Physical Chemistry Student Solutions Manual

TH EDITION Student Solutions Manual to Accompany PHYSICAL CHEMISTRY PETER ATKINS • CHARLES TRAPP CARMEN GIUNTA • MARSH...

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TH EDITION

Student Solutions Manual to Accompany

PHYSICAL CHEMISTRY PETER ATKINS • CHARLES TRAPP CARMEN GIUNTA • MARSHALL CADY

STUDENT'S SOLUTIONS MANUAL TO ACCOMPANY

PHYSICAL CHEMISTRY EIGHTH EDITION

STUDENT'S SOLUTIONS MANUAL TO ACCOMPANY

PHYSIC~l

CHEMISTRY Eighth Edition

P. W. Atkins Professor of Chemistry, University of Oxford and Fellow of Uncoln College

C. A. Trapp Professor of Chemistry, University of Louisville, Louisville, Kentucky, USA

M. P. Cady Professor of Chemistry, Indiana University Southeast , New Albany Indiana, USA

C. Giunta Professor of Chemistry, Le Mayne College, Syracuse, NY, USA

II w. H. Freeman and Company New York

Student's solutions manual to accompany Physical Chemistry, Eighth Edition © Oxford University Press, 2006 All rights reserved ISBN-13: 978-0-7167-6206-5 ISBN-lO: 0-7167-6206--4 Published in Great Britain by Oxford University Press This edition has been authorized by Oxford University Press for sale in the United States and Canada only and not for export therefrom. Library of Congress Cataloging in Publication Data Data available Second printing W. H. Freeman and Company 41 Madison Avenue New York, NY 10010 www.whfreeman.com

Preface This manual provides detailed solutions to all the end-of-chapter (b) Exercises, and to the odd-numbered Discussion Questions and Problems. Solutions to Exercises and Problems carried over from previous editions have been reworked, modified, or corrected when needed. The solutions to the Problems in this edition rely more heavily on the mathematical and molecular modelling software that is now generally accessible to physical chemistry students, and this is particularly true for many of the new Problems that request the use of such software for their solutions. But almost all of the Exercises and many of the Problems can still be solved with a modem hand-held scientific calculator. When a quantum chemical calculation or molecular modelling process has been called for, we have usually provided the solution with PC Spartan pro™ because of its common availability. In general, we have adhered rigorously to the rules for significant figures in displaying the final answers. However, when intermediate answers are shown, they are often given with one more figure than would be justified by the data. These excess digits are indicated with an overline. We have carefully cross-checked the solutions for errors and expect that most have been eliminated. We would be grateful to any readers who bring any remaining errors to our attention. We warmly thank our publishers for their patience in guiding this complex, detailed project to completion. P. W. A. e.A.T. M. P.e. e. G.

Contents PART 1 Equilibrium

1

1

The properties of gases

3

Answers to discussion questions

3 4 13 13 18 20

Solutions to exercises Solutions to problems

Solutions to numerical problems Solutions to theoretical problems Solutions to applications

5

Simple mixtures

Answers to discussion questions Solutions to exercises Solutions to problems


6

Phase diagrams


2

The First Law

22




22 23 33 33 41 47


7

Chemical equilibrium


3

The Second Law

50




4

50 51 58 58 68 74



PART 2 Structure 8

Physical transformations of pure substances


78 78 80 83 83 86 87

91 91 91 98 98 104 107

112 112 113 119 119 124 124

127 127 128 137 137 148 150

155

Quantum theory: introduction and prinCiples



157 157 158 162 162 165 172

viii

9

Contents

Quantum theory: techniques and applications

176

Solutions to theoretical problems Solutions to applications

176 176 183 183 186 195

10 Atomic structure and atomic spectra

199

Answers to discussion questions Solutions to exercises Solutions to problems Solutions to numerical problems

14 Molecular spectroscopy 2: electronic transitions


280 28 1 284 284 289 292

15 Molecular spectroscopy 3: magnetic resonance

297


Answers to discussion questions Solutions to exercises

Solutions to applications

199 200 207 207 211 218

11 Molecular structure

221

16 Statistical thermodynamics 1 : the concepts

Answers to discussion questions Solutions to exercises Solutions to problems Solutions to numerical problems Solutions to theoretical problems

Answers to discussion questions Solutions to exercises Solutions to problems Solutions to numerical problems Solutions to theoretical problems Solutions to applications

22 1 223 226 226 238 241

Solutions to problems Solutions to numerical problems Solutions to theoretical problems Solutions to applications


12 Molecular symmetry Answers to discussion questions Solutions to exercises Solutions to problems Solutions to applications

Solutions to problems Solutions to numerical problems Solutions to theoretical problems Solutions to applications

297 299 305 305 309 311

315

315 315 322 322 326 329

244

244 245 249 255

13 Molecular spectroscopy 1 : rotational 259 and vibrational spectra Answers to discussion questions Solutions to exercises

280

259 260 269 269 275 276

17 Statistical thermodynamics 2: applications

331


331 332 338 338 345 353

18 Molecular interactions

357


357 358 361

Answers to discussion questions Solutions to exercises Solutions to problems Solutions to numerical problems Solutions to theoretical problems


Contents Solutions to numerical problems Solutions to theoretical problems Solutions to applications

361 366 368

22 The rates of chemical reactions

440


440 443 450 450 455 458

Solutions to exercises

19 Materials 1: macromolecules and aggregates

Solutions to problems

370

Solutions to numerical problems Solutions to theoretical problems


370 372 375 375 379 383


23 The kinetics of complex reactions

464


464 465 468 468 471 478


20 Materials 2: the solid state

389


389 390 398 398 405 408

Solutions to numerical problems Solutions to theoretical problems

Solutions to exercises Solutions to problems Solutions to numerical problems Solutions to theoretical problems Solutions to applicati ons

ix


24 Molecular reaction dynamics

489


489 490 497 497 502 506

Solutions to exercises Solutions to problems Solutions to numerical problems Solutions to theoretical problems

PART 3 Change

411

21 Molecules in motion

413

25 Processes at solid surfaces

509


413 414 424 424 430 433


509 511 521 521 531 534

Solutions to exercises Solutions to problems Solutions to numerical problems Solutions to theoretical problems Solutions to applications


Solutions to exercises Solutions to problems Solutions to numerical problems Solutions to theoretical problems Solutions to applications

PART 1 Equilibrium

The properties of gases

Answers to discussion questions 01.1

An equation of state is an equation that relates the variables that define the state of a system to each other. Boyle, Charles, and Avogadro established these relations for gases at low pressures (perfect gases) by appropriate experiments. Boyle determined how volume varies with pressure (V ex lip), Charles how volume varies with temperature (V ex T), and Avogadro how volume varies with amount of gas (V ex n). Combining all of these proportionalities into one we find nT Vex - . p

Inserting the constant of proportionality, R, yields the perfect gas equation

V 01.3

RnT

= -p

or

pV

= nRT.

Consider three temperature regions: (1) T < TB . At very low pressures, all gases show a compression factor, Z ~ I. At high pressures, all gases have Z > I , signifying that they have a molar volume greater than a perfect gas, which implies that repulsive forces are dominant. At intermediate pressures, most gases show Z < I, indicating

that attractive forces reducing the molar volume below the perfect value are dominant. (2) T ~ TB . Z ~ I at low pressures, slightly greater than I at intermediate pressures, and significantly greater than I only at high pressures. There is a balance between the attractive and repulsive forces

at low to intermediate pressures, but the repulsive forces predominate at high pressures where the molecules are very close to each other. (3) T > TB. Z > I at all pressures because the frequency of collisions between molecules increases with temperature. 01.5

The van der Waals equation 'corrects ' the perfect gas equation for both attractive and repulsive interactions between the molecules in a real gas. See Justification 1.1 for a fuller explanation. The Bertholet equation accounts for the volume of the molecules in a manner similar to the van der Waals equation but the term representing molecular attractions is modified to account for the effect of temperature. Experimentally one finds that the van der Waals a decreases with increasing temperature. Theory (see Chapter 18) also suggests that intermolecular attractions can decrease with temperature.

4

STUDENT'S SOLUTIONS MANUAL

This variation of the attractive interaction with temperature can be accounted for in the equation of state by replacing the van der Waals a with a/ T o

Solutions to exercises E1.1(b)

(a) The perfect gas law is

pV

= nRT

implying that the pressure would be nRT P=V

All quantities on the right are given to us except n, which can be computed from the given mass of Ar.

n

=

25 g 39.95 g mol- 1

= 0.626 mol

(0.626 mol) x (8.31 x 10- 2 dm 3 bar K- 1mol-I ) x (30 + 273 K)

so P =

1.5dm

3

1

-

1

= . 10.5 bar .

not 2.0 bar. (b) The van der Waals equation is P

RT a V - b - V2

=

m

sop

=

m

10- 2 dm 3 bar K- 1mol-I ) x (30 + 273) K (1.53 dm 3 / 0.626 mol) - 3.20 x 10- 2 dm 3 mol - 1

(8 .31

X

(1.337dm 6 atmmol- 2 ) x (1.013baratm-

E1.2(b)

3

1 )

_I 10.4- bar 1 .

-

( 1.5 dm / 0.626 mo1)2

(a) Boyle's law applies: PV

= constant so

p f Vf

= Pi Vi

and Pi

prVr

= -- = Vi

(1.97 bar) x (2.14dm (2.14 + 1.80) dm 3

3

)

=

1

1.07 bar

1

(b) The original pressure in bar is Pi

E1.3(b)

= (1.07 bar)

x ( 1 atm) x (760 TOrr) 1.013 bar I atm

= I803 Torr I

The relation between pressure and temperature at constant volume can be derived from the perfect gas law pV

= nRT

so P ex: T

and

Pi

Ti

Pr Tr

THE PROPERTIES OF GASES

5

The final pressure, then, ought to be

= pjTr = Pr E1.4(b)

Tj

= 1120 kPa I

(125 kPa) x ( II + 273) K (23 + 273) K

According to the perfect gas law, one can compute the amount of gas from pressure, temperature, and volume. Once this is done, the mass of the gas can be computed from the amount and the molar mass using pV

= nRT

pV (1.00atm) x (1.013 x 105 Paatm- I) x (4.00 x 10 3 m3 ) so n - - . - RT (8 .3145J K- Imol - I) x (20+273) K

and m E1.S(b)

= ( 1.66 x

lOS mol) x (16.04 g mol-I)

x 106 g

= 12.67

X

5 10 mol

10 3 kg 1

Identifying P ex in the equation P = Pex + pgh [1.3] as the pressure at the top of the straw and P as the atmospheric pressure on the liquid, the pressure difference is P - Pex

=

pgh

= ( 1.0 x 103 kg m- 3 ) x (9.8 1 m s- 2) x (0.15 m)

=11.5 x 10 Pa 1(=1.5 3

E1.6(b)

= 2.67

= 1.66 x

X

10- 2 atm)

The pressure in the apparatus is given by

= Palm + pgh [1.3] P alm = 760 Torr = I atm = 1.013 x

P

pgh = 13.55 g cm-

P = 1.013 E1.7(b)

X

3

x

105 Pa

(/o~gg) x (1O:~m3) x

105 Pa + 1.33 x 104 Pa

= 1.146 x

0.100 m x 9.806 m s-2 = 1.33

105 Pa

X

104 Pa

= 1115 kPa I

All gases are perfect in the limit of zero pressure. Therefore the extrapolated value of pVm/T will give the best value of R. m The molar mass is obtained from PV = nRT = - RT M . mRT RT which upon rearrangement gives M = - - = p V P

P

The best value of M is obtained from an extrapolation of p / P versus P to P = 0; the intercept is M / RT. Draw up the following table

0.750000 0.500000 0.250000

0.082 00 14 0.082 0227 0.082 0414

From Figure l.l (a), (PVm) T p=O

1.428 59 1.428 22 1.427 90

= I0.082 0615 dm 3 atm K- I mol-I

I

6


[1.• ::: "'M

: ,.: .I · •.· •.

:,.!.:.• .

:I, . · ·•.

•. · •.

·· i.. ·!···!···i .. ·!· .. !··· .. · ··· ··· .. . . 'i" '!"' !"'! '" .. ,." ..... .

",,,.,,,.! ...

···'···I···I···,···j·· ·j··

~ ,,;8.202,

...

'· !.!j~!~.f !• !•• • . • . • • ••

· .r ·.• .• •. ri,.:· .•= · ·•. .•·: .!, •• .·:,i

•.• •.•

";"';";"';";"';"';" ... i ... i ... i ... ! .

··-!- ··!··-!- ·-!- ·· !···, ··· ,··!· ··.··! .. ·i .. ·!

... ............ . ! . .. ! ... ! . ..

. ~~ . ~ ...~ . .~ . .... ~ . .~ ...!... !...j ··i"!"'! "! "-!-". 'i"+" .. ·!· .. i.. ·!···!··· ..

"M

,,!"' ! '''! . ... . ,',,' ! , .. !. ,, !',, '

". :2 ";8.200' ·; .. ·;··;· ·;.. ·;.. ·:.. Y·i ...:... :... :.. :

::!:··!:.·!:::!: : !::i:::.:::·:::~.~~:· : : ·~.~~;.: ::'.:~.~~: ...;... '1.0

.:t::::ttt:::::J:::':::;:::':::':::: ::.f.!~ti:!:':::::::':::'::t:· :::i:::

From Figure 1.l(b),

(~) p

- 1.42755 g dm- 3 atm- I

p=o-

..

...:

i .4288 ..

"': ·':.. ·1.4286'

..:.

.,: ... ~ .. ':

,

.. ~ .. .., ..., .. ~ . ,! .. 'r"'~ ... ... "f " '~ .

"': .,,?

Figure 1.I(a)

..~

~

~

... ; .. ! .. ;....: .. i . . i ... .: .. j ••. ~

Figure 1.I(b) M = RT

(~) p

(0.0820615 dm3 atm mol-I K- 1) x (273.15 K) x (1.42755 g dm- 3 atm-

l

)

p=o

= 131.9987 g mol-II The value obtained for R deviates from the accepted value by 0.005 percent. The error results from the fact that only three data points are available and that a linear extrapolation was employed. The molar mass, however, agrees exactly with the accepted value, probably because of compensating plotting errors. E1.8(b)

The mass density p is related to the molar volume Vrn by

M p

where M is the molar mass. Putting this relation into the perfect gas law yields pVrn

= RT

so

pM =RT p


7

Rearranging this result gives an expression for M; once we know the molar mass, we can divide by the molar mass of phosphorus atoms to determine the number of atoms per gas molecule

+

(8.314 Pa m3 mol-I) x [(100 273) K] x (0.6388kgm- 3 ) ----------------------~----------~----

RTp

M--P -

1.60xl()4Pa

= 0. 124 kg mol- I = 124 g mol- I The number of atoms per molecule is 124g mol - I ----'------,-I = 4.00 31.0g molsuggesting a formula of ~ E1.9(b)

Use the perfect gas equation to compute the amount; then convert to mass. PV

pV

= nRT

so

n

= RT

We need the partial pressure of water, which is 53 percent of the equilibrium vapor pressure at the given temperature and standard pressure. p = (0.53) x (2.69 x 10 3 Pa) = 1.43 x 103 Pa

so n =

(1.43 x 103 Pa) x (250 m3 ) 2 = 1.45 x 10 mol (8.3145 J K- I mol-I) x (23 + 273) K

or m = (l.45 x 102 mol) x (l8.0g mol-I) = 2.61 x 103 g = 12.61 kg 1 E1.10(b)

(a) The volume occupied by each gas is the same, since each completely fills the container. Thus solving for V we have (assuming a perfect gas) V

nJRT 0.225 g =- nNe = --------=-----,I

20.18 g mol-

PJ

= 1.115x 10- 2 mol, V=

(!.l15 x

10- 2

PNe= 8.87kPa,

T=300K

mol) x (8.314 kPa K- I mol-I) x 300 K) 3 =3 .137dm 8.87 kPa dm 3

=13 .14dm3 1 (b) The total pressure is determined from the total amount of gas, n = nCH4 nCH4 =

0.320 g -2 I = 1.995 x 10 mol 16.04 g mol-

n = (1.995

+ 0.438 + 1.115)

nAr =

+ nAr + nNe.

0.175 g = 4.38 x 1O- 3 mol 39.95 g mol- I

x 1O- 2 mol = 3.548 x 1O- 2 mol

p = nRT [1.8] = (3.548 x 10- mol) x (8.314 d_m 3 kPa K- I mol-I) x (300 K) 2

3.137dm 3

V

= 128.2 kPa 1

8

E1.11(b)


This is similar to Exercise l.ll(a) with the exception that the density is first calculated.

RT M = p - [Exercise 1.8(a)] p

33.5mg =0.1340gdm- 3 , 250cm

p=

---3

M= E1.12(b)

p= 152 Torr,

T=298K

(0.1340gdm- 3 ) x (62.36dm3 TorrK- 1 mol-I) x (298K) I -II = 16.14gmol 152 Torr

This exercise is similar to Exercise 1.12(a) in that it uses the definition of absolute zero as that temperature at which the volume of a sample of gas would become zero if the substance remained a gas at low temperatures. The solution uses the experimental fact that the volume is a linear function of the Celsius temperature. Thus V = Vo

+ aVoO =

+ bO , b =

Vo

aVo

At absolute zero, V = 0, or 0 = 20.oodm3 + 0.0741 dm 3 °C- I x O(abs. zero) 3

O(abs. zero) = -

E1.13(b)

20.00 dm 3 I = t -270 °C 0.0741 dm °C-

t

which is close to the accepted value of -273°C. nRT (a) P= V

n = 1.0mol T = (i) 273.15K; (ii) 500K V = (i) 22.414dm 3 ; (ii) 150cm 3

(i)

(ii)

(1.0 mol) x (8.206 x 10- 2 dm3 atmK- I mol-I) x (273.15K) P= 22.414dm 3 = t 1.0 atm t (1.0mol) x (8.206 x 10- 2 dm 3 atmK- I mol-I) x (5OOK) P= 0.150dm 3 = t 270 atm t (2 significant figures)

(b) From Table (1.6) for H2S

a = 4.484 dm 6 atm mol- I

b = 4.34 x 10- 2 dm 3 mol- I

nRT an 2 P=V-nb-V2'

(i)

(1.0 mol) x (8.206 x 10- 2 dm 3 atm K- I mol-I) x (273. 15 K) P= 22.414 dm 3 - (1.0 mol) x (4.34 x 10-2 dm 3 mol I) (4.484 dm 6 atm mol-I) x (1.0 mol)2 (22.414 dm 3 )2

=

t

0.99 atm

t


9

( 1.0 mol) x (8.206 x 10- 2 dm 3 atm K- I mol-I) x (500K)

(ii)

p

=

0.150dm 3

(1.0 mol) x (4.34 x 10- 2 dm 3 mol I)

-

(4A84dm 6 atmmor

1 )

x (1.0mol)2

(0.150 dm 3 )2

= 185.6atm ~ 1190 atm 1(2 significant figures). E1.14(b)

The conversions needed are as follows:

Therefore,

E1.1S(b)

a

= 1.32 atm dm 6 mol- 2 becomes, after substitution of the conversions

a

= 11.34 x

b

= 0.0436 dm 3 mol - I becomes

b

= 14.36 X

10- 1 kg m5s- 2 mol- 2

~

and

10- 5 m 3 mol- 1 I

The compression factor is pVm Vm z=-=RT

V;:'

= V;:' + 0.12 V;:' = (I. 12)V;:', we have Z = [Iill IRepulsive Iforces dominate.

(a) Because Vm

(b) The molar volume is

V

= (1.12)V~ = (1.12)

x

(R;) 3

V=(1.12)x (

E1.16(b)

(a)

RT

o

V - -

p

m -

-

0.08206dm atmK- I mol-I ) x (350K») 127 d 3 I-I 1 12atm =.' m mo .

(8.314JK- I mol -I) x (298.15 K)

--------------~----~(200 bar) x (l05Pabar- l )

I

= 1.24 x 10-4 m3 mol- I = 0.124 dm 3 mol- I

I

(b) The van der Waals equation is a cubic equation in Vm. The most direct way of obtaining the molar

volume would be to solve the cubic analytically. However, this approach is cumbersome, so we proceed as in Example 104. The van der Waals equation is rearranged to the cubic form Vm3

-

(

b+

RT) V 2 + (a)p V P m

m -

abp = 0 or x

3

- (b+

RT) P x + (a) p x - ab p=0 2

10


The coefficients in the equation are evaluated as

b + RT P

=

(3.183 x 1O-2 dm3 mol - I)

= (3.183

X

+

3 2 (8.206 x 1O- dm mol- I) x (298.15 K) (200 bar) x (1.013 atm bar-I)

10- 2 + 0.1208) dm 3 mol- I

1.360 dm 6 atm mol- 2

a

-----------::- = 6.71 (200 bar) x (1.013 atm bar-I)

P

= 0.1526 dm 3mol - 1

3 3 I 2 x 10- (dm mol - )

(1.360 dm 6 atmmol- 2) x (3.183 x 1O- 2dm 3 mol- I) (200 bar) x (1.013 atmbar- I )

ab

= 2.137-

-------~--.:......---__;_------.:.

P

4 3 I 3 x 10- (dm mol- )

Thus, the equation to be solved is x3 - 0.1526x 2 + (6.71 x 1O- 3)x - (2.137 x 10- 4 ) = O. Calculators and computer software for the solution of polynomials are readily available. In this case we find x = 0.112

or

Vrn = 10.112 dm 3 mol-II

The difference is about 15 percent. E1.17(b)

The molar volume is obtained by solving Z

Vrn

= pVrn / RT [1.17] , for Vrn , which yields

ZRT

(0.86) x (0.08206dm 3 atmK- l mol- l ) x (300K)

p

20atm

= -- =

(a) Then, V

= n Vrn = (8.2

x 10- 3 mol) x (1.059 dm 3 mol- I)

= 8.7

=

3 -I 1.059dm mol

x 10- 3 dm 3 = 18.7 cm 3 1

(b) An approximate value of B can be obtained from eqn 1.19 by truncation of the series expansion after the second term, B/Vrn , in the series. Then,

B

= Vrn (p;;

- I)

= Vrn

= (1.059 dm 3 mol- I) E1.18(b)

X

(Z - I)

x (0.86 - I)

= \ -0.15 dm 3mol- 1 \

(a) Mole fractions are nN

XN=-= ntotal

Similarly, XH

(2.5

2.5 mol ~6 + 1.5) mol =~

= 10.371

(c) According to the perfect gas law Ptotal V

so

Ptotal

=

= ntotal RT ntotal RT --V-

I _ (4.0 mol) x (0.08206 dm 3 atm mol- K- I ) x (273.15 K) -140 1 - . . atm. 22.4dm 3


11

(b) The partial pressures are PN

=

XNPtot

= (0.63) x (4.0 atm) = 12.5 atm 1

and PH = (0.37) x (4.0 atm) = 11.5 atm 1 E1.19(b)

The critical volume of a van der Waals gas is Vc

= 3b

I

so b = 1 Vc = 1( 148cm 3 mol - I) = 49.3cm 3 mol- I = 0.0493 dm 3 mol- I

I

By interpreting b as the excluded volume of a mole of spherical molecules, we can obtain an estimate of molecular size. The centers of spherical particles are excluded from a sphere whose radius is the diameter of those spherical particles (i.e. twice their radius); that volume times the Avogadro constant is the molar excluded volume b

4n(2r)3 )

r=

so

b=NA ( - - 3

2

3(49.3cm 3 mol-

I

r=-

~ (~)1 /3

2 ( 4n(6.022 x

10 23

l

4nNA IP

)

I mol- ) )

=1.94 xlO- S cm =11.94 xlO- IOml

The critical pressure is a

Pc

=

27b2

so a = 27Pcb2

= 27(48.20 atm) x

(0.0493 dm 3 mol- I) 2

= 13.16 dm 6 atm mol- 2 1

But this problem is overdetermined. We have another piece of information

8a T. - - c - 27Rb According to the constants we have already determined, Tc should be

However, the reported Tc is 305.4 K, suggesting our computed alb is about 25 percent lower than it should be. E1.20(b)

(a) The Boyle temperature is the temperature at which Iim vm~oo dZ /(d(l I Vm to the van der Waals equation

a )

RT

Z

=

pVm RT

=

(

v:-=-h - ~ RT

Vm

Vm a --- - - Vm -b

VmRT

» vanishes. According

12


so

dZ d(l/Vrn )

=

( dZ) ( dVrn ) dVrn x d(l/Vrn )

= -V~ Cd:rn ) = -V~ V~b

a

(Vrn - b)2

RT

CV:~rnb)2 + Vrnl-b + vtRT)

In the limit of large molar volume, we have dZ

. I 1m

Vm ..... OO

and T

dO/Vrn )

a RT

=b-

=0

so

a RT

-

=b

a

(4.484 dm 6 atm mol- 2)

Rb

(O.08206dm 3 atmK- I mol-I) x (O.0434dm 3 mol-I)

=- =

= 11259 K I

(b) By interpreting b as the excluded volume of a mole of spherical molecules, we can obtain an estimate of molecular size. The centres of spherical particles are excluded from a sphere whose radius is the diameter ofthose spherical particles (i.e. twice their radius); the Avogadro constant times the volume is the molar excluded volume b 47r(2r)3 ) b=NA ( - - 3

so

r

= ~ (~)1 /3 2

47rNA

1/3 1 3(0.0434 dm mol-I) r =I = 1.286 x 102 (47r(6.022 x 1023 mol- ) ) 3

E1.21 (b)

9

dm

= 1.29 x

10- 10 m

= I0.129 nm I

States that have the same reduced pressure, temperature, and volume are said to correspond. The reduced pressure and temperature for N2 at 1.0 atm and 25°C are

Pr

1.0 atm = Pc -P = 33.54 = 0.030 atm

and

T __ (25 + 273) K __ 2.36 Tc 126.3 K

Tr __ _ I,

The corresponding states are (a) For H2S

= (0.030) x (88.3 atm) = 12.6 atm I T = TrTc = (2.36) x (373.2K) = 1881 K I

P = PrPc

(Critical constants of H2S obtained from Handbook of Chemistry and Physics.) (b) ForC02

= (0.030) x (12.85 atm) = 12.2 atm I T = TrTc = (2.36) x (304.2 K) = 1718 K I

P = PrPc

(c) For Ar

= (0.030) x (48.00atm) = 11.4 attn I T = TrTc = (2.36) x (l50.12K) = 1356 K I

P = PrPc


E1.22(b)

13

The van der Waals equation is

which can be solved for b

RT

-4

b = Vm - --a- = 4.00 x 10

3

_,

m mol

-

P+V2 m

= 11.3 x 10-4 m3 mol-'

(8.3145JK-' mol-' ) x (288K) 2 ) 4.0xI06 Pa+ 0.76m Pamol(4.00 X 10- 4 m3 mol-')2

(

6

I

The compression factor is

z=

4

6

3

pVm = (4.0 x 10 Pa) x (4.00 x 10- m mol- i) = 10.671

RT

(8.3145 J K-' mol-') x (288 K)

Solutions to problems Solutions to numerical problems P1.1

Since the Neptunians know about perfect gas behavior. we may assume that they will write p V = nRT at both temperatures. We may also assume that they will establish the size of their absolute unit to be the same as the oN. just as we write 1K = 1°C. Thus

pV(T,) = 28.0dm 3 atm = nRT, = nR x (T, + OON) . 0 pV(T2) = 40.0dm 3 atm = nRT2 = nR x (T, + 100 N). ° 40.0 dm 3 atm T, + 100 N= - - - nR

28.0dm 3 atm orT, = - - - - nR

T, + 1000N

40.0dm 3 atm

° . 3 = 1.429 or T[ + 100 N = 1.429T,. T, = 233 absolute UOltS. T, 28.0 dm atm As in the relationshi between our Kelvin scale and Celsius scale T = () - absolute zero(ON) so absolute zero (ON) = -233°N . Dividing.

COMMENT.

=

To facilitate communication with Earth students we have converted the Neptunians' units of

the pV product to units familiar to humans, which are dm3 atm. However, we see from the solution that only the ratio of pV products is required , and that will be the same in any civilization.

Question. If the Neptunians' unit of volume is the lagoon (L). their unit of pressure is the poseidon (P). their unit of amount is the nereid (n). and their unit of absolute temperature is the titan (T). what is the value of the Neptunians' gas constant (R) in units of L. p. n. and T? P1.3

The value of absolute zero can be expressed in terms of a by using the requirement that the volume of a perfect gas becomes zero at the absolute zero of temperature. Hence 0= Vo[l + a(}(abs. zero)].

14


1

Then () (abs. zero) = --. ex All gases become perfect in the limit of zero pressure, so the best value of ex and, hence, () (abs. zero) is obtained by extrapolating ex to zero pressure. This is done in Fig. 1.2. Using the extrapolated value, ex = 3.6637 x 1O- 3°C- I, or ()(abs.zero)=-

I 3 I =1-272.95°cl, 3.6637 x 10- °C· .

which is close to the accepted value of -273.15°C. .... _ · ..... ........ . . .... . .. .. . .. ..... ..:... -:-. ...: .. .

3.672··

~

.. . .;. ...... ; ...i···;'··; . . . . . . . . . . . ', .. !' ... ..... .. : .. . .. : ...:.. ,! .. .. ,! . . .~

~

~

~

~

~

~

~

..; .. , . . , "!

i··~··

. · .. . : .. . ': .. : .. .:.. .. :. ... ..· ...., ................. . . . . ,. .. ~ .. ;.. .~ .. :.. ~... ; .. ~ . .~

~

~

:··" 3.670'

. ,',. ,! ..

. .;.. . ! .. -:... ~ .. . . . ~ . . ! ...... ! .. -: .. ! .. ~ .. ~ •. ~ ... ~ .! ..:

....,.

'.,

. ...~ .. :.. ';' .. :.. ':' .. :.. ~ .. .~ ... ... _."

.;.: .. ~ .. ;.. ;... ; . .

;f'3. ~$ ·

'U

i

···;···i···;.. ·j· ··;···i.·,:..

~

......

..

. :. ':' . ;. ':' . ;. ;... ~ .. ... ;..... .. :...: ..:.. :... :.. ...; ••

~ G:~::rT: .~

! ~.:

"

~

~o

3.666·

: -

.: .. : ... ~ ..

~

i.·{·····I···:···I··

~

.~ .. ~ .. ~ .. :.. ~... "

.: .. ;... ~ .. ~ .. :

· ··~ ··:··7·· :

'''3.664 ' . . . . . ' .: .. .. . ':. .. .••. .

. - ..•. • ~ •. .• ••• . ~

"!"':- ":"~'

~

~

~

Figure 1.2

P1.S

p nR. p P3 . - = constant, If n and V are constant. Hence, - = - , where P IS the measured pressure at V T T3 T temperature, T, and P3 and T3 are the triple point pressure and temperature, respectively. Rearranging,

P=(~~)T. ~

~~

I

. . .

The ratio - is a constant = = 0.0245 kPa K- . Thus the change III P, t.p, IS proportIOnal to T3 273.16K the change in temperature, t.T : t.p = (0.0245 kPaK-I) x (t.T) .

I

(a) t.p = (0.0245 kPaK-I) x (LOOK) = 0.0245 kPa

I·

(b) Rearranging,p= (!.-)P3 = (373.16K) x (6.69kPa) =19.14kPal. T3 273.16K (c) Since

f

is a constant at constant n and V, it always has the value 0.0245 kPa K -I ; hence T 1 t.p = P374.15K - P373.15K = (0.0245 kPa K- ) x (1.00 K) = 0.0245 kPa

I

_ RT _ (8.206 x 10- 2 dm atm K- I mol-I) x (350 K) -1125 d 3 I-I I (a) Vrn - . . m mo . P 2.30atm RT a RT (b) Fromp = - - - 2 [1.2Ib], we obtain Vrn = ( ) + b [rearrangeI.21b]. Vrn -b Vrn + a p V2 3

P1.7

I·

rn


15

Then, with a and b from Table 1.6, (8.206 x 1O- 2 dm 3 atmK- I mol -

Vm "'"

(2.30 atm)

"'"

+ (6.260dm

3

28 .72dm mol2.34

I

l

)

x (350K) 2

6

+ ( 5.42

2

3

+ (5.42 x

2

3

atm mol - )/ (12.5 dm mol- I) )

I

I

x 10- dm mol- I "'" 12.3dm 3 mol-I . 2

3

)

Substitution of 12.3 dm 3 mol- I into the denominator of the first expression again results in Vrn = 12.3 dm 3 mol- I, so the cycle of approximation may be terminated. P1.9

As indicated by eqns 1.18 and 1.19 the compression factor of a gas may be expressed as either a virial

:m ).The virial form of the van der Waals equation is derived in Exercise 1.20(a)

expansion in p or in ( and is p =

~: { 1 + (b - RaT)

x ( :

m) + ...}

Rearranging Z = pV = I + (b - ~) x (_1_) ' RT RT Vm m

+ ...

On the assumption that the perfect gas expression for Vm is adequate for the second term in this expansion, we can readily obtain Z as a function of p .

(a)

Tc = 126.3 K .

V (R;) m =

x Z =

R; + (b - :T) + ...

(0.08206dm 3 atm K- I mol - I) x (126.3 K) 1O.Oatm

+ {(0.0387 dm3 mol - I) _

6

(

_ Z -

(l!...-) RT

x

(V) _ m -

2

) 1.352 dm atm mol(0.08206 dm 3 atm K- I mol- I) x ( 126.3 K)

= (1.036 -0.092) dm 3 mol - 1 =10.944dm 3 mol - l

}

l.

3

(1O.0atm) x (0.944dm mol-I) =0.911. (0.08206dm 3 atm K-I mol-I) x (126.3 K)

(b) The Boyle temperature corresponds to the temperature at which the second virial coefficient is zero, hence correct to the first power in p , Z = 1, and the gas is close to perfect. However, if we assume that N2 is a van der Waals gas, when the second virial coefficient is zero,

=0' (b-~) RTB TB =

(0.0387dm 3

or

a TB = - . bR

1.352 dm 6 atm mol- 2 = 426K. mol I) x (0.08206dm 3 atrnK-I mol-I )

I

10- dm mol - )

16


The experimental value (Table 1.5) is 327.2 K. The discrepancy may be explained by two considerations.

1. Terms beyond the first power in p should not be dropped in the expansion for Z. 2. Nitrogen is only approximately a van der Waals gas. WhenZ

=

I , Vrn

RT

= -, p

and using TB

= 327.2K

(0.08206 dm 3 atm K- I mol-I) x 327.2 K 10.Oatm

= 12.69dm3 mol-II and this is the ideal value of Vm . Using the experimental value of TB and inserting this value into the expansion for Vrn above, we have 0.08206 dm 3 atm K-1mol- 1 x 327.2 K

Vrn

= ---------------------------10.Oatm

+ { 0.0387 dm 3 mol- I = and Z (c) TI

6 2 1.352 dm atm mol) } 1 0.08206dm 3 atmK-l mol- x 327.2K

(2.685 - 0.012) dm 3 mol- 1 = 12.67 dm 3 mol-II

Vrn

=-

(

V;;'

=

2.67 dm 3 mol-I 2.69dm 3 mol-I

= 0.992 ~

I.

= 621 K [Table 2.9]. Vrn

0.08206dm 3 atm K-Imol- I x 621 K

= ----------~~-----------1O.Oatm + { 0.0387 dm 3 mol - I =

and Z

=

2 6 ) } 1.352 dm atm mol3 0.08206 dm atm K -I mol-I x 621 K

(

(5.096 + 0.012) dm 3 mol-I

5.11 dm 3 mol5.lOdm3

= 15.11 dm 3 mol-II

1

mol-I

=

1.002

~

I.

Based on the values of TB and TJ given in Tables 1.4 and 2.9 and assuming that N2 is a van der Waals gas, the calculated value of Z is closest to I at but the difference from the value at TB is less than the accuracy of the method.

I!i],

P1.11

(a)Vm

18.02g mol-I -10 353 d 3 I-I 1 _ molar mass _ M _ mmo . . - ..1 p 1.332 x \0 2 gdm- 3 densIty

-

_pVm _ (b) Z [1.l7b] -

RT

l 3 (327.6atm) x (0.1353dm mol- ) -1069571 I -. , 3 (0.08206 dm atm K-I mol- ) x (776.4 K)


17

(e) Two expansions for Z based on the van der Waals equation are given in Problem 1.9. They are

= 1 + (0.0305dm mol- I)3

{

(

6

2

5.464dm atmmol)} (0.08206dm3 atm K-I mol-I) x (776.4 K)

J x 0.J353dm 3 mol- 1 = 1-0.4084=0. 5916~0.59.

Z

= 1+ (RIT)

x (b - :T) x (P) + ...

I = I + -----,;--------;-----3

(0.08206dm atmK-1 mol-I) x (776.4K)

x

{

(0.0305 dm 3 mol-I) -

= 1- 0.2842 ~

(

6

2

5.464 dm atm mol)} x 327.6 atm (0.08206dm 3 atm K-I mol-I) x (776.4 K)

lo.nl. I

In this case the expansion in p gives a value close to the experimental value; the expansion in Vrn is not as good. However, when terms beyond the second are included the results from the two expansions for Z converge. P1.13

Vc

= 2b,

a Tc = 4bR [Table 1.7]

Hence, with Vc and Tc from Table 1.5, b =

~ Vc

=

~

x (118.8 cm 3 mol-I) = 159.4 cm 3 mol-I

a = 4bRTc = 2RTc Vc

= (2) x (8.206 x 10- 2 dm 3 atmK- 1 mol-I) x (289.75K) x (118.8 x 10- 3 dm 3 mol-I) = 15 .649 dm 6 atm mol- 2 1. Hence

(1.0 mol) x (8.206 x 10- 2 dm 3 atm K- I mol-I) x (298 K) (1.0dm 3 ) x exp

-

(1.0 mol) x (59.4 x 10- 3 dm 3 mol-I)

) -(l.Omol) x (5.649dm6 atmmol- 2 ) ( (8.206 x 10- 2 dm 3 atmK-1 mol-I) x (298K) x (1.0dm 6 atmmol- l )

= 26.0atm x e- O.23T = 121 atm

I.

I.

18


Solutions to theoretical problems PU5

This expansion has already been given in the solutions to Exercise 1.20(a) and Problem 1.14; the result is

· expansion .. Compare thIS wlthp

I

and hence find B = b Since C

=

B)

Vm

iT I

and

1200 cm6 mol- 2 ,

= RT(b -

a

P1.17

=

= -RT

b

= (8.206 x

( 1+ - B + - C + ... ) [1.19] Vm Vm2

Ie = b

=

2

C I/ 2

1.

= 134.6 cm 3 mol - II

10- 2 ) x (273 dm 3 atm mol-I) x (34.6 + 21.7) cm 3 mol-I

(22.40 dm 3 atm mol- I) x (56.3 x 10- 3 dm 3 mol - I)

= 11.26 dm 6 atm mol- 2 1.

The critical point corresponds to a point of zero slope that is simultaneously a point of inflection in a plot of pressure versus molar volume. A critical point exists if there are values of p , V, and T that result in a point that satisfies these conditions.

I"

th, ,ri,kal poiOl.

. Th at IS,

-RTc V; + 2BVc - 3C 2 RTcVc - 3BVc + 6C

which solve to Vc

~

=0 =0

}

= - C~2 , Tc = - - . B

3RC

Now use the equation of state to find Pc

Pc

RTc

= V; -

B C V2 + vg

=

(RB2) (B) ( B 3RC x 3C - B 3C 3

It follows that Zc P1.19

=

PcVc RTc

=

(B 27C2 ) x (3C) Ii x

(I) R

)2 + C (3CB)3 = iB3l ~.

x (3RC) fi2

For a real gas we may use the virial expansion in terms of p [1 .18] P

nRT =( I + B'p + V

RT I ... ) = p-(I + B P + . .. ) M

= III llJ·


RT P which rearranges to - = p M

RTB'

+ - - p + .... M

Therefore, the limiting slope of a plot of B'RT M

19

B'RT . 3 th limi· · I . p against p IS. ~. From Fig. I. e tmg s ope IS

p

4 2 (5.84 - 5.44) x 10 m s-2 _ 44 . (10.132 - 1.223) x IQ4Pa

~----'--...,...,..----:,-;-:::--

X

10-2 k - I 3 g m .

RT From Fig. 1.3, = 5.40 x 104 m 2 s-2 ; hence M l 3 2 , _4.4xlO- kg- m -081 10-6p- 1 B22- . x a, 5.40 x IQ4 m s-

B' = (0.81 x 1O-6 Pa-

l)

x (1.0133 x 105 Paatm- l ) =10.082atm-

l

l.

B = RTB' [Problem 1.18]

= (8.206 x 10- 2 dm 3 atm K- I mol-I) x (298 K) x (0.082 atm- I ) = 12.0dm3 mol - I

-

y

I.

= 5.3963 + 0.046074x R = 0.99549

5.9

~

5.8

I",

E

5.7

b

:§: 5.6 ,::, 5.5 5.4 0

2

4

6 p/( 104 Pa)

P1.21

8

10

12

Figure 1.3

The critical temperature is that temperature above which the gas cannot be liquefied by the application of pressure alone. Below the critical temperature two phases, liquid and gas, may coexist at equilibrium, and in the two-phase region there is more than one molar volume corresponding to the same conditions of temperature and pressure. Therefore, any equation of state that can even approximately describe this situation must allow for more than one real root for the molar volume at some values of T and p, but as the temperature is increased above Te , allows only one real root. Thus, appropriate equations of state must be equations of odd degree in Vm. The equation of state for gas A may be rewritten V~ - (RT / p) Vm - (RTb / p) = 0, which is a quadratic and never has just one real root. Thus, this equation can never model critical behavior. It could possibly model in a very crude manner a two-phase situation, since there are some conditions under which a quadratic has two real positive roots, but not the process of liquefaction.

20


The equation of state of gas B is a first-degree equation in Vrn and therefore can never model critical behavior, the process of liquefaction, or the existence of a two-phase region. A cubic equation is the equation of lowest degree that can show a cross-over from more than one real root to just one real root as the temperature increases . The van der Waals equation is a cubic equation in V rn . P1.23

The two masses represent the same volume of gas under identical conditions, and therefore, the same number of molecules (Avogadro's principle) and moles, n. Thus, the masses can be expressed as nMN

= 2.2990 g

for 'chemical nitrogen' and

for 'atmospheric nitrogen ' . Dividing the latter expression by the former yields so

X Ar

(MAr _ MN

and XA

(2.3102/ 2.2990) - 1 -

(MAr/MN) - 1

r -

I) =

2.3102 - 1 2.2990

(2.3102/ 2.2990) - I

=

(39.95gmol 1)/ (28.013gmol

I

~

- I)

= L.2:2.!..!J .

This value for the mole fraction of argon in air is close to the modem value.

COMMENT.

Solutions to applications P1.25

I t = 103 kg. Assume 300 t per day. n(S02)

V

=

300

103 kg

64 x 10- 3 kg mol

= nRT = p

P1.27

X

I

= 4.7

x 106 mol.

3 1 6 (4.7 x 10 mol) x (0.082 dm atm K- mol - I) x 1073 K 1.0atrn

= 14.1 x

108 dm 3

1.

The pressure at the base of a column of height H is p = pgH (Example 1.1). But the pressure at any altitude h within the atmospheric column of height H depends only on the air above it; therefore p = pg(H - h) and dp = -pg dh .

Since p

pM

= -~

[Problem 1.2], dp

This relation integrates to p

pMgdh .

.

dp

Mgdh

= --, LmplYIng that -p = --RT ~

= poe- Mgh/RT

For air M ~ 29 g mol- 1 and at 298 K


(a)

(b)

P1.29

h

= 15 cm.

P

= Po

x

e ( -O.15m )x( 1.l5 x lO -

4

m-

1 )

21

= 0.99998 Po ;

h = II km = 1.1 X 104 m . P = Po x e (- l.l x 1O- 4)x( 1.15 x IO-4 m -

l)

= 0.28 Po;

Refer to Fig. 1.4. F,op

1

~1

T h

Air . l (envlronment)

F bottom

Ground

77777777777

Figure 1.4

The buoyant force on the cylinder is

Fbuoy

= Fbottom -

FlOP

= A(Pbottom

- Ptop)

according to the barometric formula. Ptop = Pbotlom e

- Mgh / RT

where M is the molar mass of the environment (air). Since h is small, the exponential can be expanded in a Taylor series around h

Ptop

= Pbottom ( I

-

= 0 (e- x =

1- x

+ ~x2 + .. -). Keeping the first-order term only yields

Mgh) RT .

The buoyant force becomes

Fbuoy

(pbotto mM) g RT = Ah RT ( I - I + Mgh) M Pbottom VM) RT g =n g (

= APbottom =

n is the number of moles of the environment (air) displaced by the balloon, and nM = m, the mass of the displaced environment. Thus Fbuoy = mg . The net force is the difference between the buoyant force and the weight of the balloon. Thus

F net

= mg -

mballoon

g

= (m -

This is Archimedes ' principle.

mballoon)g

The First Law


Work is a precisely defined mechanical concept. It is produced from the application of a force through a distance. The technical definition is based on the realization that both force and displacement are vector quantities and it is the component of the force acting in the direction of the displacement that is used in the calculation of the amount of work, that is, work is the scalar product of the two vectors. In vector notation w = - f . d = - fd cos e, where e is the angle between the force and the displacement. The negative sign is inserted to conform to the standard thermodynamic convention. Heat is associated with a non-adiabatic process and is defined as the difference between the adiabatic work and the non-adiabatic work associated with the same change in state of the system. This is the formal (and best) definition of heat and is based on the definition of work. A less precise definition of heat is the statement that heat is the form of energy that is transferred between bodies in thermal contact with each other by virtue of a difference in temperature. At the molecular level, work is a transfer of energy that results in orderly motion of the atoms and molecules in a system; heat is a transfer of energy that results in disorderly motion. See Molecular interpretation 2.1 for a more detailed discussion.

02.3

The difference results from the definition H = U + PV; hence I'1.H = I'1.U + 1'1. (PV). As I'1.(PV) is not usually zero, except for isothermal processes in a perfect gas, the difference between I'1.H and I'1.U is a non-zero quantity. As shown in Sections 2.4 and 2.5 of the text, I'1.H can be interpreted as the heat associated with a process at constant pressure, and I'1.U as the heat at constant volume.

02.5

In the louIe experiment, the change in internal energy of a gas at low pressures (a perfect gas) is zero. Hence in the calculation of energy changes for processes in a perfect gas one can ignore any effect due to a change in volume. This greatly simplifies the calculations involved because one can drop the first term of eqn 2.40 and need work only with dU = Cv dT. In a more sensitive apparatus, louie would have observed a small temperature change upon expansion of the ' real' gas. louie's result holds exactly only in the limit of zero pressure where all gases can be considered perfect. The solution to Problem 2.33 shows that the louie-Thomson coefficient can be expressed in terms of the parameters representing the attractive and repulsive interactions in a real gas. If the attractive forces predominate, then expanding the gas will reduce its energy and hence its temperature. This reduction in temperature could continue until the temperature of the gas falls below its condensation point. This is the principle underlying the liquefaction of gases with the Linde refrigerator, which utilizes the louIe-Thomson effect. See Section 2.12 for a more complete discussion.

THE FIRST LAW

02.7

23

The vertical axis of a thermogram represents Cp , and the baselines represent the heat capacity associated with simple heating in the absence of structural transformations or similar transitions. In the example shown in Fig. 2.16, the sample undergoes a structural change between TI and T2, so there is no reason to expect Cp after the transition to return to its value before the transition. Just as diamond and graphite have different heat capacities because of their different structures, the structural changes that occur during the measurement of a thermogram can also give rise to a change in heat capacity.


The physical definition of work is dw

= -F dz [2.4]

In a gravitational field the force is the weight of the object, which is F

= mg

If g is constant over the distance the mass moves, dw may be intergrated to give the total work

W

=-

Zf 1

Fdz

=-

1Zf

Zi

w

E2.2(b)

mg dz

= -mg(Zf -

Zi)

= -mgh

where

h

= (Zf -

Zi)

Zi

= -(0.120 kg)

x (9.81 m S-

2)

x (50m)

= -59 J = 159 J needed 1

This is an expansion against a constant external pressure; hence w = -Pex l:!. V

[2 .8]

The change in volume is the cross-sectional area times the linear displacement:

l:!.V

so w E2.3(b)

=

(50.0cm

= -(121

2)

x (15cm) x

(~)3 = 7.5 x 1O-4m3 , 100 cm

x 103 Pa) x (7 .5 x 10- 4 m 3 )

= 1-91 J 1as 1 Pa m3 = 1 J.

For all cases l:!. U = 0, since the internal energy of a perfect gas depends only on temperature. (See Molecular interpretation 2.2 and Section 2. 11 (b) for a more complete discussion.) From the definition of enthalpy, H = U + pV, so l:!.H = l:!.U + l:!.(pV) = l:!.U + l:!. (nRT) (perfect gas). Hence, l:!.H = 0 as well, at constant temperature for all processes in a perfect gas. (a)

1

l:!.U

w

=

l:!.H

=01

= -nRT In (~) [2.11] = -(2.00 mol) x (8.3145 J K- 1 mol-I) x (22 + 273) K x In 31.7 dm: = 1-1.62 x 10 3 J 1 22.8dm

q (b)

= -w = 11.62 x

1 l:!.U

w

=

=

l:!.H

103 J 1

=0 1

-Pexl:!. V [2.8]

where P ex in this case can be computed from the perfect gas law pV

= nRT

24


(2.00mol) x (8.3145J K- Imol- I) x (22 + 273) K 3 31.7 dm

sop=

- (1.55

X

and w = q

I 3 (lOdmm - ) = 1.55

X

105 Pa

105 Pa) X (31.7 - 22.8) dm 3 _ / _ 3 / (lOdmm - I )3 _ . 1.38 x 10 J.

= -w = [ 1.38

/ ~U

(C)

X

X

10 3 J [

= ~H = 0 1

/ w = 0 1[free expansion] q =

~U -

w = 0 - 0 =@]

COMMENT. An isothermal free expansion of a perfect gas is also adiabatic.

E2.4(b)

The perfect gas law leads to PIV = nRTI P2 V nRT2

or

P2 = PI T 2 = (111kPa) X (356K) =1 143kPai TI 277 K

There is no change in volume, so 1w = 0 I. The heat flow is q=

f

Cy dT

~ Cy~T =

(2.5)

X

(8.3145J K- I mol-I)

X

(2.00mol)

X

(356 - 277) K

= [ 3.28 x 103 J [

~U = q + w =

[ 3.28

X

10 3 J [

(a)

_ - (7.7 x 103 Pa) x (2.5 dm 3 ) w--pex~V-(V) (lOdmm - I)3

(b)

w=-nRTln

_

E2.S(b)

~

6.56 g

_

~

-~

[2.11]

w=- ( I 39.95 g mol-

)

( ) (2.5 + 18.5) dm x 8.3145JK- l mol - 1 x (305K)x ln 3 18.5dm

3

= /-52.8 J /

E2.6(b)

~H = ~condH = - ~vapH = - (2.00 mol) x (35.3 kJ mol - I) = /-70.6 kJ / Since the condensation is done isothermally and reversibly, the external pressure is constant at 1.00 atm. Hence, q = qp =

~H = /-70.6 kJ /

w = -Pex ~ V [2.8]

where

~V

= Vliq -

Vvap ~ - Vvap

because

Vliq «

On the assumption that methanol vapor is a perfect gas, Vvap = nRT / p and P condensation is done reversibly. Hence, w

~ nRT =

Vvap

=

(2.00 mol) x (8.3145 J K- I mol-I) x (64 + 273) K = [ 5.60 x 103 J [

and ~U = q + w = (-70.6 + 5.60) kJ = / -65.0kJ /

Pex,

since the

THE FIRST LAW

E2.7(b)

25

The reaction is

so it liberates 1 mol of H2(g) for every 1 mol Zn used. Work at constant pressure is w =

-Pex~V

=-( E2.8(b)

= -pVgas = -nRT

5.0g _I) 65.4 g mol

X

~H .

(a) At constant pressure, q = q=

f

CpdT=

l

(8.3145JK- l mol- l ) x (23+273) K=I-188JI

l°O+273 K

[20.17+ (0.400I)T/K]dTJK- 1

0+273 K

I = [ (20.17) T + -(0.4001) x (T2)JI373K J K- I 2 K 273 K

= [(20.17) x (373 - 273) + w =

~(0.4001) x (373 2 -

273 2)] J = 114.9 x 103 J 1=

~H

-p~V = -nR~T = - (1.00 mol) x (8.3145JK- 1 mol-I) x (lOOK) = 1-831 J I

kJl

~U=q+w=(14.9-0. 831)kJ=114.1

(b) The energy and enthalpy of a perfect gas depend on temperature alone. Thus, ~H = 114.9

~U = 114.1 kJ Ias above. At constant volume, w =@] and ~U = q, so q = 1+14.1 kJ I. E2.9(b)

For reversible adiabatic expansion

V)I /C

Tr

= Tj ( V~

[2.28a]

where c

CVm

Cp,m- R

= - '- = --'-,-=-R

R

so the final temperature is Tr = (298.15 K) x

E2.10(b)

(37 .11 - 8.3145) J K-Imol- I - - - - - - - - - : - - - - ; - 1- -

8.3145J K-Imol-

3 3)

500 x 10- dm

(

2.00dm

= 3.463,

1/ 3.463

= 1200 K I

3

Reversible adiabatic work is w =

Cv~T

[2.27] = n(Cp,m - R) x (Tr - Tj)

where the temperatures are related by [solution to Exercise 2.15(b)] Tr = Tj

(~~) lie [2.28a]

where

c =

C~m

=

Cp,~-

R = 2.503

kJ Iand

26


and w = ( E2.11 (b)

3

(

= 156 K

)

2.00dm

3.12 g -I) x (29.125 - 8.3145) J K- I mol- I x (156 - 296) K = 1-325 J 1 28.0 gmol

For reversible adiabatic expansion

Pr = Pi

so

E2.12(b)

1/ 2.503

400 x 1O- 3 dm 3

So Tr = [(23.0 + 273.15) K] x

qp

= nCp,m!:!,.T

C

p,m

(Vi) -

Y

Vr

= (8.73 Torr) x

500 x 10- 3 dm 3)1.3 (

3.0dm

3

= 18.5 Torr I

[2 .24]

=~= n!:!"T

178J =153JK- I mol- I 1 1.9 mol x 1.78K

CV ,m = Cp,m - R = (53 - 8.3) J K- I mol - I = 145 JK- I mol- II E2.13(b)

!:!"H = qp = Cp!:!"T [2.23b, 2.24] = nCp,m!:!,.T !:!"H = qp = (2.0 mol) x (37.1 I J K- I mol-I) x (277 - 250) K = 12.0 x 103 J mol-I I !:!"H = !:!,.U

+ !:!,.(pV)

= !:!,.u

+ nR!:!"T

so

!:!,.U = !:!"H - nR!:!"T

!:!,.U = 2.0 x 10 J mol - I - (2.0 mol) x (8.3145 J K- I mol - I) x (277 - 250) K 3

= 11.6 x 10 3 J mol - I E2.14(b)

In an adiabatic process, q =

w

= -Pex !:!" V =

I

@]. Work against a constant external pressure is

-(78.5x1Q 3 Pa)x(4xI5-15)dm 3 _1_ 3 1 I 3 - . 3.5 x 10 J. (IOdm m- )

!:!,.U = q + w = 1-3.5 x 10 3 J I

One can also relate adiabatic work to !:!"T (eqn 2.27): w = Cv!:!"T = n(Cp,m - R)!:!"T

!:!,.T= !:!"H

=

!:!"T =

w n(Cp,m - R)

,

-3.5 x 103 J

=1-24KI. (5.0mol) x (37.11 - 8.3145)JK-I mol - I !:!,.U

+ !:!"(PV) =

= -3.5 x 103 J E2.1S(b)

so

!:!,.U

+ nR!:!"T,

+ (5.0 mol)

x (8.3145JK- I mol-I) x (-24K) =1-4.5 x 103 J I

In an adiabatic process, the initial and final pressures are related by (eqn 2.29)

!

pr V = Pi Vr

where

Cp,m Cp,m - CV,m - Cp,m- R

y----

I

I

6 20.8JK- mol---------;------;I = I. 7 I (20.8 - 8.31) JK- mol-

THE FIRST LAW

27

Find Vi from the perfect gas law: l I . _ nRTi _ (1.5mol)(8.3IJK- mol- )(315K) =0017-1 3 V,. m 230 x 103 Pa

Pi

so

Vr =

Vi

(Pi) I/ y = pr

(Om 71m3) (230 kPa) 1/ 1.67 = I 0.0205 m31. 170kPa

Find the final temperature from the perfect gas law:

Adiabatic work is (eqn 2.27) w = Cv/'o,.T = (20.8 - 8.31) JK- I mol-I x 1.5 mol x (279 - 315) K = 1-6.7 x 102 J E2.16(b)

I

At constant pressure q = /'o,.H = n/'o,.vapW = (0.75 mol) x (32.0 kJ mol-I) = 124.0 kJ

I

and w = -p/'o,.V ~ -pVvapor = -nRT = -(0.75 mol) x (8 .3145JK- I mol-I) x (260K) w = -1.6 x 103J = 1-1.6kJ /'0,. U

I

= w + q = 24.0 - 1.6 kJ = 122.4 kJ

I

COMMENT. Because the vapor is here treated as a perfect gas, the specific value of the external pressure

provided in the statement of the exercise does not affect the numerical value of the answer.

E2.17(b)

The reaction is C6HsOH(l)

+ 702(g)

~

/'o,.cW = 6/'o,.rW(C02) = [6(-393.15) E2.18(b)

6C02(g)

+

3H20(l)

+ 3/'o,.rW(H20) -

+ 3( -285 .83) -

/'o,.rW(C6HsOH) -7/'o,.rW (02)

(- 165.0) - 7(0)] kJ mol- I = 1-3053.6 kJ mol-I

We need /'o,.rW for the reaction (4)

2B(s)

+ 3H2(g) ~

reaction(4) = reaction(2) Thus,

B2H6(g)

+3 x

/'o,.rW = /'o,.rW{reaction(2)}

reaction(3) - reaction(l)

+3 x

/'o,.rW{reaction(3)} - /'o,.rW{reaction(l)}

= [-2368 + 3 x (-241.8) - (-1941)] kJ mol-I = 1-1152 kJ mol-I

I

I

28

E2.19(b)


For anthracene the reaction is

"""e cr = """ell" - """ngRT

[2.21],

"""ecr = -7061kJmol-

-

=

-7055kJmol -

1

"""ng

= - ~ mol

(-~ x 8.3 x 1O- 3 kJK- l mol- 1 x 298K)

1

3

Iql = Iqvl =

2.25 x 10- g ) In"""ecrl = ( 172.23gmol _I

c=M =

0.0922 kJ

"""T

1.35 K

( ) x 7055kJmol- 1

= 0.0922kJ

= 0.0683 kJ K- I = 168.3 J K- I I

When phenol is used the reaction is

"""ell"

"""eU

=

-3054kJmol- 1 [Table2.5]

= """eH - """ngRT,

= (-3054kJmol =

"""ng l

)

= -~

+ (~) x (8.314

X

1O- 3 kJK- 1 mol-I) x (298K)

-3050 kJ mol-I 135 x 10- ~) x (3050kJmOI- I) 94.12gmol3

Iql

=(

"""T

=C =

Iql

4.375kJ 0.0683 kJ K-I

= 4.375kJ

= 1+ 64 .1 K I

COMMENT. In this case f'>. ci f and f'>. cH" differed by about 0.1 percent. Thus, to within 3 significant figures,

it would not have mattered if we had used f'>.cH" instead of f'>. cif, but for very precise work it would.

E2.20(b)

The reaction is AgBr(s) -+ Ag+(aq) + Br-(aq) """solJr

=

"""rJr(Ag+,aq) + """rJr(Br-,aq) - """rJr(AgBr, s)

= [105.58 + (-121.55) - (-100.37)] kJ mol-I = 1+84.40 kJ mol- l

E2.21 (b)

I

The combustion products of graphite and diamond are the same, so the transition C(gr) -+ C(d) is equivalent to the combustion of graphite plus the reverse of the combustion of diamond, and

"""= 11" =

[-393.51 - (395.41)] kJ mol-I = 1+ 1.90 kJ mol-I

I

THE FIRST LAW

E2.22(b)

(a)

reaction(3)

= (-2)

x reaction(l) + reaction(2)

and

L'l.ng

29

= -1

The enthalpies of reactions are combined in the same manner as the equations (Hess's law). L'l.r~(3)

= (-2) x L'l.r~(l) + L'l.r~(2) = [(-2) x (52.96) + (-483 .64)] kJ mol-l = 1-589.56kJmOI- 1 1

L'l.rlr

= L'l.r~ - L'l.ngRT = -589.56kJmol- 1 =

(-3) x (8.314JK- 1mol- 1) x (298K)

-589.56kJ mol- 1 + 7.43 kJmol- 1 = 1-582.13 kJ mol-II

(b) L'l.fH>7 refers to the formation of one mole of the compound, so

L'l.f~(HI) = L'l.f~(H20) E2.23(b)

! (52.96 kJ mol-I)

= 126.48 kJ mol-II

= ! (-483.64kJmOI- 1) = 1-241.82kJmol- 1 1

L'l.r~ = L'l.rlr + RT L'l.ng [2.21]

=

-772.7kJmol- 1 +(5) x (8 .3145 x 1O-3kJK-lmol-l) x (298K)

= 1-760.3 kJ mol-II E2.24(b)

Combine the reactions in such a way that the combination is the desired formation reaction. The enthalpies of the reactions are then combined in the same way as the equations to yield the enthalpy of formation.

!N2(g) + !02(g) ~ NO(g) NO(g) + !CI2(g) ~ NOCl(g)

+ 90.25 -!(75.5) +52.5

Hence, L'l. f H B(NOCI , g) E2.25(b)

= 1+52.5 kJ mol-l

1

According to Kirchhoff's law [2.36]

L'l.r~(lOO°C) = L'l.r~(25 °C) +

100°C

{

L'l.rC;dT

125°C

where L'l.r as usual signifies a sum over product and reactant species weighted by stoichiometric coefficients. Because Cp,m can frequently be parametrized as Cp,m

= a + bT + c/T2

30


the indefinite integral of Cp,m has the form

Combining this expression with our original integral, we have

Now for the pieces /).rH"(25 °C) = 2( -285.83 kJ mol - I) - 2(0) - 0 = -571.66 kJ mol -

1

/).ra

= [2(75 .29) - 2(27.28) - (29.96)]JK- 1 mol - 1 = 0.06606kJK - 1 mol - 1

/).rb

= [2(0) - 2(3.29) - (4.18)] x 10- 3 J K- 2 mol - 1 = -10.76 x 10- 6 kJ K- 2 mol- 1

/).rC

= [2(0) - 2(0.50) - (-1.67)] x 105 J K mol- 1 = 67 kJ K mol- 1

/).rH"(1oo 0c) = [-571.66 + (373 - 298) x (0.06606) + x (-10.76 x 10- 6 )

-

~ (373 2 -

298 2)

(67) x (_1_ - _1_)] kJ mol- 1 373 298

= 1-566.93kJmol- 1 1 E2.26(b)

The hydrogenation reaction is

The reactions and accompanying data which are to be combined in order to yield reaction (1) and /).rW(T) are (2)

H2(g) + !02(g)

(3)

C2!it(g) + 302 (g)

-+

2H20(l) + 2C02(g)

(4)

C2H2(g) + ~02(g)

-+

H20(l) +2C02(g)

-+

H20(l)

/).cH"(2) = -285.83 kJ mol -

1

/).cH" (3) = -1411 kJ mol- 1 /).cH"(4) = -13ookJmol-

1

reaction (1) = reaction (2) - reaction (3) + reaction (4) (a) Hence, at 298 K: /).rH" = /).cH"(2) - /).cH"(3) + /).cH"(4) = [(-285.83) - (-1411) + (-1300)] kJ mol- 1 = 1-175 kJ mol-II /).r if' = /).rH" - /).ngRT

[2.21];

/).n g = -1

= -175kJmol- 1 - (-1) x (2.48kJmol- 1) =1-173kJmOI- 1 1

THE FIRST LAW

(b) At 348 K:

to rF'(348 K) = to rF'(298 K) torCp

=

+ to r C;(348 K -

L vJC;m(J) [2.37] = C; m(C2IL!,g) -

298 K)

[Example 2.6]

Cp~m(C2H2,g) - Cp~m (H2, g)

J

= (43 .56 to rF'(348K)

43.93 - 28.82) x 10- 3 kJ K- I mol- I

= (-175kJmol- l ) -

=

-29.19 x 10- 3 kJ K- I mol- I

(29.19 x 1O- 3 kJK- I mol-I) x (50K)

= 1-176kJmol- 1 1 E2.27(b)

NaCI, AgN03 , and NaN03 are strong electrolytes; therefore the net ionic equation is Ag+(aq)

+ Cl-(aq) --+

= tofF'(AgCI) -

torF'

= [(-127.D7) E2.28(b)

AgCI(s) tofF'(Ag+) - tofF'(CI-)

(105.58) - (-167.16)]kJmol- 1

= 1-65.49kJmol- 1 1

The cycle is shown in Figure 2. 1.

~,

Ionization

Ca(g) + 2Br(g) ~

Dissociation

Ca(g) + Br2(g)

Electron gain Br

~

Vaporization Br

Ca(g) + Br2(1)

Ca 2+(g) + 2Br- (g) t

~

Sublimation Ca

Ca(s) + Br2(1) ~~

-Formation

Hydration BrCa 2+(g) + 2Br- (aq)

,~

CaBr2(s) Hydration ci+

~I'

-Solution

,~

Figure 2.1

-to hyd F'(Ca 2+)

= -tosolnF'(CaBr2) -

=

+ tosubF'(Ca)

+ tovapF' (Br2) + todissF' (Br2) + toionF' (Ca) + toionF'(Ca+) + 2to eg F'(Br) + 2to hyd F'(Br-) [-( - 103. 1) - (-682.8) + 178.2 + 30.91 + 192.9 + 589.7 + 1145 + 2(-33\.0) + 2(-337)] kJ mol-I

= 11587 kJ mol-II so tohyd W (Ca2+)

tofF'(CaBr2, s)

= 1-1587kJmol- 1 1

31

32

E2.29(b)


The Joule-Thomson coefficient J-t is the ratio of temperature change to pressure change under conditions of isenthalpic expansion. So J-t - (aT) "'" 6.T = _ _-_I_O_K_ _ = I 0.48 Katm- I I ap H 6.p (1.00 - 22) atm

E2.30(b)

The internal energy is a function of temperature and volume, Urn = Um(T, Vm), so

dUm

aUm) =(aT

Vm

dT+ (aUm) -dVm aVm T

For an isothermal expansion dT = 0; hence

a 22.1 dm 3 mol-I

a 1.00dm 3 mol-I

----:;------;-+------;;------;-

2l.la 3 = 0.95475adm- mol 22.1 dm mol-

------;;-3-----;-1

From Table 1.6, a = 1.337 dm 6 atm mol-I 6.Um = (0.95475 mol dm 3 ) x (1.337 atm dm 6 mol- 2 )

= 129Pam 3 mol-

w =-fpdVm

where

1

=1129JmOI- 1 1 RT

a

Vm - b

Vm

p = --- - 2

for a van der Waals gas.

Hence,

w

= -/

(~) dVm + / Vm - b

-;'dVm Vm

=

-q+ 6. Urn

Thus

q

=

22.1 dm 3 mol / l.00dm 3 mol - 1

1

(

RT

--Vm - b

) dVm

= RTln(Vm -

122. 1 dm

b)

3 mol - 1 3

I

l.00dm mol -

2

I

22.1 - 3.20 x 10- ) = (8.314 JK- 1 mol-I) x (298K) x In ( 2 = +7 .7465kJmol - ~ 1.00 - 3.20 x 10and w = -q

+ 6.Um =

-(7747 J mol-I)

+ (l29Jmol- l )

I

= 1-76181 mol-I I = 1-7.62kJmol- 1 1

THE FIRST LAW

E2.31(b)

33

The expansion coefficient is V' (3 .7

X

1O- 4 K- 1 +2 x l.52 x 1O- 6 TK- 2 ) V

V' [3.7 x 10-4 + 2 x 1.52 x 10- 6 (T / K)] K- I = V' [O.77 + 3.7 x 1O- 4 (T / K) + 1.52 x 1O- 6 (T / K) 2]

[3.7 x 10-4 + 2 x 1.52 x 10- 6(310)] K- I 0.77 + 3.7 x 10- 4 (310) + 1.52 x 10 6(310)2

= E2.32(b)

=

I l.27 x

3

10- K-

I

I

Isothermal compressibility is so

Lip

LiV

= -VKT

A density increase of 0.08 percent means Li V /V applied is Lip 0.0008 - 2.21 x 10- 6 atm- I E2.33(b)

= 13.6 x .

= -0.0008. So the additional pressure that must be

102 at~ 1 .

The isothermal Joule-Thomson coefficient is ( 8H) 8p T

=

-J1,Cp

=

-(1.11 K atm- I) x (37.11 J K- I mol-I)

= 1-41.2J atm- 1 mol - I 1

If this coefficient is constant in an isothermal Joule-Thomson experiment, then the heat which must be supplied to maintain constant temperature is LiH in the following relationship LiH / n = -41.2Jatm-1 mol-I Lip LiH

=

so

LiH = -(41.2Jatm- 1 mol-l)nLip

-(4l.2J atm- I mol-I ) x (l2.0mol) x (-55 atm)

= 127.2 x

103 J I

Solutions to problems Assume all gases are perfect unless stated otherwise. Unless otherwise stated, thermochemical data are for 298 K.

Solutions to numerical problems P2.1

The temperatures are readily obtained from the perfect gas equation, T

TI

=

(1.00atm) x (22.4dm 3 ) (1.00 mol) x (0.0821 dm 3 atm mol- I K-l)

Similarly, T2 = 1546 K I.

~

= ~ = T3

= PV , nR

[isotherm).

34


In the solutions that follow all steps in the cycle are considered to be reversible. Step I --+ 2 W

= -Pexf!. V = -pf!. V = -nRf!.T

W

= -(1.00 mol) x (8.314J K- I mol-I) x (546 - 273) K = 1-2.27 x 103 J I.

f!.U = nCv.mf!.T = (l.OOmol) x q = f!.U -

W

[f!.(pV) = f!.(nRT)],

23 x

(8.314J K- I mol-I) x (273 K) = +3.40 x 103 J.

= +3.40 x 103 J - (-2.27 x 10 3 J) = 1+5.67 x 103 J I. 3

f!.H = qp = 1+5.67 x 10 J

I·

If this step is not reversible, then w, q, and f!.H would be indeterminate. Step 2 --+ 3

I W = 0 I [constant volume]. 3 qv = f!.U = nCv,mf!.T = (1.00 mol) x (2) x (8.314J K- I mol-I) x (-273K)

= 1-3.40 x 10 3 J I. FromH

==

U +pV

f!.H = f!.U + f!.(pV) = f!.U + f!.(nRT) = f!.U + nRf!.T

= (-3.40 x 103 J) + (1.00 mol) x (8.314JK- 1 mol-I) x (-273K) =1-5 .67 x 10 3

Jj.

Step 3 --+ 1

I

I

f!.U and f!.H are zero for an isothermal process in a perfect gas; hence for the reversible compression 3

-q =

W

= -nRT In -VI = (-1.00 mol) x (8.314JK- I mol- I) x (273K) x In (22.4dm 3 )

= 1+ 1.57

~

X

~8~

103 J I, q = 1-1.57

X

103 J I·

If this step is not reversible, then q and w would have different values which would be determined by the details of the process. Total cycle

State

p/atm

Vldm 3

TIK

2 3

1.00 1.00 0.50

22.44 44.8 44.8

273 546 273

THE FIRST LAW

35

Thermodynamic quantities calculatedfor reversible steps

Step

Process

q/kJ

w/kJ

t:.U j kJ

t:.H j kJ

1---+2 2---+3 3---+1 Cycle

p constant = Pex V constant

+5.67 -3.40 -1.57 +0.70

-2.27 0 +1 .57 -0.70

+3.40 -3.40 0 0

+5 .67 -5.67 0 0

Isothermal, reversible

COMMENT. All values can be determined unambiguously for the reversible cycle. The net result of the overall

process is that 700 J of heat has been converted to work.

P2.3

Since the volume is fixed , 1w Since t:.u t:.H

= 0 I.

= q at constant volume, 1t:.u = +2.35 kJ I.

=

t:.u

+ t:.(pV ) = t:.u + Vt:.p

[t:.V

= 0].

From the van der Waals equation [Table 1.6] so

Therefore, t:.H

= t:. U +

Rt:.T t:.p= - - Vm -b

[t:. Vm

= 0 at constant volume] .

RVt:.T

---. Vm -b

From the data, Vm

=

15.0dm 3 2.0 mol

t:.T

= (341

- 300) K

RVt:.T

(8.3141 K- I mol-I) x (15 .0dm 3 ) x (41 K)

Vm -b

7.46 dm mol-I

3

Therefore, t:.H P2.S

= 7.5 dm 3 mol-I,

=

(2.35 kJ) + (0.68 kJ)

= 41

K.

= 0.68 kJ.

= 1+3.03 kJ I.

This cycle is represented in Figure 2.2. Assume that the initial temperature is 298 K . (a) First, note that 1 w = 0 1 (constant volume). Then calculate t:.U since t:.T is known (t:.T and then calculate q from the First Law. t:.U

= nCv.mt:.T [2.16b];

t:.U

=

q

(1.00 mol) x

= qV =

t:. U - w

G)

CV,m

= Cp,m -

R

=

7

'2R - R

=

0

= 1+6.19 kJ I.

298 K)

5

'2R ,

x (8.3141 K- I mol-I) x (298 K)

= 6.19 kJ -

=

= 6.19

x 103 1 = 1+6.19 kJ I.

36


2 2.0

S '"

(a)

~

1.0

3 100

50

150

Figure 2.2

I':1.H

=

I':1.U + I':1.(PV)

= (6.19kJ) + (b) 1q

=

I':1.U + I':1.(nRT)

=

I':1.U + nRI':1.T

(1.00 mol) x (8.31 x 10- 3 kJ mol - I) x (298 K)

= 1+8.67 kJ I.

= 0 1(adiabatic).

Because the energy and enthalpy of a perfect gas depend on temperature alone, I':1.U(b)

=

-I':1.U(a)

= 1-6.19 kJ I, since I':1.T(b) =

Likewise I':1.H(b) = -I':1.H(a) = 1-8.67 kJ

=

w (c) I':1.U

1':1. U

-I':1.T (a).

I.

= 1-6.19 kJ 1[First Law with q = 0].

= I':1.H = 0 [isothermal process in perfect gas]. q = -w [First Law with I':1.U = 0];

=

V2

VI

W

=

VI

-nRT l ln - [2.11] . V3

nRTI

(1.00 mol) x (0.08206dm 3 atm K- I mol-I) x (298K)

PI

1.00atm

= -- =

where

so V3

=

_ 3 ((2) x (298 K) (24.45 dm ) x 298 K

q

10 3 J

= 1-4.29 kJ I.

= 1+4.29 kJ I·

R

)5/2= 138.3_dm 3.

w = (-1.00 mol) x (8.314J K- I mol-I) x (298K) x In

= 4.29 x

CVm

5 2

c = -'- = -

22.45 dm3) 3 ( 138.3dm

-

= 24.45 dm

3 .

THE FIRST LAW

P2.7

37

The fonnation reaction is ~fW(298 K) = -84.68 kJ mol-I.

In order to determine ~fW (350 K) we employ Kirchhoff's law [2.36] with T2

where ~rCp

=L

vJCp,m(J)

= Cp,m(C6H6)

= 350 K, TI = 298 K,

- 2Cp,m(C) - 3Cp,m(H2).

J

From Table 2.2

l

T2

T,

C

~ dT ~I p _I J K mol

= -100.83

X

(T2 - TI) +

2 - ( 1.56 x 106 K)

=-

(

(I)- (0.1079K- I) 2

(Ti -

T~)

- I - -I ) T2

100.83 x (52 K) +

TI

(~) (0.1079) (3502 -

298 2 ) K

6 ( - I - - I ) K - (1.56 x 10) 350 298

= -2.65

X

103 K.

Multiplying by the units J K-Imol - I, we obtain

r ~rCpdT = - (2.65 iT, T2

= Hence ~fW(350 K)

Cr(C6H6h(s) ~rW

--+

=

=

-2.65 x 103 J mol-I

~fW(298 K) - 2.65 kJ mol-I

84.68 kJ mol-I - 2.65 kJ mol-I

Cr(s) + 2C6H6(g) ,

= ~r~ + =

103 K) x (J K-Imol- I)

-2.65 kJ mol-I .

=P2.9

X

~ng

= 1-87.33 kJ mol-I I.

= +2 mol.

2RT, from [2 .21]

(8.0 kJ mol - I) + (2) x (8.314 J K- I mol- I) x (583 K)

= 1+17.7 kJ mol- I I.

38


In terms of enthalpies of formation /';.rlF or

=

(2) x /';.flF(benzene, 583 K) - /';.flF(metallocene,583 K)

/';.rH~(metallocene, 583 K) = 2/';.fH~(benzene, 583 K) - 17.7 kJ mol-I .

The enthalpy of formation of benzene gas at 583 K is related to its value at 298 K by /';.flF(benzene, 583 K)

=

/';.flF(benzene, 298 K)

+ (Tb -

298 K)Cp.m(l) + /';. vaplF + (583 K - Tb)Cp.m(g)

- 6 x (583 K - 298 K)Cp,m(gr) - 3 x (583 K - 298 K)Cp,m(H2 , g) where Tb is the boiling temperature of benzene (353 K). We shall assume that the heat capacities of graphite and hydrogen are approximately constant in the range of interest and use their values from Table 2.7. /';.flF(benzene,583 K)

= (49.0 kJ mol - I) +

(353 - 298) K x (136.1 J K- I mol-I)

+ (30.8 kJ mol-I) + (583 - 353) K x (81.67 J K- I mol-I) - (6) x (583 - 298) K x (8.53 J K- I mol-I) - (3) x (583 - 298) K x (28.82 J K- I mol-I)

= {(49.0) + (7.49) + = +66.8 kJ mol - I. Therefore P2.11

/';.fH~ (metallocene, 583 K) = (2

(18.78) + (30.8) - (14.59) - (24.64)} kJ mol-I

x 66.8 - 17.7) kJ mol- I = 1+ 116.0 kJ mol- 1 I.

(a) and (b). The table displays computed enthalpies of formation (semi-empirical, PM3 level, PC Spartan ProTM), enthalpies of combustion based on them (and on experimental enthalpies of formation of H20(l) and C02 (g), - 285.83 and -393.51 kJ mol-I respectively), experimental enthalpies of combustion (Table 2.5), and the relative error in enthalpy of combustion.

ClL;(g) C2 H6(g) C3 Hg(g) C4 H 10 (g) C5 H I2(g)

-54.45 -75.88 -98.84 -121.60 -142. 11

-910.72 -1568.63 -2225.01 - 2881.59 -3540.42

-890 -1560 -2220 - 2878 -3537

The combustion reactions can be expressed as:

The enthalpy of combustion, in terms of enthalpies of reaction, is

2.33 0.55 0.23 0.12 0.10

THE FIRST LAW

where we have left out l'lr1F (02) %error

=

39

= O. The % error is defined as:

l'lcHG(caIc.) - l'lcHG(expt.) G x 100%. l'lcH (expt.)

The agreement is quite good . (c) If the enthalpy of combustion is related to the molar mass by

then one can take the natural log of both sides to obtain:

Thus, if one plots In ll'lcHGI vs. In [M/(g mol-I )), one ought to obtain a straight line with slope nand y-intercept In Ikl. Draw up the following table. Compound

M/(g mol-I)

l'lcH j kJ mol-I

In M/(g mol - I)

In ll'lcHGj kJ mol - II

CH4 (g) C2 H6(g) C3 H8(g) C4H IO(g) CSHI2(g)

16.04 30.07 44.10 58.12 72.15

-910.72 -1568.63 -2225.01 -2881 .59 -3540.42

2.775 3.404 3.786 4.063 4.279

6.814 7.358 7.708 7.966 8.172

The plot is shown in Fig 2.3. 9 I

"0

E 8 :2 "SI

f

dqrev T

-

Cp,m dT - C L Tf --T- p,m n Ti

(2.00 mol) x ( -7) x (8.3l45JK- 1 mol - I) x In (l35 + 273) K 2 (25+273)K

=

18.31K- 1

and for the second f>"Sz =

where qrev

f =

dqrev = qrev

T

T

=

- w

JpdV = nRT In -Vf

= nR In -Pi =

so f>"Sz

Pf

=

f>"S

Vi

= nRT In

Pi

-

Pf

I I L50atm (2.00 mol) x (8.3 145 J K- mol - ) x In 7.00atm

(18.3 - 25.6)JK- 1

=

- 25 .61 K-

I

= !-7.31K- 1 !

The heat lost in step 2 was more than the heat gained in step 1, resulting in a net loss of entropy. Or the ordering represented by confining the sample to a smaller volume in step 2 overcame the disordering represented by the temperature rise in step I. A negative entropy change is allowed for a system as long as an increase in entropy elsewhere results in f>"Stotal > O. E3.4(b)

q

= qrev =0 [adiabatic reversible process] =

f>"S

if i

f>"U

dqrev

=

T

W

= nCv ,mf>"T = (2.00 mol )

x (27.5 J K- I mol-I) x (300 - 250) K

= 2750J = 1+2.75 kJ 1

w

=

f>"H

f>"U - q

= 2.75 kJ -

E3.S(b)

= 12.75 kJ 1

= nCp.mf>"T

Cp,m = CV ,m + R

So f>"H

0

=

=

(27.51 K- I mol-I + 8.3141 K- I mol-I)

(2.00 mol) x (35.8141 K- I mol - I) x (+50 K)

=

35.814J K- I mol - I

= 358l.41 = 13.58 kJ 1

Since the masses are equal and the heat capacity is ass umed constant, the final temperature will be the average of the two initial temperatures,

The heat capacity of each block is C

= mCs

where Cs is the specific heat capacity

so f>"H (individual)

= mCsf>"T =

1.00 x 103 g x 0.4491 K- I g-I x (±87.5K)

= ±39kJ

THE SECOND LAW

I

These two enthalpy changes add up to zero: 6. H tot

= mCs In

6.S

6.S1

(i);

200 °C

= 473.2 K; 2S °C =

6.S2 = (1.00 x 10 3 g) x (0.449 J K- I g- I)

E3.6(b)

=0I

(a)

q = 0 [adiabatic]

(b)

w

X

298.2 K; 112.5 °C

= 38S.7 K

G!~:~)

=

IIS .5J K- I

38S.7) In ( 473.2

=

-9 1.802J K- I

( 1.00 x 103 g) x (0.449J K- I g- I) x In

=

5

=

-Pex6.V

=

- ( I.Satm) x

=

-227.2 J

(.

1.0 I x 10 pa) x (lOO.Ocm 2 ) x (lScm) x ~

6.U

= q+w=0-

(d)

6.U

= nCv.m6.T

6. T

-227.2 J = -6.U- = - - - - -----:-

nCV.m

(

Im

3

-6--3

)

10~

= 1-230 J I

(c)

230 J

53

= 1-230 J I

( I.Smol) x (28.8 JK -I mol -I)

=

I-S.3K I

(e) Entropy is a state function , so we can compute it by any convenient path. Although the spe-

cified transformation is adiabatic, a more convenient path is constant-volume cooling followed by isothermal expansion . The entropy change is the sum of the entropy changes of these two steps : 6.S

= 6.S1 + 6.S2 = nCV.m In

Tr = 288. IS K - S.26 K

~

nRT

nR In

(~)

[3.19 and 3.13]

= 282.9K

( I.S mol) x (8.206 x 10- 2 dm 3 atm K -I mol-I) x (288.2 K)

= --- = ~-~-~----~------~~--~ Pi

9.0atm

= 3.942 dm

Vr

(i) +

3

= 3.942dm 3 + ( IOO cm2 )

3

x (lScm) x

1 dm 3 ) ( lOOOcm

54


/' O. (b) The volume of the container may be calculated from the perfect gas law. nRT ( 1.00 mol) x (0.08206 dm 3 atm K- I mol - I) x (373.2 K) 3 V= = = 30.6 dm 1.00atm P

At 57°C the vapor pressure of water is 130 Torr (Handbook of Chem.istry and Physics). The amount of water vapor present at equilibrium is then ( 130 Torr) x ( 1 atm ) x (30.6 dm 3 ) n=pV = 760 Torr =0.193mol. RT (0.08206 dm 3 atm K- I mol - I) x (330.2 K) This is a substantial fraction of the original amount of water and cannot be ignored. Consequently the calcu lation needs to be redone taking into account the fact that only a part, Ill , of the vapor condenses

I.

THE SECOND LAW

61

into a liquid while the remainder (1.00 mol - nl ) remains gaseous . The heat flow involving water, then, becomes q(H 2 0 ) = - nl t:. vapH + nl Cp,m(H20, 1)t:.T(H20)

+ (1.00 mol - nl)Cp,m(H20, g)t:.T(H20). Because nl depends on the equilibrium temperature through n I = 1.00 mol - p V, where p is the

RT

vapor pressure of water, we will have two unknowns (p and T) in the equation -q(H20 ) = q(Cu ) . There are two ways out of tills dilemma: (1) p may be expressed as a function of T by use of the Clapeyron equation (Chapter 4), or (2) by use of successive approximations. Redoing the calculation yields:

e=

nl t:.vapH + nlCp,m(H20, I) x 100°C + (1.00 - nl)Cp,m(H20, g) x lOO°C . mCs + nCp,m(H20, I) + ( \.00 - nl)Cp,m(H20 , g)

With nl = (1.00 mol) - (0.193 mol ) = 0.807 mol

(noting that Cp.m(H20, g) = 33.6Jmol- 1 K- 1 [Table 2.7]), vapor pressure of water is 80AI Torr, corresponding to

e=

47.2°C. At this temperature, the

nl = (1.00 mol) - (0.123 mol) = 0.877 mol.

This leads to e = SO.8°C. The successive approximations eventually converge to yield a value of 1 49.9°C = 323.1 K 1 for the final temperature. (At this temperature, the vapor pressure is 0.123 bar.) Using this value of the final temperature, the heat transferred and the various entropies are calculated as in part (a).

e=

q(Cu ) = (2 .00 X 103 g) x (0.38S J K- 1 g- I) x (49.9 K) = 138A kJ 1 = -q(H20) .

t:.S(Cu) = mC, In

~

= 1129.2 J K-

1

I.

t:.S (total) = -119.8JK- 1 + 129.2JK- 1 =19JK- 1 1. P3.5

q

w t:.U t:.H t:.S

t:.Stol t:.G

Step 1

Step 2

Step 3

+l1 .S kJ -l1 .S kJ 0 0 +19.1 J K0 -11.S kJ

0 -3.74 kJ -3.74 kJ -6.23 kJ 0 0 ?

-S .74 kJ 0 +S .74 kJ +3.74 kJ 0 +3.74 kJ 0 +6.23 kJ -19.1 J K- 1 0 0 0 ? +11.S kJ

1

Step 4

Cycle

-S.8 kJ -S .8 kJ 0 0 0 0 0

62


Step J

!':1U

w

= !':1H = @] [isothermal].

= -nRT In (~~) = nRT In (~~)

[2. I I, and Boyle's law]

= (1.00 mol) x (8.314J K- I mol-I) x (600 K) x In (1.00 atm) = 1- 11.5 kJ I. 10.0 a t m · .

q = -w = 111.5 kJ I. !':1S

= nR In ( ~~ )

[3.15]

= -nR In e~

)

[Boyle 's law]

x (8.314J K- I mol - I) x In ( I.ooatm) = 1+19.1J K- I 1. 10.Oatm· . I -!':1S (system) [reversible process] = -19.1 JK- .

= -(1.00 mol) !':1S(sur) !':1Stol

=

= !':1S(system) +

!':1G =!':1H - T!':1S

!':1S(sur)

=0-

= @].

(600K) x (19.1 J K- I)

= l-l1.5kJ I.

Step 2 q

= @] [adiabatic] .

!':1U

= nCv,m!':1T[2.16b] = (1.00 mol)

x

G)

x (8.314J K- I mol-I) x (300 K - 600 K)

= 1-3.74 kJ I·

w = !':1U = 1-3.74kJ I.

!':1H =!':1U

+ !':1(pV) = !':1U + nR!':1T

= (-3.74kJ) +

(1.00 mol) x (8.314J K- I mol-I) x (-300K)

= 1-6.23 kJ I. !':1S = !':1S(sur) = @][reversible adiabatic process]. !':1S101

= @].

!':1G = !':1(H - TS) = !':1H - S!':1T [no change in entropy). Although the change in entropy is known to be zero, the entropy itself is not known, so !':1G is

I indeterminate I·

THE SECOND LAW

63

Step 3

These quantities may be calculated in the same manner as for Step 1 or more easily as follows. f!.U = f!.H = @] [isothermal). Srev

Tc 300K qc = I - - [3 .10] = I - - - = O.Soo = 1+ - [3 .9] . Th 600K qh

qc = -O.SOOqh = -(O.Soo ) x (1I.SkJ) = -S.74kJ. qc = l-s.74kJ I,

f!.S =

qrev T

w = -qc = IS.74kJ I· 3

[isothermal] =

I

'I

-S .74 x 10 J 300K = -19.1 JK-.

f!.S (sur) = -f!.S(system) = +19. 1 J K-'.

f!.G = f!.H - Tf!.S = 0 - (3OO K) x (-19.1 JK - ' ) = I +II.SkJ I. Step 4 f!.U and f!.H are the negative of their values in Step 2. (Initial and final temperatures reversed.) f!.U

= 1+3.74 kJ I,

f!.H

= I +6.23 kJ ~

q = @][adiabatic].

w = f!.U = I +3.74kJ I·

f!.S = f!.S(sur) = @][reversible adiabatic process] . f!.Stot = @]. Again f!.G = f!.(H - TS ) = f!.H - Sf!.T [no change in entropy] but S is not known, so f!.G is I indeterminate I. Cycle f!.U = f!.H = f!.S = f!.G = @]

[f!.(state function) =

ofor any cycle].

f!.S(sur) = 0 [all reversible processes]. f!.Stot =@]. q(cycle) = ( II .S - S.74) kJ = IS .8 kJ ~ w(cycle) = -q(cycle) = I-S.8 kJ I. P3.7

~(T)

= ~(298 K) + f!.S .

f!.S =

l

T2

TJ

dT = T

Cp ,m -

lT2(a- + TJ

T

I (I I)

c ) T2 b+ 3 dT = a In - + b(T2 - T,) - -c T T, 2

2' - 2' T2

T,

.

64


(a)

S!(373 K)

= (l92.45J K- 1mol- l ) +

(29.75J K- I mol-I) x In (373) 298

+(25.IOx 1O- 3 JK- 2 mol -

l

)

x (75.0K)

+(-21) x (1.55 x 105 JK - l mol = 1200.7 J K- I mol-I

l

)

I _(298.15)2 I)

x (

(373.15)2

I.

(b)

+ (25.10 x 10- 3 J K- 2 mol-I) x (475K) +

(~) 2

x (1.55 x 105 J K- Imol - I) x

(~2 - ~2) 773 298

= 1232.0J K- I mol- I I. P3.9

""-S

=

nCp,m In

~ + nCp,m In

In the present case, Tf

t

1

= 2(500 K +

[3 .19] [Tf is the final temperature, Tf

250 K)

g)

Sm(T)

= Sm (O) +

~ (Th + Tc) ] .

= 375 K.

_ Tl _ (Th + Tc)2 ( 500 ",,-S - nCp.m In - - - nCp,m In = 3 Th Tc 4ThTc 63.54 g cm-

P3.11

=

-I-I x (24.4 J K mol )

loT Cp,mdT - [3 .18]. o

T

From the data, draw up the following table. TIK

10

15

20

25

30

50

Cpom I(J K- 2 mol-I ) T

0.28

0.47

0.540

0.564

0.550

0.428

TI K

70

100

150

200

250

298

Cp,m 1(1 K- 2 mol-I) T

0.333

0.245

0.169

0.129

0.105

0.089

Plot CpomIT against T (Fig. 3.1). This has been done on two scales. The region 0 to 10 K has been l constructed using Cpom = aT 3 , fitted to the point at T = 10K, at which Cp,m = 2.8JK - l mol- , 4 3 so a = 2.8 x 10- J K- mol- I. The area can be determined (primitively) by counting squares. Area A = 38.28J K-Imo]-I .

THE SECO ND LAW

65

0.6 0.5

, "0

0.4

E 'I :,( 0.3 -. ~

f-.. 0.2 --.. E

(~

R

T = 31OK,

_

T

~)

X

f>

so

T'

6.JH

=2

(b)

K

=-

10

- 21

Rln

(~)

= (~ _ ~)

let- = K

6.JH f> = (SS.84 kJ mol- I)

I

T'

K

_ l InK = SS.84kJmol- lnK

X

X

(In 2) = 139 kJ mol- II

6. r H f> = (SS .84kJmol- l ) X (In!) =1-39kJmol- 1 1

2

E7.11(b) p = p(NH3 )

(a)

K

10- 5 »)1 /2

K'

T' = 32SK ;

(8.3 14JK- 1 mol - I) Now 6.JH f> = « 1/ 3IOK) _ (1/ 32SK)) K

X

10- 3

X

T

(a)

(-2.SI4

10- 2)

3 2z 2(2.444 X 10- ) 1 = -- = = 1.6 0.300 0.300

XNO

X

[Z < 0 is physically impossible] so

z = 2.444

E7.10(b)

X

10- 5 = 0

X

z= =

1O- 3 )Z2

X

so

1O- 4 )z - 2.S14

X

1O- 4 )z + (1.69

X

=

+ p (HCI)

na;i

= 2p(NH3)

PJ

a(gases) = G;

[7 .16];

p

J

K = (P(

NH3» ) pf>

X

[P(NH3) = p(HCI)]

(P(HCI») = p(NH 3) 2 = pf>

pf>2

~

X

4

At 427 °C (700 K) ,

K =

~ (~: ~:) 2 = 19.241

At4S9 °C (732K) ,

K =

4'I

(!!...-) 2

X

X

(1IISkPa) 2 ~ lookPa =~

pf>

133

134


(b) I'::..,G" = -RTlnK [7.S] = (-S.3141 K-1mol - l ) x (7ooK) x (In 9.24) =1-12.9kJmo1 - 1 1 (at 427° C) (C)I'::..H "::::; R1n (K 'j K ) [7 .25 ] , (l j T - 1j T')

::::;

E7.12(b)

(S.3141 K- 1mol- l ) x In (31.0Sj9.24) I =. +161 kJmol (11700 K) - ( 1/732 K)

I

I .

The reaction is

For the purposes of this exercise we may assume that the required temperature is that temperature at which K = 1, which corresponds to a pressure of 1 bar for the gaseous products. For K = 1, In K = 0, and I'::..,G" = O.

Therefore, the decomposition temperature (when K = I) is I'::.. H " T=-'I'::..,S & CUS04 . 5H20 (s) ;=' CUS04 (s) + 5H20 (g) I'::..,H& = [(-771.36) + (5) x (-241.S2) - (-2279.7)] kJ mo1- 1 = +299.2 kJmo1- 1 I'::..,S" = [( 109) + (5) x (1SS.S3) - (300.4)] JK- 1 mol- I = 752 .8JK- 1 mol- I

Therefore, T =

299.2 x 1Q3Jmol - 1 ~ 1 1 = ~ 752.SJK- mol-

Question. What would the decomposition temperature be for decomposition defined as the state at which K

E7.13(b)

=

I j2?

PbI2 (S);=' PbI2(aq)

Ks = 1.4

X

10- 8

I'::..,G" =-RTlnKs=-(S.314JK - lmol - l) x (29S.15K) x In (1.4 x 10-8 ) = 44.S3 kJ mol -

I

I'::..fG" (PbI2 , aq) = I'::..,G& I'::.. + I'::..fG& (PbI2, s)

= 44.S3kJmol- 1 - 173.64kJmol- 1 =1-12S .S kJmol- 1 1

CHEMICAL EQUILIBRIUM

E7.14(b)

135

The cell notation specifies the right and left electrodes. Note that for proper cancellation we must equalize the number of electrons in half-reactions being combined. For the calculation of the standard ernfs of the cells we have used E~ potentials from Table 7.2. (a)

(b)

R:

AgzCr04(S) + 2e-

L:

Clz(g) + 2e-

-+

2Ag(s) + CrO~-(aq)

L:

2Fe3+(aq) + 2e-

+1.36 V

-+

R:

MnOz(s) + 4H+(aq) + 2e-

L:

Cu2+(aq) + 2e-

-+

-+

2Ag(s) + CrO~-(aq) + (CIZg)

2Fe2+(aq)

-+

-+

Sn2+(aq) + 2Fe3+(aq)

Mn2+(aq) + 2Fe3+(aq)

-0.62 V +1.23 V

Cu(s)

+0.34 V

Cu(s) + MnOz(s) + 4H+(aq)

-+ Cu 2+(aq)

+2HzO(l) COMMENT.

-0.91 V +0.15 V +0.77 V

Sn4+(aq) + 2Fe2+(aq)

Overall (R - L) :

+0.45 V

2Cl-(aq)

Overall (R - L): AgzCr04(S) + 2Cl-(aq) R : Sn 4+(aq) + 2e- -+ Sn2+(aq) Overall (R - L) :

(c)

-+

= t; - Et, with standard electrode

+ Mn2+(aq) +0.89V

Those cells for which E ~ > 0 may operate as spontaneous galvanic cells under standard

conditions. Those for which E ~ < 0 may operate as nonspontaneous electrolytic cells. Recall that E~ informs us of the spontaneity of a cell under standard conditions only. For other conditions we require E.

E7.1S(b)

The conditions (concentrations, etc.) under which these reactions occur are not given. For the purposes of this exercise we assume standard conditions. The specification of the right and left electrodes is determined by the direction of the reaction as written. As always, in combining half-reactions to form an overall cell reaction we must write half-reactions with equal number of electrons to ensure proper cancellation. We first identify the half-reactions, and then set up the corresponding cell. (a) R:

2HzO(I) + 2e-

L:

2Na+ (aq) + 2e-

20H-(aq) + Hz(g)

-+ -+

2Na(s)

- 0.83 V -2.71 V

and the cell is

I +1.88V I or more simply

I Na(s)INaOH(aq)IHz(g)IPt I (b) R: L:

Iz(s) + 2e-

-+

2H+ (aq) + 2e-

21-(aq) -+

Hz(g)

+0.54 V

o

and the cell is

I+0.54 V I or more simply

I PtIHz(g)IHI(aq)llz(s)1 Pt I

136


(c) R:

L :

2H+(aq) + 2e- ~ H2(g)

O.OOV

2H20(I ) +2e-~H2(g)+20H-(aq)

0.083 V

and the cell is 10.083 V

I

or more simply

IPtIH2(g) IH20(l)IH2(g) IPt I All of these cells have Ee. > 0, corresponding to a spontaneous cell reaction under standard

COMMENT.

conditions. If Ee. had turned out to be negative, the spontaneous reaction would have been the reverse of the one given, with the right and left electrodes of the cell also reversed.

E7.16(b)

(a)

E=E

Q

=

e.

RT vF

--lnQ

n

VJ

aJ

v=2

2 2 = aH+a Cl -

[all other activities

=

1]

J

= a~a:' = (y+b+)2

x (y_b_)2

[b

==

: e. here and below]

RT ( ) 2RT Hence, E = E e. - 2F In y1 b4 = E e. - Fin (y±b)

=

(b)

6. r G

(c)

logy±

-vFE [7.27]

=

= -(2)

x (9.6485 x 104 C mol-I) x (0.4658 V)

-I Z+Z_WI / 2[5.69]

=-

(0.509) x (0.010)1 /2 [I

y±

= 0.889

E e.

MT (y±b) = (0.4658 V) + = E + Fin

The value compares favorably to that given in Table 7.2. vFE e.

In each case In K (a)

="RT""

[7.30]

Sn(s) + CuS04(aq) ;=: Cu(s) + SnS04(aq)

L:

Cu2+ (aq) + 2e- ~ Cu(s) + 0.34 V } Sn2+ (aq) + 2e- ~ Sn(s) _ 0. 14 V + 0.48 V

In K

=

R:

(2) x (0.48 V) 25.693 mV

= +37.4,

K

= b for HCI(aq)] =

-0.0509

(2) x ( 25 .693 x ) 10- 3 V x In (0.889 x 0.010)

= I +0.223 V I

E7.17(b)

= 1-89.89 kJ mol-II

= 11.7 x

10

16

1


R:

Cu2+ (aq) + e- -+ Cu(aq) + 0.16 V } _ 0.36 V Cu+ (aq) + e- -+ Cu(s) + 0.52 V

L: InK E7.18(b)

-0.36V

2Bi3+(aq) + 6e-

L:

Bi2S3(S) + 6e-

Overall (R - L):

K (a)

-

= 25.693 mV = -14.0,

R:

In K

137

-+

-+

2Bi(s)

2Bi(s) + 3S 2-(aq)

2Bi3+(aq) + 3S 2-(aq)

vFE

~

0.268

o

0.02

0.04 0.06 (b/ b'7 )1/2

0.08

0.10

Figure 7.2

For the activity coefficients we obtain from equation (a)

r-E

b

In y± = 2RT I F -In be =

0.26843-E / V b 0.05139 -In be

and we draw up the following table.

3.0769

1.6077 -0.3465 0.9659

In y± y± P7.19

5.0403

-0.05038 -0.6542 0.9509 0.9367

7.6938

10.9474

-0.07993 -0.09500 0.9232 0.9094

The cells described in the problem are back-to-back pairs of cells each of the type Ag (s) IAgX (s) IMX (bl) IMxHg (s) . R:

M+ (bl)

+ e- ~ M xHg (s)

L:

AgX (s)

+ e-

R-L:

Ag (s)

---+ Ag (s)

F

+ X- (bJ)

+ M+ (bl ) + X-

RT E=r- -lnQ.

(Reduction ofM+ and formation of amalgam)

Hg

(b)) --+ MxHg (s)

+ AgX (s) ,

v

=

1.

CHEMICAL EQUILIBR IUM

145

For a pair of such cells back to back, Ag (5) IAgX (5) IMX (bl) IMxHg (s) IMX (b2) IAgX (5) lAg (5) ,

ER

=r

-

Rt FIn QR,

-RT

E = -

F

QL QR

- In -

EL =

RT

= -

F

In

r -

RT

Fin QL,

(a (M+) a (X-))L (a (M+) a (X- ))R

-7---7-----i----+------;~

(Note that the unknown quantity a (MxHg) drops out of the expression for E.)

With L = (1) and R = (2) we have

2RT bl 2RT Y±(1) E = - In - + - I n - - . F b2 F y±(2)

Take b2 = 0.09141 mol kg-I (the reference value), and write b =

:~ .

2RT y±- ) . E= -( Ib n - -- +ln - F 0.09141 y± (ref) For b = 0.09141, the extended Debye-Hlickel law gives

logy±(ref) =

(-1.461)

(0.09141) 1/ 2 1/2 + (0.20) x (0.09141) = -0.2735, (I) + (1.70) x (0.09141) X

Y±(ref) = 0.5328. y± ) b Then E = (0.05139 V) x ( In 0.09141 + In 0.5328 '

E b In y± = 0.05139 V - In (0.09141) x (0.05328)

We then draw up the following table.

bl

(mol/kg- I)

EIV y

0.09141

0.1652

0.2171

1.040

1.350

-0.0220

0.0000

0.0263

0.0379

0.1156

0.1336

0.572

10.5331

0.492

0.469

0.444

0.486

0.0555

A more precise procedure is described in the original references for the temperature dependence of (Ag, AgCl, CI-) ; see Problem 7.20.

r

146

P7.21


(a) From

(aG) = V [3.50] ,

ap

we obtain

T

aflrG) ( -ap- T =

flr V.

Substituting flrG = -vFE [7.27] yields

(b) The plot (Fig. 7.3) of E against p appears to fit a straight line very closely. A linear regression analysis yields

x-I-0---3-m-V-a-tm-------'1~ Slope ' I=-2-.4-8-0-

standard deviation = 3 x 10- 6 mV attn-I .

Intercept= 8.5583 mY,

standard deviation = 2.8 x 10- 3 mY.

R = 0.99999701 (an extremely good fit).

From

flrV ( -2.666 x 10- 6 m 3 mol-I) 1 x 9.6485

Since J = VC = Pam 3 ,

X

Pam 3 C=--

a?

T ,n -

(2.666 x 10- ) V 9.6485 X 104 Pa

Pa 5

6

-

V

or

V

Therefore

( aE)

104 C mol-I .

X

1.01325 x 10 Pa = 2.80 x 10- 6 Vatm- I atm = 12.80 x 10- 3 mV atm- I I.

This compares closely to the result from the potential measurements.

13

12

II

>

8 ~ 10

9

8

o

500

1000

p/atm

1500

Figure 7.3


147

(c) A fit to a second-order polynomial of the form E

= a + bp + cp2

yields a

= 8.5592 mY,

standard deviation

b = 2.835 x 10- 3 mVatm-

c = 3.02 x 10- 9 m V atm -2, R

= 0.999 997

= 0.0039 mV

standard deviation = 0.012 x 10- 3 mVatm-

l,

1

standard deviation = 7.89 x 10- 9 m V atm- I

II.

This regression coefficient is only marginally better than that for the linear fit, but the uncertainty in the quadratic term is > 200 per cent. = b + 2cp. (aE) ap T

The slope changes from to

(aE) ap

= max

(aE) ap

= min

b= 2.835 x 10- 3 mVatm- 1

b+ 2c(l 500 atm) = 2.836 x 10- 3 mVatm-

We conclude that the linear fit and constancy of

l.

(~:) are very good.

(d) We can obtain an order of magnitude value for the isothermal compressibility from the value of c. 2

a E ap2

= _~ vF

(aLlrV) ap T

= 2c. 2vcF V

2(1 ) x (3 .02 x 10- 12 Vatm- 2) x (9.6485 x 104 Cmol(KT )cell

I)

x (

82.058 cm3 atm) 8.3145J

= -----------------(--:--,----:-----:-----:-)-----'------'--(lcm3 j O.996g) x

= 13.2

18.016g I mol

x 10- 7 atm- I 1 standard deviation "'" 200 per cent

where we have assumed the density of the cell to be approximately that of water at 30°C. COMMENT. It is evident from these calculations that the effect of pressure on the potentials of cells involving

only liquids and solids is not important ; for this reaction the change is only::>:: 3 x 10- 6 V atm - 1 . The effective isothermal compressibility of the cell is of the order of magnitude typical of solids rather than liquids; other than that, little significance can be attached to the calculated numerical value.

P7.23

We need to obtain Ll rH f7 for the reaction ~H2 (g)

+ Uup+ (aq)

---+ Uup (s)

+ H+ (aq) .

We draw up the thermodynamic cycle shown in Fig . 7.4.

148


Ej(H)

13.6eV H(g) + Uup+(g)

-11.3eV t. hydH B- (H+) H+(aq ) + Uup+(g)

1 x 4.5eV

A

1H2 + Uup+(g)

-5.52eV

3.22eV

H+(aq )+ Uup(g)

1H2 + Uup+(aq)

-1.5eV

x

W(aq) + Uup(s)

B

Figure 7.4

Data are obtained from Tables 10.3, lOA, IIA, 2.7, and 2.7b. The conversion factor between eV and kJmol- 1 is 1 eV

= 96A85 kJ mol-I

The distance from A to 8 in the cycle is given by 6. r W

=x =

(3.22 eV) +

G)

x (4.5 eV) + (13.6 eV) - (11.3 eV) - (5.52eV) - (1.5 eV)

= 0.75eV. 6. r s""

= s"" (Uup, s) + s"" (H+ , aq) = (0.69) + (0) -

6. r

cr = 6. W r

G)

T6. r s""

=

-

!s"" (H2, g)

-

s"" (Uup+ , aq)

x (1.354) - (1.34) meV K- 1 = -1.33 meV K- 1 •

(0.75eV) + (298.15K) x (1.33 meVK- 1)

which corresponds to I + III kJ mol-I The electrode potential is therefore

= +1.l5 eV

I.

-~~cr , with v = 1, or 1-1.15 V I.

Solutions to theoretical problems P7.25

We draw up the following table using the stoichiometry A + 38

Mole fraction

-~

I -

~

I - ~ 2(2 -~)

=

vJ~.

C

Total

3

0

4

-3~

+2~

Initial amount Imol Equilibrium amount Imol

2C and 6. nJ

8

A

Change,6.nJ / mol

~

3(1

-~)

3(1 2(2 -

2~

~)

~

~)

2-~

2(2 -

~)


149

Since K is independent of the pressure 27

a 2 = - K , a constant. 16

Therefore (2 -

~)~ = a ( ; )

which solves to

~

=I-

x (1-

)1/2

I (

~)2,

1+ apjpB

We choose the root with the negative sign because shown in Fig. 7.5.

0.1

100

\0

~

lies between 0 and 1. The variation of ~ with p is

\000

ap/p-&-

Figure 7.S

P7.27

log y± = _AII /2 = _AC I / 2 y± =e -

2.303AC I / 2

Ks = S'(S'

+ C)

We solve S'

2

x

2 y±

= e -4606AC .

e-4.606ACI / 2

+ S' C -

Ks

2

y±

In y± = -2.303AC I / 2

=0

I /2

150


to get S

,

1

=-

(

2

2 4Ks C + -

. Therefore, since Y:f?

Yf

1/ 2 )

I Ks - - C "'" - 2 CYf

I/ 2 = e - 4606AC .

K e - 4.606AC I/2 S,""' _ 5_ _ __

C


6.C

= 6.c? + RTlnQ

[7 .11] .

In equation 7.11 molar solution concentrations are used with 1 M standard states. The standard state (G) pH equals zero in contrast to the biological standard state (EEl) of pH 7. For the ATP hydrolysis

we can calculate the standard state free energy given the biological standard free energy of - 31 kJ mol - I (impact 17.2).

= 6. c? +

6.C EB 6.G"

=

RTln(l0-7 M/ IM)

6.C EB - RT In(10- 7M/ I M)

=

-31 k1 mol - I - (8.3141 mol-I K- I)(310 K) In(10- 7)

= -31 kJmol- 1 + 41.5 kJ mol - I = +11 kJ mol- I.

This calculation shows that under standard conditions the hydrolysis of ATP is not spontaneous! It is endergonic . The calculation of the ATP hydrolysis free energy with the cell conditions pH [Pi] = 1.0 x 10 -6 M, is interesting.

= 7, [ATP] = [ADP] =

6.C = 6. G" + RT In Q = 6. G" + RT In{[ADP][P;- ][H+]/ [ATP](1 M)2 }

=

+11 k1 mol - I + (8.314 J mol - IK- I)(310 K) In(l0-6 x 10- 7)

=

+11 k1 mol - I -77 k1 mol-I

= -66 k1 mol - I. The concentration conditions in biological cells make the hydrolysis of ATP spontaneous and very exergonic. A maximum of 66 kJ of work is available to drive coupled chemical reactions when a mole of ATP is hydrolyzed. P7.31

Yes, a bacterium can evolve to utilize the ethanol/nitrate pair to exergonically release the free energy needed for ATP synthesis. The ethanol reductant may yield any of the following products.

ethanol

ethanal

elhano ic acid

The nitrate oxidant may receive electrons to yield any of the following products. NO) --+ NO; --+ nitrate

nitrite

N2 dinitrogen

--+

NH3 . ammonia


151

Oxidation of two ethanol molecules to carbon dioxide and water can transfer 8 electrons to nitrate during the formation of ammonia. The half-reactions and net reaction are: 2[CH) CH20H(I) -+ 2C02(g) + H20(l) NO)"(aq) + 9H+ (aq) + 8e- -+ NH) (aq)

+ 4H+(aq) + 4e- l + 3H20(1 )

6.r ~ = -2331 .29 kJ for the reaction as written (a Table 2.5 and 2.7 calculation). Of course, enzymes must evolve that couple thi s exergonic redox reaction to the production of ATP, which would then be available for carbohydrate, protein, lipid, and nucleic acid synthesis.

P7.33

The half-reactions involved are: R:

cytox

+e

-+

cytred

L:

Dox

+e

-+

Dred

t:;yt ~

The overall cell reaction is :

R - L = cytox

+ Dred ;=: cytred + Dox E" = t:;yt -

~

(a) The Nernst equation for the cell reaction is

At equilibrium, E

= 0; therefore

leq ) against In ([cyt] [Dox --[ ] ox ) is linear with a slope of one and an intercept Therefore a plot of In ( cyt red [Dredleq of

:T (~t - ~) .

(b) Draw up the following table. I ([Dox leq ) n [Dred leq ( [cytox leq ) In [cytredleq

- 5.882

-4.776

-3 .661

-3 .002

- 2.593

-1.436

-0.6274

-4.547

- 3.772

-2.415

- 1.625

- 1.094

-0.2120

-0.3293

152


[Doxleq ) . ( [cytox]eq ) . Th e plot ofln ( -[D] agamst In [ IS shown in Fig. 7.6. The intercept is -1 .2124. Hence red eq cyt,ed]eq

~t =

RT

F

X

(-1.2 124) + 0.237 V

=0.0257V

X

(-1.2124) +0.237 V

= 1 + 0.206 V I·

0 -\

g-

-2

1

-

B -3 ~ B -4 0

:5

-5 -6 -5

-4

-3

-2

- I

o Figure 7.6

P7.35

A reaction proceeds spontaneously if its reaction Gibbs function is negative. 6. r G

= 6. r (7 + RT In Q

Note that under the given conditions, RT = 1.58 kJ mol-I. (i)

6. r G/(kJ mol -I) = 6. r (7( I) - RTlnp H 2 0 = -23.6 - 1.58 In 1.3 x 10- 7 = +1.5.

(ii)

6. r G(kJ mol -I) = 6. r (7(2) - RTln(pH2oP HN0 3)

= -57 .2 - 1.58ln [(1 .3 x 10- 7) x (4. 1 x lO- IO (iii)

6. r G/(kJ mol - I) = 6. r (7(3) - RT In(p~20 PHN03)

= -85 .6 - 1.581n[(1.3 x 10- 7 )2 (iv)

)J = +2.0.

X

(4.1

10-

10

X

(4. 1 X 10-

10

X

)]

= -1.3 .

)]

= -3.5 .

6. r G/(kJ mol-I ) = 6. r (7(4) - RT In(p~2 0PHN03)

= -85.6 - 1.58In[(1.3 x 10- 7 )3


153

So both the dihydrate and trihydrate form spontaneously from the vapor. Does one convert spontaneously into the other? Consider the reaction

which may be considered as reaction (iv) - reaction (iii). Therefore t;.rG for this reaction is

We conclude that the dihydrate converts spontaneously to the 1trihydrate I, the most stable solid (at least of the four we considered).

PART 2 Structure

8

Quantum theory: introduction and principles


At the end of the nineteenth century and the beginning of the twentieth, there were many experimental results on the properties of matter and radiation that could not be explained on the basis of established physical principles and theories. Here we list only some of the most significant. (1) (2) (3) (4) (5)

The energy density distribution of blackbody radiation as a function of wavelength. The heat capacities of monatomic solids such as copper metal. The absorption and emission spectra of atoms and molecules, especially the line spectra of atoms. The frequency dependence of the kinetic energy of emitted electrons in the photoelectric effect. The diffraction of electrons by crystals in a manner similar to that observed for X-rays.

08.3

The heat capacities of monatomic solids are primarily a result of the energy acquired by vibrations ofthe atoms about their equilibrium positions. If this energy can be acquired continuously, we expect that the eq uipartition of energy principle shou ld apply. This principle states that, for each direction of motion and for each kind of energy (potential and kinetic), the associated energy should be ~ kT . Hence, for three directions and both kinds of motion , a total of 3 kT , which gives a heat capacity of 3 k per atom, or 3 R per mole, independent of temperature. But the experiments show a temperature dependence. The heat capacity falls steeply below 3 R at low temperatures. Einstein showed that, by allowing the energy of the atomic oscillators to be quantized accordi ng to Planck 's formula , rather than continuous, thi s temperature dependence cou ld be explained. The physical reason is that at low temperatures only a few atomic oscillators have enough energy to populate the higher quantized levels; at higher temperatures more of them can acquire the energy to become active.

08.5

If the wavefunction describing the linear momentum of a particle is precisely known, the particle has a definite state of linear momentum, but then, according to the uncertainty principle, the position of the particle is completely unknown as demonstrated in the derivation leading to eqn 8.21. Conversely, if the position of a particle is precisely known, its linear momentum cannot be described by a single wavefunction, but rather by a superposition of many wavefunctions, each corresponding to a different value for the linear momentum. Thus all knowledge of the linear momentum of the particle is lost. In the limit of an infinite number of superposed wavefunctions, the wavepacket illustrated in Fig. 8.31 turns into the sharply spiked packet shown in Fig. 8.30. But the requirement of the superposition of an infinite number of momentum wavefunctions in order to locate the particle means a complete lack of knowledge of the momentum.

158



The de Broglie relation is h h A= - = p

mv

v = 11.3 E8.2(b)

X

so

v= -

h

=

mA

6.626 X 10- 34 J s (1.675 x 10- 27 kg) x (3.0 X 10- 2 m)

-:-:--:-:=-::-----,;-;;----------,,-_

10- 5 m s-I 1 extremely slow!

The moment of a photon is 34 6.626 x 1O- J s 1 = 1.89 9 350 x 10- m

h A

p = - =

X

10- 27 kgms- I

I

The momentum of a particle is p

= mv

so

p 1.89 X 10- 27 kg m s-I v - - - -----------:---=----------:- m - 2(1.0078 x 10- 3 kg mol-I 16.022 x 1023 mol-I)

v = I 0.565 m S-I E8.3(b)

I

The uncertainty principle is

so the minimum uncertainty in position is

n

n

l.0546 x 10- 34 J s

Llx = - - = - - = -::-c:::--::----:-,:-:;-;-:--:---,:-::-:-::--,,....,-::-,------::-::--=-----::-;;-----cc:2Llp 2mLlv 2(9.11 x 10- 31 kg) x (0.000010) x (995 x 10 3 ms- I)

=15.8 xlO- 6m l E8.4(b)

he E=hv =_.

A'

he = (6.62608 x 10- 34 J s) x (2.99792 x 108 m s-I) = 1.986

X

10- 25 J m

NAhe = (6.02214 x 1023 mol-I) x (1.986 x 10- 25 J m) = 0.1196J m mol - I

Thus, E =

1.986 x 1O- 25 1m A

E(per mole) =

0.11961mmo[-1 A

We can therefore draw up the following table A

(a) 200 nm (b) 150 pm (c) 1.00 cm

Ell

E/(kJ mol-I)

0.93 x 10- 19 1.32 x 10- 15 1.99 x 10- 23

598 7.98 x 105 0.012

QUANTUM THEORY: INTRODUCTION AND PRINCIPLES

E8.5(b)

159

Assuming that the 4He atom is free and stationary, if a photon is absorbed, the atom acquires its momentum p achievi ng a speed v such that p = my. m = 4.00 x 1.6605 x 10- 27 kg = 6.642 x 10- 27 kg

f

v =

m h

P="i (a)

6.626 x 10- 34 J s 27 1 200 x 10- 9 = 3.313 x 1O- kgmsm

p=

27

v=!!...

1

= 3.313 x 1O- kgms- =10.499m s- 1 1 m 6.642 x 10- 27 kg

(b)

6.626 x 10- 34 J s 12 =4.417 x lO- 24 kgm s150 x 10- m

p=

27

v= f

= 4.417 x 1O- kg msm 6.642 x 10- 27 kg

(c)

=1665 ms - II

6.626 x 10- 34 J s = 6.626 x 10- 32 kgm s-I 1.00 x 10- 2 m

p =

6.626 x 10- 32 kg m s-I 1 -6 _I I = 9.98 x 10 m s 6.642 x 10- 27 kg

p

v= - = m E8.6(b)

1

1

Each emitted photon increases the momentum of the rocket by h/ A. The final momentum of the rocket will be Nh/ A, where N is the number of photons emitted, so the final speed will be Nh/Amrocket . The rate of photon emission is the power (rate of energy emission) divided by the energy per photon (he/A), so

N=

IPA he

and

h tP v = ( -IPA) x ( - -) = - he Amrocket emrocket

( 1O.0yr) x (365dayyr- l ) x (24 hday-l) x (3600sh- l ) x (1.50 x 1O- 3 W) (2.998 x IO s m s- l ) x ( 10.0 kg)

v

= 1158 m s - I I E8.7(b)

E8.8(b)

Rate of photon emission is rate of energy emission (power) divided by energy per photon (he/A) (0. IOW) x(700xlO- 9 m) (6.626 x 10- 34 J s) x (2.998 x lOS m s-I )

= I 352 .

(a)

rate =

-PA = he

(b)

rate =

( 1.0 W) x (700 x 10- 9 J s) I = 3.52 x lOIS s -II (6.626 x 10 34 J s) x (2.998 x lOS m S- I) . .

Conservation of energy requires Epholon

and

EK

=

= ~meV2

+ EK so

= hl! = he/A

so

2E ) 1/ 2

v

= ( m~

EK

= he/A -

x 10

17

s-

II

160


(a)

EK =

(6.626 x 1O- 34 Js) x (2.998 x 108 ms- 1) - (2.0geY) x (1.60 x 10- 19 J ey- 1) 650 x 10 9 m

But this expression is negative, which is unphysical. There is I no kinetic energy or velocity Ibecause the photon does not have enough energy to dislodge the electron. (6.626 x 10- 34 J s) x (2.998 x 10 8 m s-I) (b) EK = - (2.0geY) x (1.60 x 10- 19 J eV- 1) 195 x 10- 9 m =16.84 x 10- 19 J I 2(6.84 x 1O-19J») 1/ 2

and v = ( 9.11 x 10- 31 kg ES.9(b)

ES.10(b)

E

= 11.23 x 106 m s - I I

= hI! = h/ T, so

(a)

E = 6.626 x 10- 34 J s/ 2.50 x 10- 15 s = 12.65 x 10- 19 J = 160 kJ mol -1 I

(b)

E = 6.626 x 10- 34 J s/2.21 x 10- 15 s = 13.00 x 10- 19 J = 181 kJ mol- 1

(c)

E = 6.626 x 1O- 34 J s/ 1.0 x 10- 3 s = 16.62

X

I

10- 31 J = 4.0 x 10- 10 kJ mol- 1 I

The de Broglie wavelength is h

A= -

P

The momentum is related to the kinetic energy by p2 EK = 2m

so p = (2mEK)1 /2

The kinetic energy of an electron accelerated through 1 Y is leY = 1.60 x 10- 19 J, so h A = """(2=-m"""E=-K"",,)71/M2

(a)

A=

(b)

A

=

6.626 x 10- 34 J s If? »-

(2(9.11 x 10- 31 kg) x (100eY) x (1.60xlO- 19 Jey- I

6.626 x 10- 34 J S

----------------------:-~

(2(9.11 x 10- 31 kg) x (1.0x10 3 eY) x (1.60 x 1O- 19 Jey- 1»1 /2

= 13.9 x 10- 11 m I (c)

6.626 x 10- 34 J s

A=-------------------------,'"

(2(9.11 x 10- 31 kg) x (100 x 103 eY) x (1.60 x 10- 19 J ey-l»1 /2

= 13.88 x 10- 12 m I


ES.11 (b)

161

The upper sign in the following equations represents the math using the A+ ill operator. The lower sign is for the A - iB operator. r is a generalized coordinate.

f

± iB I1/1jdr

1/1/IA

f 1/1tIAI1/1

=

j dr

±

= {f 1/1n A11/1;dr

if 1/1tIBI1/1

j

r

±i

dr

{f 1/I/IBI1/I;dr

r

{f 1/I/IA 1/I;dr =f if 1/I/IBI1/I;dr = {f 1/I *IA =f iBl1/I;dr =

1

r

j

This shows that the

A±

r

A andB are hermitian [8.30]

iB operators are not hermitian. Lf they were hermitian, the result would be

{J 1/Ij*I A ± iBl1/I;dr j*. ES.12(b)

The minimum uncertainty in position is 1100 pm I. Therefore, since f1xf1p 34

Ii

f1p> - - 2f1x f1 v

ES.13(b)

=

=

f1p m

1.0546 x 10- 1 s 2(100 x 10- 12 m)

= 5.3 x

5.3 x 10- 25 kg m S-l 9.11 x 10- 31 kg

=

10- 25 k

gms

I5.8 x 10 .

5

~ ~Ii

-I

m s -I

I

Conservation of energy requires Ephoton

= Ebinding + ~mev2 = hv = hC/ A (6.626

x 10- 34 Js) x

(2.998 121 x 10- 12 m

and Ebinding =

so

Ebinding

= hC/ A -

~mev2

x 108 m s - I)

- ~(9.11 x 10- 31 kg) x (5.69 x 10 7 ms- I ) 2

=1l.67

xlO-

16

11

COMMENT. This calculation uses the non-relativistic kinetic energy, which is only about 3 percent less than

the accurate (relativistic) value of 1.52 x 10-

15

J . In this exercise, however. Ebinding is a small difference of

two larger numbers. so a small error in the kinetic energy results in a larger error in value is E binding = 1 .26 x 10 - 16 J. ES.14(b)

Ebinding:

the accurate

The quality QIQ2 - Q2Q I [Illustration 8.3] is referred to as the commutator of the operators

QI

and

Q2. In obtaining the commutator it is necessary to realize that the operators operate on functions ; thus, we form

QIQ2!(X) - Q2 Q U(X) Px

Ii d dx

=i

Therefore a

=

(x + Ii!)

and a t

=

(x - Ii ! )

162


Thenaatf(x)=~(x+n!) andataf(x)

= ~ (x-n!)

x (x-n!)f(x)

x (x+ n! )f(x)

The terms in x 2 and (d/dx)2 obviously drop out when the difference is taken and are ignored in what follows ; thus

"

I (- xn-dxd+ n-x d) f(x) dx d) f(x) at af(x) = -I (d xn-x - n-x dx aa 'f(x) = 2

2

d

These expressions are the negative of each other, therefore •

t

(aa ' - a a)f(x)

d . = n-xf(x)

dx

. d

- ru:-f(x) dx

( d.. d)

= njof (x)

= n dx x - x dx f(x) Therefore, (aa t - a t a)

= [KJ

Solutions to problems Solutions to numerical problems PS.1

A cavity approximates an ideal black body ; hence the Planck distribution applies,

8'Jlhe ( P =):5

I

ehc/ AkT _

I

)

[8.5].

Since the wavelength range is small (5 nm ) we may write as a good approximation I'1E = pI'1A,

A ~ 652.5 nm .

(6.626 X 10- 34 J s) x (2.998 x 10 8 m s-I) (6. 525 X 10- 7 m) x (1.38 1 x 10- 23 J K- )

he Ak

-'-------=---'-----'------=-:---.-1 =

2.

20

0 K 5 x I 4 .

8'Jlhe _ (8'Jl) x (6.626 X 10- 34 J s) x (2.998 x 108 m s-I) _ 22 07 J -4 -- _ - 4. I x 1 m . AS (652.5 x 10- 9 m)S I'1E

=

(4.221 x 10 7 J m- 4 ) x (

~ K) / T -

e(2.205x 1 3

I

1

) x (5

X

10- 9 m) .

I

0.211 JmQ4 = 1.6 x 10- 33 J m- 3 . e(2.205x I )/298 - I 3 0.211 J m(b) T = 3273 K, ll.E = Q4 = 2.5 X 10- 4 J m- 3 . e(2.205 x 1 )/3273 - I (a) T = 298 K , ll.E =

I

I


163

COM M E NT. The energy density in the cavity does not depend on the volume of the cavity, but the total energy

in any given wavelength range does, as well as the total energy over all wavelength ranges.

Question. What is the total energy in this cavity within the range 650--655 nm at the stated temperatures? PS.3

hI!

Js

X S- I

er: = T ' [Ih]= JK - I =K. In terms of er: the Einstein equation [8.7] for the heat capacity of solids is Cv = 3R ( ; ) 2 x

(e~7:2~ I) 2,

classical value = 3R.

hI!

IT « 1 as demonstrated in the text (Section 8.1). The criterion for classical behavior is therefore that 1T » er: I.

It reverts to the classical value when T

»

8E or when

hI! (6.626x I0- 34 JHC I) xv _II er:=T= 1.38I xlO- 23 JK-1 =4.798xlO (v/Hz)K.

(a) For v = 4.65 x 10 13 Hz,

er: =

(4.798 x 10- 11 ) x (4.65 x 10 13 K) =12231 KI.

(b) For v = 7.15 x 10 12 Hz,

er: =

(4.798 x 10- 11 ) x (7.15 x 1012K) =1343 KI.

Hence Cv (223IK)2 (e223T/(2X298l )2 (a) 3R = 298K x e223 I/298 _ I = 10.031

I·

Cv _ (343 K)2 (e 223T / (2X298 ) 2 _ (b) 3R x -1 0 .89 7 1. 298K e 343/298 - 1 COMMENT.

For many metals the classical value is approached at room temperature; consequently, the fail-

ure of classical theory became apparent only after methods for achieving temperatures well below 25°C were developed in the latter part of the nineteenth century.

PS.5

The hydrogen atom wavefunctions are obtained from the solution of the Schrodinger equation in Chapter 10. Here we need only the wavefunction that is provided. It is the square of the wavefunction that is related to the probability (Section 8.4).

4

3

8r = 3rrTo'

TO

= 1.0 pm.

If we assume that the volume 8r is so small that l/f does not vary within it, the probability is given by

164

STUDENT'S SOLUTION S MANUAL

l/I 2or =:34

(a) r = 0 :

l/I 2 or

(b) r = C/o:

=

(1.0)3 53 = 19.0 x 10-

6

1.

~3 (~)3 e- 2= 11.2 x 10-61. 53

Question. If there is a nonzero probability that the electron can be found at r = 0 how does it avoid destruction at the nucleus? (Hint . See Chapter 10 for part of the solution to this difficult question. ) PB.7

According to the uncertainty principle,

where !:"q and !:"p are root-mean-square deviations:

To verify whether the relationship holds for the particle in a state whose wavefunction is

we need the quantum-mechanical averages (x) , (x 2 ) , (p ), and

(x )

2a)

= (-;

1/ 2

1-0000

xe-

2m .2

dx

= 0;

/4e -ax2dx -_ 2) 1 00 (2a )1 /4e _ax2 x 2()1 -2a (X = -00 -IT IT

so!:,.q

(p )

I = ~ /2'

2a

=

1-0000

l/I '

(~dl/l) dx 1

dx

and

We need to evaluate the derivatives:

dl/l dx

= (2a) -; 1/4(-2ax)e- ax 2

(P2).

(2a)1 /21 00 xe2 -

IT

-oo

_ 2ax

2

dx ,

QUANTUM THEORY : INTRODUCTION AND PRINCIPLES

165

and

OO

So

(P)

J

=

- 00

(2a)I /4 -;

e-

ax2

(It) T

(2a)I /4

-;

(-2ax)e-

aX2

dx

2It (2a) 1/ 2 J OO xe -2{u 2dx -_ O., -_ - -;I rr -00

(p2)

=

2a) 1/2 (

(-2alt2) ( rr

2a

rr

l

/

2(2a)

2

rr

l

2

)

~ /2 = alt2; (2a)

3/ 2 -

/

and

Finally,

=

6.q 6.p

I 1/2 ------""I /2 x a

2a

It =

I -Ii.,

2

which is the minimum product consistent with the uncertainty principle.

Solutions to theoretical problems PS.9

p

Srrhe (

= ---;:s

1 e hc / AkT

I

_

)

[S.5]

As ).. increases, he j AkT decreases, and at very long wavelength he / AkT exponential in a power series. Let x = hej AkT, then e

I 2 I 3 = I +x + -x + -x + ...

x

2!

3!

Srrhe [

P=---;:S .

I

I

+x + -

2!

Srrhe [

A~mOO P = ---;:s

I

x2

+x -

]

I

+ - 1x 3 + ... 3

I

I

'

]

I

Srrhc (

= ---;:s

SrrkT )..4

Thi s is the Rayleigh-Jeans law [S.3].

'

I

I ) hej AkT .

«

1. Hence we can expand the

166


[ 00

PS.11

£

Let x

£

= 10

[ 00

P (A) dA

= Srehe 10

~

~

~IT

AkT

A kT

he

=

SrekT

[ 00

10

A2dx A5 (eX- I)

he

PS.13

(J

=

We require

10

[ 2n

0

SrekT

3dx-x (eX - 1)

[ 00 -::-_dx __

10

A3 (eX -l)

3 (re = 8rekT (kT) -

4 )

15'

he

J 1{!*1{! dr = 1, and so write 1{! = Nf and find N for the givenf·

00 r2

(a4

!

3

x (2) X (2re) = 1 if N =

)

2

= 10 [ 2n

by symmetry with

=-

[ 2n

cos I/> dl/>

10

(rea)

1/ 2

fansin3 BdB fa2n cos2 1/>dl/>

x r 2 e- r / a dr

We have used J sin 3 B dB

10

1

°O

=

I (2 re 5k4 / ISh3 e2 ) Iis the Stefan-Boltzmann constant.

= N2 (d) N 2

1)' [S.S].

= - . Then,dx = --2-dA ordA = - - - d x.

= 8rekT ( -kT) 3

where

dA A5 (ehc/ AkT _

[x

= r cos I/> sin B]

~ (cos B)(sin 2 B + 2) , as found in tables of integrals and

2 sin I/> dl/>

2 (cos I/>

+ sin 2 1/»

[ 2n

dl/> =

10

dl/> = 2re.


PB.1S

In each case form Qf. If the result is wf where w is a constant, then f is an eigenfunction of the operator Q and w is the eigenvalue [8. 2Sb]. dk dx e' ;f

(b)

d.x cos kx

d

!

k

Iyes; eigenvalue =ik.I

= ik e''kx ;

(a)

(c)

PB.17

.

= -k SIO kx;

no.

= 0; I yes; eigenvalue = 0 I·

(d)

d d.x kx

1

(e)

d ax2 d.x e = -

= k = :; kx;

no [) /x is not a constant].

2ax e -

a 2 .1

;

no [ - 2ax is not a constant].

Follow the procedure of Problem 8.15 . (a)

; 2e ikx = _k2e ikx ; yes; eigenvalue = l-k2

(b)

d.x 2 cos kx = _k 2 cos kx; yes; eigenvalue = ~.

(c)

d.x 2 k

1·

~

d2

d2

= 0;

yes; eigenvalue

= [2].

2

(d)

d d.x 2 kx

(e)

~ 2 d.x2 e- ax

= 0;

yes; eigenvalue

= [2]. 2

= (-2a + 4a 2x2 )e- ax

;

no.

d2 d2 d Hence, (a, b, c, d) are eigenfunctions of - 2 ; (b, d) are eigenfunctions of - 2' but not of - . d.x d.x d.x PB.19

167

The kinetic energy operator, T, is obtained from the operator analog of the classical equation

that is, , (jJ) 2 T=-

2m'

,

Px

Ii d d.x [8.26] ;

=i

hence

and

Then (T)

= N2

f

-~

1/1* (p; ) 1/1 dr: 2m ~

=

J l{I* (p2/2m) l{I dr: J 1/1*1/1 dr:

.

[N 2=

)]

J 1/1*1/1 dr:

.

- - J 1/I*-(e'kx cosx + e-'kx sin X)dr: 2m d.x 2 J 1/1*1/1 dr:

_1i2 2k;kx ~2::.cm-=_J_1/I_*_(-_k_)_x----;;-(e_'_·f_c_o_s_x_+_e_ - '_s_in_x_ ) _dr: = 1i2 k2 J 1/1*1/1 dr: = 11i2 k 2 1 Jl{I*1/Idr:

2mJ1/I*1/Idr:

2m '

168


PS.21 1/ 2

N

(a)

1

5

00

(r2)

= ( _1-3 )

[Problem 8.14].

32rca o

6

+ -r 2 )

e-r/ll{] dr = -1( 4 x 4! - 4 x 5! 8a0 0 ao a 8a 3 0 0

= - 13

(

4r4

-

-4r

= 142

+ 6!)a6

a~ I·

1/ 2

1J;

(b)

= Nr sin () cos t/> e- r / 2ao ,

(r)

= -132rca6

PS.23

1

00

0

N

= ( _1-5 )

[Problem 8.14].

32rcao

r 5 e- r / ao dr x -4rc 3

= -I24a6

~ x 5!a~ =~.

The superpositions of cosine functions of the form cos(nx) can be chosen with n equal to any integer between 1 and m. For convenience, x can be examined in the range between -rc/2 and rc/2. The normalization constant for each function is determined by integrating the function squared over the range of x [8.15]. Using MathCad to perform the integration, we find:

G.

G.

2 I ( 2 . cos rc . n) sin rc . n) + rc . n) cos(n·x) dx --+ - . -'----'-----'-----'.------''------"-n / 2 2 n n/ 2

1

When n is an even integer, sin(rcnI2) = 0 and, when n is an odd integer, cos(rcnI2) = O. Consequently, when n is an integer, the above integral equals rc/2 and we select (2/rc)I /2 as the normalization constant for the function cos(nx). The normalized function is t/>(n, x). The superposition, 1J;(m, x), is the sum of these cosine functions from n = I to n = m. Since the cosine functions are orthogonal, 1J;(m,x) has a normalization constant equal to (11m) 1/ 2. t/>(n,x) :=

A·

cos(n· x)

1J;(m, x) :=

(T m Y;;; . L t/>(n,x) 11=1


169

Constants and variables needed for MathCad plots and computations: -IT

N:= 1000

Xmin :=

2

IT

Xmax :="2

i :=O .. N

I

Xi := xmi n

+ N . (Xmax

- Xmin)

Examination of Fig. 8.1(a) reveals that, when the superposition has few terms, the particle position is ill-defined. There is great uncertainty in knowledge of position . However, when many terms are added to the superposition, the uncertainty narrows to a small region around x = O. A plot with m greater than 10 will further confirm this conclusion. Each function in the superposition has been assigned a weight equal to the normalization constant (l I m ) 1/ 2 [8.33] . This means that each cosine function in the superposition has an identical probability contribution to expectation value for momentum (see Justification 8.4). Each cosine function contributes with a probability equal to 11m. Furthermore, each cosine function represents a particle momentum that is proportional to the argument n because (d 2 ¢ (n , x)ldx 2 ) 1/ 2 (the differential component of the squared momentum operator) is proportional to n. The following plot [Fig. 8.1(b)] of momentum probability against momentum, as represented by n, is an interesting contrast to the plot of probability density against position. Variables needed for the MathCad plot: n :=

0..15

Prob(n, m) := if (n < m, ~ ,

0)

This plot shows that, when there are many terms in the superposition, the range of possible momentum is very broad even though the range of observed positions becomes narrow. Position and momentum are Probability density plots [Fig. 8.1(a)] for superpositions that have 1, 3, and 10 terms: 8r-----------,-----------~------------r_----------,

6

'i'(l,XI)2 'i'(3,xY

4

'i'(10,XI)2

2

......... ..... . .... ~

............ o.~

Figure 8.1(a) complementary variables. As location becomes more precise with the superposition of many function s, precise knowledge of momentum decreases. This illustrates the Heisenberg uncertainty principle [8.36a] .

170


1

Prob(n , 1)

1

0.8

-

0.6

-

Prob(n ,S)

-

Prob(n , 10) 0.4 I-

0.2

-

-

.... ...........

00

\

~

....j

IS

10

S

Figure 8.1(b)

n

The plot of probability density against position clearly indicates that the superposition is symmetrical around the point x = O. Consequently, the expectation position for all superpositions is x = O. The expectation value for position is independent of the number of terms in the superposition. The square root of the expectation value of x 2 is called the root-mean-square value of x, x rms. A plot of against m [Fig . 8.1 (c)] indicates that this expectation value depends upon the number of terms in the superposition. However, it does appear to very slowly converge to a very small (zero?) value when the superposition contains many functions.

Xrms

Xrm s(m): =

,,/2 1 (

2

x .1/I(I1l, x)2 dx

)1 /2

m:= 1 . . . 50

-,, /2

0.8

Xms(m)

0.2

10

20

30

m

40

so Figure 8.1(c)


PS.25

171

(a) In the momentum representation Px = Px x , consequently [x, Px] 1> = [x ,Px x] 1> = xPx x 1> - Px x x1> = ilup [8.39] Suppose that the position operator has the form x = a(d j dpx) where a is a complex number. Then,

d -(Px x dpx dpx -1> dpx

iii 1» - Px x (d -1> ) = -1>, dpx

a

+ Px -d1>

d1> iii - Px = -1> [Rule for differentiation ofJ(x)g(x)]. dpx dpx a

iii 1> = -1> a This is true when a = iii. We conclude that I x = ili(d j dpx) I'in the momentum representation. (b) The fact that integration is the inverse of differentiation suggests the guess that in the momentum representation

x - I1>=

(ili-dpxd)

- I

If

( 1>=:Iii

-

1f

P "

dpx ) 1>=:00 Iii

PX

1>dpx,

- 00

where the symbol J~'~ dpx is understood to be an integration operator which uses any function on its right side as an integrand. To validate the guess that x-I = (1 j ili) J~"oo dpx we need to

I

confirm the operator relationship x -I x = xx-I = integrals:

=

(f

P '

- 00

I

i. Using Leibnitz's rule for differentiation of

d1> ) +1>(Px) li~ -d de -1>(-oo)-d dp x = ddpx Px c- 00 Px Px

(f

PX

- 00

d1> dpx) -1>(-00) . -d Px

Since 1> ( - 00) must equal zero, we find that

from which we conclude that xx - I =

x - I x1> =

(

If

ill

PX

-00

dpx )

(

i.

d)

iii dpx 1> =

f

PX

-00

d1> = 1>(Px) -1>(-00) = 1>(Px) = 1>.

172


Solutions to applications PS.27

(a)

Anon-relativistic

=

h (2m e V) 1/2 e 6.626 {2 (9.109

Anon-relativistic

= 5.48

1O- 3I kg)

X

X

10- 34 J s

X

(1.602

X

10- 19 C)

X

X

10 3 V)} 112·

X

10 3 V)

(50.0

pm [8 .12 and Example 8.2]. h

Arelativistic

=

-{---(---e-v-)-}--c eV +

/2 I-=

2m e

1

2mec2

5.48 pm

Anon-relativistic

( eV) 1+-2mec2

=

1/2

5.48 pm {l+0.0489}1 /2

I

+

(1.602

{

2(9.109

X

X

10-

19

C) (50.0

10- 31 kg) (3 .00

X

JI /2

108 ms- I )2

= 1 535 m . p I·

(b) For an electron accelerated through 50 kV the non-relativistic de Broglie wavelength is calculated to be high by 2.4%. This error may be insignificant for many applications. However, should an accuracy of 1% or better be required, use the relativistic equation at accelerations through a potential above 20.4 V as demonstrated in the following calculation. Anon-relativistic -

Arelativistic

=

Anon-relativistic _

Arelativistic

1= (1 + ~) 1/2 _ 1 2mec2

Arelativistic

1(

= , + 2:

ev)

2mec2

1 ·3

+ 2 .4 .6

(

2mec

2

eV )3

2mec2

eV )

= -I ( - - 2

eV )2

1 ( - 2 . 4 2mec2

- ...

71

because 2nd and 3rd order terms are very small.

The largest value of V for which the non-relativistic equation yields a value that has less than 1% error: c2 V :::::: 2 ( 2m e ) e PS.29

(a)

C~(g) --+

X

~

(Anon-relativistic ArelativistiC) Arelatlvlstlc

C(graphite)

= 2 ( 2me

c2

) (0.01)

= 20.4 kV.

e

+ 2H2(g).

b.rG'

= -b.fG'(CH4) = -(-50.72kJmo\-I) = 50.72 kJ mo\-I

b.rW

= -b.fW(CH4) = -(-74.81 kJ mol-I) = 74.81 kJ mol- I at 'f .

at 'f

= 25°C.


173

We want to find the temperature at which !:!"rc?(T) = O. Below this temperature methane is stable with respect to decomposition into the elements. Above this temperature it is unstable. Assuming that the heat capacities are basically independent of temperature, !:!"rC;(T) "'" !:!"rC;cn

= [8.527 + 2(28.824) -:- 35.31]1 K- I mol-I l

"'" 30.8651 K- 1mol-

!:!,.rW(T) = !:!"rW(T)

f:

+

.

!:!"rC;(T)dT [2.36]

= !:!"rW('f) + !:!"rC; X a (!:!"rc?)) ( -aT -Tp

= -!:!"rW -T2

(T - T).

[3.52].

At constant pressure (the standard pressure)

fT !:!"rH~ T f T d(!:!"rc? I T) = - T -2-dT, T =

!:!"rc?(T)

!:!"rc?( T) _

T

T

fT !:!"rH~ ( T) + !:!"rC;

The value of T for which !:!"rc?(T) !:!"rc?(T)

T

X

(T - T) dT

T2

T

=

0 can be determined by examination of a plot (Fig. 8.2) of

.

agamst T.

!:!,. c?(T) r T

= 50.72 kJ mol - I1298.15 K = 0.1701 kJ K- I mol-I .

(10- k1) 3

x (298K) x

-1-

= 65.16 kJ mol - I. !:!"rC;

= (30.865JK- I mol-I)

( 10- kJ) = 0.030865kJK- I mol-I. 3

x

--J-

With the estimate of constant !:!,.rC; , 1methane is unstable above 825 K I.

174

STUDENT'S SOLUTIONS MANUAL Methane decomposition

0.20

0.15 I

"0 E

.,

0.10

~

-.

"'"

:::::::

h ........

0.05

h' ¢

0.00

"i -0.05

-0.10 200

400

600

800

1000

1200 1400

1600

T/K

(b)

Amax =

Amax

=

Figure 8.2

1(l .44cm K) 5 T [See the solution to Problem 8.10],

~(I.44cm K) 1000 K

1Amax (1000 K)

(c) Excitance ratio

=

9

-4

= 2.88 x

10

cm

(10 nm) 102 cm '

= 2880 nm I·

M (brown dwarf) M(Sun)

(1000K)4 (6000 K)4

a T,4

=

brow: dwarf

a TSun

=17.7

x 10- 4

[See the solution to Problem

8.) I]

I.

. . p (brown dwarf) Energy denSity ratio = '--'-----,:----,---'p(Sun)

-

8nhe/A 5

1)) -1))

(1/(e(hc/ AkTbrowndWarf) -

-8nhe/A5 (1/(e(hc/AkTs un ) e (hc/!..kTS un )

-

[85]

.

1

(e(hc/AkTbrowndwarf) -

I)'

The energy density ratio is a function of A so we will calculate the ratio at Amax of the brown dwarf.

he Abrown dwarfk

(6.62 x 10- 34 Js) x (3.00 x 10 8 ms- l ) (2880 x 10- 9 m) x (1.381 x 10- 23 J K-l)

= 4998K.


Energy density ratio =

175

e 4998 K j TSun - I -4""'9"'"98"""j= T - - - -

e

brown dwarf -

1

e(4998j6000) - I

1.300

e (4998j IOOO) - I

147

= 18.8

X 10-

3

1.

(d) The wavelength of visible radiation is between about 700 nm (red) and 420 nrn (v iolet). (See text Fig. 8.2.)

Fraction of visible energy density

=

- I4 11420nm p(.l..)d.l.. aT 700nm

I

[8.3 , 8.5, and solution to problem 8.11]

= - -C4 11420nm p(.l..)dA I . 4aT

700nm

As an estimate, let us suppose that peA) doesn ' t vary too drastically in the visible at lOOO K. Then, 420 nm

p(.l..)dA

I ~ p(560 nm)

x (700 nm - 420 nrn)

11 700nm ~

8n:hc ) x ( I ) x ( 280 x 10- 9 m ) ( (560 x 10- 9 m )5 e«4998K j IOOOK)x(2880nm j 560nm)) - 1

34

J s) x (3.00 x 10 8 m S- I) ( I ) 1.97 X 10- 25 m 4 e 25 .70 - I

= 8n: (6.626 x 10-

= 1.75 x

10-

10 J m - 3 .

Fraction of vis ible energy den sity ~

(3.00 x 10 8m s- l ) x (1.75 x 1O2

IO Jm- 3 )

4 (5 .67 x 10- 8 Wm - K-4) x (lOOOK)

~ 12.31

x 10- 7

1.

Very little of the brown dwarf's radi ation is in the visible. It doesn ' t shine brightly.

4

9

Quantum theory: techniques and applications


In quantum mechanics, particles are said to have wave characteristics. The fact of the existence of the particle then requires that the wavelengths of the waves representing it be such that the wave does not experience destructive interference upon reflection by a barrier or in its motion around a closed loop. This requirement restricts the wavelength to values A = 21n x L , where L is the length of the path and n is a positive integer. Then using the relations A = hi p and E = p2/2m, the energy is quantized at E = /1 2 h 2 18mL 2 . This derivation applies specifically to the particle in a box, the derivation is similar for the particle o n a ring; the same principles apply (see Section 9.6).

09.3

The lowest energy level possible for a confined quantum mechanical system is the zero-point energy, and zero-point energy is not zero energy. The system must have at least that minimum amount of energy even at absolute zero. The physical reason is that, if the particle is confined, its position is not completely un certain, and therefore its momentum, and hence its kinetic energy, cannot be exactly zero. The particle in a box , the harmonic oscillator, the particle on a ring or on a sphere. the hydrogen atom, and many other systems we will encounter, all have zero-point energy.

09.5

Fermions are particles with half-integral spin, 1/2, 3/2, 5/2, ... , whereas bosons have integral spin, 0, I , 2, .... All fundamental particles that make up matter have spin 1/2 and are ferrnions , but composite particles can be either fermion s or bosons. Ferrnions: electrons, protons, neutrons, 3He, .... Bosons: photons, deuterons.


E

= /72

2 2 /1 /7 - -2 [9.4a] 8meL

(6.626 x 10- 34 J s)2

_

.,----,-----'-..,......,;:-:-------'---,,-----,;= 2.678 X 31 9 8(9.109

X

10-

kg) x (1.50 x 1O- m)2

10- 20 J

The conversion factors required are leV

=

1.602 x 10- 19 J ;

I cm -

I

=

1.986 x 10- 23 J ;

leV

= 96.485 kJ

mol -

I

QUANTUM THEORY: TECHNIQUES AND APPLICATIONS 2

h 20 (a) E3 - EI = (9 - 1 ) - 2 = 8(2.678 x lO- J) 8m eL

I

= 12.14 x 10- 19 J 1= 11.34eV = 11.08 x 104 cm - 1 1= h2 (b) E7 - E6 = (49 - 36)--2 = 13(2.678 8me L

X

I 129kJmol- 1 I

10- 20 J)

I

I

= 13.48 x 10- 19 J = 12.17eV = 11.75 x 104 cm -1 1= 12lOkJmol- 1 1

E9.2(b)

The probability is

J1fr*1fr

P=

dx =

~

J

sin

2

C~X) dx ~ 2~X sin 2 C~X)

where t.x = 0.02L and the function is evaluated at x = 0.66 L.

E9.3(b)

(a)

For n = I

P=

(b)

For n = 2

P=

~

2(0.02L)

sin 2 (0.66JT) = ~

2(0.02L)

sin 2[2(0.66JT)] = ~

L

L

~

The expectation value is

J1fr*p1fr

(P) =

dx

but first we need p1fr A

p1fr

d (2) 1/2. nJT x sm ( L ) =

-In dx L •

=

-2innJT r ---uJo

L

A

so 1jJ) =

sin

. (2) 1/2 nJT

-In L

L

nJT x cos ( L )

(nJTX) (nJTX) Inl L cos L dx = L2.J

for all n. So for n = 2

E9.4(b)

The zero-point energy is the ground-state energy, that is, with nx = ny = n z = I: E=

(n 2 + n 2 + n 2 )h 2 3h 2 x Y 2 z [9.12b with equal lengths] = - -2 8mL 8mL

Set this equal to the rest energy me 2 and solve for L: 3h2 me 2 = - 8mL2

soL =

(~)1 /2 ~ 8

me

=

(~)1 /2 AC 8

where AC is the Compton wavelength of a particle of mass m.

177

178 E9.5(b)


1/15

= ( L2)1 /2 sin (5JTX) L

· ' " III P() dP(x) M aXlma andrrumma x correspon d to - = 0 dx d P (x) ex d1/l dx dx2 ex sin (5JT L X) cos (5JTX) L ex sin (IOJTX) ~

sin e

L

= 0, x = L 11'

E9.6(b)

= sin 2ex]

= 0 when e = 0, JT , 2JT , . . . , n'JT (n' = 0, 1, 2, . .. ) 10JTX , --=nJT

x

[2 sin ex cos ex

for

11'

< 10 -

so

n'L

x = -

10

are minima. Maxima and minima alternate, so maxima correspond to

= 1, 3, 5, 7, 9

x

=1

~ I, 1~~ I, [TI,1~~ I, 1~~ 1

The energy levels are

where £1 combines all constants besides quantum numbers. The minimum value for all the quantum numbers is 1, so the lowest energy is

The question asks about an energy 14/3 times this amount, namely 14£1 . This energy level can be obtained by any combination of allowed quantum numbers such that

The degeneracy, then, is []], corresponding to (nl , n2 , 113) = (1 , 2,3), (I , 3, 2), (2, 1, 3), (2, 3, I), (3 , 1, 2), or (3, 2, I).

E9.7(b)

£

= ~kT is the average translational energy of a gaseous molecule (see Chapter 17). 3

£ = -kT = 2

£

=

G)

8mL 2

n2h2 8mL2

x (1.381 x 10- 23 JK- 1) x (300K)

n2 = - £ 2

h

(n 2 + n2 + n2)h 2 1 2 3 8mL2

= 6.214 x

10- 21 J

QUANTUM THEORY : TECHNIQUES AND APPLICATIONS

(6.626

X

10-

34

J S)2

= 1.180 x 10- 42 J

0.02802 kg mol-I ) 2 x 100m ( 6.022 x 1023 mol-I

(8) x

21 2 6.214 x 10- J 21. n = 42 = 5.265 x 10 , 1.180 x 10- J

t:;.E = (2n

179

+ I)

x

h2

(

- -2

8mL

)

n= !7.26 x 101O!

= [(2) x (7.26 x 10 10 )

+ 1]

x

(

h

2

-2

8mL

)

=

14.52 x 1O 2

IO 2 h

8mL

= (14.52 x 10 10 ) x ( 1.180 x 10-42 J) = ! 1.71 X 10- 31 J! The de Broglie wavelength is obtained from h h A = - = - [8.12] p mv The velocity is obtained from EK

= !mv 2 = ~kT = 6.214 x 10- 21 J 6.214

v2 = 1) ( A=

2"

X

10- 21 J

= 2.671

X

105 m 2 s- 2;

v = 517 m S- I

(0.02802 kg mol-I ) x 6.022 x 1023 mol- I

6.626 x 10- 34 J s =2.75 x 10- 11 m=127.5pml (4.65 x 10- 26 kg) x (517 m S- I)

The conclusion to be drawn from all of these calculations is that the translational motion of the nitrogen molecule can be described classically. The energy of the molecule is essentially continuous,

t:;.E

E «< E9.8(b)

1.

The zero-point energy is 1

1 (k)1 /2

Eo = -!'u.v = -Ii 22m

1 = -(\.0546 2

X

10- 34 Js) x

(

285 Nm- I )1 /2 26 5.16 x 10- kg

= ! 3.92 x 10- 21 J ! E9.9(b)

The difference in adjacent energy levels is

(

k)

t:;.E = Ev+ 1 - Ev = !'u.v [9.26] = Ii;;;

so k

m(t:;.E)2

= --::-1i2

(2 .88

X

1/ 2

[9 .25]

10- 25 kg) x (3 . 17 X 10- 21 J) 2 ! _I ! ( 1.0546 X 10-34 J s)2 = 260 N m

180 E9.10(b)


The difference in adjacent energy levels, which is equal to the energy of the photon, is ~E

= Iiw = hv

,,(_k )1 /2

so "

he A

m

and A=

h; (~y /2 = 2Jre(T)

= 27T(2.998 x

1/2

8 I ((l5.9949U) x (1.66 x 1O- 27 kg U- I»)1 /2 10 ms- ) x I 544Nm-

A = 1.32 x 10- 5 m = 113.2 I1m 'l E9.11 (b)

The difference in adjacent energy levels, which is equal to the energy of the photon, is ~E

and A =

= Iiw = hv

-he (k- ) n 111

1/ 2

so

k) 1/2

he

n(-;;;

A

1/ 2 = 27Te (m) -

k

Doubling the mass, then, increases the wavelength by a factor of 21 / 2. So taking the result from Exercise 9 .1O(b), the new wavelength is A = 21 /2 (1 3.2 11m) = 118.7 11m ~E

E9.12(b)

I

= Iiw = hv

(a)

~E =

(b)

~E = Iiw = n(~) 1/2

hv = (6.626

X

10- 34 1 Hz-I ) x (33 x 103 Hz) = 12.2 x 10- 29 1

meff

I [ -m eff

= - 1 + -1 m,

m2

with ml

I

= m2 ]

For a two-particle oscillator meff , replaces 111 in the expression for w. (See Chapter 13 for a more complete discussion of the vibration of a diatomic molecule.) 2k)I /2

~E = n( -;;;-

= (1.055

X

10-

34

( 1 s) x

(2) x ( ll77Nm-l) )1 /2 (16.00) x (1.6605 x 10-27 kg)

=13 .14 x 10- 20 11 E9.13(b)

The first excited-state wavefunction has the form 1{1

= 2NIyexp (-!l

)

where NI is a collection of constants and y == x(mw/n) 1/2. To see if it satisfies Schrodinger's equation, we see what happens when we apply the energy operator to this function 2

. n2 d 1{1 1 2 X2 1{1 H1{I=---+-111W 2 2m dx

2

QUANTUM THEORY: TECHNIQUES AND APPLICATIONS

181

We need derivatives of l/f

(nut»

dl/f = dl/f dy = dx dy dx

and

:~ = ~:~

n

(:r

, = - n2m

x

X

(l _ i ) x exp (- ~i)

2

(m;)

(mw) x (y

2

So Hl/f

=

1/ 2(2NI)

2

n

x (2NI) x (-3y

- 3)1/f

1 2 = -"2 nw x (y - 3) x l/f

+

+ y3)

X

exp (

-~i) =

C:

W ) x

(i -

3)l/f

1 2 2 + -mw x l/f

2

1 2 "2 nwy l/f

3

= "2nwl/f

Thus, l/f is a solution of the SchrOdinger equation with energy eigenvalue

E9.14(b)

The harmonic oscillator wavefunctions have the form

1/fv(x )

1 )

= NvHv(y) exp ( -"2 i

with y

x

= -;;;

and a

=

(

n2 ) 1/4

mk

[9 .28]

The exponential function approaches zero only as x approaches ±oo, so the nodes of the wavefunction are the nodes of the Hermite polynomials. Hs (y )

= 32i

- 160y3

= 0, which leads to x = O. The other factor can be made into a quadratic equation

So one solution is y by letting z =

i

4 z2 - 20z + 15 so

z=

+ 120y = 0 [Table 9.1] = 8y (4l- 20i + l5)

-b ± .Jb2

=0 -

4ac

20 ± .J202 - 4 x 4 x 15

2a

2 x 4

Evaluating the result numerically yields x=10,±0.96a, or ±2.02a I.

z=

5±

.JTO 2

0.92 or 4 .08, so y = ±0.96 or ±2.02. Therefore

COMMENT. Numerical values could also be obtained graphically by plotting H5(y )'

E9.1S(b)

The zero-point energy is Eo

= ~nw = ~n(~) 1/ 2 2

2

meff

For a homonuclear diatomic molecule, the effective mass is half the mass of an atom, so I Eo = - ( 1.0546 2 Eo

= 12.3421

X

X

10- 34 J s) x

10- 20 J 1

( 2 2 9 3 . 8 N m- 1 ~(l4.0031 u) x (1.66054 X 10- 27 kgu- I )

)1 /2

182

E9.16(b)


Orthogonality requires that

f if m

1/1,;,1/1" dr

= 0

f=. n.

Performing the integration

If m

f=.

/1 , then

f

1/1* 1/1 dr

III"

2

=

27T

N e i (II_lIIl1 i(n - m) 0

2

=

N (I - 1) = 0 i(/1 - m)

Therefore, they are orthogonal. E9.17(b)

The magnitude of angular momentum is

Possible projections onto an arbitrary axis are

where

m, = 0 or ± I or ±2. So possible projections include

I0, E9.18(b)

± 1.0546 x 10- 34 J sand ±2.1109 x 10- 34 J s I

The cones are constructed as described in Section 9.7(d) and Figure 9.40(b) of the text; their edges are of length {6(6 + J)} 1/ 2 = 6.48 and their projections are mj = +6, +5 , ... , -6. See Figure 9.1 (a). The vectors follow, in units of n. From the highest-pointing to the lowest-pointing vectors (Figure 9. J(b », the values of are 6, 5, 4, 3, 2, 1,0, -I , - 2, - 3, - 4, -5 , and -6.

m,

~=====! In = + 6 ~-->..-----"~:z.. + 5

"'"->-----~z + 4 ~~------~~ + 3

~~--------~~~.+2

S~===:::::::====:::20+ I

~~------------~ - 1 ~------------~~ --23 -r------~ - 4 - 5

Figure 9.1(a)


183

Figure 9.1(b)


E

=

n2h2 8mL2 '

=

We take m(02)

E2 - E 1 =

We set E

E2 - EI

(32.000) x (1.6605 X 10- 27 kg), and find

1 39 1 (3) x (6.626 x 10- 34 J s)2 = 1.24 x 10- J. (8) x (32.00) x (1.6605 x 10- 27 kg) x (5.0 X 10- 2 m)2

n2 h 2 8mL2

=

=

3h 2 8mL2'

=

1

2kT and solve for n.

From above, h2 18mL 2 = (E2 - E I) 13 = 4.13 x 10- 40 J; then n 2 x (4. 13 x 10- 40 J)

We find n

=

G)

x (1.381 x 10- 23 JK - 1) x (300K)

= 2.07 x

10- 21J.

9 2.07 X lO- 2I J)1 /2 40 = 12.2 x 10 I· 10- J . .

= ( 4. 13 x

At this level,

h2 h2 h2 E" - E,,_I = (n 2 - (n - 1)2 ) x - -2 = (2n - I) x - -2 ~ (2n) x - -2 8mL 8mL 8mL

= P9.3

E

m 2Ji2

m 2Ji2

21

2mr

= - '-[9.38a] = - '-2

Eo = 0

EI

(4.4 x 109 ) x (4. 13 x 1O-4o J) [l

~ 11.8

x lO- 30 J 1 [or 1.1I!Jmol- I ].

= mr2 ] .

[m, = 0].

Ji2

=-= 2mr2

(1.055 x 10- 34 J s)2 (2) x (1.008) x (1.6605 x 10- 27 kg) x (160 X 10- 12 m)2

The minimum angular momentum is ±Ji I. 1

=

1

1 130 xlO- 22 J . .

184 P9.S


(a) Treat the small step in the potential energy function as a perturbation in the energy operator:

{O

H(I ) =

£

0

for :'S x :'S (l / 2)(L - a) and (l / 2)(L + a) :'S x :'S L for (l / 2)(L - a) :'S x :'S (l / 2)(L + a) .

The first-order correction to the ground-state energy, EI, is

E~I )

L

=

Ioo

(2)

= j(l /2)(L+a) (I /2)(L- a)

j(i /2)(L+a) . 2 (JtX) Sin dx L (I /2)(L- a) L

EI

= -2 £

E (I )

=

( I)

o/iO)*H(i ) o/~O) dx

1/ 2

L

£

(2) -

L

1/ 2

Jtx sin ( - ) dx, L

(JtX) . (JtX)) li / 2(L+ a) Jtx - L cos Sin , LJt L L i/2(L-a)

= -£

(

sa _ ~ cos (Jt(L + a») sin (Jt(L + a») L Jt 2L 2L

i

Jtx sin ( - ) L

+ ~ cos (Jt(L Jt

2L

a») sin (Jt(L - a)). 2L

This expression can be simplified considerably with a few trigonometric identities. The product of sine and cosine is related to the sine of twice the angle:

. (Jt(L±a») = -1. (Jt(L±a») = -1. ( Jta) , Sill Sin Jt ± cos ( Jt(L±a») Sin 2L 2L 2 L 2 L and the sine of a sum can be written in a particularly simple form since one of the terms in the sum is Jt : sin ( Jt

±

7) =

Thus E (I) = -sa I L (b) If a

. (Jta) + -Jt£Sin L

(~a) ± cos Jt sin (~a) = 'f sin (~a) . .

= L/ IO, the first-order correction to the ground-state energy is EI(I )

P9.7

sin Jt cos

£ . (Jt) = 10 + ;-£ Sin 10 = I0.1984 £.I

The second-order correction to the ground-state energy, EI, is

where H

(I) _

(0) _

- mgx'o/lI

-

. nJtx Sin - - ,

L

2 _ n2h and En - - - 2 ' 8mL

The denominator in the sum is

E(O) _ E(O) i

n

=

_h_2_ _ _n2_h_2 8mL2 8mL2

= _(1_-_n--;:2);-h_2 8mL2


185

The integral in the sum is

L 1/I (O)* H

loo

where a

/I

( I ) 1/1 (0) dx I

2mg =- loL x sinaxsinbx dx L

0

'

= nn l L and b = n i L.

The integral formulas given with the problem allow this integral to be expressed as 2mg d l o L . d (cos (a - b)x cos(a + b)X) IL - --cos axsmbxdx = -mg- --'--L da 0 da 2(a - b) 2(a + b) 0 _ -

- -2mg L

(-x Sin (a-b)X 2(a - b)

-

cos(a-b)x 2(a - b)2

L

+ xsin(a + b)x + -cos(a+b)x)I -----:02(a + b) 2(a + b)2 0 .

The arguments of the trigonometric functions at the upper limit are: (a - b)L

=

(n - I)n and (a

+ b)L = (n + I) n.

Therefore, the sine terms vanish. Similarly, the cosines are ± 1 depending on whether the argument is an even or odd mUltiple of n ; they simplify to ( _I)Il+ I . At the lower limit, the sines are still zero, and the cosines are all I. The integral, evaluated at its limits with Tt l L factors pulled out from the as and bs in its denominator, becomes mgL (-1)11+ 1 - I _ ( -1)11 + 1 n2 (n - 1)2 (n + 1)2

I)

= mgL[ (-I)n+1 - I] (n + 1)2 - (n _ 1)2) n2 (n _ 1)2(n + 1)2 '

4mgL[(-I)n+1 - l]n

n 2 (n 2

-

1)2

The second-order correction, then, is

Note that the terms with odd n vanish. Therefore, the sum can be rewritten, changing n to 2k , as

The sum converges rapidly to 4.121 x 10- 3 , as can easily be verified numerically; in fact, to three significant figures , terms after the first do not affect the sum. So the second-order correction is (2) _ -

£1

186


The first-order correction to the ground-state wavefunction is also a sum: ./. (I )

'PO

= '""" Cno/n ./. (0) ' ~

II 4mgL[(-I )II+ 1 - l]n

where c

=

rL ./. (0) 0 HI ./.(0) dx _JO 'I'll

'1'1

£,\0) _ £~O)

II

2

=_

2

2

(n - I ) «n2 _ I )h2 j 8mL2) Jt

32m2gL 3[(-l)n + I]n Jt 2h 2(n 2 - 1)3

Once again, the odd n terms vanish. How does the first-order correction alter the wavefunction ? Recall that the perturbation raises the potential energy near the top of the box (near L) much more than near the bottom (near x = 0); therefore, we expect the probability of finding the particle near the bottom to be enhanced compared with that of finding it near the top. Because the zero-order ground-state wavefunction is positive throughout the interior of the box, we thus expect the wavefunction itself to be raised near the bottom of the box and lowered near the top. In fact, the correction terms do just this. First, note that the basis wavefunctions with odd n are symmetric with respect to the center of the box; therefore, they would have the same effect near the top of the box as near the bottom. The coefficients of these terms are zero: they do not contribute to the correction. The even-n basis function s all start positive near x = 0 and end negative near x = L; therefore, such terms must be multiplied by positive coefficients (as the result provides) to enhance the wavefunction near the bottom and dimini sh it near the top.

Solutions to theoretical problems PS.9

The text defines the transmission probability and expresses it as the ratio of IA11 2j 1A1 2, where the coefficients A and AI are introduced in eqns 9.14 and 9.17. Eqns 9.18 and 9.19 list four equations for the six unknown coefficients of the full wavefunction. Once we realize that we can set BI to zero, these equations in five unknowns are: (a) A + B = C + D, (b) Cel, while the left side depends on r. The only way that the two sides can be equal to each other for all r, (), and 4> is if they are both equal to a constant. Call that constant -(li2 1(l + 1))/2m (with 1 as yet undefined) and we have, from the right side of the equation, -

li2

2mY

2 li2 1(l + I ) A Y = - - ...,-----'-

so

2m

A2y

= -1(1 + I )Y.

From the left side of the equation, we have

_~ (r2 a x + 2r ax ) _ Er2 = _ li 1(1 + I). 2

2m

2

X ar2

X ar

2m

After multiplying both sides by X / r2 and rearranging, we get the desired radial equation

2 (a X 2m ar2

_~

+ ~ ax ) + li r ar

2

1(l + I ) X 2mr 2

= EX.

Thus the assumption that the wavefunction can be written as a product of functions is a valid one, for we can find separate differential equations for the assumed factors. That is what it means for a partial differential equation to be separable.

198


(b) The radial equation with l

a2x

2 ax

= 0 can be rearranged to read

2mEX

+- - - -IL2· ar2 r ar Form the following derivatives of the proposed solution:

ax ar

= (2rcR)-1 /2 [coS(nrcr/R) (nrc) _ Sin(nrcr/R)] r

R

r2

an d -a2x -_ (2 rcR) -1 / 2 [ - sin(nrcr/R) (nrc)2 - 2cos(nrcr/R) (nrc) -

ar2

r

R

r2

R

2Sin(nrcr/R)] + ---;:--r3 .

Substituting into the left side of the rearranged radial equation yields

(2rcR) - 1/2 [_ sin(nrcr/R) (nrc)2 _ 2cos(nrcr/R) (nrc) r R r2 R

+ (2rcR)-1 /2 [2COS(nrcr/ R) r2

= -(2rcR) _1 /2sin(nrcr/R) r

+ 2Sin(nrcr/R)] r3

(nrc) _ 2Sin(nrcr/R)] R r3 (nrc)2 R

= - (nrc)2 x. R

Acting on the proposed solution by taking the prescribed derivatives yields the function back mUltiplied by a constant, so the proposed solution is in fact a solution. (c)

Comparing this result to the right side of the rearranged radial equation gives an equation for the energy

E

2 2 (.!!...)2 = nh . 2mR2 2rc 8mR2 2 2

= (nrc) 2 If-. = n rc R

2m

Atomic structure and atomic spectra


The Schrodinger equation for the hydrogen atom is a six-dimensional partial differential equation, three dimensions for each particle in the atom. One cannot directly solve a multidimensional differential equation; it must be broken down into one-dimensional equations. This is the separation of variables procedure. The choice of coordinates is critical in this process. The separation of the Schrodinger equation can be accomplished in a set of coordinates that are natural to the system, but not in others. These natural coordinates are those directly related to the description of the motion of the atom. The atom as a whole (center of mass) can move from point to point in three-dimensional space. The natural coordinates for this kind of motion are the Cartesian coordinates of a point in space. The internal motion of the electron with respect to the proton is most naturally described with spherical polar coordinates. So the six-dimensional SchrOdinger equation is first separated into two three-dimensional equations, one for the motion of the center of mass, the other for the internal motion. The separation of the center of mass equation and its solution is fully discussed in Section 9.2. The equation for the internal motion is separable into three one-dimensional equations, one in the angle ¢ , another in the ang le e, and a third in the distance r . The solutions of these three one-dimensional equations can be obtained by standard techniques and were already we ll known long before the advent of quantum mechanics. Another choice of coordinates would not have resulted in the separation of the Schrodinger equation just descri bed. For the details of the separation procedure, see Sections 10.1 and 9.7.

010.3

The selection rules are ,6,n

= ± I , ±2, ... ,

,6,[

=

± I,

,6,m{

= 0, ± I .

In a spectroscopic transition the atom emits or absorbs a photon. Photons have a spin angular momentum of I. Therefore, as a result of the transition, the angular momentum of the electromagnetic field has changed by ± I n. The principle of the conservation of angu lar momentum then requires that the angu lar momentum of the atom has undergone an equal and opposite change in angular momentum. Hence, the selection rule on ,6,{ = ± I. The principal quantum number n can change by any amount since n does not directly relate to angular momentum. The selection rule on ,6,m/ is harder to account for on bas is of these simple considerations alone. One has to evaluate the transition dipole moment between the wavefunctions representing the initial and final states involved in the transition. See Justification 10.4 for an example of this procedure.

200


010.5

See Section 10.4(d) of the text and any general chemistry book, for example, Sections 1.10-1.13 of P. Alkins and L. Jones, Chemical principles, 2nd edn, W. H. Freeman, and Co., New York (2002).

010.7

In the crudest form of the orbital approximation, the many-electron wavefunctions for atoms are represented as a simple product of one-electron wavefunctions. At a somewhat more sophisticated level, the many-electron wavefunctions are written as linear combinations of such simple product functions that explicitly satisfy the Pauli exclusion principle. Relatively good one-electron functions are generated by the Hartree-Fock self-consistent field method described in Section 10.5. If we place no restrictions on the form of the one-electron functions , we reach the Hartree-Fock limit which gives us the best value of the calculated energy within the orbital approximation. The orbital approximation is based on the disregard of significant portions of the electron--electron interaction terms in the many-electron Hamiltonian, so we cannot expect that it will be quantitatively accurate. By abandoning the orbital approximation, we could in principle obtain essentially exact energies; however, there are significant conceptual advantages to retaining the orbital approximation . Increased accuracy can be obtained by reintroducing the neglected electron--electron interaction terms and including their effects on the energies of the atom by a form of perturbation theory similar to that described in Further information 9.2 and Section 10.9. For a more complete discussion consult the references listed under Further reading .

Solutions to exercises E1 0.1 (b)

The energy of the photon that struck the Xe atom goes into liberating the bound electron and giving it any kinetic energy it now possesses Ephoion

= I + Ekinelic

I = ionization energy

The energy of a photon is related to its frequency and wavelength he

Epholon

= hv = T

and the kinetic energy of an electron is related to its mass and speed, s

he So A

I 2

he A

2

= I + -mes => I = - = (6.626

X

10-

2

8

J s) x (2.998 x 10 m S-

58.4

X

(1.79 x 106 m s -

I

I X

34

I

-mes 2

I) _

10- 9 m

~ (9.11 2

r

X

10- 31 k ) g

?

=11.94 x 10- 18 11= 12.leV E10.2(b)

The radial wavefunction is [Table 10.1] R30 = A (6 - 2p

,

+ ~9 p2) e- p / 6 where p ==

2Zr , and A is a collection of constants. ~

[Note: p defined here is 3 x p as defined in Table 10.1]

ATOMIC STRUCTURE AND ATOMIC SPECTRA

201

Differentiating with respect to p yields

d:;,O= 0= (6 A

2p

+ ~p2)

x (

-D

e-

p 6 /

+

(-2 + ~p)

Ae-

p 6 /

This is a quadratic equation

0= ap2 + bp +c

where a

I

= -54 '

b

5

= -9 '

and c

= -3.

The solution is p =

so r

-b ± (b 2 - 4ac)I / 2 2a

= (~± 3 (71 /2) ) 2

2

= 15 ±3J7

ao .

Z

= 7.65 and 2.35, so r = 111.5ao/Z Iand 13.53ao/Z \. Substituting m, r = 1607 pm 1and 1187 pm I·

Numerically, this works out to p Z

= I and ao = 5.292 x

10- 11

The other maximum in the wavefunction is at 1r = 0 I. It is a physical maximum, but not a calculus maximum: the first derivative of the wavefunction does not vanish there, so it cannot be found by differentiation. E10.3(b)

The complete radial wavefunction, R4, 1 is not given in Table 10.1; however in the statement of the exercise we are told that it is proportional to (20 - lOp

+ p 2 )p

where p

= -2Zr ao

[Note: p defined here is n x p as defined in Table 10.1]

The radial nodes occur where the radial wavefunction vanishes, namely where (20 - lOp

+ p2)p = O.

The zeros of this function occur at p =0,

and when (20 - lOp then r

+ p2) = 0,

with roots p 764a

= 2.764,

pao = 2z = 2pao = -2.- 2 -o = 1 1.382ao 1

and

orr=17.31 x 10- 11 ml and 11.917 x 1O- lO ml

and p

= 7.236

7.236a o

1

- - 2 - = 3.618 a o

1

202

E10.4(b)


Nonnalization requires

Integrating over angles yields 1= 47TN 2

=47TN 2

1 1

00

e- r / aO (2 - r/ao)2 r 2 dr

00

e- r / aO (4 - 4r/ ao

+ r2 /a6) r 2 dr = 47TN2(8a6)

In the last step, we used

I

So N = ---:== 4J27TG6 E10.S(b)

The average kinetic energy is

where 1/1

= N(2 -

p)e- p / 2 with N

=~ 4

d!"

(~) 1/2

and p

27TG6

= r 2 sinBdrdBd¢ =

==

Zr here ao

a 3 p2 sin B dp dB d¢ 0

Z3

In spherical polar coordinates, three of the derivatives in V 2 are derivatives with respect to angles, so those parts of V21/1 vanish. Thus


203

and (EK)

=

[ 00 [ "

[2"

N(2 _ p)e- p /

1o 1o 1o

2(~) 2 x (_n2 ) ao

x Ne- p / 2(-4/p

2m

+ 5/2 -

a 3 d sin e de p2 dp p/4)-'0'--_----::-_ __ Z3

The integrals over angles give a factor of 41T , so

The integral in thi s last expression works out to -2, using

10

00

e- P pn dp = n! for n = 1, 2, and 3. So

The average potential energy is (V )

f = 1° 1°"1° =

ljr*Vljrdr where V

~~? = -- = ----

41T cor

41T eoaop

3 ) N(2 - p)e- p / 2 a-'0'--_----,,-_ p2 sin e dp de _d _ 41T eoao p Z3 The integrals over angles give a factor of 41T , so and (V)

2 "

00

22 Z e

N(2 - p)e- p / 2 ( -

2 2 (V ) = 41TN2 ( Z e ) x 41T eOQ'O

(a~) [ 00 (2 _ Z 1o

p)2pe- P dp

The integral in this last expression works out to 2, using

e

2 2

(V) =41T

E10.6(b)

Z3-3 ) x ( - Z -) x ( 321Ta 41Teoao o

10

00

e- P pn dp

= n! for n =

1, 2, 3, and 4. So

3

(a-% Z

)

x (2)

=

The radial distribution function is defined as P

= 41Tr2ljr2

where p

2Zr

== -

nao

so P 3s

2Zr

= -

3ao

= 41Tr 2(Yo.oR3,0)2,

here.

But we want to find the most likely radius, so it would help to simplify the function by expressing it in terms either of r or p, but not both. To find the most likely radius, we could set the derivative of P3s equal to zero; therefore, we can collect all multiplicative constants together (including the factors of ao/Z needed to turn the initial r 2 into p2) since they will eventually be divided into zero P 3s

= C 2 p2(6 -

6p

+ p2)2 e- P

204


Note that not all the extrema of P are maxima; some are minima. But all the extrema of (P3s)I /2 correspond to maxima of P3s. So let us find the extrema of (P3s) 1/2 d(P )1 /2 3s

dp

= 0

d

= -Cp(6 dp

= C[p(6 -

6p

6p

+ p2)e- p / 2

+ p2)

X

(-!) + (6 -

12p + 3p 2)]e- p / 2

Numerical solution of this cubic equation yields p

= 0.49,

2.79, and 8.72

corresponding to r

= I0. 74ao/ Z ,

COMMENT.

4. 19ao/Z, and 13. 08ao/Z

I

If numerical methods are to be used to locate the roots of the equation which locates the

extrema, then graphical/ numerical methods might as well be used to locate the maxima directly. That is, the student may simply have a spreadsheet compute P 3s and examine or manipulate the spreadsheet to locate the maxima.

E10.7(b)

The most probable radius occurs when the radial wavefunction is a maximum. At this point the derivative of the function wrt either r or p equals zero. dR 31) ( dp max

=0=

(d

(4 -

2

p

p) pedp

/

[Table 10.1]

))

= (4 _ 4p +

max

p2) e- p / 2 2

The function is a maximum when the polynomial equals zero. The quadratic equation gives the roots p = 4 + 2.J2 = 6.89 and p = 4 - 2.J2 = 1.17. Since p = (2Z/nao)r and n = 3, these correspond to R R (PI) r = 10.3 x ao/Z and r = 1.76 x ao/Z. However, -31- = 31 (1.17) = 4.90. So, we conclude

I I I R31 (10.3) that the function is a maximum at p = 1.17 which corresponds to Ir = 1. 76ao /z·1 IR31 (pz)

E10.8(b)

I

)

Orbital angular momentum is

There are 1 angular nodes and n - 1 - 1 radial nodes

(L 2) 1/2 = 6 1/ 2/i = 12.45

x 10- 34 J s 1 ~ angular nodes

OJ radial node

(a)

n

= 4,1 = 2,

so

(b)

n

= 2, 1 = I,

so (U ) 1/ 2 = 21 / 2/i

= 11.49 x

10- 34 J s 1

OJ angular nodes @]radial nodes

(c)

n

= 3,1 = I,

so (U ) I/2 = 21 /2/i

= 11.49 x

10- 34 Js 1

OJ angular node OJ radial node

ATOMIC STRU CTURE AND ATOMIC SPECTRA

E10.9(b)

For I > 0, )

=

=I±

205

1/ 2, so

(a)

I

I, so)

= 1r-1-12-o-r3-12----,1

(b)

1=5 , so)

= 19/2 or 11/21

E10.10(b) Use the Clebsch-Gordan series in the form

+h-I ,·· ·,U,-hl

1=), +Jz,),

Then, with), 1 =

= 5 andh = 3

18, 7, 6, 5, 4, 3, 21

E10.11(b) The degeneracy g of a hydrogenic atom with principal quantum number n is g

= /1 2

The energy E of

hydrogenic atoms is

so the degeneracy is

(a)

g=-

(b)

g=-

(c)

g

he (2) 2 RH -4heRH he (4) 2 RH

-~heRH

=[] =~

= _ he(5 )2 RH = ~ -heRH

E10.12(b) The letter F indicates that the total orbital angular momentum quantum number L is 3; the superscript 3

is the multiplicity of the term, 25 + I, related to the spin quantum number 5 indicates the total angular momentum quantum number 1 . E10.13(b) The radial distribution function varies as

The maximum value of P occurs at r

= ao since

dP ex (2r _ 2r2 ) e- 2rjao = 0 at r dr

= ao

ao

P fall s to a fraction! of its maximum given by

and Pmax

= ~e-2 ao

=

I ; and the subscript 4

206


and hence we must solve for r in ? f-l / - _ r - rl ao -e

e

ao

f = 0.50

(a)

0.260

r

= ao e- rl ao solves to r = 2.08 ao = 1110 pm Iand to r = 0.380ao = 120.1 pm I

f = 0.75

(b)

0.319

=

:0

e - rl ao solves to r

= 1.63ao = 186 pm Iand to r = 0.555ao = 129.4 pm I

In each case the equation is solved numerically (or graphically) with readily available personal computer software. The solutions above are easily checked by substitution into the equation for f . The radial distribution function is readily plotted and is shown in Figure 10.1.

0.15

0. 10

.,

~ C!.

--Q..,

0.05

0.00 0.5

0.0

1.0

1.5 rl ao

2.0

2.5

Figure 10.1

E10.14(b) (a) 5d -+ 25 is ~ an allowed transition, for t../ = -2 (t../ must equal

= -I. 5p -+ 3f is 1 not 1 allowed, for t../ = +2 (t..[ must equal ± I). 6h: I = 5; maximum occupancy = ~

(b) 5p -+ 3s is 1allowed I, since t../

(c) (d)

The only unpaired electrons are those in the 3d subshell. There are three. S=

[I] and

~-

I=

For S

= ~,

Ms = ±

for S

= ~,

Ms = ±

IT],

I ~ and ± ~ I

I ~I

± I).


E10.16(b) (a) Possible values of S for four electrons in different orbitals are 12, I , and 0

207

I; the multiplicity is 2S + I ,

so multiplicities are IS , 3, and I 1respectively. (b) Possible values of S for five electrons in different orbitals are IS/2, 3/2 and 1121; the multiplicity is 2S + I, so multiplicities are 16, 4, and 21 respectively.

= I) and a d electron (l = 2) gives rise to L = 3 (F), 2 (D), and I (P) terms. Possible values of S include 0 and I. Possible values of J (using Russell-Saunders coupling) are 3, 2, and I (S = 0) and 4, 3, 2, 1, and 0 (S = I) . The term symbols are

E10.17(b) The coupling of a p electron (l

Hund's rules state that the lowest energy level has maximum mUltiplicity. Consideration of spin-orbit coupling says the lowest energy level has the lowest value of J(J + I) - L(L + I) - S(S + I). So the lowest energy level is 13F2 1.

= I and L = 2, so J =13, 2, and I 1are present. J = 3 has I2J states, with MJ = 0, ± I , ±2, J = 2 has [IJ states, with MJ = 0, ± I, or ±2; J = I has [IJ states, with MJ = 0, or ±l.

E10.18(b) (a) 3D has S

or ±3 ;

(b) 4D has S

= 3/ 2 and L = 2, so J

states, with MJ

=

= 17/2, S/2, 312, and 1121 are present. J

± 7/ 2, ±S/ 2, ±3 / 2 or ± I / 2; J

±S/ 2, ±3/ 2 or ± I / 2; J = 3/ 2 has possible states, with MJ = ± I / 2.

=

S / 2 has

0

= 7/ 2 has 0

possible

possible states, with MJ

=

0 possible states, with MJ = ±3 / 2 or ± I / 2; J = 1/ 2 has [I]

= 9/ 2 and 7/ 2 are present. J = 9/ 2 has [IQJ possible states, with 9/ 2, ± 7/ 2, ±S/ 2, ±3/ 2, or ± I / 2; J = 7/ 2 has 0 possible states, with MJ = ± 7/ 2,

(e) 2G has S = 1/ 2 and L = 4, so J MJ = ± ±S/ 2, ±3 / 2, or ±1 / 2.

E10.19(b) Closed shells and subshells do not contribute to either Lor S and thus are ignored in what follows.

(a) Sc[Ar ]3d 14s 2: S

= !,L = 2; J = ~ , ~ , so the terms are 12D5/2 and 2D3/21.

(b) Br[Ar ]3d I0 4s 2 4p 5. We treatthe missing electron in the4p subshell as equivalent to a single "electron" with l

= I , s = !.Hence L = I , S = !,and J = ~ , !, so the terms are 12 P3/2 and 2P I / 2 1.


All lines in the hydrogen spectrum fit the Rydberg formula

~A = RH (~-~) nT n~

[10.1 , with ii

=~] A

RH

= 109677cm- l .

208


Find nl from the value of Amax , which arises from the transition

n~

-Am-ax-R-H -

Since

=

nl

+ +

n2(nl 1

2nl

1

A

n~(nl

1, 2, 3, and 4 have already been accounted for, try

1)2

-1 =

+ 1) 2 =

(nl

=

+I_

nl .

+I + 1)2'

2nl

-

nl

nl

=

5, 6, .. . . With

=

6 we get

136. Hence, the Humphreys series is 1n2 _ 61 and the transitions are given by

1) (l09677cm- 1) x (1 - -~

~

,

n2

= 7,8""

and occur at 12372 nm, 7503 om, 5908 om, 5129 nm, ... , 3908 nm (at n2 nm as n2 - 00, in agreement with the quoted experimental result.

P10.3

nl

A Lyman series corresponds to nl

=

15), converging to 3282

= I; hence

Therefore, if the formula is appropriate, we expect to find that

v( I -

1

n2

)-1 .

is a constant (R Li 2+ ).

We therefore draw up the following table.

n

v/cm- I 1

V ( 1 - n2

)-1 /cm-

I

2

3

4

740747

877924

925933

987663

987665

987662

Hence, the formula does describe the transitions, and 1RLi2+

= 987663 cm -

I

I. The Balmer transitions

lie at

v=

R

.2+ LI

(~4 -~) n2

= (987 663cm-

n = 3, 4, . ..

l)

x

G-~2) =

1137175 cm-

The ionization energy of the ground-state ion is given by

II,

1185187 cm-

II, ····


209

and hence corresponds to i! = 987 663cm- l , P10.5

or

1122.5 eV

I.

The 7p configuration has just one electron outside a closed subshel1. That electron has / = I, s = 1/2, and j = 1/2 or 3/2, so the atom has L = I, S = 1/2, and J = 1/2 or 3/2. The term symbols are 12P I/2 and 2P3/21, of which the former has the lower energy. The 6d configuration also has just one

= 2, s = 1/2, and j = 3/2 or 5/2, so the atom has L = 2, S = 1/2, and J = 3/2 or 5/2. The term symbols are 120 3/ 2 and 205/21, of which the former has the lower energy. According to the simple treatment of spin-orbit coupling, the energy is given by electron outside a closed subshell; that electron has /

E/,sJ

=

!hcAUU

+ I ) - /(/ + I) -

s(s

+ I)]

where A is the spin-orbit coupling constant. So

and Ee03 /2) = !hcA[~(3/2

+ I) -

2(2 + 1) - !(1/2 + I)] = -~hcA.

This approach would predict the ground state to be 1203/21. COMMENT.

The computational study cited finds the 2P1 / 2 level to be lowest, but the authors cau-

tion that the error of similar calculations on Y and Lu is comparable to the computed difference between levels. P10.7

RH

= kJ.LH,

Ro

= kJ.Lo ,

R

= kJ.L

[10.16]

where R corresponds to an infinitely heavy nucleus, with J.L

= me.

R

Likewise, Ro

=

R

(I

+ (melmd»

where

mp

is the mass of the proton and md the mass of the deuteron .

The two lines in question lie at

~ = Ro Ao and hence

(1 - ~) = ~ 4

4

Ro

210


Then, since 1 + (melmd) 1 + (mel mp) '

and we can calculate md from

9.10939 31

1 (

+

9.1039 x 10- kg) (82259.098cm - 1)-1 1.67262 x 10- 27 kg x 82281.476cm - 1

= 13.3429 x Since I

10- 3 1 kg

X

10- 27 kg

I.

= Rhe, 82281.476cm- 1 82259.098cm- 1

P10.9

= 11.0002721·

(a) The splitting of adjacent energy levels is related to the difference in wavenumber of the spectral lines as follows :

helll! Ilv

=

IlE

= /-tBB,

so Ill!

/-tBB

(9.274 x 10- 24 JT- I )(2 T) (6.626 x 10- J s)(2.998 x 10 cm s-I)

= - - = --------:::-:------...,."...---:34 10 he

= 10.9 cm- I I·

(b) Transitions induced by absorbing visible light have wavenumbers in the tens of thousands of recip-

rocal centimeters, so normal Zeeman splitting is 1 small 1 compared to the difference in energy of the states involved in the transition. Take a wavenumber from the middle of the visible spectrum as typical :

Or take the Balmer series as an example, as suggested in the problem; the Balmer wavenumbers are (eqn 10.1):

The smallest Balmer wavenumber is

v = (l09677cm - l )

x (1/4-1/9)

= 15233cm- 1

and the upper limit is I!

= (109677cm - l )

x ( 1/ 4 - 0)

= 27419cm- l .


211


Consider t/l2p, = t/l2,I,Owhich extends along the z-axis. The most probable point along the z-axis is where the radial function has its maximum value (for t/l 2 is also a maximum at that point). From Table 10. 1 we know that

and so dR dp

=

Therefore, r* COMMENT.

(I - ~p) e4

=

p/4

=0

when

p= 4.

2;0, and the point of maximum probability lies at z

= ± 2;0 = I ±I06 pm I.

Since the radial portion of a 2p function is the same, the same result would have been obtained

for all of them. The direction of the most probable point would, however, be different.

P10.13

J

(a) We must show that 1t/l3 px 12 d r coordinates (Fig. 8.22).

{2n

= 10

I . The integrations are most easily performed in spherical

r 10r'" 1R31(P) {

10

= 2r/ao, r =

wherep

I

=

YI-I -

.j2 YII

pao/2, dr

}

=

12 r-sm(e)drded¢ [Table 1O. I, eqn 10.24] ?

(ao/2) dp .

{2" 10{" 10roo (2a0 ) 31[( 27(6)11 /2 )

= 210

( I

1 = 466 S6rr

)3/2(4 -"?/ I) pe-

1

6

p /

fa 2" cos2(¢ ) d¢ fa" sin 3 (e) de faoo (4- ~pr p 4e- /3d p p

' - - v - ' ' ' - - . - - ''

"

We must also show that

ao

3 1/2 ] \2p2 sin Ce) dp de d¢ 2 sinCe) cos(¢) [ (8rr)

X

=I

.

Thus,

f

4/ 3

t/l3px is normali zed to I.

t/l3px t/l3d.n dr

= O.

•

34992

'

212


Using Tables 9.3 and 10.1 , we find that I ( I ) 3/2 ( I) 6 o/3Px =S4(2n)1 /2 aD 4- P pe-p/ sin(8)cos(cf»

3

=

32(2~) 1/2 (~J 3/2 p 2e-

p 6 /

sin 2(8) sin(2cf»

where p = 2r I ao,r = pao/2, dr = (ao / 2) dp.

f

o/3px o/3dxy dr

fa oo p5 e- p /3 dp

= constant x

/a

2IT cos(cf» sin(2cf» dcf>, faIT sin4(8) d8

o Since the integral equals zero, o/3P., and o/3d.,), are orthogonal. (b) Radial nodes are determined by finding the p values (p = 2r I aD) for which the radial wavefunction equals zero. These values are the roots of the polynomial portion of the wavefunction. For the 3s

orbital, 6 - 6p

+ p2 = 0, when IPnode = 3 + J3 and Pnode = 3 -

J31.

The 3s orbital has these two spherically symmetrical modes. There is no node at p conclude that there is a finite probability of finding a 3s electron at the nucleus.

=

0 so we

For the 3px orbital, (4 - p) (p) = 0, when 1Pnode = 0 and Pnode = 41· There is a zero probability of finding a 3px electron at the nucleus. For the 3dX), orbital 1 Pnode (c)

(rhs

=

f

=0

2 IRJO Yool rdr

1

is the only radial node.

=

f

2 IRJO Yool r3 sin(8) drd8 dcf>

= ~ {OO (6 _2p + p2 19 )2 p 3e- p / 3 dp . 3888

/0

,

52488

(r h,

27ao

= -2- .

(d) The plot Fig. 10.2 (a) shows that the 3s orbital has larger values of the radial distribution function for r < aD. This penetration of inner core electrons of multi electron atoms means that a 3s electron experiences a larger effective nuclear charge and, consequently, has a lower energy than either a 3p or 3dxy electron. This reasoning also leads us to conclude that a 3px electron has less energy than a 3dX), electron.

ATOMIC STRUCTURE AND ATOMIC SPECTRA Radial distribution functions of atomic hydrogen

0. 12

0.1

0.08

r-? 0.06

-t

0.04

0.Q2

0 10

0

15

20

25

30

rlao

(e) Polar plots with

e=

Figure lO.2(a)

90°.

The p orbital 90

The ... orbital 120

90 120

60

60 150

30

180 180

0

300

240

210

270

'"

330

240

300 270

'"

The tI orbital

90 120

60

150

30

180

0

210

330

240

300 270

'"

Figure lO.2(b)

213

214


Boundary surface plots. s-orbital boundary surface

p-orbital boundary surface

d-orbital boundary surface

f- orbital boundary surface

Figure lO.2(c) P10.15

The general rule to use in deciding commutation properties is that operators having no variable in common will commute with each other. We first consider the commutation of Iz with the Hamiltonian. This is most easily solved in spherical polar coordinates. ,

Ii

a

lz

=i

H

=--

a¢ [Problem 9.28 and Section 9.6 and egn 9.46].

li 2

2J-L

V2

+V

[Further information 10.1]

Since V has no variable in common with Iz• this part of the Hamiltonian and Iz commute. V2

= terms in r only + terms in () only +

I r2

2

-

a22

sin () a¢

,

[Justification 9.7]

a2

The terms in r only and () only necessarily commute with lz(¢ only) . The final term in V2 contains - 2 a¢

which commutes with

a a¢' since an operator necessarily commutes with itself. By symmetry we can


215

deduce that, if H commutes with Iz it must also commute with Ix and Iy since they are related to each other by a simple transformation of coordinates. This proves useful in establishing the commutation of Z2 and H . We form

If H commutes with each of Ix ,Iy, and Iz it must commute with I;, I~ , and I~ . Therefore it also commutes with

72 . Thus H commutes with both f2 and fz.

COMMENT. As described at the end of Section 8.6, the physical properties associated with non-commuting

operators cannot be simultaneously known with precision. However, since H,P., and

/z

commute we may

simultaneously have exact knowledge of the energy, the total orbital angular momentum, and the projection of the orbital angular momentum along an arbitrary axis.

P10.17

With r = (nao/2Z)p and m = -1 , the expectation value is (r

(a)

_I

),,1

=

(nao) 2 r oo 2 2Z p IR"tI dp .

10

(r-t, = G~)2 {2 (~) 3/2}210 = I~ I

(r-I)2S = (c)

.(r-I)

= 2p

because

10

00

P e- P dp [Table 10.1]

00

p e- P dp

= 1.

[§J (a o )2 {_I_ Z 241 /2

= ~ (6) 24ao

(~) 3/2 } 2 10r oo p3 e-Pdp ao

because

10ro oo p3 e-

The general formula for a hydrogenic orbital is

P

dp

[Table 10.1]

= 6.

(r- I ) 1 = ~ . "

n2ao

216 P10.19

STUDENT'S SOLUTI ONS MANUAL

The trajectory is defined, which is not allowed according to quantum mechanics. The angular momentum of a three-dimensional system is given by II (l + I ) }1/ 2 n, not by nn. In the Bohr mode l, the ground state possesses orbital angular momentum (nn, with n = I ), but the actual ground state has no angular momentum (l = 0). Moreover, the di stributi on of the electron is quite different in the two cases. The two model s can be distingui shed experimentally by (a) showing that there is zero orbital angular momentum in the ground state (by examining its magnetic properties) and (b) examining the electron distribution (such as by showing that the electron and the nucleus do come into contact, Chapter 15).

P10.21

Justification 10.4 noted that the transition dipole moment, J.tfi , had to be non-zero for a transi tion to

be allowed. The Justification examined conditions that allowed the z component of this quantity to be non-zero; now examine the x and y components.

As in the Justification, express the relevant Cartesian variables in terms of the spherical harmonics, Y/ ,III'

Start by expressing them in spherical polar coordinates: x =r sin ecosl/>

and

y =r sin esinl/> .

Note that YI , I and YI ,_ I have factors of sin e. They also contain complex exponentials that can be re lated to the sine and cosine of I/> through the identities

These relations motivate us to try linear combinations YI .I + YI ,_ I and YI , I - Yl.-I (from Table 9.3 ; note c here corresponds to the normalization constant in the table): Yl,l + YI ._ I = -csine(e i ¢ +e- i¢) = -2csi n ecosl/> = -2cx/r,

so

x

= -(YI ,I + YI. _ I)r / 2c ;

Yl,l - Yl.-I = c sin e(e i¢ - e- i¢) = 2ic sin e sin I/> = 2icy/ r ,

so

y

= ( YI , I

-

YI ,_ I)r /2 ic.

Now we can ex press the integrals in terms of radial wavefunctions RII.I and spherical harmonics Y/ ,I11/

The angul ar integral can be broken into two, one of which contai ns YI , I and the other YI ,- I . According to the 'triple integral ' relation in Comment 9.6, the integral ll

loo

10211 0

Ytf

Yl,l Y/ j •III /. sin e de dl/>

111/

'

f

'

vanishes unless lr = Ii ± I and mf = mi ± I . The integral that contai ns YI ,_ I introduces no further constraints; it vanishes unless lr = Ii ± I and /1//f = m/ j ± I. Similarly, the y component introduces no

ATOM IC STRUCTURE AND ATOMIC SPECTRA

217

further constraints, for it involves the same spherical harmonics as the x component. The whole set of selection rules, then, is that transitions are allowed on ly if

I t..l = ± 1 and P10.23

t..m/

= 0 or ± 1 I·

(a) The Slater wavefunction [10.32] is 1/Ia(2)a(2) 1/Ia(2),8(2) 1/Ib(2)aC2)

1/I,,(3)a(3)

1/IaCN)a(N)

1/Ia( I),8(l) 1/I"C I )a(1)

1/Ia(3),8C3)

1/I,,(3)aC3)

1/I"CN),8CN) 1/I"CN)aCN)

1/Iz(l),8(1)

1/Iz(2),8C2)

1/Iz(3),8C3)

1/IzCN),8CN)

1/Ia(l)a(1)

1/I(I , 2,3, .. . ,N)

=

1

CN!)1 /2

Interchanging any two columns or rows leaves the function unchanged except for a change in sign. For example, interchanging the first and second columns of the above determinant gives:

-I

1/ICI , 2, 3, . . . , N)

= (N!) 1/2

=-

1/I,,(2)aC2) 1/Ia(2),8C2) 1/I,,(2)aC2)

1/1,,(1)0'(\) 1/1,,(1),8(1) 1/I,,(I)aCl)

1/Ia(3),8C3) 1/I,,(3)a(3)

1/I"CN)a(N) 1/I"CN),8CN) 1/I"CN)a(N)

1/Iz(2),8 (2)

1/Iz(l),8(1)

1/Iz(3),8C3)

1/IzCN),8CN)

1/Ia(3)aC3)

1/I(2, 1, 3, ... , N) .

This demonstrates th at a Slater determinant is antisymmetric under particle exchange. (b) The possibility that 2 electrons occupy the same orbital with the same spin can be explored by making any two rows of the Slater determinant identical, thereby, providing identical orbital and spin functions to two rows. Rows I and 2 are identical in the Slater wavefunction below. Interchanging these two rows causes the sign to change without in any way changing the determinant.

I 1/I( 1,2, 3, ... ,N) = N!I /2

= -1/1 (2,

1/I"Cl)a(l) 1/IaCl)aC l ) 1/I,,(1)aCI)

1/Ia(2)aC2) 1/Ia(2)aC2)

1/1,,(3)0'(3) 1/Ia(3)aC3)

1/I"CN)aCN) 1/I,,(N)a(N)

1/I,,(2)a(2)

1/I,,(3)aC3)

1/1" (N)a (N)

1/Iz(1),8(I)

1/Iz(2),8 (2)

1/Iz(3),8C3)

1/IzCN),8(N)

1, 3, . . . , N)

=

- 1/I(1,2, 3, .. . , N) .

Only the null function satisfies a relationship in which it is the negative of itself so we conclude that, since the null function is inconsi stent with existence, the Slater determinant satisfies the Pauli exclusion principle [Section 10.4 b]. No two electrons can occupy the same orbital with the same spin.

218



The wavenumber of a spectroscopic transition is related to the difference in the relevant energy levels. For a one-electron atom or ion, the relationship is

Solving for V, using the definition Ii = h/2 rt and the fact that Z = 2 for He, yields

Note that the wavenumbers are proportional to the reduced mass, which is very close to the mass of the electron for both isotopes. In order to distinguish between them, we need to carry lots of significant figures in the calculation. _

J.lHe(1.60218 x 1O- 19 C)4 - 2(8.85419 X 10- 12 J-I C2 m- I )2 x (6.62607 x 10 34 J s)3 x (2.99792 x 1010 cm s I)

\! -

X

(~-~) n~ n 2

v/cm- I = 4.81870

X

1035(J.lHe/ kg)

(~ - ~) . n n 2

l

The reduced masses for the 4He and 3He nuclei are m emnuc

J.l=----

me

where

moue

+ moue

= 4.00260 u for 4He and 3.01603 u for 3He,

4He moue = (4.00260 u) x (1.66054

X

10- 27 kg u-

3He moue = (3.01603 u) x (1.66054

X

10- 27 kg

I

or, in kg )

= 6.64648

U- I ) =

X

10- 27 kg,

5.00824 x 1O- 27 kg.

The reduced masses are 4He

27 31 _ (9.10939 x 10- kg) x (6.64648 X 10- kg) 27 31 J.l(9.10939 X 10- + 6.64648 x 10- ) kg

= 9.10814 x

10-31 kg,

3He

27 31 _ (9.10939 x 10- kg) x (5 .00824 X 10- kg) J.l(9. 10939 X 10- 31 + 5.00824 x 10 27) kg

= 9.10773

10-31 kg.

Finally, the wavenumbers for n

= 3 -+

n

X

= 2 are

4Hev

= (4.81870 x

10 35 ) x (9.10814 x 10- 31) x (1/4 - 1/9)cm- 1 = 160957.4 cm- I

I,

3Hev

= (4.81870 x

10 35 ) x (9.10773 x 10- 31) x (l/4-1/9)cm- 1 =160954.7cm-

l·

l


The wavenumbers for n

P10.27

219

= 2 ---+ n = I are ( I l l - 1/ 4)cm -

1

= 1329170cm- 1 I,

1035 ) x (9.10773 x 10- 31) x ( I l l - 1/ 4)cm-

1

= 1329155cm- 1 I·

4Hev

= (4.81870 x 1035 ) x (9. 10814 x 10- 31) x

3Hev

= (4.81870 x

(a) Compute the ratios vSlariv for all three lines. We are given wavelength data, so we can use: VSlar

A

V

ASlar

The ratios are: 438.392 nm 438.882nm

= 0.998884,

440.510 nm 44l.000nm

= 0.998889,

and

441.510nm 442.020nm

= 0.998846.

The frequencies of the stellar lines are all less than those of the stationary lines, so we infer that the star is 1receding 1from earth. The Doppler effect follows:

Vreceding

=

vf

f 2( 1 + sic)

where f

= (I

1-

SIC ) 1/2

= ( --I +slc

, so 1- f 2

s= - -2c.

- sic) ,

I

+f

Our average value off is 0.998873 . (Note: the uncertainty is actually greater than the significant figures here imply, and a more careful analysis would treat uncertainty explicitly.) So the speed of recession with respect to the earth is:

s

=

(I -

0.997747) c 1+0.997747

= 11.128 x .

10- 3 c 1= 13.381 x 105 m s- I

I.

(b) One could compute the star's radial velocity with respect to the sun if one knew the earth's speed with respect to the sun along the sun-star vector at the time of the spectral observation. This could be estimated from quantities available through astronomical observation: the earth's orbital velocity times the cosine of the angle between that velocity vector and the earth-star vector at the time of the spectral observation. (The earth-star direction, which is observable by earth-based astronomers, is practically identical to the sun-star direction, which is technically the direction needed.) Alternatively, repeat the experiment half a year later. At that time, the earth's motion with respect to the sun is approximately equal in magnitude and opposite in direction compared to the original experiment. Averagingf values over the two experiments would yieldf values in which the earth's motion is effectively averaged out. P10.29

See Figure 10.3. Trends: (i) II
2, Cs , 5C~ perpendicular to Cs , O"h , so I DSh I (c) Xe02F2: nonlinear, fewe r than 2Cn with n > 2, C2, no C~ perpendicular to C2, no O"h , 20"v, so I C2v

I

(d) Fe2 (CO)9: nonli near, fewer than 2Cn with n > 2, C3, 3C2 perpend ic ular to C3, O"h , so D3h (e) cubane (CgHg): nonlinear, more than 2CII with n > 2, i, no Cs, so@;] (f) tetrafl uoroc ubane (23): nonlinear, more than 2CII with n > 2, no i, so~.

I

I

248


E12.11(b) (a) Only molecules belonging to Cs , Cn , and C Il V groups may be polar. In Exercise 12.9(b)

Iarrha-dichlorobenzene Iand Imeta-dichlorobenzene Ibelong to C2v and so may be polar; in Exercise 12.6(b), IHF and Xe02F21 belong to C groups, so they may be polar. IlV

(b) A molecule cannot be chiral if it has an axis of improper rotation - including disguised or degenerate axes such as an inversion centre (S2) or a mirror plane (S , ). In Exercises 12.S(b) and 12.6(b), all the molecules have mirror planes, so I none Ican be chiral. E12.12(b) In order to have nonzero overlap with a combination of orbitals that spans E, an orbital on the central atom must itself have some E character, for only E can multiply E to give an overlap integral with a totally

symmetric part. A glance at the character table shows that IPx and Py Iorbitals available to a bonding N atom have the proper symmetry. If d orbitals are available (as in S03), I all d orbitals except dZ2 1cou ld have nonzero overlap. E12.13(b) The product rf x r(J1.) x Ii must contain Al (Example 12.7). Then, since Ii = BI , r(J1.) = r (y) = B2 (C2v character table), we can draw up the following table of characters

E B2 BI BIB2

C2

frv

a v'

-I

-I I -I

I -I -I =A2

-1

Hence, the upper state is I A21, because A2 x A2 = A I . E12.14(b) (a) Anthracene

H

H

H

:@: H

H

D.

H

The components of J1. span B3u(X), B2u(Y) , and Blu(z). The totally symmetric ground state is Ag . Since Ag x r

I

I

= r in this group, the accessible upper terms are B3u (x-polarized), ~ (y-

polarized), and ~ (z-polarized). (b) Coronene, like benzene, belongs to the D6h group. The integrand of the transition dipole moment must be or contain the Ai g symmetry species. That integrand for transitions from the ground state is AIgqf, where q is x,y, or z and! is the symmetry species of the upper state. Since the ground state is already totally symmetric, the product qf must also have Ai g symmetry for the entire integrand to have Aig symmetry. Since the different symmetry species are orthogonal , the only way qfcan have Ai g symmetry is if q and! have the same symmetry. Such combinations include

I

ZA2u, xElu , and yElu , Therefore, we conclude thattransitions are allowed to states with A2u or Elu symmetry.

I

MOLECULAR SYMMETRY

249

E12.15(b)

E AI A2 E sine

2C3

3uv

- I

I

-1 Linear combinations of sin e and cos e

2

case Product

0 -I -I

The product does not contain AI , so I yes Ithe integral vanishes.

Solutions to problems P12.1

I

I

(a) Staggered CH3CH3 : E, C3, C2, 3Ud ; D3d [see Fig . 12.6(b) of the text].

I

(b) Chair C6H1 2: E , C3, C2, 3Ud ; D3d I· Boat C6 H 12: E, C2, u V , (c) B2H6: E , C2,

2C~ ,

Uh ;

u~; I C2v I·

ID2h I·

(d) [Co(enh] 3+ : E, 2C3, 3C2; I D3 1·

I

(e) Crown Ss : E, C4 , C2, 4C~ , 4Ud, 2Ss ; D4d I· Only boat C6H 12 may be polar, since all the others are D point groups. Only [Co (enh ]3+ belongs to a group without an improper rotation axis (SI = u) , and hence is chiral. P12.3

Consider Fig. 12.2. The effect of Uh on a point P is to generate UhP , and the effect of C2 on UhP is to generate the point C2UhP. The same point is generated from P by the inversion i, so C2UhP = iP for all points P. Hence, 1CZUh = i I, and i must be a member of the group. p

Figure 12.2

P12.5

We examine how the operations of the C3v group affect lz = XPy - YPx when applied to it. The transformations of x,y, and z, and by analogy Px, Py, and pz, are as follow s (see Fig. 12.3)

250


E(x,y,z) -+ (x,y,z), av(x, y,z)

-+

(-x,y,z),

a~(x,y,z) -+

(x, - y,z),

a;(x ,y, z) -+ (x ,y, - z), cj(x,y,z) -+ (-!x + !J3Y, -!5x - !y,z) , C3(X,y,Z) -+ (-!X - !J3Y,!5x - !y, z) .

y

Figure 12.3

MOLECULAR SYMMETRY

251

The characters of all a operations are the same, as are those of both C3 operations (see the C3v character table); hence we need consider only one operation in each class. El z = XPy - YPx = 'z, avl z

= -XPy + YPx = -Iz

[(x,y,z) ---+ (-x,y,z)],

Ct lz = ( -!x + !.J3y ) x (-!.J3Px - !py) - (-!.J3x - !y) x (-!Px [(x, y , z) ---+ (-!x

= ~ (.J3xPr + XPy = xPv -

YPx

+ !.J3y , -

+ !.J3py)

!.J3x - !y, z) ]

3ypx - .J3ypy - .J3xPx

+ 3xpy -

YPx

+ .J3ypy )

= lz·

ct

The representatives of E , a v, and are therefore all one-dimensional matrices with characters I, -I , I respectively. It follows that I z is a basis for A2 (see the C3v character table). P12.7

The multiplication table is

ay

ax ax ax ay az

ax ay az

-iaz iay

az az -iay iax

ay jax -iaz

The matrices 1 do not form a group 1 since the products iaz , jay, jax and their negatives are not among the four given matrices. P12.9

(a) In C3v symmetry the His orbitals span the same irreducible representations as in NH3, which is Al + AI + E. There is an additional Al orbital because a fourth H atom lies on the C3 axis. In C3v, the d orbitals span A I + E + E [see the final column of the C3v character table]. Therefore, all five d orbitals may contribute to the bonding.

I

I

(b) In C2v symmetry the His orbitals span the same irreducible representations as in H20 , but one 'H20' fragment is rotated by 90° with respect to the other. Therefore, whereas in H20 the HIs orbitals span AI + B2 [HI + H2, HI - H2], in the distorted C14 molecule they span Al + B2 + Al + BI [HI + H2 , HI - H2, H3 + H4 , H3 -14]. In C2v the d orbitals span 2AI + BI + B2 + A2 [C2v

I

I

character table] ; therefore, all except A2(dxy ) may participate in bonding. P12.11

(a) We work through the flow diagram in the text (Fig. 12.4). We note that this complex with freely rotating CF3 groups is not linear, it has no Cn axes with n > 2, but it does have C2 axes; in fact it has two C2 axes perpendicular to whichever C2 we call principal, and it has a ah. Therefore, the point group is 1D2h I· (b) The plane shown in Fig. 12.4 below is a mirror plane so long as each of the CF3 groups has a CF bond in the plane. (i) If the CF3 groups are staggered, then the Ag-CN axis is still a C2 axis; however, there are no other C2 axes. The Ag-CF3 axis is an S2 axis, though, which means that the Ag atom

252


is at an inversion center. Continuing with the flow diagram, there is a Uh (the plane shown in the figure). So the point group is I C2h I. (ii) If the CF3 groups are eclipsed, then the axis through the Ag and perpendicular to the plane of the Ag bonds is still a C2 axis ; however, neither of the Ag bond axes is a C2 axis. There is no Uh but there are two Uv planes (the plane shown and the plane perpendicular to it and the Ag bond plane) . So the point group is I C2v I.

NC

Figure 12.4

P12.13

i,

(a) C2v. The functions x 2 , and Z2 are invariant under all operations of the group, and so z(5 z2 - 3r 2) transforms as Z(AI) , y(5i - 3r2) as y (B2) , x(5x 2 - 3r2) as x(B I), and likewise for z(x 2 - i ), y(x2 - Z2 ), and x(z2 - i ) . The function xyz transforms as BI x B2 X AI = A2 . Therefore, in group C2v, f ~ 12A I + A2 +2BI + 2B21.

i,

(b) C3v. In C3v, z transforms as AI , and hence so does z3. From the C3v character table, (x 2 xy) is a basis for E, and so (xyz, z(x 2 - i)) is a basis for AI x E = E. The linear combinations y(5i - 3r2) + 5y (x2 - Z2) <X Y and x(5x 2 - 3r2) + 5x(Z2 - y2) <X X are a basis for E. Likewise, the two linear combinations orthogonal to these are another basis for E. Hence, in the group C3v, f ~ IAI +3EI· (e) Td. Make the inspired guess that thef orbitals are a basis of dimension 3 + 3 + I, suggesting the decomposition T + T + A. Is the A representation A I or A2? We see from the character table that the effect of S4 discriminates between AI and A2. Under S4, x ~ y, y ~ - x, Z ~ - z, and so xyz ~ xyz. The character is X = I, and so xyz spans A I. Likewise, (x 3, y3, z3) ~ (y3, - x 3, - z3) and X = 0 + 0 - I = -I . Hence, this trio spans T 2. Finally,

resulting in X

= I, indicating T I· Therefore, in Td , f ~ 1A I + T I + T 21·

(d) Oh . Anticipate an A + T + T decomposition as in the other cubic group. Since x ,y, and z all have odd parity, all the irreducible representatives will be u. Under S4, xyz ~ xyz (as in (e)), and so the representation is X = -I (see the character table). Under S4, (x 3,y3, Z3 ) ~ (y3, _ x 3, - Z3) , as before, and X = - I, indicating Tlu . In the same way, the remaining three function s span T 2u. Hence, in Oh , f ~ I A2u + Tlu + T2u I· (The shapes of the orbitals are shown in Inorganic Chemistry, 3rd edn, D. F. Shriver, and P. W. Atkins, Oxford University Press and W. H. Freeman & Co (1999).) The f orbitals will cluster into sets according to their irreducible representations. Thus (a) f ~ A I + T I + T 2 in Td symmetry, and there is one nondegenerate orbital and two sets of triply degenerate orbitals. (b) f ~ A2u + Tlu + T2u, and the pattern of splitting (but not the order of energies) is the same.

MOLECULAR SYMMETRY

P12.15

253

We begin by drawing up the fo llowing table.

E C2

av a v'

N2s

N2px

N2py

N2pz

02px

02py

02pz

O' 2px

O' 2py

O' 2pz

X

N2s N2s N2s N2s

N2px - N2px N21Jx -N2px

N2py -N2py -N2py N2py

N2pz N2pz N2pz N2pz

0 2px -O' 2px O'2px -02p"

0 2py -O'2py -O'2py 02py

02p: O' 2pz O' 2pz 02pz

O' 2px -02px 02px -O'2p.t

O' 2py -02py -02py O'2py

O' 2p: 02pz 02p: O' 2pz

10 0

The character set (10, 0, 2, 4) decomposes into 14AI + 2BI adapted linear combinations as described in Section 12.5. l{f(A I) l{f(A I) l{f(A I) l{f(AI) l{f(BI)

= = = = =

(column (column (column (column (column

N2s N2p: 02p: + O' 2pz 02py + O' 2py N2p.t

I) 4) 7) 9) 2)

l{f( BI ) l{f(B2) l{f(B2) l{f(A2) l{f(A2)

= = = = =

+ 3 B2 + A21. We then form

02px + O' 2px N2py 02py + O' 2py 0 2p: + O'2P: 02px + O' 2IJx

2 4

sy mmetry-

(column 5) (column 3) (colum n 6) (column 7) (colu mn 5)

(The other columns yield the same combinations.) P12.17

Consider phenanthrene with carbon atoms as labeled in the stucture below. a

a'

e

f'

e'

(a ) The 2p orbitals involved in the 7T system are the basis we are interested in. To find the irreproducible representat ions spanned by thi s basis, consider how each basis is transformed under the sy mmetry operations of the C2v group. To find the character of an operation in this basis, sum the coefficients of the basis terms th at are unchanged by the operation.

E C2

av a v'

a

a'

b

b'

c

a -a' a' - a

a' -a a -a'

b - b' b' - b

b' - b b -b'

c -c' c' -c

d

d'

e

e'

f

f'

d -d' d' -d

d'

e - e' e' -e

e' -e

f -f' f' -f

f'

c' c' -c c -c'

-d d -d'

e - e'

-f f -f'

0

'" 0

g'

X

0'

14

'" - 0'

-g'"

0'

g

to

-

'"0 to

-0'

'"

0 0 -1 4

To find the irreproducible representations that these orbitals span, multiply the characters in the representation of the orbitals by the characters of the irreproducible representations, sum those products, and divide the sum by the order h of the group (as in Section 12.5(a». The table below illustrates the procedure, beginning at left with the C2v character tab le.

E AI A2 BI B2

C2

-I - I

av

a vI

-I

- I -I I

I -1

product

The orbitals span 17 A2 + B21.

E

C2

av

a v'

sumlh

14 14 14 14

0 0 0 0

0 0 0 0

-14 14 14 -14

0 7 7 0

254


To find symmetry-adapted linear combinations (SALCs), follow the procedure described in Section 12.5(c). Refer to the table above that displays the transformations of the original basis orbitals. To find SALCs of a given symmetry species, take a column of the table, multiply each entry by the character of the species' irreproducible representation, sum the terms in the column, and divide by the order of the group. For example, the characters of species A I are I, I, I, I, so the columns to be summed are identical to the columns in the table above. Each column sums to zero, so we conclude that there are no SALCs of A I symmetry. (No surprise here : the orbitals span only A2 and B I .) An A2 SALC is obtained by multiplying the characters I, I, -I , - I by the first column : I( , , ) I ( - a. ') 4a-a-a+a=2a

The A2 combination from the second column is the same. There are seven distinct A2 combinations in all : 1!(a - a' ), !(b - b'), . . . , !(g - g') I. The BI combination from the first column is ±(a + a' + a' + a) = !(a + a'). The B 1 combination from the second column is the same. There are seven distinct B I combinations in all: 1!(a + a'), !(b + b' ), ... , !(g + g' ) I. There are no B2 combinations, as the columns sum to zero. (b) The structure is labeled to match the row and column numbers shown in the determinant. The HUckel secular determinant of phenanthrene is:

a

b

c

d

g

e

g'

f'

e'

d'

c'

b'

a'

a

0

0

0

0

0

0

0

0

fJ

b

0

f3

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0 0 a-£ fJ 0 0 a-£ fJ 0 fJ a-£ 0 c 0 fJ f3 a - £ fJ d 0 0 fJ a-£ 0 0 e 0 f3 0 0 0 0 f3 0 0 0 g 0 f3

f3 0 0 a-£ f3 0 a-£ f3 f3 a-£ f3 0 f3 a- £ 0 0 f3

,/

0

0

0

0

0

f'

0

0

0

0

0

e'

0

0

0

0

0

0

0

0

fJ

0

0

0

0

0

0

0

0

0

0

0

0

0

0

'"

d'

c'

0

0

0

0

b'

0

0

0

0

0

0

0

f3

0

a'

fJ

0

0

0

0

0

0

0

0

0 0 0 f3 0 0 a -£ f3 0 a-£ f3 f3 a - £ f3 0 f3 a-£ 0 0 f3 0 0 0 f3

0 0 0 0 0

fJ a- £

This determinant has the same eigenvalues as in exercise 11 .13b(b). (c) The ground state of the molecule has A l symmetry by virtue of the fact that its wavefunction is the product of doubly occupied orbitals, and the product of any two orbitals of the same symmetry has A I character. If a transition is to be allowed, the transition dipole must be non-zero, which in tum can JL \IIj includes the totally symmetric species A I · only happen if the representation of the product Consider first transitions to another A I wavefunction, in which case we need the product A I JLA I .

\II;

MOLECULAR SYMMETRY

255

Now AlAI = AI, and the only character that returns AI when mUltiplied by AI is AI itself. The +-- A I transitions are allowed. (Note: transitions from the AI ground state to an AI excited state are transitions from an orbital occupied in the ground state to an excited-state orbital of the same symmetry.) The other possibility is a transition from an orbital of one symmetry (A2 or B I) to the other; in that case, the excited-state wavefunction will have symmetry of A 18 I = 8 2 from the two singly occupied orbitals in the excited state. The symmetry of the transition dipole, then, is A 1P.82 = P.B2, and the only species that yields A I when multiplied by B2 is B2 itself. Now the y component of the dipole operator belongs to species 8 2, so these transitions are also allowed (y-polarized).

z component of the dipole operator belongs to species A I, so z-polarized A I


The shape of this molecule is shown in Fig. 12.5. A

Q

OO

60

0

Figure 12.5

COB

(a) Symmetry elements 1E, 2C3, 3C2, ah , 253, 3av I· Point group 1 D3h I· I (b)

D (E)

=

D (C3)

(

=

0

0

1

o

0

(

0

0

I

0 1

o

5;

C~ and are counter clockwise rotations. a v is through A and perpendicular to 8-C a~ is through 8 and perpendicular to A-C. a~' is through C and perpendicular to A-B.

D(av )

=

0 0

G D· = G ~) 1

D(a~/)

0 0

~ G0I

0

D (" ;)

~).

256


(c) Example of elements of group multiplication table

D(C,)D(C,)

D(a; )D(a. )

~ G~

DG~ 1)

~ G~

D~

~

D(a:).

G ~)(: 0

0

~G

0 0

D~

0 0

D

D(c;).

D3h

E

C3

C2

ay

a y'

ah

E C3 C2 ay a y' ah

E C3 C2 ay a y' ah

C3 C'3 a y' a y' a y"

C2 a" y E £ C3 C2

ay a" y E £ C3 ay

a y' ay C3 C3 £ a y'

ah C3 C2 ay , ay £

C3

(d) First, determine the number of s orbitals (the basis has three s orbitals) that have unchanged positions after application of each symmetry species of the D3h point group.

Unchanged basis members

3

o

3

o

This is not one of the irreducible representations reported in the D3h character table but inspection shows that it is identical to A') + £ ' . This allows us to conclude that the three s orbitals span 1A;

+ £ ' I.

COMMENT. The multiplication table in part (c) is not strictly speaking the group multiplication ; it is instead

the multiplication table for the matrix representations of the group in the basis under consideration.

P12.21

(a) Following the flow chart in Fig. 12.4 of the text, note that the molecule is not linear (at least not in the mathematical sense); there is only one Cn axis (a C2), and there is a ah . The point group, then,

isl C2h I.

MOLECULAR SYMM ETR Y

b

d

~

a

.0

0

c

e

f

h

o

k'

oj

j'

k

g

e'

g'

j'

00.0 h'

c'

E C2

ail

a

a'

b

b'

c

c'

a a' , - a -a

a' a -a -a'

b b -b' -b

b' b -b -b'

c c' -c' -c

c' c -c -c'

a'

.0

.0

~

f'

d'

b'

(b) The 2pz orbitals are transformed under the symmetry operations of the

j' - j' -j

C2h

group as follows .

j'

k

k'

j'

k k' -k' -k

k' k -k -k'

-j' -j'

257

X

22 0 0 -22.

To find the irreproducible representations that these orbitals span, we multiply the characters of orbitals by the characters of the irreproducible representations, sum those products, and divide the sum by the order h of the group (as in Section 12.5(a». The table below illustrates the procedure, beginning at left with the C2h character table. E

C2

Ag All Bg BII

ah

-I I

-1 -1

product

-I - I

-1

E

C2

22 22 22 22

0 0 0 0

ah

0 0 0 0

-22 22 22 -22

sum / h 0 II II 0

The orbitals span I II Au + II Bg I. To find symmetry-adapted linear combinations (SALes), follow the procedure described in Section 12.5(c) . Refer to the table above that displays the transformations of the original basis orbitals. To find SALes of a given symmetry species, take a column of the table, multiply each entry by the character of the species ' irreproducible representation, sum the terms in the column, and divide by the order of the group. For example, the characters of species Au are I, I, I, I, so the columns to be summed are identical to the columns in the table above. Each column sums to zero, so we conclude that there are no SALes of Ag symmetry. (No surprise: the orbitals span only Au and Bg.) An Au SALe is obtained by multiplying the characters 1, I, -1 , -I by the first column: ±(a

+ a' + a' + a)

= !(a + a').

The Au combination from the second column is the same. There are I I distinct Au combinations in

!

!

all : 1 (a + a'), (b + b') , .. . ! (k 4I ( a - a, - a,

+ a)

+ k' ) I. The Bg combination from the first column is

I ( = 2"a -a ') .

The Bg combination from the second column is the same. There are I I distinct Bg combinations

!

!

in all: 1 (a - a'), (b - b'), ... !(k - k') I. There are no Bu combinations, as the columns sum to zero.

258

STUDEN T'S SOLUTIONS MANUAL

(C) The structure is labeled to match the row and column numbers shown in the determinant. The HUckel

secular determinant is: a a b

c

b

c

a-E fJ 0 a-E fJ fJ 0 a- E fJ 0

0

0

0

0

0

k

0

0

0

k

k'

j'

I

'1

c'

b'

a'

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0 0

.. . a-E fJ 0 0 0 0 a-E 0 0 0 fJ fJ ... 0 a - E fJ 0 0 fJ 0 0 a - E fJ 0 fJ 0 0 0 a-E fJ fJ .. . 0 0 0 0 a- / fJ

k'

0

0

0

j'

0

0

0

'1

0

0

0

c'

0

0

0

0

0

0

0

0

0

b'

0

0

0

0

0

0

0

0

0

a'

0

0

0

0

0

0

0

0

0

I

.. .

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

a - E fJ 0 a - E fJ fJ 0 a- E fJ

The energies of the fi lled orbitals are a + 1.98137,8 , a + 1.92583,8, a + 1.83442,8, a + 1.70884,8, + 1.55142,8 , a + 1.36511,8, a + 1.15336,8, a + 0.92013,8 , a + 0.66976,8 , a + 0.40691,8 , and a + 0.1 3648,8. The 7r energy is 27.30729,8 .

a

(d ) The ground state of the molecule has Ag symmetry by virtue of the fac t that its wavefunction is the product of doubly occupied orbitals, and the product of any two orbitals of the same symmetry has Ag character. If a transition is to be allowed, the transition dipole must be non-zero, which in tum can only happen if the representation of the product Ilit ILllij includes the totally symmetric species Ag. Consider first transitions to another Ag wavefunction, in which case we need the product AgILAg. Now AgAg = A g, and the only character that returns Ag when mUltiplied by Ag is Ag itself. No component of the dipole operator belongs to species A g, so no Ag __ Ag transitions are allowed. (Note: such transitions are transitions from an orbital occupied in the ground state to an excitedstate orbital of the same symmetry.) The other possibility is a transition from an orbital of one symmetry (Au or Bg) to the other; in that case, the excited-state wavefunction will have symmetry of AuBg = Bu from the two singly occupied orbitals in the excited state. The symmetry of the transition dipole, then, is A glLBu = ILBu , and the onl y species that yields Ag when multiplied by Bu is Bu itself. The x and y components of the di pole operator belongs to species B u, so these transitions are allowed.

13

Molecular spectroscopy 1. rotational and vibrational spectra


(1) Doppler broadening. This contribution to the linewidth is due to the Doppler effect, which shifts the frequency of the radiation emitted or absorbed when the atoms or molecules involved are moving towards or away from the detecting device. Molecules have a wide range of speeds in all directions in a gas and the detected spectral line is the absorption or emission profile arising from all the resulting Doppler shifts. As shown in Justification 13.3, the profile reflects the distribution of molecular velocities parallel to the line of sight which is a bell-shaped Gaussian curve.

(2) Lifetime broadening. The Doppler broadening is significant in gas phase samples, but lifetime broadening occurs in all states of matter. This kind of broadening is a quantum mechanical effect related to the uncertainty principle in the form of eqn 13.18 and is due to the finite lifetimes of the states involved in the transition. When r is finite, the energy of the states is smeared out and hence the transition frequency is broadened as shown in eqn 13.19.

(3) Pressure broadening or collisional broadening. The actual mechanism affecting the lifetime of energy states depends on various processes, one of which is collisional deactivation and another of which is spontaneous emission. Lowering the pressure can reduce the first of these contributions; the second cannot be changed and results in a natural linewidth. 013.3

(1) Rotational Raman spectroscopy. The gross selection rule is that the molecule must be anisotropically polarizable, which is to say that its polarizability, a, depends upon the direction of the electric field relative to the molecule. Non-spherical rotors satisfy this condition. Therefore, linear and symmetric rotors are rotationally Raman active.

(2) Vibrational Raman spectroscopy. The gross selection rule is that the polarizability of the molecule must change as the molecule vibrates. All diatomic molecules satisfy this condition as the molecules swell and contract during a vibration, the control of the nuclei over the electrons varies, and the molecular polarizability changes. Hence both homonuclear and heteronuclear diatomics are vibrationally Raman active. In polyatomic molecules it is usually quite difficult to judge by inspection whether or not the molecule is anisotropically polarizable; hence group theoretical methods are relied on for judging the Raman activity of the various normal modes of vibration. The procedure is discussed in Section 13. 17(b) and demonstrated in Illustration 13.6. 013.5

The exclusion rule applies to the benzene molecule because it has a center of symmetry. Consequently, none of the normal modes of vibration of benzene can be both infrared and Raman active. If we wish to characterize all the normal modes we must obtain both kinds of spectra. See the solutions to Exercises 13.2S(a) and 13.2S(b) for specific illustrations of which modes are IR active and which are Raman active.

260


Solutions to exercises E13.1 (b)

The ratio of coefficients Al B is (a)

A

8JThv3

B

c3

x 1O- J s) x (500 x 10 S-I) 3 I -8JT(6.626 --'----,..--:--,--,------'--,;---'----;--;;---'-= 7.73 8 34

6

(2.998 x 10 m s- I ) 3 .

10- 32 J m-3 s

X

(b) The frequency is 34 c A 8JT h 8JT (6.626 X 10- J s) 1 v = - so - = - 3 = = 6.2 x 10 2 A B A (3 .0 x 10- m) 3

E13.2(b)

28

Jm

- 3

s

1

A source approaching an observer appears to be emitting light of frequency Vapproaching

v

= - -s

.

[13.15 , Section 13.3]

1- c 1 Since vex -, AObs = (1 - sic) A A

For the light to appear green the speed would have to be

s=

(I -

AObS) c = (2.998 x 108 m s- I) x (1 - 520 nm) = 16.36 x 107 m A 660 nm

S-I

or about 1.4 x 108 m.p.h. (Since s

~

Vb o s -

c, the relativistic expression 1 + (5 I C»)1 /2 ( 1 _ (s ic) v

should really be used. It gives s E13.3(b)

= 7.02 X

107 m

1 5- .)

The linewidth is related to the lifetime r by

oiJ =

5.31 cm-

I

[13.19] so r

r i ps

5.31 cm-

o-v

=

I

ps

(a) We are given a frequency rather than a wavenumber

(5 .31 cm-

iJ = vic so r = or 11.59 ns

E13.4(b)

=

x (2.998 x 10 100 x 106 S- I

)

I I

(b) r

I

5.31 cm2.14 cm- I ps

=

I

2.48 ps

1

The linewidth is related to the lifetime r by

oiJ =

5.31 cm- I r i ps

(5 .31 em- I)c so 0 v = --'-----,-------'r i ps

10

em

S- I)

s P

= 1.59 X

103 s P

1

I .

261

MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA

(a) If every collision is effective, then the lifetime is 1/( 1.0 x 109 S- I) -

8v

10 I (5.31 cm- ) x (2.998 x 10 cm s-I) 3 1.0 x 10

=

= 1.6 x

i

08 - I s

= 1.0 x

10-9 s

= 1.0 x

103 pS

= I 160 MHz I

(b) If only one collision in 10 is effective, then the lifetime is a factor of 10 greater, 1.0 x 104 ps _

8v E13.5(b)

=

(5.31 cm- I) x (2.998 x 10 10 cm S-I) 1.0 x 104

= 1. 6 x

0 -I i 7s

= I16 MHz I

The frequency of the transition is related to the rotational constant by hv

= f+..E = hef+..F = heB[J(J + I) -

where J refers to the upper state (J

(J - I)J]

= 2heBJ

= 3). The rotational constant is related to molecular structure by

Ii Ii - 4ncl - 4n emeffR2

B-------_=_ where I is moment of inertia, meff is effective mass, and R is the bond length. Putting these expressions together yields v

liJ

= 2eBJ = 2nmeffR 2

The reciprocal of the effective mass is

= m- I + m- I =

m- I

C

eff

0

(12 U)-I + (l5.9949u) - 1 = 8.78348 x 1025 k - I 1.66054 x lO- 27 kgu- 1 g

_I

So v _ (8.78348 x 1025 kg-I) x (1.0546 x 10- 34 J s) x (3) 2n (1128 . I x 10 - 12 m) 2 E13.6(b)

_I

11

3.4754 x 10

s

1

(a) The wavenumber of the transition is related to the rotational constant by

hev

= f+..E = hef+..F = hcB[J(J + 1) -

where J refers to the upper state (J

(J - I)J]

= 2heBJ

= 1). The rotational constant is related to molecular structure by

Ii B=-4ncl

where I is moment of inertia. Putting these expressions together yields v

I

= 2BJ = ~

so 1= liJ

2ncl

ev

= 13.307 x

=

34 (1.0546 x 10- J s) x (I) 2n(2.998 x 1010 cm s I) x (16.93 cm- I )

10-47 kg m2 1

(b) The moment of inertia is related to the bond length by I

= meffR 2 so R = ( - I

meff

) 1/2

262

STUDENT'S SO LUTIONS MANUAL

m

-I

eff

-I

and R

x 1026 kg-I) x (3.307

= {(6.0494 =

E13.7(b)

(1.0078 U) - I + (80.9163 U) - I 1.66054 x 10-27 kg u I

- I

= mH + mBf =

1.414 x 1O-

lO

= 6.0494 x

10

26

kg-

I

10- 47 kg m 2)} 1/ 2

X

= 1141.4 pm 1

m

The wavenumber of the transition is related to the rotational constant by

= /' 2?) leads to cubic and icosahedral groups and therefore spherical rotors. Lf the molecule is not a spherical rotor, yes at the next question leads to symmetric rotors if the highest CII has n > 2; if not, the molecule is an asymmetric rotor. (3) CH4 : not linear, but more than two CII (/1 > 2), so 1spherical rotor

(b) CH3CN: not linear, C3 (only one of them), so 1symmetric rotor I. (c) C02: linear, so I linear rotor I.

I.

276


(d) CH30H: not linear, no CIl, so I asymmetric rotor I. (e) Benzene: not linear, C6 , but only one high-order axis, so I symmetric rotor

I.

(0 Pyridine: not linear, C2 , is highest rotational axi s, so I asymmetric rotor I. P13.25

S(v, J)

= (v +

6.S?

6.SJ

Dv

+ BJ(J + I ) [13 .61].

= v - 2B(21 = v + 2B(21 +

= I , M = -2] . 3) [6.v = I , M = +2]. I ) [6.v

The transition of maximum intensity corresponds, approximately, to the transition with the most probable value of J, which was calculated in Problem 13.24,

kT )1 /2

Jrnax

I

= ( 2hcB

2

The peak-to-peak separation is then 6.S

= 6.SLx - 6.S~,ax = 2B(2Jrnax +3) -

(-2B(21rnax - I)}

= 8B (Jrnax +!)

= 8B (~) 1/ 2 = (32BkT) 1/ 2 2hcB

hc

To analyze the data we rearrange the relation to

hC(6.S)2 B=--32kT Ii and convert to a bond length using B = - - , with I = 2mxR2 (Table 13.1) for a linear rotor. This gives 4J[c/ Ii

)1 / 2

R = ( 8J[cmx B

(

=

I ) (2kT)I /2 J[c6.S x -;;;;

We can now draw up the following table

T/K

mx / u 6.S/cm- 1 R/pm

HgCl 2

HgBr2

HgI2

555 35.45 23 .8 227.6

565 79.1 15.2 240.7

565 126.90 11.4 253.4

Hence, the three bond lengths are approximately 1230, 240, and 250 pm

I.


ctroscopy for studying the 0-0 ,.-'-::..:..:..:.:::..::..:.cc=.....:..L.C..,-,-! stretching mode because such a mode would be infrared inactive or at best only weakly active.


277

(The mode is sure to be inactive in free 02, because it would not change the molecule's dipole moment. In a complex in which 02 is bound, the 0-0 stretch may change the dipole moment, but it is not certain to do so at all, let alone strongly enough to provide a good signal .) (b) The vibrational wavenumber is proportional to the frequency, and it depends on the effective mass as follows,

_ (k) 1/ 2 -,

vex

so

meff

and JJe802) = (0.943)(844cm- l ) = I 796cm - 1 I. Note the assumption that the effective masses are proportional to the isotopic masses. This assumption is valid in the free molecule, where the effective mass of 02 is equal to half the mass of the 0 atom; it is also valid if the 0 2 is strongly bound at one end, such that one atom is free and the other is essentially fixed to a very massive unit. (c) The vibrational wavenumber is proportional to the square root of the force constant. The force constant is itself a measure of the strength of the bond (technically of its stiffness, which correlates with strength), which in tum is characterized by bond order. Simple molecular orbital analysis of 0 2, O2, and O~- results in bond orders ofl2, 1.5, and I respectively I. Given decreasing bond order, one would expect decreasing vibrational wavenumbers (and vice versa). (d) The wavenumber of the 0-0 stretch is very similar to that of the peroxide anion, suggesting

IFe~+o~- I·

(e) The detection of two bands due to 16 0 18 0 implies that the two 0 atoms occupy non-equivalent positions in the complex. Structures 7 and 8 are consistent with this observation, but structures 5 and 6 are not. P13.29

According to Problem 1O.27(a), the Doppler effect obeys I - SIC) 1/ 2

Vrecedin g

= vf where f = ( - - I + sic

This can be rearranged to yield 1- f2 s= - -2c, I +f

We are given wavelength data, so we use

f=

VSlar V

=~ . Aslar

The ratio is: 654.2 nm = 09260 706.5 nm . ,

f

=

S

= I

2

so

I

I

I - 0.9260 + 0.92602 C = 0.0768c = 2.30 x 107 m s - 1.

278


The broadening of the line is due to local events (collisions) in the distant star. It is temperature dependent and hence yields the surface temperature of the star. Eqn 13.17 relates the observed Iinewidth to temperature: OAobs

= 2A (2kTln2)1 / 2 so

T =

(C 0,A )2

m

C

m

2kln2'

2A

T = ((2.998 x 10 ms- )(61.8 x 1O- 12 m)2 [(47.95U)(1.661 x lO- 27 kg U- I )] 2(654.2 x 10- 9 ) 2(1.381 x 1O- 23 JK- I )ln2 ' 8

T = 18.34

P13.31

EJ

X

105 K

I.

= J(J + l)hcB ,

EI - Eo

I

gJ

= 2hcB = hc

= 2J + 1.

(_1___1_) . I) =:21( 1 Ashoner

1

B_l(

-:2

Ashorter -

Alonger

Alonger

I( -1- ) x (

--

Ashoner -

1 1

1

+ (/':;.A/Ashorter)

1) +

Ashoner

/':;.A

)

2

A shorter

1(

1 ) ( 1 ) ( 10 nm) 387.5 nm x 1 - 1 + (0.061 /3 87.5) x 10 2 cm '

9

= 2" B

= 12.031 cm- i I. 2(6.626

2hcB

X

10- 34 J S) x (3.00

1010 cm S-I) x (2.031 cm- I ) 1.381 x 10- 23 J K- I

k

X

= 5.847K. Intensity of J f +--- J absorption line IJ ex gJe - E} l kT

Solve for T

T

P13.33

EI - Eo) =(-

k

x (

1 In(glhshonerlgohlonger)

)

- ( = 5.847K

I In(3 x 4)

)

= I2.35 K.I

Temperature effects. At extremely low temperatures (10K) only the lowest rotational states are populated. No emission spectrum is expected for the cloud and star light microwave absorptions by the cloud are by the lowest rotational states. At higher temperatures additional high-energy lines appear because higher energy rotational states are populated. Circumstellar clouds may exhibit infrared absorptions due to vibrational excitation as well as electronic transitions in the ultraviolet. Ultraviolet absorptions may indicate the photodissocation of carbon monoxide. High temperature clouds exhibit emissions. Density effects. The density of an interstellar cloud may range from one particle to a billion particles per cm 3. This is still very much a vacuum compared to the laboratory high vacuum of a trillion particles


279

per cm 3 . Under such extreme vacuum conditions the half-life of any quantum state is expected to be extremely long and absorption lines should be very narrow. At the higher densities the vast size of nebulae obscures distant stars. High densities and high temperatures may create conditions in which emissions stimulate emissions of the same wavelength by molecules. A cascade of stimulated emissions greatly amplifies normally weak lines-the maser phenomena of Microwave Amplification by Stimulated Emission of Radiation. Particle velocity effects. Particle velocity can cause Doppler broadening of spectral lines. The effect is extremely small for interstellar clouds at 10K but is appreciable for clouds near high temperature stars. Outflows of gas from pulsing stars exhibit a red Doppler shift when moving away at high speed and a blue shift when moving toward us. There will be many more transitions observable in circumstellar gas than in interstellar gas, because many more rotational states will be accessible at the higher temperatures. Higher velocity and density of particles in circumstellar material can be expected to broaden spectral lines compared to those of interstellar material by shortening collisional lifetimes. (Doppler broadening is not likely to be significantly different between circumstellar and interstellar material in the same astronomical neighborhood. The relativistic speeds involved are due to large-scale motions of the expanding universe, compared to which local thermal variations are insignificant.) A temperature of 1000 K is not high enough to significantly populate electronically excited states of CO; such states would have different bond lengths, thereby producing transitions with different rotational constants. Excited vibrational states would be accessible, though, and rotational-vibrational transitions with P and R branches as detailed later in this chapter would be observable in circumstellar but not interstellar material. The rotational constant B for 12CI60 is 1.691 cm - I . The first excited rotational energy level, J = 1, with energy J(J + l)hcB = 2hcB , is thermally accessible at about 6 K (based on the rough equation of the rotational energy to thermal energy kT). In interstellar space, only two or three rotational lines would be observable; in circumstellar space (at about 1000 K) the number of transitions would be more like 20.

Molecular spectroscopy 2: electronic transitions


The process of the detennination of the term symbol for dioxygen, 3 ~;, is described in Section 14.1 (b) and will not be repeated here. The interpretation of the symbol follows~ the letter ~ means that the magnitude of the total orbital angular momentum about the internuclear axis is 0; the left superscript 3 means that the component of the total spin angu lar momentum about the internuclear axis is I (2 x I + I = 3) ; the subscript g means that the parity of the term is even; and the superscript - means that the molecular wavefunction for 0 2 changes sign upon reflection in the plane containing the nuclei.

014.3

A band head is the convergence of the frequencies of electronic transitions with increasing rotational quantum number, 1. They result from the rotational structure superimposed on the vibrational structure of the electronic energy levels of the diatomic molecule. See Figs 14.8 and 14.11. To understand how a band head arises, one must examine the equations describing the transition frequencies (eqns 14.5). As seen from the analysis in Section 14. 1(e), convergence can onl y arise when ternlS in both (B' - B) and (B' + B) occur in the eq uation . Since only a term in (B' - B) occurs for the Q branch, no band head can arise for that branch.

014.5

The overall process associated with fluorescence involves the following steps. The molecule is first promoted from the vibrational ground state of a lower e lectronic level to a higher vibrational-electronic energy level by absorption of energy from a radiation fie ld. Because of the requirements of the FranckCondon principle, the transition is to excited vibrational levels of the upper electronic state. See Fig. 14.22. Therefore, the absorption spectrum shows a vibrational structure characteristic of the upper state. The excited state molecule can now lose energy to the surroundings through radiationless transitions and decay to the lowest vibrational level of the upper state. A spontaneous radiative transition now occurs to the lower e lectronic level and this fluorescence spectrum has a vibrational structure characteristic of the lower state. The fluorescence spectrum is not the mirror image of the absorption spectrum because the vibrational frequencies of the upper and lower states are different due to the difference in their potentia l energy curves.

014.7

See Section 14.5 for a detailed description of both the theory and experiment involved in laser action. Here we restrict our discussion to only the most fundamental concepts. The basic requirement for a laser is that it has at least three energy levels. Of these levels, the highest lying state must be capable of being efficiently populated above its thermal equilibrium value by a pulse of radiation. A second state, lower in energy, must be a metastable state with a long enough lifetime for it to accumu late a popUlation greater than its thermal eq uilibrium value by spontaneous transitions from the higher overpopulated state.

MOLECULAR SPECTROSCOP Y 2: ELECTRONIC TRANSITIONS

281

The metastable state must than be capable of undergoing stimulated transitions to a third lower lying state. This last requirement implies not only that the metastable state must have more than its thermal equilibrium population, but also that it must have a higher population than the third lower lying state, namely, that it achieve population inversion. See Figs 14.28 and 14.29 for a description of the three- and four-level lasers. The amplification process occurs when low intensity radiation of frequency equal to the transition frequency between the metastable state and the lower lying state stimulates the transition to the lower lying state and many more photons (higher intensity of the radiation) of that frequency are created. Examples of practical lasers are listed and discussed in Further information 14.1 .

Solutions to exercises E14.1 (b)

According to Hund's rule, we expect one I JTu electron and one 2JT g electron to be unpaired. Hence S = I

[1].

and the mUltiplicity of the spectroscopic term is The overall parity is u x g = the complete core), one electron occupies a u orbital another occupies a g orbital. E14.2(b)

0

since (apart from

Use the Beer-Lambert law I log - = -&[J]I = (-327 dm 3 mol - I cm - I) x (2.22 X 10- 3 mol dm - 3) x (0.15 cm) [0

= -0.10889

-I = 10- 0.10889 = 0.778 II

The reduction in intensity is 122.2 percent 1 E14.3(b)

&

I [J]/

I

= --100 to

10

[13 .2 133] ,. - I 4

(6.67 x 10- mol dm

E14.4(b)

) x (0.35 cm) mol - I cm - I [I dm = 10cm]

= 787

x

= 17.9

x 105 cm 2 mol - I I

= 787 dm 3 mol - I cm- I

The Beer-Lambert law is I log 10 [J]

E14.S(b)

103 cm)

_

log 0.655

-3

=

= -&[J]I

so

[J]

- I

-I

I

t;/

10

= -Iog-

(323dm 3 mol - 1 cm - I x (0.750cm)

10g(1 - 0.523)

= 11.33 .

x 10- 3 mol dm- 3

1

.

Note: a parabolic Iineshape is symmetrical, extending an equal distance on either side of its peale. The given data are not consistent with a parabolic lineshape when plotted as a function of either wavelength or wavenumber, for the peak does not fall at the center of either the wavelength or the wavenumber range. The exercise will be solved with the given data assuming a triangular Iineshape as a function of wavenumber.

282


The integrated absorption coefficient is the area under an absorption peak A

=

f

edii

If the peak is triangular, this area is A

= ! (base) x (height) = ![(l99 x 1O-9 m) - 1 -6

010

=15 . x 1

E14.6(b)

(275 x 1O-9 m )-I] x (2.25 x 104 dm 3 mol- 1cm- I )

3 -I -I -I (1.56 x 109dm3m-Imol-Icm-l) x (lOOcmm- l ) dm m mol cm = 3 103 dm m- 3

Modeling the 7r electrons of 1,3,5-hexatriene as free electrons in a linear box yields non-degenerate energy levels of n2 h 2 Ell

=

8m L2 e

The molecule has six 7r electrons, so the lowest-energy transition is from n the box is 5 times the C--C bond distance R. So

= 3 to n = 4. The length of

Modelling the 7r electrons of benzene as free electrons on a ring of radius R yields energy levels of m I2 1i2

Em/

=-u

where I is the moment of inertia: I = meR2 . These energy levels are doubly degenerate, except for the non-degenerate ml = O. The six 7r electrons fill the m[ = 0 and 1 levels, so the lowest-energy transition is from m[ = 1 to ml = 2 (2 2 _ 12)h 2

-

87r 2meR2

Comparing the two shows

I

Therefore, the lowest-energy absorption willi rise in energy. E14.7(b)

The Beer-Lambert law is I

log - = -e[J]l = log T

10

so a plot (Figure 14.1) of log T versus [J] should give a straight line through the origin with a slope m of -d. So e = -mil.

MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS

The data follow [dye] / (mol dm- 3 ) 0.0010 0.0050 0.0100 0.0500

o __

T

10gT

0.73 0.21 0.042 1.33 x 10- 7

-0.1367 -0.6778 -1.3768 -6.8761

~~~~--~~~~~--C-C-,

~ 3.5384 X10- 3 -:- 137.60:X • R2 h 1:000. : : : Y

.. ...

- 2

-6 -8

om

0.00

0.02

0.03

0.04

0.05

0.06

3

Figure 14.1

[dye]/(mol dm- )

The molar absorptivity is £

E14.8(b)

I

3 = - -138dm mol - I = 522dm 3 mol-I em-I

0.250cm·

I .

The Beer-Lambert law is

=

log T £

-£[J]I

-I

so

£

= [J]/log T

-~

=

(0.0155 mol dm - ) x (0.250 em)

logO.32 = 1128dm 3 mol - I em-I I

Now that we have £, we can compute T of this solution with any size of cell T

E14.9(b)

= lO- e[J]1 = 1O-{ (128dm

3

mol - I cm- l ) x(0.0155 mol dm - 3 ) x(0.450cm) }

= @J.TI

The Beer-Lambert law is I

log -

10

=

(a)

1= -

(b)

1= -

-£[J]I

(30dm 3

so

I

I

£[1]

10

1 = --Iog-

I 1 ,---, x log - = 0.0 10 em 3 mol - I em- I) x (1.0 mol dm - ) 2

I

I

(30dm 3 mol - I em-I) x (1.0 moldm - 3 )

x logO. 10

I

= I0.033em I

283

284


E14.10(b) The integrated absorption coefficient is the area under an absorption peak

A

=

f

s dv

We are told that S is a Gaussian function, i.e. a function of the form S = smaxexp

(

_X2)

~

where x = v - vmax and a is a parameter related to the width of the peak. The integrated absorption coefficient, then, is

A

=

f

oo

-00

Smax exp

(_X2 ) ~ dx = smaxa.jJi

We must relate a to the half-width at half-height, XI 12

So

A

( 2) -X

I

2 Smax

1/ 2 = Smax exp -;;r-

In!

so

2

= SmaxXl/2 (~) 1/ 2 = (1.54 X

= 1l.39 X 108

dm 3 mol-I cm- 2

=

2 -X 1/ 2 a2

and

104 dm 3 mol- 1 em- I) x (4233 em-I) x

C:

2) 1/ 2

1

In SI base units (1.39 x 108 dm 3 mol-I cm- 2) x (lOOOcm 3 dm- 3 ) A=------------------~~----------1 lOOcmm-

9

= 11.39 x 10 m mol-II E14.11 (b)

Fi is formed when F2 loses an antibonding electron, so we would expect Fi to have a shorter bond than F2. The difference in equilibrium bond length between the ground state (F2) and excited state (Fi + e- ) of the photoionization experiment leads us to expect some vibrational excitation in the upper state. The vertical transition of the photoionization will leave the molecular ion with a stretched bond relative to its equilibrium bond length. A stretched bond means a vibrationally excited molecular ion, hence a stronger transition to a vibrationally excited state than to the vibrational ground state of the cation.

I

I


The potential energy curves for the X3 1:;- and B3 1:; electronic states of 0 2 are represented schematically in Fig. 14.2 along with the notation used to represent the energy separation of this problem. Curves for


285

the other electronic state of 02 are not shown. Ignoring rotational structure and anharmonicity we may write

voo ~

Te

I + -(v' - v) = 6.175 eV x

2

(8065.5 cmI eV

= 149 364 cm- I

I

)

+ -I (700 2

1580) cm

_I

I.

\-----7.v' = 4 ii'

--L 00 emT, = 6. 175eV

. __ .---

T---

v' = 3

v, ~=; 2

v' = 0 ' - - - , . . - - - - - - - - - 1

state 5.1147 eV

\------rv = 3 v

~

1580 em- l

o

---T----R

COMMENT. Note that the selection rule

Figure 14.2

tov = ±1 does not apply to vibrational transitions between different

electronic states.

Question. What is the percentage change in voo if the anharmonicity constants Xe V (Section 13.11), 12.0730 cm - I and 8.002 cm- I for the ground and excited states, respectively, are included in the analysis? P14.3

Initially we cannot decide whether the dissociation products are produced in their ground atomic states or excited states. But we note that the two convergence limits are separated by an amount of energy exactly equal to the excitation energy of the bromine atom : 18345 cm- I - 14660 cm- I = 3685 cm- I . Consequently, dissociation at 14660cm- 1 must yield bromine atoms in their ground state. Therefore, the possibilities for the dissociation energy are 14660cm- 1 or 14660cm- 1 - 7598cm- 1 = 7062 cm -I depending upon whether the iodine atoms produced are in their ground or excited electronic state.

286


in order to decide which of these two possibilities is correct we can set up the following Born-Haber cycle. (I)

rBr(g)

--->

~h(g)

+ ~Br2(1)

(2)

~ 12 (s)

--->

~12(g)

(3)

~ Br2(I)

--->

~ Br2 (g)

(4)

~12(g)

--->

I(g)

(5)

~ Br2 (s)

--->

Br(g)

IBr(g)

--->

I(g)

+ Br(g)

= -!::,.fH" (IBr, g) !::"H~ = ~!::"subH" ( 12' s) !::"Hr = ~!::"vapH" ( Br2' I) !::,.H: = ~!::"H(I-l ) !::"H~ = ~!::"H(Br-Br)

!::"Hf

!::"H"

= -!::,.fH" (lBr, g) + ~ !::" subH" (I 2'S) + ~!::"vapH"(Br2 , 1) + ~!::"H(I -I ) + ~!::"H(Br-Br)

!::"H"

= {-40.79 + ~

x 62.44 +

~

x 30.907

+~

x 151.24

+~

x 192.85 } kJ mol -

I

[Table 2.7 and data provided) = 177.93 kJ mol - I = 1 14874 cm - I

I.

Comparison to the possibilities 114660 cm- I 1 and 7062 cm - I shows that it is the former that is the correct di ssociation energy. P14.S

We write

S

= smaxe-x2 = s rnaxe-ii2 /2r the variable being vand r being a constant. v is measured from

the band center, at which

v = o. S = ~Smax when v2 = 2r In 2. Therefore, the width at half-height is A

-2

uV /

implying that

I 2 r=--. 81n2

Now we carry out the integration

f -= 1

00

=

A

sdv

= Smax A

Smax

-00

2 e - ii /2r dv-

2n!::,.v-2I/ 2 )1 / 2 (

8 1n 2

n

= s max(2 rn) 1/ 2

= (--) 41n 2

1/2

= 1.0645s max !::,. VI /2, with v centered on va·

Since

I/? v = -I , !::,.VI /2 ~ -!::,.A -2- [A ~ AO). A

A

= 1.0645s max

AO

(

!::" AI/ 2 ) A6 .

_

Smax!::,.VI /2

[1

00

-00

2

e _x ill• -- n 1/ 2J _

= 1.0645s max !::,.VI /2,


From Fig. 14.6 ofthe text, we find 6.A 1/2 = 38 nm with AD

287

= 290 nm and E"max ~ 235 dm3 mol - I cm- I ;

hence 7 3 I 1.0645 x (235dm mol - I cm- ) x (38 x 10- cm) -11 I 106 d 3 I- I -21 . x m mo cm . 7 (290 X 10- cm)2

A -

Since the dipole moment components transform as A I (z) , 8 I (x), and 82 (y), excitations from AI to AI , 8 I , and 82 terms are allowed. P14.7

We use the technique described in Example 13.5, the Birge-Sponer extrapolation method, and plot the 1 difference 6. vu against v + 2' We then draw up the following table.

6.i\

688.0665.1641.5617 .6591.8561.2534.0 I

1

2

3 2

5 2

7 -

9 -

11

2

2

2

13 2

502.1

465 .5

428.9

388.2

343.1

300.9

255 .0

15 2

17 2

19 2

21

23 2

25 2

27 2

v+-

6.v v 1

v+2

2

2

The data are plotted in Fig. 14.3. Each square corresponds to 25 cm- I . The area under the non-linear extrapolated line is 295 squares; therefore the dissociation energy is 7375 cm- I . The 3 I:;;- +- X excitation energy (where X denotes the grou nd state) to v = 0 is 49357.6 cm- I which corresponds to 6.12 eY. The 3 I:;;- dissociation energy for

is 7375 cm- I , or 0.91 eV. Therefore, the energy of

is 6.12 eV energy of

+ 0.91

eV

= 7.03 eV.

Since 0 * -+ 0 is -190 kJ mol - I, corresponding to -1.97 eV, the

0 2(X) -+ 20

is 7.03 eV - 1.97 eV COMMENT.

= 15.06 eV I.

This value of the dissociation energy is close to the experimental value of 5.08 eV quoted

by Herzberg [Further reading , Chapters 13 and 14], but differs somewhat from the value obtained in Problem 14.2. The difficulty arises from the Birge-Sponer extrapolation, which works best when the experimental data fit a linear extrapolation curve as in Example 13.5. A glance at Figure 14.3 shows that the plot

288

STUDENT'S SO LUTI ONS MANUAL . .

700 : . •

.

:

:

.... ..... .......... .... .. .... ...... ............... :

:

:

:

:

:

:

:

:

:

':"':"'-:"':" ':"'7"':" ':" '-:- "':" • ." '7" ':" ' ~' " '7 " ':"':'" "7" -: ••

:

..

··;···~· · · ·~ · ··;···i ... ~ ... ~ ... ~ ....~ ... ~ ... ! . ...::- . . . : .. . :.... i· ··~ .. .: ... j ... ( ....:.... .i··· (... ...:.... .j •. . {.. •• ..;••.• .i··· . . .. ..~ ...: ... ~ .. ..~ " '! .. .~ .. ..~ ... i ... . . . . :-.. . . ... ~ ...~ ... ~ ... ~ ... ~ ... ~ ... ~ ... "":'" .. . : ... ... ':" .... ..... :

600

~

.; ..i~t~~.

:;:::;.:. .,. ::~>~k:,.,." ...,. ,..;. ~ . ;.. .,.

.,' .,'

·· · f· ·~·~·~· ·

.-.': .. " ...•. .-,_ ......... . :

. .•..; •. j .. { .•. :- . . • ! ... { .

300 ;..

.... ... .: ... ...... ...:. ~

~

~

.

.

:

::

...

~ .

:

· ·1· ··'; . . ·j. · ·· 1 . . ~.

. . 0 :. ......... ........ . ......... ... . ~

'

o

4

:

:

:

.~ .. ·l··)·· .. ~ .. ·~ .. .

... ;. ... :... ~.... :- ... :.. . ... ~ ... ;... ~ ....;... ;... .. ~ ~ ::~ ~ ~ ~ !... ~ ....~ ... !.. . -, .).. .. ; ... .,........ ; ... '.. ~, : ~: . . . . . .- ~

j ... ;

~

.. . . ...

.. +... i... ~....;... !. . ... ... ... ...

xtrianpola tion.;';"+" :; ·· .....;;'..... ebeg s . . ; . . here "1".,. . ...!...

.. .. ., ... ..

... ~ .. ~ ... ~ .. !...~ ..

. . . ~ . . . ~ . .. ~ . . . . ~ .. . " .. !

· ... ~ ... Inaccurate .. ., .... .. ~ ... non-linear

.., .. i .. ~ ...,

...;... ~

. . ....:... ~....~ ... :... ~ -.~ -.. :._. ~-'.. . ... ~ ... ~....~ ;... ~....~ ... ~ ... ~.. ~~ .. ~ ... ~ . . .... : : : : ,: :

.

200

~ .. .~ 100 i....;

.• . j •.•

: : l·J{m·;~~:'~latiOn::TT .j.,...! .. ;.. ,.....::L'>~·T ·L:: :'>; ::':L::::LL:::::

... ~ . . . ! . " :

. ..., ....

:.... ~ ... ;... ~....~ ...;.. .~

:, :

~

) ... r::j:·····: :

~

500

, ...:....,...

.. .,... ...:- ... ..

8

.

.

.!.

.~

~

...

... :....:....:....:....: 12

" ~~i"'~''' ~'' . ..~ .

":' .. ~

...~ ... ~ ...j ...:. .

•. • j . •

~

.. ~ ..

.. ~ " l AX> lM X

M protons (a)

d

X protons

-j -

l AX

I I

(b)

Figure 15.3

Only the splitting of the central peak of Figure IS.3 (a) is shown in Figure IS.3(b).

E15.14(b) (a) Since all JHF are equal in this molecule (the CH2 group is perpendicular to the CF2 group), the H

and F nuclei are both chemicall y and magnetically equivalent. (b) Rapid rotation of the PH3 groups about the Mo-P axes makes the P and H nuclei chemically and magnetically equivalent in both the cis- and trails-form s. E15.15(b) Precession in the rotating frame follows

304


Since w is an angular frequency, the angle through which the magnetization vector rotates is

"" _ S0 ""' 1-

en _

-- -

g] fl-Nt

34

(n) x (1.0546 x 10- J s) 1 -4 1 940 x 10 T (5.586) x (5 .0508 x 10- 27 JT-I) x (12 .5 x 10- 6 s) .

a 90° pulse requires! x 12.51-1 s = 16.25 I-1S

=

I

(6.626 X 10- 34 J s) x (2.998 x 108 m S-I) (2) x (9.274 x 10- 24 J T-I ) x (8 x 10- 3 m) -

[iliJ 13T .

E15.17(b) The g factor is given by

hv

h

fl-B fA

fl-B

g =--;

= g

6.62608 x 10- 34 J s 9.2740 x 10- 24 JT- 1

----~,----~

1

_I

71.448mTGHz- x 9.2482GHz 330.02 mT -

= 7.1448

X

10-

11

THz- 1

= 71.448mTGHz- 1

1

2.0022

E15.18(b) The hyperfine coupling constant for each proton is 12.2 mT

I, the difference between adjacent lines in

the spectrum. The g value is given by

=~= g

fl-B fA

(71.448mTGHz- 1) x (9.332GHz) _~ 334.7mT -~

E15.19(b) If the spectrometer has sufficient resolution, it will see a signal spilt into eight equal parts at

± 1.445 ±

1.435 ± 1.055 mT from the center, namely 1328.865,330.975,331.735, 331.755,333.845,333.865, 334.625 , and 336.735 mT

I

If the spectrometer can only resolve to the nearest 0.1 mT, then the spectrum will appear as a sextet with intensity ratios of 1: I :2:2: I : I. The four central peaks of the more highly resolved spectrum would be the two central peaks of the less resolved spectrum. E15.20(b) (a) If the CH2 protons have the larger splitting there will be a triplet (1:2:1) of quartets (1:3:3:1).

Altogether there will be 12 lines with relative intensities 1(4 lines), 2(2 lines), 3(4 lines), and 6(2 lines). Their positions in the spectrum will be determined by the magnitudes of the two proton splittings which are not given. (b) If the CD2 deuterons have the larger splitting there will be a quintet (1:2:3:2:1) of septets (1 :3:6:7:6:3:1). Altogether there will be 35 Lines with relative intensities 1(4 lines), 2(4 lines), 3(6 lines), 6(8 lines), 7(2 lines), 9(2 lines), 12(4 lines), 14(2 lines), 18(2 Iines),and 21(1 line). Their positions in the spectrum will determined by the magnitude of the two deuteron splittings which are not given.

MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE

305

E15.21(b) The hyperfine coupling constant for each proton is 12.2mT I, the difference between adjacent lines in the spectrum. The g value is given by hv

h

hv

- = 71.448mTGHz

g= - - s o f$ = - , J1-B f$ J1-Bg

_ I

J1-B l

(a)

(b)

f$ = (71.448mTGHz- ) x (9.312GHz) =1332.3mTI

2.0024 f$

=

l

(71.448 mT GHz- ) x (33.88 GHz) = 11209 mT 2.0024

E15.22(b) Two nuclei of spin

II = 1 Igive five lines in the intensity ratio 1:2:3:2: 1 (Figure 15.4).

I I I II III

I II

2

2

3

I

First nucleus with I = I second nucleus with 1= 1

Figure 15.4

E15.23(b) The X nucleus produces four lines of equal intensity. Three H nuclei split each into a 1:3:3: 1 quartet. The three D nuclei split each line into a septet with relative intensities 1:3:6:7:6:3: 1 (see Exercise 15.20(a». (See Figure 15.5.)

II

Figure 15.5


g/ Bo

= -3.8260

(Table 15.2) . (6.626 x 10- 34 JHz- l ) x v

hv

= -- = g/J1-N

(-)(3 .8260) x (5.0508 x

Therefore, with v

= 300 MHz,

10- 27 JT-I)

= 3.429 x

8

10- (v/Hz) T.

306


8N Ii

gli-LNBO

~ ~

[Exercise 15.4(a)]

(-3.8260) x (5.0508 x 10-27 J r ' ) x (I0.3T)

=

(2) x (1.381 x 1O-23 JK- ') x (298K)

I

-51

= . 2.42 x 1 0 .

Since gl < 0 (as for an electron , the magnetic moment is anti parallel to its spin), the [ ] state (ml lies lower. P15.3

= -!)

The envelopes of maxima and minima of the curve are determined by T2 through eqn 15.30, but the time interval between the maxima of this decaying curve corresponds to the reciprocal of the frequency difference t. v between the pulse frequency vo and the Larmor frequency VL, that is, t. v = Ivo - VL I:

t.v

I

,

= - - = 10s- = 10Hz. O. IOs

Therefore the Larmor frequency is 1300 x 106 Hz ± 10Hz. 1 According to eqns 15.30 and 15.32 the intensity of the maxima in the flO curve decays exponentially as e- r/ T2 . Therefore T2 corresponds to the time at which the intensity has been reduced to I Ie of the original value. In the text figure, this corresponds to a time slightly before the fourth maximum has occurred, or about I 0.29 s I. P15.5

It seems reasonable to assume that only staggered conformations can occur. Therefore the equilibria are as shown in Fig. 15.6.

When R3 =

~

= H, all three of the conformations in Fig. 15.6 occur with equal probability ; hence

31HH (methyl)

=

*e + 1(

2 319)

[t

= trans, g = gauche;

CHR3 ~

= methyl].

Additional methyl groups will avoid being staggered between both R, and R2. Therefore

= !(J( + 19) 31HH (isopropyl) = 1(

31HH(ethyl)

[R3 = H, R4 [R3 = ~

= CH3],

= C H3]·

We then have three simultaneous equations in two unknowns 1, and 19 .

te1( + 2 1g) = 7.3 Hz, 3

!cJ1( 31(

+ 31g) = 8.0Hz,

= 11.2 Hz.

(I)

(2)

MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONAN CE

307

The two unknowns are overdetermined. The first two equations yield 3it = 10.1,3 i g = 5.9. However, if we assume that 3i t = 11.2 as measured directl y in the ethyl case then 3i g = 5.4 (eqn I) or 4.8 (eqn 2), with an average value of 5.1. Using the original form of the Karplus equation,

or 11.2

= A + B,

5.1 = 0.25A

+ B.

These simultaneous equations yield A = 6.8 Hz and B = 4.8 Hz. With these values of A and B, the original form of the Karplu s equation fits the data exactly (at least to within the error in the values of 3it and 3i g and in the measured values reported). From the form of the Karplus equation in the text [15 .27] we see that those values of A, B, and C cannot be determined from the data given, as there are three constants to be determined from only two values of i . However, if we use the values of A, B, and C given in the text, then

+ 5 Hz(cos 360°) = II Hz, I Hz(cos 60°) + 5 Hz(cos 120°) = 5 Hz.

it = 7 Hz - I Hz(cos 180°) i g = 7 Hz -

The agreement with the modern form of the Karplus equation is excellent, but not better than the original version. Both fit the data equally well. But the modern version is preferred as it is more generally applicable.

I

I

P15.7

The proton COSY spectrum of I-nitropropane shows that (a) the Ca-H resonance with 8 = 4 .3 shares a cross-peak with the Cb -H resonance at 8 = 2.1 and (b) the Cb -H resonance with 8 = 2.1 shares a cross-peak with the Cc - H resonance at 8 = 1.1. Off diagonal peaks indicate coupling between H's on various carbons. Thus peaks at (4,2) and (2,4) indicate that the H 's on the adjacent CH2 units are coupled. The peaks at ( 1,2) and (2, I) indicate that the H's on CH3 and central CH2 units are coupled. See Fig. 15.7.

P15.9

Refer to Fig. 15.4 in the solution to Exercise 15.20(a). The width of the CH3 spectrum is 3aH = 16.9 mT I. The width of the CD3 spectrum is 6ao . It seems reasonable to assume, since the hyperfine interaction is an interaction of the magnetic moments of the nuclei with the magnetic moment of the electron, that the strength of the interactions is proportional to the nuclear moments.

and thus nuclear magnetic moments are proportional to the nuclear g-values; hence

ao

~

0.85745

- -- x aH = 0. 1535a H = 0.35 mT. 5.5857

Therefore, the overall width is 6ao

= 12.1

mT

I.

308

STUDENT'S SOLUTIONS MANUAL abc N0 2CH 2CH 2CH 3

~ L-'=-A =® 2

0

=cw

@

=@

=@

3

4

5 4

5

P1S.11

3

2

0

. 5.7mT WewnteP(N2s) = - - 55.2mT

Figure 15.7

~ =~(lOpercentofitstime) ;

~

1.3 mT

P(N2pz) = 3.4 mT = ~ (38 per cent of its time).

The total probability is

I

(a) peN) = 0.10 + 0.38 = 0.481 (48 per cent of its time) . (b) P(O) = I - peN) = j 0.52j (52 per cent of its time). The hybridization ratio is P(N2p) = 0.38 = P(B2s ) 0.10

13.8l. ~

The unpaired electron therefore occupies an orbital that resembles an sp3 hybrid on N, in accord with the radical 's nonlinear shape. From the discussion in Section 11.3 we can write 2 1+ cos¢ a = -- 1- cos¢

b2 = 1 _ a2 = -2 cos ¢

1 - cos¢ A=

b'2 ~ a

=

-1 cos¢ A , implying that cos ¢ = - 1 + cos¢ 2+ e

Then, since A = 3.8, cos ¢ = -0.66, so ¢ = ~


309


Use eqn 15.22 and Illustration 15.2. For hydrogen itself, we have:

The only difference in wavefunction (and therefore in the expectation value of \ / r) between hydrogen and a more general hydrogenic ion is that the latter has ao/Z where the former has ao, so: e 2 /J Z _....:."'--,0, -- = 12n:m e ao

P15.15

ynJ.LOml

=

Bnuc

rearranges to R

=

4 "~R3

1.78 x 10- 5 Z .

2

(1 - 3 cos e)[ 15.28]

( gl J.LNJ.LO ) 1/3 = (5 .5857) x (5.0508 x 10-

4n:Bnuc

[ml

= +~ , e = 0,

yn

l

= gfJ.L N

27 J T- I) X (4n: x 10- 7 T2 r 1 m3»)

which

1/ 3

(4n:) x (0.715 x 1O-3 T)

= (3.946 x 10- 30 m 3 ) 1/ 3 = 1158 pm P15.17

glJ.LNJ.Lo =-4n:R3

I.

The shape of spectra l line I(w) is related to the free induction decay signal G(t) by

10

I(w) = a Re

00

iw G(t)e (dt

where a is a constant and Re means take the real part of what follows. Calculate the lineshape corresponding to an oscillating, decaying function G(t) = coswote - (IT

I(w)

= aRe 10

1 Re = -a 2

= ~a Re 2 =

1°

00

00

(e -,.~( .~

+ elL\) ( )e-(I r + ',w( dt

I

I

roo {ei(lL\)+w+i/ r )( + e-i (lL\) - w-i IT)(\dt

10

-~ aRe [ 2

iw G(t)e ( dt

I

i(WO

+ w + i/ r)

_ _ __ I __ ] iCwo - w - i/ r) .

310


When wand

wo

are similar to magnetic resonance frequencies (or higher), only the second term in

brackets is significant (because

I(w)

~

I

(wo + w)

I 2

I i(WO-w)2+I / r

I

I

«

I but

I

(wo -

w)

may be large if w

~ wo).Therefore,

-a Re----::----

= - a Re ...,--------,-,;-----,---;;2 (wo - w) 2 + 1/ r 2 ar

2 which is a Lorentzian line centered on wo, of amplitude ~A r and width - at half-height. r

P15.19

For non-weak fields in which the external magnetic field is comparable to the spin-orbit coupling field of an unpaired electron it is necessary to include a spin-orbit coupling term with coupling constant A [10.41] and apply the second-order perturbation equation [9.65b]. The Hamiltonian is H = -geYeBOsz YeBOlz + Al . s = -YeB . (ges + I) + Al . s where vector notation is used for the electron orbital and spin angular momentum. The first -order perturbation equation [9.65a] gives E ( I)

=

-YeB . (ge(s ) + (I )) + A (I . s )

=

-YegeB . (s ) - YeB · (I ) + A (s ) . (I).

The expectation value of orbital angular momentum, (I ), equals zero for real states and (sz) gives

= ms which

The second-order perturbation term is written using the ground state '0' and 'n' excited states with the energy difference !lEno = En - Eo [9.65b], which is positive.

The numerator may be expanded and simplified by discarding second-order terms in Bo and negligibly small.

E(2)

= _L n",O

(OI{-YeBolzlln )(nIAI · s lO) + (OIAI · sln)(nl{-YeBOlzl IO) !lEno

_ " -AYeBo L n",O

(Ollzln )(nl/ · slO) + (011· sln)(nllzIO ) !lE nO

Sz

as


311

The last manipulation uses the assumption that the local field is parallel to the applied field (i.e., I . S = lzsz). Combination of the first- and second-order perturbation estimates gives

E

(spin) -

-

E(I)

+ E(2)

-

-

_

~&

8

"+ 2A ~ 8 O~I'~ , , " (Ollzln)(nllzIO)

o~u

n;loO

Comparison with the effective spin Hamiltonian, _ g-ge

"

-2A~

(Ollzln)(nllzIO)

n;loO

I':.Eno

= -gYeBosz, indicates that

E(spin)

.

I':.E"o

g increases with increasing strength of the spin-orbit coupling (A) and with decreasing excitation energy

(I':.E"o). This analysis is presented on p. 434 of P.w. Atkins and R.S. Friedman, Molecular quantum mechanics, 3rd edn, Oxford University Press, 1997.


P15.21

(8 nucl)

=

-g/J1-NJ1-0m/

4nR

3

J~max (l

- 3 cos2 8) sin 8d8 J;maxsin 8 d8

-=----;Q,.-------

The denominator is the normalization constant, and ensures that the total probability of being between

o and 8max is 1.

Xmax (I X

Xmax -

If 8max

=

n (complete rotation), cos 8max

x~ax) 1

= -I

=

-g/J1-NJ1-0m/

4

nR

and (8nu cl)

3

= O.

2

(cos 8max

+ cos8max )

.

312


If emax = 30°, cos 2 emax (B nuel )

+ cos emax

= 1.616, and

_ (5 .5857) x (5 .0508 x 10- 271 T- I) x (4n x 10- 7 T2 r l m3 ) x (1.616) -

--------~----~~--~~~~~n_~--~~----~--~--~

(4n) x (1.58 x 10-

10 m)3

x (2)

= 10.58 mT I· P15.23

The desired result is the linear equation [110 =

[E~ov~v

- K,

so the first task is to express quantities in terms of [110 , [Elo, ~V , 8v , and K , eliminating terms such as [I], [EI], [E], VI , VEl , and v. (Note: symbolic mathematical software is helpful here.) Begin with v: [11

V

= [11

[Ell

+ [Ell VI + [ll + [Ell VEl

=

[110 - [Ell [110 VI

+

[Ell [110 VEl ,

where we have used the fact that total I (i.e. free I plus bound I) is the same as initial 1. Solve this so it must also be much greater than [Ell: [Ell

=

[I]o(v - VI) VEl -

VI

[Il08v ~'

where in the second equality we notice that the frequency differences that appear are the ones defined in the problem. Now take the equilibrium constant

K _ [E][Il _ ([Elo - [EI]) ([110 - [EI]) - [Ell [Ell

~

([Elo - [EI]) [110 [Ell

We have used the fact that total I is much greater than total E (from the condition that [110 it must also be much greater than [EI], even if all E binds 1. Now solve this for [Elo: E = K + [110 EI = (K + (110) ([Il08V) = (K [110 [1 [110 ~v [ 10

»

[Elo), so

+ [Ilo)8v . ~v

The expression contains the desired terms and only those terms. Solving for [110 yields:

[110=

~ -K ,

which would result in a straight line with slope P15.25

[Elo~ v

and y-intercept K if one plots [llo against 1/ 8v.

When spin label molecules approach to within 800 pm, orbital overlap of the unpaired electrons and dipolar interactions between magnetic moments cause an exchange coupling interaction between the spins. The electron exchange process occurs at a rate that increases as concentration increases . Thus the process has a lifetime that is too long at low concentrations to affect the 'pure' ESR signal. As the concentration increases, the linewidths increase until the triplet coalesces into a broad singlet. Further increase of the concentration decreases the exchange lifetime and therefore the linewidth of the singlet.


313

When spin labels within biological membranes are highly mobile, they may approach closely and the exchange interaction may provide the ESR spectra with information that mimics the moderate and high concentration signals below.

ESR spectrum of di-tert-butyl nilroxide

Low concentration

Moderate concentration

Hi gher concentration

Hi gh concentration

Figure 15.8

P15.27

Assume that the radius of the disk is I unit. The volume of each slice is proportional to (length of slice x8x ) , Fig. IS.9(a). Length of slice at x

= cos e, e = arcosx.

x

= 2 sin e.

314


x ranges from -1 to

+ l.

Length of slice at x = 2 sin(arcosx) .

•

Ox Ox ___ x

Figure 15.9(a)

2

1.5 MRI absorption

intensity f(xl

0.5

0 -1

- 0.5

0 x

0.5

Figure 15.9(b)

= 2 sin (arcos x) against x between the limits -1 and + l. The plot is shown above. The volume at each value of x is proportional tof(x) and the intensity of the MRI signal is proportional to the volume, so Fig. 15. 9(b) represents the absorption intensity for the MRI image of the disk.

Plotf(x)

Statistical thermodynamics 1. the concepts

16


Consider the value of the partition function at the extremes of temperature. The limit of q as T approaches zero, is simply gO, the degeneracy of the ground state . As T approaches infinity, each term in the sum is simply the degeneracy of the energy level. If the number of levels is infinite, the partition function is infinite as well. In some special cases where we can effectively limit the number of states, the upper limit of the partition function is just the number of states. In general, we see that the molecular partition function gives an indication of the average number of states thermally accessible to a molecule at the temperature of the system.

016.3

We evaluate {3 by comparing calculated and experimental values for thermodynamic properties. The calculated values are obtained from the theoretical formulas for these properties, all of which are expressed in terms of the parameter {3 . So there can be many ways of identifying {3, as many as there are thermodynamic properties. One way is through the energy as shown in Section 16.3(b). Another is through the pressure as demonstrated in Example 17.1. Yet another is through the entropy, and this approach to the identification may be the most fundamental. See Further reading for elaboration of this method.

016.5

An ensemble is a set of a large number of imaginary replications of the actual system. These replications are identical in some respects, but not in all respects. For example, in the canonical ensemble, all replications have the same number of particles, the same volume, and the same temperature, but not the same energy. Ensembles are useful in statistical thermodynamics because it is mathematically more tractable to perform an ensemble average to determine the (time averaged) thermodynamic properties than it is to perform an average over time to determine these properties. Recall that macroscopic thermodynamic properties are averages over the time dependent properties of the particles that compose the macroscopic system. In fact, it is taken as a fundamental principle of statistical thermodynamics that the (sufficiently long) time average of every physical observable is equal to its ensemble average. This principle is connected to a famous assumption of Boltzmann's called the ergodic hypothesis. A thorough discussion of these topics would take us far beyond what we need here. See the references under Further reading.


Ne - f3 ej 11;=---

q

whereq

=L

e- f3ej

316


Thus

.

n2

1

= -, 11£0 = 300cm- 1 2

GIven -

nl

k = (1.38066

In

X

lcm-I ) = 0.69506cm- 1 K- I 10- 23 JK- I) x ( 1.9864 x 10- 23 J

(:~) = -M/kT

T=

-11£0

11£0

kln(n2/nl)

kln(nl/n2)

· 300cm- l _ ~ = (0.69506cm- I K-I)ln(2) =622.7K~~

E16.2(b)

f3

A = h( 2rrm )

(a)

= (6.626

1/ 2

(

[16.19] = h _1_)

1/ 2

2rrmkT

10- 34 J s)

X

x C2rr) x (39.95) x (1.6605 x 1O-2;kg) x (1.381 x 1O-23JK-I) x T

)'/2

276pm (T / K) 1/2

V

q= -

(b)

A3

[16.19] =

(i) T = 300 K,

(1 00 x 10- 6 m 3) x (T /K)3/2

.

A = 1.59

(ii) T = 3000K,

(2.76 x 10- 10 m)3 10- 11 m =

X

A = 15.04 pm

I,

'11-5-.9-p-m-'~

= 4.76 x lO 22 (T /K)3/2 q = 12.47 x 1026 1,

q = 17.82 x 1027 1

Question. At what temperature does the thermal wavelength of an argon atom become comparable to its diameter? E16.3(b)

The translational partition function is qtr

so

E16.4(b)

= hV3 (2kTrrm)3 /2

qXe qHe

q=

= (mxe)3 /2 = (131.3U)3/2 =1187.91 mHe 4.003 u

L gje-/3e = 2 + 3e-/3e + 2e-/3€2 1

j

levels

f3£O

=

hey kT

1.4388(Y / cm -I)

=

T/K

STATISTICAL THERMODYNAMICS 1: THE CONCEPTS

317

Thus q = 2 + 3e-(1.4388 x I250/2000) + 2e-(1.4388 x I300/2000) = 2 + 1.2207 + 0.7850 = 14.0061 E16.5(b)

E = U - U(O) = _'!.... dq =

q df3

=

-'!....~(2 + q df3

_'!.. (-3f:le-fJE1 -2f:2e-fJE2) = q

=

3e- fJE1 + 2e- fJE2 )

Nheq (3ye-fJhCVI +2ye-fJhCVz )

(NAhe) x {3( 1250 cm- I) x (e-(1.4388X 1250/2(00») 4.006 +2(l300cm- l ) x (e-(1.4388X I300/2000» )}

= (NAhe)

x (2546cm- l )

4.006

= (6.022 x 1023 mol-I) x (6.626 x 10- 34 Js) x (2.9979 x IO lO cms- l ) x (2546cm- I)/4.006 = 17.605 kJ mol-II E16.6(b)

In fact there are two upper states, but one upper level. And of course the answer is different if the question asks when 15 percent of the molecules are in the upper level, or if it asks when 15 percent of the molecules are in upper state. The solution below assumes the fonner.

each

The relative population of states is given by the Boltzmann distribution

(-heY)

-n 2 = exp (-~E) - - = exp - nl kT kT

n2

-hey

so In- = - nl kT

-hey

ThusT= - - - kln(n2 / n l) Having 15 percent of the molecules in the upper level means 0.15 1-0.15

so

n2

= 0.088

nl

-(6.626 x 1O- 34 J s) x (2.998 x 1OIOcms-l) x (360cm- l ) and T = - - - - - - - - - - = -23- - - : : - - - - - - - - - - (1.381 x 1O- JK-I) x (in 0.088) =1213KI E16.7(b)

The energies of the states relative to the energy of the state with mt = 0 are -YN~, 0, + YN Ii = 2.04 X 10- 27 J T- I . With respect to the lowest level they are 0, YN Ii, 2YN Ii. The partition function is

q

=L

e-E"ale / kT

states

where the energies are measured with respect to the lowest energy. So in this case

q = 1 + exp

-YN~ ) + exp (-2YN kT ~ ) ( ---;;:r-

YN ~,

where

318


As ~ is increased at any given T , q decays from q = 3 toward q = I as shown in Figure 16.1 (a).

2

Figure 16.1(3) The average energy (measured with respect to the lowest state) is (£ )

=

'"

L.. slales

£

Slale e

-E"ate / kT

I

+ YN ~ exp (-YN ~/ kT) + 2YNrJ!$ exp (-2YN ~/ kT) 1+ exp (-YN ~/ kT ) + exp ( -2YN ~/ kT)

q

The expression for the mean energy measured based on zero spin having zero energy becomes YN ~

(I - exp (-2YN ~/ kT»

1+ exp (-YN ~/ kT) + exp (-2YN ~/ kT) As

~

is increased at constant T, the mean energy varies as shown in Figure 16.1(b).

Figure 16.1(b) The relative populations (with respect to that of the lowest state) are given by the Boltzmann factor

exp

-6.£) = (---u

exp

(-YN ~) kT

or

YN ~ _ (2.04 x 10- JT- i ) x (20.0 T) Note that k l.381 X 10- 23 J K - i 27

= 2 95 .

x 10- 3 K

so the populations are 3

(3)

exp (-2.95 x 101.0 K

(b)

exp (-2.95 x 10298

K) K) =

= 10.9971

3

and

10.999991 and

K)) = K)) =

2( - 2.95 x 10exp ( 1.0 K exp (2 (-2.95 x 10298

3

3

10.9941

10.999981


E16.8(b)

(a) The ratio of populations is given by the Boltzmann factor n2 nl

(-"""E) = e-25.0K/ T kT

= exp

n3 = e - 50.0 K/T

and

nl

(I) At LOOK n2 nl

= exp (-2S.0K) = 11.39 x 10- 11 1 1.00 K · .

and n3 = exp ( - SO.OK) = 11.93 x 10- 22 1 nl 1.00 K . (2) At 2S .0 K n2

-=exp nl

(-SO.OK) ~ =~ 25.0K

( -2S.0K) ~ =~ 2S .0 K

and

-

(-25 .0K) ~ =~ lOOK

and

~ -n3 =exp (-50.0K) =~ nl lOOK

n3

nl

=exp

(3) At 100 K n2

-=exp nl

(b) The molecular partition function is q =

L

e-ESla,elkT

= 1+ e- 25.0K/T + e- 50.0K/T

states

At 25.0 K, we note that e-2S.0K/T = e- I and e-SO.OK/ T = e- 2 q = I +e- I +e- 2 =ll.s031

(c) The molar internal energy is Urn = Urn (0)

_:A G;)

where/3 = (kT) - 1

So Um = Urn(O) - NA (-2S.0K)k (e-2S.0K/T + 2e-SO.O K/T) q At 25.0 K 1023 mol - I) x (-25 .0K) x (1.381 x 10- 23 J K- I) 1.503 I 2 x (e- +2e- ) (6.022

X

= 188.3 J mol-I I (d) The molar heat capacity is CV.rn =

( aa~m ) v

= NA(25.0 K )k aa T

=NA(25.0K)k x

_

(2~~2K

~

(e-25.0K/T + 2e-50.0 K/T)

(e- 2S.0 K/T +4e-SO.O K/T)

~ ( e - 2S.0 K/T + 2e -SO.OK/ T) aq ) ~ aT

319

320


where aq = _25_.0_K aT

so Cv

=

.m

(e- 2S.0 K / T

+

2e-SO.OK / T)

T2 NA(25 .0K) 2k ( (e-2S.0K/T + 2e- SO.OK/T)2) e -2S.0K / T +4e -SO.O K / T _ - - - -_ _ _ __ T 2q q

At 25 .0K Cv m ,

=

(6.022 x 1023 mol - I) x (25.0K)2 x (1.381 x 1O- 23 JK - I) (25.0K)2 x (1.503) I 2 x (e- I + 4e-2 _ (e- + 2e- )2) 1.503

-------;:::-:-::~~--;-::_::_:::'::-:---------.:..

= 13.53 J K- I mol-I I (e) The molar entropy is

At 25.0K Sm =

88.3Jmol- 1 25.0 K + (6.022

X

1023 mol-I ) x (1.38 1 x 10- 23 J K- I) In 1.503

=16.92JK- l mol - 1 1

E16.9(b)

no nl

Set -

no

I e

= - and solve for T.

In

G)

T

heB = ---,----".,.

= In3 +

(-:;B)

k( I+ln3)

6.626

X

10- 34 Js x 2.998 x lO iO cms- 1 x 1O.593cm- 1 +1.38 1 x 1O- 23 JK - 1 x (1 + 1.0986)

= 17.26 K 1 E16.10(b) The Sackur-Tetrode equation gives the entropy of a monatomic gas as

whereA=

h

~

v2kTrrm

(a) At 100 K

6.626

X

10- 34 J s

A--------------------------~=

- 12(1.381 x 1O- 23 JK-I) x (lOOK) x rr(131.3u) x (1.66054 x 1O- 27 kgu - l )jl /2 = 1.52 x 10- 11 m


_I

= (8 .3145J K

and Sm

_I mol

) In

321

I 2 23 (eS/ ( 1.381 x 10- J K- ) x ( lOOK») (1.013 x lOs Pa) x (1.52 x 10 II m)3

I

= 1147 J K- I mol - I

(b) At298.15K

1\

6.626 x 10- 34 J s

= =

/2 (1.38 1 x 10-

23

1/2 JK - I) x (298.15K) x JT(l31.3u) x (1.66054 x 1O- 27 kgu - I

)1

8.822 x 10- 12 m

and

Sm

= (8.3 145JK

_I

_I mol

)In

(e S / 2( 1.381 x 10- 23 J K- I) x (298.15 K») (1.013 x IOsPa) x (8.822 x 1O- 12 m) 3

= 1169.6J K- I mol - I I E16.11(b)

q

=

1-

e-{Je

I - e- hc{Jv

_ (1.4388cmK) x (321 cm- I) 600 K hcf3v =

Thus q

=

I - e-O.76976

=

-

= 0.76976

1.863

The internal energy due to vibrational excitation is

U - U(O)

=

N s e-{Je I _ e-{Je Nheve - hcv{J I - e-hcv{J

Sm

and hence NAk

=

NhcV ehcv{J - 1

-:--::-:;--- = (0.863) x (Nhc) x (32 1 cm- I)

U - U (0) + Inq NAkT

= (0.863)

x

( he )

kT

I

-

x (321 cm - ) + In(1.863)

_ (0.863) x ( 1.4388 Kcm) x (32Icm- l ) 600K +In(1.863)

= 0.664 +

0.62199

=

1.286

and Sm = 1.286R = 110.7 J K- I mol - I I E16.12(b) Inclusion of a factor of (N!) - I is necessary when considering indistinguishable particles. Because of

their translational freedom , gases are collections of indistinguishable particles. The factor, then , must be included in calculations on I (a) C02 gas I.

322



Number of configurations of combined system, W

= (1020)

W

x (2 x 1020)

s = kin W[16.34]; S

= k In (2

x 1040)

= 92.8 x

S,

= 12 X

1040

= kIn WI ;

=

W, W2.

I.

S2

= kin W2 .

= k{ln 2 + 40 In 10) = 92.8k

(1.381 x 10- 23 J K-' )

= 11.282 x

10- 2' J K - '

I.

s, = k In(1020) = k{20 In 10) = 46.lk = 46.1 S2

x ( 1.381 x 10- 23 J K - ' )

= k 1n(2

= 46.7 x

x 1020)

= I 0.637

X

10- 2' J K-'

I.

= k{ln 2 + 20 In 10) = 46.7k

( 1.381 x 10- 23 J K-' )

= I 0.645

X

10- 2' J K - '

I.

These results are significant in that they show that the statistical mechanical entropy is an additive property consistent with the thermodynamic result. That is, S = S, + S2 = (0.637 X 10- 2' + 0.645 x 10- 21) J K-' = 1.282 X 10- 2 ' J K-'.

S = kin W [16.34].

P16.3

Therefore,

) (auas) v = Wk (aw au v or

But from eqn 3.45

( au) as v = T . So,

Then


323

Therefore, ~w

~U

W

kT

-- ~ -

- (1.381 x 10- 23 J K-

= 2.4

1

P16.5

=

q

x 10

v

25

.1\ =

.1\3'

I)

x 298 K

1

h

(21tmkT) 1/

2

[16.19, f3 = kiT]'

and hence T

= (~)

x

21tmk

(i)2/3 V

10- 34 J s)2 ) = ( (21t) x (39.95) x ( 1.6605 x 1027 kg) x (1.381 x 10- 23 J K I) (6.626

X

10 )2/3 x ( 1.0 x 10- 6 m3 = 13.5 x 10- 15 K 1[a very low temperature].

The exact partition function in one dimension is 00

q

= Le-(1I2-I) h2P /8mL2. 11=1

For an Ar atom in a cubic box of side 1.0 em, h 2 f3

(6.626

X

10- 34 J s)2

(8) x (39.95) x (1.6605 x 10- 27 kg) x (1.381 x 10- 23 J K- 1) x (3.5 x 10- 15 K) x (1.0 x 10- 2 m)'

8mL2

= 0.171. 00

Then q

=L

-

e- O. 171 (1I

2

-I)

= 1.00 + 0.60 + 0.25 + 0.08 + 0.02 + ... = 1.95.

11=1

The partition function for motion in three dimensions is therefore q

= (1.95)3 = [iliJ.

COMMENT. Temperatures as low as 3.5 x 10- 15 K have never been achieved. However, a temperature of

2 x 10- 8 K has been attained by adiabatic nuclear demagnetization (Chapter 3).

Question. Does the integral approximation apply at 2 x 10- 8 K? P16.7(b)

(a) q

= Lgje- P' j [16.9] = L j

We use he f3 = (i) q

gje-hcPvj.

j

I I 1 at 298 K and I at 5000 K. Therefore, 207 cm3475 cm-

= 5 + e-4707/207 + 3e- 475 1/207 + 5e-10559/207

= (5) + (1.3

x 10- 10)

+ (3.2 x

10- 10 ) + (3.5 x 10- 22 )

= 15.00 I.

324


(ii) q

= 5 + e- 4707/3475 + 3e-475 1/ 3475 + 5e-10559/ 3475 = (5)

+ (0.26) + (0.76) + (0.24)

g'e- fJsj

= _1_ _ =

(b) Pj

1

Srn

[16.7, with degeneracy gj included]

q

= ~ = [QQJ at 298 K and !0.80 !at 5000 K .

Therefore, Po

(c)

!.

g 'e - hcf3vj

q

P2 =

= ! 6.26

q

3e-475 1/ 207 5.00 3e-475 1/3475

P2

=

=

Urn - Urn (0)

6.26

T

= 16.5 x

I

10- 1I at 298 K .

= [QJ3J at 5000 K.

+Nklnq [16.35].

We need Urn - Urn (0), and evaluate it by explicit summation Urn - Urn (0) = E = N A q

L gjCje-{3£j

[16.28 with degeneracy gj included] .

j

In terms of wavenumber units (i) Urn - Urn (0) = _I_tO NAhc

(ii) Um-Urn(O) NAhc

+ 4707 em-I

x e - 4707/207

+ ... } =

4.32 x 1O- 7cm- l ,

5.00

= _I_tO + 4707 em-I

x e - 4707/ 3475

+ ... } = 1178 em-I .

6.26

Hence, at 298 K Urn - Urn(O) = 5.17 x 10- 6 J mol - I

and at 5000 K Urn - Um(O) = 14. IOkJ mol - I.

It follows that (i) Srn =

5. 17 x 10- 6 J mOl - I) ( 298 k

-I + (8.314 J K- Imol ) x

(In 5.00)

= 113.38 J K - I mol -I I[essentially R In 5]. (ii) Srn

P16.9

q

=

3 (14.09 x 10 J mOl - I) 5OO0K

= Lgje-{3£j j

Pi

gi e -{3£i

[16.9]

+ (8.314 J K- I mol - I)

= Lgje- hc{3 Vj. 1

gie - hcf3vi

= -q- [16.7] = =---q

x (In 6.26)

= 1 18.07 J K- I mol - I I.


325

We measure energies from the lower states, and write q

= 2 + 2e- hc {3v = 2 + 2e-( 1.4388 x I21.1 )/(T/ K) = 2 + 2e- 174.2/ (T/ K).

This function is plotted in Fig. 16.2. (a) At 300 K,

Po

= ~ = I + e-:74.2/300 = I0.641,

PI = I - Po = I 0.361·

4

q 3

2

o

200

400

600

800

1000

T/ K

Figure 16.2

(b) The electronic contribution to U m in wavenumber units is U

I dq

- Um(O)

m -_:-"--'---'=-- NAhc

hcq dfJ

=

[16.3 Ial

2ve- hc {3v

= --q

( 121.1 cm - I) x (e-174.2/300) I + e- 174.2/300

-I

= 43.45 cm

which corresponds to 10.52 kJ mol - I I. For the electronic contribution to the molar entropy, we need q and U m as at 300 K. These are 300 K U m - Um(O)

q

0.518 kJ mol - I 3.120

500K 0.599 kJ mol- I 3.412

-

U m (0) at 500 K as well

326


Then we form Sm

=

Urn - Um(O)

+ R In q [16.35].

T

At300K

Sm=

At1500K

I 518JmOI- ) 300K +(8.3 141JK- l mol- l )x(ln3.120)=11.2JK- l mol- l

I

(

Sm=

I 518JmOI- ) 500K +(8.3141JK- l mol- 1) x (In 3.412) = 11.4JK- 1 mo1- 1 .

I

(

P16.11

]

1

At lOOK, hcf3 = 69.50cm- 1 and, at 298 K, hcf3 = 207.22cm- I ' Therefore, at 100 K, (a) q = 1 + e- 213 .30/69.50 + e- 435 .39/69.50 + e-636.27/69.50 + e-845.93/69.50 = 11.0491 and at 298 K (b) q = 1 + e- 213 .30/207.22 + e- 425.39/207.22 + e-636.27/207.22 + e-845.93/207.22 = e-hc{3v;

In each case, Pi = - - [16.7] . q

Po =

~q

= (a) I 0.9531,

(b) I 0.6451·

e- hc{3V\ PI = - - = (a)1 0.0441, q

(b) I 0.230 I·

e - hc {3V2

P2 = - - = (a)1 0.0021, q

(b)10.083

I.

For the molar entropy we need to form Um - Um(O) by explicit summation: Urn - Urn (0) -- -NA q

'~ " i

. -{3s; -_ -NA fie q

"'hcv,e-. ~ i

-hc{3v;

[1629 . , 1630] .

= 1123 J mol- I (at 100 K) 1,11348 J mol-I (at 298 K) Srn =

Urn - Urn(O) T + R In q [16.35].

(a) Sm =

1 123 Jmo1I I 1I lOOK + R In 1.049 = 1.631 K- mo1-.

(b)Srn=

1348Jmo]-1 I 1 1I 298K +Rln1.55 = 8.17JK- mo1-.


(a) W =

N! nl !n2! . . .

[16.1] =

I.

5! = 0!5!0!0!0!

OJ.

I.

[ill.

I

STATISTI CAL THERMODYNAMICS 1: THE CONCEPTS

327

(b) We draw up the following table. 0

& 2&

4 3 3 2 2 1 0

0 I 0 2 1 3 5

w=

3& 4& 5&

0 0 1 0 2 1 0

0 0 1 I 0 0 0

0 I 0 0 0 0 0

N!

111 !112!··· 5 20 20 30 30 20

I 0 0 0 0 0 0

I

I I

I

The most probab le configurations are (2, 2, 0, 1, 0, 0 I and (2, 1, 2, 0, 0, 0 I jointly.

P16.15

(a)

I1j

~

= e- fJ(£j - £o) = e - fJj£, which implies that -jl3& = In I1j -

In 110 and therefore that In I1j

= In 110 -

j

k& .

T

&

Therefore, a plot of In I1j agai nst j should be a straight line with slope - kT . Alternatively, plot In Pj againstj, since Inpj

j&

= const -

- . kT

We draw up the fo llowing table using the information in Problem 16.8.

o

j

2 0.69

4 1.39

2

3

2 0.69

o

I

[most probable configuration]

&

These are points plotted in Fig. 16.3 (full line). The slope is - 0.46 and, since he corresponds to a temperature

T

=

(50cm-

l)

x (2 .998 x 10 10cms- l ) x (6.626 x 1O- 34 J s) (0.46) x ( 1.381 x 1O- 23 JK -I)

=

= 50 cm- I, the slope

~

160K.

(A better estimate, 104 K represented by the dashed line in Fig. 16.3, is fo und in Problem 16. 17.) (b) Choose one of the weight 2520 configurations and one of the weight 504 configurations, and draw up the fo llowing table.

W

= 2520

j

0

I1j

4 1.39 6 1.79

Inl1j

W=504

I1j

In I1j

2 3 1.10 0 - 00

I 0 I 0

Inspection confirms that these data give very crooked lines.

3

4

0

I 0 1 0

- 00

1 0

328

STUDENT'S SOLUTIONS MANUAL 1.6

1.2

0.8 ;,,~

.E 0.4

0

-{).4

0

3

2

4 I

-00 .+ P16.17

dj3 =

=

Hence, efJ'

N - - [16.3Ib],

withq=

I P [16 . 12] . 1- e- ,

- ee- P' qdfJ = 1- e- P" 1 dq

dlnq

ae

dlnq dfJ

=-

(a) U - U(O)

Figure 16.3

U - U(O)

ee(-fJ')

N

1- e- fJ '

1+a-, . . that, fJ = -In 1 (1 + =Implymg e

a

I)

- . a

1 For a mean energy of e, a = I, fJ = - In 2, implying that e

T = _ e _ 1n2 = (50cm-

l

k~2

1 (b) q = 1 - e- fJ , =

1

(

) =

a

1-

x

)

(~) k~2

= 1104 K

I.

~

1+ a .

--

I+a S U - U(O) (c) - = + ln q[16.35]=afJe +ln q Nk NkT

= a ~ (I +

~) +

In(I + a)

= a In(l

+ a) - a In a + In ( 1 + a)

=1 ( I +a) ln(I +a ) -a lnal· . When the mean energy IS e, a P16.19

P=kT(alnQ)

av

= 1 and then ~ ~.

[17 .3] T .N

= kT (a In(qNIN!») av

[16.45b] T,N

= NkT (a In q )

= kT (a[N Inq -InN!])

av

T .N

av

T.N


=NkT

aIn(V /A3») av T.N

(

3

=NkT

_

329

a[lnV - InA ]) (

av

N~T

=NkT

(alnV) -av

T .N

T .N

IPV = NkT = nRT I.

or


= e-1V(r) - V(ro)} / kT

At equilibrium N(r)/V N(ro)/V

=

Since V(r)

N(oo)/V N(ro)/V

-GMm /r, V(oo)

[16 .00]

= 0 and [Note:

V(r) is potential energy, V is volume]

= eV (ro) / kT

which says that N(oo)/V <X e V(ro) / kT = constant. This is obviously not the current distribution for planetary atmospheres where lim N (r) / V = O. Consequently, we may conclude thatthe earth's atmosphere, r-> oo

or any other planetary atmosphere, cannot be at equilibrium. P16.23

Each protein binding site can be represented as a distinct box into which a ligand, L, may bind. All possible configurations are shown in the following table and the configuration count of i indistinguishable ligands n! being placed in n distinguishable sites is seen to be given by the combinatorial: C(n, i) = . . (n -I)!I! C(4, 0)

L

L

P16.25

1

L L

L L L L

C(4, 2)

L

=

4! (4 _ 2)!2!

= 6 conformations

L L

L L

L L L L

L L L

(n - i

Pi=

I conformation

L

L L

L

=

C(4,1)= (4- ~ ) !I! = 4 conformations

L

L L L

4! (4 - O)!O!

4'

L

L L L

=

4! C(4,3) = (4 _ 3)!3! = 4 conformations

1

+ I )asi q

C(4,0)

=

4! (4 _ O)!O!

= I conformation

(n- i+ l )as i

(i)

= L;~o ipi.

(a) The fraction distribution of molec ules with i coiled residues depends dramatically upon the value of the stability parameter s. When s < I , low values of i are observed but large i values are observed

330


when s > I. Thus, when s < I, the polypeptide is largely helical and, when s > I, it is more of a random coil. See Fig. 16.4(a) . Fraction of molecules with i coiled residues when

0.25

(J

=

0.050

.-------r----..-------,------,

02

0.1.5

s= 1.5 0.1

OD.5

10

1.5

L

Figure 16.4(a)

(b) The (i) plot, Fig . 16.4(b), shows that for s < 0.5 the polypeptide model is helical and little changed as s is varied. At the other extreme, for s > 1.5 the polypeptide is largely a random coil which changes only slightly with variance of s. The mean number of coiled residues changes rapidly in the middle range of 0.5 < s < 1.5 giving an overall sigmoidal dependence upon s. Dependence of the mean number of coiled residues upon the stability parameters. From left to right: s=O.OOO 1, 0.001 , 0.0 I, and 0.05

~r----~---~---~---~

1.5

(i )

10

s

Figure 16.4(b)

Statistical thermodynamics 2: applications

17


An approximation involved in the derivation of all of these expressions is the assumption that the contributions from the different modes of motion are separable. The expression qR = kT / heB is the high temperature approximation to the rotational partition function for nonsyrnmetrical linear rotors. The expression q v = kT / hev is the high temperature form of the partition function for one vibrational mode of the molecule in the harmonic approximation. The expression qE = gE for the electronic partition function applies at normal temperatures to atoms and molecules with no low lying excited electronic energy levels.

017.3

Residual entropy is due to the presence of some disorder in the system even at T = O. It is observed in systems where there is very little energy difference--or none-between alternative arrangements of the molecules at very low temperatures. Consequently, the molecules cannot lock into a preferred orderly arrangement and some disorder persists.

017.5

Equations of state can be thought of as expressions for the pressure of a gas in terms of the state functions, n, V, and T . They are obtained from the expression for the pressure in terms of the canonical partition function given in eqn 17.3. p

= kT

(aavQ) . In

T

Partition functions for perfect and imperfect gases are different. That for the perfect gas is given by Q = qN / N! with q = V / A 3 . There is no one form for imperfect gases. One example is shown in Self-test 17.1 . Another which can be shown to lead to the van der Waals equation of state is I 2rrmkT - 2- ) N! ( h

Q= -

3N / 2

(V-Nb)e

aN2/ kTV

.

For the case of the perfect gas there are no molecular features in the partition function, but for imperfect gases there are repulsive and attractive features in the partition function that are related to the structure of the molecules. 017.7

See Justification 17.4 for a derivation of the general expression (eqn 17 .54b) for the equilibrium constant in terms of the partition functions and difference in molar energy, t::;,rEO , of the products and reactants in a chemical reaction. The partition functions are functions of temperature and the ratio of partition functions

332


in eqn 17.54b will therefore vary with temperature. However, the most direct effect of temperature on the equilibrium constant is through the exponential term e -I':. ,Eo/ RT. The manner in which both factors affect the magnitudes of the equilibrium constant and its variation with temperature is described in detail for a simple R ;= P gas phase equilibrium in Section 17.8(c) and justification 17.5.


CV.m = !(3 + vR+ 2vt)R [17.35] with a mode active if T >

11M .

= !(3 + 3 + O)R = 3R [experimental = 3.7R] CV,m = !(3 + 3 + 2 x l)R = 4R [experimental = 6.3R] CV ,m = !(3 + 2 + O)R = ~ R [experimental = 4.5R]

(a) 0 3: CV,m

(b) C2H6: (c) C02:

Consultation of the Herzberg references in Further reading , Chapters 13 and 14, turns up only one vibrational mode among these molecules whose frequency is low enough to have a vibrational temperature near room temperature. That mode was in C2H6 , corresponding to the "internal rotation" of CH3 groups. The discrepancies between the estimates and the experimental values suggest that there are vibrational modes in each molecule that contribute to the heat capacity-albeit not to the full equipartition value-that our estimates have classified as inactive. E17.2(b)

The equipartition theorem would predict a contribution to molar heat capacity of !R for every translational and rotational degree of freedom and R for each vibrational mode. For an ideal gas,

Cp,m = R + CV,m' So for C02 With vibrations

Cv,m/ R =3(!)+2(!)=(3 X 4-6)=6.5

Without vibrations CV,m/ R

Experimental y =

and

= 3 (!) + 2 (!) = 2.5

37. 11 Jmol - 1K- 1 (37 . 11 - 8.3145) J mol - 1K-1

=

~

7.5 6.5

y=-=~

and

3.5 2.5

~

y=-=~

[ill 1.29

The experimental result is closer to that obtained by neglecting vibrations, but not so close that vibrations can be neglected entirely. E17.3(b)

The rotational partition function of a linear molecule is [Table 17.3]

qR

= 0.6950

x

a

T/ K (B / cm -

= I)

(0.6950) x (T / K) 2 x 1.4457

(a) At 25 °C:

qR = (0.2404) x (298) = ~

(b) At 250°C:

qR

= (0.2404)

x (523)

= @]

= 0.2404(T / K)

STATISTICAL THERMODYNAMICS 2: APPLICATIONS

E17.4(b)

333

The symmetry number is the order of the rotational subgroup of the group to which a molecule belongs (except for linear molecules, for which a = 2 if the molecule has inversion symmetry and 1 otherwi se). (a) C02 : full group Dooh; subgroup C2 ; hence a =

[I]

(b) 0 3: full group C2v; subgroup C2 ; a =

[I]

(c) S03: full group D3h; subgroup {E , C3,

C~ , 3C2}; a =

GJ

(d) SF6: full group Oh; subgroup 0; a = 1241

(e) A12Cl6 : full group D2d ; subgroup D2; a = E17.S(b)

[I]

The rotational partition function of a non-linear molecule is [Table 17.3] R 1.0270 (T / K) 3/2 1.0270 x 298 3/ 2 q = - a- (ABC / cm - 3) 1/2 = (2) x (2.02736 x 0.34417 x 0.293535) 1/2 [a = 2] = 15837 1

The high-temperature approximation is valid if T >

~,

where

hc (ABC) 1/3

eR=---k

(6.626 x 10- 34 J s) x (2.998 X 1010 cm S-I) x [(2.027 36) x (0.344 17) x (0.293535) cm- 3 jl / 3 1.381 x 10- 23 J K- I = 10.8479KI E17.6(b)

qR = 5837 [Exercise 17.5(b)] All rotational modes of S02 are active at 25 °C; therefore R

R

Urn - Urn (O)

ER

SR

rn

= -T

+R

= ~R + E17.7(b)

3 = E R = '2RT InqR

Rln(5837) = 184.57 J K- I mol- I I

(a) The partition function is

states

leve ls

where g is the degeneracy of the level. For rotations of a symmetric rotor such as CH3CN, the energy levels are EJ = hc[BJ(J + I ) + (A - B)K2] and the degeneracies are gJ.K = 2(21 + 1) if K t= 0 and 2J + I if K = O. The partition function , then, is q = 1+ t(21 + l)e - IhcBJ (J+I )/ kTl J= I

(I t + 2

K= I

e - IhC(A-B)K2/kTl)

334


To evaluate this sum explicitly, we set up the following columns in a spreadsheet (values for A 5.28 cm- I , B = 5.2412 cm - I , and T = 298.15 K) J

J(J + I )

2J + I

e - hcBJ (J+I )/ kT

0 2 3

0 2 6 12

I 3 5 7

I 0.997 0.99 1 0.982

82 83

6806 6972

165 167

4. 18 x 10- 5 3.27 x 10- 5

e - {hc(A- B)K 2/ kTi

K sum

J sum

8.832 23.64 43.88

I 0.976 0.908 0.808

2.953 4.770 6.381

9.832 33.47 77 .35

0.079 0.062

8 x 10- 71 2 x 10- 72

11 .442 11 .442

7498 .95 7499.01

Jterm

=

The column labeled K sum is the term in large parentheses, which includes the inner summation. The J sum converges (to 4 significant figures) only at about J = 80; the K sum converges much more quickly. But the sum fai ls to take into account nuclear stati stics, so it must be divided by the symmetry number (a

= 3).

= 12.50 x

At 298 K, qR

103 1. A similar computation at T

= 500 K

yields qR = 15.43 x 103 1.

(b) The rotational partition functio n of a nonlinear molec ule is [Table 17.3 with B R

q

=

1.0270 (T / K)3/2 - a - (A BC/c m - 3 ) 1/ 2

1.0270 (T / K )3/2 3 (5.28 x 0.307 x 0.307)1 /2

At 298 K, qR

= 0.485 x 298 3/ 2 = 12.50 x

At 500 K , qR

= 0.485

x 5003 / 2

= 15 .43

= C]

= 0.485

x (T / K)3/ 2

103 1

x 103 1

The high-temperature approximation is certainly valid here. E17.8(b)

The rotational partition function of a nonlinear molecule is [Table 17.3] R

1.0270

(T / K)3/2

q -

a

(ABC/cm-3) 1/2

(a) At 25 °C,

qR

qR

(b) At 100 °C ,

E17.9(b)

= 1.549 X =

1.0270 x (T / K)3/ 2 (3. 1752 x 0.3951 x 0.3505) 1/2

= 1.549 x

(T / K)3/2

(298)3/ 2 = 17.97 x 1031

1.549 x (373)3/ 2 = 11.12 x 104 1

The molar entropy of a collection of oscillators is given by Srn

=

Urn - Urn(O) +klnQ [17. 1]

T

= efJhcv _

T

Ov

hcv

where (&)

= NA(& ) +Rlnq

I

= k eOv / T

_

1 [17.28], q

and Ov is the vibrational temperature hcv/k. Thus

Srn

=

R(Ov/ T) _ RI ( I _ -Ov / T) n e - I

vv./ T eA

=

I I_

e -fJhcv

1 I _ e -Ov / T [17 .19]


335

A plot of Sm / R versus T lev is shown in Figure 17.1.

2.5 ~

'i;

2

""' 1.5

0.5

o

o

4

2

6

10

8

TIO v

Figure 17.1

The vibrational entropy of ethyne is the sum of contributions of this form from each of its seven normal modes. The table below shows results from a spreadsheet programmed to compute Sm / R at a given temperature for the normal-mode wave numbers of ethyne. T

= 298 K

T

= 500 K

ii/em - I

ev/K T /8v

Sm/ R

T/ev

Sm/ R

612 729 1974 3287 3374

880 1049 2839 4728 4853

0.216 0.138 0.000 766 0.000 00217 0.000 00146

0.568 0.479 0.176 0.106 0.103

0.554 0.425 0.0229 0.000 818 0.000 652

0.336 0.284 0.105 0.0630 0.0614

The total vibrational heat capacity is obtained by summing the last column (twice for the first two entries, since they represent doubly degenerate modes). (a) At 298 K, Sm

= 0.708R = 15.88 J mol-I

(b) At 500 K, Sm

=

1.982R

K- 1 1

= 116.48 J mol - I K- I 1

E17.10(b) The contributions of rotational and vibrational modes of motion to the molar Gibbs energy depend on

the molecular partition functions G m - Gm(O) = -RT In q [17.9; also see Comment to Exercise 17.6(a)]

The rotational partition function of a nonlinear molecule is given by

qR =

~ a

(kT)3 /2 (~)1 /2 = 1.0270 ( (T/K)3 )1 /2 he ABC a ABC/ cm- 3

and the vibrational partition function for each vibrational mode is given by

v q

=

I

1- e-(J / T

where 8

hcii

=-

k

=

1.4388 (ii/em- I)

--:=-::-:-:--

(T /K)

336


R _

At 298 K

q

-

1.0270 ( 298 3 ) - 2 - (3.553) x (0.4452) x (0.3948)

1/ 2

3

= 3.35 x 10

and C~, - C~(O) = -(8.3 145 J mol- I K- I) x (298 K) In 3.35 x 10 3

= -20.1 x 103 J mol - I = 1-20.1 kJ mol-I

I

The vibrational partition functions are so small that we are better off taking

Inq; "'" e- II .4388(IIIO)/298} = 4.70 x 10- 3 In

qi "'" e -1 1.4388(705 )/ 298} = 3.32 x 10- 2

Inqj "'" e- 11.4388(1042)/298} = 6.53 x 10- 3

so

C~ - C~(O) = - (8.3 145 J mol - IK - I) x (298 K)

x (4.70 x 10- 3 + 3.32 x 10- 2 + 6.53 x 10- 3) =-IIOJmol - 1 =1-O. llOkJmol-1 1

where g = (2S + 1) x

E17.11(b)

I {2

for 2: states for

n, ~,

...

states

The 32: term is triply degenerate (from spin), and the I ~ term is doubly (orbitally) degenerate. Hence

q=3+2e- fJ e At 400 K {3e =

( 1.4388 cm K) x (7918.1 cm- I) = 28 .48 400K

Therefore, the contribution to C m is COl - COl (0) = - RT In q [Table 17.4 for one mole 1

-RT Inq = - (8.3 14J K- I mol - I) x (400 K) x In (3 +2 x e- 2848 )

= -(8.3 14J K- I mol-I ) x (400 K) x (In 3) = 1-3.65 kJmol - 1

COMMENT. The contribution of the excited state is negligible at this temperature.

I


=

E17.12(b) The degeneracy of a species with S

Srn

=

Urn - Um(O)

T

+R

In q

337

~ is 6. The electronic contribution to molar entropy is

= R In q

(The term involving the internal energy is proportional to a temperature-derivative of the partition function, which in tum depends on excited state contributions to the partition function ; those contributions are negligible.)

Srn E17.13(b) Use Sm

= (8.3145 Jmol- 1 K- 1) In6 =[ 14.9Jmol- 1 K- 1 [ =Rlns [17.52bl

Draw up the following table 0

n:

s

1

2

6

Sm / R 0 1.8

3

4

5

0

m

p

a

b

c

0

m

p

6

6

6

2

6

6

1.8

3 1.1

6

1.8

1.8

1.8 0.7

1.8

1.8

3 1.1

6

6

1.8 0

where a is the 1,2,3 isomer, b the 1,2,4 isomer, and c the 1,3,5 isomer. E17.14(b) We need to calculate

Each of these partition functions is a product

with all qE

= 1.

The ratio of the translational partition functions is virtually 1 (because the masses nearly cancel; explicit calculation gives 0.999). The same is true of the vibrational partition functions. Although the moments of inertia cancel in the rotational partition functions , the two homonuclear species each have a = 2, so q

R 9 Br2)qR(SIBr2) qR (19BrS1 Br)2

C

= 0.25

The value of I'1Eo is also very small compared with RT, so

338



qE (a ) qE af3 v

Urn - Urn CO) = -NA CV,rn = -kf32

-

C~rn ) v

[17 .31a].

Let x = f3 £, then df3 = (1 /£) dx.

Therefore,

We then draw up the following table. TIK

(kT/hc)/mo\-1

x CV,m/ R CV.rn / (J K - I mol- I)

50

298

500

34.8 3.48

207 0.585

348 0.348

[Q;iliJ 2.9 1

10.0791 10.0291 0.654

0.244

Note that the double degeneracies do not affect the results because the two factors of 2 in

COMMENT.

q cancel when U is formed. In the range of temperature specified , the electronic contribution to the heat capacity decreases with increasing temperature. P17.3

The energy expression for a particle on a ring is

Therefore 00

00

m=- OO

m= -OO

STATISTICAL THERM ODYNAMI CS 2: APPLICATIONS

339

The summation may be approximated by an integration

q

~ .!.. / 00 e-

-00

a

U - U(O)

m

=

fn 2 / 2IkT dml

SOl

=

/ 00 e-00

x2

d.x

~ .!.. ( 2n /;T) 1/2 = .!.. (2;/) 1/ 2 a

a

ti

= _N aInq = -N ~ In.!.. ( 27r2 /)1 / 2 = !!.- = ~NkT = ~RT af3

C V. ol =

.!.. ( 2IkT) 1/ 2 a ti 2

aU m ) ( -aT

af3

ti f3

a

2f3

2

2

(N

ti f3

= NA) .

[ - I -I [ v = '2I R =4.2JK mol .

Urn - UOl(O) +Rlnq T

1 I (2nlkT) = -R + Rln- - 22 a ti

1/ 2

1 1 ( 2n) x (5.341 x 1O- 47 kgm 2) x (1.38 1 x 1O- 23 JK- I ) x (298»)1 /2 =-R+Rln2 3 ( 1.055 X 10- 34 J s)2

= ~R + 1.31R = 1.81R, orllS J K- 1 mol-I 2

P17.5

I.

The absorption lines are the values of differences in adjacent rotational terms. Using eqns 13.25, 13.26, and 13.27, we have

F(J+l)-F(J)=

E(J

+ I) -

E(J)

he

=2B(J+I )

for J = 0 , I , .... Therefore, we can find the rotational constant and recon struct the energy levels from the data. To make use of all of the data, one would plot the wavenumbers, which represent F(J + I ) - F(J) , vs. J ; from the above equation, the slope of that linear plot is 2B. Inspection of the data show that the lines in the spectrum are equally spaced with a separation of 2 1. 19 cm- I , so that is the slope: slope

= 21.19cm- 1 =2B

so

B= 10.S95cm-

l

.

The partition function is

00

q

= L (2J + l)e-fJE(J)

where

E(J)

=

heBJ(J

+ I ) [13 .25]

1=0

+ I is the degeneracy of the energy levels.

and the factor of 2J At 25°C , heBf3

00

q

=L

heB

=-

(2 J

kT

=

6.626 x 10- 34 J s x 2.998 X 10 10 cm S- I x IO.S95cm - 1 1.381 x 10- 23 J K- 1 x 298. 15 K

+ l )e-O.OSII 2J(J+ I)

1=0

= 1 + 3e-0.OSlI 2x 1x2 + Se- 0.OS I1 2x2x3 + 7e -0.OS I1 2x3 x4 + .. . = 1 + 2.708 + 3.679 + 3.791 + 3.238 + ... = [ 19.90 [.

= 0 .05112.

340 P17.7


The molar entropy is given by

Sm = Urn -

Um(O) +R(ln qm _ T NA

where Urn - Um(O) T

= -NA (aln q ) a{3

I) and

v

qm NA

= q~qRqVqE NA

The energy term Urn - Urn (0) works out to be

Translation: Te

~ = 2.561 x 1O- 2(T / K)5 /2 x (M/g mol- I)3/2 [Table 17.3) NA

= 2.561

x 10- 2 x (298)5/2 x (38.00)3/2 = 9.20 x 106

= ikT .

and (s T)

Rotation of a linear molecule:

0.6950 = __

qR

a

T/ K x - - - I [Table 17.3].

B/cm-

The rotational constant is

Ii Ii B - - - - -:------:-.;- 4rrcl - 4rrCf.,l,R2 (1.0546 x 10- 34 1 s) x (6.022 x 1023 mol - I) 4rr(2.998 x JOlOcms- l) x x 19.00 x 10-3 kgmol- l ) x (190.0 x 1O- 12 m)2

(!

= 0.4915cm - 1 so qR = 0.6950 x ~ = 210.7. 2 0.4915

= kT.

Also (sR) Vibration :

l-exp(-1.4388(ii/cm I) /( T / K»

= v

1.129. (6.626 x 10- 34 ls) x (2.998 x IO lO cms- l ) x (450.0 cm - l )

hciJ

(s )

= ehcv /kT _ = 1.149 X

I

exp(1.4388(450.0)/298) - I

10- 211.

The Boltzmann factor for the lowest-lying excited electronic state is

~p

(

I - exp( -1.4388(450.0) /298)

-(1.60geY) x (1.602 x 10- 19 leV- I») (1.381 x 10- 23 1 K- I) x (298K)

=6 x

l

0- 28


341

so we may take qE to equal the degeneracy of the ground state, namel y, 2 and (e E ) to be zero. Putting it all together yields

um -

U (0)

T

m

=

N ~ (lkT

T

)

+ kT + 1.149 x

10- 21 J

=

2.2R +

N (1.149 X 10- 2 1 J) _A_ _ _ _ __

2

= (2.5)

_I x (8.3 1451 mol

_I K

)

+

T

(6.022 x 1023 mol- I) x (1.149 x 10- 2 1 J) 298 K

= 23.11 J mol-I K- I. R (In

~:

- 1)

= (8.3 145 J mol- I K- I) =

P17.9

176.31 mol - I K- I

x (In[(9.20 x 106 ) x (2 10.7) x (1.129) x (2)] - I}

and

S~ = 1199.4 J mol -I

K- I \.

(a) The probability distribution of rotational energy levels is the Boltzmann factor of each level, weighted

by the degeneracy, over the partition function (21

L

+ l)e- hcBJ (1+I) / kT + l)e-hcBJ(J+I) / kT

-=-------;-;::-;-;-:---:-:-;=

(2J

[17 .13] .

J=O

It is conveniently plotted agai nst J at several temperatures usi ng mathematical software. This distribution at 100 K is shown in Fig. 17.2(a) as both a bar plot and a line plot.

Rotational distributions 0. 15 r - - - , - - - - , - - - - - , - - - - , - - - - - , - - - - - - , - - - r - - - - - - ,

lOOK 0.1 rI}(T)

Figure 17.2(a)

The plots show that higher rotational states become more heavily populated at higher temperature. Even at lOOK the most populated state has 4 quanta of rotational energy; it is elevated to 13 quanta at 1000 K. Values of the vibrational state probability distribution, y

p" (T)

e --

-0.8

-I

200

0

400

600

Temperature/ K

800

1000

Figure 17.2(b)

Energy cbange contributions

T /K

Figure 17.2(c)

The derivative dU v / dT may be evaluated numerically with numerical software (we advise exploration of the technique) or it may be evaluated analytically using eqn 17.34:

Cv

(}y

v

{(}y (

dT -

T

- dU --R-

V,m -

e-Ov /2T )} 2 l-e-Ov/ T

where = hev / k = 3 122 K. Fig. 17 .2(d) shows the ratio of the vibrational contribution to the sum of translational and rotational contributions. Below 300 K, vibrational motion makes a small , perhaps negligible, contribution to the heat capacity. The contribution is about 10% at 600 K and grows with increasing temperature.

344

STUDENT'S SOLUTIONS MANUAL Relative contributions to the heat capacity 0.2 , - - - - - , - - - - - - , - - - - - - - , - - - - - - , - - - - - - ,

C~. m

0I .

C~. m+ C~. m

o

L -_ _ _ _ _ _

~

o

____

~~~

200

_ _ _ _ _ _ _ _L __ _ _ _ _ __ i_ _ _ _ _ _

400

600

~

800

1000

TI K

Figure 17.2(d)

The change with temperature of molar entropy may be evaluated by numerical integration with mathematical software. 6.S(T)

= SeT) -

S(100 K)

=

I

T

lOOK

= (T ilOOK

=

6.S(T)

I

T

CV,m(T) T -27

R

C (T)dT p,m [3.18] T

+ R dT [2.48]

+ C~.meT) dT. T

lOOK

= 2R In (_T_) + (T C~,m(T) dT 2

lOOK

T

ilOOK

~ /),.ST +R (T)

'~----~---~ /),.SV (T)

Fig. 17 .2( e) shows the ratio of the vibrational contribution to the sum of translational and rotational contributions. Even at the highest temperature the vibrational contribution to the entropy change is less than 2.5% of the contributions from translational and rotational motion. The vibrational contribution is negligible at low temperature.

Relati ve contributions to the entropy change 0.03 , - - - - - - . - - - - - , - - - - - - - , - - - - - - - , - - - - - - ,

0.02

0.01

o L -_ _ _

o

~

_____

200

~~

____

400

~

_ _ _ _ _ L_ _ _ _

600

TI K

800

~

1000

Figure 17.2(e)


P17.11

+ DCI ;=! HDO + HCI.

H20 K

345

= q"(HDO) q" (HCI) e-{3Mo [17.54; NA factors cancel] . q" (H20 )q" (DCI)

Use partition function expressions from Table 17.3. The ratio of translational partition functions is

q~(HDO)q~(HCI) = (M (HDO)M (HCI ) ) q~(H20)q~(DCI)

3/ 2 = (19.02

X 36.46) 3/2 18.02 x 37.46

M(H20)M(DCI)

= 1.041 .

The ratio of rotational partition functions is qR (HDO)qR(HCI)

=

qR(H20)qR(DCI)

a (H20 ) (A(H20 )B(H 20 )C(H20 ) / cm-3) 1/2B(DCI)/cm- 1 - 1 - (A(HDO)B(HDO)C(HDO)/cm- 3) 1/2B(HCI) /c m- 1

=2 x

(27.88 x 14.51 x 9.29)1 /2 x 5.449 (23.38 x 9.102 x 6.417)1 /2 x 10.59

= 1.7

07

(a = 2 for H20 ; a = I for the other molecules).

The ratio of vibrational partition functions (call it Q) is q v (HDO)q v (HC!)

q(2726. 7)q( 1402.2)q(3707.5 )q(299 I )

=

q v (H20 )q v (DCI)

where q(x) I:!.Eo he

q(3656.7)q(l594.8)q(3755 .8)q(2 145 )

=Q

I

=

I_

e- 1.4388x/ (T / K ) ·

= ~ {(2726.7 + 1402.2 + 3707 .5 + 2991) 2

= -162cm-

l

(3656. 7 + 1594.8

+ 3755 .8 + 2 145)}cm- I

.

So the exponent in the energy term is -f3I:!.Eo

Therefore, K

I:!. Eo

he

kT

k

= -- = - =

I:!. Eo

x -

1.041 x 1.707 x

Q

he

I x -

T

=-

x e 233 / (T/ K)

=

1.4388 x (- 162) T/K

233

= +--. T/K

1.777 Qe 233/ (T / K).

We then draw up the following table (using a computer)

T/ K K

100 18.3

200 5.70

and specifically K

300 3.87

400 3.19

500 2.85

600 2.65

700 2.51

800 2.41

900 2.34

1000 2.29

= 13.891 at (a) 298 K and [IiI] at (b) 800 K.


(a) Bv and BR are the constant factors in the numerators of the negative exponents in the sums that are the partition functions for vibration and rotation. They have the dimensions of temperature, which

346


occurs in the denominator of the exponents. So high temperature means T » (}y or OR and only then does the exponential become substantial. Thus (}y and ~ are measures of the temperature at which higher vibrational and rotational states, respectively, become significantly populated:

~ _ hcf3 _ (2.998 x IOlO cm -

k

-

hcv k

=

x (6.626 x 10- 34 ) s) x (60.864cm- l ) ( 1.38I x lO- 23 )K - I )

S- I)

= 187.55 K 1

and (}y

=-

(6.626

X

10- 34 )

s)

x (4400.39 cm - I ) x (2.998 x 10 10 cm (1.381 x 1O- 23 )K- I )

s-I)

=

1 1 6330 K .

(b) and (e) These parts of the solution were performed with Mathcad 7.0 and are reproduced on the following pages.

Objective: To calculate the equilibrium constant K(T) and Cp(T) for dihydrogen at bigh temperature for a system made with n mol H 2 at I bar. H 2(g) ;=: 2H(g)

At equilibrium the degree of dissociation, a, and the equilibrium amounts of H 2 and atomic hydrogen are related by the expressions nH2 = ( I - a)n

and

nH

= 2a n.

The equilibrium mole fractions are XH2 XH

= (l - a)nl{(l - a)1I + 2an) = (l - a)/(l + a), = 2an/{(I - a)n + 2an} = 2a/(I + a) .

The partial pressures are

PH 2

= (l -

a)p/(1

+ a)

and

PH

= 2ap/(1 + a).

The equilibrium constant is

4a 2 - - - - ;?,-

(I - a-)

where p

= p" = I

bar.

The above equation is easily solved for a :

1a = (KI(K + 4»1 /21· The heat capacity at constant volume for the equilibrium mixture is

The heat capacity at constant volume per mole of dihydrogen used to prepare the equilibrium mixture is Cv

= Cv(rnixture)/n = {nHCv,m(H) + nH2CV,m(H2)}/n = I2aCv,m(H) + (l - a)CV,m(H2) ,.


347

The formula for the heat capacity at constant pressure per mole of dihydrogen used to prepare the equilibrium mixture (Cp ) can be deduced from the molar relationship Cp,m Cp

= CV,m + R .

= (nHC".m(H) + nH2Cp,m (H2) lin = nH (Cv.m(H) + RI + nH2 ( CV,m(H2) + RI n

n

= nHCv.m(H) +nnH2C v.m(H 2)

+R(nH:nH2)

=Cv +R( I+a ).

Calculations J = joule mol = mole h = 6.62608 x 10- 34 J s R = 8.31451 J K - 1 mol- I

s = second = gram e = 2.9979 x 10 8 m s- I NA = 6.022 14 X 10 23 mol- I g

kJ = 1000 J bar=l x lO5 Pa k = 1.38066 X 10- 23 J KpG = I bar

I

Molecular properties of H2: v = 4400.39 cm- I ,

B = 60.864 cm- I ,

D = 432.1 kJ mol-I.

I g mol- I he\!

ev = k '

heB

~ -- k .

Computation of K (T ) and a (T ) N = 200,

i = 0, ... , N

T; = 500 K

+

i x 5500 K N

qv ; = I _ e- (&v I T;) ' Keq; a;

=(

Keq;

) 1/ 2

+4

See Fig. 17.3(a) and (b). Heat capacity at constant volume per mole of dihydrogen used to prepare the equilibrium mixture is (see Fig. 17.4(a» C v (H) =

[ill,

C V(H2') =

,

2 .5R +

ev [ -T;

e-o

=

2rr.NA {r3 _ 3 I

= 2rr.NA {r3 _

3

1

E(r~

- r

i)} _T 2 rr.NA

{ E(r~

3

kT

2E(r~ -

ri) } = b _

kT

~. RT

The Joule-Thomson coefficient itself is [2.55] /1-

= _ /1-T = 2rr.NA {2E(r~ Cp

3Cp

kT

ri) _ ri}

- r

kT2

= b - ~~ . Cp

i)}


P17.17

(a) Ethene belongs to the D 2h point group, whose rotational subgroup includes E and 3 C2 elements around different axes. So a = 4. The rotational partition function of a non-linear molecule is [Table 17.3] R

=

q

1.0270 (T / K) 3/2 - a - (ABC/c m-3) 1/2

(b) Pyridine belongs to the R

q P17.19

351

1.0270

e2v

=

1.0270 x 298 .15 3/ 2 (4) x (4.828 x 1.0012 x 0.8282)1 /2

group, the same as water, so a

(T /K)3/2

= -a- (ABC/cm -3 ) 1/ 2 =

~

=~.

= 2.

1.0270 x 298.15 3/ 2 (2) x (0.2014 x 0.1936 x 0.0987)1 /2

=

1

41

4 26 x 10 . .

The partition function of a system with energy levels e (J) and degeneracies g(J) is q

= Lg(J)e - P£(J) J

The contribution of the heat capacity from this system of states is Cv

= -kfJ2 (au) afJ

[17.3Ia] v

= -N C~;q)v = -~

where U - U(O)

G;)v·

Express these quantities in terms of sums over energy levels

u-

_'!.. (- L

U(O) =

q

g(J)e (J)e-P£(J») =

'!.. Lg(J)e (J)e-P£(J) q

J

J

and Cv ) -kfJ-

= (au) = '!.. (_ L afJ

v

= _'!.. L q

q

J

q g(J )e 2 (J)e - P£(J») _ N '"' g(J)e(J)e- P£(J) ( a ) q2 afJ

g(J)e2(J)e - P£(J)

7

+~ q

J

(I)

L g(J)e(J)e-P£(J) L g(J')e(J')e-P£(J'). J'

J

Finally a double sum appears, one that has some resemblance to the terms in l;(fJ). The fact that l;(fJ) is a double sum encourages us to try to express the single sum in Cv as a double sum. We can do so by multiplying it by one in the form (L: g(J ')e - Pd J')/q, so J'

~ = _ q2 N -kfJ 2

'"' g(J)e2(J)e-P£(J) '"' g(J')e-P£(J' ) ~ ~ J

+ q2 N

J'

'"' g(J)e(J)e-P£(J) '"' g(J')e(J')e-P£(J' ). ~ ~ J

J'

Now collect terms within each double sum and divide both sides by -N :

~ kNfJ2

=

~ '"' g(J)g(J') e 2(J )e - P[£(J)+£(J' )1 - ~ '"' g(J)g(J')e(J)e(J')e-P[£(J)+£(J') ]. q- ~ q2 ~ J,./'

J,./'

352

STUDENT' S SO LUTIONS MANUAL

Clearly the two sums could be combined, but it pays to make one observation before doing so. The first sum contains a term £2(1 ), but all the other factors in that sum are related to 1 and l ' in the same way. Thus, the first sum would not be changed by writing £2 (1') instead of £2(1) ; furthermore, if we add the sum with £2 (1' ) to the sum with 8 2(1), we would have twice the original sum. Therefore, we can write (fi nall y combining the sums)

Recogni zing that £2(1)

+ 82 (1 ' ) -

28 (1 )£ (1' )

= [£(1 ) -

8(1 ' )f , we arrive at

kNfJ 2

= -2- ~(fJ ).

Cv

For a linear rotor, the degenerac ies are g(1) 8(1)

so fJ £(1 )

= hcBl(1 +

I)

= ()R kJ (1 +

= 2J + I. The energies are

I)

= ()R l(1 + I ) / T .

The total heat capacity and the contributions of several transitions are plotted in Fig. 17.5 . One can evaluate CV.m / R using the following ex pression, deri vable fro m eqn (I) above. It has the advantage of using single sums rather than double sums.

(fJ) is defined in such a way that J and J' each run independently from 0 to infinity. Thus, identical terms appear twice. (For example, both (0, 1) and (1,0) terms appear with identical value in (fJ ). In

COMMENT.

the plot, though, the (0,1) curve represents both terms.) One could redefine the double sum with an inner sum over J ' running from 0 to J - 1 and an outer sum over J running from 0 to infinity. In that case, each term appears only once, and the overall factor of 1/2 in Cv would have to be removed .

P17.21

All partition fun ctions other than the electronic partition function of atomic I are unaffected by a magnetic field; hence the relative change in K is due to the relati ve change in qE . E _ '"'

q -L e

- g/ls fJ l3Mj

,

M __ 1 _ ! +! + 1 . - ~ J 2 ' 2' 2 ' 2, g - 3'

MJ

Since gf-t BfJ B qE

=L

«

1 for normall y attainable fields, we can expand the exponentials

{ 1 - gf-tBfJBMJ

+ ~(gf-tBfJBMJ)2 + ... }

MJ

~ 4+ ~(gf-t BfJB)2

L MJ

MJ [LMJ

0

= 0] = 4(1 + 1 9(f-tBfJ B )2 )

[g= ~l

MJ

This partition function appears squ ared in the numerator of the equil ibrium constant expression. (See solution to E 17 .14(a).) Therefore, if K is the actual equilibrium constant and KOis its value when B 0,

=

STATISTICAL THERMODYNAM ICS 2: APPLICATIONS

353

1.2

0.8

-u'" c:.:::

E

0.6

0.4 1,3 0.2

----=::::::: -

0,3

0 2

0

4

3

5

Temperature , T IeR

Figure 17.5

we write

For a shift of I per cent, we req uire

Hence B ~

0.067kT

=

(0.067) x ( 1.381 x 1O- 23 JK-

J.i.B

9.274 x 10- 24 J T-

I)

x (lOOOK)

I

~

~

lOOT .


S

= k In W [16 .34] .

so S = k In 4N = Nk In 4

=

(5 x 10 8 ) x ( 1.38 x 1O- 23 J K-

I

)

x In4

= 19.57 x

10-

15

JK

- I

I.

Question. Is this a large residual entropy? The answer depends on what comparison is made. Mu lti ply the answer by Avogadro's number to obtain the molar residual entropy, 5.76 x 10 9 J K- I mol- I, surely

354


a large number-but then DNA is a macromolecule. The residual entropy per mole of base pairs may be a more reasonable quantity to compare to molar residual entropies of small molecules. To obtain that answer, divide the molecule's entropy by the number of base pairs before multiplying by NA. The result is 11.5 J K- 1 mol- I, a quantity more in line with examples discussed in Section 17.7. P17.2S

The standard molar Gibbs energy is given by

c: - c:

G

(0) = RTln qrn NA

Translation (see table 17.3 for all partition functions) :

=

(2.561 x 10- 2) x (2000)5/2 x (38.90)3/2 = 1.111

X

109 .

Rotation of a linear molecule: q

R

kT 0.6950 =--=--x ahcB

a

T /K B / cm - I

.

The rotational constant is Ii B - -- -

Ii -;-----=:;;-

- 4ncl - 4ncmeffR2

where meff

mBmSi

= ---mB

B -_ so qR

=

+mSi

4n(2.998 x

IO lO

0.6950 x 2000 0.5952 1

(l0.81) x (28.09) 1O- 3 kg mol- I -----::----:-.,...-::--::-- x 10.81 + 28 .09 6.022 x 1023 mol- 1

cms-

l)

= 1.296 x

1O- 26 kg.

34 1.0546 x 1O- J s -I -_ 05952 . cm x (1.296 x 10- 26 kg) x (190.5 x 1O- 12 m)2

= 2335.

Vibration: qv

= __-;--::-;-;-;;:1-

e-hcv/kT

-1.4388(ii / cm- I») 1- exp ( T/ K

1 - exp (

-1.4388(772) ) 2000

= 2.467. The Boltzmann factor for the lowest-lying electronic excited state is exp (

- (1.4388) x (8000») _ 3 2 10-3 2000 - . x .

The degeneracy of the ground level is 4 (spin degeneracy = 4, orbital degeneracy excited level is also 4 (spin degeneracy = 2, orbital degeneracy = 2), so qE

= 4(1 + 3.2 x

10- 3)

= 4.013 .

= 1), and that of the


355

Putting it all together yields

= (8.3145 J mol- I K- I) x

cr: - cr: (0)

x In[(1.111 x

= 5.135 x P17.27

109 )

(2000 K)

x (2335) x (2.467) x (4.013)]

105 J mol - I

= 1513.5 kJ mol- I I·

The standard molar Gibbs energy is given by

NA

where qm NA

= RTln qm

cr: - cr: (0)

T

= ~qRqV qE [17.53] NA

See Table 17.3 for partition function expressions. First, at 10.00 K T

Translation : ~ NA

x 1O- 2(T / K)5 /2(M / g mol - I )3/2

= 2.561 = (2.561

x 10- 2) x (10.00)5/2 x (36.033)3/2

= 1752.

Rotation of a nonlinear molecule:

qR

= ~ (kT) 3/2 (~)1 /2 = 1.0270 a

hc

ABC

X

a

(T/K)3 /2 . (ABC / cm-3) 1/2

The rotational constants are

B= -Ii-

so

4ncJ

ABC= ( - Ii 4nc

)3- - , IA/B/C

1.0546 x 10- 34 J s )3 ABC = ( 4n(2.998 x IO lO cm s-I) (loIDA m- 1)6

x

= R

so q

------------------------~~---------------------

(39.340) x (39.032) x (0.3082) x (u A2) 3 x (1.66054 101.2cm - 3

1.0270

= -2-

X

( 10.00)3/2 (101.2)1 /2

X

10- 27 kg u- I )3

= 1.614.

Vibration : for each mode

v q

I

=

I -

e - hcv/ kT

1 - exp (

-1.4388(ii/ cm - I)) T/ K

1 - exp (

-1.4388(63.4) ) 10.00

= 1.0001 Even the lowest-frequency mode has a vibrational partition function of 1; so the stiffer vibrations have q v even closer to I. The degeneracy of the electronic ground state is 1, so qE = 1. Putting it all together yields cr: - cr:(0)

= (8 .3145J mol- I K- I) x = 1660.8 J mol - I I.

(lO.OOK) In[(1752) x (1.614) x (I) x (l)]

356


Now at 1000 K q

Translation: ~ = (2.561 x 10- 2) x (1000)5/ 2 x (36.033)3/2 = 1.752 X 10 8 .

NA

1.0270

( 1000)3/2

Rotation:

qR -- - 2 - X (101.2) 1/2 -- 1614.

Vibration:

q'(

I = ---,------,--,,------,,..,....----,-- = 1 I 47

I-exp ( -

qi =

.

,

1

-----:-(---:-(:-l.4-:-:3:-::8-::-:8)- X---:-(1:-::2-=-24-:-.-::-:5)....,--) = 1.207, 1 - exp - - - - I0'--0-0---

v q3 =

( 1.4388) x (63.4») 1000

1

(

(1.4388) x (2040») = 1.056, 1 - exp - ---1-0-00---

qV = (11.47) x ( 1.207) x (1.056) = 14.62 .


c:, - c:,(0) =

(8.3 1451 mol- I K- I ) x ( IOOOK) x 10[( 1.752 x J08 ) x ( 1614) x (14.62) x (I )]

=2.415 x 105 1 mol - I =1 241.5 k1 mol- I I.

18

Molecular interactions


Molecules with a permanent separation of electric charge have a permanent dipole moment. In molecules containing atoms of differing electronegativity, the bonding electrons may be displaced in such a way as to produce a net separation of charge in the molecule. Separation of charge may also arise from a difference in atomic radii of the bonded atoms. The separation of charges in the bonds is usually, though not always, in the direction of the more electronegative atom but depends on the precise bonding situation in the molecule as described in Section 18.1 (a). A heteronuc1ear diatomic molecule necessarily has a dipole moment if there is a difference in electronegativity between the atoms, but the situation in polyatomic molecules is more complex . A polyatomic molecule has a permanent dipole moment only if it fulfills certain symmetry requirements as discussed in Section 12.3(a). An external electric field can distort the electron density in both polar and nonpolar molecules and this results in an induced dipole moment that is proportional to the field. The constant of proportionality is called the polarizability.

018.3

Dipole moments are not measured directly, but are calculated from a measurement of the relative permittivity, Sr (dielectric constant) of the medium. Equation 18.15 implies that the dipole moment can be determined from a measurement of Sr as a function of temperature. This approach is illustrated in Example 18.2. In another method, the relative permittivity of a solution of the polar molecule is measured as a function of concentration. The calculation is again based on the Debye equation, but in a modified form. The values obtained by this method are accurate only to about 10%. See the references listed under Further reading for the details of this approach. A third method is based on the relation between relative permittivity and refractive index , eqn 18.17, and thus reduces to a measurement of the refractive index. Accurate values of the dipole moments of gaseous molecules can be obtained from the Stark effect in their microwave spectra.

018.5

If the A-H bond in the A-H · .. B arrangement is regarded as formed from the overlap of an orbital on A, l/IA , and a hydrogen Is orbitall/lH , and if the lone pair on B occupies an orbital on B, 1/1'8 , then, when the two molecules are close together, we can build three molecular orbitals from the three basis orbitals:

One of the molecular orbitals is bonding, one almost nonbonding, and the third antibonding. These three orbitals need to accommodate four electrons, two from the A-H bond and two from the lone pair on B.

358


Two enter the bonding orbital and two the nonbonding orbital, so the net effect is a lowering of the energy, that is, a bond has formed. 018.7

A molecular beam is a narrow stream of molecules with a narrow spread of velocities and, in some cases, in specific internal states or orientations. Molecular beam studies of non-reactive collisions are used to explore the details of intermolecular interactions with a view to detennining the shape of the intermolecular potential. The primary experimental information from a molecular beam experiment is the fraction of the molecules in the incident beam that is scattered into a particular direction. The fraction is normally expressed in terms of dI , the rate at which molecules are scattered into a cone that represents the area covered by the 'eye' of the detector (Fig. 18.14 of the text). This rate is reported as the differential scattering crosssection, a , the constant of proportionality between the value of dI and the intensity, I, of the incident beam, the number density of target molecules, N , and the infinitesimal path length dx through the sample: dI

= alNdx.

The value of a (which has the dimensions of area) depends on the impact parameter, b, the initial perpendicular separation of the paths of the colliding molecules (Fig. 18.15), and the details of the intermolecular potential. The scattering pattern of real molecules, which are not hard spheres, depends on the details of the intermolecular potential, including the anisotropy that is present when the molecules are non-spherical. The scattering also depends on the relative speed of approach of the two particles: a very fast particle might pass through the interaction region without much deflection, whereas a slower one on the same path might be temporarily captured and undergo considerable deflection (Fig. 18.17). The variation of the scattering cross-section with the relative speed of approach therefore gives information about the strength and range of the intermolecular potential. Another phenomenon that can occur in certain beams is the capturing of one species by another. The vibrational temperature in supersonic beams is so low that van der Waals molecules may be formed, which are complexes of the form AB in which A and B are held together by van der Waals forces or hydrogen bonds. Large numbers of such molecules have been studied spectroscopically, including ArHCI, (HClh, ArC02 , and (H20h. More recently, van der Waals clusters of water molecules have been pursued as far as (H20k The study of their spectroscopic properties gives detailed information about the intermolecular potentials involved.


A molecule that has a center of symmetry cannot be polar. S03(D3h) and XeF4(D4h) cannot be polar.

ISF41 (see-saw, e 2v ) may be polar. E18.2(b)

+ M~ + 2MiM2 cose)i /2 [18.2a] = [(1.5)2 + (0.80)2 + (2) x (1.5) x (0.80)

M = (Mf

x (cos 109S)]i /2

D= II.4D I

MOLECULAR INTERACTION S

E18.3(b)

359

The components of the dipole moment vector are J.Lx = I>iXi = (4e) x (0)

+ (-2e) and J-Ly =

L

+ (-

2e) x ( J62 pm)

x (l43pm) x (cos 30°) = ( -572 pm )e

+ (-2e)

qiYi = (4e) x (0)

x (0)

+ (-2e)

x (l43 pm ) x (sin 30°) = (- 143 pm )e

i

The magnitude is J-L = (J-L~

+ J-L;') 1/ 2 =

« -570)2

+ (-143)2) 1/2 pm e =

(590 pm )e

= (590 x 10- 12 m) x ( 1.602 x 10- 19 C) = 19.45 x 10- 29 C m 1 and the direction is

e=

tan - I J-Ly = tan - I - 143 pm e = 1194.0° 1from the x-axis (i.e. 14.0° below the J-Lx

-572pm e

negative x-ax is). E18.4(b)

The molar polarizati on depends on the polari zability through

Thi s is a linear eq uati o n in T -

so

J.L =

I

with slope

9E:okm) 1/2 = ( --;:;;-

? (4.275 x 1O- -9 Cm) x (m/(m 3 moJ - i K»i /2

and with y-intercept

Since the molar polarization is linearly dependent on T - I , we can obtain the slope m and the intercept b

m=

P m .2 - Pm.1

TI

and

I -

T-

I

(75.74 - 71.43) cm 3 mol - I 3 3 I (320.0K)- I _(42 1.7K)- 1 =5 .72 x 10 cm mol- K

2

b = Pm - mT- 1 = 75.74cm 3 mol- I - (5.72 x 10 3 cm 3 mol- I K) x (320.0 K) - 1 = 57.9cm 3 mol - I

It follows that J-L = (4.275 x 10- 29 C m) x (5.72 x 10- 3 ) 1/ 2 = 13.23 x 10- 30 C m 1

and

360

E18.5(b)


The relative permittivity is related to the molar polarization through

er - I pPm --=-=C er +2 M

C

=

( 1.92gcm- 3 )

r

X

= 0.726

I

(0.726) + I = 18.971 1-0.726

The induced dipole moment is

= ae = 4JTeoa ' e

J.L*

= 4JT(8.854

12

10-

X

rl

C2 m - ' )

X

(2.22

X

10- 30 m 3 )

X

(15 .0

X

10 3 V m-')

x 1O- 36 C m 1

= 13.71 E18.7(b)

(32. l6cm 3 mol-')

X

85 .0gmol

e = 2

E18.6(b)

2C + 1 er = - - , I-C

so

If the permanent dipole moment is negligible, the polarizability can be computed from the molar polarization

3eoP NA

m a=--

so

and the molar polarization from the refractive index

pPm

M

a-

=

er - l 6 r +2

3 X (8.854

-

(6.022

= 13.40 E18.8(b)

X

3eOM(n;-I)

n;-I

= 1l~+2 X

X

a=

so

10- 12 r l C2 m- I )

1023 mol - ')

X

n} +2

NAP

(2.99

X

(65.5gmol - ' )

X 1Q6

gm - 3)

X

(1.622 2

10- 40 C 2 m2 J- ' 1

The solution to Exercise 18.7(a) showed that

M)

= (3 eO

a

pNA

X

(~)

or

+2

n~

a

= (3M) 4JTpNA

I

X

which may be solved for nr to yield

n. __ r

1

(f3 + 2a. ' ) f3' - a '

' /2

,

(3)

f3 = (4JT) nr

with

X _

= (

(0.865

X

33.14+2 x 2.2 33.14 - 2.2

X

f3 = -3M -I

4JTpNA

(72.3 g mol- I)

106 g m- 3 ) ) 1/ 2

X

(6.022

=~

X

1023 mol

-

I)

1.6222 +2

(11;-1)

n}+2

MOLECULAR INTERACTIONS

E18.9(b)

361

The relative permittivity is related to the molar polarization through I

Cr -

pPm

- - = - - == C so Cr

+2

M

Cr

2C + I 1- C

= --

The molar polarization depends on the polarizability through so

C

pNA =-

3eoM

2 ( 4n coa , +1L- )

3kT

(1491 kgm - 3) x (6.022 x 1023 mol - I) C = ---~--:-::-~~--;---=-=-::-:-~:-:;--:------:=.-:3(8.854 x 1O- 12 J- I C2 m- I) x ( 157.01 x 10- 3 kg mol I) x (4n(8.854 x 10- 12 r I C2 rn- I) x (1.5 x 10- 29 m 3) (5.17 x 1O- 30 Cm)2 ) +3 (1.381 x 1O- 23 JK -I) x (298K) C = 0.83

E18.10(b)

M

p

Vm =

and

er =

2(0.83) + 1 I."Zl I _ 0.83 = L..!iJ

18.02gmol- 1 -5 3 _I = 999.4 x 10 3 gm- 3 = 1.803 x 10 m mol 2(7.275 x 1O- 2 Nm- l ) x (1.803 x 10-5 m 3 mol-I) (20.0 x 10- 9 m) x (8.314J K- I mol I) x (308.2K)

2yVm

rRT

-

= 5.119 x 10

-2

P = (5.623 kPa) e00511 9 = 1 5.92 !cPa E18.11(b)

y =

~pghr= ~ (0.9956 g cm- 3 )

1

x (9.807ms- 2) x (9.11 x 1O- 2 m)

3 x (0.16 x 1O- m) x

C~c~_~-3)

=17.12 x 1O- 2 Nm - 1 1 E18.12(b)

Pin - Pout

2y (2) x (22.39 x 1O- 3 N m- I) I 5 = ~ [18.38] = 2.20 x 10 7 m = 2.04 x 10 p~

I


The positive (H) end of the dipole will lie closer to the (negative) anion. The electric field generated by a dipole is g =

(~) x (~)[18.21] 4n cO r (2) x (1.85) x (3.34 x 10- 30 Cm) (411:) x (8.854 x 10- 12 J I C2 m-l) x r3

l.ll x 10- 19 Vm- I

l.ll x 108 Vrn- I

(rlm)3

(rlnm)3

362


(a) rff =

11.1

(b) rff = (c) rff = P18.3

X

1. I I

X

1. I I

X

10 8 Y m-'

1

when r = 1.0nm.

I

108 Y m - , 0.3 3 =4x 109 Ym - '

I forr=0.3nm .

10 8 Y m - \ = 14kYm-'1 forr=30nm. 303 . .

The equations relating dipole moment and polarizability volume to the experimental quantities fr and pare P

m -

x (fr-l) -(-M) P fr + 2

411 , NAJ-L 2 [\8.14] and Pm = -NACf + - 3 9fokT

[18.15, with a = 411foO"].

Therefore, we draw up the following table (with M = 1I9.4 g mol - ').

e;oe

-80

-70

-60

-40

-20

0

20

TIK

193 5.18 3.1

203 4.93 3.1

213 4.69 7.0

233 4.29 6.5

253 3.95 6.0

273 3.66 5.5

293 3.41 5.0

0.41

0.41

0.67

0.65

0.63

0.60

0.57

1.65 29.8

1.64 29.9

1.64 48.5

1.61 48.0

1.57 47.5

1.53 56.8

1.50 45.4

1000 / (T / K)

Cr cr- I -f r +2 p i g cm - 3 Pm / (ern 3 mol-I) Pm

is plotted against 1IT in Fig. 18.1. 50

o

2

3 4 5 103/(T /K) m.p!.

6

Figure IS.1

The (dangerously unreliable) intercept is ~ 30 and the slope is ~ 4.5

X

103 . It follows that


363

To determine fJ- we need fJ-

Ok) 1/2 = ( 9& NA x

=

_ 2 (slope x cm 3 mol I K)I /

I

(9) x (8.854 x 10- 12 r l C2 m- I ) x ( 1.381 x 10- 23 J K 6.022 x 10- 23 mol- I

x (,lope x om' mol-' K)'I'

I

mol )1 / 2

= (4.275

x 10- 29 C) x ( K m

= (4.275

X

I

)

1/2

)

x (slope x cm3 mol-I K) I/ 2

10- 29 C) x (slope x cm 3 m- I )I /2

= (4.275 x 10- 32 C m) x (slope) 1/ 2 = (1.282 x 10- 2 D) x (slope) 1/ 2 = (1.282 x 10- 2 D) x (4.5 x 103)1 / 2 =

I 0.86 D I.

The sharp decrease in Pm occurs at the freezing point of chloroform (-63°C), indicating that the dipole reorientation term no longer contributes. Note that Pm for the solid corresponds to the extrapolated, dipole-free, value of Pm , so the extrapolation is less hazardous than it looks.

P18.5

4n 3

Pm = -NM:t

,

2

NAfJ+- [18 .15, with a

9&okT

,

= 4n&oa].

Therefore, draw up the following table.

T/ K

292.2

309.0

333.0

387.0

413 .0

446.0

1000/( T / K) Pm /(cm 3 mol-I )

3.42 57.57

3.24 55.01

3.00 51.22

2.58 44.99

2.42 42.51

2.24 39.59

The points are plotted in Fig. 18.2. The extrapolated (least squares) intercept lies at 5.65 cm 3 mol - I (not shown in the figure), and the least squares slope is 1.52 x 104 cm 3 K - I mol - I. It follows that

a'

3 x 5.65 cm 3 mol - I 4n x 6.022 x 1023 mol-I

3P m (at intercept) 4nNA

fJ-

= \ 2.24 x

10- 24 cm 3

= 1.282 x

10- 2 D x ( 1.52 x 104 ) 1/ 2 [from Problem 18.3]

\.

The high-frequency contribution to the molar polarization, refractive index:

p:n = (M) p

X

p:n, at 273 K may be calculated from the

(&+- 2I) [18 . 14] = (M) - I) . - x (n2 -;-n, + 2 - ' --

&,

p

= 11 .58 D I.

364


3.0 1000 KJT

2.0

4.0

Figure 18.2

Assuming that ammonia under these conditions ( 1.00 atm pressure assumed) can be considered a perfect gas, we have

p M

and p

=

pM RT

RT

=-

p

=

Then P~, = 2.24

82.06 cm 3 atm K- I mol-I x 273 K 1.00 atm X

= 2.24

4 3 I x 10 cm mol- .

I

I

( 1.000379)2 - I } 104 cm 3 mol - I x { 2 = 5.66 cm 3 mol - I . (1.000379) + 2

If we assume that the high-frequency contribution to Pm remains the same at 292.2 K then we have

N f-L2

= Pm - Pm' = (57.57 - 5.66) cm 3 mol-I

_A_

9t:okT

Solving for f-L we have

The factor

(

Therefore f-L

9~:k )

1/2

has been calculated in Problem 18.3 and is 4.275 x 10- 29 C x (mol / K m) 1/2.

= 4.275

x 10- 29 C x

= 5.26

10- 30 C m

X

The agreement is exact'

(~I)

1(2

= 11.58 D I.

x (292 .2 K) 1/2 x (5.191 x 10- 5 ) 1/2(m 3 / mo!) 1/2


P18.7

365

(a) The depth of the well in energy units is f: = hcD e = 11.51 x 10-

23

J

I·

The distance at which the potential is zero is given by Re =2 1/ 6ro

so

ro =ReT I/6 = T I/ 6 (297 pm) = 1265pm l.

(b) In Fig. 18.3 both potentials were plotted with respect to the bottom of the well, so the Lennard-Jones

potential is the usual L-J potential plus f:. 10

8

....

u

~

;:J

6

6

-

Morse

'-'

~ 4

2

0 200

400

300

500

r/pm

Figure 18.3

Note that the Lennard-Jones potential has a much softer repulsive branch than the Morse. P18.9

Neglecting the permanent dipole moment contribution, N A fY.

Pm = -

3f:o

[18 .15]

.

(6.022 X 1023 mol - I) x (3.59 x 10- 40 r I C2 m 2 ) 3(8.854 x 10- 12 r l C 2 m- I) = 8.14 x 10- 6 m 3 mol - I = 18. 14 cm) mol - I I. f: - I

_r_ _

=

f: r +2

-

[18.16]

M =

f: r

pP

~

(0.7914 gcm- 3 ) x (8. 14cm 3 mol32.04gmol -

I = 0.201 f: r

nr = f: : /

2

[18.17]

+ 0.402 ;

1

l)

= 0.201.

1f: r = 1.76 1.

= ( 1.76) 1/ 2 = [Lill.

The neglect of the permanent dipole moment contribution means that the results are applicable only to the case for which the applied field has a much larger frequenc y than the rotational frequency. Since red light has a frequency of 4.3 x 10 14 and a typical rotational frequency is about I x 10 12 Hz, the results apply in the visible.

366



Exercise 18.7 showed

a= (3epNAoM) x (n;nt +- 2I) n; - I Therefore, - 2- nr + 2

=

nt+2'

4npNA

) 1/ 2

(

=

( I _ 4na pNA 3M

8na'p) 1/ 2 1+-3kT I _ 4na'p 3kT

[( 1+-8na'p) x ( 1+-4na'p)] 1/ 2 3kT

~ (

= I + const.

PNA + 8na' 3M ,

~ ~

3kT

12na'p

1+--+ ··· 3kT

x p

)1 / 2

A

2na'p

~ I+--

kT

~ with constant =

3M) (n; - I) a, = (4rr.N P x n; + 2 = P19.13

, = (-3M - - ) x (n;-I) --

3M

I

I

a

4na'NA P

Solving for n r , nr =

Hence, nr

or

-.E]

M M [for a gas, P = = Vm RT

~ I+X] [ _I I-x

[(l+X) I /2~1+~X].

2n~

IT' From the first line above,

(~ (n; - I) ;;r+2 . 3kT)

x

Consider a single molecule surrounded by N - 1(~ N) others in a container of volume V. The number of molecules in a spherical shell of thickness dr at a distance r is 4nr2 x (N IV) dr . Therefore, the interaction energy is

u=

l

a

R4 rr.r 2 x (N) x (-C6) - - dr V

r6

-4nNc6 1R -dr

=

V

a

r4

where R is the radius of the container and d the molecular diameter (the di stance of closest approach). Therefore,

u

4n) =(3

x

(N) V

(C6) x

(

I

I)

R3 - d3

~

-4nNC6 3Vd 3

because d « R. The mutual pairwise interaction energy of all N molecules is U = !Nu (the! appears because each pair must be counted only once, i.e. A with B but not A with Band B with A). Therefore,

U= 2

For a van der Waals gas, -/l 2a = V

(au) av

2nN2C6

-

2 3

T

3V d

and therefore a

=


P18.15

367

The number of molecules in a volume element dr isN dr IV = Ndr. Tbeenergy of interaction of these molecules with one at a distance r is V N dr . The total interaction energy, taking into account the entire sample volume, is therefore

u

f

=

VN dr = N

f

V dr

[Vis the interaction energy, not the volume].

The total interaction energy of a sample of N molecules is ~Nu (the ~ is included to avoid double counting), and so the cohesive energy density is

U

2

- V = 2rrN

1

00

C6

a

dr r4

Nlc 6

2rr

=3

X

~.

However, N = NAP 1M, where M is the molar mass; therefore

P18.17

Once again (as in Problem 18.16) we can write

8(v) =

I

. ( rr - 2 arCSIn 0 R,

b

+ R2(V)

)

b ::: R,

+ R2(V)

b > R, +R2(V)

but R2 depends on v

Therefore, with R, (a)

8(v)

= ~R2 and b = ~R2'

= rr-2arcsin( 1 + 2e1 v/ . ) . v

(The restriction b ::: R, + R2(V) transforms into This function is plotted as curve a in Fig. 18.4. The kinetic energy of approach is E (b)

8(E)

~R2

:::

= ~mv2 , and so

= rr - 2 arcsin (1 + 2 e~(E/E*)1/2 ) with E* =

in Fig. 18.4.

~R2 + R2e- v/ v', which is valid for all v.)

~mv*2. This function is plotted as curve b

368



(a) The energy of induced-dipole-induced-dipole interactions can be approximated by the London formula (eqn 18.25): V

C = -= -3a;a; -- Ilh - - = -3a'21 -6 6 r6

2r

II

+h

4r

160

120

40

2

4 (a) v/ v*

and

6 8 (b) £ / £*

10

Figure 18.4

where the second equality uses the fact that the interaction is between two of the same molecule. For two phenyl groups, we have

V=

-

3(1.04

X

10- 29 m 3 )2(5 .0eY)(1.602 x 10- 19 Jey-I) 4(1.0 x 10- 9 m )6

= 6.6 x

-?3

10 - J

or l-39 Jmol - l l· (b) The potential energy is everywhere negative. We can obtain the distance dependence of the force by

taking

dV

6C

F=--=-dr r7 .

This force is everywhere attractive (i.e. it works against increasing the distance between interacting groups). The force approaches zero as the distance becomes very large ; there is no finite distance at which the dispersion force is zero. (Of course, if one takes into account repulsive forces , then the net force is zero at a distance at which the attractive and repulsive forces balance. ) P18.21

(a) The dipole moment computed for trans-N-methylacetamide is i-L = (3.092 D) x (3 .336 X 10- 30 C m D- I) =

I 1.03

X

10- 29 C m I

(semi-empirical, PM3 level, PC Spartan Pro™). The dipole is oriented mainly along the carbonyl group. The interaction energy of two parallel dipoles is given by eqn 18.22: V

=

i-L li-L"zf (e)

4rreor

3

where/eel

= 1-

3 cos

2

e


369

and r is the distance between the dipoles and () the angle between the direction of the dipoles and the line that joins them . The angular dependence is shown in Fig. 18.5. Note that V (() is at a minimum for () = 0 0 and 1800 while it is at a maximum for 90 0 and 270 0 .

20

r

0

o

E

~

::::- -20

-40

o

50

150

100

200

250

350

300

O/ deg

Figure 18.5

(b) If the dipoles are separated by 3.0 nm, then the maximum energy of interaction is: 29

Vrnax =

(1.03 1 x 10- C m) 2 4rr(8.854 x 10- 12 r I C2 m - I ) x (3 .0 x 10- 9

I -

= 3.55 x 10

m )3

- 23

I

J.

In molar units Vrnax = (3.55

X

10- 23 J) x (6.022

X

1023 mol-I) = 21 Jmol- I = 2.1 x 10- 2 kJmol -

l.

Thus, dipole--dipo1e interactions at this distance are dwarfed by hydrogen bonding interactions. However, the typical hydrogen bond length is much shorter, so this may not be a fair comparison. P18.23

Here is a solution using MathCad.

(8)

Data:=

7.36 3.53 ( 1.00

logA := (DataT )(0)

8.37 8.~ 4.24 4.09 1.80 1.70

S:= (DataT )(1)

info := regress(Mxy, togA,1)

(b)

W:= 1.5 S := 4.84 Given

7.47 3.45 1.35

7.25 2.96 1.60

6.73 8.52 7.87 2.89 4 .39 4.03 1.60 1.95 1.60

W:= (DataT )(2)

Mxy:= augment(S,W) b=

0.957) bo 0.362 bl ( 3.59 b2

W := Find(W)

W = 1.362

b := submatrix(info, 3,5,0,0)

Estimate for Given/Find Sotve Bank togA:= 7.60 togA = bo + b1 . S + b2 . W

7.53) 3.80 1.60

Materials 1: macromolecules and aggregates

19


Number average is the value obtained by weighting each molar mass by the number of molecules with that mass (eqn 19.1 )

- = IVI"

~N;M;.

Mn

;

In this expression, N; is the number of molecules of molar mass M; and N is the total number of molecules. Measurements of the osmotic pressures of macromolecular solutions yield the number average molar mass. Weight average is the value obtained by weighting each molar mass by the mass of each one present (eqn 19.2)

LN;M;

- = -I" Mw

m

~m;M; i

= -=,i = , --L.,N;M;

[19.3].

In this expression, m; is the total mass of molecules with molar mass M; and m is the total mass of the sample. Light scattering experiments give the weight average molar mass. Z-average molar mass is defined through the formu la (eqn 19.4)

LN;M; Mz

=

2'

LN;M;

The Z-average molar mass is obtained from sedimentation equilibria experiments . 019.3

Contour length: the length of the macromolecule measured along its backbone, the length of all its monomer units placed end to end. This is the stretched-out length of the macromolecule, but with bond angles maintained withi n the monomer units. It is proportional to the number of monomer units, N , and to the length of each unit (eqn 19.30). Root mean square separation: one measure of tbe average separation of tbe ends of a random coil. It is the square root of the mean value of R2 , where R is the separation of the two ends of the coil. This mean val ue is calculated by weighting each possible value of R2 with the probability, f (eqn 19.27), of that value of R occurring. It is proportional to N 1/ 2 and the length of each unit (eqn 19.31).

MATERIALS 1: MACROMOLECULES AND AGGREGATES

371

Radius of gyration: the radius of a thin hollow spherical shell of the same mass and moment of inertia as the macromolecule. In general , it is not easy to visualize this distance geometrically. However, for the simple case of a molecule consisting of a chain of identical atoms this quantity can be visualized as the root mean square distance of the atoms from the center of mass. It also depends on N I / 2, but is smaller than the root mean square separation by a factor of (1/6) I / 2 (eqn 19.33). 019.5

For a molecular mechanics calculation, potential energy functions are chosen for all the interactions between the atoms in the molecule; the calculation itself is a mathematical procedure that locates the energy minima (local and global) of the molecule as a function of bond distances and bond angles. Because only the potential energy is included in the calculation, contributions to the total energy from the kinetic energy are excluded in the result. The global minimum of a molecular mechanics calculation is a snapshot of the molecular structure at T = O. No equations of motion are solved in a molecular mechanics calculation. The structure of a macromolecule (or any molecule, for that matter) can, in principle, be determined by solving the time independent Schr6dinger equation for the molecule with methods similar to those described in Chapter II. But, due to the very large size of macromolecules, these methods may be impractical and, due to approximations to make them tractable, inaccurate. In a molecular dynamics calculation, equations of motion are integrated to determine the trajectories of all atoms in the molecule. The equations of motion can, in principle, be either classical (Newton's laws of motion) or quantum mechanical. But, in practice, due to the very large number of atoms in a macromolecule, Newton's equations of motion are used. Quantum mechanical methods are too time consuming, complicated, and at this stage too inaccurate to be popular in the field of polymer chemistry.

019.7

A surfactant is a species that is active at the interface of two phases or substances, such as the interface between hydrophilic and hydrophobic phases. A surfactant accumulates at the interface and modifies the properties of the surface, in particular, decreasing its surface tension. A typical surfactant consists of a long hydrocarbon tail and other non-polar materials, and a hydrophilic head group, such as the carboxylate group, -C0 2, that dissolves in a polar solvent, typically water. In other words, a surfactant is an amphipathic substance, meaning th at it has both hydrophobic and hydrophilic regions. How does the surfactant decrease the surface tension? Surface tension is a result of cohesive forces and the solute molecules must weaken the attractive forces between solvent molecules. Thus molecules with bulky hydrophobic regions such as fatty acids can decrease the surface tension because they attract solvent molecules less strongly than solvent molecules attract each other. See Section 19. IS(b) for an analysis of the thermodynamics involved in this process .

019.9

A Langmuir- Blodgett (LB) film is a monolayer or multilayer film that has been placed upon a substrate by transferring a surface film from a liquid to the substrate. A Langmuir trough, shown in Fig. 19.1 (a), is designed to perform the transfer. A surface film of water-insoluble, fi lm-forming molecules is assembled upon the water by mechanical compression. Dipping and withdrawing the substrate affects monolayer transfer. Repeated dipping produces multi layers. Weak van der Waals forces hold the monolayers together. Self-assembled monolayers (SAMs) do not require assembl y by mechanical compression. SAMs form from charged materials that have adsorption--desorption properties that promote self-assembly as shown in Fig . 19.1(b). The substrate is si mply immersed in a dispersion of the charged materials, withdrawn, and rinsed. Films are held together with either strong ionic bonds or covalent bonds.

372

STUDENT'S SOLUTIONS MANUAL Substrate

Figure 19.1(a)

self-assembly ~

-

F;oo",19.1(b)

Both methods yield well-organized monolayers but LB films upon water provide better organizational control than is possible with spontaneous self-assembled films. However, not requiring mechanical compression, SAMs are much more versatile. The strong bonding of SAMs gives long-lasting, stable films in contrast to the less stable van der Waals LB films.


The number-average molar mass is (eqn 19.1)

Mn =

~ LN;M; =

[3 x (62)

+ 2 ~ (78)] kg mol-

I

I

= 68 kg mol - I

I

The mass-average molar mass is (eqn 19.3)

_ LN;Ml _ 3 x (62)2 Mw LN;M; 3 x (62) E19.2(b)

+2 X +2 x

(78)2 k I-I - 169k I- I g mo g mo (78)

I

For a random coil , the radius of gyration is ( 19.33)

Rg = L(N / 6) 1/ 2 so N = 6(Rg / L)2 = 6 x (18.9 nm / 0.450 nm )2 = 11 .06 x 104 1 E19.3(b)

(a) Osmometry gives the number-average molar mass, so

(m l / MI)MI (ml / Md

+ (m2/ M2) M2 + (m2/ M 2)


373

(b) Light-scattering gives the mass-average molar mass, so

Mw = mIMI +m2 M 2 = (25) x (22) +(75) x (22/3) kgmol- I =lllkg mol-II ml+m2 25+75 E19.4(b)

The formula for the rotational correlation time is

r =

4rra 31/ 3kT

1/(H20, 20 0c) = 1.00 x 10- 3 kgm- I s-I[CRC Handbook] I 3 r = 4rr x (4.5 x 1O- 9 m)3 x 1.00 x 10- kgm- s-I = 19.4 x 10- 8 s I 3 x 1.381 X 10- 23 J K- I x 293 K E19.5(b)

The effective mass of the particles is meff = bm = (I - pl!s)m [19.14] = m - pl!sm = l!pp - l!p = l!(pp - p) where l! is the particle volume and Pp is the particle density. Equating the forces meffrw2 =/s = 6rr1/as [19.15, 19.12] or l!(pp - p)rw2 = 1rra3(pp - p)rw2 = 6rr1/as Solving for s yields 2a 2(pp - p)rw2 s= - - ' - - - - 91/

. .. S2 a~(pp - ph (a2)2 (pp - ph Thus, the relatIve rates of sedimentatIOn are - = 2 = . Sl a l (pp - P)I al (pp - P)I The value of this ratio depends on the density of the solution. For example, in a dilute aqueous solution with p = 1.0 I g cm -3 , the difference in polymer densities matters in that the factor involving densities is significantly different than I: S2 2 ( 1.10 - 0.794) - = (8.4) SI ( 1.18 - 0.794)

~ =~

In a less dense organic solution, for example a dilute solution in octane with p = 0.71 gcm - 3, the density difference has a smaller effect, for the factor involving densities is closer to 1: S2 2 (1.10 - 0.71h - = (8.4) SI (1.18 - 0.71)1

~ =~

In both cases, the larger particle sediments faster. E19.6(b)

The molar mass is related to the sedimentation constant through eqns 19.19 and 19.14: -

SRT

SRT

M = -bD- = -:-(I- --pl!s-:-)D-

374


where we have assumed the data refer to aqueous solution at 298 K. (7.46 x 1O- 13 s) x (8.314SJK- 1 mol-I) x (298K)

--

------------~----~~----~--~--~--~----~~~ [I - (lOOOkgm- 3 ) x (8.01 x 1O-4 m 3 kg- I)] x (7.72 x 10- 11 m 2 s- l )

Mn=

= 1120kgmOI- 1 1 E19.7(b)

See the solution to Exercise 19.5(b). In place of the centrifugal force meff r 2 we have the gravitational force meffg. The rest of the analysis is similar, leading to

2a2 (pp - p)g ~-

• -

(2) x (lS.S x -

9T]

-

=11.47 x E19.8(b)

m ) 2 x (12S0 - 1000) kgm- 3 x (9.81 m s- 2 ) (9) x (8.9 x 1O- 4 kgm IS- I)

1O-6

------------------------~--~~~------------

1O-4

m s-11

The molar mass is related to the sedimentation constant through eqns 19.19 and 19.14: --

SRT

SRT

M - --- - - - - - - - bD - (I - pvs)D

Assuming that the data refer to an aqueous solution, (S.l x 1O- 13 s) x (8.314SJ K- I mol-I) x (293K) 1 6 _I 1 = S kg mol [ 1 - (0.997 gcm- 3 ) x (0.72 1 cm 3 g- I)] x (7 .9 x 10- 11 m 2 s- I)

-M = E19.9(b)

In a sedimentation experiment, the weight-average molar mass is given by (eqn 19.20) Mw

=

2RT (ri -

C2

rf)bw 2

In --

so

CI

C2

In -CI

=

Mw(r~ - r? )bw 2 2RT

-----=---'---

This implies that M r 2 bw2

~RT

In C =

so the plot of In

C

Mwbw2

In

+ constant

versus r2 has a slope m equal to --

= ----- and M w = 2RT

2RTm - - -2-

bw

(8.3 14SJK- 1 mol- I) x (293K) x (821 cm - 2 ) x ( IOOcmm - I)2 Mw = [I _ (lOOOkgm- 3 ) x (7.2 x 1O- 4 m 3 kg I)] x [(1080 s- l ) x (2JT)]2

_

2

X

=13 . 1 x 10 3 kg mol - I 1 E19.10(b) The centrifugal acceleration is

2 2 a = rw so a / g = rw / g

a/

3 = (S. SO cm) x [2JT X ( 1.32 x 10 s- I) ] 2 = 13.86 x 105 1 I g ( IOOcm m- ) x (9.81 m s 2) . .


375

E19.11(b) For a random coil, the rms separation is [19.31)

Rrms = N 1/ 2 l = ( 1200) 1/2 x (1.1 25 nm ) = 138.97 nm 1 E19.12(b) Polypropylene is - (CH (CH3)CH2) - N, where N is given by

N

=

174kgmol- 1 0 ---..::::...-- -----:-I = 4.13 x I 3 3 42.1 x 10- kg mol-

Mpolymer Mmonomer

The repeat length l is the length of two C-C bonds. The contour length is [19.30) Rc = Nl = (4.13 x 103 ) x (2 x 1.53 x IO- Io m) = 11.26 x 10- 6 m 1

The rms seperation is [19.31) Rrms = INI /2 = (2 x 1.53 x 10- 10 m) x (4.13 x 103 ) 1/2 = 11.97

X

10- 8 m 1= 19.7 nm


s S = - 2 [19.16) . rev

. dr s I dr dIn r Since s = - - = - - = - dt ' r

r dt

dt

and, if we plot In r against t , the slope gives S through 1 dIn r

S= -ev 2 dt .

The data are as follows t/ min

15.5

29.1

36.4

58.2

r/cm

5.05 1.619

5.09 1.627

5.12 1.633

5.19 1.647

In(rlcm )

The points are plotted in Fig. 19.2. The least-squares slope is 6.62 x 10- 4 min - I, so S=

P19.3

6.62 x 10- 4 min - I (6.62 x 10- 4 min-I ) x (I min / 60 s) -1 3 ~ 2 = 2 =4.97 x 10 s or~. 4 ev (2IT x 4.5 x 10 / 60 s)

[1)) = lim (1) / 1)0 - I) [19 .23). c-+o

C

376


1.64

c ..... , ... ... , ... .

Eu

---.E ~

1.62

1.60

o

40

20

60

I/ min

Figure 19.2

We see that the y-intercept of a plot of the right-hand side against c, extrapolated to c begin by constructing the following table using 1)0 = 0.985 g m- I s-I.

1)/1)0 c

I)

3

/ (dm g-I)

1.32

2.89

5.73

9.17

0.0731

0.0755

0.0771

0.0825

= 0, gives [1)]. We

(

The points are plotted in Fig. 19.3 . The least-squares intercept is at 0.0716, so [1)]

8.2

,.

8.0

OIl

Ma ~

,.---

7.8

):!..

7.6

>?

---

S 7.4 00 7.2

Figure 19.3

= 10.0716 dm3 g-I

I.

MATERIALS 1: MAC ROMOLECULES AND AGGREGATES

P19.5

377

We follow the procedure of Example 19.5. Also compare to Problems 19.3 and 19.4. [17]

= lim

c-o (

17 / 170 -

I)

[ 19.23]

c

and

[17]

= K~

[19.25]

with K and a from Table 19.4. We draw up the following table using 170

17 / ( 10- 3 kg m -

I S- I )

= 0.647

x 10- 3 kg m- I s- I.

o

0.2

0.4

0.6

0.8

1.0

0.647

0.690 0.332

0.733 0.332

0.777 0.335

0.821 0.336

0.865 0.337

« 17 / 170 - 1)/c) /(100cm 3 g- l )

The values are plotted in Fig 19.4, and the y-intercept is 0.330 .

.,

0.336

eo

~E

u

8 0.334

~ Ie

~

0.332

0.330 0

0.2

0.4

0.6

0.8

cI(g1 I00 em 3)

Hence [17]

and

=

M

--y-

g mol- I

(0.330) x ( 100cm 3 g-I)

=

(33 0

3 - I

. cm g 8.3 x 10- 2 cm 3 g- I

That is, M = G58 kg mol - I P19.7

Figure 19.4

= 33.0cm 3 g- I ) 1/ 0.50

= 158 x 10 3

.

I.

The empirical Mark-Kuhn-Houwink-Sakurada equation [19.25] is

As the constant a may be non-integral the molar mass here is to be interpreted as unitiess, that is, as M y I(g mol - I). The units of K are then the same as those of [17]. We fit the data to the above equation and obtain K and a from the fitting procedure. The plot is shown in Fig 19.5.

1

K = 0.0117 cm 3 g- I

1

and

1a = 0.7171·

378


y

= 0.01167xo.7 1661 ,

R

= 0.99983

250

200

,. Oil

ISO

M e ~

----.£

100

50.0

0.00

'2L.J,---,-----'--L.......L-..L.......,'---'-----'---..L......L-..I....-L.......1----,---L-,---,---L.....J

o

0.4

0.2

0.8

0.6

1.0

Figure 19.5

(Many plotting programs can fit a power series directly. If not, the equation can be transformed into a linear one In[1)]

= InK +alnMy

so a plot of In [1)] versus In M v will have a slope of a and a y-intercept of In K.) COMMENT. This value for a is not much different from that for polystyrene in benzene listed in Table 19.4.

This is somewhat surprising as one would expect both the K and a values to be solvent-dependent. THF is not chemically similar to benzene . On the other hand, benzene and cyclohexane are very much alike, yet the values of K and a as determined in Example 19.5 are markedly different from those in Table 19.4 for polystyrene in cyclohexane.

P19.9

See Section 5.5(e) and Example 5.4.

h

RT

BRT

e

pgMn

pgMn

- = -=- +

-2'

e [Example 5.4].

We plot hie against e. Draw up the following table. c/(g/lOO cm 3 )

0.200

0.400

0.600

0.800

1.00

hlcm h - / (100 cm 4 g- l) e

0.48

1.12

1.86

2.76

3.88

2.4

2.80

3.10

3.45

3.88

The points are plotted in Fig. 19.6, and give a least-squares intercept at 2.043 and a slope 1.805. Therefore, RT I pgMn - _ Mn -

=

(2.043) x (100 cm 4 g- I)

=

2.043

X

10- 3 m4 kg-I and hence

(8 .314 J K - 1 mol-I) x (298 K) _ (0.798 x 10 3 kg m- 3 ) x (9.81 m s-2) x (2 .043 x 10- 3 m4 kg I)

=

1

k I-I 1 . 155 g mo

MATER IALS 1: MACROMOLE CU LES AN D AGGR EG ATES

379

4

,.

01)

"uE

8

3

~ ~

0.2

0.4

0.6

0.8

cI(gli ()() em 3)

Figure 19.6

From the slope, 4

BRT? = ( 1.805) x

pgM ~

(~OO cm g~ l)

= 1.805

X

104 cm 7 g- 2 = 1.805

X

10- 4 m7 kg- 2

",/(100 cm )

and hence

B=

(p!~n)

x Mn x ( 1.805 x 10- 4 m 7 kg- 2 )

(155 kg mol - I) x ( 1.805 x 10- 4m7 kg- 2)

2.043 x 10- 3 m4 kg- I = 113.7 m 3 mol- I I.


See the discussion of radius of gyration in Section 19.8(a). For a random coi l Rg ex N I / 2 ex MI / 2. For a rigid rod, the radius of gyration is proportional to the length of the rod, which is in turn proportional to the number of polymer units, N , and therefore also proportional to M. Therefore, poly(y-benzyl-L-glutamate) is rod-like whereas polystyrene is a random coil (in butanol ).

P19.13

Call the constant of proportionality K , and evaluate it by requiring that L etM-M = (2y)I / 2X

and N =

10

00

so

1dN =

N.

dM=(2y)I /2dt

Ke -(M- A1)l / 2y dM = K (2y) 1/2

1: e-

x2

dx where a = M / (2y)I /2.

Note that the point x = 0 represents M = M, and x = - a represents M = O. In a narrow distribution, the number of molecules with masses much different than the mean falls off rapidly as one moves away

380


from the mean; therefore, dN

Hence, K =

-Mn

N

(2ny)

= -1 N

=

f

~

0 at M ::: 0 (that is, at x ::: -a) . Therefore

1/2 . It then follows from turning eqn 19.1 into an integral that

MdN

I

1 1

(2ny) I/2

00

1

(2ny)I /2 2y

= (-) n

1/2

1

00

=

[(2y)I /2x

0

00

(

-a

_x2

xe

- ? / 2Y dM Me-(M-M)-

0

+ M]e-

X2

(2y)I /2dM

M _x2 )

+1 /2 e (2y)

dx.

Once again extending the lower limit of integration to -

P19.15

00

adds negligibly to the integral, so

(a) Following Justification 19.4, we have

.

a

3

2 _ a2R2

wlthf=4n(n l / 2 ) R e 2

Therefore, Rrms =4n

= Hence, Rrms =

-

3

2a

(

a )3

n l/ 2

3

,a= ( 2Nl2 )

{oo Re4

Jo

_ a2 R2

1/2 [19 .27].

(

a )3

dR=4n n l / 2

x

(

3)

"8 x

(n) 1/2 aiD

2 2 =Nl .

11N I / 2 1.

(b) The mean separation is

(c) The most probable separation is the value of R for which! is a maximum, so set d! / dR = 0 and solve for R.


381

Therefore, the most probable separation is

= 4000 and 1 = 154 pm, R rm s = 19.74 nm I; (b) Rmean = 18.97 nm I;

When N (a) P19.17

(c) R*

= 17.95 nm I·

We use the definition of the radius of gyration given in Problem 19.19, namely,

R~ = ~ LRJ. j

(a) For a sphere of uniform density, the center of mass is at the center of the sphere. We may visualize

the sphere as a collection of a very large number, N, of small particles distributed with equal number density throughout the sphere. Then the summation above may be replaced with an integration. 2

Rg =

I N

J; r 2P(r)dr

N---=~::7.~':-P-(-r)-d-r-

P(r) is the probability per unit distance that a small particle will be found at distance r from the center, that is, within a spherical shell of volume 4Jtr 2dr. Hence, P(r) = 4Jtr2dr. If P(r) were normalized, the integral in the numerator would represent the average value of r2 , so N times that integral replaces the sum. The denominator enforces normalization. Hence

Rg

_(3)

-"5

1/ 2

a.

(b) For a long straight rod of uniform density the center of mass is at the center of the rod and P( z) is constant for a rod of uniform radius; hence,

2

Rg

COMMENT.

rl / 2 Z2d Z 2 Jo

13 (1/) 2 3

I

2

= 2Jci/2 dz = ~ = 12 1 ,

~ 1 Rg

= 2.J3

.

The radius of the rod does not enter into the result. In fact, the distribution function is P(r ,z),

the probability that a small particle will be found at a distance r from the central axis of the rod and Z along that axis from the center, that is, within a squat cylindrical shell of volume 2nrdrdz. Integration radially outward from the axis is the same in numerator and denominator.

For a spherical macromolecule, the specific volume is

so

3vsM ) a= ( - 4JtNA

1/ 3

382


and

= (~) 1/2 x (3 Vs M) 1/ 3

R

5

g

4rrNA

( ~ ) 1/2 x

=

(3 vs/ em 3 g-I) x em 3 g-I x (MIg mol- I) x g mOl - I) 1/ 3 (4 rr ) x (6.022 x 1023 mo l- l )

5

=

(5.690 x 10- 9 ) x (v s/e m 3 g- I) 1/ 3 x (M /g mol- I) 1/ 3 em

=

(5.690 x 1O-

ll

m) x {(vs/em 3 g-I) x (M/g mol- I)}1 /3 .

That is Rg/ nm = [ 0.05690 x When M

=

1(v s/e m 3 g- l ) x (M /g mol - I )1 13) [.

100 kg mol- I and Vs

= 0.750 em3 g-I ,

R g/ nm = (0 .05690) x {O.750 x 1.00 x 1O5} 1/3 = 12.40 I.

For a rod, Vmol

Rg

_

-

= rra 2l, so

Vmo l

2rra 2y'3

_ vsM -- x

NA

I 2rra 2y'3

(0.750 em 3 g- I) x ( 1.00 x 105 g mol - I) (6.022 x 10 23 mol - I) x (2 rr ) x (0.5 x 10- 7 em)2 x y'3

= 4.6 COMMENT.

~=

x 10- 6 em

= 146 nm I.

Rg may also be defined through the relation

L:m;rl iL,mi '

Question. Does this definition lead to the same formulas for the radii of gyration of the sphere and the rod as those derived above? P19.19

Refer to Fig. 19.7.

....

~---=--

j

Figure 19.7 The defi nition in the text (eqn 19.32) is

so

R 2g

=

1~hij " 2 2N2 ij

=

I " " 2' 2N2 ~~hij . j


383

The scalar quantity hij can be written as the dot product hij . hij . If we refer all our measurements to a common origin (which we will later specify as the center of mass), the interatomic vectors hij can be expressed in terms of vectors from the origin: hij = Rj - R; . (If this is not apparent, note that R; + h ij = Rj .) Therefore

R2 = _1- " "(R - R ) . (R - R) g 2N2 L L } I } I ;

j

=~ " " ( R j . Rj + R;· R; 2N- L L I

2R;· Rj)

}

=~ L L(Rj + R; - 2R;· Rj ). 2N . . I

}

Look at the sums over the squared terms:

j

j

If we choose the origin of our coordinate system to be the center of mass, then

" L R; " = L Rj

=0

and

1LRj' "2 Rg2 = N

j

j

for the center of mass is the point in the center of the distribution such that all vectors from that point to identical individual masses sum to zero. P19.21

Write I

= aT, then

( ~) aT

=a

(au) at

and, using PI9.20,

I

T

=1-aT=O

.

Thus the internal energy is independent of the extension. Therefore

t

= aT =

T(~) aT = I-T (as) at I[p19.20] I

T

and the tension is proportional to the variation of entropy with extension. Extension reduces the disorder of the chains, and they tend to revert to their disorderly (non-extended) state.


The center of the sphere cannot approach more closely than 2a; hence the excluded volume is

vp

where

4

(4

= "3rc(2a) 3= 8 "3rca 3) = ~ ~ Vrnol

is a molecular volume.

384


The osmotic virial coefficient, B (see eqn 5.41), arises largely from the effect of excluded volume. If we imagine a solution of a macromolecule being built by the successive addition of macromolecules to the solvent, each one being excluded by the ones that preceded it, then the value of B turns out to be (PI9.18)

where

vp

is the excluded volume due to a single molecule. 1 = -NA x 2

B(BSV)

= B(Hb) =

32 -rca 3 3

(l~rc)

(I~rc)

16 3 = -rca NA 3

x (6.022 x 1023 mol-I) x (14.0 x

x (6.022 x

10-

1023 mol-I) x (3 .2 x 1O-

9

m)3 = 128 m3 mol- II.

9 m)3

= I 0.33 m3 mol- I I.

.

Smce n = RT [J] +BRT[Jf + .. . [5.41], if we write n ° = RT[J] then

,

n - n°

no

~

BRT[J]2 RT[J]

= B[J] .

For BSV, g [J]=(1.0 ) x (lOdm- 3) = M

3 IOgdm=9.35 x 1O- 7moldm-3 1.07 x 107 g mol- I = 9.35 x 1O- 4 molm- 3

and

n-n°

no

For Hb, [J] = and P19.25

n-n°

n0

= (28 m3 mol- I) x (9.35 10 g dm- 3

66.5 x 103 g mo]-

X

10- mol m-3) = 2.6 x 10- 2 corresponding to 12.6 per cent I. 4

I = 0.15 mol m- 3

= (0.15 mol m- 3) x (0.33 m3 mol-I) = 5.0 x 10- 2 which corresponds tol5 percent I.

(a) We seek an expression for a ratio of scattering intensities of a macromolecule in two different conformations, a rigid rod or a closed circle. The dependence on scattering angle e is contained in the Rayleigh ratio Re. The definition of this quantity, in eqn 19.7, may be inverted to give an expression for the scattering intensity at scattering angle

e

Ie

sin 2 ¢

= ReIo--2-, r

where ¢ is an angle related to the polarization of the incident light and r is the distance between sample and detector. Thus, for any given scattering angle, the ratio of scattered intensity of two conformations is the same as the ratio of their Rayleigh ratios: Prod

P ee

The last equality stems from eqn 19.8, which related the Rayleigh ratios to a number of angleindependent factors that would be the same for both conformations, and the structure factor (Pe)


385

that depends on both conformation and scattering angle. Finally, eqn 19.9 gives an approximate value of the structure factor as a function of the macromolecule's radius of gyration Rg , the wavelength of light, and the scattering angle: 3)..2 -161[2R~ sin2 (~e) 3)..2 The radius of gyration of a rod of length 1 is Rrod =

1/ (12)1 /2 [Section 19.8(a)].

For a closed circle, the radius of gyration, which is the rms distance from the center of mass [PI9.19], is simply the radius of a circle whose circumference is I:

[ = 21[Ree

[

so

Ree = - . 21[

The intensity ratio is: Irod

3)..2 _11[2[2 sin 2 (~e)

Icc

3)..2 - 4[2 sin 2 (~e)

Putting the numbers in yields:

er lrod / lee

20 0.976

45 0.876

90 0.514

(b) I would work at a detection angle at which the ratio is smallest, i.e. most different from unity, provided I had sufficient intensity to make accurate measurements. Of the angles considered in part (a), 1900 1is the best choice. With the help of a spreadsheet or symbolic mathematical program, the ratio can be computed for a large range of scattering angles and plotted (Fig. 19.8).

u

0.5

::::u

J 0.0

-0.5

L-.~~-'--~~---'--~~-'--~~--'

o

45

90 (JIO

135

180

Figure 19.8

A look at the results of such a calculation shows that both the intensity ratio and the intensities themselves decrease with increasing scattering angle from 0° through 180°, that of the closed circle conformation changing much more slowly than that of the rod. Note: the approximation used above yields negative numbers for Prod at large scattering angles; this is because the approximation, which depends on the molecule being much smaller than the wavelength, is shaky at best, particularly at large angles.

386

P19.27


The molar mass is given by eqn 19.19 Mn

=

SRT bD

=

(1

SRT D [19.14, for b] - pV s )

_ (4.5 x 1O- 13 s) x (8.3 141 K- I mol- I) x (293 K) _I _ II (1 - 0.75 x 0.998) x (6.3 x 10- 11 m2 S-I) -. 69 kg mol . Now combine!

= 6nary [19.12] with! = kT / D [19.11]:

kT

a P19.29

= -6nryD =

(1.381 x 10- 23 1 K - I) x (293 K) I (6n) x (1.00 x 10- 3 kgm- S- I) x (6.3 x 10- 11 m 2 s- l )

= 13.4 nm I·

The isoelectric point is the pH at which the protein has no charge. At that point, then, its drift speed under electrophoresis, s, vanishes. Plot the drift speed against pH and extrapolate the line to s = O. The plot is shown in Fig. 19.9. o~--~------~--------------~

-D. I

-D.3

-D.4+-----+-----+-----+-----+---~

4

3.5

4.5

5.5

5

pH

6

Figure 19.9

Isoelectric pH is the x-intercept on the graph, that is, the value of x at which y solving the fit equation: s/(l1m/s) so pH

=

- 0.17pH + 0.655

= O. One can find this by

=0

= 13.851.

COMMENT.

One could obtain the result to about ±O.05 pH by reading the value directly from the

graph. P19.31

(a) The data are plotted in Fig. 19.10. Both samples give rise to tolerably linear curves, so we estimate the melting point by interpolation using the best-fit straight line. The best-fit equation has the form Tm/ K = m! + b, and we want Tm when! = 0.40: 10- 2 mol dm - 3 :

Csalt

= 1.0 x

Csalt

= 0.15 mol dm- 3 :

Tm

Tm

= (39.7 x 0.40 + 324) K = 1340 K I·

= (39.7 x 0.40 + 344) K = 1360 K I·


375

365

::.:

f-.=

•

..

370

387

• 0.01 • 0.15 -Linear (0.01)

···· ··· ···f ~· · ·· · ···

360 355 350 345 340 335 0.3

0.4

0.6

0.5

0.7

f

0.8

Figure 19.10

(b) The slopes are the same for both samples. The different concentrations of dissolved salt simply offset the melting temperatures by a constant amount. The greater the concentration, the higher the melting point. This behavior is not what is typically observed with small molecules, where the presence of dissolved impurities disrupts freezing and depresses the freezing point. The dissolved ions can interact with charged regions of the macromolecule that might otherwise experience unfavorable intramolecular interactions. For example, if two regions bearing negative charge would have to approach each other in the absence of dissolved salts, the incorporation of a cation very close to each region and an anion in between tbem would turn an unfavorable interaction into a favorable one. (See Fig. 19. 11).

Figure 19.11

The melting points are greater at both larger fractions of G--c base pairs and at larger salt concentrations. Tm increases with the number of G--c base pairs because this pair is held togethar with the three hydrogen bonds in the double helical structure, whereas the A -T pair is held with two hydrogen bonds (see Section 19.11). The f:!"Hm contribution is greater for the G--c pair. Low salt concentrations destabilize the double helix by inadequately contributing to the attractive forces between the solution and the sugar-phosphate backbone of the double helix. This makes it easier for a base to rotate out from the center of the double helix. P19.33

The peaks are separated by 104 g mol- I , so this is the molar mass of the repeating unit of the polymer. This peak separation is consistent with the identification of the polymer as polystyrene, for the repeating group of CH2CH(C6Hs) (8 C atoms and 8 H atoms) has a molar mass of 8 x (12 + I) g mol-I = 104 g mol-i . A consistent difference between peaks suggests a pure system and points away from different numbers of subunits of different molecular weight (s uch as the I-butyl initiators) being incorporated into the polymer molecules . The most intense peak has a molar mass equal to that of n repeating groups plus

388


that of a silver cation plus that of terminal groups: M(peak)

= nM(repeat) + M(Ag+) + M(terrninal) .

If both ends of the polymer have terminal t-butyl groups, then

= 2M(t-butyl) = 2(4

M(terminal) and n

P19.35

=

x 12 + 9) g mol- 1

M(peak) - M(Ag+) - M(terrninal) M(repeat)

=

=

114 g mol-I

25598 - 108 - 114 104

= 12441 .

The procedure is that described in Problem 19.7. The data are fitted to the Mark-Kuhn-HouwinkSakurada equation. [1)]

= KM".,

[19.25].

The values obtained for the parameters are K

= 12.38

X

10- 3 cm 3 g-I 1 and

a

= I0.9551·

This K value is smaller than any in Table 19.4 or that in Problem 19.7. The value for a is quite close to 1. When a = 1 exactly, the molar mass, M v corresponds to the weight average molar mass, M w · COMMENT.

The magnitude of the constant a reflects the stiffness of the polymer chain as a result of

rr-orbital interactions between heterocyclic rings.

20

Materials 2: the solid state


Lattice planes are labeled by their Miller indices h, k, and I , where h, k, and 1 refer respectively to the reciprocals of the smallest intersection di stances (in units of the lengths of the unit cell , a, b, and c) of the plane along the x , y, and z axes.

020.3

If the overall amplitude of a wave diffracted by planes (hk/ ) is zero, that plane is said to be absent in the diffraction pattern. When the phase difference between adjacent planes in the set of planes (hkl) is n , destructive interference between the waves diffracted from the planes can occur and this wi ll diminish the intensity of the diffracted wave. This is illustrated in Fig. 20.21 in the text. The overall intensity of a diffracted wave from a plane (hkl) is determined from a calculation of the structure factor, Fhkl , which is a function of the positions (hence, of the Miller indices) and of the scattering factors of the atoms in the crystal (see eqn 20.7). If Fhkl is zero for the plane (h kl) , that plane is absent. See Example 20.3.

020.5

The majority of metals crystalli ze in structures that can be interpreted as the closest packing arrangements of hard spheres. These are the cubic close-packed (ccp) and hexagonal close-packed (hcp) structures. In these models, 74% of the volume of the unit cell is occupied by the atoms (packing fraction = 0.74). Most of the remaining metallic elements crystallize in the body-centered cubic (bcc) arrangement, which is not too much different from the close-packed structures in terms of the efficiency of the use of space (packing fraction 0.68 in the hard sphere model). Polonium is an exception; it crystallizes in the simple cubic structure, which has a packing fraction of 0.52. See the solution to Problem 20.24 for a derivation of all the packing fraction s in cubic systems. If atoms were trul y hard spheres, we would expect that all metals would crystalli ze in either the ccp or hcp close-packed structures. The fact that a significant number crystallize in other structures is proof that a simple hard sphere model is an inaccurate representation of the interactions between the atoms. Covalent bonding between the atoms may influence the structure.

020.7

Because enantiomers give almost identical diffraction patterns it is difficult to distinguish between them. But absolute configurations can be obtained from an analysis of small differences in diffraction intensities by a method developed by J.M . Bijvoet. The method makes use of extra phase shifts that occur when the frequency of the X-rays approaches an absorption frequency of atoms in the compound . The phase shifts are called anomalous scattering and result in different intensities in the diffraction patterns of different enantiomers. See Section 23.7(b) of the 7th edition of this text for an explanation of the origin of this anomalous phase shift. The incorporation of heavy atoms into the compound makes the observation of the extra phase shift easier to observe, but with very sensitive modern diffractometers this is no longer strictly necessary.

390

020.9

STUDENT'S SOLUTI ONS MANUAL

The Fenni-Dirac distribution is a version of the Boltzmann distribution that takes into account the effect of the Pauli exclusion principle. It can therefore be used to calculate the population, P, of a state of given energy in a many-electron system at a temperature T: P

=

1

+I'

-'-;:'---'C7,':;:---

e (E-/.L )/ kT

In this expression, /l- is the Fermi energy, or chemical potential, the energy of the level for which P = 1/ 2. The Fenni energy should be distinguished from the Fermi level, which is the energy of the highest occupied state at T = O. See Fig. 20.54 of the text. From thermodynamics (Chapter 3) we know that dU = -pdV + TdS + /l-dn for a one-component system. This may also be written dU = -pdV + TdS + /l-dN, and this /l- is the chemical potential per particle that appears in the F-D distribution law. The term in dU containing /l- is the chemical work and gives the change in internal energy with change in the number of particles. Thus, /l- has a wider significance than its interpretation as a partial molar Gibbs energy and it is not surprising that it occurs in the F-D expression in comparison to the energy of the particle. The Helmholtz energy, A, and /l- are related through dA = - p dV - S dT + /l- dN, and so /l- also gives the change in the Helmholtz energy with change in number of particles. To fully understand how the chemical potential /l- enters into the F-D expression for P, we must examine its derivation (see Further reading) which makes use of the relation between /l- and A and of that between A and the partition function for F- D particles.


(1 , 0,1) is the midpoint of a face. All face midpoints are alike, including

G,!, 0) and (0, !,!) .

There are six faces to each cube, but each face is shared by two cubes. So other face midpoints can be described by one of these three sets of coordinates on an adjacent unit cell. E20.2(b)

Taking reciprocals of the coordinates yields ( I, ~, -I) and

I

yields the Miller indices (3 13) and (643) E20.3(b)

(!, ~, ! ) respectively. Clearing the fractions

I

The distance between planes in a cubic lattice is a dhkl

=

(h2

+ k2 + [2) 1/ 2

This is the distance between the origin and the plane which intersects coordinate axes at (h la, k i a , [I a). d

-

121 -

(I

523pm

+ 22 + I) 1/ 2

- \ 214 m\ P

MATERIALS 2: THE SOLID STATE

E20.4(b)

391

The Bragg law is /1A

= 2d sine

Assuming the angle given is for a first-order reflection, the wavelength must be A = 2(128.2 pm) sin 19.76° = 186.7 pm 1

E20.S(b)

Combining the Bragg law with Miller indices yields, for a cubic cell sin ehkl

= ~ (h2 + k 2 + p)I / 2 2a

In a face-centered cubic lattice, h, k, and l must be all odd or all even. So the first three reflections would be from the ( I I I ), (2 0 0), and (2 2 0) planes. In an fcc cell , the face diagonal of the cube is 4R, where R is the atomic radius. The relationship of the side of the unit cell to R is therefore

4R

so

a=-

../2

Now we evaluate

A A - = -- = 2a 4../2R

154pm

4../2 (144 pm)

=0.189

We set up the following table

E20.6(b)

e

hkl

sin

III 200 220

0.327 0.378 0.535

19.1 22.2 32.3

38.2 44.4 64.6

In a circular camera, the di stance between adjacent lines is D = R6. (2e), where R is the radius of the camera (distance from sample to film) and is the diffraction angle. Combining these quantities with the Bragg law (A = 2d sin relating the glancing angle to the wavelength and separation of planes), we get

e

e,

d)

1 A D = 2R6.e = 2Rt{sin- 2

= 2(5 .74cm ) x ( sinE20.7(b)

I

I

96.035 95.401 pm ) - sin - I = 0.054cm 2(82.3 pm) 2(82.3 pm )

I

The volume of a hexagonal unit cell is the area of the base times the height c. The base is equivalent to two equilateral triangles of side a. The altitude of such a triangle is (/ sin 60°. So the volume is

v=

2

Ua x asin 60° ) c = (/2 c sin 60° = (1692.9 pm)2 x (506.96 pm) x sin 60°

= 1.2582 x 109 pm 3 = 11.2582 nm 3 1

392

E20.8(b)


The volume of an orthorhombic unit cell is V = abe= (589 pm) x (822 pm) x (798 pm) =

3.86 x 108 pm 3 10 3 = 3.86 x 1O- 22 cm 3 (10 pmcm- I)

The mass per formula unit is m=

135.01 gmol- I = 2.24 x 10- 22 g 6.022 x 1023 mol-I

The density is related to the mass m per formula unit, the volume V of the unit cell, and the number N of formula units per unit cell as foHows

Nm

3 22 N = P V = (2.9 g cm- ) x (3.86 x 10- cm 3) = Isl m 2.24 x 10- 22 g L::J

so

p=-

V

A more accurate density, then, is p =

E20.9(b)

22 5(2.24 x 10- g) 1 -31 = 2.90gcm 3.86 x 10- 22 cm 3

The di stance between the origin and the plane which intersects coordinate axes at (hla, klb, lie) is given by h2

k2

l2 )-1 /2

dhkl = ( - 2 + - 2 + - Z a b c d322

= 1182pm

=

(32 (679 pm)2

+

22 (879 pm)z

22 )-1 /2 + ----;:(860 pm)z

I

E20.10(b) The fact that the III reflection is the third one implies that the cubic lattice is simple, where all indices give reflections. The III reflection would be the first reflection in a face-centered cubic cell and would

be absent from a body-centered cubic. The Bragg law

can be used to compute the cell length a=

A 2sinehkl

(h2+k2+p)I /2=

137pm (I2+12+12)1 /2=390pm 2sin 17.7°

With the cell length, we can predict the glancing angles for the other reflections expected from a simple cubic ehkl = sin-I elOO

(~ (h 2 + k Z + P)I / Z)

= sin- I (0.176(h 2 + k Z + p)I /Z)

= sin-I (0.176(1 2 +0+0)1 /2) = 10.1° (checks)

ello = sin-I (0.176(12 + 12 +O)I / Z) = 14.4° (checks)

ezoo = sin - I (0.176(2 2 + 0 + O)I /Z) = 20.6° (checks)

MATERIALS 2: THE SOLID STATE

393

These angles predicted for a si mple cubic fit those observed, confirming the hypothesis of a simple lattice; the reflections are due to the ( 100), ( 110), (III), and (200) planes.

I

I

E20.11(b) The Bragg law relates the glancing angle to the separation of planes and the wavelength of radiation

A = 2d sin

e

so

e=

A sin - \ 2d

The distance between the origin and plane which intersects coordinate axes at (hla, klb, lie) is given by

So we can draw up the following table

hkl 574.1 796.8 339.5

100 010 III

4. 166 3.000 7.057

E20.12(b) All of the reflections present have h + k + I even, and all of the even h + k + I are present. The unit cell,

I

then, is body-centered cubic

I

E20.13(b) The structure factor is given by

All eight of the vertices of the cube are shared by eight cubes, so each vertex has a scattering factor off/8. The coordinates of all vertices are integers, so the phase
I [22.46]. The activation energies for the parallel reactions are equal in case n and, consequently, the two products appear at identical rates. If the reactions are irreversi ble, [P2l/ [Pll = k2 / kl = I at all times. The results are very different for reversible reactions. The activation energy for PI -+ R is much larger than that for P2 -+ R and PI accumulates as the more rapid P2 -+ R -+ PI occurs. Eventually the ratio [P2l/ [P ll approaches the equilibrium value for which

[P2l ) ( [PIl

= e-(6~-6GIl/RT

< I.

eq

This is thermodynamic control. Case III represents an interesting consecutive reaction series R -+ PI -+ P2. The first step has relatively low activation energy and PI rapidly appears. However, the relatively large activation energy for the second step is not available at low and moderate temperatures. By usi ng low or moderate temperatures and short reaction times it is possible to produce more of the thermodynamically less favorable PI . This is kinetic control. High temperatures and long reaction times will yield the thermodynamically favored P2. The ratio of reaction products is determined by relative reaction rates in kinetic controlled reactions. Favorable conditions include short reaction times, lower temperatures, and irreversible reactions. Thermodynamic control is favored by long reaction times, higher temperatures, and reversible reactions. The ratio of products depends on the relative stability of products for thermodynamically controlled reactions.

THE RATES OF CHEMICAL REACTIONS

022.9

443

The primary isotope effect is the change in rate constant of a reaction in which the breaking of a bond involving the isotope occurs. The reaction coordinate in a C-H bond breaking process corresponds to the stretching of that bond. The vibrational energy of the stretching depends upon the effective mass of the C and H atoms. See eqn 13.50. Upon deuteration, the zero point energy of the bond is lowered due to the greater mass of the deuterium atom. However, the height of the energy barrier is not much changed because the relevant vibration in the activated complex has a very low force constant (bonding in the complex is very weak), so there is little zero point energy associated with the complex and little change in its zero point energy upon deuteration. The net effect is an increase in the activation energy of the reaction. We then expect that the rate constant for the reaction will be lowered in the deuterated molecule and that is what is observed. See the derivation leading to eqns 22.51-22.53 for a quantitative description of the effect. A secondary kinetic isotope effect is the reduction in the rate of a reaction involving the bonded isotope even though the bond is not broken in the reaction. The cause is again related to the change in zero point energy that occurs upon replacement of an atom with its isotope, but in this case it arises from the differences in zero point energies between reactants and an activated complex with significantly different structure. See Illustration 22. 3 for an example of the estimation of the magnitude of the effect in a heterolytic dissociation reaction. If the rate of a reaction is altered by isotopic substitution it implies that the substituted site plays an important role in the mechanism of the reaction. For example, an observed effect on the rate can identify bond breaking events in the rate-determining step of the mechanism. On the other hand, if no isotope effect is observed, the site of the isotopic substitution may play no critical role in the mechanism of the reaction .


E22.2(b)

v

= _ d[A] = _~ d[8] = d[C] = ~ d[D] = l.00moldm -3 s-1 so dt

3 dt

dt

2 dt

Rate of consumption of A

= 11.0 mol dm- 3 S- I 1

Rate of consumption of B

= 13.0 mol dm - 3 S- I 1

Rate of formation of C

= 11.0 mol dm- 3 S- I 1

Rate of form ation of D

= 12.0 mol dm- 3 s- I 1

. of 8 Rate of consumptIOn Rate of reaction

'

d[8] = I 1.00 mol dm - 3 s - ~ = - dr

I

= - ~ d[8] = 10.33 mol dm- 3 s- I 1= d[C] = ~ d[D] = _ d[A] 3 dt

Rate of formation of C

= 10.33 mol dm - 3 S- I I

Rate of formation ofD

= 10.66 mol dm- 3 s-I I

Rate of consumption of A = 10.33 mol dm - 3 S- I I

.

dt

2 dt

dt

444

E22.3(b)


The dimensions of k are dim of v

amount x length- 3 x time- I

(dim of [AJ) x (dim of [BJ)2

(amount x length - 3 )3

= length 6

x amount- 2 x time- I

In mol, dm, s units, the units of k are Idm 6 mol - 2 S- I I

E22.4(b)

(a) v

= - d[A] = k[A][Bj2 so

(b) v

=-

d[A]

dt

d[C]

dt

d[C] so -

dt

dt

= -k[A][Bj2

= k[A][B] 2

The dimensions of k are amount x length- 3 x time- I

dim of v

- - - ----==--- - - : ; - -

dim of [A] x dim of [B] x (dim of [CJ)-I

amount x length - 3

= time- I

The units of k are ~

v E22.S(b)

= d~~] = I k[A][B][C] - 1 I

The rate law is

v = k[AY' ex p"

= {Po(1 -

f)}a

where a is the reaction order, andl the fraction reacted (so that I -lis the fraction remaining). Thus VI

{Po(1 - !J)}(I {Po(l - h))"

=

(1 -/1)a h

and

a

1-

= _In--;-(v,.--I_I_v2:-):- = In

(_I-_II) 1-

E22.6(b)

(I - 0.100)

= 12 00 I

.

The half-life changes with concentration, so we know the reaction order is not 1. That the half-life increases with decreasing concentration indicates a reaction order < 1. Inspection of the data shows the half-life roughly proportional to concentration, which would indicate a reaction order of 0 according to Table 22.3. More quantitatively, if the reaction order is 0, then (I)

tl / 2

ex p

and

t 1/2

-;m 1/ 2

PI P2

We check to see if this relationship holds ( I) tl / ?

340 s

t (2)

178 s

-- = - - = 1/ 2

1.91

so the reaction order is E22.7(b)

h

In (9.71 /7 .67) In 1 - 0.200

@] .

The rate law is 1 d[A] v = - - - =k[A]

2 dt

and

~ P2

= 55 .5 kPa = 1.92 28.9kPa

THE RATES OF CHEMICAL REACTIO NS

445

The half-life formula in eqn 22.13 is based on the assumption that

= k[A].

_ d[A] dt

That is, it would be accurate to take the half-life from the table and say

In 2

tl / 2

= k'

where k' = 2k . Thus t

1/ 2

=

In 2

2(2.78 x 10- 7 S- I)

=

I1.80

X

106 s

I

Likewise, we modify the integrated rate law (eqn 22. 12b), noting that pressure is proportional to concentration:

(a) Therefore, after 10 h, we have 4 p = (32. 1 kPa) exp[-2 x (2.78 x 1O- 7 s- l ) x (3.6 x \0 s) ] = 13 1.5 kPal

(b) After 50 h, p = (32.1 kPa)exp[-2 x (2.78 x 10- 7 S-I) x ( 1.8 x 105 s) ] = 129.0 kPa E22.8(b)

From Table 22 .3, we see that for A

kt

+ 2B

--+

I

P the integrated rate law is

- 2[P])] = [B]o - I 2[A]0 In [[A]O([B]O ([A]o - [P])[B]o

(a) Substituting the data after solving for k

k =

I [ (0.075 x (0.080 - 0 .060) ] x In (3.6 x 103 s) x (0.080 - 2 x 0.075) x (mol dm- 3 ) (0.075 - 0.030) x 0.080

= \ 3.47 x 1O- 3 dm 3 mol - l s- l \ (b) The half-life in terms of A is the time when [A) = [A]0/2 = [P], so

(A) tl / 2

I I [[A]O([B]0-(2[A]0 / 2»] = k([B]o _ 2[A]0) n ([A]0[B]0/2)

which reduces to (A) -

t 1/ 2

-

=

I I (2 2[A]0) k([B]o _ 2 [A]0) n - [B]o I x ln ( 2 - 0.150) -(3.47 x 10- 3 dm 3 mol- I S- I) x (-0.070 mol dm- 3 ) 0.080

=856fs=~

446


The half-life in terms of B is the time when [8] [A]o

1

(1 /2(B)

=

k([B]o _ 2[A]o) In

= [B]o/2 and [P] = [B]o/4:

([B]o - 2[B]O)]

[ ( [A]o -

[B]o) [B]o 4

which reduces to

1/2

E22.9(b)

1

(B)

t

I (

= k([B]o _ 2[A]o) n

[A]o/2 ) [A]o - [B]o/4

=

I 0.075/2 ) x In ( 0.Q75 - (0.080/4) (3.47 x 10- 3 dm 3 mol - I s- I) x (-0.070 mol dm- 3 )

=

1576 s

= I0.44 h I

(a) The dimensions of a second-order constant are

dim ofv (dim of [A))2


- - --

---"'----=---

(amount x length- 3 )2

= length 3 x

amount- I x time-I

I

In molecule, m, s units, the units of k are m3 molecule-I s- I

I

The dimensions of a third-order rate constant are dim of v (dim of [A])3


-

-

-

-----3,,-----

(amount x length- )3

= length

6

x amount

I

In molecule, m, s units, the units of k are m6 molecule- 2 s- I

- 2. - 1 x lIme

I

COMMENT. Technically, "molecule" is not a unit , so a number of molecules is simply a number of individual

objects, that is, a pure number. In the chemical kinetics literature, it is common to see rate constants given in molecular units reported in units of m 3 S - 1 , m 6 S - 1 , cm3 S -1 , etc.

(b) The dimensions of a second-order rate constant in pressure units are dim of v (dim of p )2

press ure x time(pressure)

i

~-:----:-;;-2-

= pressure - Ix .lIme _ I

In SI units, the press ure unit is N m- 2

=

I

Pa, so the units of k are Pa- I S- I

The dimensions of a third-order rate constant in pressure units are dim of v (dim of p)3

press ure x time-I

=----:-- --,--:;-3-

(pressure)

= pressure

-2.

x tIme

In SI pressure units, the units of k are I Pa- 2 s- I I. E22.10(b) The integrated rate law is

k( =

I In [A]o( [B]o - 2[C)) [Table 22.3] [Blo - 2[A]o ([A]o - [C))[B]o

_I

I


447

Solving for [C] yields. after some rearranging [A]o[B]o{exp[kt([B]o - 2[A]0)] - I} [C] = -[B-]'-o-ex-p--=-[k:-t(-C:: [B=-=]:-o-----::2--=-[A-:-:]'--o)--=-]-----::2=-=-[A-:-"C]=--o [C] so mol dm - 3 =

(0.025) x (0.150) x (eO. 21x (0.lOO)x r/ s - I) (3.75 x 10- 3 ) x (eO.021x r/s - I) (0.150) x e O.2Ix (0.lOO)x r/s - 2 x (0.025) = (0.150) x eO.02I xr/s - (0.050)

0-3 (0.21 I) 3 ) x e [C] = (3 .75 x I rna I d m -3 =.165 ' x 10- rna I d m -31. (0. 150) x e0 21 - (0.050) 3 126 [C] = (3.75 x 10- ) x (e - I) mol dm - 3 = 1 0.025 mol dm- 3 1 (0. 150) x e 12.6 - (0.050)

(a)

(b)

E22.11 (b) The rate law is

v

= _~ d[A] =

k[A] 3

2 dt

which integrates to

2kt

t

I(I I)

= 2'

= (4(3 .50 X = 11.5

[A]~

[A]2 -

~m6 moI- 2 s-I Jx CO.02I m~1 dm -

10- 4

x 106 s

3 )2

- (0.077

m~I dm- )2 ) 3

I

E22.12(b) A reaction nth-order in A has the following rate law

_ d[A] = k[Al"

so

dt

d[A] [A]"

= -k dt = [Ar" d[A]

Integration yields [A]I - n - [A]I - II _ _ _ _--'0'--

= -kt

l-n Let tl /3 be the time at which [A]

d

(~ [A]O)I-II - [A]b- II

[A]b - "[(V-n - I]

I-n

1-n

= ~---,---~

so -kt1 /3

an

= [A]0 / 3.

tl / 3 =

3"- 1 - I [A]I-n ken _ I)

°

E22.13(b) The equilibrium constant of the reaction is the ratio ofrate constants of the forward and reverse reactions:

K

= -kr kr

so

kr

= Kk r .

The relaxation time for the temperature jump is (Example 22.4): r = {kr

+ kr([B] + [C])}-I

so

kr = r - I - kr([B]

+ [CD

448


Setting these two expressions for kr equal yields so

k r -

T(K

I

+ [B] + [CD

Hence I

kr = ~~--~~----------~--------~----------------~ (3.0 X 10- 6 s) x (2.0 X 10- 16 + 2.0 X 10- 4 + 2.0 x 10- 4 ) mol dm- 3 = 18.3 x 10 8 dm 3 mol - I s-I

I

E22.14(b) The rate constant is given by

k

= A exp (

-Ea ) RT

[22.31]

so at 24°C it is 1.70

X

10- 2 dm 3 mol - I S-I

= A exp (

-Ea

(8.31451K- I mol-I) x [(24 + 273) K]

)

and at 37 °C it is 2.01 x 10- 2 dm 3 mol - I S- I

= A exp (

-Ea

(8.31451 K-I mol - I) x [(37

+ 273) K]

)

Dividing the two rate constants yields 1.70 2.01 so

2

X X

1010- 2

= exp

2 (1.70 X 10- ) In 2.01 x 10- 2

=

[(

(

-Ea ) 8.31451K - I mol - I x

(1

1)]

297K - 310K

-Ea ) (1 I) 8.31451K- Imol - 1 x 297K-31OK

and Ea = _ ( __1_ _ _1_)-1 In (1.70 x IO-~ ) x (8.31451 K- I mol - I) 297 K 310 K 2.0 I x 10-

= 9.9

x 10 3 1 mo]-I

= 19.9 kJ mol - I I

With the activation energy in hand, the prefactor can be computed from either rate constant value A = kexp

Ea )

( RT

2 3 I I ( = (1.70 x 10- dm mol- s- ) x exp

= I0.94 dm 3 mol- I s-I

I 3 9.9 x 10 J mol) I (8.3145JK-I mol- ) x (297K)

I

E22.15(b) (a) Assuming that the rate-determining step is the scission of a C-H bond, the ratio of rate constants

for the tritiated versus protonated reactant should be [22.53 with hciJ

= fiw = Ii(kj /1-) 1/ 2]


449

The reduced masses will be roughly I u and 3 u respectively, for the protons and 3H nuclei are far lighter than the rest of the molecule to which they are attached. So A~

(1.0546

2 x (1.381 x 1O- 23 JK- I) x (298K)

(~ /2 - ~ /2 ) (l u) (3 u)

x ~

10- 34 I s) x (450Nm- I)I / 2

X

~-----:-.,------,---=-:~-;--C"":"""":"-:-::-:c-

27

I kgu - )- 1/2

2.8

~ e- 2S = 10.06 ~

so ::

x ( 1.66 x 1O-

1/ 16 1

(b) The analogous expression for 16 0 and ISO requires reduced masses for C- 16 0 and C-I SO bonds. These reduced masses could vary rather widely depending on the size of the whole molecule, but in no case will they be terribly different for the two isotopes. Take 12CO, for example:

/-L 16 = _(1_6_.0_u_)_x_(1--:2_.0_u_) (16.0+ 12.0) u

( 1.0546

= 6.86 u

and

/-LI S =

( 18.0u) x (12.0u) (18.0 + 12.0) u

= 7.20 u

10- 34 J s) x (1750 N m- I) 1/ 2

X

A =---=--:-:-::-::-:--:-::--;;-;;-:-=--;-:---=-::-:-::-=:23

2 x ( 1.381 x 1O-

1K- I) x (298K)

1 1/2 1 1/2 ) x ( 1.66 x 1O- 27 kgu - I) - 1/2 (6.86 u) (7.20 u)

x (

= 0.12 so kl s kl 6

= e- O.12 = 10.891

At the other extreme, the 0 atoms could be attached to heavy fragments such that the effective mass of the relevant vibration approximates the mass of the oxygen isotope. That is, /-L16 ~ 16 u and /-L IS ~ I 8 u so A ~ 0. 19 1

E22.16(b)

k'a

-k = k k a b

kl s = e- O.19 = 10.831

so

kl6

1

+ -kaPA

[analogous to 22.67]

Therefore, for two different pressures we have

so

ka =

(~ - ~) (~ - ~) - I P

P'

k

k'

1

2.2

X

10- 4 s-I

)-1

450



A simple but practical approach is to make an initial guess at the order by observing whether the half-life of the reaction appears to depend on concentration. If it does not, the reaction is first-order; if it does, it may be second-order. Examination of the data shows that the first half-life is roughly 45 minutes, but that the second is about double the first. (Compare the 0 ~ 50.0 minute data to the 50.0 ~ 150 minute data.) Therefore, assume second-order and confirm by plotting I/[Al against time. If the reaction is second-order, it will obey

1 [Al

I

= kt + [Ala

[22.15bl .

We draw up the following table (A

tlmin m(urea)/g m(A)/g

[AJ /(mol dm - 3 ) [Ar l /(dm 3 mol -I)

0 0 22.9 0.381 2.62

= NH4CNO).

20.0 7.0 15.9 0.265 3.78

50.0 12.1 10.8 0.180 5.56

65.0 13.8 9.1 0.152 6.60

150 17.7 5.2 0.0866 11.5

The data are plotted in Fig. 22.2 and fit closely to a straight line. Hence, the reaction is / second-order /. The rate constant is the slope: 1k

= 0.0594 dm 3 mol - I min- I I.

12

10 t

0 E

8

E

~

--

~

-I ~

6

'--"'

4

2

0

20

40

60

80 t/ min

100

120

140

Figure 22.2

To find [AJ at 300 min, use eqn 22.15c: [Ala [Al

= I + kt[Ala

0.382 mol dm - 3 = 0.0489 mol dm -3. 1 + (0.0594) x (300) x (0.382)


451

The mass of NH4CNO left after 300 minutes is

P22.3

The procedure adopted in the solutions to Problems 22. 1 and 22.2 is employed here. Examination of the data indicates the half-Life is independent of concentration and that the reaction is therefore 1first-order I. That is confirmed by a plot of In (A = nitrile). t/( 103 s) [Al /(mol dm-

3

)

0

2.00

4.00

6.00

8.00

10.00

12.00

1.10

0.86

0.67

0.52

0.41

0.32

0.25

0.78

0.6 1

0.47

0.37

0.29

0.23

-0.246

-0.496

-0.749

-0.987

-l.235

-1.482

ill [Alo

In

(l&)

(l&) against time (eqn 22. 12b). We draw up the following table

0

-slope

A least-squares fit to a linear equation gives k coefficient of 1.000. P22.5

11.23

X

10- 4

S- i

1 with a correlation

As described in Example 22.5, if the rate constant obeys the Arrhenius equation [22.29] , a plot of In k against I / T should yield a straight line with slope -Ea/ R . However, since data are available only at three temperatures, we use the two-point method, that is,

-R In (576/ 2.46)

4-1

Ea= «(1 / 313K)-(i / 273K)) =9.69 x 10 Jmol

.

For the pair () = 20°C and 40°C, -Rln (576/45. 1)

4

1

Ea= ((l/313K)-(1/293K)) =9.71 x 10 Jmol-.

The agreement of these values of Ea indicates that the rate constant data fits the Arrhenius equation and that the activation energy is 19.70 x 104 J mol-I I. P22.7

The data for this experiment do not extend much beyond one half-life. Therefore the half-life method of predicting the order of the reaction as described in the solutions to Problems 22.1 and 22.2 cannot be used here. However, a si milar method based on three-quarters lives will work. For a first-order reaction, we may write (analogous to the derivation of eqn 22.13)

~[Alo

3 4

4 3

kt3 /4 = - In - - = - In - = In - = 0.288

[Alo

or

0.288

t3 /4

= -k-'

452


T hus the three-quarters life (or any given frac tional life) is also independent of concentration for a firstorder reaction. Examination of the data shows that the first three-quarters life (time to [A] = 0.237 mol dm - 3) is about 80 min and by interpolation the second (time to [A] = 0.178 mol dm- 3) is also about 80 min. Therefore the reaction is first-order and the rate constant is approximately

k

0.288 0.288 ~ --t3/4 80 min

= --

=

36 x 10- 3 min-I. .

A least-squares fit of the data to the first-orde r integrated rate law [22.12b] gives the slightly more accurate result, k = 13.65 x 10- 3 min - I I. The half-life is

rl /2

In 2

In 2

= - k = 3.65

x 10- 3 min - I

= 1190 min I..

The average lifetime is calculated form [A] [A]o

= e - kt

[22.12b].

which has the form of a distribution functio n. The ratio ~ is the fraction of sucrose molecules that have lived to time t. The average lifetime is then

The denominator ensures normalization of the distribution function . COMMENT.

The average lifetime is also called the relaxation time. Compare to eqn 22.28. Note that the

average lifetime is not the half-life. The latter is 190 minutes. Also note that 2 x 13/ 4

P22.9

-I

11 / 2 .

The data do not extend much beyond one half-life; therefore, we cannot see whether the half-life is constant over the course of the reaction as a preliminary step in guessing a reaction order. In a first order reaction, however, not only the half-life but any other similarly defined fractional lifetime remains constant. (T hat is a property of the exponential funct ion .) In this problem, we can see that the ~ -life is not constant. (It takes less than 1.6 ms for [CIO] to drop from the first recorded val ue (8.49 /-Lmol dm - 3) by more than of that value (to 5.79 !-Lmol dm - 3 ) ; it takes more than 4 .0 more ms for the concentration to drop by not even of that value (to 3.95 !-LmOI dm- 3 ). So our working assumption is that the reaction is not first-order but second-order. Draw up the following table.

1

1

rims

[CIO]/(!-Lmol dm- 3)

( II[CIO])/(dm3!-Lmol-l)

0. 12 0.62 0.96 1.60 3.20 4.00 5.75

8.49 8.09 7.10 5.79 5.20 4 .77 3.95

0. 11 8 0. 124 0.141 0. 173 0. 192 0.210 0.253

THE RATES OF CHEM ICAL REACTIONS

453

The plot of [CIO] vs. t in Fig. 22.3 yields a reasonable straight line; the linear least squares fit is: (I / [CIO)) / (dm\lmol - I)

= 0.118 + 0 .0237(t / ms)

R2 = 0.974.

The rate constant is equal to the slope

The half-life depends on the initial concentration (eqn 22.16): '1 / 2

= __1 - = k[ClO]o

I (2.37 x 10- 7 dm 3 mol- I s-I )(8.47 x 10- 6 mol dm- 3 )

= 14.98

x 10- 3 s I.

0.30

,.

0.25

"0

E

:::l

ME

0.20

"0

:::::'

-I~

0.15 0.10 2

0

P22.11

A+B

~

d[P]

P,

3 rIm s

4

5

6

Figure 22.3

= k[A]'"[BJ"

dt

and, for a short intervall3t, 8[P]

~

k[A]'"[B]"8t

Therefore, since 8[P]

= [P]f

- [P]o

=

[P]I>

[Chloropropanel · . d d f [P ] . I· h [Propene] IS III epen en! 0 ropene , Imp ylllg t at m [Chloropropane] [HCI]

= {P(HCI)

10 0.05

7.5 0.03

= I.

5.0 0.01

These results suggest that the ratio is roughly proportional to p (HCI)2, and therefore that m A is identified with HC!. The rate law is therefore d[Chloropropane]

-"---d"":t---'-"'::'

I

= k[Propane][HCI]

I

3

I

I

and the reaction is first-order in propene and third-order in HC!.

= 3 when

454 P22.13


2HCI ;=; (HCI)z,

[(HClh ] = KI [HClf

KI

HCI + CH3CH=CH2 ;=! complex (HClh + complex rate

--+

[complex]

= K2[HCI][CH3CH =CH2]

CH3CHCICH3 + 2HCI k

d[CH3CHCICH3]

=

K2

dt

= k[(HClh][complex].

Both (HClh and the complex are intermediates, so substitute for them using equilibrium expressions: rate

= k[(HClh][complex] = k(KI [HClf)(K2[HCI][CH3CH=CH2]) = 1 kKIK2[HCI] 3[CH 3CH=CH2] 1

which is third-order in HCI and first-order in propene. One approach to experimental verification is to look for evidence of proposed intermediates, using infrared spectroscopy to search for (HClh , for example. P22.1S

We can estimate the activation energy of the overall reaction by proceeding as in P22 .5: -R In (keff /k~ff)

I

-R In 3

-I

I

E - -18 kJ mol a,eff- ((l / T) - (l / T'» - (I / 292K)-(l / 343K ) - '

To relate this quantity to the rate constants and equilibrium constants of the mechanism (P22.13), we identify the effective rate constant as keff = kKI K2 and apply the general definition of activation energy (eqn 22.30): ?

d In keff

= RT-~ = RT

Ea,eff

keff d( l i T) d(l I T) ~

2 din

din keff

= -R d(l I T)'

This form is useful because rate constants and equilibrium constants are often more readily differentiated when considered as functions of liT rather than functions of T, as in this case: In keff

=

In k + In KI + In K2

__ Rdlnkeff --R dink _RdlnKI _RdlnK2 =E. +t:-..H +t:-..H so Ea,eff d(l / T) d(l I T) d(l / T) d(l l T) a r 1 r 2 dinK since - d(l / T)

-t:-..rH . =- [van't Hoff equation, 7.23b]. R

Hence Ea = Ea,eff - t:-..rHI - t:-..rH2 = (-18 + 14 + 14) kJ mol - I P22.17

-I = _ k'a k

kakb

= 1+10 kJ mol- I I·

I + _ [analogous to 22.67] . kaP

We expect a straight line when

~ k

is plotted against

~ . We draw up the following table.

P

plTorr

84. 1

11 .0

2.89

0.569

0.120

0.067

1/(plTorr)

0.012 0.336

0.091 0.448

0.346 0.629

1.76 1.17

8.33 2.55

14.9 3.30

1O- 4 /(kl s- I)

THE RATES OF CHEM ICAL REACTIONS

455

These points are plotted in Fig. 22.4. There are marked deviations at low pressures, indicating that the Lindemann theory is deficient in that region.

4

.,...... '"1;

~

..,.......

3

2

I

0

4

8

12

l/ (p/ Torr)

P22.19

Figure 22.4

The reasoning that led to eqn 22.46 holds as long as the rate laws for the two products have the same reaction orders:

Then, si nce Ea . 1 > Ea.2, the exponent in the exponential function is negative, and it gets less negati ve as the temperature increases. Therefore, the exponential function itself increases and the product concentration ratio also increases. COMMENT. A qualitative argument can be made that leads to the same conclusion, provided one under-

stands that the activation energy is a measurement of the strength of a reaction's temperature dependence. (See eqn. 22 .30.) Since Ea.1 > Ea.2, the rate of reaction 1 increases faster with increasing temperature than does the rate of reaction 2.


A ;=" B d[A] dt

= -k[A] + k'[B]

At all times, [Al Therefore, [B]

d~~]

+ [Bl

= [Alo

and

=

-k' [Bl

[AJ.

+ k' {[Alo + [Blo -

[All = -(k

To solve, one must integrate

f

(k

d[Al k' ([Alo

+ k' )[A] -

+ k[Al .

+ [Blo·

= [Ala + [Blo -

= -k [A ]

d[B] dt

+ [Blo) = -

f

dt.

+ k' )[Al + k'( [A]o + [B]o) .

456


· . [Al Th e so IutlOn IS

k'([Alo + [Blo) + (k[Alo - k' [Blo)e-(k+k')r = --------:------:-,-------=........:..:'-'---k+k'

The final composition is found by setting t

=

[Aloo

and [Bloo

=

C~ k') [Alo

= 00:

x ([Alo + [Blo).

+ [Blo -

=

[Aloo

(_k_)

x ([Alo

k +k'

+ [Blo) .

Note that

P22.23

d[Al dt

= -2k[Aj2[Bl

(a) Let [Pl = x at

I,

'

2A

+ B --+

then [Al = Ao - 2x and [Bl = Bo - x = ~ - x. Therefore,

dx -d[Al = -2= -2k (Ao dr dr

-dx = k(Ao dt

P.

(I

- 2x) 2 x

2

2x) x (Bo - x) ,

)= I

-Ao - x

2

-k(Ao - 2x) 3

2

'

1 = 10r (Ao -dx2r) 3 = 4I x [( Ao -12x )2 - (IAo )2] .

"2kt

Therefore, kt (b) Now Bo

2x(Ao - x)

= ,

Aii(Ao - 2x)

2

= Ao , so

dx 2 2 - = k(Ao - 2x) x (Bo - x) = k(Ao - 2x) x (Ao - x) , dt kt _

r ..,.-_--:--,dx_,-----_--:x (Ao - x) .

- 10 (Ao -

2x)2

We proceed by the method of partial fractions (which is employed in the general case too), and look for the values of ex, fJ, and y such that ex (Ao - 2x)2

(Ao - 2x)2 x (Ao - x)

+ _fJ_ + _y_ . Ao - 2x

Ao - x

This requires that ex(Ao - x)

+ fJ(Ao -

2x) x (Ao - x)

+ y(Ao -

2x)2

= I.

Expand and gather terms by powers of x: (Aoex

+ A6fJ + A6Y) -

(ex

+ 3fJAo + 4yAo)x + (2fJ + 4y)x 2 = I.


This must be true for all x; therefore

+ 3Ao,8 + 3Aoy = 0,

a

2,8

+ 4y = 0. 2

=-

These solve to give a

Ao

,,8

-2

I

Ao

Ao

= -2' and y = 2·

Therefore, kt

r(

=

(2/Ao) _ (2/A6) (Ao-2x)2 Ao-2x

10

+ (i/A6))

I

(i/Ao)

dx

Ao-x x

I

- + - In(Ao - 2x) - - In(Ao - x) ) (Ao - 2x A2 A2 0 0 0 I

P22.2S

d[Al The rate law - dt kt =

n- I

tl /2

kt3 /4

=

t3 /4

v

=

2n -

= k([Alo

=

1 -

I

x)([Blo

I

+ x).

- x) - k([Blo

[Blo

[E22.12(a)l.

4 ),,-1 - (I[Alo )"-1] . (n _I 1 ) [( 3[Alo

dv The extrema correspond to dx [Alo - x

[Al3-

C~ I) [C~lOr-1 - c~lor-1

(3"4),,-1 -

= k([Alo -

dv dx

I_I)

[Al"-

ktl /2 =

= t3 /4,

Hence, -

P22.27

I integrates to

(_1_) x (_1_1 _ _

Att = tl / 2,

At t

= -k[Al" for n f.

+x

or

+ x).

= 0, or 2x

=

[Alo - [Blo

or

x=

[Alo - [Blo 2

457

458


Substitute into v to obtain

Since v and x cannot be negative in the reaction, 1[Blo::: [Alo I·

To see the variation of v with x , let [Blo v

or

=

k([Alo - x)([Alo + x)

k[:16 =

(I - [~~6)

Thus we plot -vk[Aol

= (I-

2

= k([A16 -

(1 +

=

= [Alo . The rate equation becomes

[:10)

x )

(1-

X2) = (l -

-2

[Aol

=

k[A16 - kx 2

[:10).

X2) against -x-

[Aol

= X from X = o.

x x . The plot is shown in Fig. 22.5 in which X = - - . - - < I corresponds to realIty . [Alo [Alo-

1.0

r----===-~-----:---------:-----__:---_.

0.8

0.6 N

> 6 ~

'I'

5.5

E u

01)

c:

"5'

-=

5 4 .5 4 3.5 3

0

100

200

300 tlmin

400

500

Figure 22.6(a)

The data are plotted in Figs . 22.6(a) and (b). The first-order plot fits closely to a straight line with just a hint of curvature near the outset. The second-order plot, conversely, is strongly curved throughout. Hence,

460

STUD ENT'S SOLUTIONS MANUAL 0.05

~

~ R2 = 0.8403:

0.04

•

~ 0.03

'7 E u

OIl

.5 0.02

0.0 1

0 0 tlmin

Figure 22.6(b)

the reaction is 1first-order I. The rate constant is the slope of the first-order plot: k 10.459 h-

I

=

1

0.00765 min -

I

1

=

I·

The half-life is (eqn 22.13) In2

tl / 2

=

k

In 2 = 0.459 h- I

~

=~=191 min I·

COMMENT. As noted in the problem, the drug concentration is a result of absorption and elimination of the

drug, two processes with distinct rates . Elimination is characteristically slower, so the later data points reflect elimination only, for absorption is effectively complete by then . The earlier data points, by contrast, reflect both absorption and elimination. It is, therefore, not surprising that the early points do not adhere so closely to the line so well defined by the later data.

P22.33

(a) For the mechanism k.

hhhh ...

~

hehh . . .

hehh . ..

kb ~

ecce . ..

k~

the rate equations are d[hhhh ... J --d-t--

I

= -ka[hhhh . .. J + ka[hehh . .. J,

d[hehh . .. J - - - = ka[hhhh .. .J dt d[ccee .. .J """:""-d-:-t- -"

= kb[hchh . .. J -

I

I

ka[hehh . .. J - kb[hehh . .. J + kb[eeee . .. J, I

kb[eeee . . .J.


461

(b) Apply the steady-state approximation to the intennediate: , ' 0 ka[hchh . . .] - kb[hchh . .. ] + kb[cccc . .. ] =

d[hchh . .. ] - - - = ka[hhhh . .. ] dt

so [hchh . .. ] Therefore,

ka[hhhh . . .] + k~[cccc . .. ] = -----~-- k~ + kb

d[hhhh . .. ] ~

kakb k~k~ ---[hhhh . .. ] + --[cccc . . .].

=

~+~

~+~

This rate expression may be compared to that given in the text [Section 22.4] for the mechanism k

A -.::= B . k' keff

Here hhhh ... -.::= cccc ... with k~ff

(c) It is difficult to make conclusive inferences about intennediates from kinetic data alone. For example,

if rate measurements show fonnation of coils from helices with a single rate constant, they tell us nearly nothing about the mechanism. The rate law d[cccc . . .] dt

= k[hhhh ... ]

is consistent with a single-step mechanism, with a two-step mechanism with a rate-determining second step, and with a two-step mechanism with a steady-state intennediate. Even if kinetic monitoring of the product shows production with two rate constants, the rate constants could belong to competing paths or to steps of a single reaction path. The best evidence for an intennediate's participation in a reaction is detection of the intennediate, or at least detection of structural features that can belong to a proposed intennediate but not reactant or product. P22.35

We assume a pre-equilibrium (as the initial step is fast), and write K

=

[unstable helix] [A][B] ,

implying that [unstable helix]

= K[A][B].

The rate-determining step then gives v

=

d[double helix] dt

,----, [B) I [k

= k2[unstable helix] = k2 K [A][B] = Ik[A]

The equilibrium constant is the outcome of the two processes kl

k,

A + B -.::= unstable helix,

K=k'I

k;

.

Therefore, with v

P22.37

= k[A][B],

E!J k

= - 1k2 . k'I

The Arrhenius expression for the rate constant is k

= Ae- Ea / RT

[22.31] so Ink

= InA - Ea/RT [22.29].

= k2K].

462


A plot of In k versus l i T wiU have slope -Ea / R and y-intercept InA . The transformed data and plot (Fig. 22.6) follow. 295 3.55 15 .08 3.39

T IK 1O- 6 kl(dm 3 mol-I s-I) lnk l(dm 3 mol-I S- I) 10- 3 K I T

15.5 15 14.5

"'" .:

14

.

223 0.494 13.11 4.48

I

'"

218 0.452 13 .02 4.59

213 0.379 12.85 4.69

206 0.295 12.59 4.85

200 0.241 12.39 5.00

195 0.217 12.29 5.13

I

In k = - 1642 KIT + 20.585 R'=O.9937

~

13.5

~

13

"; ~

12.5

~

12

0.003

0.0035

0.004

0.0045

0.005

KJT

0.0055

Figure 22.7

SoEa = -(8.3145JK- 1 mol - I) x (-1642K) = 1.37 x 104 Jmol- 1 =1 13.7 kJ mol-I 1 and A = e20 .585 dm 3 mol- I S-I =18.7 x 108dm 3 mOI- l s- 1 1 P22.39

The rate constants are:

k=Aexp(~~a) 9

[22.31]. 3

I

I

kl = ( 1.13 x 10 dm s- mol- ) exp

(

1 ) -14.1 x 103JmolI I (8.3145J K- mol -) x (298 K)

Compared to reaction I, reaction 2 shows a significant kinetic isotope effect whereas reaction 3 shows practically none. This difference should not be surprising: in reaction 2 a C-D bond is broken, whereas


463

in reaction 3 the D atom is simply along for the ride already attached to the 0 atom. Compare the measured isotope effect of 0.13 to that expected in reaction 2.

(1 1))

l2 I72 - I72 -k2 = exp (hkf --kl

We take

k2 kl

2kBT

Ji,CH ~ mH

= ex

/I

""CH

and Ji,CD

[E22.15(a»).

/I

""CD

~ mD ~ 2mH,

so

34 (( -(1.0546 x 10- J s) x (500 kg s-2) P 2(1.381 x 1O-23JK-I) x (298K)

23 x (6.022 x 10 mOl- I) 1 x 10- 3 kg mol- I

1/2)

=~ in agreement with the experimental value.

1/2) ( x

1) 1- 21 /2

23

The kinetics of complex reactions


(a)

(l)AH~A·+H·

(2) A· ~ B· +C (3) AH + B· ~ A· + D (4) A · + B·

(b)

~

P

(l)A 2 ~A·+A·

(2) A· ~ B· + C (3) A· +P ~ B· (4)A · +B·~P

Initiation [radicals formed] Propagation [new radicals formed] Propagation [new radicals formed] Termination [non-radical product formed] Initiation [radicals formed] Propagation [radicals formed] Retardation [product destroyed, but chain not terminated] Termination [non-radical product formed]

023.3

The Michaelis-Menten mechanism of enzyme activity models the enzyme with one active site that, weakly and reversibly, binds a substrate in homogeneous solution . It is a three-step mechanism . The first and second steps are the reversible formation of the enzyme-substrate complex (ES). The third step is the decay of the complex into the product. The steady-state approximation is applied to the concentration of the intermediate (ES) and its use simplifies the derivation of the final rate expression. However, the justification for the use of the approximation with this mechanism is suspect, in that both rate constants for the reversible steps may not be as large, in comparison to the rate constant for the decay to products, as they need to be for the approximation to be valid. The simplest form of the mechanism applies only when kb » k~ . Nevertheless, the form of the rate equation obtained does seem to match the principal experimental features of enzyme-catalyzed reactions; it explains why there is a maximum in the reaction rate and provides a mechanistic understanding of the turnover number. The model may be expanded to include multi substrate reaction rate and provides a mechanistic understanding of the turnover number. The model may be expanded to include multisubstrate reactions and inhibition.

023.5

The primary quantum yield is associated with the primary photochemical event in the overall photochemical process which may involve secondary events as well. An example that illustrates both kinds of events is the photolysis of HI described in Section 23.8(a). The primary quantum yield is defined as the ratio of the number of primary events to the number of photons absorbed (eqn 23.28) and its value can never exceed one. However, in reactions described by complex mechanisms, the overall quantum yield, which is the number of reactant molecules consumed in both primary and secondary processes per photon absorbed, can easily exceed one. Experimental procedures for the determination of the overall

THE KINETICS OF COMPLEX REACTIONS

465

quantum yield involve measurements of the intensity of the radiation used, defined here as the number of photons generated and directed at the reacting sample, and of the amount of product formed . This ratio is the overall quantum yield. See Example 23 .5. In addition to chemical reactions, the concept of the quantum yield enters into the description of other kinds of photochemical processes, such as fluorescence and phosphorescence, and in each case there are techniques specific to the process for the determination of the quantum yield. 023.7

The Forster theory of resonance energy transfer examines the interaction between an induced oscillating dipole moment in chromophore S, the energy donor, with a second chromophore Q, the energy acceptor. The oscillating dipole moment of S is induced by incident electromagnetic radiation and the chromophores are separated by distance R. S transfers the excitation energy of the radiation to Q via a mechanism in which its oscillating dipole moment induces an oscillating dipole moment in Q. Resonance energy transfer can be efficient when R is short (typically less than about 9 nm) and when the absorption spectrum of the acceptor overlaps with the emission spectrum of the donor. Fluorescence resonance energy transfer (FRET) experiments commonly use the fluorescent spectrum and relaxation times of the Forster donor and acceptor chromophores to find the distances between fluorescent dyes at labeled sites in protein, DNA, RNA, etc. FRET is a type of spectroscopic ' ruler' . The computation uses either experimental quantum yields or relaxation lifetimes to calculate the efficiency of resonance energy transfer ET . ET

=

I- -

r/Jf

=

T

I- -

r/Jf,O

[23 .37] .

TO

ET is used to calculate R.

or

R

IE) ' /6

= Ro ( ~T

T

[23 .38].

Solutions to exercises In the following exercises and problems, it is recommended that rate constants are labeled with the number of the step in the proposed reaction mechanism and that any reverse steps are labeled similarly but with a prime. E23.1(b)

The intermediates are NO and N03 and we apply the steady-state approximation to each of their concentrations k2 [N02] [N03] - k3 [NO] [N20S]

=0

k, [N20 S] - k; [N02] [N03] - k2 [N02] [N0 3] Rate = _~ d [N 20 S] 2 dt d [N?Os] d~ = -k, [N20 S]

I

=0

+ k, [ N02][N03] - k3[NO][N20S]

466


From the steady-state equations k3 [NO] [N20S]

[N20 S] k; + k2

= kl

[NO] [NO] 2

= k2 [N02] [N03]

3

Substituting,

E23.2(b)

Apply the steady-state approximation to both equations

+ k3 [R'] k3 [R' ] = 0

2kl [R2] - k2 [R] [R2]

k2 [R] [R2] -

The second solves to [R']

=

2k4 [R]2 = 0

k2 -[R][R2] k3

k ) and then the first solves to [R] = ( k~ [R2]

E23.3(b)

1/ 2

(a) The figure suggests that a chain-branching explosion I does not occur I at temperatures as low as 700 K. There may, however, be a thermal explosion regime at pressures in excess of 106 Pa. (b) The lower limit seems to occur when

There does not seem to be a pressure above which a steady reaction occurs. Rather the chainbranching explosion range seems to run into the thermal explosion range around log (pjPa) E23.4(b)

= 4.5

so

p

= 104.s Pa = 13

The rate of production of the product is

x 104 pa 1


467

HAH + is an intermediate involved in a rapid pre-equilibrium

= ~ so [HAH+ ] =

[HAH+ ] [HAl [H+ ] and d[BH + ] dt

k;

= kl~2

kl [HA1[H+ ]

k;

[HAl [H+ ]rBl

kl

L -_ __ _ __

---'

This rate law can be made independent of [H+l if the source of H+ is the acid HA, for then H+ is given by another equilibrium 2 [H+ )[A - l = K = [H+ l so [H+l = (K [HAl) 1/2 a [HAl a [HAl and d[BH + l dt E23.5(b)

= klk2K~ /2 [HAl3/ 2[Bl k;

L -_ __ _ __

-'

A2 appears in the initiation step only.

Consequently, the rate of consumption of [A2l is first order in A2 and the rate is independent of intermediate concentrations. E23.6(b)

The maximum velocity is kb [Elo and the velocity in general is

v

V

E23.7(b)

= k [Elo =

Olax

=

kb [Sl [Elo KM + [Sl

so

V max

= kb [Elo =

KM

+ [Sl

[Sl

v

3 (0.042 + 0.890) mol dm - (2.45 x 10- 4 mol dm -3 s -l) 0.890 mol dm - 3

= 12.57

X 10- 4 mol dm - 3 s-I 1

The quantum yield tells us that each mole of photons absorbed causes 1.2 x 102 moles of A to react; the stoi chiometry tells us that I mole of B is formed for every mole of A which reacts. From the yield of 1.77 mmol B, we infer that 1.77 mmol A reacted, caused by the absorption of 1.77 x 10- 3 mol / ( 1.2 x 102 mol Einstein-I) = 11.5 x 10- 5 moles of photons

I

E23.8(b)

The quantum efficiency is defined as the amount of reacting molecules nA divided by the amount of photons absorbed nabs. The fraction of photons absorbed Jabs is one minus the fraction transmittedflrans; and the amount of photons emitted nphoton can be inferred from the energy of the light source (power P times time t ) and the energy of the photons (he/A) . nAheNA =----( I - flrans) APt

(0.324mol) x (6.626 x JO-34JS) x (2.998 x J08 m s-l) x (6.022 x 1023 mol-I) ( I -0.257) x (320 x JO-9 m) x (87.5 W) x (28.0 min ) x (60 s min - I)

=[D]

468


Solutions to problems Solutions to numerical problems

These equations serve to show how even a simple sequence of reactions leads to a complicated set of nonlinear differential equations. Since we are interested in the time behavior of the composition we may not invoke the steady-state assumption. The only thing left is to use a computer and to integrate the equations numerically. The outcome of this is the set of curves shown in Fig. 23.1 (they have been sketched from the original reference). The similarity to an A -+ B -+ C scheme should be noticed (and expected), and the general features can be analysed quite simply in terms of the underlying reactions.

3.0

~

2.0

u

"0

E........ ~

1.0

0.0

o

2

4

6 t/ ms

8

10

Figure 23.1


P23.3

The roles are

+0

(1)

N20 ---+ N2

(2)

o + SiH4 ---+

(3)

(4) (5)

(6)

SiH3 + OH OH + SiH4 ---+ SiH3 + H2O SiH3 + N20 ---+ SiH30 + N2 SiH30 + Si~ ---+ SiH30H + SiH3 SiH3 + SiH30 ---+ (H3 SihO

initiation, propagation [or transfer], propagation [or transfer], propagation, propagation, termination.

The rate of silane consumption is

Steady-state approximation (SSA) for 0

SSA forOH

so [OH]

=

kl [N20 ] . k3[SiH4]

SSA for SiH30 and SiH3

+ ks[SiH30][SiH4] = 2kl [N20] -

k6[SiH30][SiH3]

k4[SiH3][N20]

+ ks[SiH30HSiH4] -

k6[SiH30HSiH3] ~

Adding these expressions together yields

and subtracting them gives

Solve for [SiH30]:

0= 2k [N 0] - k 1k4[N2 0 ]2 I

2

k6[ SiH 30 ]

= 2klk6[SiH30][N20] -

+ ks[SiH3 O][S1.'4 ·u 0]

klk4[N20]2

+ ksk6[SiH30f[SiH4].

o.

469

470

STUD ENT'S SOLUTIONS MANUAL

If kl is small, then

[SiH 30 ] "" kl [N 20] (k4kS[SiH4]) 1/ 2 kS[S IH4] k, k6

= [N2 0 ] (

klk4 ) ksk6[SIH4]

1/ 2


P23.5

In the steady-state approximation for [I ·],

Substitution of (2) into ( I) gives 2kbka[h][H2]

d[HI]

k~

dl

+ kb[H 2]

This simple rate law is observed when step (b) is rate-determining so that step (a) is a rapid equilibrium and [I·] is in an approximate steady state. This is equivalent to kb[H2] « k~ and hence,

P23.7

(a)

!.i~ = e-

I

/

ro [23.3 1]

or

In

(!.i) = _!...-. ~ ~

A plot of In(lr 110) against t should be linear with a slope equal to - l i To (i.e. TO = -l/slope) and an intercept equal to zero. Consequently, we make the plot to determine whether it is linear. If it is linear (it is), we do a linear regression fit with a zero intercept and use the regression slope to calculate TO . See Fig. 23.2. Alternatively, average the experimental values of (l I t ) In Urllo) and check that the standard deviation is a small fraction of the average (it is). The average equals - l i TO (i.e. TO = - I1average). Slope

= -0.150 ns- l , - (-0.150 ns- I )- I ,

TO

=

TO

= 16.67 ns I·


(b)

kf

= rpflTo [23.34] = 0.70/(6.67 ns)

kf

= 10.105 ns- I I·

471

0 y = -D.ISOlx

-D.S

R2 =0.997

-I ~

~

:::: -=

-I.S -2 -2.S -3 -3.S

20

IS

10

S

0

tIns

P23.9

2S

Figure 23.2

Since Jr = kf[S*]t = kf[S*]O e t / ro , we surmise that a graph of In (If / 10) against t should be linear with a slope equal to -l i To in the absence of a quencher. The plot is in fact linear with a regression slope equal to -1.004 x 105 S-I, TO

1

= 1.004 x 105s- I = 9.96I1- s.

In the presence of a quencher, a graph of In(Jr / 10) against t is still linear but with a slope equal to -l / T. This plot is found to be linear with a regression slope equal to -1.788 x lOSs- I. 1

T = 1.788 x 105 s-I

1 T

I TO

- = - + kQ[Q]

= 5.59I1-s.

[23 .36].

(0.08206dm 3 atmK- I mol-')(300K)(1.788 - 1.004)105 s-I 9.74 x 10-4 atm kq = 11.98 x 109 dm 3 mol-' s-I

I.


d[CH3CH3] dt

=

-ka[CH3CH3] - kb[CH3][CH3CH3] - kd[CH3CH3][H]

+ ke[CH3CH2][H].

We apply the steady-state approximation to the three intermediates CH3, CH3CH2 , and H. d[ CH 3] -d-t- = 2ka[CH3CH3] - kb[CH3CH3][CH3] = 0

· h Imp ' l'les th at [C H3] whIC

2ka. =kb

472


d[ CH3CH ? ) dt -

= kb[CH3)[CH3CH3) + kd[ CH 3CH 3)[H) -

kdCH3CH2) ke[CH3CH2)[H)

= O.

These three equations give

0' [CH, CH, I - /

('';;J + [ (,~)' + (~~)

rI

[CH, CH, i

which implies that

If ka is small in the sense that only the lowest order need be retained,

The rate of production of ethene is therefore

The rate of production of ethene is equal to the rate of consumption of ethane (the intermediates all have low concentrations), so

d[

CH CH 3 3) dt

= -k[CH 3 CH 3,)

Different orders may arise if the reaction is sensitized so that ka is increased.

P23.13

-

(M )N

= -MI-p

-

[eqn 23. 8a with (M )N

= (n)M ).

The probability Pn that a polymer consists of n monomers is equal to the probability that it has n - 1 reacted end groups and one unreacted end group. The former probability is pll- I; the latter I - p.


473

Therefore, the total probability of finding an n-mer is

P"

= p"-I (1

We see that (n 2 )

- p).

= (i~~2

_

and that (M2 )N - (M)N2

= M2

( 1+ P I ) = (l-p)

2 -

( I-p)

2

pM2 (l-p)

2'

pi /2M

Hence, 8M

= -- . I-p

The time dependence is obtained from p

=

kt[Alo

I I

and - -

I-p

+ kt[Alo

= 1 + kt[Alo

pl / 2

I-p

P23.15

(kt[Alo(\

+ kt[Alo)) 1/ 2

= I M{kt[Alo(l + kt[Alo)) 1/ 2 I.

In termination by disproportionation, the radicals do not combine. The average number of monomers in a polymer molecule equals the number in the radical, the kinetic chain length, v. (n)

P23.17

[23.8bl.

= pl / 2(l + kt[Alo) =

Hence - and 8M

[23 .7l

= v = Ik[.M][Il - I/21 [23. 14l·

(a) A + P -+ P + P autocatalytic step, v = k[A][Pl. Let [Al = [Alo - x and [Pl = [Plo

+ x.

We substitute these definitions into the rate expression, simplify, and integrate.

d~~l

v= -

= k[A][Pl

d([Alo - x) dt = k([Alo - x)([Plo

-

dx ([Alo - x)([Plo

-::-:--=---~:-::---

[Alo

I

+ [Plo

(

+ x)

+ x).

= k dt.

I [Alo - x

+

I ) dx [Plo + x

= k dt.

474


t

[Alo

+ [Plo

1

Jo

(_1_ [Alo -

[Alo

+ [Plo

1

{In (

x

+

1 ) dx [Plo +

x

+ In ([Plo

[Alo ) [Alo - x

=k

t dt.

Jo

+x)} -_ . kt

[Plo

X) }= k([Alo + [Plo)t.

[Al o ) ( [Plo + In { ( [Plo [Alo _ x

In { In {

(~~i~) ([Alo + ~~~o _ [Pl) } = k([Alo + [Plo)t

G)

CAlo

+ ~~~o _

_ _ _[_P_l_ _ [Alo + [Plo - [Pl [Pl

0+ beat)[Pl = [Plo

+

I)

e

I

= at

where

a

= k([Alo + [Plo)

and

[Plo [Alo

b=- .

= beal .

= ([Alo + [Plo)beal -

[Pl _ (b [Plo -

[Pl) }

(1

+

beal [Pl. [Alo) beal [Plo

= [Plo (1 + ~) beal = [Plo(b + l)eal . b

al

+ beal

(b) See Figure 23.3(a). Autocatalytic process 12 10

b=O.1

8 [P] / [P]o

b =0.2 6

4 b= 1.0 2 0

0

2

4

6 at

8

10

Figure 23.3a

The growth to [Pl reaches a maximum at very long times. As t -+ 00, the exponential term in the denominator of [Pl 1[Plo = (b + 1) (e al 1( I + be al )) becomes so large that the denominator becomes beal. Thus, ([Pl / [Plo)max = (b+ 1)(eal I be al ) = (b + 1) l b where b = [Plo / [Alo and this maximum occurs as t -+ 00. The autocatalytic curve [Pl / [Plo = (b + I )(e al 10 + beal)) has a shape that is very similar to that of the first-order process ([Pl / [Alo) = 1 - e- kl . However, [Pl max = [Alo at t -+ 00 for the first-order


475

0.8 0.6 [PJ/[AJa

4

3

2

5

kt

Figure 23.3b

process whereas [Pl max = (1 + 1/ b) [Plo for the autocatalytic mechanism. In a series of experiments at fixed [Alo and assorted [Plo, only the autocatalytic mechanism will show variation in [Pl max . Another difference is that the autocatalytic curve is initially concave up, which gives an overall sigmoidal curve, whereas the first-order curve is concave down. See Fig. 23 .3(b). (c) Let [Pl vmax be the concentration of P at which the reaction rate is a maximum and let tmax be the corresponding time. v

+ x)

= k[A][Pl = k([Alo -

x)([Plo

= k{[Alo[Plo + ([Alo

- [Plo)x - x 2 ).

dv

- = k([Alo dt

[Plo - 2x).

The reaction rate is a maximum when dv / dt

x

=

[Pl V

max

- [Pl

0

=

[Alo - [Plo 2

or

= O. This occurs when [Pl vmax [Plo

Substitution into the final equation of part (a) gives [Pl vmax [Plo

= b+1 =

(b

+ I)

2b

Solving for t max ,

atmax

(max

= In(b- I) =

= I-~ In(b) I.

-In(b) .

ealmax 1 + beGlmax

b+l 2b

476 (d)


d[P] dt

= k[Af[p].

= Ao -x,

[A]

[P]

= Po +x, -d[P] = -dx = k(Ao dt dt

t

-:-_---;;dx;--_ _ Jo (Ao - X)2(PO + X)

2

X) (PO

+ X)

= kt

Solve the integral by partial fractions

__------::--__ = (Ao - X)2(PO

a (Ao - X)2

+ x)

a(Po

a

+ _f3_ + _y_ Ao - x Po + x

+ x) + f3(Ao

I

Poa + A oPof3 + A6Y = I + (Ao - PO)f3 - 2AOY = 0 -f3 + y = 0

- x) (PO

(Ao - X)2(PO

+ x) + y(Ao + x)

X)2

.

This set of simultaneous equations solves to

a

I a=---, Ao +PO

f3=y=--. Ao +Po

Therefore,

kt=

(Ao~pJ fox

= (Ao ~ = (AO

pJ {

(Ao

~x)

-

+

(AO

~ po) [(Ao(A: - x»)

Therefore with y

Ao(Ao

(AO~Po) (Ao~X + Po~x) Jdx (~J + ~ [In (AoA~ J+ In ( + ~ pJ In G~~p~ ~;~)] .

[(Aol-xr

x

=-

Ao

+ Po )kt =

and p

(Ao

Po

=- , Ao

(-y-) + (_1_) In ( + y) . I - Y

I - P

p p(l - y)

The maximum rate occurs at dvp

dt = 0,

Vp

= k[A]2[p]

and hence at the solution of 2k (d[A]) [A][P] dt

k[A]([A] -

+ k[Af d[P] = O. dt

+ k[A]2vp = 0 2[p])vp = o.

- 2k[A][P]vp

Po)

[as VA

= -vp] .

Pop: X) ] }


That is, the rate is a maximum when [A]

Ao -x = 2Po + 2x,

or

= 2[P], which occurs at

x = !(Ao - 2Po);

!(l- 2p).

Y=

Substituting this condition into the integrated rate law gives

Ao(Ao + Po)ktmax = or (Ao 1

(e)

(_1_) (~(l - 2p) + In~) +p 2p I

2

+ Po)2ktmax = ! - p -In 2p I·

d[P] dt

= k[A][P]2.

-dx = k(Ao -x)(Po +x) 2 [x = P - Po). dt

kt =

foX

(Ao _

x~Po + x)2·

Integrate by partial fractions (as in part (d))

kt

= (Ao

~ Po) fox {(Po ~ J + (Ao ~ Po) [Po ~ x + Ao ~ x J} dx 2

= (Ao ~ Po) = (Ao

{

(~o - Po ~

J+

~ Po) [In ( Pop: X) + In (AoA~

(Ao

~ Po) [(Po(p: +X)) + (Ao ~Po) In (~~A:~~n]·

x Po Therefore, with y = and p = - , [A]o Ao

Ao(Ao

+ Po)kt = (

p(p

y

+ y)

)

+

(_1_) + p(lp +1

p

In (

y ) . y)

The rate is maximum when dvp dt

= 2k[A][P] (d[P]) + k (d[A]) [pf dt

= 2k[A][P]vp That is, at[A]

dt

k[pfvp

= k[P](2[A] -

[P])vp

= O.

= ![P].

On substitution of this condition into the integrated rate law, we find

Ao(Ao

or (Ao

+ Po)ktmax = (

2-p ) 2p(1 +p)

2 -p

2

2p

P

+ Po)2ktmax = - - + In -

.

+ ( -1- ) 1 +p

2 in-

p

J] }

477

478


A ~ 2R

P23.19

I.

A+ R

~

R +B

R +R

~

R2

k2.

k3.

d[A] dt = I-I -

k2[A][R]I,

The latter implies that [R]

d[A]

-

dt

d[B]

-

dt

=

I ) 1/2

= ( k3

' and so

( 1 ) 1/ 2 -I - k2 [A] ,

k3

= k2[A][R] = k2

(/)1 /2[A]. k3

~~2

Therefore, only the combination

may be determined if the reaction attains a steady state.

k3 COMMENT. If the reaction can be monitored at short enough times so that termination is negligible compared

to initiation, then [RJ ~ 21t and ~ ~ k21t [AJ. SO monitoring B sheds light on just k2·

_d[=--C_r(_C_O.;..:)s:..:.]

P23.21

dt

=1_

Hence, [Cr(CO)s]

=

d[Cr(CO)sM] --d-t--'---

k2[Cr(CO)s][CO] - k3[Cr(CO)s][M]

+ k4[Cr(CO)sM] = 0 [steady state].

1 + k4[Cr(CO)sM] k2[CO] + k3[M] .

= k3[Cr(CO)s][M] -

k4[Cr(CO)sM] .

Substituting for [Cr(COh] from above, d[Cr(CO)sM] dt . Iff

=

=

k3 / [M] - k2 k4[Cr(CO h M [CO] k2[CO] + k3[M]

= -f[Cr(CO)

k2 k4[CO] k2[CO] + k3[M]

and we have taken k3/[M]

«k2~[Cr(CO)sM][CO].

Therefore,

I I k3[M] -= -+-:--::'-'::-:::-'::-:c f k4 k2k4[CO]

and a graph of I If against [M] should be a straight line.


(a) The mechanism considered is ka

k,

k~

k~

E + S ;=:(ES) ;=: P + E.

M] s


We apply the steady-state approximation to [(ES)]. d[ES] dt

= ka[E][S]

_

= [E]o -

Substituting [E]

k~[(ES)] -

kb[(ES)]

+ k~[E][P] = o.

[(ES)] we obtain

ka([E]o - [(ES)])[S] - k~[(ES)] - kb[(ES))

+ ka[E]o[S] -

(-ka[S] - k~ - kb - k~[P])[(ES)]

[(ES)]

Then d[P] 'dt

ka[E]o[S] ka[S] + k~

=

=k

+ k~([E]o -

[(ES)])[P] = O.

k~[E]o[P]

= O.

+ k~[E]o[P] + kb + k~[P]

[(ES)] _ k' [P][E] b b

[EJo[S] + (kVka) [E]o[P] - k' [P] b KM + [S] + (k~/ka) [P] b

=k

x ([E] _ [E]o[S] + (~/ka) [E]0[P]) o KM + [S] + (k~/ka) [P]

kb [[EJo[S]

+ (k~/ka) [E]o[P]] - k~[EJo[P]KM KM + [S] + (kVka) [P]

Substituting for KM in the numerator and rearranging d[P] dt

=

kb[E]o[S] + (k~k~/ka) [E]o[P] KM + [S] + (k~/ka) [P]

[v = d[P] ] . dt

(b) For large concentrations of substrate, such that [S]

»

KM and [S]

»

[P],

d[P]

-dt = kb[E]o which is the same as for the unmodified mechanism. For [S] d[P] _ k E {[S] - (k/kb)[P] } dt - b[ ]0 [S] + (k/k~)[P]

For [S] - +0, where kp =

d[P] _ _

dt

+ kb - - ,- . k~

kb

-k~k~[E]o[P] k~

+ kb + k~[P]

_

_

k'k'

k=~

ka

»

KM, but [S]

~

[P]

479

480


COMMENT. The negative sign in the expression for d[P]/dt for the case [S] ~ 0 is to be interpreted to

mean that the mechanism in this case is the reverse of the mechanism for the case [P] and S are interchanged.

->-

O. The roles of P

Question. Can you demonstrate the last statement in the comment above? P23.25

(a)---------------------------------[ATPJ/(I-Lmol dm - 3 ) v/(I-Lmol dm- 3 S- l) v/[ATP]/ S-l

0.60 0.81 1.35

0.80 1.4 0.97 1.30 1.21 0.929

2.0 l.47 0.735

3.0 1.69 0.563

[23.21]. Taking the inverse and mUltiplying by Vrnax v, we find that

Thus,

v

= Vrnax -

v . KM (Eadie-Hofstee plot) [S]o

or

v

Vrnax

v

[S]o

(b) The regression slope and intercept of the Eadie-Hofstee data plot of v against v/[S]o gives -KM and vrnax , respectively. Alternatively, the regression slope and intercept of an alternative form of the Eadie-Hofstee data plot of v/[S]o against v gives - 1/ KM , and vrnax/ KM, respectively. The slope and intercept of the latter plot can be used in the calculation of KM and Vrnax . (c) We draw up the following table, which includes data rows required for a Eadie-Hofstee plot (v against v/ [S]o). The linear regression fit is found for the plot as seen in Fig. 23.4.

Vrnax

= 12.30 I-Lmol dm - 3 s- l I

and

KM

= 11.10 I-Lmol dm- 3 1·

2 1.6 I

'7'"

e

1.2

y = -1.1015x + 2.303 1

"0

"0 0.8 e

R2 =0.998

::1.

;.

0.4 0 0

0.5

1.5

Figure 23.4


P23.27

481

When using reaction rates v, the Lineweaver-Burk plot without inhibition [23 .22] has the fom :

where the intercept and slope are simple functions of Vmax and KM. When using reaction rates relative to a specific, non-inhibited rate (Vrel = V/ Vreference), the Lineweaver-Burk plot without inhibition has the same basic form:

I

I

Vrel

= vmax,rel +

(KM)

1

vmax,rel

[S]o

The linear regression fit of the non-inhibited Lineweaver-Burk data plot is

-

I

vrel

= 0.797 + (2.17)

1

[CBGP]o/l0-2 mol dm-

3'

R2

= 0.980.

= l / intercept = 1/ 0.797 = l.25 and = slope x vmax.rel = (2.17 X 10- 2 mol dm- 3 ) x (l.25) = 2.71

Consequently, Vmax.rel KM

x 10- 2 mol dm- 3 .

The Lineweaver-Burk plot with inhibition has the basic form

I

vrel

=

a' vmax,rel

+

(aKM)

I

vmax,rel

[S]o '

The linear regression fit of the Lineweaver-Burk data plot for phenylbutyrate ion inhibition is

-

I

vrel

=

1.02 + (6.01)

I

[CBGP]o/l0-2 mol dm-

3'

R2

= 0.972.

Therefore, a' = intercept x vmax,rel = 1.02 x 1.25 = 1.28 and a = slope x vmax ,reI/KM = (6.01 X 10- 2 mol dm- 3 ) x (l.25) / (2.71 x 10- 2 mol dm- 3 ) = 2.77 . Since both a > I and a' ~ 1 (see Section 23.6(c», we conclude that phenylbutyrate ion is a competitive inhibitor of carboxypeptidase. The linear regression fit of the Lineweaver-Burk data plot for benzoate ion inhibition is

-

1

vrel

= 3.75 + (3.01)

I [CBGP]o / 1O- 2 mol dm- 3

,

R2

= 0.999.

Therefore, a' = intercept x vmax,rel = 3.75 x l.25 = 4.69 and a = slope x vmax ,reI/KM = (3.01 x 10- 2 mol dm- 3 ) x (l.25)/(2.71 x 10- 2 mol dm- 3 ) = 1.39. Since both a ~ 1 and a' > 1, we conclude that benzoate ion is an I un competitive Iinhibitor of carboxypeptidase.

482


4>r = 4>r,o

P23.29

T

ET = I - -

R6 ET = __0_ R8 +R6

I- -

[23.37].

TO

[23.38].

Equating these two expressions for ET and solving for R gives

R6

T

+ R6

TO

_ _0 _ = 1 _ _

R8

Rg +R6 ~

I-(T / To)

-,---:--_-,- _ I = _--,T/,--T...::.O_ l-(T/To) I-(T / TO) T/ TO

P23.31

= 10 ps/ lO3 ps = O.OlO

and R

or

R

= Ro

T/ TO ( 1 - (T/TO)

)1/6

0.010 )1 / 6 = I2.6 nm.I = 5.6 nm ( 1-0.010

Hypothesis: The 1270 nm emission band is the emission of the first excited state of 02(a l ~

E(a) < E(b and c)

(vi) E (Cr, C~+ ) = (-0.74 - 0.12) V = -0.86 V < E(a, b, and c). (vii) E(Co, Co2+) = (-0.28 - 0.15) V = -0.43 V < E(a, b, and c). Therefore, the metals with a thennodynamic tendency to corrode in moist conditions at pH = 7 are

IFe, AI, Co, Cr Iif oxygen is absent, but, if oxygen is present, all seven elements have a tendency to corrode. (b) A metal has a thermodynamic tendency to corrosion in moist air if the zero-current potential for

the reduction of the metal ion is more negative than the reduction potential of the half-reaction 4H+ + 0 2 + 4e- --+ 2H20, E" = 1.23 V. The zero-current cell potential is given by the Nemst equation

We are asked if a tendency to corrode exists at pH 7 ([H+] = 10- 7 ) in moist air (P(02 ) ~ 0.2 bar) , and are to answer yes if E :::: 0 for a metal ion concentration of 10- 6 , so for v = 4 and 2+ cations E=1.23V-~-

0.02569 V (10- 6 )2 In 74 =0.983V-~. v (1 x 10-) x (0.2)

In the following, z = 2. For Ni: E" = 0.983 V - (-0.23 V) > 0 1corrodes I. For Cd: E" = 0.983 V - (-0.40 V) > 0 1corrodes I· For Mg: E" = 0.983 V - (-2.36 V) > 0 1corrodes I. For Ti: E" = 0.983 V - (-1.63 V) > 0 1corrodes I. For Mn: E" = 0.983 V - (-1.18 V) > 0 1corrodes I. P25.39

Corrosion occurs by way of the reaction

The half-reactions at the anode and cathode are: Anode: Fe

--+

Fe 2+ + 2e - ,

Cathode: 2H+ + 2e-

--+

H2 .

t:..¢corr = (-0.720 V) + (0.2802 V) = -0.440 V, t:..¢eorr = 71(H) + t:..¢e(H) [Justification 25.1] , t:..¢e(H) = (-0.0592 V) x pH = (-0.0592V) x 3 = -0.1776V, jeorr I 71(H) = --In - - . af jo(H)

PROCESSES AT SOLID SURFACES

Then I'leorr jeorr

and In -. -

jo(H)

=

1

jeorr

-

-0.440 V = - - I n - - - 0.1776 V ex! jo(H)

= (0.262 V)

x ex!

= (0.262 V)

x (18 V

_I

)

= 4.716.

Faraday's laws give the amount of iron corroded leorrt

n

= -- =

m

=n x

ZF

=

(1.12 x 10- 5 Acm- 2 ) x (8.64 x 104 sd- l ) (2) x (9.65 x

(55.85 g mol- I)

I0.28 mg cm- 2 d- I I·

= (5.0 x

104

C mol- I)

5.0 x 10

-6

mol cm

-2 _I

d

10- 6 mol cm -2d- l ) x (55 .85 x 103 mg mol-I)

.

539

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