First Edition, 2009
ISBN 978 93 80168 56 2
© All rights reserved.
Published by: Global Media 1819, Bhagirath Palace, Chandni Chowk, Delhi-110 006 Email:
[email protected] Table of Contents 1. The Equilibria 2. Chemical Equilibrium 3. Surface Phenomenon 4. Bioinorganic Chemistry 5. Bio-organic Chemistry 6. The Photochemistry 7. Quantum Chemistry 8. Analytical Chemistry 9. Nuclear Chemistry 10. Organic Name Reactions 11. Reagents in Organic Synthesis 12. Kinetics of Reactions 13. The Macromolecules 14. Fast Reactions 15. Conformational Analysis
The Equilibria 1 The Equilibria The E°cell of an aluminium-air battery is 2.73 volts and it involves a 12 electron process. The free energy change (DG°) of the battery in kJ. Calculated : DG = _ nFE° = _ 12 × 96500 × 2.73 J = _ 3161340 J = _3161.34 Id For the following reaction— H2(g) + C12 (g)
2HCl (g).
DG° is _ 262 kJ. Calculate the equilibrium constant (K) for the reaction at 298 K. DG° = _ 262000 J = _2.303 RT log K
log K =
= 45.918
K = 8.279 × 1045 Hydrogen-oxygen Fuel Cell Two half-cells of hydrogen-oxygen fuel cell under basic conditions can be depicted as OH_/O2 (g)/ Pt and OH_/H2 (g)/ Chemistry : Basic Elements Pt and their standard electrode potentials at 25°C are 0.4009 and _ 0.8279 V respectively. Write the half cell reactions and the complete cell reaction. Depict the complete cell and the e.m.f. of the cell. Calculated : _Pt
|H2(g)| OH_| O2(g)| Pt +
At anode, the reaction is H2 + 2OH_ ® 2H2O + 2e_
At cathode, the reaction is 2e_ +
The cell reaction is H2 +
O2 + H2O ® 2OH_
O2 ® H2O
The e.m.f. of the cell is = 0.4009 _ (_0.8279) = 1.2288 V The reduction potentials of Ag+/Ag and Fe+3/Fe+2 are 0.799 and 0.771 V respectively. The equilibrium constant of the reaction. Ag + Fe+3
Fe+2 + Ag+
Ag | Ag+ || Fe+3 · Fe+2| Pt(+) E°cell = 0.771 _ 0.799 = 0.028 V At (_ ), Ag ® Ag+ + e_ At (+), Fe3+ + e_ ® Fe+2 Cell Reaction : Ag + Fe3+ ® Ag+ + Fe2+, Here n = 1
log K =
=
=
= _ 0.47329
K = 0.33628 The EMF of the cell, Pb | PbCl2 || AgCl | Ag at 300 K is 0.50 V. If the temperature coefficient of EMF is _ 2 × 104 volt deg_1, DH and DS for the cell reaction. Calculated : Pb + 2AgCl ® PbCl2 + 2Ag Pb | PbCl2| Cl_ | AgCl | Ag + The cell reaction is Pb + 2AgCl ® PbCl2 + 2Ag The Equilibria
DG = _ nFE = _ 2 × 96500 × 0.5 J = _ 96,500 J or _ 96.5 kJ
DS = = 2 × 96500 (_2 × 10_4) = _ 38.6 JK_1 DG = DH _ TDS, _ 96,500 = DH _ 300 (_38.6) DH = _ 96,500 _ 300 × 38.6 = _ 108,080 J or, _108.08 kJ The potential of pentane/oxygen fuel cell given that the standard free energies of formation (in kJ/mole) at 298 K are _ 8.2, _ 237.2 and _ 394.4 for pentane, H2O (1) and CO2 (g) respectively. Calculated : The chemical reaction that takes place in the pentane-oxygen fuel cell is C2H12 (g) + 8 O2 (g) ® 5 CO2 (g) + 6H2O (1) DG° for the reaction = [5(_ 394.4) + 6 (_ 237.2)] _ [_ 8.2] = _ 1972 _ 1423.2 + 8.2 = _ 3387.0 kJ The electrode reactions are C5H12 + 10 H2O ® 5 CO2 + 32 H + + 32 e_ 8 O2 + 32 H+ + 32 e_ ® 16 H2O
DG° = _ n FE°, DG° = _ 3387 × 103 J, n = 32
E° =
= 1.0968 volts
Potential of Hydrogen-electrode Taking the case of hydrogen electrode, consisting of H2 gas Chemistry : Basic Elements in equilibrium with H+ ions, the electrode reaction, written as reduction reaction is
H + + e_ ®
H2 (n = 1)
Applying Nernst equation, the electrode potential of the hydrogen electrode is given by
EH+, H2 = E°H+, H2 _ If H2 gas is at 1 atmospheric pressure, aH2 = 1
\ EH+, H2 = E°H+, H2 Replacing the activity of H+ ions by its molar concentration, we have
E°H+, H2 = E°H+, H2 + Since the standard electrode potential of hydrogen electrode is taken as zero. E°M+, H2 = 0 Hence
EH+, H2 =
= 0.0591 log [H+] at 25° C
The following cell is used to measure the mean activity coefficient (Y+) of HCl Pt | H2 (g) | HCl (aq) | AgCl (s) | Ag (s) (i) The cell reaction, (ii) activities of H+ and Cl_ ions express the emf of the cell, (iii) the activities in terms of molality of HCl and the mean ionic activity coefficient, obtain an expression for In Y+ in terms of the emf. The cell is Pt | H2 (g) | HCl (aq) | AgCl (s) | Ag (s) i.e., The Equilibria it consist of hydrogen and silver-silver chloride electrodes in HCl as the electrolyte. The cell reaction will be
H2(g)+AgCl (s) ® Ag (s) + H+ (aH+) + Cl_ (aCl_) and the EMF of the cell is
Ecel = E°Ag|AgCl _
=
... (1)
where a is the activity of HCl as a whole. Now as the activity (a) of HCl at any molality m is related to the mean activity coefficiency Y+ by the expression a = m2Y2+ , substituting this value in eq. (1) we get
Ecel =
Ecel =
or, Ecell +
=
... (2)
All the quantities on the left hand side of equation (2) are known experimentally. Hence to calculate Y±, the value of E°Ag|AgCl is required. To determine E°Ag|AgCl the quantity is plotted against m and the result is extrapolated to m = 0, when m = 0, Y+ = 1, therefore from equation(2).
E°Ag|AgCl =
, i.e.,
equal to the value of the ordinate. The mean ionic activity coefficient of 0.1 molar HCl at 25° C given that the EMF of the cell. Calculated :
Chemistry : Basic Elements H2 (1 atm) | HC1 (a), AgCl (s) | Ag is 0.3524 V at 25° C and that the standard electrode potential of Ag-AgCl is 0.2224 V at 25°C. For the given cell
=
or,
=
Putting T = 298 K, R = 8.314 JK_1 mol_1 and F = 96500 C, this equation becomes Ecell + 0.1183 log m = E°Ag|AgCl _ 0.1183 log Y+ Substituting Ecell = 0.3524 V
E°Ag|AgCl = 0.2224 V and m = 0.l mol kg_1 0.3524 + 0.1183 log 0.1 = 0.2224 _ 0.1183 log Y±
or log Y± =
= _ 0.0989
or Y+ = 0.796 The standard reduction potentials of the electrodes Fe+3 | Fe and Fe+2 | Fe are _ 0.035, _ 0.440 V respectively. It is easy to oxidise Fe to Fe+2 or Fe to Fe+3. Since the standard reduction potentials are given the standard oxidation potentials will be + 0.036 V and + 0.44 V. The standard oxidation potential of Fe | Fe+2 is more positive than that of Fe|Fe+3 electrode. So it is easy to oxidise Fe to Fe+2. (a) Write sell electrode for the following reaction Cu (OH)2 (s) ® Cu2+ + 2OH_ (b) Write the cell reaction for the following cell Pt | H2 | HCl | Hg2Cl2 | Hg | Pt The Equilibria (a) Half-cell reactions are Cu(OH)2 (s) + 2e- ® Cu + 2OH_ Cu2+ + 2e_ ® Cu (s) and electrodes are Cu2 + | Cu and Pt | Cu(OH)2 | OH_ (b) The electrode reactions are 2 H+ + 2 e_ ® H2(g) Hg2Cl2 + 2e_ ® 2Hg (1) + 2Cl_
The cell reaction is Hg2Cl2 (s) + H2 ® 2 Hg (1) + 2H+ + 2Cl_ The standard electrode potentials of the electrodes Cu2+ | Cu and Ag+ | Ag are 0.337 V and 0.7991 V. The concentration of Ag+ in a solution containing 0.06 M of Cu2+ ion such that both the metals can be deposited together ? The activity coefficients are unity and both silver and copper do not dissolve among themselves. Assumed : The individual reactions are Cu2+ + 2e_ ® Cu (s) and Ag+ + e_ ® Ag (s) The electrode potentials given by Nernst equation E (Cu2+ | Cu) = E° _
= 0.337 _ 0.036 = 0.301
= 0.337 _
E(Ag+ |Ag) = 0.7991 _
=
= 8.428
Chemistry : Basic Elements
= 108.428 or, [Ag+] = 10_8.428 = 0.37 × 10_8mol dm_3 The Ksp of AgI by forming proper cell. Give E°, I _ AgI (s)|Ag = _ 0.151 V and E°Ag+|Ag = 0.7991V. Calculated :
The cell can be written as Ag | Ag+; I_ | AgI | Ag At left electrode Ag (s) ® Ag+ + e_ E° = 0.7991V At right electrode AgI (s) + e_ ® Ag (s) + I_, E° = _ 0.151 V AgI (s) ® Ag+ + I_ The standard emf of the cell is E° = E°R _ E°L = _ 0.151 _ 0.7991 = _ 0.9501 V
We know that log Ksp =
= _ 16.1
Ksp = 10_16.1 = 7.94 × 10_17 At 25° C, (¶E°/¶T)p = _1.25 × 10_3 VK_1 and E° = 1.36 V for the cell Pt|H2(g)|HCl(aq)|Cl2|Pt The enthalpy and entropy for the cell reaction calculated. The cell reaction is : H2 (g) + Cl2 (g) = 2HCl (aq)
DS° = nF DS° = (2 mol) (96,485 C mol_1) (_1.25 × 10_3 VK_1) = _ 241JK_1
We also know that DH° = _ nF DH° = (2 mol) (96,485C mol_1) [(1.36 V _ 298.15 K) (1.25 × 10_3 VK_1)]
DH° = _ 191 kJ The Equilibria The fact that two electrodes must have the same potential when equilibrium is attained to calculate K for the familiar reaction. Utilized : Zn + Cu+2
Zn+2 + Cu
If (aZn+2) and (aCu+2) represent the activities of ions when equilibrium is attained, the potentials of Zn | Zn+ 2 and Cu | Cu2+ electrodes, which must then be equal are given by
EZn = = + 0.763 _ 0.0296 log (aZn+2)
ECu = = _ 0.337 _ 0.0296 log (aCu+2) Equating these potentials, it is seen that + 0.763 _ 0.0296 log (aZn+2)e = _ 0.337 _ 0.0296 log (aCu+2)
= K = 1.7 × 1037 Finely divided metallic lead and tin, shaken with solutions containing stannous and plumbous perchlorates until the equilibrium in the reaction Sn (s) + Pb+2
Sn+2 + Pb (s)
reached; the ratio of the concentrations of stannous and plumbous ions at equilibrium i.e., (CSn+2/CPb+2) was found to be 2.98 at 25°C. The standard oxidation potential of the Sn | Sn+2 electrode at 25°C, E°Pb is 0.126 volt at 25°C.
If the ratio of the concentrations is equal to the ratio of the activities in terms of molalities, as is probably the case if the solutions are dilute,
Chemistry : Basic Elements
\K=
= 2.98
The oxidation and reduction processes in the above reaction is Sn (s)
Sn+2 + 2e-
and Pb+2 + 2e_
Pb (s)
so that the complete reaction as written in the question above, takes place in the reversible cell Sn | Sn+2 || Pb+2| Pb for the passage of two faradays i.e., n is 2. In this case E°cell = E°Sn _ E°Pb E°cell = E°Sn _ 0.126
=
= 0.0140
E°Sn = +0.140 volt The standard potential of the Sn | Sn+2 electrode is thus + 0.140 volt at 25° C. Nature of the Electrode Process In an alkali-chlorine cell a saturated (about 6 N) solution of sodium chloride is electrolyzed, at ordinary temperatures, between a steel cathode (hydrogen overvoltage 0.2) and a graphite anode (oxygen overvoltage 0.6 volt; chlorine overvoltage negligible). The nature of the electrode process. Explained : At Cathode : The cations present in the solution are H+ and Na+; the concentrations are 10_7 g (for a
neutral solution) and 6 g ion per liter, respectively. The standard oxidation potentials are 0.0 and + 2.71V respectively; hence the reversible potentials in the given electrolyte are : EH = 0 _ 0.059 log 10_7 = + 0.41 volt ENa = + 2.71 _ 0.059 log 6 = + 2.66 volt The Equilibria Since hydrogen overvoltage is 0.2, the potential for the discharge of hydrogen ions and the evolution of hydrogen gas is + 0.41 + 0.2 = 0.61 volt, this is much below required for Na+ ion discharge, only by raising the potential to 2.66 volt the discharge of Na+ ions become possible. The removal of H+ ions by discharge leaves an excess of OH_ ions in the solution and this accounts for the formation of NaOH. At Anode : The anions present are OH_ and Cl_; the concentrations being 10_7 and 6 g ion per litre as for H+ and Na+ respectively. The standard oxidation potentials of O2 and C12 are _ 0.40 and _1.36 respectively and hence the reversible potentials in the given electrolyte are EO = _ 0.40 + 0.059 log 10_7 = _ 0.81 volt ECl = _ 1.36 + 0.059 log 6 = _1.31 volt Allowing for the overvoltage (0.6 volt) the oxygen evolution potential resulting from the discharge of OH_ ions is _ 0.81 _ 0.6 = _ 1.41 volt and hence discharge of chloride ions and the formation of chlorine gas, will take place in preference. By increasing the anode potential, oxygen evolution would tend to occur. Four types of fuel cells. They are superior for : Depending on the kind of fuel used, the types of fuel cells are— 1. Hydrogen-oxygen fuel cell 2. Hydrocarbon-oxygen fuel cell 3. Carbon monoxide-oxygen fuel cell 4. Solid coal-oxygen fuel cell. Superiority of Fuel Cells :
1. They possess very high efficiency (75.90%). In heat engines efficiency is around 40% or less. 2. The individual cells can be stacked and connected in series to generate higher voltage.
Chemistry : Basic Elements 3. They are also very light. 4. The fuel cells do not create pollution problems. 5. These cells play an important role in manned space flights. The temperature dependence of the emf of an electrochemical cell can often be written in the form : E = (a + bT + cT2 + dT3) volt where a, b, c and d are constants. A certain commercially suitable battery was found to have a = 1.19237, b = _ 1.537 × 10_4, c = 2.73 × 10_8 and d = 1.78 × 10_11. DG, DH and DS for this cell calculated at 27° C if n = 3. DG = _ n EF, where n = 3, F = 96,487 coulombs is equal to 23.06 kcal volt_1 equiv_1 DG = _ 3(23.06) [1.19237 _ (1.537 × 10_4) (300) + (2.73 × 10_8)(300)2 + (1.78 × 10_4)(300)3] = _ 79.52 k cal
DS = nF
= nF (0 + b + 2cT + 3dT2)
DS = 3 (23.06) [_ 1.537 × 10_4 + 2(2.73 × 10_8)T] + 3 (1.78 × 10_11) T2 DS = _ 9.17cal deg_1
DH = DG + TDS DH = _79.52 + 300(_ 9.17 × 10_3) DH = _ 82.27 k cal A voltaic cell is constructed using Al and Al+3 in one half cell and Ag and Ag+ in the other half-cell (a) What total reaction will occur ? (b) What half reaction will occur at each electrode? (c) What is the anode and which is cathode ? The Equilibria (d) How many volts will the cell produce if [Al+3] and [Ag+] are 1.0 M : (a) The reduction potential for Ag is positive, relative to hydrogen and the reduction potential for aluminium is negative. Therefore silver will be reduced and Al will be oxidized. The balanced equation for the reaction is Al + 3Ag+
Al3+ + 3Ag
(b) The half reactions are Al ® Al3+ + 3e_ 3Ag+ + 3e_ ® Ag (c) Since the oxidation occurs at the Al electrode this is called the anode. Reduction occurs at the Ag electrode and is called the cathode. (d) The voltage or potential difference, produced by the cell is just the total difference between the standard potentials for Al and Ag. The standard reduction potential for Al is _1.66 V and for Ag it is + 0.80 V. The total difference is 2.46. This is the voltage produced by the cell. With the help of electrochemical series. Which substance can be used to oxidize fluorides to fluorine. Shown : From the values of standard potentials in electrochemical series
F2 + 2e_
2F_ E° = 2.87 V
This half cell has the highest reduction potential. This implies that fluorides cannot be oxidized chemically by any substance listed in electrochemical series. They can be oxidized only electrolytically. Tin is (II) stable towards disproportionation in non-complexing media ? Given E°Sn+2Sn = _ 0.15 V,
Chemistry : Basic Elements E°Sn+4,Sn+2 = 0.15 V The disproportionation reaction is 2 Sn+2(aq)
Sn+4 (aq) + Sn (s)
which results from the following half-reactions (i) Sn+2 (aq) + 2e_ (ii) Sn+2(aq)
Sn (s) E°el = _ 0.15 V
Sn+4 (aq) + 2e_ E°el = _ 0.15 V
Adding (i) and (ii) 2Sn+2 (aq)
Sn+4 (aq) + Sn (s) E° = _ 0.30 V
Since E° is negative, the disproportionation reaction is not spontaneous. Hence tin (II) is stable. The galvanic cell for each of the following reactions and write down the corresponding expression for the cell potential. Constructed : (1) Zn(s) + H2SO4 (aq) (2) Fe(s) + Cl2(g)
ZnSO4 (aq) + H2(g)
FeCl2(aq)
(1) Zn is oxidized to Zn2+ and H+ is reduced to H2. Thus we have RHC, 2H+(aq) + 2e_
H2(g) ... (1)
LHC, Zn2+ (aq) + 2e_
Zn (s) ... (2)
Subtracting eq. (2) from eq. (1) we have Zn (s) + 2H + (aq)
H2(g) + Zn2+ (aq)
The cell would be Zn (s) | Zn2 + (aq) || H+ (aq) | H2 (g) | Pt and the cell potential will be given by
Ecell = where E°cell = E°(H+,H2 | Pt) _ E° (Zn2+ |Zn) The Equilibria = _E°(Zn2+ |Zn) (2) Fe (s) + Cl2(g)
FeCl2 (aq)
Fe is oxidized to Fe2+ and Cl2 is reduced to Cl_. Thus we have RHC, Cl2 (g) + 2e_ LHC, Fe2+ (aq) + 2e_
2 Cl_ (aq) ... (1) Fe (s) ... (2)
Subtracting eq. (2) from eq. (1), we have Fe (s) + Cl2 (g)
Fe2+(aq) + 2Cl_ (aq)
The cell would be Fe | FeCl2 (aq) | Cl2 (g) | Pt and the cell potential is given by
Ecell = with E°cell = E° (Cl_ |Cl2| Pt) _ E° (Fe2+ | Fe) In acid solution the following are true under the standard conditions (a) H2S will react with oxygen to give H2O and Sulphur. (b) H2S will not react in the corresponding reaction with Selenium and Tellurium. (c) H2Se will react with Sulphur giving to H2S and Se but not react with Tellurium. The hydrides H2O, H2S, H2Se and H2Te in order of their tendency to lose electrons to form the corresponding elements. Arranged.
(a) The reaction is H2S
O2 ® H2O + S
The two partial reactions are 2H+ + H2S
O2 + 2e_
H2O (Reduction)
2H+ + S + 2e_
Chemistry : Basic Elements (Oxidation) A cell corresponding to the given reaction can be constructed with the oxidation reaction at anode and reduction reaction at cathode. The cell potential would be E°cell = E°RHC _ E°LHC = E° (H+ | O2 | Pt) _ E° (H+ | H2S | S) Since the cell reaction is spontaneous it follows that
E° (H+ |O2| Pt) > E° (H+ | H2S | S) (b) Similarly we have E°(H+|H2S|S) > E° (H+ | H2Se | Se) E° (H+ | H2S | S) > E° (H+ | H2Te | Te) (c) Similarly we have E° (H+ | H2S | S) > E° (H+ | H2Se | Se) E° (H+ | H2Se | Se) > E° (H+ | H2Te | Te) Arranging the above reduction potentials in the, decreasing order, we get E° (H+ | O2 | Pt) > E° (H+ | H2S | Pt) > E° (H+ | H2Se | Se) > E°(H+ | H2Te | Te) The above-order implies that the tendency for the reduction reaction
2H+ +
Sn + 2e_
H2X
to take place is greatest for oxygen and least for Te. The tendency for the reverse reaction i.e., the oxidation of hydride to the corresponding element will be greatest for H2Te and least for H2O. Determined at 298 K for the cell Pt | Q·QH2, H+ |1 M KCl, Hg2Cl2 (S) | Hg | Pt (a) its emf when pH = 5.0 (b) the pH when Ecell = 0 The Equilibria
(c) the positive electrode when pH = 7.5. For the given cell, we have Half cell Reaction Potential RHC Ecalomel = E°calomel (1 mol dm_3 calomel)
LHC
Q + H+ + e_
QH2 EQ,QH2, H+|Pt =
(quin hydrone) The emf of the cell is Ecell = E°calomel _ EQ,QH2,H+|Pt = E°calomel _ E°Q,QH2, h+ | Pt + at 298 K
We have Ecell = = (0.280 V _ 0.6996 V) + (0.05913 V) pH = _ 0.4196V + (0.05913V) pH Thus we have (a) Ecell = _ 0.4196 V + (0.05913 V) × 5 = _ 0.1239 V
(b) PH =
=
= 7.1
(c) Ecell = _ 0.4196 V + (0.05913 V) × 7.5 = 0.0239 V The following equations for a cell reaction. Considered
A + B 1 C + D, 2A + 2B 2 2C + 2D) How are E° and K for the two reactions related ?
The E° values are same K2 =
= (K1)2
In aqueous solution Cobalt(III) ion is able to oxidize under Chemistry : Basic Elements Co3+ + e_ ® Co2+, E° = 1.842V. The formation constant for [Co(NH3)6]3 + is 5 × 103 and for [Co(NH3)6]2+, 1 × 105. An aqueous solution of [Co(NH3)6]3+ in 1 M NH3 does not oxidize water. Shown : Co(NH3)26+
6 NH3 + Co2 +,
K = 1 × 10_5 = [(Co(NH3)6)2+] = [Co2+] × 10_5, [Co(NH3)3+6 ] = [Co3+l (5 × 1033) Co3+ + e_ ® Co2+
E=
=
=
=
= The Equilibria
E= This equation is exactly in the form of Nernst equation for the reduction of Co(NH3)3+6 to Co(NH3)2+6 with an E° value of 0.143 V. This small E° value is not sufficiently positive for the overall cell reaction with water to have a positive E value. Proved that for two half reactions having potentials E1 and E2 which are combined to yield a third half-reaction, having a potential E3,
E3 = DG3 = DG1 + DG2 _ E3n3F = -E1n1 F - E2n2F Division by - n3F yields
E3 = Standard reduction potentials of the electrodes Cu (s) | CuCl (s) | Cl_ and Cu (s) | Cu+
are + 0.137 V and + 0.518 V. The arrangement of the cell for the reaction : CuCl ® u + + Cl_ The standard cell EMF. The cell reaction spontaneous. The solubility product of CuCl. Calculated : The electrode reaction for the half cell Cu | CuCl | Cl_ is CuCl + e_ ® Cu + Cl_ (Reduction) At the Cu | Cu+ Cu ® Cu+ + e_ (Oxidation) _____________________________________________________________
Overall cell reaction: CuCl ® Cu+ + Cl_ The concerned cell is
Chemistry : Basic Elements Cu (s) | Cu+ || Cl_ | CuCl (s) | Cu+(s) Standard EMF of such a cell = E°rhs _ E°Ihs = 0.137 _ 0.518 = _ 0.381 V The cell reaction is not spontaneous under standard conditions. The equilibrium constant for the cell reaction, K is given by the relation
In K = _ 0.381 Here n = l
\
= _ 0.381
0.0592 log K = _ 0.381 log K = log K = _ 6.4358 = 7_.5642 K = 3.666 × 10_7 Solubility product = 3.666 × 10_7
Chemical Equilibrium 2 Chemical Equilibrium The average bond energies of S8, H2 and H2S are 264, 436 and 338 kJ mol_l respectively. The enthalpy of the reaction, estimated : S8 + 8H2
8H2S
DH of the above reaction = (Total bond energies of the reactants) _ (Total bond energy of the product) DH = 264 + (8 × 436) _ (8 × 338) kJ = 1048 kJ Ten moles of a gas is allowed to expand from a state, A, at 10 atm and 300 K, to a state, B, at 100 atm and 600 K. If the value of Cp is 6.955 cal deg_1, the entropy change of this process, calculated : S = Cp In T _ R In P + S0
DS = Cp In
_ R In
= 6.955 In
In
Chemistry : Basic Elements = 6.955 × 0.69315 _ 1.987 × 2.3026 = 4.8209 _ 4.5753 = 0.2456 cal K_1 mol_1 For 10 moles of the gas DS = 2.456 cal K_1
Cp is always greater than Cv and find the relation between Cp and Cv. If the volume of the system is kept constant when the heat is added to a system then no work is done by the system. Thus the heat absorbed by the system is used up completely to increase the internal energy of the system. Again if the pressure of the system is kept constant when the heat is supplied to the system; then some work of expansion is also done by the system in addition to the increase in internal energy. Thus if at constant pressure, the temperature of the system is to be raised through the same value as at constant volume, then some extra heat is required for doing the work of expansion. Hence Cp > Cv.
Cv =
and Cp =
Cp _ Cv =
_
But H = E + PV and PV = RT H = E + RT Differentiating this equation w.r.t. T, we get
=
or,
+R
_
=R
C p _ Cv = R Thus, Cp is greater than Cv by the gas constant R, i.e., approximately 2 calories or 8.314 joules. Chemical Equilibrium Joule-Thomson Coefficient Joule Thomson coefficient m may be defined as the temperature change in degrees produced by a drop of one atmospheric pressure when the gas expands under conditions of constant enthalpy. It is expressed as
m= (1) For cooling m will be positive (because dT and dP both will be negative). (2) For heating m will be negative (because dT is positive while dP is negative). (3) If m = 0, the gas gets neither heated up nor cooled on adiabatic expansion because m = 0, only if dT = 0 for any value of dP. Two moles of gas at 1 bar and 298 K are compressed at constant temperature by use of a constant pressure of 5 bar. If the compression is driven by a 1000 kg mass, it will fall in the earths gravitational field to a certain extent : w = _ P2 (V2 _ V1)
= _ P2
=_
= _2 × 8.314 × 298(1 _ 5) = 19820 joules
h=
=
= _ 2.022 m The molar heat capacity at constant volume of O2 (g) is given by C_v = a + bT + gT2 where a = 17.23 JK_1 mol_1, b = 13.61 × 10_3 JK_2 mol_1 and g = 42.55 × 10_7 JK_3mol_1. The change in molar internal energy when oxygen is heated from 298 to 500 K :
Chemistry : Basic Elements DU_ =
= 17.23(500 _ 298) + (13.61 × 10_3/2) (5002 _ 2982) _ (42.55 × 10_7/3) (5003 _ 2983) = 4437 J mole-1 The Joule-Thomson coefficient for a real gas is not zero in the limit of zero pressure : For a real gas the compressibility factor at low pressures can always be represented by the viral equation PV = RT(1 + BP + C¢P2+ ...... ) where B and C¢ are functions of temperature only. Thus,
= R (1 + BP + C¢P2 + .....)
+ The Joule-Thomson coefficient is therefore given by
mjT = Thus, not all the properties of a real gas approach those of an ideal gas when the pressure of the real gas is reduced to zero. Inversion Temperature Every gas has a definite temperature (at a particular pressure) at which m = 0. Below this temperature m is positive and above this temperature m is negative. This temperature (at a particular pressure) at which m = 0, i.e., the gas neither cooled down nor heated upon adiabatic expansion and below which Chemical Equilibrium m is positive adiabatic expansion and above which m is called the inversion temperature.
Deduced relationship m =
.
Enthalpy H is a state function. Taking temperature T and pressure P as the independent variables. H = f (T, P) ... (1)
\ dH =
... (2)
In Joule-Thomson effect, dH = 0
\
= 0 ... (3)
Dividing this equation throughout by dP and imposing the condition of constant enthalpy, we get
\
But
= 0 ... (4)
= Cp, heat capacity at constant pressure and
= m, Joule-Thomson coefficient,
Putting these values in equation (4), we get
+ mCp = 0 or, m = 2Al + 6NaCl is _ 256.8 k cal. The heat of The enthalpy change of the reaction, Al2 Cl6 + Na formation of solid NaCl is _ 98.2 k cal. Heat of formation of solid Al2Cl6, Calculated: The enthalpies of the reactants and products may be written as follows—
Chemistry : Basic Elements Al2Cl6 + 6 Na 2Al + 6 NaCl x 0 0 6(_ 98.2) where x is the enthalpy and hence the heat of formation of Al2Cl6. For the whole reaction DH is _ 256.8 k cal. Consequently _ 256.8 = [0 + 6(_ 98.2)] _ (x + 0) x = _ 332.4 k cal. In case a non-linear triatomic gas obeys the principle of equilibration of energies, the molar heat capacity, calculated: In a triatomic gas, there will be nine degrees of freedom, three translational, three rotational and three vibrational, contribution to the energy by each degree of translational and rotational motion is ½RT per mol, whereas that of vibrational mode is RT per mol of gas. Thus,
E_T =
C_V =
= 6 RT
= 6R = 6 × 8.314 J mol_1 deg_1
= 49.884 J mol_1 deg_1 If the contribution to the internal energy due to vibrational mode is
E_(vib) = then the value of C_v would be :
Ans. E_(vib) =
C_V (vib) =
= Chemical Equilibrium
=
C_V =
The Joule-Thomson coefficient for Van der Waals gas can be written as mj.T = dm6 atm mol_2, b = 0.0318 dm3 mol_1. The value of DH for the isothermal (T = 300 K) compression of 1 mol of oxygen from the pressure of 1 to 200 atm, calculated : Combining equation
mjT =
=
and the given relation we can write
mj.T =
DH =
=
a = 1.36
At constant temperature
DH =
\ DH = = _ 15.46 atm dm3 mol_1 = _ (15.46 at dm3 mol_1) (101.325 dm_3 atm_1J) = _1556.5 J mol_1
Chemistry : Basic Elements Heat of neutralisation of HCN and NaOH is _12.13 kJ mol_1, DH°ionization of HCN : DH° = DH°ioniz + DH°neutr \ _12.13 = DH°ioniz _55.90 DH°ioniz = + 43.77 kJ mol_1 Various properties into Intensive and Extensive variables, listed : Temperature, Heat capacity, Pressure, Dielectric constant, Density, Boiling point, Viscosity, Concentration, Refractive index, Enthalpy, Entropy, Gibbs free energy, Molar enthalpy, Chemical potential, Molality, Volume, Mass, Specific heat. No. of moles. Free energy per mole. Intensive Variables : Temperature, Pressure, Dielectric constant, Density, Boiling point, Viscosity, Concentration, Refractive index, Molar enthalpy. Chemical potential, Molality, Specific heat, Free energy per mole. Extensive Variable : Heat capacity, Enthalpy, Entropy, Gibbs free energy, Volume, Mass, No. of moles.
Matched : (a) Isothermal process (1) when P = constant (b) Adiabatic process (2) when heat capacity of body = constant (c) Isobaric process (3) when dE = dH = 0 (d) Polytropic process (4) when T = constant (e) Quasistatic process (5) when both E and H = constant (f) Isochoric process (6) when q = constant (g) Cyclic process (7) when V = constant (a) 4, (b) 6, (c) 1, (d) 2, (e) 3, (f) 7, (g) 5. A psychologist develops a theory that states of mind (anger, suspicion, greed) are thermodynamic states of a region of the brain that can be considered a system. Chemical Equilibrium The brain is a thermodynamic system. The mind is a thermodynamic single valued function because it depends only on the states of the system. It means that mind is a function of anger, suspicion, greed. Mathematically Mind = f (anger, suspicion, greed) Hence (1) Mind is a state function. (2) Differential of mind is an exact differential. Shown mathematically that the magnitude of the work involved in a reversible expansion of an ideal gas from volume V1 to V2 is larger than the corresponding work involved in an irreversible expansion against a constant pressure of p2.
Wrev = nRT In
= nRT In
Expanding the logarithmic term, we have
Wrev =
=
+ higher terms
= p1 (V2 _ V1) + higher terms and Wirrev = p2 (V2 _ V1) Therefore Wrev _ Wirrev = {p1 (V2 _ V1) + higher terms} _ p2 (V2 _ V1) = (V2 _ V1) (p1 _ p2) + higher terms Since, in expansion V2 > V1 and p1 > p2, therefore Wrev _ Wirrev = positive i.e., the magnitude of the work involved in a reversible expansion is larger than the corresponding work involved in an irreversible expansion.
Chemistry : Basic Elements The enthalpy of fusion, DHf , of ice at _ 10°C, from the following data, calculated DHf = 6.02 kJ mol_1 at 0°C,
Cp(ice) = 37.66 JK_1 mol_1, Cp(Water) = 75.31 JK_1 mol_1. Using the Kirchoffs equation we have
DHf, 263 =
= 6020 J mol_1 + = 6020 Jmol_1
+ = 6020 J mol_1 + 37.65 JK_1 mol_1 (263 K _ 273 K) J mol_1 = 6020 J mol_1 _ 376.5 J mol_1 = 5643.5 J mol_1 = 5.64 kJ mol_1
Shown that Cp _ Cv = compressibility :
, where a and b signify the usual coefficient of expansion and
Since V = f (P, T)
\ dV = If volume is constant (i.e.,dV = 0), then
0=
or,
=
Chemical Equilibrium Dividing by dT,
=_
or,
=
... (1)
But we know that
Cp _ CV =
(Maxwells relation) ... (2)
Combining equations (1) and (2), we get
C p _ Cv =
Multiplying numerator and denominator by
, we get
Cp _ Cv =
... (3)
But coefficient of expansion,
a=
a2 =
... (4)
and coefficient of compressibility,
b=
... (5)
Chemistry : Basic Elements Combining equations (3), (4) and (5) we get
C p _ Cv =
or Cp _ CV =
Chemical potential is defined as mi =
We know that mi =
= G_i
i ¹ j. Shown that
= _S_i
Differentiating the above equation with respect to temperature at constant pressure and composition of other components
= We know that dG = VdP _ SdT But at constant P and composition of system
= _S_i Differentiating this equation w.r.t. ni, maintaining other variables constant
=
= _S_i
Since dG is a perfect differential, hence
= Chemical Equilibrium
\
= _S_i
This equation gives the variation of the chemical potential of a constituent in a mixture with respect to temperature at constant pressure and composition of the system. S_i is the partial molal entropy of ith component of the mixture. The entropy of mixing of 1 mole of N2 and 2 moles of O2, assuming the gases to be ideal calculated.
The result in S.I. units expressed :
DSmix (per mole) =
= = 2.303 × 8.314 (0.1590 + 0.1174) JK_1 = 5.29 JK_1 \ DSmix for 3 moles of the mixture = 3 × 5.29 = 15.87 JK_1 DH, DS and DG for the vapourisation of 2 moles of benzene at its boiling point (80.2°C) calculated. Latent heat of vapourisation = 101 cals. gm_1 : DH = 101 cals gm_1
cals gm_1
DS =
DG = 0, since DGT, P = 0, for the equilibrium between liquid and vapour. For 2 moles, DH = 101 × 2 × 78 cals = 15,750 cals K-1
For 2 moles, DS =
= 44.6 cals K_1
Chemistry : Basic Elements For a certain reaction, DG = _a + bT In T where `a' and `b' are constants. DH as a function of temperature, expressed:
=
=
=
=
\ _ DH = a + bT
Proved that for an ideal gas
=0=
By definition H = E + PV = (E + RT) for 1 mole of an ideal gas
\
=
=
=
_P=
=
=0
=0
The molar entropy change for the reversible process, calculated H2O(I)
H2O(g), DH = 2269.4 Jg_1
1 atm, 100° C 1 atm, 100° C This is an equilibrium process \ DG = 0 = DH _ TDS
\ DS =
=
= 109.5 J deg_1 mole_1
Shown that
=
The Gibbs Helmholtz equation is DG = Chemical Equilibrium It may be rearranged to give
= _DH Dividing by T2, we have
=
\
=
For a certain reaction the equilibrium constant does not change with temperature. The value of DH° for the reaction is:
= It is given that K does not change with temperature.
\
=0
\ DH° is zero. If an exact differential, dz is given in the form dz = M dx + N dy, then by the rules of partial
differentiation i.e.,
=
. Four Maxwell relations.
Using the above technique for the equation
dE = TdS _ pdV
we get
=
... (1)
Chemistry : Basic Elements Similarly from dH = TdS + VdP, we get
=
... (2)
dA = _pdV _ SdT gives
=
... (3)
and dG = VdP _ SdT gives
=
... (4)
These four equations are Maxwell relations. A gas obeying the equation of state P(V _ B) = RT undergoes a change from the initial state T1, V1 to a final state T2, V2. An expression derived for the entropy change of this gas. The variation of heat capacity at constant volume is given by Cv = a + bT + cT2.
dS =
and dE =
give dS =
... (1)
Thermodynamic equation of state
=
, when applied to equation (1) yields
dS =
... (2)
Chemical Equilibrium For the given gas, P = RT/(V _ B)
and
=
... (3)
Substituting equation (3) into (2) we get
... (4)
dS = ...
CV = a + bT + cT2
\
=
... (5)
dS = On integrating for proper limits, we get
=
\ DS = One mole of toluene is vapourized at its boiling point 111° C. The heat of vapourization at this temperature is 363.3 Jg_1. The maximum work done against 1 atm, q, DH, DE, DG and DS for the vapourization of 1 mole toluene calculated : C6H5CH3 (liquid) ® C6H5CH3 (vapour) The process is isothermal, isobaric and reversible T = 111+273 = 384 K _ W = PDV = P(VV_ V1) » PVV = RT (... VV>> VI) = (8.314 JK_1 mol_1) × (384 K) = 3193 J mol_1
Chemistry : Basic Elements qp = DHvap = (363.3 Jg_1) × (92 g mol_1) = 33423.6 J mol_1 DE = DH _ PDV = (33424 J mol_1 _ 319.3 J mol_1 = 30231 J mol_1
DG = òV dP = nRT In
= nRT In
=0
DS =
=
=
= 87.04 JK_1 mol_1 Percentage T1 is of T2 for a 10 per cent efficiency of a heat engine. Equating the efficiency equal to 0.1 we have
h=
\
=
= 0.1
× 100 = (1 _ 0.1) × 100 = 0.9 × 100 = 90
T1 = 90 per cent of T2 Without doing calculations, the signs of Dp, DV, DT, DE, DH and DS for one mole of an ideal gas calculated taken through each of the following four steps of a Carnot cycle. Cp and CV as constants assumed : Step (1) Isothermal reversible expansion p1, V1, Tl ® p2, V2, T2 Step (2) Adiabatic reversible expansion p2, V2, T1 ® p3, V3, T2 Step (3) Isothermal reversible compression p3, V3, T2 ® p4, V4, T2 Step (4) Adiabatic reversible compression p4, V4, T2 ® p1, V1, T1 Chemical Equilibrium Step (1): DV = +ve as expansion takes place
DT = 0 as process is isothermal Dp = _ve as increase in volume at constant temperature will decrease the pressure (Boyle's law) DE = 0 since for an ideal gas E = f (T) & DT = 0 DH = 0 DH = DE + D(pV) and DE = 0 and D(pV) = D(RT) = 0
DS = +ve since DS = nR In V2 > V1
and
Step (2): DV = +ve as expansion takes place DS = 0 as adiabatic process is reversible DE = _ ve since DE = w & w is _ ve (expansion process) DT = _ ve since DE = Cv, m (DT) DH = _ ve since DH = DE + D(pV) = DE + RDT DP = _ ve Step (3) : Here the signs will be just opposite to those listed in step (1). Step (4) : Here the signs will be just opposite to those listed in step (2). 1 mole of helium is mixed with 2 moles of Neon, both at the same temperature and pressure. Calculate DS for this process if the total volume remains constant calculated : VNe = 2 VHe Now the volume of the gas after mixing will be Vtotal = VNe + VHe = 2VHe + VHe = 3VHe
Chemistry : Basic Elements Each of the two gases will suffer entropy change due to the volume change. Thus
DSHe =
= (1 mole)
(8.314 JK_1 mol_1) × 2.303 log = 9.136JK_1
DSNe =
= (2 mole)
(8.314 JK_1 mol_1) × 2.303 1og = 6.743 JK_1 Thus DSmix = DSHe + DSNe = 9.136 JK_1 + 6.743 JK_1 = 15.879 JK_1 Shown that for an ideal gas undergoing isothermal reversible expansion, DG = DA. The function G and A are given by G = H _ TS A = E _ TS Function, H = E + PV \ G = (E + PV) _ TS = (E _ TS) + PV or, G = A + pV Thus, for a process
DG = DA + D(pV) Now for an isothermal expansion of an ideal gas, we have D(pV) = D (nRT) = 0, as DT = 0 so that DG = DA The vapour pressure of liquid Hg at 433 K is 4.19 mm Hg. The free energy change accompanying the expansion of one Chemical Equilibrium mole of Hg vapour in equilibrium with liquid at 433 K to a pressure of 1 atm at the same temperature assuming the vapour behaves like an ideal monoatomic gas, calculated: The transformation of Hg (1) to Hg (g) at temperature 433 K and at the corresponding equilibrium pressure 4.19 mm Hg will be reversible in nature. Thus DG for this process will be equal to zero. But there will occur a change of free energy when the pressure of Hg (g) is changed from 4.19 mm Hg to 760 mm Hg which can be calculated using the expression DG = nRT In p2/p1 Here p2 = 760 mm Hg, p1 = 4.19 mm Hg
\ DG = (1 mole) (8.314) (433 K) 2.303 log = 18725.4 J mole_1 Trouton's rule concerning entropy of vapourization. Trouton's rule states that for most normal liquids the entropy of vaporization per mole 21 e.u. By a normal liquid, we mean which is not associated. In general association in the liquid state may be expected when intermolecular forces of a dominant type operate. Dipole moments, hydrogen bonding etc., lead to this situation. Abnormally high boiling points are a consequence of molecular association in the liquid state. An ideal gas is carried through a Carnot cycle. The diagram of the cycle drawn using T and H as variables. Carnot cycle consist of four steps—
(1) Isothermal and reversible expansion T = constant, DH = 0 (2) Reversible adiabatic expansion DH = _ve Temperature decreases, (3) Reversible isothermal compression T = constant, DH = 0 (4) Reversible adiabatic compression T=positive, DH= positive
Chemistry : Basic Elements
Therefore Shown that entropy change in a binary mixture for isothermal process is maximum when x1 = x2 = 1/2. DSsyst = _ (Rx1 In x1 + Rx2 In x2) (For binary mixture) This can be rewritten as DSsyst = _[R (1 _ x2) In (1 _ x2) + Rx2 In x2] For maximum condition
=0
=
i.e.,
=0
= 0 = In 1
= 1, x2 =
Thirdly xl = 1 _ x2 = 1 _
=
The entropy of CO crystals at absolute zero, calculated : The crystal of CO are the mixture of two kinds of molecules. It is assumed that mole fraction of each kind is 1/2 the entropy of the crystal of CO. Chemical Equilibrium
\ S_0 = _ R Sxi In xi = = (8.314 JK-1 mol-1) In 2 = 5.76 JK-1 mol-1 = 8.76 EU
Using equation, m = m0 + RT In f shown that From Gibbs-Duhem relation i.e., n1dm1 = _ n2dm2 or, x1dm1 = _ x2 dm2 At constant temperature and pressure equation
=
m = m° + RT In f becomes dm = RTd In f So x1 RT d In f1 = _ x2 RT d In f2 Divide both sides by RT dx1 at constant T and P
=Now x1 = 1 _ x2, i.e., dx1 = _ dx2 The above relation becomes
= 0.5 mole of an acid (R—COOH) and 2.0 moles of an alcohol (R¢OH) were mixed and allowed to attain equilibrium at 25°C. The amount of the ester at equilibrium if DG° for the reaction calculated : R—COOH + R¢OH ® R—COOR¢ + H2O is 440 cal.
Chemistry : Basic Elements _ RT In K = DG°
\ log K =
= _ 0.3226 = 1_.6774
K = 4.758 × 10_1 0.5 mole 2 moles 0 mole 0 mole R — COOH + R¢OH
R — COOR¢ + H2O
(0.5 _ x) (2 _ x) x x
K=
= 0.4758
0.5242 x2 + 1.1895x _ 0.4758 = 0 On solving quadratic equation, the admissible value of x = 0.35. Using the Gibbs-Helmholtz equation obtained Clapeyron equation for the equilibrium solid
liquid.
Clapeyron equation deals with the change in the equilibrium pressure with a change in the equilibrium temperature. Suppose T and P are temperature and pressure at equilibrium. At equilibrium, free energies of the two phases are equal. Gsolid = Gliquid (at temperature T and pressure P) ...(1) Suppose the above equilibrium is disturbed by changing the equilibrium pressure by dP, then the equilibrium temperature should be changed by an amount dT, so that equilibrium is attained once again. Then Gsolid + dGsolid = Gliquid + dGliquid ... (2) (at temperature T + dT and pressure P + dP) From equations (1) and (2) dGsolid = dGliquid ... (3) The Gibbs-Helmholtz relation for any phase is dG = VdP _ SdT Chemical Equilibrium Combining this with equation (3) we have
Vsolid dP _ Ssolid dT = Vliquid dP _ Sliquid dT dT (Sliquid _ Ssolid) = dP (Vliquid _ Vsolid)
= , where DS and DV correspond to the forward step (of the concerned equilibrium) namely Solid ® Liquid. In the present case, DS is the entropy of fusion per mole and DV is the volume change per mole.
DS = where Diffusion is molar enthalpy of fusion and Tmelt is the melting' point of the solid.
Thus,
=
From the two laws of thermodynamics, Maxwell's relations, written down : (1) When T and V are independent variables
= (2) When T and P are independent variables
= (3) When S and V are independent variables
= (4) When S and P are independent variables
=
Chemistry : Basic Elements (5) When P and V are independent variables
=1 (6) When T and S are independent variables
=1
=
Using Maxwells relation shown that
=
We know that,
Cp =
=
Differentiating with respect to P at constant T
=
... (1)
From Maxwells relation
Differentiating with respect to T
=
=
... (2)
Combining equations (1) and (2)
=
The EMF of the cell Ag | AgCl (s), KCl (aq); Hg2Cl2 (s) | Hg involving the cell reaction Ag + Hg2Cl2 (s) = AgCl (s) + Hg is + 0.0455 volts at 25° C. The temperature coefficient is Chemical Equilibrium + 3.38 × 10_4 volt/degree. The heat content change at the given temperature. What is the heat of the reaction taking place in the cell calculated :
E=
or, DH = It is given that,
= 3.38 × 10_4 volt/degree E = 0.0455 volts, n = 1, T = 25 + 273 = 298 K, F = 96500 coulombs So DH = 1 × 96500 × 298 × 3.38 × 10_4 _ (1 × 96,500 × 0.0455) = 1,273 cals \ Heat content change = 1,273 cals Heat of reaction = _1,273 cals Using the third law of Thermodynamics shown that
(1) limT ® 0 (¶P/¶T)V = 0 (2) limT ® 0(¶V/¶T)P = 0 (1) We know that A = E _ TS and dA = _ SdT _ PdV and Euler's theorem for exact differentials gives (¶S/¶V)T = (¶P/¶T)V But from third law of thermodynamics for (¶S/¶V)T = 0 as T ® 0
Hence
(¶P/¶T) V = 0
Chemistry : Basic Elements (2) Again dG = _ SdT + VdP Euler theorem gives _ (¶S/¶P)T = (¶V/¶T)P But according to third law of thermodynamics (¶S/¶P)T = 0 as T ® 0
\
(¶V/¶T)p = 0
Shown variation of chemical potential with pressure :
We know that
= mi
Differentiating it with respect to pressure, we get
=
... (1)
On differentiating equation
= V, with respect to ni, we get
=
= V_i ... (2)
Combining equations (1) and (2) we get
= V_i where V_i is partial molar volume of the ith constituent. The pressure on liquid benzene at 25°C is raised from 1 to 11 bar. The change in molar enthalpy is. For liquid benzene under these conditions a = 1.237 × 10_1 K_1 and the density is 0.879 g cm_3
= V_ (1 _ aT) Chemical Equilibrium
= = 5.61 × 10_5 m3 mol_1
DH_ = (5.61 × 10_5 m3 mol_1) DP = (5.61 × 10_5 m3 mol_1 (10 bar) (105 Pa bar_1) = 56.0 J mol_1 Proved that in a reversible process net entropy change for the system and surroundings is zero. If qrev is the heat absorbed by the system reversibly, then the heat lost by the surroundings will also be qrev. If the process takes place isothermally at the absolute temperature T, then
(1) Entropy change of the system is DSsystem =
(2) Entropy change of the surroundings is DSsurr = _ Hence the total entropy change for the combined system and the surroundings will be
DSsystem + DSsurr =
_
=0
The terms fugacity, activity and activity coefficient defined: Fugacity may be defined as a substitute for pressure to explain the behaviour of real gas and activity may be defined as the substitute for the concentration to explain the behaviour of a non-ideal solution. Thus in a mixture of real gases, the chemical potential mi of any constituent i is given by mi = m0i + RT In fi ... (1) where fi is the fugacity of the constituent i in the mixture. Similarly, in a non-ideal solution the chemical potential of any component; is given by
Chemistry : Basic Elements mi = m0i + RT In ai ... (2)
where ai is the activity of component i in the solution. Thus to explain the behaviour or real gases, the pressure has to be corrected by multiplying with a suitable factor to get the fugacity i.e., f = Y × P ... (3) where Y is called the activity coefficient of the gas. From equation (3), we have Y = f/P ... (4) Hence activity coefficient of a gas may be defined as the ratio of the fugacity of the gas to the pressure of the gas in the same state. For ideal gas Y = 1, for real gas Y is always less than 1. Similarly in a non -ideal solution the concentration has to be corrected to give the activity as a = Y × C ... (5) \ Y = a/C Hence activity coefficient of any component in the solution may be defined as the activity of that component in the solution to the concentration of the same component in the solution. For an ideal solution (i.e., for very dilute solution), Y = 1 (so that a = C). For other solutions, greater is the departure in the value of Y from 1 more non-ideal is the solution. Nernst's Heat Theorem The Gibbs Helmholtz equation is
DG _ DH = T
If
is a finite quantity, then
at T = 0° K, we have
Chemical Equilibrium DG _ DH = 0 In other words, limT®0(DG) = limT®0 (DH)
This means that at absolute zero, DG and DH are equal. Nernst showed experimentally that
decreases decreases with decrease of temperature. He suggested that in a condensed system with lowering of temperature asymptotically and becomes equal to zero at absolute zero (0° K). This statement is known as Nernst's heat theorem. Mathematically this theorem is expressed as
=
=0
Residual Entropy The third law of thermodynamics says that the entropy of pure, perfect crystalline substance is zero at absolute zero. But, in actual practice, it has been found that certain chemical reactions between crystalline substance, do not have DS = 0 at 0°K, which indicates that exceptions to third law exist. Such exceptional reactions involve either ice, CO, N2O or H2. It means that in the crystalline state these substances do not have some definite value of entropy even at absolute zero. This entropy is known as Residual Entropy. At 0°K the residual entropies of some crystalline substances are Ice—3.3 JK_1, Solid CO—5.8 JK_1 Solid N2O—5.8 JK_1, Solid H2—6.2 JK_1 m _ m0 for 1 mole of an ideal gas at 25°C and 400 mm of pressure calculated : For an ideal gas mi = mi0 + RT In pi (pi = partial pressure if ith gas in atmosphere)
Chemistry : Basic Elements \ mi _ mi0 = RT In pi = 2.303 RT log pi = 2.303 × 8.314 × 298 log (400/760) = _ 1590 J The criteria for phase equilibrium for multicomponent system. Let us consider the simplest case consisting of only two phases A and B. Suppose a number of components (C1, C2, ..., C3) are distributed between them. (See Figure)
Further suppose that at constant pressure and temperature a small amount dn1 of component 1 is transferred from phase A tophase B. If (m1)A and (m2)B represent the chemical potentials of component 1 in phases A and B respectively, we have Decrease in the free energy of component 1 in phase A = (m1)A dn1 Increase in the free energy of component 1 in phase B = (m1)B dn1 Since the system is a closed one, (DG)T, P = 0 \ _ (m1)A dn1 + (m1)B dn1 = 0 (minus sign to represent the decrease) or (m1)A = (m1)B It means for a system consisting of two phases in equilibrium, the chemical potential of any given
component is same in both the phases. Similarly if a system consists of three phases, A, B and C in equilibrium, we will have (m1)A = (m2)B = (m1)C Chemical Equilibrium \ For a multiphase equilibria containing a number of components distributed among them, the chemical potential of any component is the same in all the phases. The criteria for two-phase equilibrium for one component system. For a system consisting of one pure substance only which may exist in two different phases in equilibrium the criterion is obtained in a simple manner in terms of free energy. In such a case, if a certain amount of the substance is transferred from one phase to the other, the molar free energy of one phase decreases while that of the other phase increases by an equal amount. Hence the net result is that there is no change in free energy. i.e., DG = 0 This implies that if at equilibrium, G1 is the molar free energy of phase I and G2 that of the phase II, we will have G2 _ G1 = 0 or, G1 = G2 Thus whenever two phases of the same single substance (one component system) are in equilibrium, at a given temperature and pressure, the molar free energy is the same for each phase. An iceberg is floating in the lake. If one considers the lake, iceberg and atmosphere as a one system; number of phases are: Iceberg is a solid form of water, lake is liquid water and atmosphere contains water vapours and other gases. So there are three phases (solid, liquid and vapour). The number of phases, components and degree of freedom in the following systems calculated: (1) A mixture of nitrogen and oxygen gases contained in a vessel.
(2) Rhombic sulphur in equilibrium with monoclinic sulphur.
Chemistry : Basic Elements (3) N2O4 (g) 2NO2 (g) (4) Solid carbon in equilibrium with gaseous CO, CO2 and O2 at 100° C. Is it possible to have a quadruple point in a phase diagram for one component system? (1) No. of phases = 1 (Gases form a single phase) No. of component = 2 Degree of freedom is given by F=C_P+2=2_l+2=3 (2) No. of phases = 2 No. of components = 2 Degrees of freedom is F = C _ P + 2 = 2 _ 2 + 2 = 2 (3) No. of phases = 1 (Gases mix uniformly to give one phase) No. of components = 1 Degree of freedom, F = C _ P + 2 = 1 _ 1 + 2 = 2 (4) No. of phases = 2 (1 for solid carbon and 1 for gases) No. of components = 2 Degree of freedom, F = C _ P + 2 = 2 _ 2 + 2 = 2 Quadruple point is a point where four different phases meet. Thus P = 4, substituting the values in the phase rule equation F=C_P+2
or, F = 1 - 4 + 2 or, F = _1 This is absurd, therefore one component system cannot have a quadruple point in the phase diagram. A substance `A' exists in two solid modification A1 and A2 as well as liquid and vapour under one atmospheric pressure. Chemical Equilibrium A1 is stable at lower temperature while A2 at higher temperature and the transition from A1 to A2 is accompanied by increase in molar volume. Both A1 and A2 are denser than the liquid phase. If no metastable equilibria are observed, a P-T phase diagram for the system drawn. Each area of the P-T diagram labelled : For the transition, A1 ® A2
=
=
But as per problem SA2 > SAl and VA2 > VAl, hence SA2 _ SAl = + ve, VA2 _ VAl = +ve
\
= +ve
That is, the line representing the equilibrium between A1 and A2 will slope away from the pressure axis. Again for the transformations A1 (s) ® liquid and A2 (s) ® liquid
both DS and DV are positive, therefore for both transformations
= + ve This means that the lines indicating the equilibria A1 (s)
Liquid
and A2 (s)
Liquid
will slope away from the pressure axis. On the basis of above information, the phase diagram (P-T) will be
Chemistry : Basic Elements
A pure substance A exists in three modifications A1, A2 and A3. At the triple point Sa1 > Sa2 > Sa3 and Va1 > Va2 > Va3, S and V are entropy and volume respectively are there. A suitable P-T diagram on the basis of informations given above and also label the region, line and points constructed : The entropy and volume of A1-phase are greater that the two other phases A2 and A3. Therefore phase A1 must be stable in the region of low pressure and high temperature. The entropy and volume of phase A3 are the lowest among all the three phases, hence A3 must be stable in the region of high pressure and low temperature. The phase A2 must be stable for intermediate case.
Now,
=
Hence for the following equilibria For A1
A2,
DS = Sa2 _ Sa1 = _ve DV = Va2 _ Va1 = _ve
\
=
=
= + ve
For A2
A3
Chemical Equilibrium
=
=
= + ve
For A1
A3
=
=
= + ve
The +ve values of DP/DT means that the lines representing the above equilibria will slope away from the pressure axis. On the basis of above informations, the phase diagram (P-T) is
A schematic phase diagram for CO2 drawn :
The phase diagram for CO2 shows that there is equilibrium between solid and gas at 1 bar andat _ 78° C. Liquid CO2 is produced only above 5.1 bar. That is why solid CO2 sublimes at atmospheric pressure.
Chemistry : Basic Elements The equation for the molar enthalpy of an ideal solution derived :
=
=
Thus
=
Thus, H_ = SxiH_i°
=
Proved thermodynamically that at a given temperature, the vapour pressure of metastable phase is greater than that of the stable phase : Since the transformation Metastable phase ® Stable phase is a spontaneous process, DG = ms _ mms = _ ve where ms and mms are chemical potentials of stable and metastable phases. Since in both stable and metastable region, the substance is in equilibrium with its vapour, we will have ms = mvs ± mv0 + RT In ps mms = mvm = mv0 + RT In pms where mvs and mvm are the chemical potentials of the vapour phase in stable and metastable regions respectively, hence
DG = ms _ mms = RT In Since DG is negative, it follows that pms > ps that is vapour pressure of metastable phase is larger than that of the stable phase at the same temperature. Chemical Equilibrium Matched the following :
(a)_(3), (b)_(1), (c)_(2). The phase diagram of a one component system having solid phases (a, b) are given below. Starting from point A. The phases observed if the system is heated at constant pressure. The degrees of freedom at A and B calculated :
(1) Cooling curve of a liquid mixture of a two component system (2) Supercooling phenomenon (3) Cooling curve of a pure component.
Chemistry : Basic Elements It is a one component system consisting of four phases i.e., solid a and b, liquid and vapour phase. Curve AB : It is sublimation or vapour pressure curve of solid a. Along this curve, two phases a and vapour are in equilibrium. F=C_P+2=1_2+2=1 System is univariant. Point B : At this point a changes to b. It is known as transition temperature. Thus there are three phases [Two solids (a & b) and one vapour) F=C_P+2=1_3+2=0 Thus B is a triple point. Curve BC : It is vapour pressure or sublimation curve for b. On this curve two phases b and vapour are in equilibrium. F=C_P+2=1_2+2=1 Point C : It represents the melting point of b. This is another triple point. At this point three phases (b, vapour and liquid) are in equilibrium. Curve CF : Along this curve liquid and vapour are in equilibrium. F=1_2+2=1 Curve BD : Both phases a and b are solids. Curve CD : b and liquid are in equilibrium. This also has one degree of freedom. Distinguished between triple point and the freezing point of a pure substance. The triple point is the point where solid, liquid and vapour are in equilibrium with one another, with no other substance present. The freezing point is the point at which solid and liquid are in equilibrium under 1 atm total pressure. There must be some other substance present to achieve 1 atm pressure. Chemical Equilibrium
Proved that the liquid vapour (L-V) equilibrium line in a one component system must always have a positive slope on a pressure-temperature diagram :
=
=
=
Since Sv > Sl and Vv > Vl
\
=
= + ve
Hence proved. The phases present in the numbered areas of figure given below identified :
From the figure congruently melting compound is formed at XB = 0.667 which means that there are two moles of B for every mole of A so that the compound has the empirical formula AB2. Horizontal `tie lines' indicate that in area ¹ 1, Solid A and liquid are in equilibrium, in areas ¹ 2 and ¹ 4 liquid and solid AB2; in area ¹ 3 solid A and solid AB2, in area ¹ 5 solid B and liquid and in area ¹ 6 solid B and solid AB2 are in equilibrium.
Surface Phenomenon 3 Surface Phenomenon Relation for Calculating Contact Angle : That the surface of the drop makes with solid surface. YS = YS . liq + Yliq cos q where YS and Yliq are the surface tension of solid and liquid respectively, YS . liq is the Interfacial tension between the solid and liquid face. Variation of Surface Tension of an Adsorbent : Related to the surface excess concentration of a gas by the relation. Y _ Y0 = a _ bé where a and b are constants. Show that Gibbs adsorption equation when applied to gas adsorption leads to Freundlich adsorption equation. The Gibbs equation when applied to gas adsorption has a form
é=
... (1)
é ® Excess concentration of solute per unit area at the surface.
Chemistry : Basic Elements p ® Pressure of the gas. where the surface excess concentration of the gas at the surface is given by
é=
...(2)
Vm ® Molar volume of the gas. V ® Volume of gas adsorbed per unit mass of solid.
s ® Surface area of the solid per unit of mass. It is given that Y _ Y0 = a _ b é Substituting for é, we get
Y _ Y0 = a _ b
Hence dY = Substituting dY from the above realtion and é from equation (2) in equation (1) we get
=
or,
=
or,
=
= d In p1/n
where 1/n = RT/b. Integrating the above relation In V = In p1/n + In K where In K ® constant of integration. The above expression can be written as In V = In (Kp1/n) or, V = Kp1/n which is required Freundlich adsorption equation.
Surface Phenomenon Matched : (a) Streaming Potential (1) The pressure difference per unit potentialdifference when the flow of matter is zero. (b) Electro osmosis (2) The potential difference per unit pressure difference at zero electric current (c) Electro osmotic pressure (3) The flow of matter per unit electric current when the pressure is uniform. (d) Streaming current (4) The current flow per unit matter flow.at zero potential difference. (a)_(2), (b)_(3), (c)_(1), (d)_(4). Factors which Influence the Absorption of Gases by Solids : (1) Nature and surface area of the adsorbent. (2) Nature of gas being adsorbed. (3) Temperature. (4) Pressure. If a molecule dissociates on being adsorbed, the process is referred to as dissociative adsorption. The Langmuir adsorption isotherm for dissociative adsorption, derived : When dissociation occurs on the surface two sites are required and so the probability of sticking is proportional to the pressure and to the availability of adjacent sites, K (1 _ q)2 P. The probability of desorption is proportional to the probability that adjacent sites are occupied rq2, since recombination has to take place on the surface prior to desorption. At equilibrium rq2 = K (1 _ q)2 P
=
\q= where K = (K/r)1/2
Chemistry : Basic Elements `STM' which makes it possible to see "a single molecule adsorbed on a surface." Scanning Tunneling Microscopy. Peaks expected in Auger Electron Spectroscopy (AES) spectrum of the following elements. (a) Sodium, (b) Carbon, (c) Hydrogen. (a) Sodium ® Two (b) Carbon ® One (c) Hydrogen ® None The technique for investigating electronic properties used to find the oxidation state of an adsorbed oxygen species. The information expected to obtain : We can use Ultraviolet Photo Electron Spectroscopy (UPES) and Auger Electron Spec-troscopy (AES). UPES will give us information about chemical shift and finger print, and AES will give us finger print information. The dissociative adsorption of oxygen on tungsten described by the Langmuir isotherm with K = 0.35 KPa_1. The fractional surface coverage at a pressure of 1 KPa. Calculated :
Fractional covered q =
=
=
= 0.2593
Langmuir method used to derive expressions for the fraction qA and qB of a surface covered by adsorbed molecules A and B, assuming that the molecules compete for the same sites.
The Langmuir adsorption isotherm can be written as
P=
or, q =
\ From this equation we can write Surface Phenomenon For Pure A, rAqA = KA (1 _ q) PA, For Pure B, rBqB = KB (1 _ q) PB when gases A and B are both present, these equations become rAqA = KA(1 _ qA _ qB) PA rBqB = KB (1 _ qA _ qB)PB Solving these two simultaneous equations for qA and qB yields
qA =
=
qB =
= where KA = KA/rA and KB = KB/rB The Micelles
Micelles can be spherical or laminar or cylindrical. Micelles tend to be approximately spherical over a fairly wide range of concentrations above CMC (critical micelle concentration) but often they are marked transitions to larger, non spherical liquid crystal structures at high concentrations. For straight chain ionic surfactants, the number of monomer units per micelle ranges between 30 and 80. Micelle formation effects the conductance of Surfactant solutions. The Reason : The total viscous drag on the surfactant molecules is reduced on aggregation but counter ions become kinetically a part of Chemistry : Basic Elements the micelle, thus reducing the number of counter ions which carry the current. Also the retarding influence of the ionic atmosphere of unattached counter ions on the migration of the surfactant ions is greatly increased on aggregation. Micro Emulsions They are emulsions with droplet diameters in the range 0.01 to 0.1 pm. It is formed when the oil-water interfacial tension approaches zero. Micro emulsions represents an intermediate state between micelles and ordinary emulsions. They are usually of low viscosity. The change in properties of solutions of surfactants at CMC shown below in the diagram as 1, 2, 3. Labelled :
(1) Molar conductivity. (2) Surface Tension. (3) Osmotic Pressure. Kraft Temperature
Miscelles from ionic surfactants can be formed only above a certain temperature called the Kraft temperature. Types of Surfactants. Examples : Surfactants are of three types— Cationic : They dissociate in water to yield positively charged ions e.g., octadecyl ammonium chloride (C18H37N+H3 Cl_). Surface Phenomenon Anionic : They are sodium salts of higher fatty acids e.g., sodium palmitate (C15H31COONa), Sodium stearate (C17H35COONa). Non-ionogenic : They are those whose molecule cannot undergo dissociation e.g., when alcohol having a high molar mass reacts with several molecules of ethylene oxide, a non-ionogenic surfactant is produced; CnH2n + 1 (OCH2CH2)mOH Stabilisation is Important. Here : Solubilisation is important in the formulation of pharmaceutical drugs containing water insoluble ingredients, in detergency (removal of oily soil), in emulsion polymerisation and in micellar catalysis. For n = 1, B.E.T. Isotherm gives the Langmuir Isotherm: For multilayer adsorption of gases on a solid, the B.E.T. adsorption isotherm can be written in a slightly different notation as
v= where n is the number of layers of the adsorbate molecules, P° is the vapour pressure of the pure liquid and c = exp {[DHads (mono) _ DHads]/ RT} where DHads (mono) is the heat of vaporization for the first monolayer, assumed to be different from the heat of adsorption of the liquid, DHads. For n = 1, we have
v=
=
Chemistry : Basic Elements
=
=
(where b = c/P°)
This is Langmuir Isotherm. In ESCA chemical shifts can be observed for every element in the periodic table except for two elements. Named : Hydrogen and Helium. The Occlusion This term is restricted to the sorption of gases by metals only. ESCA being utilized for finding the possible structure of doubly oxidised cystine. Cystine is Cy—S—S—Cy and doubly oxidised cystine is Cy—SO2—SCy. There are two possible structures for doubly oxidised cystine. O O || (a) Cy— S —S—Cy (Thiosulphonate structure) || O
Or (b) Cy—S—S—Cy (Disulphoxide structure) || || OO Thiosulphonate structure (a) is expected to give two peaks of equal intensity corresponding to the two oxidation states in the molecule whereas the disulphoxide structure should give only one peak. ESCA spectrum shows that the thiosulphonate structure applies to the doubly oxidised cystine because two peaks of equal intensity is observed. Surface Phenomenon The time for 10 per cent of the sites on a (100) surface of a certain substrate to be covered with nitrogen at 298 K when the pressure is 0.30 mPa and the sticking probability is 0.55. The substrate, taken to have a b.c.c. structure with lattice constant 316 pm. First the number of atoms on the surface should be calculated. The distance between lattice sites on the (100) face is 316 pm; \ each atom accounts for (316 pm)2 of the surface. The number of atoms in 1.00 m2 is therefore (1.00 m2)/(316 pm)2 = 1.00 × 1019 so the surface density of sites is 1 × 1019 m_2. \ From equation Zw = p/(2p mKT)1/2 = 8.6 × l015 m_2s_1 where Zw is the number of collisions per unit area per unit time when the pressure is p. Therefore the time required for 10 per cent coverage is
t=
= = 1.1 × 102 s The decomposition of phosphine (PH3) on tungsten is first order at low pressures and zero order at high pressures. Accounted for these observations : If the rate is supposed to be proportional to the surface coverage, then
v = Kq = where `p' is the pressure of phosphine, when the pressure is so low
Chemistry : Basic Elements Kp > 1 v=K \ Decomposition is zero order. Following types of experimental Physical adsorption Isotherms. Matched :
(a)
(b)
(c) Surface Phenomenon
(d)
(e) (l)_(d), (2)_(e), (3)_(b), (4)_(a), (5)_(c).
Bioinorganic Chemistry 4 Bioinorganic Chemistry Active Site Structure of Rubredoxin : There are several non-heme iron-sulphur proteins that are involved in electron transfer. They contain distinct iron-sulphur clusters composed of iron atoms, sulphydryl groups from cysteine residues and inorganic or labile sulphur atoms or sulphide ions. The labile sulphur is readily removed by washing with acid. The cysteine moieties are incorporated within the protein chain and are thus not labile. The simplest type of cluster is bacteria rubredoxin, (Cys-S)4 Fe (often abbreviated Fe1S0 where S stands for inorganic sulphur), `and contains only non labile sulphur. It is a bacterial protein of uncertain function with a molecular weight of 6000. The single iron atom is at the centre of a tetrahedron of four cysteine ligands (Fig.).
Ferredoxin : Two common forms of ferredoxin are photosynthetic ferredoxin and cubane-like ferredoxin.
Chemistry : Basic Elements The cluster in the ferredoxin molecule associated with photosynthesis in higher plants is thought to have the bridged structure Fe2S2 as shown in figure. It is known as photo-synthetic ferredoxin.
The cluster found in certain bacterial ferredoxins involved in anaerobic metabolism. It consists of a cubane-like cluster of four iron atoms, four labile sulphur atoms, thus Fe4S4, and four cysteine ligands (Fig.).
Fe4S4 in cubane like ferredoxin The Structure of the Proposed Model for the Active Site of Nitrogenase : Nitrogenase : The best known nitrogen fixing bacterium is Rhizobium. This contains the metalloenzyme nitrogenase. The active enzyme in nitrogen fixation is nitrogenase. Nitrogenase contains two proteins, molybdoferredoxin and azoferredoxin. Molybdoferredoxin is brown, air sensitive, contains two Mo atom, 2436 Fe atoms and 2436 S atoms together with a protein d2B2 tetramer with a and is a molecular weight of about 225000. Azoferredoxin is yellow, air sensitive, is a derivative of ferredoxin Fe4S4(SR)4 and has a molecular weight in the range 5000070000. Bioinorganic Chemistry
It seems likely that the active site for dinitrogen binding involves the molybdenum atom. It has been established by extended X-ray Absorption Fine Structure (EXFAS) measurements that the coordination sphere consists of several sulphur atoms at distance of about 235 pm. An Mo=O double bond, so common in complexes of Mo (IV) and Mo (VI) is not present. There are other heavy atoms, perhaps iron, nearby (~ 270 pm). The iron-sulphur clusters probably act as redox centres. It is not certain exactly how nitrogen fixation occurs. It is thought that N2 bonds to the Mo in molybdoferredoxin. If the Mo is sufficiently reduced it can probably bond to the antibonding orbitals on N2. Then the Fe in azoferredoxin is reduced by free ferredoxin. An electron is transferred from reduced azoferredoxin to molybdoferredoxin, possibly via the Fe, then from the Mo to the. N2-. Protons are then added to N2 giving NH3. A.T.P. is required to provide the necessary energy. Nitrogenase also reduces alkynes to
alkenes, and cyanide. Four Steps Involved in Oxidation of Water : The photosynthetic process in green plants consists of splitting the elements of water, followed by reduction of carbon dioxide. 2H20 ® 4[H] + O2
CO2 +
4[H] ® (CH2O)x +
O2
where 4[H] does not imply free atoms of hydrogen but a reducing capacity of four equivalents. In all di-oxygen Chemistry : Basic Elements producing orgnisms, there are two coupled photosynthetic systems PS-I and PS-Il. The two differ in the type of chlorophyll present and in the acessory chemicals for processing the trapped energy of photon. The primary product of PS-I is reduced carbon, and the primary product of PS-II is energy in one form of two moles of A.T.P. with molecular oxygen as a chemical by product. The oxidised form of PS-II is a strong enough oxidizing agent to oxidize water. However, the reactions involves the net transfer of four electrons and is achieved only by a complex 2H2O(l) ® 4H + (aq) + O2 (g) + 4e series of redox reactions of a manganese based enzyme which transfers electrons to the photo-chemically active centre. Experiments with successive light flashes show that the enzyme system is oxidized in four one-electron steps before accomplishing the four-electron oxidation of water to O2. If the redox reservoir involves the reduction of Mn (IV) to Mn (II), the enzyme must contain two Mn atoms to accommodate four electrons. The four steps of oxidation which precede the release of O2 in the release of O2 in oxygen evolving system of PS-II is shown in fig.
Oxygen uptake and equilibrium in Myoglobin and Hemoglobin : The function of both hemoglobin and myoglobin Bioinorganic Chemistry is to bind oxygen but their physiological roles are very different. Hemoglobin picks up oxygen in the lungs and carries it to the tissues via the circulatory system. Cellular oxygen is bound by myoglobin molecules that store it until it is required for metabolic action, where upon they release it to other acceptors. Hemoglobin has a additional function, however, and that is to carry CO2 back to the lungs; this is done by certain amino acid side chains, and the heme groups are not directly involved. Because the circumstances under which Hb and Mb are required to bind and release O2 are very different, the two substances have quite different binding constants as a function of O2 partial pressure (Fig.). Hemoglobin is not simply a passive container of oxygen but an intricate molecular machine. This may be appreciated by comparing its affinity for O2 to that of myoglobin. For myoglobin (Mb) we have the following simple equilibrium Mb + O2 ® MbO2
If f represents the fraction of myoglobin molecules bearing oxygen and P represents the equilibrium partial pressure of oxygen then
This is the equation for the hyperbolic curve shown in Fig. Hemoglobin with its four subunits has more complex behaviour; it approximately follows the equation
n » 2·8 where the exact value of n depends on pH.
Chemistry : Basic Elements
Thus, for hemoglobin (Hb) the oxygen binding curves are sigmoidal as is shown in the above figure. The fact that n exceeds the unit can be ascribed physically to the fact that attachment of O2 to one heme group increases the binding constant for the next O2, which in turn increases the constant for the next one and so on. This is called cooperativity. Just as a value of n = 1 would represent the lower limit of no cooperativity a value of n = 4 would represent the maximum cooperativity, since it would imply that only Hb and Hb(O2)4 would be participants in the equilibrium. Although Hb is about as good an O2 binder as Mb at high O2 pressure, it is much porer at the lower pressure prevailing in muscle and, hence, passes on its oxygen to the Mb as required. Moreover, the need for O2 will be greatest in tissues that have already consumed oxygen and simultaneously have produced CO2. The CO2 lowers the pH, thus causing the Hb to release even more O2 to the Mb. The pH-sensitivity (called the Bohr effect), as well as the progressive increase of the O2 binding constants in Hb, is due to interactions between the subunits; Bioinorganic Chemistry Mb behaves more simply because it consists of only one unit. It is clear that each of the two is essential in the complete oxygen transport process. Important Structural Features of the Active Sites of Metal Centres Deoxyhemoglobin, Cytochrome c, Vitamin B12.
Deoxyhemoglobin : The oxygenated form of hemoglobin is called oxyhemoglobin and the reduced form is called deoxyhemoglobin. In oxyhemoglobin the Fe2+ is in the low spin state and is diamagnetic. It is just the right size to fit in the hole at the centre of the porphyrin ring. The porphyrin is both planar and rigid. In deoxyhemoglobin the Fe2+ is in the high spin state and is paramagnetic. The size of Fe2+ increases by 28% when it changes from low spin to high spin i.e., from 0·61 Å to 0·78 Å. Thus in deoxyhemoglobin the Fe2+ is too large to fit in the hole at the centre of the porphyrin, and is situated 0·70·8 Å above the ring, thus distorting. the bands round the Fe. Thus the presence of O2 changes the electronic arrangement of Fe2+ and also distorts the shape of the complex. The iron atom in deoxyhemoglobin has square pyramidal coordination.
Cytochrome c : There are many cytochhomms, which differ in slight detail, but these are broadly grouped together as cytochrome a, cytochrome b and cytochrome c. The active centre of the cytochrome is the heme group. It consists of a porphyrin ring chelated to an iron atom. The porphyrin ring Chemistry : Basic Elements consists of a macrocyclic pyrrole system with conjugated double bonds and various groups attached to the perimeter. Cytochrome c is an important biological intermediate in electron transfer. This metalloprotein, found in all cells, has a molecular weight of approximately 12,800 and contains 104 amino acids (in vertebrates). It is an electron carrier for oxidative phosphorylation, transferring electrons to O2. The energy released in this process is used to synthesize A.T.P. The heme group of cytochrome c lies near the surface of the protein. The iron atom is six coordinate, with bonds to five nitrogen atoms (four from porphyrin, one from a histidine nitrogen) and one sulphur atom from a cysteine. Since all six coordination sites are occupied, direct electron transfer to iron is not possible, and the electron must pass through the surrounding protein bridge work. The importance of cytochrome c in photosynthesis and respiration indicates that it is probably one of the oldest of the chemicals involved in biological processes.
The iron porphyrin unit in chytochrome c. Vitamin B12 : Vitamin B12 is an important cobalt complex. The vitamin was isolated from liver after it was found that eating large quantities of raw liver was an effective treatment for pernicious anaemia. The term vitamin B12, refers to cyanocobalamin. Vitamin B12 is a coenzyme, and serves as a Bioinorganic Chemistry prosthetic group which is tightly bound to several enzymes in the body. It is called a biological Grignard reagent. Vitamin B12 is unique in that it appears to be synthesized only by microorganisms especially anaerobic bacteria. The structure of vitamin B12 complex consists of four principal components
1. A cobalt(III) ion. 2. A macrocyclic ligand called the corrin ring, which bears various substituents. It resembles the porphine ring, but differs in various ways, notably in the absence of one methine, =CH, bridge between a pair of pyrrole rings. 3. A complex organic portion consisting of a phosphate group, a sugar, and an organic base, the latter being coordinated to the cobalt atom.
Chemistry : Basic Elements 4. CN ligand coordinated to the cobalt ion. The Co atom is bonded to four ring N atoms. The fifth position is occupied by another N from a side chain (a-5, 6-dimethylbenzimadazole) and this is also attached to corrin ring. The sixth group which makes up the octahedron is the active site, and is occupied by a CN group in cyanocobalamin. The CN is introduced in isolating the coenzyme, and is not present in the active form in living tissues. This position is occupied by OH in hydroxocobalamin, by water or by an organic group such as CH3 (methylcobalamin) or adenosine. This shows that a metal to carbon s bond can be formed. The cobalamins can be reduced from CoIII to CoII and CoI in neutral or alkaline solutions both in laboratory
and in vivo (in the living body). The Col complex is strongly reducing. Its structure is five-coordinate i.e., the site usually occupied by CN or OH is vacant. Vitamin B12 is converted to co-enzyme B12 by extracts from microorganisms supplemented with ATP Coenzyme B12 is associated with many biochemical reactions1. Conversion of methylmalonyl CoA to succinyl CoA. 2. Isomerisation of dicarboxylic acids e.g., glutamic acid into b-methyl-aspartic acid. 3. Dismutation of vicinal diols to the corresponding aldehydes e.g., propane-1, 2-diol into propionaldehyde. 4. Methylation of homocysteine to form methionine. The Metals in I with the Appropriate Biomolecule in I II (A) Zn (i) phosphotransferase (B) Cu (ii) peroxidase (C) Fe (iii) carboxypeptidase (iv) haemocyanin (v) nitrogenase Bioinorganic Chemistry Matched : I II (A) Zn (i) carboxypeptidase (B) Cu (iv) haemocyanin (a copper containing protein) (C) Fe (ii) peroxidase (a heme enzyme) and
(v) nitrogenase (molybdenoferridoxin) A few metalloproteins (enzymes) in which porphyrin ligands are involved. The structural features of the ligand systems, responsible for the enzyme activity. The three major types of metalloproteins (enzyme) in which porphyrin ligands involved are peroxidases, catalases and cytochrome P450 enzymes. Peroxidases catalyse the oxidation of a substrate by peroxides, mainly H2O2. Catalases catalyse decomposition of H2O2 (and some other peroxides) to H2O and O2. They have many similarities, both in structure and in aspects of their mechanisms. They both have high-spin ferric heme groups lodged deeply in rather large protein molecules, with a histidine nitrogen atom constituting the fifth ligand. The sixth position may be occupied by a water molecule in the resting enzyme. There is growing evidence that a (proph) FeIV=O species is a key intermediate in peroxidases and catalases. The catalytic activity of peroxidase entails the following steps, where P represents the resting enzyme and AH a reducing substrate P + H2O2 ® compound I + H2O compound I + AH ® compound II + A. compound II + AH ® P + A. Compound I has an oxidation level two units higher than P, but it reverts to P in two one-electron steps. There is evidence to suggest that compound I and II can be more specifically formulated as follows
Chemistry : Basic Elements The R·+ represents a p-cation radical formed on the porphyrin ligand. Cytochrome P450 enzymes are heme containing oxygenases they catalyse the introduction of oxygen atoms from O2 into substrates. Of the many possible substrates the most important are molecules in which CH groups are converted to COH groups. The catalytic cycle entails a species in which the iron atom attains a high (IV or V) oxidation number. The coordination sphere of the iron atom includes,'in addition to the porphyrins, one sulphur atoms, but whether the sixth position is occupied by water molecule or is vacant in the resting enzyme is still uncertain.
In step (1) of the catalytic cycle of Cytochrome P450 the substrate is bound, at or near the iron atom, and in step (2) reduction is affected by another enzymatic system. In step (3) the (RH)FeII species binds an O2 molecule. In step (4) another one electron reduction occur to give an iron(III) peroxo complex. In step (5), this complex loses H2O to give the crucial high oxidation state (RH)FeIV=O.
A cagtalytic cycle for cytochrome P450 enzyme. Cytochrome c a Redox Protein : The active centre of Cytochrome c is the heme group. It consists of a porphyrin ring chelated to an iron atom. The heme group in Cytochrome c has a polypeptide chain attached and wrapped around it. This chain contains a variable number of amino acids ranging from 103 to 112. Å nitrogen atom from a histidine segment and a sulphur atom from a methionine segment of this chain are coordinated to the fifth and sixth coordination site. Thus, unlike Bioinorganic Chemistry the iron in hemoglobin and myoglobin there is no position for further coordination. Cytochrome c therefore cannot react by simple coordination but must react indirectly by an electron transfer mechanism. It can reduce the dioxygen and transmit its oxidising power towards the burning of food and release of energy in respiration. The oxidation state of iron may be either + 2 or + 3 and the importance of the Cytochrome lies in their ability to act as redox intermediates in electron transfer. Myoglobin is a protein of molecular weight of about 17,000 with the protein chain containing 153 amino acid residues folded about the single heme group. This restricts access to the iron atom (by a second heme) and reduces the likelihood of formation of a hematin-like Fe(III) dimer. The micro environment is similar to that in Cytochrome c, but there is no sixth ligand (methionine) to complete the coordination sphere of the iron atom. Thus there is a site to which a dioxygen molecule may reversibly bind. Myoglobin contains iron(II) in the high spin state. The function of both hemoglobin and myoglobin is to bind oxygen, but their physiological roles are very different. Hemoglobin picks up oxygen in the lungs and carries it to tissues via, the circulatory system. Cellular oxygen is bound by myoglobin molecule that
store it until it is required for metabolic action, whereupon they release it to other acceptors. Role of Metal Complexes in Nitrogen Fixation : The discovery that dinitrogen was capable of forming stable complexes with transition metals led to extensive investigation of possibility of fixation via such complexes. For example, phosphine complexes of molybdenum and tungsten containing dinitrogen readily give ammonia in acidic media [MOCl3(thf)3] + 3e + 2N2 + excess dppe ® [Mo(N2)2 (dppe)2] + 3Cl [Mo(N2)2 (dppe) 2] + 6H+ ® 2NH3 + N2 + MoIV products.
Chemistry : Basic Elements where thf = tetrahydrofuran and dppe = 1, 2-bis (diphenylphosphino) ethane, Ph2PCH2CH2PPh2. Both reactions takes place at room temperature and atmospheric pressure. There are several bacteria and blue-green algae that can fix molecular nitrogen in vivo. Molybdenum is present in the catalysts of nitrogen fixing bacteria. The best known nitrogen fixing bacterium is Rhizobium. This contains the metalloenzyme nitrogenase. Nitrogenase contains two proteins, molybdoferredoxin and azoferredoxin. Molybdoferredoxin contains two Mo atoms, 24-36 Fe atom and 24-36 Sz atoms together with a protein and has a molecular weight of about 225000. Azoferredoxin is a derivative of ferredoxin Fe4S4(SR)4 and has a molecular weight in the range of 50000-70000. It is thought that N2 bonds to Mo in the molybdoferredoxin. If the Mo is sufficiently reduce it can bond to antibonding orbitals on N2. Then the Fe in azoferredoxin is reduced by free ferredoxin Fe4S4(SR)4. An electron is transferred from reduced azoferredoxin to molybdoferredoxin, possibly via the Fe, then from the Mo to the N2. Protons are then added to N2, eventually giving NH3. Adenosine triphosphate ATP is required to provide the necessary energy for the process. The Ionophores Ionophores (ion carriers) are lipophilic substances, capable of binding and carrying specific cations through the biological membranes. They differ from the uncouplers in that they promote the transport of cations other than H+ through the membrane. The behaviour of valinomycin is typical of a group known as ionophore antibiotics. The antibiotic forms a lipid-soluble complex with K+ which readily passes through the inner mitochondrial membrane, whereas K+ alone in the absence of valinomycin penetrates only very slowly. Valinomycin binds K+ more strongly than Na+. Thus, valinomycin interferes with oxidative phosphorylation in mitochondria by making them
Bioinorganic Chemistry permeable to K+. The result is that mitochondria use the energy generated by electron transport to accumulate K+ rather than to make ATP.
Schematic representation of the action of ionophores on membranes. M+ = Metal ion Molecular mechanism of ion transport across membrane: Cells are enclosed by a membrane of about 70 Å thickness and composed of double layers of protein separated by lipids. Cation cannot pass through the lipid layer without encapsulation and thus the enclosed cation presents an organic, lipid soluble surface to the membrane. Most animal cells have higher concentration of K+ ions inside the cell membranes and a higher concentration of Na+ outside the cell. Maintenance of these concentrations gradients requires a sodium and potassium pump. The immediate source of energy required to maintain this is ATP. In some cells, each ATP molecule hydrolysed transports 3Na+ out of the cell and 2K+ into the cell. The K+ is required in the cell for glucose metabolism, protein synthesis and activation of some enzymes. The transport of glucose and amino acids into the cell is coupled with Na+ transport, which is favoured by great concentration gradient. The Na+ entering the cell in this way must be pumped out again.
Chemistry : Basic Elements
A valuable clue to the manner in which iron pump functions, came in 1957 when an enzyme of molar mass 110 kg/mol was discovered by Jens Skou which hydrolyses ATP only if Na+ and K+ ions are present in addition to Mgt+ required for all ATPases (enzymes catalysing the hydrolysis of ATP). The activity of this enzyme correlates quantitatively with the extent of ion transport. Another important clue was provided by the observation that this ATPase is phosphorylated at an aspartase site only in the presence of Na+ and Mg2+ ions. The phosphorylated product is hydrolysed if K+ ions are present. It has been also observed that the enzyme undergoes a conformational change when it is phosphorylated. Bioinorganic Chemistry At first step ATP and three Na+ ions bind to the inside of the membrane and the enzyme is phosphorylated. The product of the reaction undergoes a conformational change called eversion and brings the Na+ ions to the outside of the cell membrane. There three Na+ ions are replaced by two K+ ions. The attachment of the K+ ions induces dephosphorylation and hydrolysis of ATP to ADP produces a conformational change that carries two K+ ions to the interior of the cell where they are released. The process builds up a charge gradient across the membrane because three Na+ ions
are released for two K+ ions incorporated, and the outer surface becomes relatively positively charged. The outstanding feature of the active transport is the selectivity of the complexation process, between K+ and Na+. The enzyme is selective first to Na+ and then to K+ in its different conformations. Selectivity is governed by coulombic forces and differences in ionic radii. The ion channels which permit transport of ions, are constructed from protein polymers in helical conformations. These helices form the wall of cell channels and their precise conformations result in specific geometrical arrangements of a number of potential ligand-metal binding sites. There are striking differences between the channels, selective towards either Na+ or K+ ions. The Photosynthesis The photosynthetic process in green plants consists of splitting the elements of water, followed by reduction of carbon dioxide 2H2O ® [4H] + O2
x CO2 +
[4H] ® (CH2O)x +
O2
where [4H] does not imply free atoms of hydrogen but a reducing capacity of four equivalents. In all dioxygen-producing organisms ranging from cyanobacteria to algae to higher plants, Chemistry : Basic Elements there are two coupled photosynthetic systems, PS-I and PS-II. The two differ in the type of chlorophyll present and in the accessory chemicals for processing the trapped energy of the photon. The primary product of PS-1 is reduced carbon, and the primary product of PS-II is energy in the form to two moles of ATP with molecular oxygen as a chemical by product.
In addition to chlorophyll molecules at the reaction centers of PS-I and PS-II, there are several other pigments associated with light-harvesting complex. Among these are cartenoid, open-chain tetrapyrrole pigments and others are important. These serve dual roles of protecting the cell from light radiation and at the same time harvesting much of it for photosynthesis. Some of these compounds are arranged in antenna-like rods that gather the light energy and funnel it into the reaction centers. The energy of an absorbed photon in either PS-I or PS-II initiates a series of redox reactions. PS-I produces a moderately strong reducing species (REDI) and a moderately strong oxidising species (OXI). PS-II provides a stronger oxidising agent (OXII) but a weaker reducing agent (REDII). OXII is responsible for the production of molecular oxygen in photosynthesis. Å manganese complex, probably with four atoms of manganese, is attached to a protein molecule. It reduces OXII which is recycled for use by another excited chlorophyll molecule in PS-II. In the redox reaction the manganese shuttles between two oxidation states with each manganese atom Bioinorganic Chemistry increasing its oxidation state by one unit, but it is not known with certainity what these oxidation are [three Mn(II) and one Mn(III) or three Mn(III) and one Mn(IV)]. Butterfly cluster of Mn4 configuration have been suggested along with others.
Copper Metalloenzymes and Metalloprotein : The copper enzymes are mostly oxidases, that is enzymes which catalyse oxidations. Examples are : Ascorbic acid oxidase : (MW = 1,40,000; 8 Cu). It is widely distributed in plants and micro-organisms. It catalyses oxidation of ascorbic acid (vitamin C) to dehydro ascorbic acid. Cytochrome Oxidase : The terminal electron acceptor in the oxidative pathway of cell mitochondria. This enzyme also contains heme. Tyrosinase : Various tyrosinase catalyse the formation of pigments (melanins) in a host of plants and animals. Copper Metalloprotein (Hemocyanin) : In the phyla Mollusca and Arthropoda the oxygen carrying molecule is a copper containing protein is hemocyanin. They are very large of very high molecular weight, but they are consist of subunits. Each subunit contains a pair of Cu atoms, and they can bind one O2 molecule per pair of Cu atoms. It generally display cooperativity in O2 binding.
The Metalloenzymes Metalloenzymes-Enzymes are large protein molecules so built that they can bind at least one reactant (substrate) and catalyse a biochemically important reaction. Some enzymes incorporate one or more metal atoms in their normal structure they are called metalloenzymes. The metal not merely participate during the time that the enzyme-substrate complex Chemistry : Basic Elements exists,but a permanent part of the enzyme. The metal atom or at least one of the metal atoms when two or more are present, occurs at or very near to the active site and plays a role in the activity of the enzyme. Ca, Mn, Fe, Cu, Zn and Mo have been found in metalloenzymes. Metalloenzymes are extremely efficient as catalyst, typically causing rates to increase 106 times or more. Zinc Metalloenzymes : Two of the most important zinc metalloenzyme are the following : Carbonic Anhydrase : (MW2 = 30,000, 1 Zn) This enzyme occurs in red blood cells and catalyse the dehydration of bicarbonate ion and hydration of CO2. OH- +CO2
HCO3-
In the absence of this enzyme the reaction is not compatible with physiological requirements. Carboxy Peptidase : (MW = 34,300, 1 Zn) This enzyme in the pancreas of mammals catalyses the hydrolysis of the peptide bonded at the carboxyl end of a peptide chain. Natural Oxygen Carrier The two most important metalloprotein, other than hemoglobin and myoglobin are hemerythrins and hemocyanin which act as natural oxygen carrier. Hemocyanin is discussed in previous question. Hemerythrins : In a great variety of worms (marine) the oxygen-carrying molecules are iron bearing proteins, but they do not contain porphyrins, called hemerythrins. Its essential structural unit is a single chain protein of 113 amino acid residues, which contains two atoms of iron. Hemerythrins differ functionally from hemoglobin in two major ways. They show little or no cooperativity, and two iron atoms are employed to bind one O2 molecule. Each dioxygen binding site contain two iron(II) atoms and the reaction takes place via a redox reaction to form iron(III) and peroxide (O22-).
Bio-organic Chemistry 5 Bio-organic Chemistry The Concept Primary, Secondary, Tertiary and Quaternary Structure of a Protein. The Bonding, Greatly Responsible for the Secondary structure. Primary structure of a protein is simply amino acids sequence of the peptide chain. The secondary structure is a result of the different conformations that the chain can take. The tertiary structure refers to the three dimensional shape that results from twisting, bending and folding of protein helix. The quaternary structure refers to the way in which these amino acid chains of a complex protein are associated with each other (oligomer, dimers, trimers, etc.). (a) The H-bonding between N—H of one amino acid residue and O=C of another properly situated amino acid residue. (b) The peptide sequence is coiled into a right handed spiral in the a-helix, with the R groups positioned on Chemistry : Basic Elements the outside of the spiral. Each amide H—N bonds to the O=C on the next turn of the coil, four residues away by H-bons, stabilizing this arrangement. In the pleated sheet of b-structure, the peptide chains lie side by side in an open structure, with interchain amide H-bonding holding the chains together. Parallel pleated sheets have chains running in the same direction, all with their N-terminal residues, starting at the same end. Antiparallel pleated sheets have their chains running in opposite directions. The a C's rotate slightly out of plane of the sheet to minimize repulsions between their bulky R groups, giving rise to the pleats. In both cases, the R groups alternate positions above and below the sheets. The random coil structure has no repeating geometric pattern; encompassed within it are sequences in a helical conformation, a pleated conformation, and regions that appear to have no discernible repeating structure, but are actually not random conformation. Bonding Responsible for the Structure of Proteins : The following are the attractive forces responsible for the tertiary structure—
(i) Ionic : bonding between COO_ and NH3+ at different sites; (ii) H-bonding : mainly between side chain nH2 and COOH, also involving OH's (of serine, for example) and N—H of tryptophane; (iii) Weakly hydrophobic Vander Waals attractive forces engendered by side chain R groups; (iv) Disulphide cross linkages between loops of the polypeptide chain. The same kind of attractive and repulsive forces responsible for the tertiary structure operate to hold together and stabilize the subunit of the quaternary structure. Bio-organic Chemistry Amino Acids given in the List B with that of Type given in List A. Matched : List A List B (1) Mercapto (i) Valine (2) Hydroxy (ii) Proline (3) Carboxamide (iii) Methionine (4) Unsubstituted (iv) Tyrosine (5) Aromatic (v) Serine (6) Heterocyclic (vi) Cysteine (7) Thioether (vii) Glutamine (1)—(vi), (2)—(v), (3)—(vii), (4)—(i), (5)—(iv), (6)—(ii), (7)—(iii). An elemental analysis of cytochrome c, an enzyme involved in oxidation reduction process, gave 0.43% Fe and 1.48% S. The minimum molecular weight of enzyme and The Minimum Number of Iron Atoms per Molecule of Sulphur Atom. If one Fe atom per molecule
0.43% Fe = m.w. = 13,000 If one S atom per molecule
1.48% S = m.w. = 2170 The minimum m.w. 13,000; 1 Fe and 6 S. Nucleic Acids Nucleic acids : Nucleic acids are bio molecules which are found in the nuclei of all living cells in form of nucleoproteins. Chemistry : Basic Elements They are biopolymers in which the repeating structural unit or monomeric unit is a nucleotide (polynucleotides). Each nucleotide consists of three components— (i) A pentose sugar i.e., D(_) ribose or 2-deoxyribose (furanose form). (ii) A heterocyclic nitrogenous base, two types— (a) purines : Adenine (A) and Guanine (G). (b) pyrimidines : Uracil (U), Thymine (T), Cytosine (C). (iii) Phosphoric acid : Nucleotides are the phosphoric acid esters of nucleosides (nucleosides contain only pentose sugar and a nitrogen base). Two types of nucleic acids are : Deoxyribonucleic Acid (DNA) : Sugar 2-deoxyribose, a phosphate group and a heterocyclic base. The base is either purine A or G and pyrimidine C or T.
Ribonucleic Acid (RNA) : Sugar is ribose, a phosphate group and a heterocyclic base. The base is either purine A or G and pyrimidine C or U. Thus the RNA differ from DNA in sugar (ribose) and the pyrimidine base U replaces T. The primary structure of nucleic acids refers to the sequence in which the four nitrogen bases (A, G, C and T in DNA and A, G, C and U) in RNA are attached to sugar phosphate backbone of the nucleotide chain. Secondary structure of DNA consists of two strands of polynucleotides coiled around each there in the form of double helix. The backbone of each strand is sugar-phosphate unit and the base unit of each strand are pointed into the interior of the helix and are linked through H-bonds. G and C are held by three H-bonds, A and T are held by two bonds. Unlike DNA, RNA has a single strand. Change in Base Sequence in DNA Causing Mutation : A change in single base pair in DNA can cause substitution of Bio-organic Chemistry a different amino acid in the protein being synthesized (as is seen in sickle cell hemoglobin). In a silent mutation, the substitution does not significantly change the normal activity of protein. A lethal mutation is characterised by production of a defective harmful protein or failure to produce an essential protein. The H-bonds between the paired bases and the Van der Waals attractive forces that hold the staked basepairs, aligned one pair on top of the other, are disrupted. The ordered conformation breaks and the double helix dissociates into random, disordered coils. The viscosity of the solution drops sharply. The double helix structure of DNA has the hydrophobic bases pointing to the centre of the helix in an almost planar arrangement. These base-pairs are closely stacked perpendicular to the long axis of the chain, and are attracted to each other by Van der Waals forces. The hydrophilic phosphates are negatively charged at the pH of the cell and point to the outside.
Double-helix structure of DNA.
Chemistry : Basic Elements The two sugar-phosphate chains are aligned in opposite directions. The 5¢OH linkage is at the top of the one strand and 3¢OH linkage at the top of the other strand. In the figure the number of dashes between the base symbols indicate the number of H-bonds. S is deoxyribose and P is O=P—O_. DNA directing Protein Synthesis in a Cell : The DNA directed protein synthesis occurs in two steps— Transcription : It involves copying of DNA base sequences into a mRNA. A small portion of DNA double helix unwinds and are of the two DNA strands act as the template for the synthesis of mRNA. Ribonucleotides assemble along the uncoiled template in accordance with the base pairing principle. For example, U in the RNA being formed appears opposite A of DNA; C opposite G; A opposite T and G opposite C of DNA. Then occurs the bond formation between the various nucleotides thus assembled. When mRNA strand is synthesized, the DNA-RNA double helix separates. The mRNA migrates to the cell cytoplasm while DNA returns to its normal double helix structure. Similarly tRNA and rRNA are synthesized. Translation : The mRNA directs protein synthesis in the cytoplasm of the cell with the help of rRNA and tRNA. This process is called translation. The mRNA synthesized above gets attached to very small ribosome particles (60% rRNA and 40% protein). On the ribosome, mRNA serves as the template for protein synthesis.
The specific ribonucleotide sequence in the mRNA forms a code that determines the order in which the different amino acid residues have to be joined. The four bases, i.e., A, C, G and U have been shown to act in the form of triplets. Since a particular amino acid codes a specific amino acid, therefore, these triplets are called codons. Since there are four bases, therefore, 43 = 64 triplets or codons are possible. Out of these 61 for specific amino acids while the remaining 3 (UUA, UGA and UAG) are known to code for chain termination (nonsense Bio-organic Chemistry or stop codons). Since there are only 20 different amino acids, therefore, more than one triplet combinations code for the same amino acid, e.g., GGU, GGA, GGC, GGG code for glycine and two triplets UUU and UUC code for phenylanine. mRNA recognise the amino acid with the help of tRNA so there are 61 different tRNA. All the tRNA have a CCA triplet which is attached to amino acid by an ester linkage. At one end of the loop of tRNA there is a ribonucleotide triplet called anticodon which is complementary to a codon on mRNA. Each codon of mRNA is read in a serial order by an anticodon of tRNA and matched. If matching occurs, the tRNA transfers the desired amino acid to the growing polypeptide chain on the ribosome. When the synthesis of a specific protein is completed, a stop codon signals the end and the synthesized protein is released from the ribosome. On hydrolysis of DNA, it is found that although the ratio of bases vary from one DNA to another, the ratios of C : G and A : T are always 1 : 1. The C and A is one strand always matches the G and T respectively in other strand. This matching is called base-pairing. (b) Base pairing occurs by H-bonding with maximum efficiency between a pyrimidine and a purine base, specifically between A and T and between C and G.
This suggest that RNA helixes are single stranded not double helixes like DNA.
Chemistry : Basic Elements An error of a single base in each strand of DNA can bring about the amino acid residue error that
causes sickle-cell anemia. A change from C—T—T to C—A—T or a change from C—T—C to C—A—C. Tautomeric structures of adenine, guanine, cytosine, thymine and uracil are—
(Cytosine)
(Thymine)
(Uracil) Bio-organic Chemistry The successive hydrolytic degradation products of a nucleoprotein. outlined. Nucleoprotein
nucleic acid (+ proteins)
nucleotides
nucleosides (+ phosphate ion)
pentose + heterocyclic base. Base-pairs in the gene, needed to code for the enzyme lysozyme (129 amino acids) found in egg white. Three base-pairs code for one amino acid, and two more triplets are required to start and stop signals, or 3 × 129 + 3 + 2 = 393 base pairs. HNO2 converts —NH2 to —OH, i.e., cytosine to uracil and guanine to xanthine via diazotization, thereby changing the genetic code and leading possible mutation.
Cytosine Uracil
Guanine Xanthine Replication : The process by which a single DNA molecule produces two identical copies of itself is
called replication (cell division). Replication is a enzyme catalyzed process. It begins with partial unwinding of the double helix through breaking of H-bonds between pairs of bases leading to the formation of two isolated strands. Each strand then act the template for the synthesis of two new strands. Because of the base pairing principle, each new strand is complementary to its old template strand. As a result, two identical copies of DNA from the Chemistry : Basic Elements parent DNA are produced. Each copy is then passed on the two new cells resulting from cell division. In this way hereditary effects are transmitted from one cell to the other. ATP Regarded as Store House of Energy : Adenosine triphosphate (ATP) in a nucleotide consists of purine base adenine, a pentose sugar ribose and three molecules of phosphate. It contains two oxygen to phosphorus bonds between two phosphate units. These phosphorus bonds are called high energy phosphatic bonds. ATP is an energy rich molecule. This is because of the presence of four negatively charged oxygen atoms very close to each other. These four negatively charged O-atoms experience very high repulsive energy. When ATP is hydrolysed, the hydrolysis of bond results in decrease in repulsive forces and consequently, a large amount of energy is released. They hydrolysis of ATP proceeds as— ATP
ADP + Pi
Adenosine diphosphate ADP
AMP + Pi
The structure of ATP, ADP and AMP. Bio-organic Chemistry The energy producing reactions are coupled to carry out some other chemical reactions which are otherwise not energetically possible. ATP, ADP and AMP occur not only in cell cytosol but also in mitochondria and the nucleus. During phosphate transfer, ATP breaks down to give AMP or ADP and not adenosine and PPPi. The structure of a mononucleotide. ATP is an energy rich molecule because its triphosphate unit contains two phosphoanhydride bonds. A large amount of free energy is liberated when ATP is hydrolysed to ADP or AMP. ATP + H2O
ADP + Pi + H+ DG° = _ 7.3 k cal/mol
ATP + H2O
AMp + PPi + H+ DG° = _ 7.3 k cal/mol
ATP is not breakdown to adenosine and PPPi because ATP, AMp and ADP are interconvertible by the enzyme myokinase. ATP + AMP
ADP + ADP
Moreover DG° value of hydrolysis of yield adenosine and phosphate is much lower (about _ 3.4 k cal/mol). Thus the phosphate group of AMP (i.e., the a-phosphate group of ATP) is in the lower energy class.
Nucleotides are the phosphoric acid ester of nucleoside, while nucleosides are compounds in which nitrogenous bases (purines and pyrimidines) are conjugated to the pentose sugar (ribose or deoxyribose) by a b-glycosidic linkage. AMP, ATP, ADP, GMP, CMP, UMP are the examples of mononucleotides. Glyceraldehyde phosphate and dihydroxyacetone phosphate react in the presence of aldolase to yield fructose diphosphate as the only product under physiolocial conditions. By the reaction of D-glyceraldehyde and 1,3-dihydroxypropane (both as monophosphate ester), Dfructose as the 1,6-diphosphate ester is formed. The process is readily reversible and is catalyzed by an enzyme known as aldolase. Chemistry : Basic Elements This is a type of aldol addition (known as biological aldol addition) and is one of the reaction in the metabolism of carbohydrates by the glycolic pathway.
(glyceraldehyde 3-phosphate) (D-fructose-1, 6-diphosphate) It seems unlikely that this reaction could occur in quite the same way as in the laboratoryaldol reaction, because the enolate anion of the donor molecule (dihydroxypropanone) is not expected to be formed in significant amount of the pH of living cell. In fact there is strong evidence that the enzyme behaves as amino (ENH2) compound and reacts with carbonyl group of dihydroxy propanone to form an imine. This implies that the imine form of dihydroxy propanone is a key intermediate in the overall aldol-type addition.
+ H2N—E
+ H2O
The Photochemistry 6 The Photochemistry An example of carbonyl compound which illustrate Photodecarbonylation and example of Abstraction of hydrogen atom to form alcohol. Cis-trans isomerization due to absorption of light, illustrated :
+ CO Photodecarbonylation
(C6H5)2C=O + C6H5CHOHC6H5 Benzophenone Benzhydrol Benzpinacol The photochemical cis trans isomerization is effected by direct irradiation or in the presence of a sensitiser or a catalyst.
e.g., trans stilbene cis stilbene
Chemistry : Basic Elements
cis-2-butene trans-2-butene
Intramolecular Hydrogen Abstraction Intramolecular hydrogen abstraction from a g carbon occurs when certain ketones are irradiated. For e.g., vapours of Hexan-2-one on photolysis gives an alkene and a ketone (via the enol form). It is referred to as Norrish type II cleavage.
Matched : Reaction Type of Photoisome- rization reaction (a) cis MLnXm
trans MLnXm Optical
n = 2 or 4; m = 2 n = 3; m = 3 (b) (d)M(L—L)3 (c) LnM—XY
(l)M(L—L)3 Linkage LnM—Y—X Geometrical
(a) Geometrical, (b) Optical, (c) Linkage. The Photochemistry
The reactions completed :
Chemistry : Basic Elements
(e)
(f) (a) Drawn and labelled Jablonski diagram : (b) The product of the following reaction :
(Norrish type I section)
(a)
(b) Drawn diagram for the relative energies of the various types of p orbitals of butadiene (which has two bonding and two antibonding p orbitals) to illustrate the ground state and the S1, T1, S2 and T2 excited states. The Photochemistry
The following equations completed :
(e) MeCH2CH2COMe
(h) CH3CH2CH2CH2 N3
(a cyclobutane is major product)
(a cyclic compound)
Chemistry : Basic Elements
(a) (b) Me2CHONO
Me2CHO + NO
Me2C=O + HOCHMe2
(c)
The Photochemistry
A mechanism for the reaction suggested : 2PhCH2CH2ONO
(PHCH2NO)2 + 2CH2O
PhCH2—CH2O—NO
NO + PhCH2—CH2O.
CH2O + NO + PhCH2.
| Dimer ¬¾ PhCH2NO¬¾ The Norrish type I process is not important for the photolysis of diaryl ketones. A reason suggested : Norrish type I process is—
R1CO—R2
+ R2—
+ CO
If diaryl ketones followed this route we would have
Ar1—CO—Ar2
+ Ar2—
+ CO
Now aryl free radicals are extremely unstable, they are not stabilised by resonance. Since the stability of a product can act as a driving force for the reaction to proceed along that path then we can say that the more unstable the product is, the less likely that path will be followed. Hence Norrish type I process is energetically unfavourable for diaryl ketones. (a) A synthesis of the following compound, planned :
(b) The structures of the photosensitized irradiation of the following compounds :
Chemistry : Basic Elements (c) Rationalize the following reaction :
(a) The cyclobutane ring can be constructed via photochemical cycloaddition of the alkene using acetophenone sensitizer.
(b) (I) (c) The photochemical conversion of I to II involves the coupling rather than the fragmentation of the Norrish type II biradical to the cyclobutanol.
1,4-cineole and ascaridole are the naturally occurring oxide and peroxide respectively. There are three isomeric terpinenes. The following out of these is a suitable isomer for photochemical conversion into either 1,4-cineole or ascaridole:
Formation of cyclic peroxides by conjugated dienes is a general reaction and although ultraviolet light often initiates the reaction, better results are achieved by carrying out the The Photochemistry irradiation of a-terpinene in the presence of sensitisers, e.g., chlorophyll, dyes etc. Sesquiterpenoids Caryophyllene The sesquiterpenoids caryophyllene and its geometrical isomer occur naturally in the oil of cloves. Plan a synthesis of a suitable starting material containing a four membered ring :
Compounds A and B undergo photodecarboxylation resulting in the formation of I and II (from A) and III and IV (from B). Give their structures and the mechanism of their formations:
Chemistry : Basic Elements The likely products from the vapour phase irradiation of trans 2,6-dimethyl cyclohexanoneare: A mixture of cis and trans-1,2-dimethyl cyclopentanes will be ormed as the stereochemical integrity of the molecule would be destroyed on opening of the ring.
The products you expect to be formed in the irradiation of:
(i) H2C: + C6H5CH=CH2
(ii)
+
(iii) CH2
CH2
singlet triplet
Cyclopropane Cyclopropane The Photochemistry (Singlet state either decays to triplet and then react with alkene to form cyclopropanes or directly react with alkene to give cyclopropanes) Suggest mechanism for the following reactions—
(i)
(ii)
(i) The starting 1,2,3-thiodiazole is formally equivalent to a diazoketone and the first part of the reaction is the same as the photocatalyzed Wolf rearrangement.
Chemistry : Basic Elements
(ii) The observed products correspond to the formation of the most stable intermediate by the addition of the photoexcited triplet (diradical) state of acetophenone to the alkene. The Products Predicted :
The Photochemistry (a) Mentioned two reasons as to why benzophenone is a good triplet sensitizer.
(b) The following compounds will quench benzophenone triplets at a diffusion controlled rate: (a) Sensitizers should be chosen to produce high yields of triplets for maximum efficiency, benzophenone is a good example of this category because— (i) benzophenone forms triplets when irradiated in fluid solution at room temperature; (ii) benzophenone can efficiently transfer its triplet energy to the triplet state of other molecules. Its inter system crossing efficiency is almost equal to1. (b) Benzophenone has ET = 69. Thus, butadiene (ET = 60), anthracene (ET = 42) and 2-acetonaphthone (ET = 59) will quench benzophenone triplets at the diffusion control rat. Whereas benzene (ET = 85) and acetophenone (ET = 75) will not. The following compound is expected or not to undergo efficient photoreduction :
No, because the molecule carries a built in quencher around with it. Each benzophenone triplet formed will be quenched at a rate much faster than hydrogen abstraction from solvent. The major products from the photosensitized dimerization of isoprene predicted. The ratio of diene dimers vary as the sensitizer is varied or not :
Chemistry : Basic Elements
(1)
(2)
(a) The ratio of dimers will not vary because stereoisomeric triplets are not possible.
Quantum Chemistry 7 Quantum Chemistry
The wave function y = A sin
for a particle in one dimensional box of length a normalised :
A wave function is normalised if the probability is unity i.e.,
=1 Substituting the values of limits and the wave function, the integral becomes
1=
= (A2/2) a
=
A=
Chemistry : Basic Elements Thus, y = Motion of Wave Packet The particles momentum is equal to its mass times the group velocity vg
vg =
... (1)
where w is the circular frequency (w = 2pv). But this type of wave pattern is inadequate to a general case where the shape of the wave packet changes as the packet moves along. \ We are interested in the movement of the centroid of the wave packet as the latter moves. This implies .
that we want Let
be some operator. Then we have a general equation
= If [
... (2)
is the Hamiltonian operator then we have to find , x]. We will write [
, x] as
...(3) The only non-vanishing term is
=
=
=
=
... (4)
Quantum Chemistry On comparing the identity
= With equation (2) we see that
=x
Hence
or,
=
=
=
The possible values of total angular momentum resulting from the additions of angular momentum with quantum numbers :
(a) j1 = 2, j2 = 4 (b) j1 =
, j2 =
(a) The maximum to minimum values of the quantum number J are given by J = (j1 + j2), (j1 + j2 _ 1), (j1 + j2 _ 2) ... | j1 _ j2 | \ For j1 = 2 & j2 = 4, we have Jmax = 2 + 4 = 6 Jmin = |2 _ 4| = 2 Hence possible values are J = 6, 6 _ 1, 6 _ 2, 6 _ 3, 6 _ 4, \ J = 6, 5, 4, 3, 2.
(b) Jmax =
=
=2
Jmin =
=
=1
\ J = 2, 1
Chemistry : Basic Elements The correct alternative with explanation : The system for which energy (E) increases quadratically with the quantum number (n) is Particle in one dimensional box.
Because E =
where n is the principal quantum number.
If and denote the average potential energy, the average kinetic energy and the average total energy of a particle. If `x' represents the positional coordinate and the potential energy V is proportional to xb, `b' being a constant, then assume that for (1) a harmonic oscillator and (2) hydrogen atom, calculated : For the harmonic oscillator Potential energy (V)
=_
kx2 = (constant) x2
= (constant) xb (given) \b=2 For the hydrogen atom
The ratios
and
Potential energy (V) = (constant) r_l = xb Identifying x with r, we have b=1 It is given that = Quantum Chemistry For the harmonic oscillator b=2 \
=
i.e., . That is average kinetic energy is equal to the average potential energy. This is known as the virial theorem, thus the ratio
= 1 and
=2
For the harmonic oscillator For the hydrogen atom b = _1, But \
= =_
That is average potential energy (which is negative) is numerically equal to twice the average kinetic
energy. Thus
=
. The total energy of the electron in the hydrogen atom is numerically equal to
half of the average potential energy. Total energy = PE + KE i.e.,
=
i.e.,
=
\
=
=
=
Note : In the above discussion for the hydrogen atom, the average potential energy that _
is positive.
The total energy
Since
is negative so
=
is also negative,
Chemistry : Basic Elements Wave function in quantum mechanics represents : A state of the system. A 1s orbital in quantum mechanics refers to ...... A one electron wave function. The zero point energy of a simple harmonic oscillator is:
E0 = Note : Classical mechanics predicts that the zero point energy of the oscillator is zero whereas quantum
mechanics predicts that the zero point energy is non zero. An electron is confined to a two dimensional square box of side 1 nm. The (a) its minimum energy, and (b) the minimum excitation energy are : For a particle in a two dimensional box the expression for energy is
E= where nx and ny are integers excluding zero.
(a) Emin =
=
=
J
= 1.2048 × 10_19 J (b) For the next higher energy level
E=
=
= 3.012 × 10_19J Quantum Chemistry \ DE = (3.012 _ 1.2048) × 10 _19J = 1.8072 × 10_19J For a particle in a one dimensional box with its potential energy zero, the de Broglie relation from its energy expression, deduced: The length of the box can accommodate only an integral number of l/2. i.e.,
a=n
E=
where n is an integer
=
=
But E, which is the kinetic energy of the particle in the box is equal to
\
or p = which is the de Broglie's equation. The following wave functions are acceptable in quantum mechanics over the range when x goes from 0 to 2p. (a) sin x and (d) cos x + sin x are acceptable wave functions whereas (b) tan x and (c) cosec x are not since tan x tends to infinite at x = p/2 and cosec x tends to infinite at x = 0. Hermite Polynomial The Hermite polynomial for v = 2 and v = 1 are H2 (x) = 4x2 _ 2 and H1 (x) = 2x. The wave functions of harmonic oscillator in the stationary states 1 and 2 are orthogonal, shown:
Chemistry : Basic Elements The wave functions for harmonic oscillator are y1 = N1 exp (_x2/2)2x = 2N1x exp (_x2/2) y2 = N2 (4x2 _ 2) exp (_ x2/2)
where x = The above wave functions are orthogonal if the integral
=0 Substituting the values of wave functions and x
=
= Both the integrands are odd functions so the values of both the integrals are zero. Thus the wave functions. y1 (x) and y2 (x) are orthogonal. The allowed wavelengths for a particle in a box is : According to de Broglie principle
l=
... (1)
We also know that boundary conditions are also satisfied by sine functions if
= np ... (2)
From equation (1) and (2), we have
= np and l =
Therefore, the wavelength of the particle in a box can take the values
l= Quantum Chemistry The normalized wave functions for a particle in one dimensional box are : In order for the functions to be normalized, we must have
=
º1
Evaluation of integral after changing the variable, Let
= x,
Then,
= dx
=
=
Hence the constant is given by A = + Therefore, the normalized functions for a particle in a box are
yn = The eigen values i.e., energy of each rotational level, of the rigid rotator problem is givenby
EJ = where I is the moment of inertia and J = 0, 1, 2, ...... the energies of the first three levels, calculated, the energy difference between each level found :
As EJ =
E0 = 0, E1 =
Chemistry : Basic Elements E2 =
, E3 =
The energy difference will be
E1 _ 0 =
E2 _ 1 =
E3 _ 2 = Write the Hamiltonian operator of a free particle moving in one direction under the influence of zero potential energy. In classical mechanics, Hamiltonian is the sum of kinetic and potential energy
H= Here P.E. = 0,
\H=
=
where px is the linear momentum along x-axis. For linear momentum the operator is
, so
=
=
=_ Quantum Chemistry
The operators x and
commute with each other? The condition for the two commuting operators is
Let f(x) be a function
Now,
=
=
=
... (1)
Thirdly
=
=
= f(x) + xf ¢ (x) ... (2)
From equations (1) and (2) it is clear that relation given in the question is not satisfied i.e., these two operators do not commute. A particle in a box with the lowest energy (n = 1). What is the probability of the particles being
between (L/2 + 0.01 L) and (L/2 _ 0.01 L), considered : From the limits we have dx = 0.02 L
The probability density at x =
=
is given by
=
\ Probability =
=
= 0.04
The average position of a particle in a one dimensional box, calculated :
Chemistry : Basic Elements The operator for the position of the particle is
. Hence
=
Let
=
Substituting for x and dx in the above expression, we get
=
=
=
The operators d^/dx and d^ldy commute, shown :
Let f = x2y, then
=
and
= 2x
=
= 2x
Hence
=0 The n = 1 and n = 2 eigen functions of a particle in a box are orthogonal, shown : Substitute the values of n = 1 and n = 2 functions into equation ò y1y2 dv = 0 and evaluation of the resulting integral shows Quantum Chemistry
=
= =0 The functions are therefore orthogonal. A Hamiltonian for a freely moving particle and hence the corresponding Schrodinger equation: If m is the mass of the particle and moving with a velocity vx in the direction of x, then
K.E. = T =
=
Potential energy is zero for freely moving particle. So the classical Hamiltonian is
H = K.E.+ P.E. =
The operator for momentum is
H^ =
i.e.,
=
Substituting the values of H^ in equation H^y = Ey
= Ey
=0 The standard deviation in the velocity of an electron if the uncertainty in its position is 100pm:
Chemistry : Basic Elements
D px >
>
> 5.272 × 10_25 kg ms_1
Dvx =
>
> 5.79 × 105 m/s Normalizing a wave function. Given that a particle is restricted to the region _ a < x < a and has a wave function y proportional to cos
, normalize the wave function :
We must find a constant N such that
= 1, where Y = N cos
Thus 1 =
= N2a
= or N = (1/a)1/2
The eigen functions and eigen values of the operator d/dx:
= kf (x)
= k dx In f (x) = kx + c f(x) = ec ekx = c¢ ekx where c and c¢ are constants. For each different value of k there is an eigen function c¢ ekx or to put it another way, the eigen function c¢ekx has the eigen value K. Quantum Chemistry The ground state energy for an electron that is confined to a potential well with a width of 0.4 nm :
E=
= = 3.765× 10_19 J
= = 226.7 kJ mole_1 The most probable translational energy for an atom in a gas at temperature T is equal to 3/2 KT where K = R/NA is Boltzmann's constant. Calculate the degeneracy of the most probable energy level for an argon atom at 300 K and 1 bar pressure, assuming that the atom can be treated as a particle in a three dimensional box. The volume of the gas at these conditions is 0.022 m3: As volume is 0.022m3, the side of the box `a' can be taken as (0.022)1/3 m.
We have E =
or, n2 =
= n2 = 6 × 1020 \ n = 2.25 × 1010
= 3/2 KT
(This is the most probable value) The degeneracy is of the order 6 × 1020 for this level.
Chemistry : Basic Elements The reduced mass and moment of inertia of H 35Cl. The equilibrium internuclear distance Re = 127.5 pm. The values of L, Lz and E for the state with J = 1 are : Atomic masses of H is 1.007825 & Cl is 34.96885 amu. Reduced mass
m= = 1.62665 × 10_27 kg Moment of inertia I = mRe2 = (1.62665 × 10_27 kg) (127.5 × 10_12 m)2 = 2.644 × 10_47 kg m2
L=
=
= 1.491 × 10_34 Js
Lz = = _ 1.054 × 10_34 Js, 0, 1.054 × 10_34 Js.
E=
=
= 4.206 × 10_22 J The equation for a standing wave in a string has the form y (x, t) = y (x) cos (wt), (a) Calculate the time averaged potential energy and kinetic energy for this motion. (b) Show that E µ y2(x). Quantum Chemistry P.E. = _ òFdy,
F = ma, a = = _ w2y (x) cos wt = _w2y (x, t) or, P.E. = _ òFd y = _ ò_ mw2y (x, t) dy
=
mw2 y2 (x, t)
=
mw2 y2 (x) cos2 wt
K.E. =
=
=
Hence E = K. E. + P. E. = The following functions are acceptable as wave functions: Acceptable as y ® 0 as x ® + ¥ The expressions for the following operators are :
=
=
=
=
or,
=
Chemistry : Basic Elements
=
=
=
=
=
= The operator Aop is said to be a Hermitian operator if it satisfies the expression òyn* Aop ymdt = ò ym A*op yn* dt where ym and yn are the two eigen functions of Aop. The eigen value of a Hermitian operator are all real, proved : Let yn be one of the eigen function of the operator Aop and let an be the corresponding eigen value. We will have Aopyn = anyn ...(1) Multiplying both of sides of equation (1) by y*n and then integrating, we get òy*n Aop ynn dt = an ò yn* yn dt .. .(2) The complex conjugate of equation (2) must also be valid. Thus we have òyn A*op y*n dt = a*n òyn y*n dt ...(3) Since Aop is Hermitian the left hand sides of equation (2) and (3) are equal. Thus we have an ò y*n yn dt = a*n ò yn y*n dt or, (an _ an* ò y*n yn dt = 0 Since ò y*n yn dt is non zero, we must have an = an* which will be applicable only when the eigen value a is real. Quantum Chemistry
Quantum Mechanical Operator
The quantum mechanical operator for the linear momentum in one direction is d/dx. The operator applied on the eigen functions of a particle in a one dimensional box and thus shown that these functions are not eigen functions of the momentum operator and suggest a possible reason for this. The expression of eigen function for one dimensional box is given by the expression
yn =
\
=
=
¹ (constant) yn it follows that yn is not an eigen function of the momentum operator. The particle in the box has only kinetic energy which is related to the momentum p by E = p2/2m. Thus p = ± Thus for a given value of E, two directions of the momentum operator are possible and hence it is not uniquely defined. Making use of the expression Dp Dx > , shown that for a free particle dE Dt > h/4p where dE and Dt are the minimum uncertainties in energy and duration of measurement of velocity respectively.
Chemistry : Basic Elements For a free particle the energy expression is
mv2 =
E=
The uncertainty in energy will be given by
dE = where dp is the uncertainty in p. Now p/m which is velocity, may be measured by determining the distance Dx traversed in time Dt, such that
= If Dx is the maximum uncertainty in position of the particle during the measurement of p, then the minimum uncertainty in the value of p will be
dp >
Hence dE =
=
>
or, dE Dt > The ground state energy of an electron in a one dimensional box of 300 pm length is about 4 eV. The radius of a hydrogen atom is about 100 pm and suppose that the electron in hydrogen atom is thought of as being in three-dimensional cubic box of 100 pm on the side. The energy of the hydrogen electron on this basis, estimated : The ground state energy in one dimensional box is
E=
, where l = 300 pm
Quantum Chemistry
\ E1 =
Hence,
= 4eV
= (4eV) (300 pm)2
The energy expression in a three dimensional box is
E¢1 = For the ground-state electron in the hydrogen atom nx = ny = nz = 1 and it is given that l1 = l2 = l3 = 100 pm
Thus E¢1 =
=
= (4 eV) (300 pm)2 = 4 × 9 × 3eV = 108 eV The function y(x) = A exp (ikx) represents a de-Broglie wave, shown : As x changes to x + l (l = wavelength of the de-Broglie wave) y (x + l) = A exp (ik (x + l))
= A exp (ikx) exp (ikl) = A exp (ikx) exp (ikh/p) = A exp (ikx) exp (2pi)
= A exp (ikx) (cos 2p + i sin 2p) = A exp (ikx) = y
Chemistry : Basic Elements That is the function y(x) repeats after an interval of l which is the wave length.
and
If two operators and
is also Hermitian if and only if
commute, then shown :
Since
=
, the integral
=
=
means operating on the function
(Because Since
are Hermitian then their product
is Hermitian, the integral is equal to
=
Therefore,
=
(because
is Hermitian)
is also Hermitian.
The energy required for a transition from the nx = ny = nz = 1 to nx = ny = 1, nz = 2 state for an Argon atom (atomic mass = 39.95 g) in a cubic container with a 1.0 cm side, determined:
The mass of one argon atom = = 6.633 × 10_26 kg
E1, 1, 1 =
= = (8.274 × 10_39 J) (3) = 2.482 × 10_38 J Quantum Chemistry E1, 1, 2 = (8.274 × 10_39 J) (12 + 12 + 22) = (8.274 × 10_39 J) (6) = 4.964 × 10_38 J Hence, DE = E1, 1, 2 _ E1, 1, 1 = (4.964 × 10_38 J _ 2.482 × 10_38 J) = 2.482 × 10_38 J The value of x at which the ground state wave function of the harmonic oscillator exhibits the maximum, determined: The ground state wave function is
y0 = Making use of the mathematical criterion of maximum (i.e. d y0/dx = 0), we get
=
=0
The above expression gives x = 0 at which the function y0 will exhibit a maximum.
Shown that
= _ 1.
By definition
= Let y (x) be the operand, then operating on y (x) by the right hand side expression, we have
=
=
= _y
Chemistry : Basic Elements Removing y from both sides of the equation, we have
=_1 If the walls of the one dimensional box are suddenly removed, then it happens : When the walls of the particle in one dimensional box are suddenly removed, it becomes free to move without any restriction on the value of the potential energy. If the potential energy is taken to be zero, then the solution of Schrodinger equation
= 0 ...(1)
is given by the equation (2) We know that y = A cos Kx = B sin Kx as A must be zero y = B sin Kx Since y = 0 at x = 0 and at x = 0 we have B sin Ka = 0 or, sin Ka = 0 or Ka = np
so that K =
. ...(2)
However the arbitrary constants A or B and K in above equation can have any values which may be assigned to them. Thus, the energy eigen values given by E = are not quantized. Hence we conclude that a freely moving particle (which may be an electron) without any restriction has continuous energy spectrum. Quantum Chemistry The magnitude of the angular momentum of an electron that occupies the following atomic orbitals, calculated : We know that
= [l (l + 1)]1/2 h/;
l = 0, 1,2,3 (h/ = l.055 × 10_34Js)
...
h/ =
For 1s electron :
l = 0;
=0
For 2s electron :
= [1(1 + 1)]1/2 h/
l = 1;
= 1.414 h/ = 1.49 × 10_34 Js For 3d electron l = 2;
= [2 (2 + 1)]1/2 h/ =
h/ = 2.58 × 10_34 Js
The average distance of the electron from the nucleus in the ground state of hydrogen atom calculated: given that the normalized ground state wave function is :
y1s =
.
In quantum mechanics, the average (or expectation) value of dynamical variable r is given by = òyn ryn dt In the present case, the average distance of the electron from the nucleus is given by = òy1n ry1s dt
Chemistry : Basic Elements Since dt = r2 dr sin q dq df, hence integrating between appropriate limits, we get
=
=
=
The Stationary State is : A state is said to be in a stationary state if the function explaining the state does not include time. Thus the function f (x, y, z) is a stationary state whereas function y(x, y, z, t) is not. Matched :
(1) Linear Operator (a) Ñ2 =
(2) Commutator (b) Ñ = (3) Vector Operator (c) (4) Laplacian (d)
=0
(f + g) =
f+
g
(1)_(d), (2)_(c), (3)_(b), (4)_(a). If two operators
and
commute then they have the same set of eigen functions :
Suppose, the operator has eigen function yi, then yi = ai yi where ai is the eigen value. Since yi =
yi =
Quantum Chemistry
ai yi = ai
and yi
commute
This shows that if (
yi is an eigen function of the operator
with the eigen value a1. This is possible only
yi) is a multiple of yi, i.e.
yi = bi yi In other words, yi is also an eigen function of
.
A projection operator has only two eigen values 0 and 1, shown : Let f be the eigen function, of
with the eigen value l. Thus.
f = lf 2f
=
f = lf = l2f
We then have l2f = lf l(l _1) f = 0 i.e., l = 0, or, l = l Although the `s' orbital in an atom has a (small but) non zero overlap with the nucleus, the electron normally never gets trapped within the nucleus, accounted : The size of the nucleus » 10_13 cm Take this as Dx
Then Dpx =
» 10_14g cm sec_1
px must at least be equal to this. The kinetic energy corresponding to this for an electron is
=
Chemistry : Basic Elements = 5.6 × 10_2erg = 5.6 × 10_2 × 0.63 × l012eV = 3.53 × 1010eV This is so enormously large that no reasonable potential well in the nucleus can retain the electron. The mass of a particle which must provide a binding for the nucleus within the nucleus is: Assume the velocity of the particle to be » c (the velocity of light). If Dx is the size of the nucleus,
then Dt = Further
DE × Df » h/
\
» h/
If the mass of the particle is m, then DE = mc2
\
» h/
\m»
=
=
gm
Dividing this by the electron mass i.e., 9 × 10_28 gm.
We have
=
= 370
Quantum Chemistry A harmonic oscillator (1 dimension) has a zero point energy: Consider the equilibrium position of the oscillator as the origin. Then the total energy, E = Kinetic energy + Potential energy
= Since both the kinetic energy of the potential energy are > 0, E > 0. Suppose E = 0, then px = 0 and x = 0. Thus both px and x are exactly known and this is not possible according to the uncertainty principle. Thus E > 0 for an oscillator even at the lowest energy level.
The wave function of a particle is y (x) = A exp momentum is: The expectation value of the momentum is
= The value
. The expectation value of its
= = + ah/ y.
Hence the expectation value is
=
= ah/
Since
=1 The zero point energy of a particle in a one dimensional box at a infinite height is: The occurrence of zero point energy in accordance with the Heisenberg uncertainty principle is :
Chemistry : Basic Elements Since En = the particle.
, the energy corresponding to n = 1, i.e., E1 =
is called the zero point energy of
Since this energy is finite it means that the particle inside the box is not at rest even at 0K. This being so, the position of the particle cannot be precisely known. Again since only the expectation value of the kinetic energy is known, the linear momentum of the particle is also not precisely known. Thus occurrence of zero point energy implies uncertainty in the position, Dx and also certainty in the x component of the linear momentum Dpx. This means that occurrence of zero point energy is in keeping with the Heisenberg uncertainty principle.
Analytical Chemistry 8 Analytical Chemistry On a 1000 cm wall coated tabular column of 0·25 mm bore, the helium carrier gas velocity is 40 cm/sec. The retention time tR, for decane is 1·28 min; peak width at half height is 0·88 sec. The retention time for a non-retained compound is:
The average linear velocity, u of the mobile phase is where `L' is the column length and tm is the retential time for a non retained compound.
\
sec or 0.41 min
The partition ratio, where t´R is adjusted retention time which is equal to tR tm.
\
Chemistry : Basic Elements Substances A and B were found to have retention times of 17·30 and 19·92 minutes respectively on a 25·0 cm column. The widths (at the base) for A and B were 1·10 and 1·22 minutes respectively. The average number of plates in the column and the plate height calculated resolutions. Column resolution,
Rs
where WA and WB are the widths at the base for A and B respectively. DZ is the difference in time of their arrival at the detector. DZ = (19·92 17·30) min = 2·62 min WA = 1·10 min and WB = 1·22 min
Number of plates,
Thus
and
Average Basis of Classification of Chromatographic Methods : Classification is based on the phenomenon involving the process of either partition or adsorption. Exclusion chromatography involves separation of sample components according to the molecular size. Recent techniques are gel permeation and sieving separation. Analytical Chemistry The Rf Value : Rf value is the relative rate of solute and solvent.
The substance having same Rf value can be separated by two dimensional paper chromatography. Descending technique is better because compounds with low Rf values which cannot be completely separated by ascending method can be separated by descending technique. The Rf values of individual functional groups in a chemically related compounds are very close. So the term RM is additive and is composed of the partial RM values of the individual functional groups or other groupings of atoms in the molecule. Mathematically RM = log (1/Rf 1). On a column 125 cm long, operated at 160° C, these retention times (in minutes) were obtained : air peak, 0·90, heptane, 1·22 and octane, 1·43. The base widths of the bands were 0·16 for heptane and 0·22 for octane. The relative retention and the resolution for these bands are :
The degree of separation or resolution of two adjacent bands is defined as the distance between band peaks divided by the average band width. i.e.,
Chemistry : Basic Elements Chromatography Matched : (1) Packed Column (a) Tungsten rhenium alloy (2) Capillary column (b) Volatile sulphur or phosphorus compounds (3) Thermal conductivity detector (c) Rubidium silicate bead (4) Flame Ionization detector (d) Ultraviolet radiation
(5) Thermionic Emission detector (e) Fused Silica (very high purity glass) (6) Electron capture detector (f) Electrolytical conductivity (7) Flame photometric detector (g) Stainless steel, nickel or glass (8) Photoionization detector (h) High energy electrons (Beta particles) (9) Electrolytic conductivity detector (i) Pure hydrogen/Air flame (1)(g), (2)(e), (3)(a), (4)(i), (5)(c), (6)(h), (7)(b), (8)(d), (9)(f). The problems arising from using a poorly packed HPLC column. Resettling of the packing with use that can create a void at the top of the column that would lead to broad peaks with poor symmetry. At high mobile phase velocities the equation for reduced plate height becomes approximately h/v = c. Under conditions required to obtain the maximum number of plates, the pressure drop tends to infinity. Measuring column efficiency in HPLC Column efficiency could be measured in terms of plate number, N i.e., N = 16 (V/w)2 \ Narrower the peak (low w) the higher will be the value of N and so more efficient the column will be. Analytical Chemistry Main advantages of TLC (Thin layer chromatography) are ® Simple equipment ® Short development time ® Separation of micrograms of the substances
® Highly sensitive ® Early recovery of separated components and easy visualisation of separated compounds. ® The technique may be employed for adsorption, partition or ion exchange chromatography. The cation exchanger in the H + form required to separate Ca+2 ions from 1 dm3 of 0·1 N CaC12 are : The adsorption effect of an ion is characterised by the distribution coefficient
... (1) where Cr and Cs are the equilibrium concentrations of ions in the respective phases and m is the mass of the exchanger in grams. V is the volume of aqueous phase in cm3. Therefore from equation(1)
The quantity C1/D is the static exchange capacity of the cation exchanger, hence
The mixtures of Zn (II) and Mg (II) usually separated employing anion exchanger rather than cation exchanger are: Several metal ions (e.g., Fe, Al, Zn, Co, Mn, etc.) can be absorbed from hydrochloric acid solutions on anion exchange Chemistry : Basic Elements resins owing to the formation of negatively charged chloro complexes. Each metal is absorbed over a well defined range of pH, and this property can be used as the basis of a method of separation. Zinc is absorbed from 2M acid while Magnesium is not thus by passing a mixture of Zn (II) and Mg (II) through a column of anion exchange resin a separation is effected. Zn (II) is eluted with dilute nitric acid. A cation exchange resin column is saturated with copper(II) ions recovered from rinse waters from plating operations. It is desired to recover the copper and to convert the resin
to the H+ form for reuse. Normally this might be done by washing the column with 3M H2SO4 but 6M HCI is found to be superior. 6 M HCl is found to be superior in this case because it forms tetra chlorocuprate (II) (2 ) and trichlorocuprate(II) (1 ) species with copper(II) which will not bind to the cation exchange sites. In multiple extraction of solute from aqueous to organic phase, if the volume of the organic solvent used each time remains constant and equal to the volume of aqueous phase, the weight unextracted after n such operations is :
and weight extracted in n operations is
Suppose we have a solution containing `m' gm of a substance in `V' ml of solution. This solution is repeatedly extracted using every time `v' ml of organic solvent which is immiscible with the first. Suppose `ml' be the amount of solute that remains unextracted at the end of first operation. Then the distribution coefficient may be written as Analytical Chemistry
... (1) or m1 = m (KV/KV + v) ... (2) Exactly in the similar manner, amount m2 that remains unextracted at the end of second extraction,
... (3) or, m2 = m (KV/KV + v)2 ... (4)
Therefore the weight that remains unextracted at the end of n operations will be m2 = m (KV/KV + v)n and the weight extracted in `n' operations will be given by m mn = m [1 (KV/KV + v)2] Note : The efficiency of extraction increases by increasing the number of operations using only a small quantity of the extracting liquid at each time. Synergistic Extraction The phenomenon in which two reagents when used together, extracted a metal ion with enhanced efficiency compared to their individual action is called synergism. A common form of synergistic extraction is that in which a metal ion, Mn + is extracted by a mixture of an acidic chelating reagent, HR and an uncharged basic reagent, S. The joint action of the reagents is especially pronounced in those cases where the coordination capacity of the metal ion is not fully achieved in the MRn chelate, then the extractant S gives a mixed complex, MRn Sx which is extracted with much greater efficiency than the parent chelate. An example is the synergistic influence of zinc in the extraction and AAS determination of trace cadmium in water.
Chemistry : Basic Elements The weight of Fe(III) left unextracted from 100 ml of a solution having 400 mg of Fe3+ in 6M HCl after two extractions with 25 ml of diethyl ether : D = 150, Calculated : Amount left unextracted,
where, W = amount of solute, Vw = volume of aqueous solution, D = 150, V0 = volume of organic phase.
\
In the extraction of cerium(IV) with 2-thenoyl trifluoroacetone in benzene, the distribution ratio was 999. The volume of organic phase was 20 ml and that of aqueous phase 50 ml, what was the percentage extraction : D = 999, V0 = 20 ml, Vw = 50 ml, Assume E = x is percentage of extraction
The electronic transitions will have the lowest energy : n ® p* transition will generally have the lowest energy. The effect of addition of Co2+aq to the equilibrium reaction is : Co2+(aq) + 4NCS ® [Co(NCS4)]2 (Pale Pink) (Colourless) (Blue) on the % transmittance of the solution. As Co2 + is added to the equilibrium mixture, the concentration of the complex ion in the mixture increases. The percentage of absorption by the complex increases and transmittance decreases. Analytical Chemistry The Matched : Energy changes involved Region in electromagnetic spectrum (1) Nuclear (a) Infrared. (2) Inner shell electrons (b) Radio waves. (3) Valence electrons (c) Visible.
(4) Molecular vibrations (d) Gamma rays. (5) Spin orientation (e) X-rays. (1)(d), (2)(e), (3)(c), (4)(a), (5)(b). By the use of potential energy curves for ground and excited electronic states, the sequence of steps leading to phosphorescence and fluorescence shown :
Raman and Rayleigh scatter can be identified when included in a fluorescence emission spectrum. When excitation wavelength is varied, Rayleigh Scatter will vary exactly as the excitation, whereas Raman shift is a constant energy shift. Any fluorescence peaks vary only in intensity with the wavelength maxima remaining unchanged.
Chemistry : Basic Elements After insertion or injection of the sample into the electrothermal atomizer, the temperature profile with ramp heating drown :
See the Figure Two advantages and a major limitation of each of the following methods-(a) Flame AAS, (b) electrothermal AAS, (c) FES and (d) AFS, stated : Technique Advantages Limitation 1. Flame AAS (i) Relatively inexpensive. Analyte must be in (Atomic absorption, solution. spectroscopy) (ii) Fast technique. 2. Electrothermal AAS (i) Lower detection limits. Less precision, slower. (ii) Handles solid samples directly. 3. FES (Flame emission (i) It complements AAS. It is limited to alkali spectroscopy) and alkaline earth metals. (ii) It is fast and inexpensive. 4. AFS (Atomic fluore- (i) Good for trace analysis. No instrumentation
scence spectrometry) is made commercially. (ii) It is faster than electrothermal AAS. Acetylene-nitrous oxide flame is suitable for elements such as Al, Be and rare earths: These elements form stable oxides. The temperature of Analytical Chemistry acetylene-nitrous oxide flame is about 3000° C and it has an advantage that its burning velocity is almost same as that of air-acetylene. Consequently it can be easily and safely handled in the laboratory. Aqueous solution of a drug (0·10 milli molar) shows a percentage transmission of 50 in a 1 cm cell at 250 run. The molar absorptivity, calculated : According to Beer's law
A = log
= ecl
where T is transmittance, c is concentration in moles per litre, l is path length in cm, e is molar absorptivity.
\
= e × 1 × 104 moles/lit × 1 cm
0·3010 = e × 1 × 104 × 1 e = 0·3010/1 × 104 = 3010 The ratio of number of sodium atoms in 3p excited states to the number in the ground state at 2500 K calculated. The average wavelength of the two sodium emission lines involving the 3p to 3s transition is 5892 Å : = 1/5892 Å × 108 cm = 1.698 × 104 cm1 Ej = 1·698 × 104 cm1 × 1·986 × 1016 erg cm1
= 3·372 × 10 12 ergs (since h × c = 1·986 × 1016 erg cm1) 3s and 3p levels have 2 and 6 quantum states respectively. Pj / P0 = 6/2 = 3.
Hence
or,
=1·7 × 104
Chemistry : Basic Elements The temperature at which plasma sources operate stated. The region where plasma source produces a great number of excited emitted atoms is : Between 7000 and 15000 K. In the ultraviolet region plasma source produces a great number of excited emitted atoms. The difference between emission and excitation fluorescence spectra is : In emission, excitation is carried out at a fixed wavelength and the emission intensity records as a function of wavelength. Excitation fluorescence spectra are obtained by measuring the fluorescence intensity at a fixed wavelength while the wavelength of the exciting radiation is varied. In the determination of selenium in urine by AAS it was found that Beer's law applied up to the 2 ppm range. If the 2 ppm standard was set at 100 and a blank set at zero. The selenium concentration of the sample to read at 73 would be:
Selenium concentration of urine = The advantages of molecular fluorescence methods compared to the corresponding absorption spectrophotometric methods and how do they arise are : Fluorescence is caused by the absorption of radiant energy and re-emission of some of this energy in the form of light. The advantages of molecular fluorescence methods over absorption spectrophotometric methods are
(1) The sensitivity of fluorescence methods is generally 10 to 103 times more than the sensitivity of absorption methods. (2) These methods have greater specificity and selectivity than absorptiometric techniques. This is due to the fact that in luminescence both the wavelength of excitation and emission are selected whereas in absorptiometry only the wavelength for absorption is selected. Analytical Chemistry The g-value of the methyl radical whose centre of ESR spectrum appears at 329.4 mT in a spectrometer operating at 9·233 G. Hz (h = 6·627 × 1034 Js, electron Bohr Magneton = 9·27 × 1024 JT1) is. The number of lines in the ESR spectrum of this methyl radical predicated:
or,
Hence g = 2·0048, The interaction of an unpaired electron with three equivalent protons as in CH3 radical gives four equally spaced lines according to (n + 1) rule. The Polarizability The case with which the electron distribution in a molecule or atom may be distorted by the application of an electric field. The basis of turbidimetric and nephelometric analysis is: Measurement of the intensity of the transmitted light as a function of the concentration of the dispersed phase is the basis of turbidimetric analysis and measurement of the intensity of the scattered light (at right angles to the direction of incident light) as a function of the concentration of dispersed phase constitute the basis of nephelometric analysis.
In the determination of manganese by atomic absorption method, a solution containing an aliquot of unknown strength gives a metre reading of 23. Another sample containing this solution with added 50 ppm of manganese solution gives a metre reading of 92·5. Each reading was corrected for the background. The concentration of manganese in the original solution calculated :
Chemistry : Basic Elements As absorbance is directly proportional to the concentration of manganese. . . \ 23 = mx, where `x' ppm is concentration of Mn. 92·5 = m (x + 50) 92·523 = 50 m or, m = 1·39
and x =
= 16·5 ppm.
Amperometric titration is a better method than polarographic method for quantitative analysis. Amperometric titration is a quick, accurate and convenient method similar to potentiometric and conductometric titration and at the equivalence point there is sharp change in diffusion current. The galvanometer used need not be calibrated. The specific characteristic of capillary does not influence the titration. No polarising unit is used, suitable half cell can be easily used for the purpose. The Match: List I List II Quantity Measured & Variable Controlled Name of method (1) E, i = 0 (a) Linear potential sweep voltammetry (2) E vs volume of titrant, i = 0 (b) Amperometric titrations (3) i vs E, concentration (c) Ion selective potentiometry (4) i vs E, t (d) Conductometric titrations
(5) i vs volume of titrant, E (e) Voltammetry (6) 1/R vs volume of titrant (f) Potentiometric titrations (1)(c), (2)(f), (3)(e), (4)(a), (5)(b), (6)(d). The amperometric titration of iodine with sodium thiosulphate using two indicator electrons. (a) The amperometric titration curve and correlate the several regions on this curve with the corresponding regions on the current potential curves sketched. Analytical Chemistry (b) This titration curve change if excess iodide ion were initially present would :
(a)
(b) Using equivalent conductance values, the general form of the titration curve in (a) titration of Ba(OH)2 with HCl, (b) titration of NH4Cl with NaOH sketched :
Chemistry : Basic Elements
Titration of electroactive substance A with another electroactive substance B was carried out at the potential at which both A and B are reducible at cathode. The current vs volume of titrant B curve and label the diagram accordingly sketched : In this titration the current will drop to the end point and there will be increase again to give a V-shave titration curve.
The residual and limiting current is : Residual or condenser current flowing in a polarographic cell is the sum of the Faradic current (if) and the charging current (ic). Thus Residual current (ir) = if + ic Analytical Chemistry The limiting current (il) is the sum of the diffusion current (id) and the residual current (ir). Thus il = id + ir The merits of cyclic voltammetry are : Merits are in the realm of qualitative or diagnostic experiments. A cyclic voltammogram is to the electrochemist what the frequency domain spectrum is to the spectroscopist. It gives a picture of a substance electrochemical behaviour while varying the voltage (energy) of an indicator electrode. Cyclic voltammetry is capable of rapidly generating a new oxidation state during the forward scan and then probing its fate on the reverse scan. Peak height is related to both concentration and reversibility of the reaction. Thus it yields information about reaction reversibilities and also offers a very rapid means of analysis for suitable systems. This method is particularly valuable for the investigation of stepwise reactions and in many cases direct investigation of reactive intermediates is possible. By varying the scan rate, systems that exhibit a wild range of rate constants can be studied and transient species with half lives of milliseconds are readily detected. Matched : (1) U.V. spectroscopy (a) DC Arc
(2) FT. IR (b) Interoferometer (3) AES (c) Xenon flash lamp (4) AAS (d) Thermal conductivity detector (5) Gas chromatograph (e) Hollow cathode lamp (6) Fluorescence & (f) Deuterium discharge lamp Phosphorescence spectro photometry (1)(f), (2)(b), (3)(a), (4)(e), (5)(d), (6)(c). The full form of the techniques mentioned below and the quantity measured by them given DSC, DTA, TGA, TMA, DMA, EGA.
Chemistry : Basic Elements (1) DSCDifferential Scanning Calorimetry (Heats and temperatures of transitions and reactions). (2) DTADifferential Thermal Analysis. (Temperature of transitions and reactions) (3) TGAThermo gravimetric analysis. (weight change) (4) TMAThermomechanical analysis. (Dimension and viscosity changes) (5) DMADynamic mechanical analysis. (Modulus, damping and viscoelastic behaviour) (6) EGAEvolved gas analysis (Amount of gaseous products of thermally induced reaction) The probable thermal decomposition patterns of CaC2O4·H2O deduced using the thermogram (obtained in air) given below (Atomic weights of Ca, C and O may be assumed to be 40, 12 and 16 respectively).
The probable thermal decomposition of CaC2O4·H2O according to above thermogram will be CaC2O4·H2O ® CaC2O4 + H2O CaC2O4 ® CaCO3 + CO CaCO3 ® CaO + CO2 Analytical Chemistry A purity of a compound can be checked by Differential Scannig calorimetry (DSC) this way: In cases where Van't Hoff equation is applicable and material is pure about 97 to 100% a single endotherm is obtained using a very small sample and heating at the rate of about 1·25° C/min. The curve so obtained is integrated to the vertex by parts and the temperatures are tabulated with the partial areas upto that temperature. The following equation is employed,
where Ts is the instantaneous temperature of the sample in K, To·melting point of the infinitely pure sample (solvent) in K, R = gas constant, DH = Heat of fusion of sample, F = fraction of the total sample melted at Ts. Now a plot of temperature as a function of the reciprocal of the fraction melting, 1/F to that temperature is made which should give a straight line of slope KT02/DH with an intercept of T0. AH can be determined from the DSC curve. The three types of automatic chemical analyzers with respect to the mode of operation, advantages and disadvantages compared :
Analyzer Mode of Operation Advantages Disadvantages (1) Discrete Batch, serial proce- Runs selected test on Expensive prepackaged ssing of sample. limited number of reagents and cells, samples, quick start up. mechanically complex. (2) Continuous Continuous pro- Fast, simple equipment, Difficult to vary test flow cessing of large reagent cost minimised. doing a series of number of sam- samples, tubing is ples for same critical. series of tests. (3) Centrifugal Sample mixed Fast, all samples and Loading samples is with reagent by controls measured under time consuming, can centrifugal force. the same conditions. run only a limited number of tests.
Nuclear Chemistry 9 Nuclear Chemistry Order of Radioactive Disintegration Process : Radioactive disintegration is similar to a chemical reaction of the first order, that is, a chemical reaction in which the rate of the reaction varies as the concentration of one molecular species only. Suppose such a reaction is representation by the equation A®B Let the number of atoms of A originally present be N0 and number of atoms of A, Present at any given time t, be N. Then, the rate of change of A into B is
=
…(1)
= Disintegration constant or decay constant Thus, _ldt = dN/N …(2) If, dt = 1s
l=
Chemistry : Basic Elements On Integtrating equation (2)
=
…(3)
or In N = -lt + c …(4) where c is the integration constant.
At t = 0, N = N0 \ c = In N0 Substituting in equation (4) we have In N =
…(5)
or, In (N/N0) = -lt …(6)
or,
…(7)
Relation between half-life and decay constant : Suppose, after time t1/2 half of the atoms of the radioactive substance have disintegrated , that is N = N0/2 Substituting in equation (7) we have
= or, In 0.5 = -lt1/2 or, 2.303 log 0.5 = -lt
Half-life period of a given radioactive substance is independent of the amount of substance present initially. It depends only on disintegration constant of the element. Nuclear Chemistry Half-life nuclide 220Rn is 54.5 s
l= We know, 1 mci = 3.7 × 107 disintegration per sec.
= lN
N= Mass of
= Number of Rn nuclei
mass of one Rn nucleus
= or, = An accident occurs in a laboratory in which a large amount of radioactive material with a known t1/2 of 20 days becomes embedded in floor. Tests show that the level of radiation is 32 times the permissible limit. The laboratory can be safely occupied after : Since the initial level of radiation is 32 times the permissible limit, we have to calculate the time t in which the activity drops to 1/32 of it initial value, i.e.,
= Disintegration constant,
l=
Chemistry : Basic Elements Suppose the laboratory becomes safe for use after t days. Than
=
=
=
t= Radioactive Equilibrium Radioactive equilibrium : If in a disintegration series A, B, C, D etc. are some of the intermediates consecutive atoms i.e., a stage may come when the amounts of A, B, C, D etc. become constant which is so because equal. If then
=
But -
=
=
etc., represent the number of atoms of A, B, C, D etc., at equilibrium,
Hence
=
or,
= ...
=
Nuclear Chemistry Thus the amounts present at equilibrium are inversely proportional to their disintegration constant or directly proportional to their half-lives. Radioactive equilibrium may be of two types— Secular equilibrium : It is limiting case of radioactive equilibrium, in which the half-life of the parent is many times greater than the half-life of the daughter) usually by a factor of 104 or greater). After a sufficient long time
NB = Transient equilibrium : It is similar to secular equilibrium but differs in that the half-lives differ only by a small factor. After a sufficiently long time
NB = No. of atoms of 226Ra in 1 g of
= At equilibrium =
or,
=
=
= =
Chemistry : Basic Elements Reason Having more Biochemical Importance is biochemically more important because of its shorter half-life period. Shorter halflife ensures attainment of transient equilibrium faster. barrier height for an a- particle inside the
nucleus
Barrier height =
= A thin sample of gold was irradiated in a thermal neutrons flux of 1012 neutrons cm2 sec_1 for 25.6 hrs. In the reaction the nuclide 198 Au is produced with a half-life of 64 hours. If the thermal neutron absorption cross section is 98 barns, the specific activity of the sample is:
A= Suppose 1g of the gold was irradiated,
then N = f= s= t1/2 = Substituting these values, we get
A= = Nuclear Chemistry A sample of 100 mg of a radioactive nuclide decay to 81.85 mg of the same in exactly 7 days. The decay constant for this disintegration and the half life of the nuclide. Calculated: Disintegration constant,
l= t=
l=
Half-life, t1/2 = A sample of river water was found to contain 8 × 10_18 tritium atoms,
per atoms of ordinary
hydrogen. Tritium decomposes radioactively with a half-life of 12.3 Y. The ratio of tritium to normal hydrogen atoms 49 Y after the original sample was taken if the sample is stored in a place where additional tritium atoms cannot be formed. It would be : 49 years is almost exactly four half-lives (4 × 12.3 = 49.2Y), so we can avoid the log formula. The fractions of the tritium atoms remaining after four half-lives would be (1/2)4 = 1/16. The normal hydrogen atoms are not radioactive, and therefore, their number does not change. Thus the final ratio of tritium to normal hydrogen will be
= Using the approximate equation for the radius of the nucleus, calculate the density of the nucleus of and density of metallic silver (10.5 g/cm3). Compared : The approximate equation for the radius is r = R0 A1/3
Chemistry : Basic Elements where r is the radius of nucleus of mass number A and Ro is a constant whose value is equal to 1.5 × 10_13 cm So r = 1.5 (107)1/3 × 10_13 = 7.12 × 10-13 cm The volume, mass and density of a single nucleus are calculated from the radius by the following equations
V=
m=
d= The nuclear density is 1.18 × 1014 times that of silver metal. The ratio of the nuclear radii of the s - particle and the proton, estimated : is a-particle Radius equation = R0 A1/3
=
= 41/3 = 1.587
Suppose M represents the mass of last bound neutron from
and M1 that of
Let E be the energy needed to remove the
From the law of conservation of energy
E=
E= where mn is the mass of neutron
=
= Nuclear Chemistry A nucleus splits into nuclei, which have mass numbers in the ratio of 2 : 1. Assuming R = 1.5 f, The ratio of radii of nuclei formed : Following the equation
r = R0 A1/3 R0 is given 1.5 f For the two nuclei formed r1 = R0A11/3; r2 = R0A21/3 Since the mass numbers of nuclei formed are in the ratio of 2 : 1, therefore, r1 = 1.5 f × (2 × 235/3)1/3 = 8.056 f r2 = l.5f × (l × 235/3)l/3 = 6.418 f The average mass of chlorine is 35.5. The mass numbers of two isotopes are 35 and 37. The proportion to which two isotopes are present in ordinary chlorine is : Let the proportion of CI-35 and Cl-37 be 1 : x.
Then the average mass would be But this is given to be 35.5
So
or,
= 35.5
35 + 37x = 35.5 (1 + x) 35 + 37x = 35.5 + 35.5x 1.5x = 0.5 x = 1/3
Therefore, the ratio is 1 : 1/3 or 3 : 1.
Chemistry : Basic Elements Justified Statement"We need a very powerful crane of lift microscopic nuclear mass." Suppose volume of nuclear mass of microscopic size to be lifted is 10_4 cm3. Since the mass of a nucleon is approximately 1 amu, hence the mass (m) of the nucleus of mass number 1.66 × 10_27kg. The nuclear radius is given by r = R0 A1/3 where, R0 = 1.5 × 10_15m Nuclear volume, V = 4/3 pr3 = (4/3)p(R0 A1/3)3 = (4/3)p(1.5 × 10_15)3Am3 Hence, the nuclear density,
r=
= Thus we see the nuclear density is very high and is independent of the mass number of the nucleus. Therefore, all nuclei have approximately the same density. Now volume of the nuclear mass = 10_4 cm3 = 10_10 m3 (Q 1 m3 = 106 cm3) Mass = volume × density
= 10_10m3 × 1.17 × l0I7 kg m_3 = 1.17 × l06 kg Such a high mass can only be lifted only by a powerful crane. Cross Section The probability of nuclear process is generally expressed in terms of cross-section, which is denoted by s that has the Nuclear Chemistry dimensions of an area. This originates from the simple picture that the probability for the reaction between a nucleus and impinging particle is proportional to the cross sectional target area presented by the nucleus. The cross section for a particular nuclear process is defined by the equation Ri = lnx si This equation applies when there is a well defined beam of particles incident on a target. where, R, = number of processes of the type under consideration occurring in the target per unit time, I = number of incident particles per unit time, n = number of target nuclei per cubic centimetre of target, si = cross section of the specified process, expressed is square centimetres. x = target thickness in cm. The total cross section for collision with a fast particle is never greater than twice the geometrical crosssectional area of the nucleus and therefore, fast particle cross-sections are rarely much larger than 10_24 cm2 (radii of the heaviest nuclei are about 10_12 cm). Hence a cross-section of 10_24 cm2 is considered as big as barn and 10_24 cm2 has been named the barn (b) and used in expressing cross-sections. The millibarn (mb = 10_3 b), microbarn (mb = 10_6 b) and nanobarn (nb = 10_9 b) are also commonly used. Cross Section and Reaction Rate : For a sample containing N nuclei in a flux of f particles per square centimetre per second, the rate of reaction of type i, which have a cross section a, is given by Ri = f Nfi
This is the situation when a sample embedded in a uniform flux of particles incident on it from all directions and exactly this happens in a nuclear reactor.
Chemistry : Basic Elements How long a 60 mg piece of Co wire has to be placed in a flux of 5 × 1013 thermal neutrons per square centimetre per second to make 1 mci (1 millicurie = 3.7 × 107 dis s_1) of 5.22 Y 60Co, calculated. The cross-section for the reaction 59Co (x, y) 60Co is 3.7 b. Following the equation Ri = f Nsi
R= = 1.13 × 1012 atom s_1
From equation
we calculate that 1 mci of 60Co corresponds to 8.87 × 1015 atoms.
Thus it will take 8.87 × 1015/1.13 × 1012 = 7.85 × 103 sec or, approximately 2.2 hr to produce 1 mci of 60Co. The effective neutron capture radius of a nucleus having a cross section by 1.0 barn, calculated : l.0 barn = l.0 × l0_24cm2 The area of a circle is given by A = pr2 Hence r =
=
= 5.6 × 10_13 cm.
The ratio of number of neutrons to protons decides the so called `belt of nuclear stability'. Showing this by drawing a graph and using this, explain the decay process involving :
Nuclear Stability : n/p ratio : The stability of nucleus depends on the number of neutrons and protons present. Light Nuclear Chemistry nuclei upto atomic number 20 form most stable nuclei when the number of protons p is equal to the number of neutrons n i. e., n/p = 1. For elements with higher atomic number n/p ratio increases progressively to about 1.6. In such cases the number of neutrons is greater than that of protons. With increase in the number of protons the force of repulsion between them increases and this tends to lower the stability of nucleus.
By plotting a graph between number of neutrons and protons for the nuclei of various elements it has been found that most stable nuclei (non-radioactive nuclei) lie within the shaded area which is called the zone or belt of stability because it contains the stable nuclei. Nuclei that fall above or below this belt are unstable. Nuclei that fall above the stability belt have more neutrons while those lying below have more protons. Such unstable nuclei would attain stability by undergoing change that would produce a nucleus with n/p ratio within the stability zone. Examples :
(1) Nuclei lying above the belt of stability are richer in neutrons and hence they disintegrate is such a manner that one of their neutrons is converted into a protons i.e.,
Chemistry : Basic Elements ® such nuclei emit p-particle. For example, ® Radioactive Stable or, 6p + 8n ® 7p + 7n + n/p = 8/6 = 1.33 ® n/p = 1/1= 1.00 (2) The nuclei lying below the zone of stability are deficient in neutrons and hence disintegrate in such a way that one of their protons is converted into a neutrons. The conversion of a proton into a neutron can be done by any of the following two ways (a) Emission of a positron ® (b) Electron-capture process ® ® or, 12p + 11n ® 11p + 12n + n/p = 11/12 = 0.91 ® n/p = 12/11 = 1.09
(3) Other nuclei having higher number of protons or neutrons
disintegrate by
(a),
decay. More unstable of each of the following pairs and in each case, that type of process could the unstable nucleus undergo: (a) 16C , 16N (b) 18F, 18Ne has the ratio of neutron to protons farther above the belt of stability. It would emit a (a) to set n/p ratio back to the stable range.
particle
Nuclear Chemistry (b) has a lower n/p ratio than into the range of stability.
It could emit a positron or capture a K electron to get that ratio
Each of the following nucleide classified as `probably stable', `beta emitter' or `positron emitter'
The nucleide near the belt of stability are probably stable, those above the belt are beta emitters, those below, positron emitters. Thus the stable hucleides are 208Pb and 120Sn; the beta emitters are 40Ca, 3OA1 and 94Kr, the positrons emitters are l95Hg, 8B and l50Ho. The binding energy per nucleus of 16O is 7.97 MeV and that of I7O is 7.75 MeV. The energy needed to remove a neutron from 17O, calculated : I7O
® 16O +
Energy needed to remove a neutron from 17O is given by E = Energy equivalent of (mass of i6O + mass of neutron _ mass of l7O) = (8mp + 8mn _ 7.97 × 16) + mn _ (8mp + 9mn _ 7.75 × 17)
where mp = mass of proton, mn = mass of neutron = (7.75 × 17 _ 7.97 × 16) MeV = 4.23 MeV The approximate mass of uranium that must undergo fission to produce the same energy as 105 kg of coal, calculated. Assumed that heat of combustion of coal = 8000 cal g -1 and one fission of uranium releases 200 MeV. Mass of coal = 105 kg = 108 g Heat of combustion of coal = 8000 cal g_1 Total heat produced on combustion of 108g coal = 8000 × l08 = 8 × 1011 cal = 8 × 4.184 × 1011J = 35.5 × 1011 J
Chemistry : Basic Elements Thus, we have to calculate the mass of uranium that will undergo fission to produce 33.5 × 1011J of energy. Energy released per fission = 200 MeV = 200 × l06 × l.6 × l0_19J = 3.2 × 10_11 J
Number of U-235 atoms needed to produce this amount of energy =
\ Mass of 1023 atoms of U-235 = `K-electron Capture' When an electron from the nearest orbital i.e., K-shell orbital is absorbed by the nucleus to convert a proton into a neutron without emitting any particle, the process is known as orbital electron capture. Since it is most usually a K-electron which is captured by the nucleus, the process is also known as Kelectron capture. Usually an electron from higher energy level L, M, N etc. drop back to fill the vacancy
in K-shell.
® ® ® Spallation Reactions In these reactions, a high speed projectile having high energy (400 MeV) chips a fragment off from a heavy nucleus and a large number of light particles are emitted. The daughter nucleus has atomic number 10-20 units less and mass number about 30 units less than the parent nucleus. For example, ®
Nuclear Chemistry Nuclear Reactions completed by writing equations (i)
(ii)
(iii) 3H (a, n) ... (iv) (vi) .... (D, T)O18
(v)
(viii) 30P (..., p) 33S
(vii) (i) (ii)
® ®
(iii)
®
(iv)
®
(v)
® ®
(vi) (vii)
®
(viii)
®
If an atom of , after the absorption of slow neutrons, undergoes fission to produce 139Xe and 94Sr. Other particles are also produced. The fission reaction is ® Thus, three neutrons are emitted. Nuclear reactions balanced : (i)
® X + 5a + 3b
(ii)
® X + 6a + 4b
Chemistry : Basic Elements (iii) (iv)
® X + 8a + 6b ® X + 6a + 4b
(i)
®
(ii)
®
(iii)
®
(iv)
®
Q-value of Nuclear Reactions : All types of nuclear reactions are accompanied by either absorption or release of energy. The amount of energy released or absorbed in a nuclear reaction is called Q-value of the nuclear reaction. Thus the following two cases arises (i) When the sum of the masses of the reactants (mR) is less than that of the product (mP) (mR < mP), energy is absorbed i.e., when there is an increase in mass, energy is absorbed. Q-value (Energy absorbed) = Increased in mass in amu × 931 MeV = (mP _ mR) × 931 MeV When Q is absorbed it is given negative sign and the reaction is known as endoergic. For example, ® (ii) When the sum of the masses of the reactants is greater than that of the product (mR > mP), energy is released i.e., when there is a decrease in mass; energy will be released. Q-value (Energy released) = Decrease in mass in amu × 931 MeV = (mR - mP) × 931 MeV Nuclear Chemistry Q-value which is released is generally represented by a positive sign and the nuclear reaction is known as
exoergic. For example, ® Q-value of the following nuclear reactions calculated : (i)
®
(ii)
® ®
(i)
The change in mass in atomic mass unit (amu) can be found as follows Reactant Particles Product Particles = 4.00387
= 17.00450
= 14.00753
= 1.00814
Total = 18.01140 = mR Total = 18.01264 = mP Increase in mass = (mP _ mR) amu = 18.01264 _ 10.01140 = 0.00124 amu \ Q-value = 0.00124 × 931 = 1.15 MeV Since mR < mP, energy is absorbed in this reaction and hence Q-value is negative i.e., Q-value = _1.15 MeV. (ii)
®
Here, the mass values are as under Reactant Particles Product Particles = 1.00814
= 4.00387
= 7.01822
= 4.00387
Total = 8.02636 = mR Total = 8.00774 = mP
Chemistry : Basic Elements Decrease in mass = (mR _ mP) amu = 8.02636 _ 8.00774 = 0.01862 amu \ Q-value = 0.01862 × 931 MeV = 17.3 MeV Since mR > mP, energy is released in this reaction and hence Q-value is positive i.e., Q-value = + 17.3 MeV. 14C
is believed to be made in upper atmosphere by an (n, p) process on I4N. Q for this reactionis: The reaction (n, p) is
=
Reactant Product I4N
= 14.00753 1H = 1.00814 = 1.008665 14C = 14.00324
Total = 15.016195 = mR Total = 15.01138 = mP
Decrease in mass = mR _ mP = 15.016195 _ 15.01138 = _0.004815 amu Q = 0.004815 × 931 = 4.482 MeV Q-value for the 7Li (p, n) 7Be reaction, evaluated : Reactant Product 7Li
= 7.01822 7Be = 7.019465
1H
= 1.00814 1n = 1.008665
Total = 8.02636 = mR Total = 8.02813 = mP Increase of mass = mP _ mR = 8.02813 _ 8.02636 = 0.00177 amu Q = 0.00177 × 931 = 1.64 MeV Nuclear Chemistry Q mR < mP, energy is absorbed and hence Q-value = _ 1.64 MeV. This energy is supplied as kinetic energy of the bombarding proton and is part of the acceleration requirement for the proton supplied by the particle acceleration. The Q-value for the 3He (n, p) reaction is 0.76 MeV. The nuclidic mass of 3He is : The reaction is Reactants Products
®
3He
= x 1H = 1.00814
= 1.00865 3H = 3.01605 Total = x + 1.00865 Total = 4.02419 The mass lost must be = 0.761931 = 0.0008164 Then (x+ 1.00865) _ 4.02419 = 0.0008164 x = 3.01634 Nuclear fission. Explain why is
not suitable for a chain reaction, defined :
Nuclear Fission : The splitting of heavier atom like that of U-235 into a number of fragments of much smaller mass, by suitable bombardment with sub-atomic particles with liberation of huge amount of energy (due to mass defect) is called nuclear fission. For example, + Energy The neutrons released from the fission of first uranium atom can hit other uranium atoms and which again release neutrons, each of which can further hit are uranium atom. In this way, a chain reaction is set up resulting into the liberation of a tremendous amount of energy. not suitable for chain reaction : neutrons is
nuclei do not break up unless the energy of the bombarding
Chemistry : Basic Elements above 1.2 MeV. Such neutrons are called fast neutrons. On the other hand by slow neutrons or thermal neutrons having low energy of value 0.25 MeV slow neutrons form 239U and 239Pu as follows
®
nuclei can be break up when bombarded by
® Nuclear Reactor Nuclear reactor is an arrangement in which the energy produced in a nuclear fission can be used in a controlled manner to produce steam which can run the turbine and produce electricity. Nuclear reactor consists of following parts Fuel Rods : The fuel or fissionable material used in enriched uranium-235 (in the form of U3O8). Control Rods : The energy can be controlled by controlling the fission which, in turn can be controlled by controlling the number of neutrons released during fission. To control the fission cadmium or boron rods are used which absorb the neutrons ® ® Moderator : The material used to slow down the neutrons without absorbing them is called moderator. Graphite and Heavy water are good moderator. Coolant : The material used to remove the liberated heat is called coolant. Generally heavy water is used as coolant. Breeder reactor is different from nuclear reactor : Breeder Reactor : In such type of reactors fissionable nucleides are formed by the nonfissionable nucleides. Nuclear Chemistry Plutonium-238 and uranium-233 are suitable fissionable nucleides and are produced from more abundantly available uranium-238 and thorium-232. For example, (fast) ® ®
® Nuclear fusion reaction takes at very high temperature or why nuclear fusion reactions are known as thermo nuclear reactions : The nuclear reactions in which tighter nuclei fuse together to form a heavier nuclear are called nuclear fusion reactions. Such reactions, occur at very high temperature (of the order of > 106 K) which exist only in the sun or interior of stars therefore, such reactions are also called thermonuclear reactions. Nuclear fusion can take place at high temperature only because lighter nuclei combining together do not have sufficient kinetic energy to overcome the mutual repulsion. ® ® ® Attempts to make fusion reactors are not successful : Controlled fusion can also be carried out; however, attempts are not successful because of the difficulty of producing extremely high temperature in the laboratory and more over which vessel will withstand such high temperatures. Red giant stars, which are coolar than the sun, produce energy by means of the reaction: ® From the nuclidic mass 9Be (9.0154) and 6Li (6.01702), the energy released in MeV calculated compared with the energy released in carbon cycle (30 MeV) and in solar-hydrogen cycle (26.6 MeV). (4He = 4.0015; 1H = 1.00728)
Chemistry : Basic Elements The total mass of product and reactant nuclei are Products = (6.01702 + 4.0015) = 10.0185
Reactants = (9.01504 + 1.00728) = 10.0223 Mass difference = 0.0038 Energy released = 0.0038 × 931.5 = 3.5 MeV This value is much lower than the energy released in carbon cycle or solar helium-hydrogen cycle. 13N
decays by b+ emission. The maximum kinetic energy of b+ is 1.20 MeV. The nuclidic mass of 13N
is : ® Mass difference = [M (nucleus) for 13N] _ [M (nucleus) for 13C] - M (e) = [M (13N) _ 7M (e)] _ [M (t3C) _ 6M(e)] _ M (e) = M(13N) _ M(13C) _ 2M(e) = M(13N) _ 13.00335 _ 2 (0.00055) = M(13N) _ 13.00445 This mass equal to mass expressions of the kinetic energy of the b+
= 0.00129 Then 0.00129 = M (13N) _ 13.00455 or, M(13N) = 13.00574 Nuclear Reactions Completed :
(i)
(ii) (iii) Nuclear Chemistry (i)
(ii)
(iii)
The energy released in a positron decay process calculated: ® (11C = 11.011443; 11B = 11.009305) Mass defect = Mass of reactants _ Mass of product = 11.011443 _ 11.009305 = 0.00213 \ Energy released = 0.00213 × 931 = 1.984 MeV (Positron carries no mass)
A mixture is to be assayed for penicillin. Added 10.4 mg of penicillin of specified activity 0.405 mc mg_1. For this mixture are isolated only 0.3 mg of pure crystalline penicillin and determined its specific activity to be 0.035 mc mg_1. The penicillin content of the original sample was : From the given data, m´ = 1.0 mg, S´ = 0.405 mc mg_1 S = 0.035 mc mg_1 Substituting these values in following equations
m=
=
Chemistry : Basic Elements Tracer and Tracer Technique Most of the element occurs as mixture of isotopes, the proportion of which remains constant throughout the course of physical, chemical and biological changes. Some of these isotopes are radioactive. The radioactive isotopes of a given element behave chemically in same way as non-radioactive isotopes. Hence, if a radioactive isotope is added to its non-radioactive variety the mixture will behave the same way chemically. Since radioactive isotopes can always be detected easily, they may be used as indicators or tracers in various fields. The process in which a radioactive isotope is mixed with non-radioactive variety and the activity of radioisotope is noted after physical, chemical or biological changes, is called radioactive tracing or tracer technique. The success of radioactive tracing lies in the selections of a suitable nuclide. Neutron activation Analysis It is one of the most sensitive and specific methods available for the determination of trace quantities of a wide range elements in such diverse media as terrestial, lunar and meteoritic materials, marine sediments, air borne particles, natural water, biological material, hair, drugs, coal, petroleum, blood, etc.
The essential basis of this analysis is that element to be determined is made radioactive usually by slow neutrons irradiation and then this radioactivity (radiation emitted, on, y) is a measure of the mass of the element originally present. When an element is made radioactive by placing it in a homogenous flux of energetic charged particles of neutrons, the activity produced in the element is given by A = N f s (1 = elt)
or A = where N = the number of atoms of the nuclide in the sample capable of forming radio-isotope in question Nuclear Chemistry f = the neutron flux (i.e., no. of neutrons cm_2 s_1) s = the neutron capture cross section in cm2 for the given target nuclide. t = time of irradiation t1/2 = half-life of nuclear species produced l = decay constant in s_1 of the radioactive product The cross sectional area of the nucleus of a 10B atom assuming a spherical shape calculated. Their value to its cross section for neutrons 3.99 × 10_21 cm2 compared. r = l.4A1/3 × l0_13 = l.4 × (10)1/3 × 10_13 = 1.4 × 2.15 × 10_13 = 3.0 × 10_13 cm Area = pr2 = 3.14 × (3.0 × 10_I3)2 = 2.9 × 10_25 cm2 Compare to cross section 3.99 × 10_21 cm2, in capturing neutrons 10B acts as if it has a much larger cross section than its geometric cross-section i.e., a direct hit is unnecessary.
The fission of
represented by overall reaction considered
® Calculate the total energy released in the fission reaction. The masses in atm of various nucleides are as follows : 236U
= 236.0457, 94Zr = 93.90610, 140Ce = 139.9054,
= 0.00055,
= 1.00867
Mass of the reactants = 236.0457 Mass of the product = 93.90610 139.9053 0.00055 × 6 1.00867 × 2 235.83204
Chemistry : Basic Elements Mass defect = Mass of the reactants _ Mass of the products = 236.045 _ 235.83204 Energy released = 0.21296 × 931.5MeV = 199 MeV Matched : (i)
®
(A) Hydrogen bomb
(ii)
®
(B) Positron emission
(iii)
®
(iv)
®
(D) cc-decay
®
(v) (vi)
(C) Fission reaction
®
(E) Spallation reaction (F) K-electron capture
(G) Exo-ergic reaction (H) Capture reaction (i) (H), (ii) (F), (iii) (E), (iv) (G), (v) (A), (vi) (B). The nucleus and atom can be assumed to be spherical. The radius of the nucleus of mass number A is given by 1.25 × 10_13 × A1/3 cm. The atomic radius of atom is 1 Å. If the mass number is 64, the ratio by the atomic volume that is occupied by the nucleus would be :
= where r and R the radii of the nucleus and atom respectively.
= = 1.25 × 10_13 Nuclear Chemistry If two radioactive nuclide A and B have half-life of t and 2t respectively. If we start an experiment with one mole of each of them, Their mole ratio after a time interval of 6t
would be:
Number of half-lives of A = nA =
Number of half-lives of B = nB Initial amount of A and B = 1 mole each.
NtA =
NtA =
…(i)
Similarly NtB =
Dividing (i) by (ii)
…(ii)
=
Heat energy of the sun is generated by the fusion of four nuclei into one stock of hydrogen in the sun is consumed, then energy will be obtained from :
nucleus. If the
When all the jH nuclei are exhausted by their conversion into nuclei, the temperature an pressure in the sun will decrease and the sun will shrink in its size under its own gravitational attraction. When the sun has considerably shrunk in its size, the temperature in the sun will again became so hig that nuclei will fuse together to form
isotope and a tremendous amount of energy will be liberated
In this way sun will regain its lost sun shine and energy.
Chemistry : Basic Elements An element with mass number 15 and isotropic mass 15.00486 amu has mass defect of 0.124043 amu.
If mass of a proton and a neutron is 1.008145 amu and 1.008986 amu respectively, the atomic number of the element. Mass of the electron can be neglected would be : Let the element be represented as _ a) neutrons.
, thus the nucleus of the atom of this element has a proton and (15
Now, M´ = Mass of protons + Mass of (15 _ a) neutrons = [a × 1.008145 + (15 _ a) × 1.008986] amu M = 15.00482 (given) Mass defect = M´ _ M or, 0.124043 = 1.008145a + 1.008986 × (15 _ a) _ 15.00486 On solving this equation we get a = 7
Organic Name Reactions 10 Organic Name Reactions Favorskii Rearrangement The reaction of a-haloketones (chloro, bromo or iodo) with alkoxide ions to give rearranged esters is called Favorskii rearrangement. For example, 2-chlorocyclohexanone is converted to the methyl ester of cyclopentane carboxylic acid by treatment with sodium methoxide in ether.
The Favorskii rearrangement is an example of a migration to an electron rich carbon atom. Its mechanism with 2-chlorocyclohexanone can be represented as follows—
Chemistry : Basic Elements
(ii)
Completed Reactions :
(i)
(ii)
(i)
Organic Name Reactions Explanation : Both PhCH2COCH2Cl and PhCHCICOCH3 form PhCH2CH2COOH when reacted with OH-, followed by acidification. A common intermediate is required in this Favorskii rearrangement to set the same product. Removal of an aH by OH- is followed by an SN2 displacement of C1- to give a cyclopropanone ring. Ring opening occurs to give the more stable benzylic carbanion.
Favorskii Reaction : The reaction of aldehydes with terminal acetylenes is known as Favorskii reaction.
Sodium acetylides are the most common reagents, but lithium, magnesium and other metallic acetylides have also been used. A particularly convenient reagent is lithium acetylide-ethylene diamine complex. Alternatively, the substrate may be treated with the alkyne itself in the presence of a base, so that the acetylide is generated in situ. 1,4-Diols can be prepared by treatment of aldehyde with dimetalloacetylenes.
The Stork Enamine Reaction with its mechanism : Stork Enamine Reaction : Aldehydes and ketones react with secondary amines to form compounds called enamines. The general reaction for enamine formation can be written as
Chemistry : Basic Elements Water is removed as an azeotrope or by a drying agent. Enamine formation is also catalyzed by the presence of traces of an acid. The secondary amines commonly used are pyrrolidine, piperidine and morphine.
Cyclohexanone, for example, reacts with pyrrolidine in the following way
Enamines are good nucleophiles and have both a nucleophilic nitrogen and a nucleophilic carbon.
The nucleophilicity of the carbon of enamines makes them particularly useful reagents in organic synthesis because they can be acylated, alkylated, and used in Michael addition. Organic Name Reactions When an enarnine reacts with an acyl halide or acid anyhydrides the product is the C-acylated compound. The iminium ion that forms hydrolyzes when water is added, and overall reaction provides a synthesis of b-diketones.
Although N-acylatinn may occur in this synthesis, the N-acyl product is unstable and can act as an acylating agent itself.
As a consequence, the yields of C-acylated products are generally high. Alkylation of enamines may lead to the formation of N-alkylated product, which on heating is converted to C-alkyl compound (This rearrangement is common with allylic halide, alkyl halide or a-haloacetic ester.
Chemistry : Basic Elements Michael Addition Reaction When a new carbon-carbon bond is produced by nucleophilic addition to conjugated systems, the process is called Michael addition. The generalised process involves an a, b-unsaturated compound and a compound containing an active hydrogen attached to a carbon atom (e.g., malonic ester, acetoacetic ester, nitrocompounds, aldehydes, ketones etc.) These are condensed in the presence of a base. The overall reaction and its mechanism can be represented as follows : Mechanism : Overall reaction
Mechanism : Step 1
Step 2
Organic Name Reactions Step 3
Other examples are as follows :
Chemistry : Basic Elements The products of the following Michael addition, predicted: (i) Ethyl crotonate + Malonic ester ® (ii) Ethyl acrylate + ethylacetoacetate ® (iii) Methyl vinyl ketone + malonic ester X. (iv) Benzalacetophenone + acetophenone ® Y.
(v) Acrylonitrile + allyl cyanide ®
The acrolein, H2C=CHCHO is epoxidised much more readily with a basic solution of H2O2 than with a peroxyacid : (b) What is Robinson annelation reaction ? (c) Give the steps for this transformation Organic Name Reactions
(a) Base converts H2O2 to conjugate base, HOO-, that undergoes a Michael addition with acrolein to give an a-carbanion that then displaces HO- from the HOO- group leaving epoxide.
The acid catalyzed epoxidation goes by the typical electrophilic attack by HO+ (from H2O2) on C=C. (b) Robinson annelation reaction involves a Michael addition to an a, b-unsaturated ketone immediately followed by an aldol addition. For example,
(c) The transformation can be achieved by Michael addition using methyl vinyl ketone (Robinson annelation).
The sequence that follows illustrates how a conjugate aldol addition (Michael addition) followed by a simple aldol condensation may be used to build one ring onto another. This procedure is known as the Robinson annulation (ring forming) reaction (after the English chemist Sir Robert Robinson, who won the Nobel Prize in Chemistry in 1947 for his research on naturally occurring compounds).
Chemistry : Basic Elements
The mechanism of the following reaction :
The reaction is a Michael type of addition to a base-induced dehydration product of the (4-hydroxy phenyl) methanol.
The mechanism of addition of cyciohexanone to C6H5CH = CHCOC6H5 : Conjugate addition of enolate anion to a, b-unsaturated carbonyl compounds is known as Michael addition.
Organic Name Reactions The Mannich Reaction with its Mechanism : Mannich Reaction : This reaction takes place between an amine, an aldehyde (or ketone) and a highly nucleophilic carbon atom. The reaction is the addition of a nucleophilic carbon to an imonium ion intermediate. (The reaction is usually carried out in acid solution, but may also be base catalysed. This is the condensation between aldehydes, ammonia or a primary or secondary amine and a compound containing at least one active hydrogen atom e.g., ketones, b-ketoesters, b-cyanoesters, nitroalkanes, alkynes with CºH). For example, C6H5COCH3 + CH2O + HN(CH3)2 Mechanism : Step 1
Step 2
C6H5COCH2CH2N(CH3) 2 + H2O
The products of the Mannich reaction (Mannich bases) are useful intermediate on organic synthesis e.g.
(1) PhCOCH2CH2NHR2+Cl _PhCOCH==CH2 (2) PhCOCH2CH2MR2+
PhCOCH2CH2CN
Chemistry : Basic Elements Completed reactions :
(i) C6H5COCH3
(ii) (iii) NMe2 CHNO2+HCHO+EtNH2
(iv)
(v)
PhCOCH2CH3
(vi)
(i)
(ii)
(iii)
Organic Name Reactions
(iv)
(v)
(vi)
The Sharpless Asymmetric Epoxidation with an example: Sharpless Asymmetric Epoxidation : This is a method of converting allylic alcohols to chiral epoxy alcohols with very high enantioselectivity (i.e., with preference for one enantiomer rather than formation of racemic mixture). It involves treating the allylic alcohol with tert-butyl hydroperoxide, titanium(IV) tetra isopropoxide [Ti(O—iPr)4] and a specific stereoisomer of tartaric ester. For example,
The oxygen that is transferred to the allylic alcohol to form epoxide is derived from tert-butyl hydroperoxide. The enantioselectivity of the reaction results from a titanium complex among the reagents that includes the enantiomerically pure tartrate ester as one of the ligands. The choice whether to use (+) or (-) tartrate ester for stereochemical control depends on which enantiomer of epoxide is desired. The stereochemical preferences are such that it is possible to prepare either enantiomer of a chiral epoxide in high Chemistry : Basic Elements enantiomeric excess, simply by choosing the appropriate (+) or (-) tartrate stereoisomer as the chiral ligand.
The Ene reaction with example. How an ene reaction is related to Diels-Alder reaction: Ene reaction (Hydro-allyl addition)-A reaction of an allylic compound with an olefin which resembles both a cycloaddition and a [1, 5]-sigmatropic shift of hydrogen. In this reaction an alkene having an allylic hydrogen atom reacts thermally with a dienophile (enophile) C = C, C = O, N = N, N = O, C = S
etc. with the formation of a new s bond to the terminal carbon of the allyl group. The interaction of a hydrogen atom with the HOMO of the allyl radical and the LUMO of the enophile is a symmetry allowed process.
As related to Diels-Alder reaction this also represents a 6p electron electrocyclic reaction. Like DielsAlder reaction the ene reaction is reversible. For example, the produce) 1-pentene of the ene reaction between ethene and propene, gives back ethene and propene on decomposition at 400° C.
Organic Name Reactions A reaction with structure between .malefic anhydride and propene under thermal conditions:
Propene Maleic anhydride It is an ene reaction. Barton Reaction The photolysis of nitrites which do not contain g-hydrogen atoms usually results in the elimination of nitric oxide and the formation of hydroxy and carbonyl compounds. When g-hydrogen atoms are present than the product is an oxime or the corresponding nitroso dimer. This reaction is known as Barton S reaction.
The other example of Barton reaction is :
Chemistry : Basic Elements
By Barton reaction a methyl group in the d position to an OH group can be oxidised to a CHO group. The alcohol is first converted to the nitrite ester. Photolysis of the nitrite results in conversion of the nitrite group to the OH group and nitrosation of the methyl group. Hydrolysis of the oxime tautomer gives the aldehyde. The Hofmann-Loffler-Freytag reaction : Hofmann-Loffler-Freytag Reaction : It is a photochemically initiated decomposition of N-haloamines in acidic solution. First d-halo-amines are formed, these are then converted into pyrrolidines by intramolecular nucleophilic substitution, which involves a hydrogen abstraction. The first step of the reaction is a rearrangement, with the halogen migrating from the nitrogen to the 4 or 5 position of the alkyl group and the second step, the ring closure takesplace.
A similar reaction has been carried out on N-haloamides, which give
-lactone.
Organic Name Reactions A similar reaction has been carried out on N-haloamides, which give g-lactone.
The Shapiro Reaction with its Mechanism : Shapiro Reaction : Treatment of the tosylhydrazone of an aldehyde or a ketone with a strong base leads to the formation of an olefin, the reaction being formally an elimination accompanied by a hydrogen shift. This reaction is called Shapiro reaction.
The most useful method synthetically involves treatment of the substrate with at least two equivalent of an organolithium compound (MeLi) in ether, hexane, or tetramethylenediamine. Tosylhydrazones of a, bunsaturated ketones give conjugated dienes. The mechanism of the reaction has been formulated as
Mechanism of Baeyer-Villiger Oxidation : Both aldehydes and ketones are oxidized by peroxy acids. This reaction, called the Baeyer-Villiger oxidation, is especially useful with ketones, Chemistry : Basic Elements because it converts them to carboxylic esters. For example, treating acetophenone with a peroxy acid converts it to the ester, phenyl acetate.
The oxidation may be carried out by Caro's acid (per monosulphuric acid, H2SO5) or with perbenzoic, peracetic or monoperphthalic acid. The products of this reaction show that a phenyl group has a greater tendency to migrate than a methyl group. Had this not been the case, the product would have been C6H5COOCH3 and not CH3COOC6H5. This tendency of a group to migrate Organic Name Reactions is called its migratory aptitude. Studies of the Baeyer-Villiger oxidation and other reactions have shown that the migratory aptitude of groups is H > phenyl > 3° alkyl > 2° alkyl > 1° alkyl > methyl. In all cases, this order is for groups migrating with their electron pairs, that is, as anions. Completed Reactions :
Chemistry : Basic Elements
The product is omega-caprolactone, a seven membered lactone that cannot be made by an intramolecular ring closure.
Chichibabin Reaction Pyridine and other heterocyclic nitrogen compounds an be aminated with the alkali-metal amides by chichibabin reaction. The attack is always in the 2 position unless both such positions Organic Name Reactions are filled, in which case the 4 position is attacked. Substituted alkali-metal amides have also been used.
A similar reaction of this type is
Completed Reactions :
(i)
(ii)
(iii)
(iv)
(v)
Chemistry : Basic Elements
(vi)
(vii)
(viii)
(ix)
(i)
(ii)
Reaction (i) and (ii) are Baeyer-Villiger reaction.
Organic Name Reactions
(The reaction (vii)—(ix) are examples of Mannich reaction) The product and name the reaction :
Chemistry : Basic Elements
Robinson synthesis of tropinone (example of Mannich reaction in synthesis)
Completed sequence of reaction :
(i) A reasonable meachnism for the following reaction :
(ii) The way one may link together, phenol, formaldehyde and a secondary amine. Organic Name Reactions (iii) The structure of the product Indole + CH2O + (C2H5)2NH (iv) Starting from (I and II), one can get III (4,4-dimethyl cyclohexanone)
(ii) This is Mannich reaction, the active hydrogen compound being phenol (the hydrogens in the o- and ppositions are active) and the result is amino alkylation in the ortho position which is preferred.
(iii) Indole undergoes Mannich reaction with CH2O and (C2H5)2NH to give 3-N, N´diethylamino-ethylindole.
(iv) The first reaction is the Michael addition of the enolate anions to the methyl vinyl ketone followed by intramolecular aldol condensation.
Chemistry : Basic Elements
The following canverted :
The structures of the major product in the following reactions :
This is an example of Barton reaction.
Organic Name Reactions This is an example of Shapiro reaction.
The transformations carried out :
Chemistry : Basic Elements
Reagents in Organic Synthesis 11 Reagents in Organic Synthesis Complete the Reactions :
In terms of the catalysts used, The two reactions classified. Shown the steps in the mechanism of reaction : Product in both the reactions is CH3CH2CH3. Both reactions are catalytic hydrogenations (addition of H2). (i) is heterogeneous catalytic hydrogenation and (ii) is homogeneous catalytic hydrogenation. In step-1 an H2 adds to the rhodium complex and one Ph3P ligand (L) is lost, resulting in a five coordinate rhodium complex, A (L = Ph3P). In this oxidative addition, the Rh changes oxidation state from + 1 to + 3. In step-2 the alkene reacts with A to form a p complex, B, which undergoes rearrangement (step-3) of an H to one of the C's of the double bond, the other C forming a s bond to the Rh (C).
Chemistry : Basic Elements
In the last step, a second H is transferred to the other C, and the alkane is lost with simultaneous regeneration of the metal catalyst—
The reagents for carrying out the following transformations, indicated :
Reagents in Organic Synthesis Crown Ethers Crown ethers are cyclic polyethers and their structure permits a conformation with certain sized `holes' in which cations can be trapped by co-ordination with the lone pair electrons on the oxygen atoms. These are used as phase transfer catalysts. The cyclic polymers of ethylene glycol (OCH2CH2)n are named as X-crown-Y. X refers to total number of atoms in the ring and Y to the total number of oxygens in the ring. The ability of a crown ether depends to complex a cation: Why the nucleophilic reactions
under the reaction conditions give almost same quantitative yields ? On the size of the cavity in the crown ether which can be tailored to allow for the selective binding of only certain cations, i.e., whose ionic radius is best accommodated by the polyether. Thus 18-crown_6 shows a high affinity for K+ (ionic diameter 2.66 Å) and 15-crown-5 for Na+ (ionic diameter 1.80Å).
The crown ether, 18-crown-6 acts as a phase transfer catalyst and gets the anion into the organic phase. On coordination with a metal cation the crown ethers convert the metal ion into a species with a hydrocarbon like exterior. The crown ether, Chemistry : Basic Elements 18-crown-6, e.g., coordinates effectively with potassium ions as the cavity size is correct and because the six oxygens are ideally situated to donate their electron pairs to the central ion. Crown ethers render many organic salts soluble in non-polar solvents. Therefore salts like CH3COOK, KCN can be transferred into aprotic solvents by using catalytic amounts of 18-crown-6. In the organic phase the relatively unsolvated anions of these salts bring about nucleophilic substitution reaction on an organic substrate. Phase Transfer Catalyst A compound whose addition to a two-phase organic water system helps to transfer a water soluble ionic reactant across the interface to the organic phase where a homogeneous reaction can take place is called a phase transfer catalyst. These catalysts enhance the rate of a reaction. A quaternary ammonium halide R4N+ X_ e.g., tetrabutylammonium halide is phase transfer catalyst. It can cause the transfer of the nuclepphile for example CN_ as an ion pair (Q+ CN_) into the organic phase; since the cation (Q+) of the ion pair with its four bulky alkyl groups resembles a hydrocarbon even thoughit has a positive charge. It is thus lipophilic, i.e., it prefers a non-polar environment to an aqueous one.
Prevost and Woodward Reactions Both the reactions are essentially the additions of iodine carboxylate (formed in situ) to an alkene, i.e., the reaction of an alkene with iodine and silver salt. The Prevost procedure employs iodine and silver carboxylate under dry conditions. The initially formed transiodocarboxylate (b) from a cyclic iodonium ion (a) undergoes internal displacement to a common intermediate acylium ion (c). The formation of the diester (d) with retention of configuration provides an example of neighbouring group participation. The diester on subsequent hydrolysis gives a trans-glycol. Reagents in Organic Synthesis
In the Woodward procedure, water is present which intercepts the intermediate (c) to give a cishydroxy ester (e) and a cis-glycol on subsequent hydrolysis.
Umpolung : The reversal of polarity of the carbonyl carbon atom is termed umpolung (German: for polarity reversal). Normally the carbonyl carbon atom of an aldehyde (or a ketone) is partially positive i.e., electrophilic and therefore it reacts with nucleophiles. When the aldehyde is converted to a dithiane by reaction with 1,3-propanedithiol and reacted with butyl lithium the same carbon now becomes negatively charged to react with electrophiles. This reversed polarity of the carbonyl carbon is termed umpolung which increases the versatility of the carbonyl group in synthesis. The sulphur atoms stabilize
carbanions because sulphur has the capacity to utilize 3d orbitals for bonding and to occur in valence states higher than 2.
Chemistry : Basic Elements
Dithianes : These are thioacetals which can be prepared by treating an aldehyde with 1,3-propanedithiol in the presence of trace of acid.
A 1,3-dithiane is a weak proton acid (pKa = 32) which can be deprotonated by strong bases such as nbutyllithium. The resulting carbanion is stabilized by the electron withdrawing effect of the two sulphur atoms.
The reactions completed by writing the appropriate product:
Reagents in Organic Synthesis
Reaction completed :
Peterson's Synthesis : In the Peterson's synthesis the b-hydroxysilanes are converted to alkenes in either acidic or basic solution.
Chemistry : Basic Elements
Reagents Used in Transformations
(2) (1) (CH3)2CuLi (Lithium dimethyl cuprate). (2) (CH3)3SiCl (Trimethylsilyl chloride). Lobry de Bruyn-van Ekenstein Rearrangement
(1)
(2)
(3) Reducing sugars on treatment with dicyclohexyl carbodiimide (DCC) undergo rearrangement to form a mixture Reagents in Organic Synthesis of other reducing sugars e.g., glucose forms a mixture of fructose and mannose.
Suitable Mechanism for the Reaction :
(1)
(ii) DCC is dicyclohexylcarbodiimide.
A carboxylic acid cannot be directly converted to an acid amide by the action of an amine. Hence the need of a dehydrating agent like DCC. The acid is converted to a compound with a better leaving group.
Chemistry : Basic Elements
Difference in behaviour of RMgX and R2CuLi: Since the C—to—Mg bond has more ionic character than the C—to—Cu bond, its R group is more like R : _ and is much more reactive. A set of reagents is listed under column A and their uses under column B. The reagent with its most appropriate use—
Column A Column B (1) (Ph3P)3RhCl (a) Hydroboration (2) 18-crown-6 (b) Epoxidation (3) Bu4N+ Br_ (c) Alkylation (4) Disiamyl borane (d) Hydrogenation (5) DDQ (e) Dehydration (6) DCC (f) Dehydrogenation (7) CH3CO3H (g) Cation complexing (8) LDA (h) Phase transfer catalysis (1)—(d), (2)—(g), (3)—(h), (4)—(a), (5)—(f), (6)—(e), (7)—(b), (8)—(c). Reagents in Organic Synthesis Formulas for A through D in the following reactions
LDA rather than NaOEt is used as the base in the abvoe reaction :
Structure for A, B and C
(a)
Tributylstannane (C4H9)3SnH reduces an alkyl halide to the corresponding alkane by a free radical mechanism. The initiator is an azo compound, (CH3)2C(CN)—N = N—C(CN)(CH3)2 which breaks down to N2 and a radical. A possible mechanism : The initiator gives the carbon radical (CH3)2C(CN),(Radical) which abstracts an H from the tin compound. Radical + (C4H9)3Sn—H —® (C4H9)3Sn + Radical—H
Chemistry : Basic Elements The tributyltin radical abstracts a halogen atom from the alkyl halide and the chain is propagated as follows— (C4H9)3Sn · + R—X —® (C4H9)3Sn—X + R · then R · + (C4H9)3Sn—H —®R—H + (C4H9)3Sn · Mathced :
(1) Wilkinsons catalyst (a) Cyclic polyethers
(2) DDQ (b)
(3) Prevost reagent (c) (4) Crown ethers (d) RhCl[(Ph3P)3] (5) 1, 3-dithiane (e) (i-Pr)2N _ Li + (6) disiamyl borane (f) 2,3-dichloro-5,6-dicyano-1, 4-benzoquinone (7) LDA (g) Mixture of silver salt of an acid and iodine. (1)_(d), (2)_(f), (3)_(g), (4)_(a), (5)_(b), (6)_(c), (7)_(e). Application of Phase Transfer Catalysts In water : Na +: CN_ + n-Bu4N+Cl_ _® n-Bu4N+: CN_ + Na + Cl_ In non-polar solvent : n-C8H17Cl + n-Bu4N + : C_ _® n-Bu4N + Cl_ + n-C8H17CN n-Bu4N + Cl _ goes back into H2O and the two steps repeat. Activation of the C-terminus of an AA, followed by coupling with a second AA, is accomplished with the reagent, DCC, Reagents in Organic Synthesis
C6H11—N=C=N—C 6H11. Give the structure of the product of reaction of DCC with RCOOH :
The Merrifield solid-phase process for synthesizing peptides:
The solid phase is beads of polystyrene whose mer is mainly . However occasional benzene rings have p-CH2C1 substituents projecting out to the surface. The solid phase may be indicated as. [P]—CH2C1 (A), where [P] is the polystyrene backbone. The peptide chain is started at the Cterminus by bonding the BOC protected AA to the solid phase through benzyl ester formation, followed by removal of the BOC group.
A second AA (BOC-protected so that it will not self couple) is added, along with DCC. The two steps of addition of a BOC-protected AA and regeneration of NH2 with CF3COOH are repeated as many times as required. In between each step, the [P]-growing peptide chain is washed with suitable solvents to remove excess reagents and undesirable products. Thus intermediates do not have to be isolated and purified and the yields are high. Reactant BOC-AA´s, DCC and CF3COOH are added through an automated system. Removal of the completed peptide from the polymer is accomplished with anhydrous HF, which also removes the last BOC group. The Reactions Completed :
(i)
Chemistry : Basic Elements
(ii)
(iii)
(iv)
(i)
(ii)
(iii)
(iv) Use of SeO2
Reagents in Organic Synthesis
SeO2 is used as oxidising agent, dehydrogenating agent and carbonylating agent. Selenuous acid is formed when water reacts with SeO2.
Chemistry : Basic Elements The following reactions :
Reagents in Organic Synthesis
Baker Yeast
Saccharomyces cerevisiae (Baker Yeast) are micro organisms and are used for the reduction of carbonyl group to hydroxyl group and for reduction of double bond.
Products Formed in the Reactions
Chemistry : Basic Elements In Prevost
With OsO4
In Woodward-Prevost
The Reactions Completed :
(1)
(3)
(2)
(4) E Me3Sil.
Kinetics of Reactions 12 Kinetics of Reactions Methods to Determine Rate Laws Rate law can be determined by the following methods — Differential Method : The reactants are mixed and the concentration of reactant `c' (any one) is measured at regular intervals and a graph is drawn between concentration and time. The slope of the curve is measured. If the order of the reaction is V then the value of V can be calculated from the equation
Slope (rate) =
= Kcn, where `K' is the rate constant.
Integration Method or Hit and Trial Method : Here known quantities of standard solutions of reactants are mixed in a reaction vessel and the progress of the reaction is determined by determining the amount of reactant consumed after different intervals of time. These values are then substituted in the equations of first, second, third order and so on. The order of the reaction is the order corresponding to that equation which gives a constant value of K.
Chemistry : Basic Elements Fractional Change Method : From the equations of half life tor reactions of various orders except first order reaction, time required to complete a definite fraction of the reaction is inversely proportional to a0n_1 where `n' is the order of reaction and `a0' is initial concentration.
tµ Graphical Method : This is an extension of integration method, for a reaction of nth order, the rate of reaction
is equal to K (a0 _ x)n where K, a0 and x are the rate constant, initial concentration of the
reactant and the concentration of the reactant consumed at time `t'. If the plot of straight line than the order of the reaction is `n'. The value of
vs. (a0 _ x)n is a
(rate of reaction) is obtained from the
slope of the plot between the concentration of the reactant various time intervals`t'. In a certain reaction, time, t varies arithmetically while the concentration varies geometrically time 0 t 2t 3t ... Conc. a aa a2a a3a ... Shown that reaction is of the I order. For the first order
K= Substituting values
K=
= Kinetics of Reactions In a second calculation
K=
=
= In a thud calculation
K=
=
Thus we get the same value of K, which confirms I order. Shown that t3/4 = (2n _ 1 + l) t1/2 where `n' is the order of the reaction. Using this in the reaction of pyrolysis of CH3—CHO to determine `n'. t1/2 = 420 secs. t3/4 = 1220 secs in a particular kinetic run : For a reaction of order, `n' we have
Kt =
\ Kt1/2 =
since c =
when t = t1/2
Kt3/4 =
since c =
Dividing,
when t = t3/4
=
= (2n _ 1 + 1)
\ t3/4 = (2n _ 1 + 1) t1/2
Chemistry : Basic Elements In the pyrolysis of CH3_CHO,
=
» 3 = 2n _ 1 + 1
\n=2 On the action of Br2 on fumaric acid, following data were obtained _ First Experiment: (t) min. Concentration 0 8.87 95 7.87
Mean concentration = 8.37 and
= 0.0106
Second Experiment : (t) min. Concentration 0 3.81 1.32 3.51
Mean concentration = 3.66 and
= 0.00227
The order of reaction, found out :
n=
=
= 1.87
n=2 It is a second order reaction. The rate law for the following reaction. Assuming that the rate of formation and disappearance of NOC12 are equal in the first step, derived
Kinetics of Reactions K1 NO + C12
NOCl2(Fast) ... (1)
K2 K3 NO + NOC12 ® 2NOCl (Slow) ... (2)
[NOC1] = K3 [NO] [NOC12] ... (3)
From equation (2), But from equation (1),
K1 [NO][C12] = K2 [NOC12]
or, [NOC12] =
[NO] [Cl2] ... (4)
Substituting the value of (4) in (3), we have
[NO]2 [Cl2]
[NOCl] =
Overall reaction being 2NO + C12 ® 2NOC1 The rate of the reaction is
=
[NO]2 [Cl2]
=
[NO] = _
[C12]
For the Bromide catalysed decomposition of hydrogen peroxide (at pH < 7) 2H2O2 (aq) ® 2H2O (aq) + O2 (g) a mechanism proposed involves three steps (I) a pre-equilibrium step, (II) a reaction between Br_ and [HOOH2]+ to give HOBr and H2O which is considered to be the slowest step and (III) a fast step involving the release of O2 (g) ?
Chemistry : Basic Elements It is further observed experimentally that the rate of this reaction depends on pH and Br_concentration. The complete mechanism and obtain the rate law : K1 Mechanism: H2O2 + H+
[H2OOH]+ ... (1)
K_1 (protonation equilibrium) Br_ + [H2OOH]+ ® HOBr + H2O ... (2) K2 Fast HOBr ® H+ + Br_ + ½ O2 ... (3) Adding H2O2 ® H2O + ½O2 Applying steady state approximation to the unstable intermediate [H2OOH]+ K1 [H2O2][H+]_K_1 [H2OOH]+_K2 [Br_] [H2OOH]+= 0
[H2OOH]+ =
From equation (2)
= K2[Br_] [H2OOH]+
= The step (3) involving the formation of O2 is not shown in the above expression, this step has been omitted, since it is a fast step, subsequent to the rate determining step. The expression shows that the rate of decomposition of H2O2 depends on the concentrations of the catalyst (Br_) and the hydrogen ions. A decrease in pH increases the rate. In the following reaction mechanism the expression for the rate of formation of P : Kinetics of Reactions K1 R1 + R2
I
K2 K3 I is the reactive intermediate. I_ ® P Writing the differential equation for I and equating it to zero.
= K1 [R1][R2] _ K2[I] _ K3 [I] = 0
[I] =
Now
= K3 [I] =
[R1] [R2] = K[R1] [R2]
where K is the effective rate constant.
K= The isomerization of cyclopropane follows the Lindemann mechanism and is found to be unimolecular. The rate constant at high pressure is 1.5 × 10_4 s_1 and that at low pressure is 6 × 10_6 torr_1 s_l. The pressure of cyclopropane at which the reaction changes its order, found out: K1 Lindemann mechanism : A + A ® A + A (Activation) K2 A* + A ® A + A (deactivation) K3 A* ® B (product)
Chemistry : Basic Elements For the steady state concentration of A*
= Suppose the required transition pressure is pA. If the actual pressure is much greater than pA, the reaction is of the first order, i.e.,
= K (pressure)1
If the actual pressure is > K_1, calculated :
r= For [S] >> Km, r = Vmax = K2 [E]0 when r is half of this maximum value, then by definition [S] = Km = (K_1 + K2)/K1 = K_1/K1] (assuming that K2