg (NO :;h . T he ;vlgH ion is smalle r th a n th e Na ' ion (Fig u re 7.12 ), and the mag ne sium ion h a s a 2 + ch arge (a s oppo se d to 1+ fo r so d iu m ) . Both of th ese effects lead to stro ng e r ion-io n forces of attraction in m agn e sium ni tra te . (c) NiCI2 • T h e nickclfl l ) ion h as a large r charge than Rh " an d is co ns id e ra b ly sm alle r. Roth cfIccts mea n that there arc stronger ion-ion forc e s of att ractio n in nic kel ( lI ) chlo ride .
12.11 '1 = + 90. 1 kJ
12,1 3 (a) Wa ter vap o r pressure is about 150 mm JIg at 60 °C. (A p pe n d ix G give s a value of 149 .4 m m H g a t 60 "C .) (b) 6 00 mm H g at abo u t 9 3 °C (c ) AI 70 DC , eth anol h a s a vapo r pressu re o f ab o u t 520 mm H g , whereas th at of wa te r is a bo ut 225 m m Hg. 12,1 5 At 30 °C. the va por pl"l'SSUH' o f e th e r is a bo u t 590 m m H g . (T h is pressure req u ires 0.23 g of e th e r in the va po r ph ase a t th e g iven condition s, so the re is sufficient ether in th e flask .) A t 0 °C. th e vapor pressu re is about 160 m m Ilg , so so m e ether co n d e ns es whe n the te m p e ra tu re declines.
12. 17 (a) 0 , ( - 18 3 ' C) (b p of N, ~ - 1% ' C) (b) SO, (- 10 ' C) (CO , su blim es at - 78 ' C ) (c) Il F ( + 19.7 ' e ) (Il l, - 35.6 "C) (tI ) e-n , (- 9 0.0 ' C ) rsm., - 111.8"C)
12.19 (a ) (I»
(c) (d) (e)
12.21
CS~ , abou t 62 0 mm Hg; C H sNO~, abou t 80 m m Hg CS ~ , induced dipol e-j uduced d ipole fo rces; C H :~NO ~ , di po le-d ipole fo rces CS';,! , abou t 46 °C; CII sK O ';,! , abou t 100 °C Abou t :l9 °C About 34 -c
Using th e eq uatio n for th e stra ig h t lin e in th e plot
In
(a ) 80. 1 · C (b) At ab ou t 4S °C , the liqu id ha s a vapor p ress u re o f 250 mm II g.
T he vapor pressu re is 650 nun II g a t 75
-c .
(c) 33 .5 kJ /nw l (fro m slo pe of plo t) 12.23 No , CO ca n no t be liquefi ed a t ro o m temperatu re because th e criti cal tem pera tu re is low er th an room tem pera ture .
12.25 Ar < CO , < C H,.
16.45 (a ) Fracti on dis so ciated = 0.15 (b) Fract ion di ssoc iated = 0 . 189 . If th e pressure d ecrea se s, t he e q u ilib riu m sh ifts to th e rig ht, in crea sing the fra c tion o f N t O _1 d issoci ated.
= 0.67
16.4 7 [N il , ] [~ ". al
=
M; [N,] 1HO a tm
16.49 (a ) [N I1,] (h) [N il, ]
~ ~
= 0.57
M; [11 2 ]
[ II ,S] ~ 0.0 13 M 0.027 M and [ II,S]
~
=
1.7 M;
0.0067 M
K.., = K, = 56. Because 2 mol o f reactan ts give s 2 mo l of product , A n d o e s not ch an ge and K.,. ~ K (see p age 730) .
(b , c) In itial P(I -It ) 5 .2 auu
= P(l t ) = 2.6 a nn
and
~ula l
=
At equi libriu m , [ )(H t ) = [ )(11 ) = (>..~4 a tm a n d P(III ) = 4. 1 a tm. T herefo re, [~ u. al = 5.2 a tm. Th e in itial tOLaI p re ssu re an d th e e q u ilib riu m to ta l p re s~ sure a re the sa m e owi ng to the re a ct io n sto ichi omctr y. Pe rce nt d isso ciatio n = 69 %
16.55 P(CO ) 16.5 7 1.7 X
~
(1.0010 a im
10 11'10
16.6 5 (a ) As more K~CN is added , Lc Chan-Her's p rinciple pre dicts more o f th e red co m p lex ion [Fe ( II,o ), (SCN ) + w ill fo rm . (h ) Addi ng Ag + io n s lead s to a p re ci p itate of AgSCN , thus rem ovin g SCN - io n s from so lu tio n . The e q u ilib riu m sh ifts left. dropping th e con cc ruratio n o f th e red co m plex io n . 16.67 (a) False . Th e magn it ude of Kis always d e p en d ent o n te m pe ratu re , (h) True (c ) False. T ile e qu ilib r-i um con stan t fo r a re action is the re ci p roca l of th e va lue o f K for it" reve rse . (d ) TrlTe (e ) Fa lse . D. n ~ 1 so
16.61 (a) K." = 0.20 (b ) Wh en initial [N t 0 4 ] = 1.00 atm , th e eq uil ibrium pressures are [N tO j ] = 0.8 0 a tm an d [l'~'() 2] = 0.40 a tm . Wh e n in it ial [N';,104] = 0 ,10 an n, til e e q uili briu m pressures a re [N to. j ] = 0 .050 atm a n d [NOt ] = 0 .10 a tm . T he p e rce nt d isso ciatio n is n ow :)0.%. T h is is in accord with Lc C ha tclier's p rin ciple: If th e in iti al pn'ssul"l' o f th e re actan t
Appendix 0
I
K."
~
Answers to Selected Study Questio ns
K (RT) I 1 1
16.7 1 T h e syste m is n o t at eq u ilib rium becau se it con t inues to gain e Iler6'}' from th e surroundin gs . Tile temperatu re o f th e water/ ice m ixt u re will remain at 0 °C un til all th e ice is mel ted , then th e u-m pcra ture will rise a s more e ne rg y is ga in ed . Only if the be aker of wa te r / ice were m ove d to a p e rfe ctl y in su la te d co rnpa rt m en t, also a t 0 "'C , wo u ld it a ttain eq ui libriu m a t o CC . III this case , it wo u ld h e a dy n amic equ ilib ri um wit h wa ter m o lecules m ovin g from th e so lid to th e liqu id ph a se and fro m th e liqu id to th e so lid ph ase . T h e qua nt ity o f ice wo u ld n ot ch a nge , If a 1)';,1 0 ice cub e was added to so m e H 2 0 (f) , a n equi lib rium would b e obtained. Til e a m o u n t of 1)20 in th e liq u id ph ase wo u ld increase d ue to the con tin u ing molcc uI'l l' e xchange . Th e water cou ld th en be sa mpled for th e prese n ce o f D 20 .
ato m s
16.5 9 Glvcerin co ncentration sho u ld b e 1.7 M
A -100
= 69%
16.69 (a ) P roduct-favored , K » (b ) Rea ct ant-favo red . K « (c) P ro du ct-favored , K »
16.51 P(N (\ ! ) = 0 .379 a un and P( NtO ,j) = 0 .960 au n : P(tot al) = 1.339 atm 16.53 (a )
16.63 (a ) T ill' flas k co ntai nin g (1I 3N ) R(CH 3h will h aw th e larg-est p ar tial p re ssu re o f R(CH 3 ) 3' (h) 1' [I\ (CII 3 ),] ~ 1'(1\113 ) ~ 2. 1 and 1' [ ( H , N) 11 (CH ,) ;.] ~ !. O aim
Pe rce n t d isso ciatio n
(a) T he equilibrium will sh ift to th e left on a d d ing more Cit . (b ) K is calcul ated (from the q u anti tie s o f re ac ta nts a n d produ cts a t equi librium } to be 0 .0470. Afte r Cit is add e d , til e co nc e n tra tio ns a re ; [PC I:;] 0.0 199 M, [PCl 3 ] ~ 0.0231 M, and [CI, ] ~ 0.0403 M.
16.4 3 K."
de creases, th e e q u ilib riu m sh ifts to th e rig ht, iu creasin g t he frac tio n o f th e reactan t d isso ciated . Sec also Q ues tion 16 .45 .
CHAPTER 17 17.1
(a ) C N-, cyan id e ion (h) 50/ - , su lfa te ion (c) F- , fluori d e ion
17. 3
17.5
(a ) H 3 0 +(aq ) + N0 3 - (aq ) ; H :\O +(aq ) is the co nj uga te ac id o f H 20 , a nd NO:1- (aq ) is the co nj uga te base of 1-1 1'\° 3 , (h) H 3 0 +(aq ) + SO/- (aq ) ; H 3 0 +(aq ) is th e conjugate acid o f H 2 0 , a nd SO / - (aq ) is the co nj ug a te base of I-I SO j - . (c) H 20 + I-IF; H 2 0 is th e co nj uga te base of I13 0 +. a nd II F is the co nj uga te acid of F- . Bronstcd ac id : HC 20 j - (aq ) Bren su-d base: II C20~ - ( aq )
+ H 20 (l')
~
H 30 + (aq)
+ C, O .,' - (a q )
+ 112 0 (f)
~
H,C,O, (ag ) + Ofr (aq) 17.7
Acid (A)
Base
Conjugate Acid of B
17. 35 (a ) Left; N ll j and HBr arc th e stro nger base a nd ac id , re sp ecti vel y. (b ) Left ; PO l - a nd Cll jC0 2 fl are the str onger base a nd acid , res pe ct ivel y. (c) Right; [Fe( H 2 0 )lY + a nd 1-I C 0 3 - arc the stro ng er acid a nd ba se , re spe ctively. 17.37 (a ) OH - (ag )
II , O (C)
pn.:
Conjugate Base of A
(a) HCO, H
H, O
HC0 2 -
H]O+
(h ) H, S
NH ,
HS-
NH/
17.39 (a) CH"CO , H (a g) + HI'O,'- (ag ) q CII ,CO , - (ag) + II ,PO.,- (ag ) (b) C lI j(X) 2H is a str o nger ac id th a n H 2P o.1- , so the e q u ilib riu m will lie to th e right.
(c ) HSO, -
OW
50 4 2 -
H,O
17.41 (a) 2. 1 X 10-' M; (b) K, 17.43 K;,
~
1.8 X 10 - '
~1 ;
[11,0+]
17.11
HCI is a stro ng ac id , so [ H 3 0 +] = co nce n tr a tio n o f the acid . [ H, O +] ~ 0.0075 ~1 and [O H - ] ~ 1.3 X 10 - 12 M . pH ~ 2.12.
aci d ic
17.13 Ba (OH h is a stro ng base , so [O H - ] tration of the base . [O l l "]
pH
~
= 3.0 X 10- ' M; p Oll 11.48
=
2 X co nce n-
= 2.52 ; and
17.15 (a ) Th e stro n ge st acid is HC0 2H (la rge st K..) a nd th e weake st acid is C 1,II:;CH I (sm a lle st KJ . (h ) T he stro nges t ac id ( H C0 2H ) has th e weak e st conj uga te base . (c) T h e we a kes t a cid (CljH[)O H ) has th e st ro ng es t co nj uga te base . 17.17 (c) H CIO , th e weakest ac id in thi s list (Tab le 17.3) , has the stro n gest co nj uga te base .
17.19 CO / - lag)
+
H,O (C) -> HCO , - (a g )
+
O H - (ag )
H igh e st p H , Na2S; lowe st pH, AICl3 (wh ich g ives th e \v'ca k acid [AI( H 20) liP + in so lu tio n) ~
17.23 1'1\,
17.25 K,
~
4.HI
3.0
X 10 - 1
17.27 2-C h lo ro he nl oic ac id has the sma lle r pK" value .
= 7.09 X 10- 12 K" = 6.3 X 10 - ' CH,CO , H (ag ) + H CO , - (a g ) q
17.29 K;,
17.31 17.33
+ PO,3- (ag )
(h) O H - is a stro nge r b ase th a n PO j 3- , so th e e q uilib ri um will lie to th e righ t. (T h e predomi na n t spec ies in so lu tio n is PO l - , so th e so lu tio n is l - is th e co nj ug ate likel y to be ba sic be cause base o f a weak acid .]
(B)
17.9
17.21
+ HPO ,' - (a q ) q
CH 3CO , - (aq) + H,CO , (aq ) Eq ui libri u m lies pred om in a n tly to th e rig h t because C113C 0 2 ' J is a stro nger acid th an H 2C O j .
~
6.6
X
~
3.5 X 10-'
10-'
17.45 (a ) [11,0+] ~ 1.6 X 10-' M (b) Mod e rately we ak; K" = 1.1 X 17.47 [CII , CO , - ] = [11,0 +] [C II "CO , II ] ~ 0.20 M
=
IO - ·~
1.9 X 10- ' M a nd
17.49 [H "O +] ~ [C N- ] ~ 3.2 X 10- 1; M ; [H Cl\:] 0.025 M ; I'" = 5.50 17.51 [Nil" ] ~ [0 11- ] ~ 1.64 X 10- 3 M; [Nil ,,] 0 .15 M; pH ~ 11.22 17.53 [0 11- ]
= O.lJI 02 M; I'"
12.!J1 ; 1'0"
=
~
1.99
17. 55 p H = 3.25 17. 57 [11,0+]
=
1.1 X 10- ' M ; I'"
= 4.98
17.59 [ HCN] ~ [O H - ] ~ 3.3 X 10- ' M ; [ H,o +] 3.0 X 10- 12 M ; [N a +] ~ 0.441 M 17.61 [ H, O +]
~
~
1.5 X 10-" M ; pH
8.8 1
17.63 (a ) Th e re actio n prod uce s ace ta te io n , th e co nj ugat e base o f ace tic acid , T he so lu tio n is weak ly basic. p H is greate r th au 7. (b ) Th e reac tio n p ro d uce s Nl l , " , the co nj uga te acid of NH 3 . T he so lu tio n is weakly acid ic. p H is less th an 7. (c ) Th e reaction m ixe s eq ual m ol a r amou nts of stro ng ba se an d s tron g ac id. T he solu tio n will he neutral. pH will be 7, 17. 65 (a ) I'"
=
1.17; (b ) [50 ,'- ]
17.67 (a ) [0 11- ]
= [N,I V
]
= 6.2
= 9.2
X 10 - < M
X 10- ·; M; [N, II r ]
8.9 X lO - lh M
(h) pH
~
9.96
17.69 (a) Lewis ba se (h ) Lewis ac id
Appendi x 0
I
Answers t o Selected Study Quest ions
A-101
(c) Lewis base (ow ing to lo ne pa ir of e lectrons on t he N a tom)
17.71 CO is a Lewis base in its reaction s wit h tra nsitio n m e tal a to ms. It dona tes a lon e p air of e lec tr o ns o n th e C ato m . 17.73 I l o e N sh o uld he a stro ng er ac id th an II e N because th e II a tom in 1I0e N is a uac lu-d 10 a h igh ly elec trone gati ve 0 a to m . Th is induce s a positive ch arge o n the H a tom , m ak in g it mo re readily remove d by a n interact ion with wa ter. 17.75 T he S a to m is surro u nded by four h ighly el ect ronegati ve 0 atoms. T he indu ctive eff ect o f th e se a toms indu ce s a positi ve charge o n the" a tom , m ak in g it susce p tible to removal by water, 17.77 pll
~
2.1;7 1
17.79 Both Ba (OH h and Sr( O H h di ssol ve co m ple te ly in wa ter to p ro vide ~1 H and O H - ions. 2.50 g s.tom, in 1.00 1. of water g ives [5 r H ] = (1.021 M a nd l O H - ] = 0.04 1 M. Th e co ncentra tion o fOH is refl e cted in a pl I of 12.fil. 17.81 H,S (aq ) + C H,CO , - (aq) !::::::j C II ,CO , II (a'l)
+
IIS - (aq )
Th e eq uilibriu m lies to th e left a nd favors the reactan ts.
17.83 [X - J ~ [ H,O ' J pll ~ 2.52 17.85
1\.
17.87 pH
~ ~
~
3.0 X 10- ' M: [H XJ
~
~
~
X 10-"
17. 109 (a ) Ani line is bot h a Bron stcd a n d a Lewis base. As a proto n acce ptor it gives Cj ; H ,,;'\a--l :~ + , T he N a to m can also d on ate a n elec tro n pai r to give a Lewis acid-base ad d uc t, F:\B 0(-- N H 2Cj ; H .~. (h) pH ~ 7.97 17. 111 Wat e r ca n bot h acc e p t a p roton (a Bronstcd base) an d don at e a lone pa ir (a Lewis base}. Water can also do na te a p roton (Bronstcd acid ), but it ca n no t accep t a pa ir of (..h-ctrou s (a n d a ct as a Lewis aci d ). 17. 113 (a) HOCI is the strong es t ac id (s mallest p R:, ami largest R:,), a nd HO I is the weakest ac id . (h) CI is m ore el ectron egative th an I, so the O C lan ion is more stab le than th e 0 1- a nion . 17.11 5 (a) II CIO , + II ,SO , !::::::j CIO, - + II ,SO/ (h ) T he 0 ato ms on su lfuric acid ha ve lo ne pai rs of ele ctrons th at ca n he used to bind to an H + io n . :0 :
.. I .. H -O- S-O-H .. I 17. 117 (a )
5.84
7.GG
17.97 Fo r th e reac tion HC0 2H (a'1) + OH - (aq ) ~ II ,O (C) + II CO, - (a q ), /\'
17.95 K,wl = K.d X K..t = 3.8 X lO- ti
A- 10 2
17. 107 K...
;0 :
1.4 X 10-' ; p K,
17.93 Acid ic: Na HSO ,j , N H tBr, Fe CI3 Neu tra l: KCI0 4 • NaN0 3 , LiBr Basic: Na 2CO ;\, (N I 14)2S, Na 2I IPO _1
>
A = 11+ solu tion; B = N I-I 4 + sol u tio n : C = Na + solution ; Y = Cl " so lu tio n; Z = 0 11- so lutio n
0.00 7 M:
17.89 (a) Ethylamin c is a stro nge r base than eth an olamin e . (h) For ct h ylamin c , th e pll of the so lu tion is 11.82. 17.91 pH
17. 105 T he po ssible ca tion-anion com bina tio ns a rc NaCI (ne u tra l) , Na O H (basic ), N H'ICI (aci d ic) , :":H IO H (basic}, II CI (acid ic) , a nd 11 20 (ne u tra l) .
~
K...,
~
K,K., KJ
~
K I K...t;,
(b ) Su bstitute ex pressio ns fo r the eq ua tion.
K.,~
Ka, an d
Ku
(b) [N a +] = 0.0323 M , l OIn = 1.5 x 10- ' M , [H :iO +] = 6.5 X 10- 12 !\.1, a nd [C(;H "O - ] = 0.030 8 M (e) pH ~ II.I !)
into
[H 0 ' ] [OW ]( [HP ' j [N H'd ) 3 [I' ll, ' ] [O W ][ HCN ] [CW ]
In a solu tio n ofN I I ~C N , we h ave I N I I~ + ] IC N - ] and [N H,] ~ [HCN ]. Wh en these and [ O H - ], are can celed from the e xp re ssion , we sec it is eq ua l to [H :lO+] . (c) I' ll = 9.33 (d ) K.l for N H 1+ and 1 Ag +( aq ) + CN - (aq) , K" = [Ag +] [CW ] (b ) NiC O , (s) -> Ni H (aq ) + CO ,' - (aq ) , K,,, = [N iH ] [CO/ - ] (e) AuBr,,(s) -> Au 3 +(aq ) + :, Br - (a q ) , 1
18.71 (a ) Na h rf aq ) (h ) 2 KC I(a'l ) 18.73 Q >
K.I"
+
NaNO,(a'l) I'h (NO, ), (a'l ) ---> 2 KNO,,(aq )
+
AgBr (s )
+
I'hCI, (s)
so RaS0 4 precip itates.
18.75 [H ,O ' ] ~ 1.9 X 10- '1> M; pH ~ 9.73 18.77 lIa CO, < Ag,CO " < Na,CO, 18.79 O rig in a l pH
2.1 X 10 6; yes , AgI form s
18.87 (a) [F - ] ~ 1.3 X 10- ' M ; (h ) [CaH ] 2.9 X 10- " M
18.59
Zn( O H ),( s)
=
=
8 .62 ; d ilu tio n will no t affec t th e pH .
18 .8 1 (a ) 0. 100 ~f ace tic aci d has a pl l of 2.87 . Ad ding sodi u m ac eta te slow ly raises th e pll . (b ) Ad di ng NaNO s to 0.1 00 M HN0 3 has no effec t o n the p H.
A - 104 Appendix 0
I
Answers to Selected Study Questions
18.91 Wh en [CO/-] ~ 0.050 M . [CaH ] ~ 6.8 X 10- ' M . Th is m eans on ly 6.8 X 10- 1 % o f th e io ns re m ai n , o r th at essen tia lly all of the calcium ions ha ve bee n re moved. 18.93 (a ) Add H 2SO ,I, preci pita t ing RaS0 4 an d leavi ng Na ' (aq) in so lu tio n . (h ) Add I-ICl or ano ther so urce o f ch lori d e ion . Pb Cl 2 will p re cipitat e , hut NiCl2 is wa ter- sol uble . 18.95 (a ) BaSn l will p recipitate fi rst. (h ) [Ila " ] = 1.8 X 10- 7 M p H ~ 2.8 1 pH at eq uiva le nce p o int = 8. 72 pl l at the mi d po int = p K~ = 4.62 Ph e n o lph th ale in «0) Aft e r 10.0 ml. , I'll = 4.39. After 20.0 mL, pH = 5 .0 7, After 30.0 m l., I' l l = 11.84. (f) A pl ot of pi I versus vo lu me o f Na O I I ad ded wo u ld begin a t a pH o f2 .8 1, rise sligh tly to th e midpoi nt at pH = 4.62 , an d th e n hegin to rise more stee p ly as th e eq uiva lence po int is a pproac h ed (wh e n th e volum e o f l\'a O I I ad d e d is 27. 0 mL) . T he pH rises ver tica lly th rough th e e quivalence po int , and tl u-n heg in s to level o ff above a pH o f about 11.0.
18.97 (a) (b) (c ) (d )
18.99 T hc K, value for ct hyla min c (4. 2 7 X 10- 4) is found in Ap pe ndix I. (a ) pH ~ 11.89 (h) Midpoin t I'll = 10.63 (c) pH ~ 10.15 (d) pH = 5.93 a t the eq u ivale nce po int «0) I'll ~ 2.13
(c) 1\, ~ I X 10 - 3 (d) 10% (e ) I' ll a t half-way p o in t = P K:, = 3.0; pl I a t cq uivalc n cc p oi nt = 7.;l
(1) T itration cu rve
14 12 ___
10 ;r:
~---
CHAPTER 19
8
c. 6
19.1
(a) Fo r a g iven su bstan ce a t a give n tempe ra tu re , a g-as a lways h as a g reate r e n tro py th an th e liqu id . M att er an d cll ergy arc mort' d isp erse d. (h) Liq u id wa ter a t 50 °C (c ) Rub y (rl) One m o le of N 2 a t 1 bar
19. 3
(a) d rS" = + 12. 7 .J/ K· mo l-rxn . En t ro py in creases. (h) d rS o = - 102.55 .1/ K . mo l-rxn . Signi fican t de crease in e n tro py. (c ) d r '~' o = + 93.2 J/ K · m o l-rx n . En tr o py increases. (d ) 8 r S o = - 129.7 .1/ K . m ul- rxn . T h e so lu tion h as a smaller e n tropy (with 11+ fo r m in g I1;jO + an d hydrogen bon ding oc currin g ) th an H e l in th e gaseous sta te .
19.5
(a)
4
'-----
2
oo!---=_-;:;------;-::----=------:-=----:-:::---:-::;20 40 60 80 100 120 140 Tit ran t Volume (mL)
(g ) Aliza rin o r b ro m creso l purple (se e Figu re 18.10)
18.101 110 m l , Na O Il
18.10 3 Add dil u te H C!, say 1 M H CI, to a so lu tio n o f th e sal ts. Both Ag-Cl a n d PbC1 2 will p recip ita te , b u t Cu2+ io n s will stay in so lu tion (as Cu C I2 is wa rc r-so lu hlc }. Deca nt o il t he co p p er-c o n ta in ing so lu tio n to leave a preci p ita te of wh itt' Ag e l a nd PbC12 . Leadf l l ) ch lo ride ( /(p = 1.7 X 10- ;') is m u ch m ore sol u bl e th an Ag Cl (K il = 1.8 X 10 - 10 ) . "Varm in g th e preci pi ta tes in wa te r will di ssol ve th e P bC12 an d leave th e AgCl as a wh ite so lid .
18.111 (a) C-C-C ang le . 120 "; O - C=O. 120"; C- O - ll . l(J9o; C-C- ll , 120 " ( b ) Roth th e ring C ato ms a n d th e C in C0 2 H arc Sp 2
+ 9 .3 .J/ K . ITl Ol-r XIl; (1))
~, S"
=
19.7
(a) d rS = - 507.3 .I/ K . m ol-rx n ; e n tropy d ecli n e s as a gaseo us re act a n t is in corp orat ed in a so lid co mpou nd . (b) 6. r '~· o = +3 13.25 .1 / K . m ol-r xn ; e n trop y in cre ase s as five mole cu le s (two o f th e m in th e gas p h ase ) form six m olecule s of p ro ducts (a ll gases ).
19.9
d sysSO= - 134. 18 J / K · mo l-rxn : d syJr = - 662.75 kJ/mol-rxn ; 6, ,,,rr.\'o = + 2222.9 J/ K . mo l-rxu : d \ll1 i1'So = + 2088.7 .1/ K . rno l-rx n
Q
18.107 W hen Ag)}o.l di ssolves slig h tly, it p ro d u ces a sm a ll
18.109 (a) Base is a d d ed to in cre ase th e pH . T h e added base re ac ts with ace tic aci d to fo rm more acetate ion s in th e m ixt ure , T h u s, th e frac tio n of ac id d ecl in e s an d th e fract ion of co nj uga te hasc rises (i.e., the ra tio [CII ;jC0 2 11l/[ CII 3C0 2 - ] d ecreases) as the p H rise s. (h ) At pH = 4 , acid p redomi n a te s (85 % ac id ami 15 % ace ta te ions ) . At p H = 6, ace ta te io n s p redo minate (95% ac eta te ion s a n d 5 % ac id) . (c) At th e point th e line s cross , [C,H :l C0 2 H ] = (CIJ;jC0 2 - ] . At thi s p o int p l I = PK:" so pl\, for ace t ic ac id is 4.75 .
=
- 293.97 .1/ K · mol-rx u
18.105 Cu (O H h will di sso lve in a no noxidizi n g acid su ch as H CI, whereas Cu S will n o t. con ce n tration of th e ph o sph at e io n , 1'0/ - . This io n is a strong b ase a n d hyd rol yzes to ur o,> . As th is reac tion rem oves th e 1'0 / - ion from e q u ilibriu m with Ag 3 P0.1, th e equ ili bri um sh ifts to th e right , p ro d u cin g m o re 1'0. 13 - and Ag " ion s. T hu s, AKl P0 4 di ssolv es to a g re a te r e x tent th an m igh t he ca lcu la ted fr om a " I' valu e (un less th e K,I' value was act u all y d e te rmi n ed ex perim en ta lly) .
~ ,S "
o 19.11 d mS = + l fi :~ . 3 .1/ K · mol-rxn: l:1'i's l/" = + 285. 83 kJ / mol-rxn ; f},," rrS o = - 95s' 68 .11K ' m o l-rxn : d llni,.SO =
- 795.4 ] IK T h e re act io n is no t spon ta neo us , b ecaus e th e ove rall e n tro py c h ange in th e u n iverse is n ega tive . T he re actio n is di sfavored hy e n ergy d ispersal.
19.13 (a) Type 2. T h e rea cti on is e n thalpy-favo red b ill e n tropy-d isfavored . It is more favo rable at low
tempe ra tures . (h) Type 4. T hi s e n dother mic re ac tion is not favored b y th e en th alp y ch an g e n or is it favor ed h y th e e n tr o p y change . It is not sp o n ta neo us un de r a ny con d ition s. ~ , ll" = - 438 k.J/lIlo l-rxn ; ~ , S " = - 20 1.7 J/ K . mol-rxn : ~ rG o = - ;l 7H ~I / m o l- rx n . T he rea ct io n is produ ct-favo red an d is en th alpy-drive n . (h ) ~ ,W ~ - 81l.1l1 kJ/ mol-rx n ; ~ ,S " ~ - 7!JA J/ K . mol-rxn ; ~ rG o = - 6 2.9 ~ J /mo l-rx n
19.15 (a)
hvbrid tzcd . Appendix 0
I
Answe rs t o Selected Study Quest ions
A-105
T he react io n is p roduct-favored. Th e c m hal pv change favo rs th e re ac tio n.
19. 17 (a) a,H' ~ + 116.7 kJ/ mo l-rxn; a,s' ~ + 168.0 J/ K mol-rxn; ~l; o = + 66.6 kJ /mol (b) a,1r = - 425.93 kJ/ mol-rxlI; a ,S" = - 154.6 J/ K · mol-rxn ; ~IG o = -;l79.82 kJ/ m ol (c) a,H' ~ + 17.51 kl/ mo ]-rx n; a,s' ~ + 77.95 J/ K m ol-rx n ; 19. 19 (a) (h) (c )
19.21
~fG o
=-
5 .73 kJ/m ol
19.29 ~ rG o = - 10 0.24 kJ/ mol-rxn an d ~) = 3.64 X 10 17 . Both the free e nergy c hange and K ind ica te a p ro duct -favo red p ro n 'ss.
19.31 (a ) Il l1 r (b) l'o: H ,CI (aq) (c ) C, H ,(g)
(d) l'o:aCI (g )
19.33
~, G O = ~T G o ~I Go
- 8 17.54 kJ/ mo l-rx n ; spo llla neous = + 256.6 q /m ol-rxn ; n ot spon taneous = - 1101. 14 kJ / mol-rxn ; spon tan eo us
ap " [l1aCO, (s)]
19.23 (a) a,H'
~
~,Go
m ol-rx n;
=
mo l-rxn ;
~ r Go =
~
mol-rxn: a,s'
- 283.99
=
+ 179.1 J/ K .
19.37 (a) The reactio n oc cu rs spon ta neo usly and is productfavore d. Th erefo re , ~ lTll i"S o is positive an d ~ r G o is negat ive . T he re acti on is likel y to he e xo thermic, so !J.,Ho is negat ive , an d ~ "'TT S O is po sitive . ~ s~'sS o is e xpec ted \0 he n egative becau se two moles of gas fo rm one m ole of so lid. T he calc ula ted val ues arc as follows:
~ l /mol -rx n
~
4 CO , (g ) + Ii II , O (g ) (a) No t on ly is this all exot herm ic co m bustio n reacLion , h UI th e re is a lso a ll increase in the nu m be r of mo le cu les from reac ta n ts to prod ucts. Th erefo re . we wo uld predi c t a n increase in ~ S o for both the svsn-m a nd t ilt' surroundings and th u s fo r the universe as well. ( h) T he exothermic re action has li TH " < O. Co m hincd with a pos itive li""So, the value of A,G" is Iwgatiw . (c) TI ll' value of ~, is likely to he much greate r tha n I. Further, 1H.'c ;:H1 St.' Ii""s ois positivt,. the val ue of A;, will be even la rge r at a hig her tempcra tu rc . (Set' th e left pa nel o f Fig-u re 19. 12.)
= -80.7 .1 / K · mol-t-xu ~ : Ii ,,\;'; = - 16U)O.1 1K · m ol-rxn
19.73 Rea ctio n 1:
19.55 ti rS" = - 459 .0 .l/ K . mol-rx n : !i ,H o = - 79:~ kJ / mol-rx n ;
Rea ctio n
Re actio n :l:
~ , Sj
Ii , .\~j
=
- ~ 4 2 . :-\ .J / K
. mol-rx n
d ,(; o = - 65 7 kJ /mol-rxll T he reac tio n is spon ta neous a nd c rulr al py-drivcu . 19. 57
( 2 Cr (O fl ), (s) + 3 5° 4' - (aq ) + 4 ( )II - (aq )
+ '1 c "
(a ) Cr (s) -> Cr3+(aq )
+ 1I,o ( /') + 2,' - ]
(c) Zn (s)
Cr is a red uc ing agent; this is an ox idatio n reacti o n .
+ 4 ()II - (aq ) -> Zn (O II ).,' - (aq) + 2
C u (O H),(s) Zn (s)
+2
+2
c " -> Cu(s )
Ol-l Iaq) t
+ 20 H - (a q )
+ Cu (O H ), (s) -> Zn( O fl)/-( aq)
AsH j is a reduci ng agen t; this is an oxi datio n actio n . (c ) VO, - (aq)
!"L '.
(d) 3[IIS- (aq )
V" (aq ) VO :~ -
(aq ) is an oxi dizing age nt; this is a red uctio n reaction.
'1 HS- (a q )
+ 2 ()II -(aq ) ->
(d ) 2 Ag (s)
Ag, O (s)
+ II , O (f) + Zc"
20.7
Silver is a redu cing age nt; this is an ox ida tion reaction . 20 .3
(a ) Ag (s) -> Ag +(aq)
+ c"
(h )
+
NO, - (aq )
+
2[ ~nO , - (aq)
+
2 H +(aq ) -> Ag +(aq ) + NO , (g )
H H 't Iaq)
+ II , O (/')
+ 5 e - -> + 4 II , O (f) ]
~1n" (aq )
5[ 1I50,,- (aq )
+ 1I, 0 ( f) -> 50 ,' -( aq)
+ 3 H +(a q ) + 2 c - ]
2 MllO ,- (aq) + H ' (aq ) + 5 H50 , - (aq )-> 2 Mn " (aq ) + 3 1I,o (l) + 5 SO , ' -(aq ) (c ) 4[Zn (.s ) -> Zn " (aq ) 2 NO, - (aq)
+
+ 2 ,,- ]
10 H f faq}
+
H c - ->
N,O (g ) 4 Zn (s )
3 c"
+ 3 e-
+ N0 3 - (aq) + 4 H +(aq ) -> NO (g )
Cr (s) 20 .5
+ 5 1I, 0 (f)
+ 2 N0 3 - (aq ) + 10 1I+(a q ) -> 4 Zn ' +(aq) + N, O(g ) + 5 H, O (f )
(d) Cr (s) -> Cr3+(aq )
+ 2 H,o (/')
+ NO, - (aq ) + 4 H ' (aq ) -> Cr3+(aq ) + NO (g ) + 2 II , O (/')
(a) 2 [A1 (s)
+ 4 O II - (aq ) -> AI (O H ), - (aq )
3 [2 H, O (f) 2 AI (s)
+ 3 c- ]
+ 2 e - -> H, (g ) + 20W (aq ))
+ 2 O H -( aq ) + 6 H, O (f) -> 2 AI(O fl) 4- (aq )
A -10B
Appendix 0
I
Electrons flow from the Cr e lec trode to the Fe clcctro rk-. :'\ega tive ions move via the salt bridge fro m the Fe/Fe2+ h al f-ce ll to the Cr/Cr H half-cell (and positive ions move in the opposite direc tion} .
+ 3 1I, (g )
Answers to Selected Study Questions
---?
Cr'l+(aq)
Ca thode (re ductio n): FeH (aq ) + 2 c "
20 .9
+ 6 ( )II -(aq )
+ CIO, - (aq )-> 3 5 (s) + C I- (aq ) + 3 O H - (aq)
Anode (o xidatio n) : Cr(s)
e - + NO, - (aq ) + 2 W (aq) -> NO,(g ) + H P (f ) Ag (s)
+ H P (f ) + 2 e - ]
CI0 3 -(aq) + :J H, O (f) + 6 e - -> CJ- (a q)
+ 3 H, O (f )
+ Cu (s)
+ Orr (aq ) -> 5 (s)
+ 6 1I +(aq ) + 3 e-->
c"
+ 3 e" ---?
Fc fs)
(a) Ox idation: Fe (s) + 2 c" Red uction : O , (g ) + 4 1I+(aq ) + 4 e " -> 2 II P (f ) Overall : 2 Fe (s) + O ,(g ) + 4 l l t faq) -> 2 Fe" (a q ) + 2 II ,O ( /') ---? Fc2 +(aq)
(b ) Anode, o xidatio n: Fe( s) ---? FeH (aq ) + 2 e " Cathode, red uction : ()2(g) + 4 H + (aq) + 4 c " ---? 2 H, O (f ) (c) Elec trons flo w fro m the ne ga tive anode (Fe) to the positive cathode (site of the O 2 h al freac tion ). Negative ion s m ove through the sal t bridge from the cathode compart m en t ill wh ich the 0 :1 re ductio n o ccu rs to the an ode com partment in which Fe ox idatio n o ccurs (and pos itive ion s move in the o pposite direc tion) . 20. 11 (a ) All arc primary bat te ries , not rec hargeable . (h) Dry ce lls and alka lin e ba tte rie s have Zn anodes . Ni-Cd ba tteries have a cadmium anode . (c) Dry ce ll... have an acidic en vironment, whe reas the en vironme nt is alkaline for alkalin e ami NiCd cells. 20 ,1 3 (a) R On _II = (b) EO n ' lI = (c) EO u -II = (d) EO n ' lI =
- 1.298 V;
1I0 t
prod uct-favored
- 0.5 1 V ; no t produc t-favo red
- 1.023 V; not product-favo re d + 0.029 V; produc t-favored
20 . 15 (a ) SIl" (ag ) + 2 Ag (s) --> SIl (S) + 2 Ag ' (ag ) E On _II
= - 0.94 V; not product-favored
( h) 3 SIlH (a q ) + 2 AI(s) --> :1 Sn 2' (a q) + 2 AI" (aq ) f ; O(.-l1
= + 1.8 1 V: prod uct-favo red
(e) 2 C IO , - (ag) + 10 Ce 3 ' (ag ) + 12 1I ' (ag ) --> CI 2(ag ) + 10 C" " (ag ) + 6 11 20 ({)
=-
E"'(f'1I
0. 1-1 V; not product-favore d
(d) 3 C u rs) + 2 :\"03 - (aq} + 8 H ' (a q} --> 3 Cu' +(ag ) + 2 :\"O (g ) + 4 11 20 ({)
EO cr tl
=
+ 0.62 V; product-favored
20 . 17 (a) Al (h) Zn a nd Al (e) Fe" (ag ) + Sn (s) --> Fe( s) + S n" (ag ) ; rca ctan t-favorcd (d) Zn 2+(aq ) + Sn (s) ~ ZIl (S) + Sn 2+(aq ) ; rc act an t-favored
20.4 7 Time
= 250 h
20.49 (a ) C O / (a g ) + 4 Il ' (ag ) + e - --> U' +( C I- (a g ) + 20W ( C, H,CO , H(aq) + 2 H ' (aq ) + 2 c : 2 A~ ' (aq)
+ C"H, C H O(aq) + 11,o (£) -> CGH ,C O , H( aq) + 2 H 't Iaq) + 2 Ag f s)
(b ) 3[C H 3C H,o H(aq)
2[C r, O,'-( aq) + 14 H ' (aq ) + 6e --> 2 CrH (aq ) + 7 H,o «( ) ] 3 C H 3C H,o H (aq ) + 2 C r, O ,'-(aq) + l fi H ' (aq ) -> 3 C H 3C O, H(aq) + 4 Cr3 +(aq) + II H, O «( ) 20.73 (a) 0 ,974 kJ/ ~ (h) 0, 60 k J / ~ (c) The silve r-zinc h atn- ry produces mo re e n e!"!.,'. )' per gram of reactan t". 20. 75 (a ) 2 N O ,,-(aq) + 3 Mn " (aq ) + 2 H,o (£) -> 2 N O(~) + 3 ~l nO , ( s ) + 4 1I' (aq ) ~~
Mn0 2 (s)
+
4 11+(aq) + 2 NI r ~ + (a q) ~ N,(g) + 3 ~ln" ( aq ) + 6 H,O «( )
20.79 (a)
2 Cu' +(a'l)
r: 2 Li,o (s)
Li,O (s)
+ H,o (O -> 2 l.iOH (a q )
2 Cars) + O , (g ) -> 2 Ca O(s) Ca O(s)
+ H, O «( ) -> Ca (O H ), (s)
21.3
These are th e ele m e n t" of Group 3A: boron , B; a lu m inum , AI; ga llium, Ca : in d ium , In; and thall iu m . TI.
21.5
2 Na(s) + Cl, (g ) -> 2 NaCI(s)
21.7
T he pro du ct , NaCI, is a co lo rl ess so lid an d is so lu ble in wa te r. Other a lka li metal c h lorides have sim ilar p rope rt ie s.
21.9
Ca lciu m will n o l exist in the e arth 's crust be cau se the metal reacts wit h wa te r.
;';-j 2+(aq )
Ca th o d e
21. 11 In cr e asing ba sicity: C O 2 < Si0 2 < SIl0 2 2 1.13 (a) (h) (c) (d)
2 Na (s) + fir , (£) -> 2 Nafir (s) 2 Mg (s) + O , (g ) -> 2 ~l g0 ( s ) 2 A1(s) + :1 F, ( ~ ) -> 2 AIF3 (s) C (s) + O , (g ) -> CO,(g)
2 1.15 2 H, (g ) + O , (g ) -> 2 H,O (g ) H, (g ) + CI,(g) -> 2 HCI (g ) :l H, (g )
Appendix 0
+4
20.87 29 0 h
~i
(h) Anode : Cells) -> Cel"(aq) + 2 c " Ca thode : N i2 +(a q) + 2 c " ~ N i( s) Ne t: Cells) + Ni"(aq) -> Cel" (aq ) + Ni (s) (c ) The ano de is negati ve a nd th e ca th o de is positive.
A-110
=
T he reac tion is ex o the rm ic a nd the product is io n ic. See Figure 104.
~
..... :-': ° :1
E Oallnd C H,CO, H(aq) + 4 H ' (a q ) + 4 e-]
E On _II
Answers t o Selected Study Questions
+ N,(g) -> 2 NH 3(g)
21.17 C H, (g) + HP (g ) --->
co (g)
+
~
H, (g )
tJ.,H' ~ + 206.2 kJ ; tJ., S ' ~ + 2 14.7 .J1K; tJ., G' ~
+ 142.2 kJ (at 298. 15 K) . 21.19 Step I ; 2 SO , (g) + 4 IJP (f) + 2 1, (, ) ---> 2 H,SO , (f) + 4 HI (g) Ste p 2: 2 H,SO, (£) ---> 2 1I, 2 AI" (a q) + Ii CI- (aq} + 3 1I, (g ) 2 AI(s) + 3 CI, (g ) ---> 2 AICI,,(s) 4 AI (, ) +
Ste p S: 4 HI (g) ---> 2 H, (g ) + 2 1, (g)
~
21. 37 2 AI(s) + 2 O l-l Iaq) + ti H ,O (£) ---> 2 AI(O Il ), - (aq ) + 3 11,(g) t
Net : 2 H,O (( ) ---> 2 H, (g ) + O , (g) 21.2 1 2 Na (, ) + F, (g) ---> 2 NaF (s)
Volum e o f H ';! o btai ned fro m 13.2 g Al = 18.4 I.
2 Na (, ) + CI, (g) ---> 2 NaCI(, )
21.39 AI,O, (s) +
2 Na (, ) + Ilr , (£) ---> 2 Na ll r (, )
~
H,SO, (aq ) ---> AI, (SO ,), (s) +
The a lkali m e ta l hal id es a rc wh ite , crysta lline so lids . T he y h aw h igh m eltin g an d boili n g p o in ts, an d a rc so luble i ll wa te r;
21. 23 (a) 2 CI- (aq ) + 2 II , O (£) ---> CI, (g) + 1I, (g) + 2 0 1I- (aq ) (I» If th is were the o n ly pro cess used to produce c h lorin e , th e ma ss of e Ll re p orted for in dust ria l p rodu cti on wo u ld be 0 .88 tim e s the mass o f Na O H produced (2 mol NaC l, 117 g , wo uld yiel d 2 m o l Na O Il , 1)0 g. a n d I mo l Cit. 70 g). T h e a moun ts quo ted in d ica te a CI\Clo..NaO H mass ra.. tio 0.96 . C h lo ri ne is p resu m ab ly also prep a re d by oth er ro ut es th a n thi s one.
21.43 T h is structu re h as a six-m e mber ring o f Si a to ms wit h 0 a tom b rid ge s. Eac h Si also h as two 0 a to ms a ttach ed. The b asic u nit is SiO / - . an d th e ove rall c h a rge is - 12 in [ (S i0 3) 6] - IZ. (Ele ctron lo n e pa irs a re ornincd in the foll owin g str uctu rc. ) -0
-0
c!\
-0
CaCO ,(s) + 1I, Ca"' (aq ) + 2 I1C(},- (aq ) 21.29 1.4 X lOGg SO , :0:
I
I
_.. / B . . . . . .. / B.. . . . . ... _ :() () () :
..
:0:
:0:
I .. I
:0 .. -1l-0 .. -1l-0.. :
."
/ \
0 -
!
Si - O -
\)
!
- O - Si
2 1.27 CaC 0 3 is used in agri cu ltu re to n eu tralize aci d ic so il, to p repa re Ca O for use in mortar, and in steel pro.. duction .
I
0 -
o-'::',Lo\ \!
- O - Si
Mg (' ) + N, (g) ---> Mg, N, (, )
. ........... B.. . . . .. ;0 0 :
H, O (C)
21.41 Pyrox e n es h ave as t h e ir b asic stru ctu ral li n it a n exte nde d ch ai n of lin ked SiO j te tr ah e d ra . T h e rati o of Si to 0 is 1 to 3 .
21.25 2 Mg (s) + O , (g ) ---> 2 MgO (, )
21. 31
~
Mass o f H ';!SOI requir ed = 860 g" a n d m ass of AI';!03 re q ui red = 298 g.
2 Na (, ) + 1, (, ) ---> 2 Na l (s)
~
O , (g) ---> 2 AI, O , (,)
Si- O -
/ \
O -Si-O
-d "0-
o
21.45 Co ns id er the ge n e ra l d e co mposition rea ct io n : N ~O "~---7
XI';! N z + .lI z O 2
T h e valu e of d (;" can h e o b ta in ed fo r all N xO ~ m olecu les b e cause dJ; o = - ill ;" . T h ese da ta sh o~\' th a t the d ecompos iti on reactio n is spo ntaneo us for all o f th e n itroge n ox id e s. All a rc unstable with re spect to de co m p os itio n to th e clem ents. Compound
-dlGo (kJj mol)
NO(g)
- 86.58
- 51.23 - 104.20
21.33 (a) 2 1l)1, (g ) + 12 O , (g ) ---> 5 IIp,,(, ) + 9I1 P (g ) (b) Enth alp y of combu sti on of B:;H !, = - 434 1.2 ~ J / m o l. Th is is m o re than do uble th e e n thal py o f co m b ustio n of B';!H 6 .
- 97.7 3
Appendix 0
I
Answers t o Selected Study Question s
A-111
21.47 :1Jr = - 114.4 kJ; exoth ermi c :1,Go = - 70.7 ~ I, product-favored at equil ibrium
(c) Cae:t 3 is planar tri gon al ; AsC lj is pyram idal. .. ...CI : :C I-Ga . .
21.49 (a) N, HI( aq) + O , (g ) -? N,(g) + 2 H,O (f) (h) 1.32 X 10' g
21.51
(a) O xid ati on number = +:~ (h ) Diph o sphoni c ac id ( H 1P20 ,,) should he a d ip ron e aci d (los ing the two II a to ms a ttac h ed 10 ato m s) .
°
:0:
:0:
I
I
H -P-O-P-H
I .. I
H-O .. :
:O.. H
"'CI : 2 1.67 (a) 2 KCIO , (s) -? 2 KCI (s) + 3 0,(g) (b) 2 H,S (g ) + 3 O , (g ) -? 2 H,O (g ) + 2 SO ,(g) (c) 2 Na (,) + O , (g ) -? Na , O , (s) (d) P, (s) + 3 KOH (aq ) + 3 H,o (f)-? PII , (g ) + 3 KII ,PO, (a'l ) (e) N H INO ,I(s) -? N, O(g) + 2 H ,O (g ) (t) 2 In (s) + 3 I\r, (C) -? 2 Inl\r 3 (s) (g) S nC I I «( ) + 2 H,O (C) -? S nO ,(s) + 4 HCI (aq )
2 1.69 1.4 X 10" metric ton s 2 1.53 (a) 3.5 X 10' kg SO , ( h) 4. 1 X 10' kg Ca( O Il) ,
2 1.71 Mg :
b-q
2 1.55
di sulfide leu
21.57
E O{ Zn"(aq) + 2 c "
UO r
(aq)
+
uo,« (aq ) +
4 H' (aq)
+
2 c " ---> U4+ (a q) + 2 H,o (C)
4 Il +(aq ) + Zn (s) ---> U4+(aq ) + 2 Il ,O (C) + Zn"(aq )
(c) 5[U' +(aq) + 2 1l,o (C) ---> UO r (aq ) + 4 fj +(a q ) + 2 e - ] 2[ ~ln () , - ( a q)
+ H H t taq) + 5 e ----> Mn " (aq ) + 4 Il ,o (f )]
5 U4+(aq) + 2 Mn O ,- (aq ) + 2 1I,0 (f ) ---> 5 UO r(aq ) + 4 fj +(aq ) + 2 Mn' +(aq) 22.69
Ion
K'o, mation
[0 2+
1.3 X 10'
Nj2+
5.5 X 108
Cu 2 +
2.1 X lOB
Zn 2+
2.9 X 109
22,71 Wilkin son 's ca ta lyst and th e EAN rule . Slfj) I: Rhod iu m-containing reacuu u: Eac h PR3 liga nd do nates two elec tr o ns as d o cs th e CI- liga nd . T he Rh " ion has B d e lec t ro ns . Tot al ele ct ron s = 16. Slfj) I: Rhnd iu m-contni ni ng [ntxl u ct. Assum in g that th e H liga nd is H - , each dona te s two e lec trons , as do CI- a n d the two PR 3 lig an d s. Th e m e tal ce n ter is Rh H , a ri(i m etal ion . To tal electrons = I R
Slrp 2.' Th e p rod uc t has 16 electro ns be cau se o n e PR 3 lig and h as been d isso cia ted . Appendix 0
I
Strj) 4: T h e p ro d uct h as 16 e lectron s. T h ere are three a nio n ic, two-el e ctron do no r ligan d s (CI- , II - , an d CII 3C Il t - ) , two two-el ectro n donor PR 3 lig and s, and an Rh H io n (d (i). Slrj) 5.' Th e p roduct has 14 el e ct ro ns. It is an Rh " complex ((p') with two PR 3 ligan d s ( two electrons eac h ) an d o ne CI - (two elec trons). (I t is likely a so lve n t m o lec ul e fi lls th e vacan t site he re to give a n-an sic n t, Ifi-clcc tro u co m plcx.) Strj) 6.' Th e p ro duct is on ce aga in the active , l fi-ek-ctro n ca talyst.
CHAPTER 23
nr ; (d) 1~Tc ;
23.11 (a) ~~Ni ; (h) I,ll ; (c) (1)
'k
(e)
-'lf3 ;
(pos itro n)
23.1 3 (a ) - Y!3; (h) ~iR h; (c ) ~a; (ct) 'imRa; (e )-Y!3; (I) jtNa
(ammine compLexes)
T he d ata fo r th ese h exa mmin c co m plex es d o in d e ed , veri fy th e Irvin g-Willia ms se ries. In t he book Ch('misfr)' f!f flu' Elements (N. N. G ree nwo o d a nd A. Ea rnsh aw: Zn d ed itio n, P: 90 B, O xfo rd , En gland , Bu tt erwo rth-H e in e m an n , 199 7) , it is state d : "the stahili tics o f co r respon d in g com ple xes o f th e hivalcnt io ns of th e first t ran sition se ries, irr espe ct ive o f th e part icu la r ligand involve d, usu ally vary in the IrvingWill ia m s o rd er, ... , wh ich is the re verse o f th e 01'dcr fo r th e ca tion rad ii. T h ese ob se rvati on s a re ( o nsistc n t with th e view tha t, a t leas t fo r me tals in o xid ati o n sta tes +2 a nd + 3, th e coord ina te bo nd is larg el y electrosta tic. T hi s was a maj or facto r in the ac ceptan ce o f th e crysta l fiel d th eo ry."
A -116
SI'1) 3: T h l~ prod uct has 1B electrons . T he twoel ectron do nor lig and C H 2C H 2 has rep lace d the d isso ci at ed PR 3 liga nd .
Answers to Selected Study Qu estions
2ijST h 4
2ij\ Pa
+ - Yf3
2ij \ Pa
4
2~&Ac
+
2~~Ac 4
2~lT h
+ _jlf3
2~l.T h 4
t~~ Ra
+ Ju
~u
tk:jPo
4
Ju 2 ~1 Po + Ju tMPh + Ju
2M Ph
4
2~~ B i
+ _jlf3
t h~Bi 4
2k~ P o
+ _jlf3
2 k~ Po 4
2~~ Ph
2~~ Ra -) 2MtRn 2k~R n 4
23.17 (a ) J¥fAu
4
+
+
Ja
l ~tHg
+ -\'f3
(b) 2~~ Rn 4 2 k~ r o + Ja (c) 1t,!,CS 4 1\f,Ba + _Vf3 (d ) 1~~ l n 4
23,19 (a)
1 ~~C d
+ Ve
~g R r
has a h igh neutro n /proto n ra tio o f 45/ 35 . Beta d ecay will al low th e ra tio to d ecrease: ~g B r 4 ~~ K r + -Y f3 . So m e ilo"'Rr decays by ga m m a e mi ssion. (h) Alp h a decay is likel y: 2~~C f 4 2i1:lCm + Ja (c) C o bal t-fil h as a high nil) rat io so be-ta d ec ay is likel y: t~Co 4
tkKr
+ - \'f3
(d) Carbon-L l h as only 5 ne u trons so Kccapturc or pos it ro n em ission may o ccur:
ILc + _\'e -)
I ~R
23.21 (;ene ra lly b eta d ecay will occur when th e nil) ratio is hi gh , whe reas positron e m issio n will o ccur whe n th e nil) ra tio is lo w. (a) Be ta d e cay: ';! \W -) T~Ne + _\I{3 ~H e
1H 4
+ -!i{3
TiiNe
+ \'(3 =
23.23 Bindi n g e n ergy p er n ucleon for liB
6.7 0 X
Hr' kJ
Bindi n g energy p e r nucl e on fo r HIR = 6.26 X 10'" k ]
23.27 7. 700 X 10'" kJ/ nu cleo n 23.29 0. 78 1 micrograms 23. 31 (a ) I ~~I 4 1 ~~X e + -Y{3 (b) 0 .07 5 m icrog ram s 23. 33 9 .5 X 10- -1 mg -7 2 ~~ po
=
+
~a
8.8 7 d
23.37 (a) 15.8 y; (h) 88 % 23.39 2~\: pu 23.41
~~Ca
23.43 (a ) 23.45
+ +
+
4
!lln
2~?,A m
(I» ill e
4
+ IH +
WiUuq +
2~* pU 4
'.:;Co
Ign +
~a
~Li
+
2
23.59 130 mL 23.61 Energy obtain ed from 1.00 0 lb (452.6 g) of 235 U = 4.05 X 10 - 1() ~J
23.63 27 fish tagg ed fi sh o u t of 5250 fish caugh t repre sents 0 .5 1 % of th e fish in th e lak e . Th e refore , 1000 fish put in to the lak e rep re sent 0 .51 % of th e fi sh in th e lake , o r 0 .5 1% of 190 ,000 fish .
23.67 (a) 231Pa isotope helon g s to th e 235 U d ec ay series (se e Questio n 23.65h) . (b) 2ij~U -) 2~!,T h + Ja
' lll
3 lIn (c) Ja
(d) ~~Cu
Ja
23.47 Tim e = 4.4 X lO w y 23.49 If t l /2 = 14 .28 d . th en h = 4 .854 X 10- 2 .r'. If the o riginal d isint egrati on rat e is 3.2 X 10(; dpm , th e n (fro m th e inn-gran -d fi rst order rate equation} , til l' rate afte r 365 d is 0.0 65 d pm. The plot will re semble Figu re 23.5 . 23.51
23.57 T ime = 1.9 X 10 \1 Y
23.65 (a ) Th e m ass d ecrease s by 4 units (with a n Ja c m issio n) OJ' is un ch an g ed (with a _Vf3 em issio n) so th e o n ly masses possibl e are 4 uni ts apar t. (b) 2:l2T h series , 111 = 4 n; n ',U series 111 = -in + 3 (c ) n !;Ra a n d ~ IO B i. 4 n + 2 series: 215,\ (. 4n + 3 se ries ; 22HT h, 4 n se ries ) (d) Eac h series is head ed by a lo ng-l ived iso to pe (in th e o rd er o f 10!! yea rs, th e ag e of the Ea rt h .) Th e 4n + I series is mi ssin g beca use the re is n o lo n g-lived iso to pe in thi s se ries. O ve r geo log ic tim e . all th e members o f th is se ries h ave decaye d co m p le te ly.
23.25 H,256 X 10'" kJ/nucleo n
23. 35 (a ) ';!~~ R n (b ) Ti me
23.55 Plot In (activity) versus ti me . T he slo p e o f the p lot is - k, th e ra te cons ta n t for d ec ay. H ere , k = 0 .00 50 FeS (s) + 2 1I +(a'l ) 2 lIi3+(a'l ) + 3 H, S (aq ) -> lIi,S3(s) + 6 ll t Iaq) Ca2+ (aq) + SO ,' - (aq ) -> Ca SO ,(s)
1. Step 1: Ca lcu late th e am ount of 12 in so lut io n fr om titra tio n d at a : Amo unt I, = (0.042:'; m o l S, 0 3' - / L) (0 .0253 L) ( I mo l 1, / 2 mol S,OJ' - ) = 5.38 X 10-' mo l I,
Case Study: Killing Bacteria with Silver 1. 100 X 10'" Ag + ions(l mo l/6J) 22 x IW' ion s) = 2 X 10 - 7 mol Ag+
Ste p 2: Ca lcu la te th e amou n t o f N'aClO prese n t based o n th e a mo u n t o f 12 for med, and fr om th ai value ca lcu la te th e mass of NaC IO .
2. 2 X 10- 7 mo l Ag +( I0 7.9 gAg +/ 1 mol Ag +) = 2 X 10 - 5 g Ag + ions
Mass NaCIO = 5.38 X 10- ' mo l 12 ( 1 mo l I ICIO /I mol 1, ) (1 mo l NaC IO / I mo l H CIO ) (74.44 g Na CIO / I mo l NaC IO ) = 0 .0400 g NaCIO
CHAPTER 4
Puzzler: CHAPTER 5
1. O xid at io n-reduct ion reactio n s.
Puzzler:
2. O xid at io n o f Fe gives Fe20 3; oxidat ion o f AI g ives A 120 3 .
Ste p I: Ca lcu late mass o f air in th e ball o on Mass air
A p p en d i x Q
I
=
1100 m 3( 1.200 g / m 3)
=
1.3 X 10 6 g
An swers to Chapter Opening Puzzler and Case Study Questions
A-123
Step 2: Calcu late e nerg y as heat need ed to raise the temperature of air in the ballo on . Energy as heat = ex m X tl T = ( 1.0 I I/ g . K) ( 1.3 X l OGg)(383 K - 295 K) = 1.2 X 10' J ( = 1.2 X 10:' kJ) Step 3: Calculate e nthalpy change for the ox idation of 1.00 g prop ane from e nthalpy o f fo rmation data. Assume formatio n of wa te r vapo r, H2 0 (g ) , in this reactio n. C, H,(g)
+5
O ,(g) -> 3 CO 2(g)
+ 4 H,O (g )
tl,J r = (2 mol CO 2 / mo l-rxn) [-393.5 kJ/m ol CO,] + (3 mol H, O /m ol-rxn )[ - 24 1.8 I
tl ,ll' = (2 mol CO';mol-rx n ) [tl/ H '(CO ,)] + (3 molll ,O / m ol-rxn )[tlr!r(II,O )] - ( I mo l C 2H 50 H / mo l-rxn) [tl/{'(C, H:,O H )] A -124
Appendix Q
I
For oc ta ne, per liter: 'l = - 44.40 kl/g (699 g / I.) = 3. 10 X 10 ' 1 KCl (s)
+4
2. There is a dou b le bo n d be tween N a n d the tcrm in al O .
M gO (s)
3. Reso n a n ce struc tu res are no t n eed ed to d escribe the bo n d in g in th is io n .
CHAPTER 7
PuzzLer: CHAPTER 9
1. Cr : [Ar ]3 d '4s 1; Cr:l+: [Ar]3(j3; CrO .,'(chro rniu m rVl l} : [Ar]
PuzzLer:
2. C r H is p a ram a gn eti c with th ree u npai red e lec tro ns.
1. Xe F2 is line a r. T he electro n pai r geom e try is trigo nal hi pyrami d al. Three lo n e p ai rs are lo cat ed in th e equ a tori a l pl an e , a n d th e two F a tom s a re lo cat ed in the ax ia l positions. T hi s sym me tr ica l struc tu re will no t h ave a dipo le .
3. Ph in PbC r0-t is pr esen t. as Ph 2 +: [Xe]4f 145 d I06s'
2. The Xe atom is sp\l h ybrid ized . Xe -F bon d s: ove rla p o f Xc sp 3 d orb ita ls with F 2p o rbita l. :\ lone pairs in Xe sp:ld o rb itals.
Case Study: MetaLs in Biochemistry and Medicine 1. Fc2+ : [Ar1 3d "; Fe:l+: [Ar] 3d '
3. T hi s :\f}-e lec tro n m ol ecule h as a be nt rnolccu la r stru c tu re .
2. Bot h ir on ion s a re pa ra m agn e tic.
.
3. Cu " : [Ar]3d lO ; Cu 2 +: [Ar]3 d"; Cu 2 + is paramagn e tic; Cu " is d iam agn e tic. 4. T he sligh tly la rge r size o f Cu co m pa red to Fe is rel a ted to g re at e r e lec tr o n-elec tr o n re pu lsio n s.
.x e. ..
. . " F:
Case Study: Two ChemicaL Bonding Mysteries 1. Eig h t two-el ectron bo n d s, wh ic h wou ld req uire l f elec tr o ns, four more than the 12 ava ila b le.
5. Fe 2 + is larg er th an Fe H a n d will fit less we ll into th e str uc tu re. As a re su lt, so m e di sto rtion o f the rin g struc tu re fro m p la n a rity oc c urs. A p p en d i x Q
.; O~
.. ; Xc : : F' ··
I
An swers to Chapter Opening Puzzler and Case Study Qu estions
A-12 5
2. The com poun d has Sp :l h ybridi zed B a nd N at o ms a n d is p o lar. The R a tom has a - 1 form al c h a rge a n d N ha s a + 1 fo rm charge. All bond ang les a re a bou t 109 °. II
I k ,\
/
Case Study: Biodiesel, a Fuel for the Future 1. C 12 H"CO ,CH ,( t) 14 11 ,0 (g ) 2.
II
N - I1 \ "II II H
/
3. (54. :' X 10 - 3 g AgBF,)(l mol/ 194.7 g) = 2.79 X 10 - 4 mol T he a mou n t of Ag(C2I1") .tB F.t tha t must have d ecompo sed is 2.79 X 10 - " mol. Mol ar m ass of unkn own = (62. 1 X 10- 3 g) / (2.79 X 10 - 4 mol ) = 223 g / mo l Th e co m pou nd Ag(C2 H ..) BF.. where x = 1 ha s a m ol ar m ass of 223 g / mo l.
+ 20 0 2(g )
---7
14 CO 2 (g)
+
( 14 m ol CO 2 /mo l-rx n)[i'1r!r(C02 ) ] + (14 mol H 20 /m ol-rxn)[i'1JH O(H 20 ) ] - (I mol C I2H 2,; C0 2CH , / mol-rx n) [i'1 rH O(C I2H"CO,C H,)]
s.u: «
i'1 cH o = (1 4 mol CO,/ m o l-rx n) [ -3935 kJ/m o l CO 2 ] + ( 14 mol II , O /m ol-rxn )[ - 24 1.8 ~ I / m ol H 20 ] - (I m ol C I2H 2"CO ,CH.,/ m o l-rxn) [ -77 1.0 kJ/m ol C I2H ",C0 2CH 3 ) ] = - 8 123.2 ~I /m ol-rxn I m ol m e th yl m yrisrat c pe r mo l-rxn , so q p e r m ol = - 8 123.2 kl /mol
3. Burn ing h e xade can e: C Il1 H 3 .. (f ) ---7 16 CO 2(g) + 17 H 20 (g )
+ 24.5
0 2(g )
i'1 cH o = (16 mol CO 2/ m o l-rx n) [i'1 rH O(C0 2)] + ( 17 m o l 1120 /m ol-rxn ) [i'1 rIlO( II ,O)] - ( I m ol C\
1:\9 pm
x
2. Bond an gle s: H - C-H and H - C-S, 1095 ' ; C- S- II som ewhat less th an 109°. 3. Pol ar 4. It shou ld beh ave as a n ideal gas a t moderate pre ssure s and temperature s we ll a bove its boiling point.
2. LOO l.tn] = l. OO X 10- ' em an d 240 pm = 2.4 X 10-' em .
5. H 2S (with th e lowest molar m ass) will d iffu se fa stest.
The number of C ti rin gs spa nn ing I ILm is l.OO X 10- ' em /2 .40 X IO- Rem = 4. 17 X 10" rings
CHAPTER 12
3. Graph en e is d escribed as be ing on e carb o n a to m th ick so th e thi ckness is twice th e radiu s o f a ca rbon a to m o r 154 pm .
Puzzler: 1. In ice , wate r m o lec u les are n o t packed as closely as th ey are in liquid water. This str uc-
Case Study: The World's Lightest Solid
ture re sults so th at h ydro gen bonding interactions bet ween water m o lec u les a re m a ximi zed.
1. The mass of 1.00 ern" o f aerogel is 1.00 m g ( LOO X 10- " g) and 0.2% of this, 2.00 X 10- 6 g, is the mass of the polyme r. The number o f silico n atoms in 1.00 cru' = 2.00 X 10- fi g ( I mo l (C 211,O),SiO / 134.2 g)( 1 mol Si/I mol (C211 ,O),SiO) (6.022 X 1023 ato ms Si/mo l Si) = 9.0 X IO"j ato ms Si.
A p iece of ice floa ts at a leve l where it will di sp lace its we ight of seawate r. Mo st of th e vo lum e of a n iceberg is below th e wa te r line. 2. A loon em" piece o f ice (m ass = DI7 g) wou ld float with [0.9 17/ L026]( IOO%) = 89.4 % belo w th e su r fac e o r 10.6 % a bove th e su r fac e .
2. Volume between glass pan es = 1:">0 em X ISO em X 0.2 em = 5.400 em"
Case Study: The Mystery of the Disappearing Fingerprints
De n sity o fa e ro gel = 1.00 X 10- 3 g /c m cl , so the m ass of ae rog el needed = 5400 em'( l. OO X 10- ' g /e m") = 5.4 g.
1. The che m ica l co m pou n ds in a child 's fin gerpr ints a re m ore vo lat ile because th ey have lo wer mo lecul ar weig h ts than co m pou n ds in ad u lts' fin gerprints.
A p p en d i x Q
I
Answers to Chapter Opening Puzzlerand Case Study Questions
A-127
CHAPTER 14
2.
--l .OO
:1. 75
Puzzler:
3.:)0
3.23 3.00 2.7:)
1. "Like di ssol ves like ." Both liquids are pol a r, a n d both are ca pa b le o f str ong hydrogen bonding. 2. Ethyle ne g lyco l, a n onvolatil e so lu te , lowers th e fr eezin g p oin t. It is no t co rros ive. Th e liqu ids a re mi scible in a ll p roportio ns. 3.
(" yen' =
2 .30 2 .2:) 2 .00 ~
1.75
0:::
1.50
"
CO 2(g )
1. (a) O xidiz e part o f th e N H, to H :-.1 0
CHAPTER 18 Puzzler:
+ H 20 (g ) ---> CO , (g ) + 1I, (g )
1. CaC O " ( h.p fo r CaC O , = 3.4 X 10- 9 ~ln CO, = 2.3 X 10- 11)
!1,II' = (I mol C0 2/ mo l-rxn )[!1/ I ' (C O , )] ( I mol CO / mo l-rxn )[!1; II' (CO )] - (I mo l H20/ mol-rxn )[!1;II' (H 20 )]
> h.p for
2. PbS
!1,II' = ( I mo l C0 2/mo l-rxn )( - 393.5 " j/ mo l) - ( I mol CO / mo l-nm )( - 110.5 " l/ mo l) ] - ( I mol 11 20 /mol-rx n )( - 2-1 1.8 kl/ mo l) ] = - -1 1.2 kj /mol-rx n (exo the rmic)
3. CaF, (,) Ca' +(aq) ( K,p = 5.3 X 10- " )
+ 2 F- (aq)
Define sol ubi lity (mo I/ L) a, x; th e n [Ca"" ] = x and [F - j = 2x 5.3 X 10- " = [Ca' +][ F- j' = [x][2xj' x
A p p en di x Q
I
= 2.4
X 10- '; solubi lity
= 2.4
X 10- ' rnol /L
Answers to Chapter Op ening Puzzler and CaseStudy Questions A · 129
eRIrIf! o xid at ion , free enerb'1' cha nge: dJ:;o = d rH o - T d rSo. (I1,H Ofro m ab ove . \Ve need to calc u la te a lS o fo r thi s re ac tio n.)
Case Study: Take a Deep Breath! 1. pH = pK;, + 10g [HI'O ,' -j / [H ,I'O, - j
7.4 = 7.20
+ 10g[HI'0,' - ] / [H ,I'O, - ]
/!" ,S' = (S mol CO , / mo l-rx n)S' (CO , )j + (9 mol I I,O / m o l) [S ' (I I, O)] - ( 1 mol C,II ,,/ mo l)[S' (C,H '8) ] - (12.5 mo l O , )[ S' (O , ) ]
[ H I'O ,' -j /[ H, PO , - j = 1.6 2. Assign x = [HPO/ - ], then [H , PO,- j = (0.020 - x)
/!" ,S' = (S mo l CO , / mo l-rx n) [21 3.74j / K · mo l CO ,] + (9 m o l II ,O/mol-r xn ) [ ISS.S4J!K· mol 11 ,0 ] - ( I mo l CRII ,, / mo l-rxn ) [36 1.2 kJ/K ' mo l C,H '8)] - (12.5 mol O g/m o l-rxn ) [20.0.07 .I/K . mol 0 , ] = .oB7.5 J! K . mol-rxn ( = 0 .5S7.o ~J /K . mol-rxn )
l .f = x/ (0.020 - x); x = 0.01 2 [HPO ,' -j [H , PO,- j
= x = 0.01 2 mo l/L = 0.20 - x = O.OOBO
mo l/L
/!" ,G' = /!",1l' - T /!" ,S ' = - .0 .0 70 kJ/mo l 29S.2 K(O.5S7S kj /K . mo l-rxn ) = - .0 ,24.0 kJ/ rnol-r xn
CHAPTER 19 Puzzler: 1. Ethanol oxidation : C, H, O H(f) 2 CO , (g) + 3 11,0 (g)
+ 3 O , (g )
---7
The enthalpy change per gr a m , - 26.S0 kJ/ g, was ca lcu lated for th e Ca se Study in Ch apter ;) (sec pa ge A-124 ) . From thi s, th e enthalpy cha nge p e r kg is - 2.6S0 X 10' kJ/kg. C,II " oxidati on : C, H ,,(A) S CO , (g) + 9 11 ,0 (g)
+
12..0 O , (g )
---7
Th e e n tha lpy change per g ra m , - 44.4 kj / g , was ca lcu la te d for th e Case Study in Ch a p te r 5 (see page A-124) . From th is, th e e n th alpy c ha nge per kg is 4.44 X I0' ~J /kg 2. Eth anol o xid at ion, free en ergy c ha n ge : /!" ,G' = (2 mol CO ,jm o l-rx n) [/!"/i' (CO,)] + (3 mol H ,O/mol-rxn )[/!"/ i' (H ,O)] - ( I mol C, I1501 I/mo l-rxn ) [/!"/Ci' (C, I I, OI I) J /!" ,G' = (2 mol CO , / m o l-rx n) [-394.4 kJ/m o l CO ,] + (3 mol II ,O/mo l-rxn ) [ - 22S.6 ~J /m ol H ,O] - ( L mo l C, H "O H / lllo l-rx n)[ - 174.7 kJ/ mo l C, 1I 5 0 Il)] = - 1300 . kJ/lllol-r xn
I mol e than o l per I ruol-rxu : th e refore !1J: pe r mol: - 1300 . ~J /Illol /!" ,G' per kg: - 1300 . kj /mo l ( 1 mo l C, H 5 0 H / 46 .07 g C, H 5 0 H) (1000 g / kg) - 2.S22 X 10' kl/kg C, lI"O Il
/!,.J;' per kg = (- .0,27.0 kJ/m ol-rxn ) ( I mol -rxn / mo l C, II IR) ( I mo l/ 114.3 g) (I kg /1000 g) = - 4.59 X 10' kj /kg 3. For th e oxidation o f ethanol , entropy ch anges in crease th e ene rgy availab le 10 do useful wo rk. For oxid ation of h yd roge n, th e opposite is tru e .
4. Th e differen ce in values o f 6.l'Ho a n d 6.,(; 0 re sn it because e ne rgy is ex pe n d ed or ac qu ired to ac hi eve a hig he r di sp e rsion of e nerbT)' in the uni verse (syste m a n d su rro u n d ings ) .
Case Study: Thermodynamics and Living Things 1. C re atine phosphate + H 20 ~ Creatine + 1]+ /!" G' = - 43 .3 kJ/lllol
Adenosin e + HP i ~ Ad en osin e rnonophosph at e + 11 ,0 /!,.G o = + 9.2 kJ/mol
Ne t reac tio n (su m of th e two re actions): Crea tine phosph ate + Ad enosine ~ Creatine + Adenosine monophosphat e For thi s /!" G' = - 43 .3 ~J /mol + 9.2 kJ/mol = - 34. 1 kJ/mo l; th e negative value in dicates th at th e tra nsfe r of phosp h at e from c rea tin e phosphate to ad enosine is product-favored.
2. /!" G" = /!" G' + RT In[C][H ,O +] / [A][B] = M i' + (S.3 1 X 10-' ) (29S) In ( 1)[ I X 10- ' ]/ [ 1][1 ] /!" G" = /!"(i' - 34 .2 kj /mol
A - 130 A p p en di x Q
I
+ HP i
Answe rs to Chapter Opening Puzzle r and Case Study Questions
CHAPTER 20
[; ,Ir = (I mol CO 2/ mo l-rxn ) (-393.5 kJ/rn ol ) - (I mol CH ,/ mo l-rxn) (- 74.87 kl/m ol )] - (2 mol II ,O / mol-rxn ) ( - 285 .8 kl /m ol )] = 252 .4 I;I/m ol-rxn
Puzzler: Possible reactio ns are:
1) Cn2+(aq ) 2) Cn(s )
+ Ni( s)
-> Cn(s)
+ Ni2+(aq)
For Sil L: assume s JI 2 0 (f) is the reactant
+ Ni2+(aq) -> Cn 2+ (aq ) + Ni( s) .
[; ,11° = (I mol Si0 2/m ol-rxn ) [[;/ r(SiO ,)]
- ( I mol SifL z'mol-rx n ) [[;/H O(SiH , )] - (2 mol II 20 /m ol-r xn )[[;t1r(II ,O )]
A reactio n g iven by eqnatio n ( 1) would produce a n electri c cu rre nt of 0 .:)9 vo lts. I };' n'" = f:' (cath oc!e ) - RO(anode ) = 0.34V - (- 0.25 V) ]
[; ,1I " = ( I mo l SiO , )/ mo l-rxn ) ( - 9 10.86 IV mol ) - (I mol SiII,/ mo l-rxn )(34.3 1 kJ / mo l) ] - (2 mol H 20 /mol-rxn ) (- 285.8 ~J / m o l)] =
- 373.6 kf/mol-rxn
Case Study: Manganese in the Oceans 1. Catho de reac tion: ~'1 n 3+
Anod e reaction : Mn :VlnO , + 4 H + + e -
H
+ e-
-t
+ 2 1120
Net reaction: 2 ~'l n :H + 2 H 2 0 :Vl nO , + Mn 2+ + 4 II + f:' cdl
3. Ele ctronegativities: C 2.5, Si, 1.9 , II 2.2. From this we co nclude that polarities o f C- IJ and Si- H bon ds arc in the oppo site direct ion s: in Slf L , the II has a sligh t negative charge (it is hydriodi c) and in e ll" the II has a slig h t po si-
!vl n2+
-t
~
tive charge . 4. Ge neral o bse rvatio n fro m the se exam ple s:
= f:'(cathode ) - f:' (anod e )
=
Carbo n often bonds to o ther ato ms via double bo nds, whereas Si does not. \Ve wo uld 1I0 t expect H2Si= SiH 2 to exist as a mo lecular spec ies; instead a po lymer ic struc tu re [ - SiII 2SiII2 - ] x is predicted .
1.50 V -
0.95 Vj = 0.55 V Th e positive value asso cia ted with d isproportion ation (the ne t react ion ) is pos itive, ind icating a prod uct-favore d reactio n. 2. (a) MilO ,
: 0:
+ 112S + 2 11 + -> Mn 2+ + S + 2 11,0
(b) 2 Mn' + + O 2 + 2 11 20 -> 2 Mn02 + 4 11+ 3. Cathode reactio n: O 2 (F: = 1.229 V)
+ 4 II + + 4 e : - t 2 H 20
Anode reac tio n: !\'1n 2+ + 2 1120 -t :VlnO, + 4 H + + 2 e - (I,' = 1.23 V, fro m Ap pe ndi x M)
Total CaO = 115.3 mg (2 significan t figur es)
+ 2 11,0 (£) -> CO, (g) + 4 112(g )
n
1120 ( C) -> Si0 2(s)
eac h of CaCO :~ and !\lgC0
+ 4 1I, (g )
= ( I mo l CO , / mo l-rxn ) [ [; / 10( C0
+ 209 .9 mg
= 330 mg
\\.Te get 2 mol CaCC):l per mol e CaN and I mol 3
per mol Mg2+
CaC0 3 fro m Ca2+ react io n: (0.150 g Ca2+ ) (I mo 1/40.08 g Ca 2+) (2 mol CaC O , / 1 m ol Ca2+ ) ( IOO. 1 g CaC O , / 1 mo l CaC O ,) = 0.749 g
2. Fo r CH 4 : assumes II 20 (t') is the reactant
s.ir
j
For Ca2+: (50 mg ) ( I m mol Ca' +/ 40.0R mg ) ( I mmol CaO / m mo l Ca H ) (56.07 mg Ca O / I m mol Ca O ) = 209.8 mg CaO
Puzzler:
+2
I .. ell
1. For Mg 2+: (50 mg ) (I mmo l Mg 2+/ 24.3 1 mg ) (I mmol Ca O / m mo l Mg 2+) (56 .07 mg CaO / I mmo l Ca O) = 115.3 mg Ca O
CHAPTER 21
Sill , (g)
CI H'..t
Si-O
Case Study: Hard Water
f-:' cdl = E'{cath ode ) - f:' (anode) = 1.229 V 1.23 V = 0 v.
1. CII , (g )
-t
II H,C- C - CH,
2) ]
( I mo l CH,/ m o l-rxn ) [ [;j H O(C H,) ] - (2 mo l H 20 /m ol-rxn ) [ [; /H O(H ,O)]
Appe ndi x Q
I
Answers to Chapter Opening Puzzlerand Case Study Questions
A- 131
CaC O :l from ~f g2 + reactio n: (0.050 g ~l g2 + ) (I mo 1/ 24.31 g MgH ) (I mol CaC O ,,/ 1 mol Mg 2 + ) ( 100 . 1 s CaC0 3/ I mol CaCO,) = 0.0.206 s YI gCO" from Mg' + reaction: (0.050 s Mg' +) (I mo 1/ 24 .31 g MgH ) (I mol MgCO,,/1 mol YI g 2 + ) (84 .3 1 s CaCO, / I mol CaCO,) = 0.173 s Total mass of so lids = 0.747 g + 0.206 g 0.173 g = 1.1 g (2 significa n t figures)
+
Case Study: Lead, a Mystery Solved 1. 50 ppb is 50 g in I X 10" g o f blood . Assu me th e d en sity of blood is 1.0 g /m L. In 1.0 X 10" mL (i.e ., 1.0 L) o f blood , th ere will be 50 X 10- 6 g o f Pb . From thi s:
50 X 10 - 6 g (I mol I'b /20 7.2 s I'b ) (6.022 X 10' " at o ms Pb /mol Ph ) 1.5 X 101i ato ms Pb
+ 4 Ol-l r taq) + 3 O , (a'1) + 6 lI ,O (f)
---7
Ne t: 6 NO ,,'(a'1) + 5 CH, O H ---7 3 N,(g) 5 CO,(a'l) + 6 Ol I T aq ) + 7 11,O (C)
2. Calcu late d d e nsity: Mass of one uni t ce ll is th e mass of on e T i and one Ni atom = (47.87 g/ mo l)(I mo l/ 6.022 X 102:\ ato ms Ti ) + (58.69 g/ mo l) ( I mol / 6.022 X 102:\ atoms Ti ) = 1.77 X 10- 22 g
ass u m e d .
+
3. H C0 3 - is th e predo minan t species. Recall th at whe n aci d and base co nce n tra tion s are eq ua l, pH = pK", If H, CO" and H CO,,- are present in eq ual concentra tions, the pH wo u ld be abo u t 6.4 . If II CO:~ - an d CO:l2- are p resent in eq ua l co nce n tra tio ns , th e pH wo uld be 10.2. For th e p II to be abo u t 8 (in a salt water aq uariu m ) , [ HCO:~ -l wo uld have to be high er tha n e ither of th e oth er carbo na te species. 4. Co ne. o f N in ppm (rug N/ L) = [( 1.7 X 10' kg N0 3 - ) ( 10 6 mg NO ,,- / kg N0 3 - ) ( 14.0 mg N / 62 .0 mg NO , ')] /(2.2 X 10 7 L) = 1.7 X 10' mg/L = 1.7 X 10' ppm Co ne. ofNO, - in ppm (mg / L) = (1.7 X 10' kg ) ( 1()6 mg/kg )/ (2 .2 X 107 L) = 770 mg /L
I
1. Defin e len gth of th e sid e of th e cu be as x, th en th e len gth o f th e diagonal across the cu be is xV3. T his is set equa l to 2 r T i + 2 r :-.li , i.e., xV:) = 2 rTi + 2 r :\' i = 540 pm ; x = 3 12 pm (a = b = c = 3. 12 X 10- ' em )
The ag ree me n t is no t very good , p robabl y beca use ato ms don 't pack togeth er as tight ly as is
O xid at ion half-reaction: ClI ,O ll(a'1) + 6 O W(aq) ---7 CO ,(a'1) + 5 H,O (C) + 6 e -
Appendix Q
Puzzler:
Ca lcula te d d ensity = 1.77 X 10 - 22 g / 3.04 X 10- ' " em ' = .'}.82 g/c m"
2. Reduction hal f-reaction : 2 NO:\-(aq) + 611,O (f) + 10 e : ---7 N,(g) + 12 OW (a'1)
A- 13 2
CHAPTER 22
Volume o f the unit ce ll is i~ = (3. 12 X 10- 8 em )? = 3.04 X 10- " em'
Case Study: A Healthy Aquarium and the Nitrogen Cycle 1. 2 N H / (aq) 2 NO , - (a'1)
COliC. o f NO,,' in m o l/ L = [( 1.7 X 10' kg ) (10 1 g / kg) (I mo 1/ 62.0 g)] / (2.2 X 107 1.) 0.01 2 mo l/ I.
3. As fre e ato ms, both T i and Ni are paramagn eti c.
Case Study: Accidental Discoveries 1. First order kin etic s: In [xl xul = - kr In[ x/IO mg ] = - 7.0 X 10-"
S- I
[24 h X 36 00 s/ h] x/ IO mg = 1.4 X 10' ''; x =
1.4
X
10- 2 mg remain
2. Usc Hcndc rson-H assclbalch eq ua tio n fo r thi s aci d d issociation eq ui libriu m .
I'll = pK" + log [base/ acid ] 7.4 = 0.6 + log{[P tCI(NI I, j,OIl J/ [PtC I(N I1"j, (11,0 )
n
[PtC I(N ll , y,0 1l] /[ PtC I(N Il ,,),( 1l20)] = 6.3
Case Study: Ferrocene 1. Fc H in fcrroccnc ha s an elec tro n co nfigu ratio n [Ar ] 3d" and is presen t in th e low spin state .
Answers t o Chapter Opening Puzzle r and Ca se Study Questio ns
2. Cr (0 ). Cr( O) in th is com po un d is assu m ed to have a n e lec tro n co n fig urat io n [Ar] 3d 1i a nd is p resent in the low spi n state.
Case Study: Nuclear Medidne and Hypothyroidism: 1.
3. Bo th are in acco rd th e 18 electron rule.
1311 ~ U 1X e
+ ~ 1 {3
2. Calcul a te the fra ction ( = / ) of each re m ain ing after 7 d ays
4. Selec t oxidizing agents from Table 20. 1 ( page 920) based on the north east-sou thwest rule (a bove r.~ = 0 0400 v) . Common ox id izing age n ts include th e ha loge n s, 1-1 2 0 2 a n d
Fo r 123 1: k = 0.693/ Iv, = 0.693/ 13.3 h 0.05 2 1 h - I
~ lnO , - .
In (f)
5. ;'\iCI, + 2 ;'\a [C, H, ] --> :-li(rrC, II , ), + 2 NaC!. Nick elocene is predicted to h ave 2 unpaired elect rons (N iH , with a d " co nfigu ra tio n , in a n octa h ed ral envi ronmeut. )
J=
=-
=
0.052 1 h - l[i d(2 4 hid )]
1.6 X 10 -'
Fo r 131 1: k = 0.693/ tv, = 0.693/ 8.04 d 0.0862 d - I
=
In (f) = - 0.0862 d - l (7 d )
CHAPTER 23
J = 0.55
Puzzler:
Ra tio of a mo u n ts remain ing, [ 13 11]/ [ l23I] = 0.55 / ( 1.6 X 10- ' ) = 3400
1. 23r,U: 92 proto ns, 235 - 92 = 1 4 ~~ ne u trons 2:~ U :
2. (a )
T he a m o u n t of th e 13 11 iso tope is 3400 times g reate r th a n th e amoun t of 123 1.
92 proto ns, 146 n eu tro ns
2:lRU
+ ~n ~
239U
+ ~1 {3
( b) 2:l9U ~ 2:l9N p
+ ~ 1 {3 + ~a
2:wN p ~ 2:l9p u
(c ) 2:l!l P U
~
n ", U
Appe ndix Q
I
An swers to Ch apte r Opening Puzzler and Case Study Ouestions
A-133
