544 Pages • 160,782 Words • PDF • 120.4 MB
Uploaded at 20210924 16:04
This document was submitted by our user and they confirm that they have the consent to share it. Assuming that you are writer or own the copyright of this document, report to us by using this DMCA report button.
TH EDITION
Student Solutions Manual to Accompany
PHYSICAL CHEMISTRY PETER ATKINS • CHARLES TRAPP CARMEN GIUNTA • MARSHALL CADY www.elsolucionario.net
STUDENT'S SOLUTIONS MANUAL TO ACCOMPANY
PHYSICAL CHEMISTRY EIGHTH EDITION
www.elsolucionario.net
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
www.elsolucionario.net
Student's solutions manual to accompany Physical Chemistry, Eighth Edition © Oxford University Press, 2006 All rights reserved ISBN13: 9780716762065 ISBNlO: 0716762064 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
www.elsolucionario.net
Preface This manual provides detailed solutions to all the endofchapter (b) Exercises, and to the oddnumbered 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 handheld 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 crosschecked 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.
www.elsolucionario.net
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
Solutions to numerical problems Solutions to theoretical problems Solutions to applications
6
Phase diagrams
Answers to discussion questions
2
22
The First Law
Solutions to exercises Solutions to problems
Answers to discussion questions Solutions to exercises Solutions to problems
Solutions to numerical problems Solutions to theoretical problems Solutions to applications
22 23 33 33 41 47
Solutions to numerical problems Solutions to theoretical problems Solutions to applications
7
Chemical equilibrium
Answers to discussion questions
3
50
The Second Law
Solutions to exercises Solutions to problems
Answers to discussion questions Solutions to exercises Solutions to problems
Solutions to numerical problems Solutions to theoretical problems Solutions to applications
4
50 51 58 58 68 74
78
Answers to discussion questions Solutions to exercises Solutions to problems
Solutions to numerical problems Solutions to theoretical problems Solutions to applications
PART 2 Structure 8
Physical transformations of pure substances
Solutions to numerical problems Solutions to theoretical problems Solutions to applications
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
Answers to discussion questions Solutions to exercises Solutions to problems
Solutions to numerical problems Solutions to theoretical problems Solutions to applications
www.elsolucionario.net
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
Solutions to numerical problems Solutions to theoretical problems Solutions to applications
280 28 1 284 284 289 292
15 Molecular spectroscopy 3: magnetic resonance
297
Answers to discussion questions Solutions to exercises Solutions to problems
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
280
Solutions to problems Solutions to numerical problems Solutions to theoretical problems Solutions to applications
Answers to discussion questions Solutions to exercises 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
12 Molecular symmetry Answers to discussion questions Solutions to exercises Solutions to problems Solutions to applications
244 245 249 255
13 Molecular spectroscopy 1 : rotational 259 and vibrational spectra Answers to discussion questions Solutions to exercises Solutions to problems Solutions to numerical problems Solutions to theoretical problems Solutions to applications
259 260 269 269 275 276
17 Statistical thermodynamics 2: applications
331
Solutions to applications
331 332 338 338 345 353
18 Molecular interactions
357
Answers to discussion questions
357 358 361
Answers to discussion questions Solutions to exercises Solutions to problems Solutions to numerical problems Solutions to theoretical problems
Solutions to exercises Solutions to problems
www.elsolucionario.net
Contents Solutions to numerical problems Solutions to theoretical problems Solutions to applications
361 366 368
22 The rates of chemical reactions
440
Answers to discussion questions
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
Answers to discussion questions Solutions to exercises Solutions to problems Solutions to numerical problems Solutions to theoretical problems Solutions to applications
370 372 375 375 379 383
Solutions to applications
23 The kinetics of complex reactions
464
Answers to discussion questions
464 465 468 468 471 478
Solutions to exercises Solutions to problems
20 Materials 2: the solid state
389
Answers to discussion questions
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
Solutions to applications
24 Molecular reaction dynamics
489
Answers to discussion questions
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
Answers to discussion questions
413 414 424 424 430 433
Answers to discussion questions
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 applications
Solutions to exercises Solutions to problems Solutions to numerical problems Solutions to theoretical problems Solutions to applications
www.elsolucionario.net
PART 1 Equilibrium
www.elsolucionario.net
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.
www.elsolucionario.net
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 1molI ) 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 1molI ) 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)
1 )
3
_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
www.elsolucionario.net
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 molI)
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 s2 = 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 molI
www.elsolucionario.net
I
6
STUDENT'S SOLUTIONS MANUAL
[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 molI K 1) x (273.15 K) x (1.42755 g dm 3 atm
l
)
p=o
= 131.9987 g molII 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
www.elsolucionario.net
THE PROPERTIES OF GASES
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 molI) x [(100 273) K] x (0.6388kgm 3 ) ~~
RTp
MP 
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 molI) x (23 + 273) K
or m = (l.45 x 102 mol) x (l8.0g molI) = 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 molI) 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 molI) x (300 K) 2
3.137dm 3
V
= 128.2 kPa 1
www.elsolucionario.net
8
E1.11(b)
STUDENT'S SOLUTIONS MANUAL
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 molI) 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 molI) x (273.15K) P= 22.414dm 3 = t 1.0 atm t (1.0mol) x (8.206 x 10 2 dm 3 atmK I molI) 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=VnbV2'
(i)
(1.0 mol) x (8.206 x 10 2 dm 3 atm K I molI) x (273. 15 K) P= 22.414 dm 3  (1.0 mol) x (4.34 x 102 dm 3 mol I) (4.484 dm 6 atm molI) x (1.0 mol)2 (22.414 dm 3 )2
=
t
0.99 atm
t
www.elsolucionario.net
THE PROPERTIES OF GASES
9
( 1.0 mol) x (8.206 x 10 2 dm 3 atm K I molI) 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 molI ) x (350K») 127 d 3 II 1 12atm =.' m mo .
(8.314JK I mol I) x (298.15 K)
~~(200 bar) x (l05Pabar l )
I
I
= 1.24 x 104 m3 mol I = 0.124 dm 3 mol 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+
www.elsolucionario.net
RT) P x + (a) p x  ab p=0 2
10
STUDENT'S SOLUTIONS MANUAL
The coefficients in the equation are evaluated as
b + RT P
=
(3.183 x 1O2 dm3 mol  I)
= (3.183
+
3 2 (8.206 x 1O dm mol I) x (298.15 K) (200 bar) x (1.013 atm barI)
10 2 + 0.1208) dm 3 mol I
X
1.360 dm 6 atm mol 2
a
:: = 6.71 (200 bar) x (1.013 atm barI)
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 molII
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
www.elsolucionario.net
THE PROPERTIES OF GASES
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
www.elsolucionario.net
» vanishes. According
12
STUDENT'S SOLUTIONS MANUAL
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 + Vrnlb + 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 molI) x (O.0434dm 3 molI)
= =
= 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 molI) 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
www.elsolucionario.net
THE PROPERTIES OF GASES
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 104 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)].
www.elsolucionario.net
14
STUDENT'S SOLUTIONS MANUAL
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 =1272.95°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 kPaKI) x (t.T) .
I
(a) t.p = (0.0245 kPaKI) 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 molI) x (350 K) 1125 d 3 II 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
www.elsolucionario.net
THE PROPERTIES OF GASES
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 molI . 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 molI) =0.911. (0.08206dm 3 atm KI molI) 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 atrnKI molI )
www.elsolucionario.net
I
10 dm mol  )
16
STUDENT'S SOLUTIONS MANUAL
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
= ,
and using TB
p
= 327.2K
(0.08206 dm 3 atm K I molI) x 327.2 K 10.Oatm
= 12.69dm3 molII 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 K1mol 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 atmKl mol x 327.2K
(2.685  0.012) dm 3 mol 1 = 12.67 dm 3 molII
Vrn
=
(
V;;'
=
2.67 dm 3 molI 2.69dm 3 molI
= 0.992 ~
I.
= 621 K [Table 2.9]. Vrn
0.08206dm 3 atm KImol 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 molI x 621 K
(
(5.096 + 0.012) dm 3 molI
5.11 dm 3 mol5.lOdm3
= 15.11 dm 3 molII
1
molI
=
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 molI 10 353 d 3 II 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 KI mol ) x (776.4 K)
www.elsolucionario.net
THE PROPERTIES OF GASES
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 KI molI) x (776.4 K)
J x 0.J353dm 3 mol 1 = 10.4084=0. 5916~0.59.
Z
= 1+ (RIT)
x (b  :T) x (P) + ...
I = I + ,;;3
(0.08206dm atmK1 molI) x (776.4K)
x
{
(0.0305 dm 3 molI) 
= 1 0.2842 ~
(
6
2
5.464 dm atm mol)} x 327.6 atm (0.08206dm 3 atm KI molI) 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 molI) = 159.4 cm 3 molI
a = 4bRTc = 2RTc Vc
= (2) x (8.206 x 10 2 dm 3 atmK 1 molI) x (289.75K) x (118.8 x 10 3 dm 3 molI) = 15 .649 dm 6 atm mol 2 1. Hence
(1.0 mol) x (8.206 x 10 2 dm 3 atm K I molI) x (298 K) (1.0dm 3 ) x exp

(1.0 mol) x (59.4 x 10 3 dm 3 molI)
) (l.Omol) x (5.649dm6 atmmol 2 ) ( (8.206 x 10 2 dm 3 atmK1 molI) x (298K) x (1.0dm 6 atmmol l )
= 26.0atm x e O.23T = 121 atm
I.
www.elsolucionario.net
I.
18
STUDENT'S SOLUTIONS MANUAL
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 molI) x (34.6 + 21.7) cm 3 molI
(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
www.elsolucionario.net
= III llJ·
THE PROPERTIES OF GASES
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 s2 _ 44 . (10.132  1.223) x IQ4Pa
~'...,...,..:,;:::
X
102 k  I 3 g m .
RT From Fig. 1.3, = 5.40 x 104 m 2 s2 ; hence M l 3 2 , _4.4xlO kg m 081 106p 1 B22 . x a, 5.40 x IQ4 m s
B' = (0.81 x 1O6 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 molI) 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 twophase 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 twophase situation, since there are some conditions under which a quadratic has two real positive roots, but not the process of liquefaction.
www.elsolucionario.net
20
STUDENT'S SOLUTIONS MANUAL
The equation of state of gas B is a firstdegree equation in Vrn and therefore can never model critical behavior, the process of liquefaction, or the existence of a twophase region. A cubic equation is the equation of lowest degree that can show a crossover 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
www.elsolucionario.net
THE PROPERTIES OF GASES
(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 IO4 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 firstorder 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 
mballoon)g
This is Archimedes ' principle.
www.elsolucionario.net
The First Law
Answers to discussion questions 02.1
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 nonadiabatic process and is defined as the difference between the adiabatic work and the nonadiabatic 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 nonzero 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 louieThomson 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 louIeThomson effect. See Section 2.12 for a more complete discussion.
www.elsolucionario.net
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.
Solutions to exercises E2.1(b)
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 crosssectional area times the linear displacement:
l:!.V
so w E2.3(b)
=
(50.0cm
= (121
2)
x (15cm) x
(~)3 = 7.5 x 1O4m3 , 100 cm
x 103 Pa) x (7 .5 x 10 4 m 3 )
= 191 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 molI) x (22 + 273) K x In 31.7 dm: = 11.62 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 www.elsolucionario.net
24
STUDENT'S SOLUTIONS MANUAL
(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 molI)
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 ) wpex~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 molI) x (64 + 273) K = [ 5.60 x 103 J [
and ~U = q + w = (70.6 + 5.60) kJ = / 65.0kJ /
www.elsolucionario.net
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=I188JI
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 molI) x (lOOK) = 1831 J I
kJl
~U=q+w=(14.90. 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
(37 .11  8.3145) J KImol I
Cp,m R
=  ' = ',=R
         :     ;  1 
R
8.3145J KImol
so the final temperature is Tr = (298.15 K) x
E2.10(b)
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
www.elsolucionario.net
kJ Iand
26
STUDENT'S SOLUTIONS MANUAL
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 = 1325 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 molI) x (277  250) K = 12.0 x 103 J molI 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(4xI515)dm 3 _1_ 3 1 I 3  . 3.5 x 10 J. (IOdm m )
!:!,.U = q + w = 13.5 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
=124KI. (5.0mol) x (37.11  8.3145)JKI mol  I !:!,.U
+ !:!"(PV) =
= 3.5 x 103 J E2.1S(b)
so
!:!,.U
+ nR!:!"T,
+ (5.0 mol)
x (8.3145JK I molI) x (24K) =14.5 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
www.elsolucionario.net
THE FIRST LAW
27
Find Vi from the perfect gas law: l I . _ nRTi _ (1.5mol)(8.3IJK mol )(315K) =00171 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 molI x 1.5 mol x (279  315) K = 16.7 x 102 J E2.16(b)
I
At constant pressure q = /'o,.H = n/'o,.vapW = (0.75 mol) x (32.0 kJ molI) = 124.0 kJ
I
and w = p/'o,.V ~ pVvapor = nRT = (0.75 mol) x (8 .3145JK I molI) x (260K) w = 1.6 x 103J = 11.6kJ /'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 = 13053.6 kJ molI
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 molI = 11152 kJ molI
www.elsolucionario.net
I
I
28
E2.19(b)
STUDENT'S SOLUTIONS MANUAL
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 molI) x (298K)
3050 kJ molI 135 x 10 ~) x (3050kJmOI I) 94.12gmol3
Iql
=(
"""T
=C =
Iql
4.375kJ 0.0683 kJ KI
= 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 molI = 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 molI = 1+ 1.90 kJ molI
www.elsolucionario.net
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 moll = 1589.56kJmOI 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 = 1582.13 kJ molII
(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 molI)
= 126.48 kJ molII
= ! (483.64kJmOI 1) = 1241.82kJmol 1 1
L'l.r~ = L'l.rlr + RT L'l.ng [2.21]
=
772.7kJmol 1 +(5) x (8 .3145 x 1O3kJKlmoll) x (298K)
= 1760.3 kJ molII 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 moll
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 www.elsolucionario.net
30
STUDENT'S SOLUTIONS MANUAL
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
= 1566.93kJmol 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 = 1175 kJ molII /).r if' = /).rH"  /).ngRT
[2.21];
/).n g = 1
= 175kJmol 1  (1) x (2.48kJmol 1) =1173kJmOI 1 1
www.elsolucionario.net
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 molI) x (50K)
= 1176kJmol 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
= 165.49kJmol 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) 
=
tofF'(CaBr2, s)
+ 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 molI
= 11587 kJ molII so tohyd W (Ca2+)
+ tosubF'(Ca)
= 11587kJmol 1 1 www.elsolucionario.net
31
32
E2.29(b)
STUDENT'S SOLUTIONS MANUAL
The JouleThomson coefficient Jt is the ratio of temperature change to pressure change under conditions of isenthalpic expansion. So Jt  (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 molI
a 1.00dm 3 molI
:;;+;;;
2l.la 3 = 0.95475adm mol 22.1 dm mol
;;3;1
From Table 1.6, a = 1.337 dm 6 atm molI 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 molI) x (298K) x In ( 2 = +7 .7465kJmol  ~ 1.00  3.20 x 10and w = q
+ 6.Um =
(7747 J molI)
+ (l29Jmol l )
I
= 176181 molI I = 17.62kJmol 1 1
www.elsolucionario.net
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 104 + 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 104 + 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 JouleThomson coefficient is ( 8H) 8p T
=
J1,Cp
=
(1.11 K atm I) x (37.11 J K I molI)
= 141.2J atm 1 mol  I 1
If this coefficient is constant in an isothermal JouleThomson experiment, then the heat which must be supplied to maintain constant temperature is LiH in the following relationship LiH / n = 41.2Jatm1 molI Lip LiH
=
so
LiH = (41.2Jatm 1 moll)nLip
(4l.2J atm I molI ) 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 Kl)
~
= ~ = T3
Similarly, T2 = 1546 K I.
www.elsolucionario.net
= PV , nR
[isotherm).
34
STUDENT'S SOLUTIONS MANUAL
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 molI) x (546  273) K = 12.27 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 molI) 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 molI) x (273K)
= 13.40 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 molI) x (273K) =15 .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 = 11.57
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
www.elsolucionario.net
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 molI) x (15 .0dm 3 ) x (41 K)
Vm b
7.46 dm molI
3
Therefore, t:.H P2.S
= 7.5 dm 3 molI,
=
(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.
www.elsolucionario.net
298 K)
5
'2R ,
x (8.3141 K I molI) x (298 K)
= 6.19 kJ 
=
= 6.19
x 103 1 = 1+6.19 kJ I.
36
STUDENT'S SOLUTIONS MANUAL
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)
= 16.19 kJ I, since I':1.T(b) =
Likewise I':1.H(b) = I':1.H(a) = 18.67 kJ
=
w (c) I':1.U
1':1. U
I':1.T (a).
I.
= 16.19 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 molI) x (298K)
PI
1.00atm
=  =
where
so V3
=
_ 3 ((2) x (298 K) (24.45 dm ) x 298 K
q
10 3 J
R
)5/2= 138.3_dm 3.
w = (1.00 mol) x (8.314J K I molI) x (298K) x In
= 4.29 x
CVm
= 1+4.29 kJ I·
= 14.29 kJ I.
www.elsolucionario.net
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 molI.
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 KImol  I, we obtain
r ~rCpdT =  (2.65 iT, T2
= Hence ~fW(350 K)
Cr(C6H6h(s) ~rW
+
=
=
2.65 x 103 J molI
~fW(298 K)  2.65 kJ molI
84.68 kJ molI  2.65 kJ molI
Cr(s) + 2C6H6(g) ,
= ~r~ + =
103 K) x (J KImol I)
2.65 kJ molI .
=P2.9
X
~ng
= 187.33 kJ molI I.
= +2 mol.
2RT, from [2 .21]
(8.0 kJ mol  I) + (2) x (8.314 J K I mol I) x (583 K)
www.elsolucionario.net
= 1+17.7 kJ mol I I.
38
STUDENT'S SOLUTIONS MANUAL
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 molI .
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 molI)
+ (30.8 kJ molI) + (583  353) K x (81.67 J K I molI)  (6) x (583  298) K x (8.53 J K I molI)  (3) x (583  298) K x (28.82 J K I molI)
= {(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 molI
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 (semiempirical, 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 molI 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
www.elsolucionario.net
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 molI )), one ought to obtain a straight line with slope nand yintercept In Ikl. Draw up the following table. Compound
M/(g molI)
l'lcH j kJ molI
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 molI) 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 molI + 8.3141 K I molI)
(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 gI x (±87.5K)
www.elsolucionario.net
= ±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
63
)
10~
= 1230 J I
(c)
230 J
53
= 1230 J I
( I.Smol) x (28.8 JK I mol I)
=
IS.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 constantvolume 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 molI) 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
www.elsolucionario.net
54
STUDENT'S SOLUTIONS MANUAL
/'
0, V
c> 00 ,
a
c>
0, and T
c> 00 .
The fact that lfT > 0 ( because a > 0) is consistent
with a representing attractive contributions, since it implies that
(au) av
.

> 0 and the Internal energy T
rises as the gas expands (so decreasing the average attractive interactions).
www.elsolucionario.net
THE SECOND LAW
P3.33
If S
73
= S(T,p)
then dS
=
( as ) dT+ ( as) dp , aT p ap T
=T (
Td S
as ) dT aT p
=_ T
dp . T
x C
p
[
(~~)
p
= T,
Problem 3.25] ,
( av ) [Maxwell relation]. aT P
= CpdT 
Hence, TdS
as ) ap
(~ ) ( aH)  ~ aH p aT p  T
Use ( as) _ aT p 
as ) ( ap
+T (
G~) p dp =
T
1
CpdT  aTVdp I·
For reversible, isothermal compression, TdS = dqrev and dT = 0; hence dqrev qrev
= aTVdp .
=
l
Pf
aTV dp
= laTV t:..p I[a and Vassumed constant].
Pi
For mercury qrev = (1. 82 P3.35
In!/>
=
10 4 K I) x (273 K) x (1.00
X
(Z;
foP
104 m 3) x (1.0 x 108 Pa)
= 10.50 k.J I·
I) dp [3 .60],
8
C
Vm
Vm
Z=I++2 8 with 8 ' =  , RT
X
C' =
,, 2
=1+8p+Cp
+ ...
C8 2 R2T 2 [Problem U8].
Z I , ,   =8 +Cp+ ···. p
{P
{P
1
8p
Therefore, In!/>
= 10 8' dp + 10 C'pdp + .. . = B'p + "2 C'p2 + . . . = lIT +
8p For argon, RT
=
(C  8 2)p2
2R2T2
+ .. .
(21.1 3 x 10 3 dm 3 mol  l ) x (l.00atm) _ 4 2 3 I I  9.43 x 10 , (8.206 x 10 dm atm K mol ) x (273 K)
(C  8 2)p2 _ {( 1.054 x 10 3 dm 6 mol  2)  (2 1.13 x 10 3 dm 3 moll)2) x (l.OOatm)2 2R2T2 (2) x {(8.206 x 10 2 dm 3 atm K I molI ) x (273 K) }2
=
6.05
X
10 7
www.elsolucionario.net
74
STUDENT'S SOLUTIONS MANUAL
Therefore, In¢ = (9.43 x 10 4 )
+ (6.05
x 10 7 ) = 9.42 x 104 ; ¢ = 0.9991.
Hence,f = (1.00atm) x (0.9991) = 10.9991 atm
I.
Solutions to applications P3.37
6. r G [3.38] .
wadd, max =
6. r G"(37°C) = r 6. rG"(Tc)
= (
+ (1 r)6. rfF(Tc)
310K ) x (6333kJmol l ) 298.15K
[problem 3.17, r =
+ (1
~J
310K ) x (5797kJmol l ) 298.l5K
= 6354kJmol l . The difference is 6. r G"'(37°C)  6. r G"'(Tc) = {6354  (6333») kJmol 1 = 121 kJ molI I. Therefore an
additional 21 kJ mol I of nonexpansion work may be done at the higher temperature. As shown by Problem 3.16, increasing the temperature does not necessarily increase the
COMMENT.
maximum nonexpansion work. The relative magnitude of t:. rGB and t:. rW is the determining factor.
P3.39
The relative increase in water vapor in the atmosphere at constant relative humidity is the same as the relative increase in the equilibrium vapor pressure of water. Examination of the molar Gibbs function will help us estimate this increase. At equilibrium, the vapor and liquid have the same molar Gibbs function. So at the current temperature Gm,liq(To) = Gm,vap(TO)
so
~, liq(To) = ~,vap(TO)
+ RTo In PO,
where the subscript 0 refers to the current equilibrium and p is the pressure divided by the standard pressure. The Gibbs function changes with temperature as follows
and similarly for the vapor. Thus, at the higher temperature
Solving both of these expressions for C:,liq (To)  ~,vap (To) and equating them leads to (6.T)(s;rq  S:p)
+ R(To + 6.T) Inp =
RTo Inpo.
Isolating p leads to Inp
(6.T)(S!p 

p=exp (
s;rq) +ToInpo 
R(To+6.T)
To+6.T'
s;rq»)
(6.T)(S!p R(To+I'!T)
(To /( To + t.T) )
Po
.
www.elsolucionario.net
THE SECOND LAW
7S
1 I _ (2.0K) x (188 .8369.91 ) Jmol K ) 0089 (290K / (290+2.0)K) SO p  exp I x ( . 1 ) , (8.3145Jmol KI) x (290+2.0)K
= 0.0214 which represents a 113 per cent 1increase.
p
P3.41
The change in the Helmholtz energy equals the maximum work associated with stretching the polymer. Then
d WmaJ<
= cIA =
Jdt.
For stretching at constant T
Assuming that (aU / al)T =
J=T
o(valid for rubbers)
(asat ) =T (a) at T
T
{3kBt2 + C} 2Na 2
=T{_3kBt} =_(3kBT)t. Na 2 Na 2
This tensile force has the Hooke's law formJ P3.43
= kHt with kH = 3kBT /Na 2.
The Otto cycle is represented in Fig. 3.5. Assume one mole of air.
P
C (2) B
D
(4)
v Figure 3.5 c = Iwlcycle [3 .8].
Iq21 Wcycle q2 C
=
=
=
WI + W3
~U2
=
=
~UI
+
~U3
[ql
= q3 = 0] = CV(TB
Cv(Te  TB)·
ITB  TA + To  Te l ITeTBI
=
1 _ (To  TA ).
Te TB
www.elsolucionario.net
 TA) + Cv(To  Tc) [2.27] .
76
STUDENT'S SOLUTIONS MANUAL
We know that
T
~ =
and
(~)I /C
.£ Vo
Te
[2.28a].
. TA To TATe SInce VB = Ve and VA = Vo ,  =  , orTo =   . TB Te TB
7 5 2 Given that Cp,m = "2R, we have CV ,m = "2R [2.26] and c = S. V For ~ VB
=
10,
£
=
I 
( I
10
)2/5=10.471.
~Sl = ~S3 = ~Ssur. 1 = ~Ssur,3 =
@]
[adiabatic reversible steps].
~S2 = CV ,m In (;:). At constant volume (;:) =
~S2 =
G)
~Ssur,2
=
X
e:)
= 5.0.
(8.3 14JK 1 molI) x (In 5.0) =1+33JK I I·
~S2
= 133 J K I I.
llS4=llS2 [Te = TB]=1_33JK I. ' To TA llSsur,4
P3.45
= llS4 = 1+33 J K I I·
1.00
11.00 1
In case (a), the electric heater converts kJ of electrical energy into heat, providing kJ of energy as heat to the room. (The Second Law places no restriction on the complete conversion of work to heatonly on the reverse process.) In case (b), we want to find the heat deposited in the room Iqhl :
Iqhl = Iqcl + Iwl
so
where
Iqcl = Iwl

Tc Th  Tc
c = [ImpactI3.I]
_ ~ _ 1.00 kJ x 260 K = 4 kJ (295 _ 260)K 7. .
Iqcl  Th _ Tc 
www.elsolucionario.net
THE SECOND LAW
77
The heat transferred to the room is Iqlrl = Iqcl + Iwl = 7.4 kJ + 1.00 kJ = 18.4 kJ I. Most of the thermal energy the heat pump deposits into the room comes from outdoors. Difficult as it is to believe on a cold winter day, the intensity of thermal energy (that is, the absolute temperature) outdoors is a substantial fraction ofthat indoors. The work put into the heat pump is not simply converted to heat, but is ' leveraged' to transfer additional heat from outdoors.
www.elsolucionario.net
Physical transformations of pure substances
Answers to discussion questions 04.1
Consider two phases of a system, labeled ct and 13. The phase with the lower chemical potential under the given set of conditions is the more stable phase. First consider the variation of J.t with temperature at a fixed pressure. We have
= ( aJ.ta) aT p
S
a
and
= Sf3. ( aJ.tfJ) aT p
Therefore, if Sf3 is larger in magnitude than Sa , the 13 phase will be favored over the ct phase as temperature increases because its chemical potential decreases more rapidly with temperature than for the ct phase. We also have
aJ.ta) m,a ( ap T v
and
(
aJ.tf3) ap T = Vm .f3.
Therefore, if Vm ,a is larger than V rn ,f3 the 13 phase will be favored over the ct phase as pressure increases because the chemical potential of the 13 phase will not increase as rapidly with pressure as that of the ct phase. See Example 4.1. 04.3
Refer to Fig. 4.1 below and Fig. 4.5 in the text. Starting at point A and continuing clockwise on path peT) toward point B, we see a gaseous phase only within the container with water at pressures and temperatures peT). Upon reaching point B on the vapor pressure curve, liquid appears on the bottom of
the container and a phase boundary or meniscus is evident between the liquid and less dense gas above it. The liquid and gaseous phases are at equilibrium at this point. Proceeding clockwise away from the vapor pressure curve the meniscus disappears and the system becomes wholly liquid. Continuing along peT) to point C at the critical temperature no abrupt changes are observed in the isotropic fluid. Before point C is reached, it is possible to return to the vapor pressure curve and a liquidgas equilibrium by reducing the pressure isothermally. Continuing clockwise from point C along path peT) back to point A, no phase boundary is observed even though we now consider the water to have returned to the gaseous state. Additionally, if the pressure is isothermally reduced at any point after point C, it is impossible to return to a Liquidgas equilibrium. When the path peT) is chosen to be very close to the critical point, the water appears opaque. At near critical conditions, densities and refractive indices of both the liquid and gas phases are nearly identical. Furthermore, molecular fluctuations cause spatial variations of densities and refractive indices on a scale large enough to strongly scatter visible light. This is called critical opalescence.
www.elsolucionario.net
PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
79
Vapor pressure curve of water Explorati on path peT)
22 1.2 bar p
647.4 K
T
04.5
Figure 4.1
Vapor pressure curve of water
The supercritical fluid extractor consists of a pump to pressurize the solvent (e.g. C02), an oven with extraction vessel, and a trapping vessel. Extractions are performed dynamically or statically. Supercritical fluid flows continuously through the sample within the extraction vessel when operating in dynamic mode. Analytes extracted into the fluid are released through a pressuremaintaining restrictor into a trapping vessel. In static mode the supercritical fluid circulates repetitively through the extraction vessel until being released into the trapping vessel after a period of time. Supercritical carbon dioxide volatilizes when decompression occurs upon release into the trapping vessel.
Advantages
Disadvantages
Current uses
Dissolving power of SCF can be adjusted with selection of T and p
Elevated pressures are required and the necessary apparatus expensive
Extraction of caffeine, fatty acids, spices, aromas, flavors, and biological materials from natural sources
Select SCFs are inexpensive and nontoxic. They reduce pollution
Cost may prohibit large scale applications
Extraction of toxic salts (with a suitable chelation agent) and organics from contaminated water
Thermally unstable analytes may be extracted at low temperature
Modifiers like methanol (110%) may be required to increase solvent polarity
Extraction of herbicides from soil
The volatility of SCC02 makes it easy to isolate analyte
SCC02 is toxic to whole cells in biological applications (C02 is not toxic to the environment)
SCH20 oxidation of toxic, intractable organic waste during water treatment
SCFs have high diffusion rates, low viscosity, and low surface tension
Synthetic chemistry, polymer synthesis and crystallization, textile processing
02 and H2 are completely miscible with SCC02. This reduces mUltiphase reaction problems
Heterogeneous catalysis for green chemistry processes
www.elsolucionario.net
80
04.7
STUDENT'S SOLUTIONS MANUAL
Firstorder phase transitions show discontinuities in the first derivative ofthe Gibbs energy with respect to temperature. They are recognized by finite discontinuities in plots of H, V, S, and V against temperature and by an infinite discontinuity in Cpo Secondorder phase transitions show discontinuities in the second derivatives of the Gibbs energy with respect to temperature, but the first derivatives are continuous. The secondorder transitions are recognized by kinks in plots of H, V, S, and V against temperature, but most easily by a finite discontinuity in a plot of Cp against temperature. A Atransition shows characteristics of both first and secondorder transitions and, hence, is difficult to classify by the Ehrenfest scheme. It resembles a firstorder transition in a plot of Cp against T , but appears to be a higherorder transition with respect to other properties. At the molecular level firstorder transitions are associated with discontinuous changes in the interaction energies between the atoms or molecules constituting the system and in the volume they occupy. One kind of secondorder transition may involve only a continuous change in the arrangement of the atoms from one crystal structure (symmetry) to another while preserving their orderly arrangement. In one kind of 8transition, called an orderdisorder transition, randomness is introduced into the atomic arrangement. See Figs. 4.18 and 4.19 of the text.
Solutions to exercises E4.1(b)
Assume vapor is a perfect gas and p* /:;.vapH In=+p R
/:;. vapH
is independent of temperature
( ~ ~) T
T*
1 1 R p* =+InT
T*
p
/:;. vapH
1 8.314J K I mol I (S80) =+ xln  ' 293.2K 32.7 x 10 3 Jmol 1 66.0 = 3.378 x 10 3 K I T =
= 296 K = 123 °C 1
3.378 x 10 3 KI
E4.2(b)
assuming
/:;.fusS
_
and
/:;.fus V
independent of temperature.
3
/:;.fus S (lS2.6cm mol
= (l0.6crn 3 rnol 
I _
5 II x (1.2 x 106 Pa)  (1.01 x 10 Pa) 142.0crn rno) 429.26K427. ISK
l
(~)
)
x
3
x (S.21 x 105PaKI)
= S.S2Parn 3 K I molI =IS.SJKIrnol II
www.elsolucionario.net
PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
I1rusH
=
Trl1S
=
(427.15 K) x (5 .521 K I molI)
= 12.4 kJ molII E4.3(b)
Use
f
d Inp
=
f 11~;2H
= constant 
In p
dT
6. vapH RT
Terms with I / T dependence must be equal, so _ I1 vapH
3036.8 K T/K
I1vapH
RT
= (3036.8 K)R = (8 .3141 K I molI)
x (3036.8 K)
= 125.25 kJ molII E4.4(b)
(a)
logp
= constant 
I1vapH /(RT(2.303))
Thus
I1 vapH
x (8.3141 K I molI) x (2.303)
= (1625 K)
= 131.11 kJ mol  I I
(b) Normal boiling point corresponds to p log(760) 1625 T/K
T/ K Tb
E4.5(b)
6. T
= 8.750 
= 8.750 =
1625 T/ K
log(760)
1625 8.750  log(760)
11 rusS
x I1p
=
Tr 6. f.US V x I1p 6.rusH
[Tr
= 3.65 + 273 .15 = 269.50 K]
6.T
=
Tr I1pM =I1rusH
x 11 ( I ) P
(269.50 K) x (99.9 MPa)M x ( I _ 1 ) 8.68kJmol  1 0.789gcm 3 0.801 gcm 3
= (3 . 1017 x
I1T

= 276.87
= 1276.9 K 1
I1rus =V
=
= 1.000 atm = 760 Torr
106 K Pari mol ) x (M) x (+0.01899cm 3 / g) x
(+ 5.889 x 10 2 K Pa m 3 rig I mol)M
= (46.07 g mol  I)
=
= 269.50K +
x (+5 .889 x 10 2 KgI mol)
2.71K
10 cm
(+ 5.889 x 10 2 KgI mol)M
= +2.71K Tr
(~) 6 3
= 1272 K 1
www.elsolucionario.net
81
82 E4.6(b)
STUDENT'S SOLUTI ONS MANUAL
dm dn  = dl dt
X
MH, O
dn
dq / dt
dt
!:!"vapH

q where n =  !:!" vapH
(0.87 x 103 Wm 2) x ( 104 m2) 44.0 x 103 J molI = 197.7 J s I rl mol = 200 mol sI
dm I I  = (197.7mol s ) x (l8.02gmol  ) dt
=13 .6kgs  1 1 E4.7(b)
The vapor pressure of ice at 5 °C is 0.40 kPa. Therefore, the frost will sublime. A partial pressure of 0.40 kPa or more will ensure that the frost remains.
E4.8(b)
(3) According to Trouton 's rule (Section 3.3(b), eqn 3.16)
(b) Use the ClausiusClapeyron equation [Exercise 4.8(a)]
At T2 = 342.2 K, P2 = 1.000 atm; thus at 25 °C In I = P
29.T x 103JmOII) ( 8. 314J K I mol I
X
(I 298.2 K

I) = 1.509 342.2 K

PI = 10.22 atm 1= 168 Torr
At60 °C, In
PI
= 
29.T x 103JmOI  I) x (1 I) = 0.276   ( 8.3 14JKImol 1 333.2K 342.2K
PI = 10.76 atm 1= 576Torr E4.9(b)
!:!,.T=Tfus(lOMPa) Tfus(O. IMPa) = Tfu s!:!"pM !:!,. !:!,. fusH
(I) 
p
[See Exercise4.5 (b)]
!:!"fusH = 6.01kJmol  1
!:!,.T=
(273.15K) x (9.9 x 106 Pa) x (18 x 1O3 kgmOI  I) } { 6.01 x 103 J mol I x {9.98
X
1~2 kg m3 
9.15 x
1~2 kg m3 }
= 0.74K Tfus(lO MPa) = 273.15 K  0.74 K = 1272.41 K 1
www.elsolucionario.net
PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
E4.10(b)
fl vapH = fl vapV + fl vap(PV) fl vapH = 43.5 kJ mol  I flvap(P V) = pfl vap V = p(Vgas  Vliq) = pVgas = RT [per mole, perfect gas] fl vap(PV) = (8.3 141 K I molI) x (352 K) = 29271 mol  I . flvap(P V) 2.927kJmol 1 FractIOn = = I fl vap H 43.5 kJ mol= 16.73 x 10 2 1= 6.73 percent
Solutions to problems Solutions to numerical problems P4.1
At the triple point, T3, the vapor pressures of liquid and solid are equal; hence 1425.7 K =8.3186; T3 = 1196.oK I. T3 T3 1871.2 K log(P3/ Torr) = 196.0 K + 10.5916 = 1.0447; P3 = 111.1 Torr I·
10.5916
P4.3
(a)
(b)
1871.2 K
dp fl vapS fl vapH  =  = [4.6, Clapeyron equation] dT fl vapV hfl vapV I 3 14.4 x 10 1 molI II = = +556 kPa K(180 K) x (14.5 X 10 3  1.15 x 10 4 ) m3 mol  I .
$]
fl ~H ~ = 2 x p [ 4.II,withdlnp=dT RT P =
(14.4 x 103 1 mol  I) x (1.013 x 105 Pa) (8.3141 K I mol I) x (180 K) 2
The percentage error is 12.5 per cent P4.5
(a)
_(BILBp(S» ) ( BIL(1») Bp T
I = +5 .42 kPa K .
I.
= Vm (I)  Vm (s) [4.13] = Mfl = (18.02 g molI) x ( = 11.63 cm 3 mol  I
(b)
( BIL(g» ) Bp
_ T
(BIL(I») Bp
(~) P
T
= Vm (g) 
I _ _ _1_.".) 1.000 g cm 3 0.917 g cm  3
I.
Vm(l)
T
= (18.02 g molI) x ( = 1+30.1 dm 3 mol  I
I _ _ _ _1:_0) 0.598 g dm 3 0.958 x 103 g dm 3
I.
www.elsolucionario.net
83
84
STUDENT'S SOLUTIONS MANUAL
At 1.0 atm and 100o e, JL(I) (as in Problem 4.4)
= JL(g); therefore, at 1.2 atm and 100o e , JL(g) 
JL(l) "" b. Vvapb.p =
(30.1 x 10 3 m 3 molI) x (0.2) x (1.013 x 105 Pa) "" 1+ 0.6 kJ mol I I. Since JL (g) > JL (I), the gas tends to condense into a liquid. P4.7
The amount (moles) of water evaporated is ng = The heat leaving the water is q
AT __
u
_
2_
RT
= nb. vapH.
The temperature change of the water is b. T
Th eref ore,
PH oV
= ~, nCp,m
n
= amount of liquid water.
P H20 V b.vapH RTnCp,m  (3.17 kPa) x (50.0 dm 3 ) x (44.0 x 103 J mol I) (8.314kPadm 3 K I molI) x (298.15 K) x (75.5 J K  I molI) x
=
C8.0~5g0~OI
I)
2.7K.
The final temperature will be about 122°e I. P4.9
(a) Follow the procedure in Problem 4.8, but note that Tb
=
1
227.5°e
1
is obvious from the data.
(b) Draw up the following table.
Or e TI K 1000 KI T Inpl Torr
57.4
100.4
133.0
157.3
203 .5
227.5
330.6 3.02 0.00
373.6 2.68 2.30
406.2 2.46 3.69
430.5 2.32 4.61
476.7 2.10 5.99
500.7 2.00 6.63
The points are plotted in Fig. 4.2. The slope is 3
6.4 x 10 K, so implying that b.vapH P4.11
b.vapH R
= 6.4
3
x 10 K,
= 1+53 kJ mol 1 I.
(a) The phase diagram is shown in Fig. 4.3 . (b) The standard melting point is the temperature at which solid and liquid are in equilibrium at 1 bar. That temperature can be found by solving the equation of the solid liquid coexistence curve for the temperature 1 = P3/bar + 1000(5.60 + 11 .727x)x.
www.elsolucionario.net
PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
85
i:::::·;··:::::I::::::f::::::r:::::I::::::r:::::r::::r:::::I::::::f:::::j
6
i·· ·· · ·~·· · ··· ; · ·
..;. ...... ! .......; ...... ~ .......: ...... .;. ... .. .;...... .; ..... ~
! 41·····+·····1······j······: · ····[·····+···+····+·····1··· ·+···1
;'IFlll l iiF!l , o :......L. ... L .... i. ....L .....~ ...... L.. ....t. ..... L.....L .... : .... J 2.0
2.2
2.4
2.6
2.8
3.0
3
Figure 4.2
(10 /T)K
2
. . · ·:i· · · · ~· ···+·····
o ::::::!::::::! ::::::;::: ·r~~;~··r:··::r····;
j
i
2 4
:::~~:;J:::::I::: : f : : t:· ·:f···::F::t:::l::::r~: l: : : : ::::::C:::L:::::::: :r·.c::L::L::L::t:::::L:::L:::
6 ••
S
• Liquid...Vapor • SolidLiquid
iF.:•.i • •tJ.·• FIFEL ~
1
o
100
;
;
l
200
j.
~
300 1}K
.~
400
l
.;
l
500
600
Figure 4.3
So 11727x2 + 5600x + (4.362 x 10 7

I)
= o.
The quadratic formula yields 5600 ± 1(5600)2  4 (11727) x (I)} 1/ 2 X =
1 ±
2 (11 727)
/I + (4(l1727)/56002)}1 / 2 2 ((11727)/5600)
The square root is rewritten to make it clear that the square root is of the form {I a
«
I; thus the numerator is approximately I
+ (1 + ~a )
~a,
+ a} 1/2,
with
and the whole expression
reduces to x ;;:,; 1/5600
=
1.79 x 104 .
Thus, the melting point is T
=
(I +x) T3 = ( 1.000179) x (178.15 K)
= 1178.18 K I.
(c) The standard boiling point is the temperature at which the liquid and vapor are in equilibrium at
I bar. That temperature can be found by solving the equation of the liquidvapor coexistence curve
www.elsolucionario.net
86
STUDENT' S SOLUTIONS MANUAL
(d) The slope of the liquidvapor coexistence curve is given by !:;. vapHG T!:;. vapVG
dp dT so
!:;. vapW
=
dp dT
T !:;.vap ~ 
The slope can be obtained by differentiating the equation for the coexistence curve. dp dT
d In p dy
d In p
= P;rr = Pd.Y dT '
2dp = (10.413 dT y
x
15.996 + 2(14.015)y  3(5.0120)i

 (1.70) x (4.7224) x (1 y )070)
(:J.
At the boiling point, y = 0.6458, so dp = 2.851 x 10 2 bar K 1 = 2.851 kPa KdT
1
3
and
!:;. vapW
= (383.6 K) x
30.3  0.12) dm mOlI) ( _ I) 3 x 2.851 kPa K ( 1000 dm m 3
= 133.0 kJ mo]  I I.
Solutions to theoretical problems P4.13
a!:;'G) =(aGfJ ) ( ap T ap
T
_( aCet ) ap
=VfJ Vet .
T
Therefore, if VfJ = Vet , !:;.C is independent of pressure. In general, VfJ f= Vet , so that !:;'C is nonzero, though small, since VfJ  Vet is small. P4.1S
Amount of gas bubbled through liquid = pV
RT
(p
=initial pressure of gas and emerging gaseous mixture). .
Amount of vapor earned away
m = .
M
. f· . mj M M oIe f ractIOn 0 vapor III gaseous mixture =       (m j M)
.
Partial pressure of vapor = p =
mj M (m j M)
+ (PV j RT)
+ (PV j RT)
xp =
p (mRT / PVM)
mPA
= , (mRT j PVM) + 1 rnA + 1
RT A = . PVM
For geraniol, M = 154.2 g mol  I, T = 383 K, V = 5.00 dm 3 ,p = 1.00 atm, and m = 0.32 g, so A=
(8.206
X
10 2 dm 3 atmK
(1.00 atm) x (5 .00
dm3 )
1
mol I) x (383 K)
x (154.2 x
10 3
kg molI)
www.elsolucionario.net
I =40.76kg .
PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
87
Therefore p
P4.17
=
(0.32 x 10 3 kg) x (760 Torr) x (40.76 kgI) (0.32 x 103 kg) x (40.76kg')
+1
=
I98 Torr I .
.
In each phase the slopes are given by
= Sm [4. 1] . ( al1) aT p The curvatures of the graphs of 11. against T are given by 2
a 11) ( aT2
p
td
=  (aSm) aT = T p
x C
p,m [Problem 3.26].
Since Cp,m is necessarily positive, the curvatures in ail states of matter are necessarily negative. Cp,m is often largest for the liquid state, though not always, but it is the ratio Cp,m/T that determines the magnitude of the curvature, so no precise answer can be given for the state with the greatest curvature. It depends upon the substance. P4.19
Sm
= Sm(T,p).
dSm
= ( aSm ) aT
p
dT
+ (aSm) ap
dp. T
asm ) Cp ( aT p = T [Problem 3.26] , dqrev Cs
= T dSm = Cp dT 
= ( aq ) = Cp aT s
m) (aS ap
=  (aVm) T
aT
p
[Maxwell relation].
aVm) T(dp. aT p
p TVma ( a ) aT s
= Cp aVm x
trs
!:1 H [4.7]. !:1 trs V
Solutions to applications P4.21
(a) The Dieterici equation of state is purported to have good accuracy near the critical point. It does fail badly at high densities where V m begins to approach the value of the Dieterici coefficient b. We will use it to derive a practical equation for the computations.
Substitution of the Dieterici equation of state derivative (apr/aTr )vr reduced form of eqn 3.48 gives
www.elsolucionario.net
= (2 + Tr Vr)pr/T? Vr into the
88
STUDENT'S SOLUTIONS MANUAL
Integration along the isotherm T, from an infinite volume to V, yields the practical computational equation.
I'1.U,(T" V,)
=_
2 (T,
1
00
p,
Trconstant
~)
" , dV,. T, V,
The integration is performed with mathematical software. (b) See Fig. 4.4(a). (c) 8(T" V,) = JPcI'1.Ur/Vr where Pc = 72.9 atm. Carbon dioxide should have solvent properties similar to liquid carbon tetrachloride (8 when the reduced pressure is in the approximate range 10.85 to 0.90 when T,
(a)
4 l;U r
\ Tr =1
5
Tr= 1.2
T,= 1.5
6
7 1.5
0.5
(b)
2
2.5 p,
3
3.5
4
4.5
Solubility parameter of carbon dioxide
20
15
.L atm!!>
10
5 0.75
0.8
0.9
0.85
0.95
Pr
www.elsolucionario.net
8
~
9)
= 1 I. See Fig. 4.4(b).
2
3
~
Figure 4.4
PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
P4.23
C (graphite) ;= C (diamond), D.r~
89
= 2.8678 kJ molI at T . T = 25 °C.
We want the pressure at which D.rG = 0; above that pressure the reaction will be spontaneous. Equation 3.50 determines the rate of change of D.rG with p at constant T.
(1)
(aD.rG)
ap
T
= D.rV = (Vo  VG)M
where M is the molar mass of carbon; Vo and VG are the specific volumes of diamond and graphite, respectively. D.rG( T , p) may be expanded in a Taylor series around the pressure pdlnr +dl/> . Subtract din r from both sides, to obtain as din r
= (I/> 
I)d Inr+dl/>
(I/>  I)
= dr+dl/>. r
. . and notmg . that In (as) Then, by mtegratlOn r r=O
A(s)
PS.23
;='
r
r=O
= In (YS)r=O = In I = 0,
A(I) .
J1.~(s)
=
J1.~(I) + RTlnaA
and ~fus G
=
J1.~ (I)  J1.~ (s)
Hence, In aA
= In (YSXS) 
=
 RT In aA·
~fu s G
= ~.
din aA I d (~fus G) ~fusH  = T =  [GibbsHelmholtzeqn]. dT R dT RT2 For
~T
= T( 
T , d~T
=  dT and
www.elsolucionario.net
106
STUDENT'S SOLUTIONS MANUAL
Therefore, dlnaA
MA
MAdt:.T
dt:.T
Kf
Kf
  =   and dlnaA =    According to the GibbsDuhem equation
which implies that
nB
and hence that d In aA =   d In aBo nA
din aB nAMA I Hence    =   =  'dt:.T
nBKf
bBKf
We know from the GibbsDuhem equation that XA d In aA + XB d In aB = 0
and hence that J d In aA = Therefore In aA = 
f
f
XB

d In aBo
XA
xB d In aB. XA
The osmotic coefficient was defined in Problem 5.21 as I
XA
r
XB
¢ =   In aA =   In aA . Therefore, ¢ =XA xB
f
XB I dlnaB=
lob
1 bdlnaB=bob
xA
lob 0
lob
bdlnyb=1 bdlnb+1 bob
lob
=1+1 bdlny . b 0 From the DebyeHiickellimiting law,
In y
= A ' b I /2
[A'
= 2.303AJ.
1
Hence, dIn y = 2A' b1 / 2db and so
¢
= 1+ bI ( 2I A ') 10{b b 1/ 2 db = I  21 (AI) b
COMMENT.
x _2b3/ 2
3
11
For the depression of the freezing point in a 1, 1electrolyte
www.elsolucionario.net
_IAII /21.
3
lob bdlny 0
SIMPLE MIXTURES
(1 1) (~ 2.) _
D.fusH and hence rl/> = R _ D.fusHxA
Therefore, I/> 
R
Xs
T
  T T*
_
T*
.
D.fusH xA

R
Xs
(T*  T) "" IT*
D. T R T*2
D.fuSH xA
Xs
D.fusHD. T
"" vRbsT*2MA where v
MRT*2
= 2. Therefore, since Kf =  , H D.fus
T
~
1/> 2bsKf .
Solutions to applications PS.2S
In thi s case it is convenient to rewrite the Henry 's law expression as
(1) At PN2
= 0.78
x 4.0 atm
= 3.1
= 3.1 atm
mass ofN2
atm,
x lOOgH20 x 0. 18 ~gN2 / (g H20atm)
= IS6~gN2 1.
(2) At PN 2 = 0.78 atm, mass of N2 = 114 ~g N21. (3) In fatty tissue the increase in N2 concentration from 1 atm to 4 atm is
PS.27
(a) i = I only, N,
= 4, K, = 1.0 X
107 dm 3 mol  ' ,
4 x IOdm 3 ~mol  '
v
[AJ
I + IOdm3 ~mol' x [A(
The plot is shown in Fig. S.6(a) . (b)
i = I; N, = 4, N2 = 2; K, = 1.0 X 105 dm 3 mol' = 0.10 dm 3 ~mol',
K2
= 2.0
v
[AJ
X
10 6 dm 3 mol  '
= 2.0 dm 3 ~mol  '.
4 x 0.IOdm3~mol'
2 x 2.0dm3 ~mol'
1+ 0.IOdm 3 ~mol' x [AJ + 1+ 2.0dm3
~mol'
The plot is shown in Fig. S.6(b).
PS.29
By the van ' t Hoff equation [S.40], cRT fl=[BJRT=  . M
www.elsolucionario.net
x [A(
107
108
STUDENT'S SOLUTIONS MANUAL
40.0 r                ,
30.0
Al (OHh + 3HCI AICl) ;==> AI3+ + 3CI H20 ;==> H+ +OHand one condition of electrical neutrality
Hence, the number of independent components is C
=7
(3 + I )
= [IJ
www.elsolucionario.net
116
STUDENT'S SOLUTIONS MANUAL
NH 4CI(s)
E6.7(b)
;='
+ HCI(g)
NH3(g)
= I 1[Example 6.1] and 1P = 21 (s and g). added before heating, 1C = 21 (because NH4CI, NH3
(3) For this system 1C (b) If ammonia is
are now independent) and
1P = 21 (s and g). E6.8(b)
(3) Still 1C
= 21 (Na2 S04 , H20) , but now there is no solid phase present, so 1P = 21 (liquid solution,
vapor).
0.
(b) The variance is F = 2  2 + 2 = We are free to change any two of the three variables, amount of dissolved salt, pressure, or temperature, but not the third. If we change the amount of dissolved salt and the pressure, the temperature is fixed by the equilibrium condition between the two phases. E6.9(b)
See Figure 6.4.
+ 10  10
 30
50
 70
 90
Figure 6.4 E6.10(b)
See Figure 6.5. The phase diagram should be labeled as in figure 6.5. (a) Solid Ag with dissolved Sn begins to precipitate at QI , and the sample solidifies completely at Q2. (b) Solid Ag with dissolved Sn begins to precipitate at bl, and the liquid becomes richer in Sn. The peritectic reaction occurs atb2, and
(a)
b
(b)
a
Liquid
800
L + Ag solid contaminated with Sn
U
~ Bi
460 °C b2
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 StefanBoltzmann constant.
= N2 (d) N 2
1)' [S.S].
=  . Then,dx = 2dA 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.
www.elsolucionario.net
QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
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 = lk2
(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:
www.elsolucionario.net
2mJ1/I*1/Idr:
2m '
168
STUDENT'S SOLUTIONS MANUAL
PS.21 1/ 2
N
(a)
1
5
00
(r2)
= ( _13 )
[Problem 8.14].
32rca o
6
+ r 2 )
er/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
= ( _15 )
[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
www.elsolucionario.net
QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
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 illdefined. 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] .
www.elsolucionario.net
170
STUDENT'S SOLUTIONS MANUAL
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 rootmeansquare 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
www.elsolucionario.net
40
so Figure 8.1(c)
QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
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>=
(ilidpxd)
 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 xI = (1 j ili) J~"oo dpx we need to
I
confirm the operator relationship x I x = xxI = 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>.
www.elsolucionario.net
172
STUDENT'S SOLUTIONS MANUAL
Solutions to applications PS.27
(a)
Anonrelativistic
=
h (2m e V) 1/2 e 6.626 {2 (9.109
Anonrelativistic
= 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
=
{(ev)}c eV +
/2 I=
2m e
1
2mec2
5.48 pm
Anonrelativistic
( eV) 1+2mec2
=
1/2
5.48 pm {l+0.0489}1 /2
I
+
(1.602
{
2(9.109
X
10
19
C) (50.0
10 31 kg) (3 .00
X
X
JI /2
108 ms I )2
= 1 535 m . p I·
(b) For an electron accelerated through 50 kV the nonrelativistic 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. Anonrelativistic 
Arelativistic
=
Anonrelativistic _
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 nonrelativistic equation yields a value that has less than 1% error: c2 V :::::: 2 ( 2m e ) e PS.29
(a)
C~(g) +
X
~
(Anonrelativistic 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 molI) = 74.81 kJ mol I at 'f . www.elsolucionario.net
at 'f
= 25°C.
QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
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 molI 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 2dT, 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 molI .
(10 k1) 3
x (298K) x
1
= 65.16 kJ mol  I. !:!"rC;
= (30.865JK I molI)
( 10 kJ) = 0.030865kJK I molI. 3
x
J
With the estimate of constant !:!,.rC; , 1methane is unstable above 825 K I.
www.elsolucionario.net
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 Kl)
= 4998K.
www.elsolucionario.net
QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
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  K4) 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.
www.elsolucionario.net
4
9
Quantum theory: techniques and applications
Answers to discussion questions 09.1
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 zeropoint energy, and zeropoint 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 zeropoint energy.
09.5
Fermions are particles with halfintegral 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.
Solutions to exercises E9.1(b)
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
www.elsolucionario.net
= 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 zeropoint energy is the groundstate 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.
www.elsolucionario.net
177
178 E9.5(b)
STUDENT'S SOLUTIONS MANUAL
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)
(n 2 + n2 + n2)h 2 1 2 3 8mL2
n2h2 8mL2
x (1.381 x 10 23 JK 1) x (300K)
= 6.214 x
8mL 2
n2 =  £ 2
h
www.elsolucionario.net
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 molI ) 2 x 100m ( 6.022 x 1023 molI
(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 1042 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 molI ) 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 zeropoint 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 1034 J s)2 = 260 N m
www.elsolucionario.net
180 E9.10(b)
STUDENT'S SOLUTIONS MANUAL
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 HzI ) 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 twoparticle 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 ( ll77Nml) )1 /2 (16.00) x (1.6605 x 1027 kg)
=13 .14 x 10 20 11 E9.13(b)
The first excitedstate 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
www.elsolucionario.net
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 
20 ± .J202  4 x 4 x 15
4ac
2a
5±
2 x 4
Evaluating the result numerically yields x=10,±0.96a, or ±2.02a I.
z=
.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 zeropoint 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
( 2 2 9 3 . 8 N m 1 ~(l4.0031 u) x (1.66054 X 10 27 kgu I )
10 20 J 1
www.elsolucionario.net
)1 /2
182
E9.16(b)
STUDENT'S SOLUTIONS MANUAL
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"
27T
2
=
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 highestpointing to the lowestpointing 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)
www.elsolucionario.net
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
183
Figure 9.1(b)
Solutions to problems Solutions to numerical problems P9.1
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 1O4o 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
www.elsolucionario.net
=
1
1 130 xlO 22 J . .
184 P9.S
STUDENT'S SOLUTIONS MANUAL
(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 firstorder correction to the groundstate 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(La)
= £
(
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 firstorder correction to the groundstate energy is EI(I )
P9.7
sin Jt cos
£ . (Jt) = 10 + ;£ Sin 10 = I0.1984 £.I
The secondorder correction to the groundstate 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
www.elsolucionario.net
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
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 (ab)X 2(a  b)

cos(ab)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 secondorder 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 secondorder correction is (2) _ 
£1
www.elsolucionario.net
186
STUDENT'S SOLUTIONS MANUAL
The firstorder correction to the groundstate 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 firstorder 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 zeroorder groundstate 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 evenn 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.
www.elsolucionario.net
198
STUDENT'S SOLUTIONS MANUAL
(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
www.elsolucionario.net
Atomic structure and atomic spectra
Answers to discussion questions 010.1
The Schrodinger equation for the hydrogen atom is a sixdimensional 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 onedimensional 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 threedimensional 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 sixdimensional SchrOdinger equation is first separated into two threedimensional 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 onedimensional equations, one in the angle ¢ , another in the ang le e, and a third in the distance r . The solutions of these three onedimensional 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.
www.elsolucionario.net
200
STUDENT'S SOLUTIONS MANUAL
010.5
See Section 10.4(d) of the text and any general chemistry book, for example, Sections 1.101.13 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 manyelectron wavefunctions for atoms are represented as a simple product of oneelectron wavefunctions. At a somewhat more sophisticated level, the manyelectron wavefunctions are written as linear combinations of such simple product functions that explicitly satisfy the Pauli exclusion principle. Relatively good oneelectron functions are generated by the HartreeFock selfconsistent field method described in Section 10.5. If we place no restrictions on the form of the oneelectron functions , we reach the HartreeFock 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 electronelectron interaction terms in the manyelectron 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 electronelectron 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]
www.elsolucionario.net
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
and p
= 7.236
7.236a o
1
  2  = 3.618 a o
orr=17.31 x 10 11 ml and 11.917 x 1O lO ml
www.elsolucionario.net
1
202
E10.4(b)
STUDENT'S SOLUTIONS MANUAL
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
www.elsolucionario.net
ATOMIC STRUCTURE AND ATOMIC SPECTRA
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)
Z33 ) 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 www.elsolucionario.net
204
STUDENT'S SOLUTIONS MANUAL
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
www.elsolucionario.net
ATOMIC STRU CTURE AND ATOMIC SPECTRA
E10.9(b)
For I > 0, )
=
=I±
205
1/ 2, so
(a)
I
I, so)
= 1r112or312,1
(b)
1=5 , so)
= 19/2 or 11/21
E10.10(b) Use the ClebschGordan series in the form
+hI ,·· ·,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
and Pmax
= ~e2
P fall s to a fraction! of its maximum given by
www.elsolucionario.net
ao
=
I ; and the subscript 4
206
STUDENT'S SOLUTIONS MANUAL
and hence we must solve for r in ? fl /  _ 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 www.elsolucionario.net
± I).
ATOMIC STRUCTURE AND ATOMIC SPECTRA
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 RussellSaunders 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 spinorbit 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.
Solutions to problems Solutions to numerical problems P10.1
All lines in the hydrogen spectrum fit the Rydberg formula
~A = RH (~~) nT n~
[10.1 , with ii
=~] A
RH
= 109677cm l .
www.elsolucionario.net
208
STUDENT'S SOLUTIONS MANUAL
Find nl from the value of Amax , which arises from the transition
n~
AmaxRH 
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
II,
1185187 cm
The ionization energy of the groundstate ion is given by
www.elsolucionario.net
II, ····
ATOMIC STRUCTURE AND ATOMIC SPECTRA
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 spinorbit 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 spinorbit 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
(1  ~) = ~ 4
4
Ro
and hence
www.elsolucionario.net
210
STUDENT'S SOLUTIONS MANUAL
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 sI)
=   = ::::...,."...: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 (1/41/9)
= 15233cm 1
and the upper limit is I!
= (109677cm  l )
x ( 1/ 4  0)
= 27419cm l .
www.elsolucionario.net
ATOMIC STRUCTURE AND ATOMIC SPECTRA
211
Solutions to theoretical problems P10.11
Consider t/l2p, = t/l2,I,Owhich extends along the zaxis. The most probable point along the zaxis 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
=
YII 
.j2 YII
pao/2, dr
}
=
12 rsm(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.
www.elsolucionario.net
•
34992
'
212
STUDENT'S SOLUTIONS MANUAL
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 pep/ 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.
www.elsolucionario.net
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
'" www.elsolucionario.net
Figure lO.2(b)
213
214
STUDENT'S SOLUTIONS MANUAL
Boundary surface plots. sorbital boundary surface
porbital boundary surface
dorbital 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
2JL
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 www.elsolucionario.net
ATOMIC STRUCTURE AND ATOMIC SPECTRA
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 noncommuting
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
(rt, = G~)2 {2 (~) 3/2}210 = I~ I
(rI)2S = (c)
.(rI)
= 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 ePdp ao
because
10ro oo p3 e
The general formula for a hydrogenic orbital is
P
dp
[Table 10.1]
= 6.
(r I ) 1 = ~ . "
n2ao
www.elsolucionario.net
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 threedimensional 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 nonzero for a transi tion to
be allowed. The Justification examined conditions that allowed the z component of this quantity to be nonzero; 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
www.elsolucionario.net
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.
www.elsolucionario.net
218
STUDENT'S SOLUTIONS MANUAL
Solutions to applications P10.25
The wavenumber of a spectroscopic transition is related to the difference in the relevant energy levels. For a oneelectron 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 JI 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
1031 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
1031 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/41/9)cm 1 =160954.7cm
l·
www.elsolucionario.net
l
ATOMIC STRUCTURE AND ATOMIC SPECTRA
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 sunstar 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 earthstar vector at the time of the spectral observation. (The earthstar direction, which is observable by earthbased astronomers, is practically identical to the sunstar 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 <
h < 13 because of decreased nuclear shielding as each successive electron is removed.
www.elsolucionario.net
220
STUDENT'S SOLUTIONS MANUAL
First three ionization energies of group 13 40
>
'"
35 30
;>.,
~
::? 25
'"c: '"c:0
'Zl to
20
•
•
15
N
10
~
5
'2
•
.
0 0
20
40
60
80
Atomic number, Z
100
Figure 10.3
(ii) The ionization energies of boron are much larger than those of the remaining group elements because
the valence shell of boron is very small and compact with little nuclear shielding. The boron atom is much smaller than the aluminum atom. (iii) The ionization energies of AI , Ga, In, and Tl are comparable even though successive valence shells are further from the nucleus because the ionization energy decrease expected from large atomic radii is balanced by an increase in effective nuclear charge.
www.elsolucionario.net
Molecular structure
Answers to discussion questions 011.1
Our comparison of the two theories will focus on the manner of construction of the trial wavefunctions for the hydrogen molecule in the simplest versions of both theories. In the valence bond method, the trial function is a linear combination of two simple product wavefunctions, in which one electron resides totally in an atomic orbital on atom A, and the other totally in an orbital on atom B. See eqns Il.l and 11 .2, as well as Fig. 11 .2. There is no contribution to the wavefunction from products in which both electrons reside on either atom A or B. So the valence bond approach undervalues, by totally neglecting, any ionic contribution to the trial function. It is a totally covalent function. The molecular orbital function for the hydrogen molecule is a product of two functions of the form of eqn 11.8, one for each electron, that is,
1/1
= [A(l) ± B(I)][A(2) ± B(2)) = A(l)A(2) + B(1)B(2) +A(1)B(2) + B(l)A(2).
This function gives as much weight to the ionic forms as to the covalent forms. So the molecular orbital approach greatly overvalues the ionic contributions. At these crude levels of approximation, the valence bond method gives dissociation energies closer to the experimental values. However, more sophisticated versions of the molecular orbital approach are the methods of choice for obtaining quantitative results on both diatomic and polyatomic molecules. See Sections 11.611.8. 011.3
Both the Pauling and Mulliken methods for measuring the attracting power of atoms for electrons seem to make good chemical sense. If we look at eqn 11 .23 (the Pauling scale), we see that if D(AB) were equal to 112[D(AA) + D(BB)) the calculated electronegativity difference would be zero, as expected for completely nonpolar bonds. Hence, any increased strength of the AB bond over the average of the AA and BB bonds, can reasonably be thought of as being due to the polarity of the AB bond, which in turn is due to the difference in electronegativity of the atoms involved. Therefore, this difference in bond strengths can be used as a measure of electronegativity difference. To obtain numerical values for individual atoms, a reference state (atom) for electronegativity must be established. The value for fluorine is arbitrarily set at 4.0. The Mulliken scale may be more intuitive than the Pauling scale because we are used to thinking of ionization energies and electron affinities as measures of the electron attracting powers of atoms. The choice of factor 112, however, is arbitrary, though reasonable, and no more arbitrary than the specific form of eqn 11.23 that defines the Pauling scale.
www.elsolucionario.net
222
011.5
STUDENT'S SOLUTIONS MANUAL
The Hi.ickel method parameterizes, rather than calculates, the energy integrals, IX and (J , that arise in molecular orbital theory. They are considered to be adjustable parameters; their numerical values emerge only at the end of the calculation by comparison to experimental energies. The overlap integral is neglected, set equal to zero. Three other rather drastic approximations, listed in Section 11.6(a) of the text, eliminate many terms from the secular determinant and make it easier to solve: all diagonal terms of the determinant are set equal IX  E ; nearestneighbor terms (that is, between bonded atoms) all have the same value, {J; and all other terms are zero. (Ease of solution was important in the early days of quantum chemistry before the advent of computers, and without the use of these approximations, calculations on polyatomic molecules would have been difficult to accomplish.) The simple Hi.ickel method is usually applied only to the calculation of rr electron energies in conjugated organic systems. The simple method is based on the assumption of the separability of the a and rr electron systems in the molecule. This is a very crude approximation and works best when the energy level pattern is determined largely by the symmetry of the molecule. (See Chapter 12.)
011.7
The ground electronic configurations of the valence electrons are found in Figures 11.3111.33 and 11.37.
Oz
z 1agz 1auz lrr42a u g lazlaz2a21rr4)rrz g u g u g
NO
laz2a23a2lrr42rr I
Nz
b=3
2S + 1= 0
b=2
2S+ I = 3
b= 2!
2S+I=2
The following figures show HOMOs of each . Shaded vs. unshaded atomic orbital lobes represent opposite signs of the wavefunctions. A relatively large atomic orbital represents the major contribution to the molecular orbital.
Nz
2a molecular orbital
NO
2rr
Dinitrogen with a bond order of three and paired electrons in relatively low energy molecular orbitals is very unreactive. Special biological, or industrial, processes are needed to channel energy for promotion of 2a electrons into high energy, reactive states. The high energy Irrg LUMO is not expected to form stable complexes with electron donors. Molecular nitrogen is very stable in most biological organisms, and as a result the task of converting plentiful atmospheric N 2 to the fixed forms of nitrogen that can be incorporated into proteins is a difficult one. The fact that Nz possesses no unpaired electrons is itself an obstacle to facile reactivity, and the
www.elsolucionario.net
MOLECULAR STRUCTURE
223
great strength (large dissociation energy) of the N2 bond is another obstacle. Molecular orbital theory explains both of these obstacles by assigning N2 a configuration that gives rise to a high bond order (triple bond) with all electrons paired. (See Fig. 11 .33 of the text. ) Dioxygen is kinetically stable because of a bond order equal to two and a high effective nuclear charge that causes the molecular orbitals to have relatively low energy. But two electrons are in the high energy IJTg HOMO level, which is doubly degenerate. These two electrons are unpaired and can contribute to bonding of dioxygen with other species such as the atomic radicals Fe(lI) of hemoglobin and Cu(Il) of the electron transport chain. When sufficient, though not excessively large, energy is available, biological processes can channel an electron into this HOMO to produce the reactive superoxide anion of bond order I ~. As a result, 0 2 is very reactive in biological systems in ways that promote function (such as respiration) and in ways that disrupt it (damaging cells). Although the bond order of nitric oxide is 2~, the nitrogen nucleus has a smaller effective nuclear change than an oxygen atom would have. Thus, the one electron of the 2JT HOMO is a high energy, reactive radical compared to the HOMO of dioxygen. Additionally, the HOMO, being antibonding and predominantly centered on the nitrogen atom, is expected to bond through the nitrogen. Oxidation can result from the loss of the radical electron to form the nitrosyl ion, NO+ , which has a bond order equal to 2. Even though it has a rather high bond order, NO is readily converted to the damagingly reactive peroxynitrite ion (ONOO) by reaction with 0; without breaking the NO linkage.
Solutions to exercises E11.1(b)
E11.2(b)
E11.3(b)
Use Figure I 1.23 forH;, I l.33for N2 , and 1l.31 for 0 2. 110 220* 1 1 b = 0.5
(a)
H; (3 electrons) :
(b)
N2 (10 electrons) :
110 220*21JT 430 2 1 b=3
(c)
02(l2electrons) :
110 2 20*230 2 1n 4 2n*21 b=2
CIF is isoelectronic with F2, CS with N2 . 110 220 *230 2 In 42n*41
(a)
CIF(14electrons):
(b)
CS(IOelectrons):
110 220*21 n 430 2 1 b = 3
(c)
0;(13 electrons) :
110 220*230 2 In 42n*3 1 b
b
=
I
= 1.5
Decide whether the electron added or removed increases or decreases the bond order. The simplest procedure is to decide whether the electron occupies or is removed from a bonding or anti bonding orbital. We can draw up the following table, which denotes the orbital involved
(a) ABChange in bond order (b) AB + Change in bond order
N2
NO
02
2JT *
2n*
 1/2
112
30
2n*
2n*
 1/2
+112
+1/2
C2
F2
CN
2n*
30
40*
1/2
+1/2 In  1/2
 1/2
2JT*
30+112 30
+1/2
1/2
www.elsolucionario.net
224
STUDENT'S SOLUTIONS MANUAL
I
I
(a) Therefore, C2 and CN are stabilized (have lower energy) by anion formation.
I
(b) NO, 02 and F21 are stabilized by cation formation; in each of these cases the bond order increases. E11.4(b)
Figure 11 .1 is based on Figure 11.31 of the text but with Cl orbitals lower than Br orbitals. BrCl is likely to have a shorter bond length than BrCl; it has a bond order of 1, while BrCl has a bond order of 112.
3p
3s
Figure 11.1 E11.S(b)
0i (11 electrons) :
10" 220" *230" 21 JT 42JT *1
b = 5/ 2
02 (12 electrons) :
J0"220"*230"21 JT 42JT*2
b=2
0; (13 electrons) :
J0"220"*230"21JT 42JT*3
b = 3/2
O~ (14 electrons) :
Each electron added to
I0i, E11.6(b)
02, 0;,
10"220"*230"21JT 42JT*4
0i is added to an anti bonding orbital, thus increasing the length. So the sequence
o~I has progressively longer bonds.
1
1/f2 dr = N 2 2
1
= N (l
(1/fA
+ A1/fB)2 dr
+ A2 + 2AS) 1
Hence N E11.7(b)
b= 1
= 1= N
[I
2
1
1/fA1/fsdr
(1/fl
=
+ A21/f~ + 2A1/fA 1/fs) dr
= 1
s]
) 1/ 2
= ( 1 + 2AS + A2
We seek an orbital of the form aA + bB, where a and b are constants, which is orthogonal to the orbital N(0.145A + 0.844B) . Orthogonality implies
1+ 1 (aA
N
bB)N(0.145A
[0. 145aA2
+ 0.844B) dr = 0
+ (0.145b + 0.844a)AB + 0.844bB2] dr = 0
www.elsolucionario.net
MOLECULAR STRUCTURE
The integrals of squares of orbitals are I and the integral
0= (0.145
+ 0.844S)a + (0.145S + 0.844)b
so
JAB d r a=
225
is the overlap integral S, so
0. 145S + 0.844  0.145 + 0.844S b
This would make the orbitals orthogonal, but not necessarily normalized. If S = 0, the expression simplifies to a
= _ 0.844 b 0.145
and the new orbital would be normalized if a = 0.844N and b = 0.145N. That is
IN(0.844A 
0.145B)
I
E11.8(b)
The trial function 1/1 = x 2 (L  2x) does not obey the boundary conditions of a particle in a box, so it is I not appropriate I· In particular, the function does not vanjsh at x = L.
E11.9(b)
The variational principle says that the minimum energy is obtajned by taking the derivatjve of the trial energy with respect to adjustable parameters, setting it equal to zero, and solving for the parameters: Etrial
=
3an _ e (~) 1/2 2J.L eo 2JT 3 2
2
so
Solving for a yields:
Substituting this back into the trial energy yields the minimum energy:
E11.10(b) Energy is conserved, so when the photon is absorbed, its energy is transferred to tbe electron. Part of
it overcomes the binding energy (ionization energy) and the remainder is manifest as the now freed electron 's kinetic energy.
so
Epholon
= I + Ekinelic
Ekinelic
=
Epholon  /
he (6.626 x 10 34 1 s) x (2.998 X 10 8 m sI) =  / =  4.6geV A (584 x 1O 12 m) x (1.602 x 1O 19 1eV I)
=1211gevl=13.39 x 10 16 11 E11.11 (b) The molecular orbitals of the fragments and the molecular orbitals that they form are shown in Figure 11.2. E11.12(b) We use the molecular orbital energy level diagram in Figure 11.41 . As usual, we fill the orbitals starting
with the lowest energy orbital, obeying the Pauli principle and Hund 's rule. We tben write
www.elsolucionario.net
226
STUDENT'S SOLUTIONS MANUAL 7r *
Figure 11.2
(a) C6H6 (7 electrons) :
E
= 2(a +
2,8) + 4(a +,8) + (a ,8)
(b) C6 H t(5 electrons):
E
= 1 7a +
7,8 I
Ia~Ue~g I
= 2(a + 2,8) + 3(a + ,8) = 1 5a + 7,8 I
E11.13(b) The secular determinants from E 11.l3(a) can be diagonaJized with the assistance of generalpurpose mathematical software. Alternatively, programs specifically designed for Hiickel calculations (such as the one at Australia's Northern Territory University, http://www.smps.ntu.edu.auimodules/mod3/ interface.html) can be used. In both molecules, 14 nelectrons fill seven orbitals. (a) In anthracene, the energies ofthe filled orbitals are a + 2.414 21,8, a + 2.00000,8, a + 1.41421,8 (doubly degenerate), a + 1.00000,8 (doubly de enerate), and a + 0.41421,8, so the total energy is 140: + 19.313 68,8 and the n energy is 19.313 68,8 . (b) For phenanthrene, the energies of the filled orbitals are a + 2.43476,8, a + 1.95063,8, a + 1.51627,8, a+ 1.30580,8, a+ 1.14238,8, a +0.76905,8, a+0.60523,8,sothetotalenergyis 140: + 19.44824,8 and the n energy isI19.44824,8 I.
Solutions to problems Solutions to numerical problems P11.1
VrA = cos kx measured from A, 1/rB Then, with Vr
= cos k' (x 
R) measuring x from A.
= VrA + Vrs,
Vr = coskx + cosk'(x  R) = cos kx + cos k'R cos k'x + sin k'R sin k'x
= cos a cos b + sin a sin b]. Jt k = k' = ~. cos k'R = cos ~ = O' sink' R = sin 2" = I. 2R' 2' [cos(a  b)
(a)
Jtx
Vr = cos 2R + sin
Jtx
2R'
www.elsolucionario.net
MOLECULAR STRUCTURE
I
"2R,
For the midpoint, x =
. I 1/1 ( I R) = cos I Jl + SIn  Jl =
so
2
4
interference (1/1 > 1/1A, 1/IB)· Jl I 3Jl I 3Jl (b) k=2R' k =2R; coskR=cosT=O, JlX 2R
4
21 / 2
and there
227
. . IS constructive
sink'R = 1.
3Jlx
1/1 =cos  si n  . 2R
I
interference P11.3
(1/1
I
"2R,
For the midpoint, x =
= cos  Jl 4
. 3 SIn 
4
Jl = 0 and there is destructive
1/IA, l/Is).
<
The s orbital begins to spread into the region of negative amplitude of the p orbital. When their centers coincide, the region of positive overlap cancels the negative region. Draw up the following table. R/ao
0
S
0
00429
2
3
4
5
6
0.588
0.523
0.379
0.241
0.141
7 0.D78
8 0.041
9
10
0.021
0.01
1.0
0.8
5 0.6
1
l'y
(
0.4
/ 0.2
o
5(\5, 2pz)
"\. 't
/
II o
"\ ~ ........
2
4
6
8
R/ao
'10
Figure 11.3
The points are plotted in Fig. II .3. The maximum overlap occurs at 1 R = 2. lao I. P11.S
We obtain the electron densities from
p+ = 1/Ii and p_ =
1/1~ with 1/1+ and 1/1 as given in Problem 1104.
www.elsolucionario.net
228
STUDENT'S SOLUTI ONS MANU AL
We evaluate the factors preceding the exponentials in 1fr+ and 1fr.
N+
(

I
:n:al
Likewise, N _
)
I~
= 0.561 x
(~) 1/ 2 :n:ao
1/ 2
I
(:n:
1216pm 3/ 2 ·
x (52.9 pm)3 )
621 pm 3/ 2 ·
Then and The 'atomic ' density is
=
p
~(1frls(A)2 + 1frls(B)2) = ~ 2
x
2
e  2rA / aO
9.30 x
+ e 21ll /ao
e2lzl/ao
105 pm 3
(_1_) n~
(e  2rA/aO
+ e  21ll /aO)
+ e  2Iz  RI/ao
9.30 x 105 pm 3
The difference density is op±
= P± 
p.
Draw up the following table using the information in Problem 14.4. z/pm
p+ x 107 / pm 3 p_ x 107 / pm 3 p x 107 / pm 3 op+ x 107 / pm  3 op_ x 107 / pm 3
100
80
60
40
20
0
20
40
0.20 0.44 0.25 0.05 0.19
0.42 0.94 0.53 0.11 0.41
0.90 2.01 1.13 0.23 0.87
1.92 4.27 2.41 0.49 1.86
4.09 9.11 5.15 1 .05 3.96
8.72 19.40 10.93 2.20 8.47
5.27 6.17 5.47 0.20 0.70
3.88 0.85 3.26 0.62 2.40
100
120
140
160
180
200
7.42 14.41 8.88 1.46 5.54
5.10 11 .34 6.40 1.29 4.95
2.39 5.32 3.00 0.61 2.33
1.12 2.50 1.41 0.29 1.09
0.53 1.17 0.66 0.14 0.51
0.25 0.55 0.31 0.06 0.24
z/pm
60
p+ x 10 7 / pm  3 p_ x 10 7 / pm 3 p x 107 / pm 3 op+ x 107 / pm  3 op_ x 107 / pm  3
80
3.73 0.25 3.01 0.70 2.76
4.71 4.02 4.58 0.13 0.56
The densities are plotted in Fig. 11.4(a) and the difference densities are plotted in Fig. 11.4(b). P11.7
P
=
11fr1 2dr ~ 11fr1 2 or ,
or
=
1.oopm 3
(a) From Problem 11 .5
www.elsolucionario.net
MOLECULAR STRUCTUR E
20
15 ME Q.
'0 10 X
.
5 0  100
100
0
200
z/pm
Figure 11.4(a)
10 8 6
1:
Q.
.0
x .
"0
4
2 0 2 4 6 100
200
100
0 z/pm
Figure ll.4(b)
Therefore, the probability of finding the electron in the volume
P = 8.6
X
or at nucleus A is
10 7 pm  3 x 1.00pm 3 = 18.6 x 10 7 1.
(b) By symmetry (or by taking (c) From Fig. 11.4(a),
z=
1/Ii (~R) =
106pm) P
=
18.6 x 10 7 1
3.7 X 10 7 pm  3 , so
P=
rI3 .7  xI0 7'1
(d ) From Fig. 11 .5, the point referred to lies at 22.4 pm from A and 86.6 pm from B.
0.0 pm
Figure 11.5
A _'''..,. B
e 22.4/ 52.9
Therefore,
1/1
=
1/1 2 = 4.9
+ e  86.6/ 52.9
12 16pm X
10 7 pm  3 ,
3/ 2
so
P
_0._65_ +_0:;.1=9 _ 4 3/ 2 12 16pm 3 / 2  6.98 x 10 pm .
= 14.9 x 10 7 1.
www.elsolucionario.net
229
230
STUDENT'S SOLUTIONS MANUAL
For the antibonding orbital, we proceed similarly. (a) 1/I~ (z
= 0) = 19.6 X 10 7 pm 3 [Problem 1l.5], so
(b) By symmetry, P
(c)
GR) =
1/1~
(d) We evaluate
=
0,
12.0 x 106 1.
so
=
P
_ 0.65  0.19 _
1/1~ = 5.49 EH
= El =
[QJ.
1/1 at the point specified in Fig. 1l.5.
1/1  621 pm 3/ 2
P11.9
12.0 x 10 6 1.
P

 4
10 7 pm  3 ,
X
 3/ 2
7.41 x 10 so
pm
.
= 15.5 x 10 7 1.
P
hcRH [Section 1O.2(b)].
Draw up the following table using the data in question and using
?
?
?
~
~
=x= x4nEoR 4nEoao R 4nEo x (4nEon2 / mee2) R
so that
(e2 / 4nEoR) Eh
ao
=. R
Draw up the following table. R / ao
0
2 (e / 4n EoR) / Eh
00
(VI  V2) / Eh
0
(E  EH) / Eh
00
0.007 l.049
2
3
4
0.500 0.031 0.425
0.333 0.131 0.132
0.250 0.158 0.055
00
0 0 0
The points are plotted in Fig. 11.6. The contribution V2 decreases rapidly because it depends on the overlap of the two orbitals. P11.11
The internuclear distance (r )n ~ n2ao , would be about twice the average distance (~ l.06 x 106 pm) of a hydro genic electron from the nucleus when in the state n = 100. This distance is so large that each of the following estimates is applicable. Resonance integral,
f3
Overlap integral, S
~ E
~
8 (where 8
(where
E ~
~
0).
0).
www.elsolucionario.net
MOLECULAR STRUCTURE
231
0.5
0.4
..c:
0.3
... .. .!
~
:c L; ::':L::::LL:::::
... ~ . . . ! . " :
. ..., ....
:.... ~ ... ;... ~....~ ...;.. .~
:, :
~
) ... r::j:·····: :
~
500
, ...:....,...
.. .,... ...: ... ..
8
.
.
.!.
.~
~
...
... :....:....:....:....: 12
" ~~i"'~''' ~'' . ..~ .
":' .. ~
...~ ... ~ ...j ...:. .
•. • j . •
~
.. ~ ..
.. ~ " l AX> lM X
M protons (a)
X protons
j 
d
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 MoP axes makes the P and H nuclei chemically and magnetically equivalent in both the cis and trailsform s. E15.15(b) Precession in the rotating frame follows
www.elsolucionario.net
304
STUDENT'S SO LUTIONS MANUAL
Since w is an angular frequency, the angle through which the magnetization vector rotates is
"" _ S0 ""' 1
en _
 
g] flNt
34
(n) x (1.0546 x 10 J s) 1 4 1 940 x 10 T (5.586) x (5 .0508 x 10 27 JTI) x (12 .5 x 10 6 s) .
a 90° pulse requires! x 12.511 s = 16.25 I1S
=
I
(6.626 X 10 34 J s) x (2.998 x 108 m SI) (2) x (9.274 x 10 24 J TI ) x (8 x 10 3 m) 
[iliJ 13T .
E15.17(b) The g factor is given by
hv
h
flB fA
flB
g =;
= g
6.62608 x 10 34 J s 9.2740 x 10 24 JT 1
~,~
_I
1
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
flB 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.
www.elsolucionario.net
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$ =  , J1B f$ J1Bg
_ I
J1B 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
Solutions to problems Solutions to numerical problems P15.1
g/ Bo
= 3.8260
(Table 15.2) . (6.626 x 10 34 JHz l ) x v
hv
=  = g/J1N
()(3 .8260) x (5.0508 x
Therefore, with v
10 27 JTI)
= 3.429 x
= 300 MHz,
www.elsolucionario.net
8
10 (v/Hz) T.
306
STUDENT'S SOLUTIONS MANUAL
8N Ii
gliLNBO
~ ~
[Exercise 15.4(a)]
(3.8260) x (5.0508 x 1027 J r ' ) x (I0.3T)
=
(2) x (1.381 x 1O23 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(
(I)
(2)
+ 31g) = 8.0Hz,
= 11.2 Hz. www.elsolucionario.net
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 Initropropane shows that (a) the CaH resonance with 8 = 4 .3 shares a crosspeak with the Cb H resonance at 8 = 2.1 and (b) the Cb H resonance with 8 = 2.1 shares a crosspeak 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 gvalues; hence
ao
~
0.85745
  x aH = 0. 1535a H = 0.35 mT. 5.5857
Therefore, the overall width is 6ao
= 12.1
mT
I.
www.elsolucionario.net
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
Figure 15.7
~ =~(lOpercentofitstime) ;
. 5.7mT WewnteP(N2s) =   55.2mT
~
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 ¢ = ~
www.elsolucionario.net
MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
309
Solutions to theoretical problems P15.13
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 1O3 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 )( + ei (lL\)  wi IT)(\dt
10
~ aRe [ 2
iw G(t)e ( dt
I
i(WO
+ w + i/ r)
_ _ __ I __ ] iCwo  w  i/ r) .
www.elsolucionario.net
310
STUDENT'S SOLUTIONS MANUAL
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(WOw)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 halfheight. r
P15.19
For nonweak fields in which the external magnetic field is comparable to the spinorbit coupling field of an unpaired electron it is necessary to include a spinorbit coupling term with coupling constant A [10.41] and apply the secondorder 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 secondorder 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 secondorder 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
www.elsolucionario.net
Sz
as
MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
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 secondorder 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, _ gge
"
2A~
(Ollzln)(nllzIO)
n;loO
I':.Eno
= gYeBosz, indicates that
E(spin)
.
I':.E"o
g increases with increasing strength of the spinorbit 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.
Solutions to applications
P15.21
(8 nucl)
=
g/J1NJ10m/
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/J1NJ10m/
4
nR
and (8nu cl)
3
= O.
www.elsolucionario.net
2
(cos 8max
+ cos8max )
.
312
STUDENT'S SOLUTIONS MANUAL
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 yintercept 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.
www.elsolucionario.net
MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
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 ditertbutyl 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
= 2 sin e.
= cos e, e = arcosx.
x
www.elsolucionario.net
314
STUDENT'S SOLUTIONS MANUAL
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)
www.elsolucionario.net
Statistical thermodynamics 1. the concepts
16
Answers to discussion questions 016.1
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.
Solutions to exercises E16.1(b)
Ne  f3 ej 11;=
q
whereq
=L
e f3ej
www.elsolucionario.net
316
STUDENT'S SOLUTIONS MANUAL
Thus
.
n2
1
= , 11£0 = 300cm 1 2
GIven 
nl
k = (1.38066
In
X
lcmI ) = 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 KI)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 1O2;kg) x (1.381 x 1O23JKI) 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,
'115.9pm'~
= 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
www.elsolucionario.net
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:lefJE1 2f:2efJE2) = q
=
3e fJE1 + 2e fJE2 )
Nheq (3yefJhCVI +2yefJhCVz )
(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 molI) x (6.626 x 10 34 Js) x (2.9979 x IO lO cms l ) x (2546cm I)/4.006 = 17.605 kJ molII 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 10.15
so
n2
= 0.088
nl
(6.626 x 1O 34 J s) x (2.998 x 1OIOcmsl) x (360cm l ) and T =           = 23   : :           (1.381 x 1O JKI) 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
eE"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
www.elsolucionario.net
YN ~,
where
318
STUDENT'S SOLUTIONS MANUAL
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
www.elsolucionario.net
3
3
10.9941
10.999981
STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
E16.8(b)
(a) The ratio of populations is given by the Boltzmann factor n2 nl
("""E) = e25.0K/ 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
eESla,elkT
= 1+ e 25.0K/T + e 50.0K/T
states
At 25.0 K, we note that e2S.0K/T = e I and eSO.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 (e2S.0K/T + 2eSO.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 molI 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
~
(e25.0K/T + 2e50.0 K/T)
(e 2S.0 K/T +4eSO.O K/T)
~ ( e  2S.0 K/T + 2e SO.OK/ T) aq ) ~ aT
www.elsolucionario.net
319
320
STUDENT'S SOLUTIONS MANUAL
where aq = _25_.0_K aT
so Cv
=
.m
(e 2S.0 K / T
+
2eSO.OK / T)
T2 NA(25 .0K) 2k ( (e2S.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 + 4e2 _ (e + 2e )2) 1.503
;::::::~~;::_::_:::':::.:..
= 13.53 J K I molI I (e) The molar entropy is
At 25.0K Sm =
88.3Jmol 1 25.0 K + (6.022
X
1023 molI ) 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 SackurTetrode 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 JKI) x (lOOK) x rr(131.3u) x (1.66054 x 1O 27 kgu  l )jl /2 = 1.52 x 10 11 m
www.elsolucionario.net
STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
_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  eO.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  ehcv{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.
www.elsolucionario.net
322
STUDENT'S SOLUTIONS MANUAL
Solutions to problems Solutions to numerical problems P16.1
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
www.elsolucionario.net
STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
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(1I2I) 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
gjehcPvj.
j
I I 1 at 298 K and I at 5000 K. Therefore, 207 cm3475 cm
= 5 + e4707/207 + 3e 475 1/207 + 5e10559/207
= (5) + (1.3
x 10 10)
+ (3.2 x 10 10 ) + (3.5 x 10 22 ) = 15.00 I. www.elsolucionario.net
324
STUDENT'S SOLUTIONS MANUAL
(ii) q
= 5 + e 4707/3475 + 3e475 1/ 3475 + 5e10559/ 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
3e475 1/ 207 5.00 3e475 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) UmUrn(O) NAhc
+ 4707 emI
x e  4707/207
+ ... } =
4.32 x 1O 7cm l ,
5.00
= _I_tO + 4707 emI
x e  4707/ 3475
+ ... } = 1178 emI .
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
www.elsolucionario.net
x (In 6.26)
= 1 18.07 J K I mol  I I.
STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
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 (e174.2/300) 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
www.elsolucionario.net

U m (0) at 500 K as well
326
STUDENT'S SOLUTIONS MANUAL
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 + e636.27/69.50 + e845.93/69.50 = 11.0491 and at 298 K (b) q = 1 + e 213 .30/207.22 + e 425.39/207.22 + e636.27/207.22 + e845.93/207.22 = ehc{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 molI (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.
Solutions to theoretical problems P16.13
(a) W =
N! nl !n2! . . .
[16.1] =
I.
5! = 0!5!0!0!0!
OJ.
www.elsolucionario.net
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.
www.elsolucionario.net
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
= kT (a[N Inq InN!])
av
= NkT (a In q )
T .N av T.N www.elsolucionario.net
STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
=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
Solutions to applications P16.21
= e1V(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
~,
where
hc (ABC) 1/3
eR=k
(6.626 x 10 34 J s) x (2.998 X 1010 cm SI) 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
e  IhC(AB)K2/kTl)
K= I
www.elsolucionario.net
334
STUDENT'S SOLUTIONS MANUAL
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 hightemperature 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/cm3) 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
=
I I_
e fJhcv
and Ov is the vibrational temperature hcv/k. Thus
Srn
=
R(Ov/ T) _ RI ( I _ Ov / T) n e  I
vv./ T eA
www.elsolucionario.net
1 I _ e Ov / T [17 .19]
STATISTICAL THERMODYNAMICS 2: APPLICATIONS
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 normalmode 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 molI
(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)
www.elsolucionario.net
336
STUDENT'S SOLUTIONS MANUAL
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 = 120.1 kJ molI
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 =1O. llOkJmol1 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 molI ) x (400 K) x (In 3) = 13.65 kJmol  1
COMMENT. The contribution of the excited state is negligible at this temperature.
www.elsolucionario.net
I
STATISTICAL THERMODYNAMICS 2: APPLICATIONS
=
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 temperaturederivative 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
www.elsolucionario.net
338
STUDENT'S SOLUTIONS MANUAL
Solutions to problems Solutions to numerical problems P17.1
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
www.elsolucionario.net
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 e00
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 molI 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)efJE(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 )eO.OSII 2J(J+ I)
1=0
= 1 + 3e0.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 [.
www.elsolucionario.net
= 0 .05112.
340 P17.7
STUDENT'S SOLUTIONS MANUAL
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 103 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 :
lexp(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 lowestlying 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
www.elsolucionario.net
STATISTICAL THERMODYNAMICS 2: APPLICATIONS
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 molI 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)ehcBJ(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
v
{(}y (
dT 
T
 dU R
V,m 
eOv /2T )} 2 leOv/ T
(}y
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.
www.elsolucionario.net
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
www.elsolucionario.net
800
~
1000
Figure 17.2(e)
STATISTICAL THERMODYNAMICS 2: APPLICATIONS
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 ) / cm3) 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.
Solutions to theoretical problems P17.13
(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
www.elsolucionario.net
346
STUDENT'S SOLUTIONS MANUAL
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 )
sI)
=
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) ,. www.elsolucionario.net
STATISTICAL THERMODYNAMICS 2: APPLICATIONS
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 s = second = gram e = 2.9979 x 10 8 m s I NA = 6.022 14 X 10 23 mol I
J = joule mol = mole h = 6.62608 x 10 34 J s R = 8.31451 J K  1 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 molI.
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;
eo
=
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
 r
kT2
~. RT
The JouleThomson coefficient itself is [2.55] /1
= _ /1T = 2rr.NA {2E(r~ Cp
3Cp
kT
ri) _ ri}
= b  ~~ . Cp
www.elsolucionario.net
i)}
STATISTICAL THERMODYNAMICS 2: APPLICATIONS
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 nonlinear molecule is [Table 17.3] R
=
q
1.0270 (T / K) 3/2  a  (ABC/c m3) 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)eP£(J») =
'!.. Lg(J)e (J)eP£(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)eP£(J) L g(J')e(J')eP£(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)eP£(J) '"' g(J')eP£(J' ) ~ ~ J
+ q2 N
J'
'"' g(J)e(J)eP£(J) '"' g(J')e(J')eP£(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')eP[£(J)+£(J') ]. q ~ q2 ~ J,./'
J,./'
www.elsolucionario.net
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 gft BfJ B qE
=L
«
1 for normall y attainable fields, we can expand the exponentials
{ 1  gftBfJBMJ
+ ~(gftBfJBMJ)2 + ... }
MJ
~ 4+ ~(gft BfJB)2
L MJ
MJ [LMJ
0
= 0] = 4(1 + 1 9(ftBfJ 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,
=
www.elsolucionario.net
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 JK9.274 x 10 24 J T
J.i.B
I)
x (lOOOK)
I
~
~
lOOT .
Solutions to applications P17.23
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
www.elsolucionario.net
354
STUDENT'S SOLUTIONS MANUAL
a large numberbut 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
ehcv/kT
1.4388(ii / cm I») 1 exp ( T/ K
1  exp (
1.4388(772) ) 2000
= 2.467. The Boltzmann factor for the lowestlying electronic excited state is exp (
 (1.4388) x (8000») _ 3 2 103 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 .
www.elsolucionario.net
= 1), and that of the
STATISTICAL THERMODYNAMICS 2: APPLICATIONS
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 / cm3) 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 sI) (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 lowestfrequency 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
(lO.OOK) In[(1752) x (1.614) x (I) x (l)]
= 1660.8 J mol  I I. www.elsolucionario.net
356
STUDENT'S SOLUTIONS MANUAL
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
Iexp ( 
qi =
.
,
1
:(:(:l.4::3:::8:::8) X:(1:::2=24:.:::5)....,) = 1.207, 1  exp     I0'00
v q3 =
( 1.4388) x (63.4») 1000
1
(
(1.4388) x (2040») = 1.056, 1  exp  1000
qV = (11.47) x ( 1.207) x (1.056) = 14.62 .
Putting it all together yields
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.
www.elsolucionario.net
18
Molecular interactions
Answers to discussion questions 018.1
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 AH bond in the AH · .. 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 AH bond and two from the lone pair on B.
www.elsolucionario.net
358
STUDENT'S SOLUTIONS MANUAL
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 nonreactive 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 nonspherical. 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 crosssection 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.
Solutions to exercises E18.1(b)
A molecule that has a center of symmetry cannot be polar. S03(D3h) and XeF4(D4h) cannot be polar.
ISF41 (seesaw, 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
www.elsolucionario.net
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 JLy =
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 JL = (JL~
+ JL;') 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 JLy = tan  I  143 pm e = 1194.0° 1from the xaxis (i.e. 14.0° below the JLx
572pm e
negative xax 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 yintercept
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 JL = (4.275 x 10 29 C m) x (5.72 x 10 3 ) 1/ 2 = 13.23 x 10 30 C m 1
and
www.elsolucionario.net
360
E18.5(b)
STUDENT'S SOLUTIONS MANUAL
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 10.726
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 =   , IC
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
X
gm  3)
(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
(11;1)
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
=~
www.elsolucionario.net

I)
1.6222 +2
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 105 m 3 molI) (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
Solutions to problems Solutions to numerical problems P18.1
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 ml) x r3
l.ll x 10 19 Vm I
l.ll x 108 Vrn I
(rlm)3
(rlnm)3
www.elsolucionario.net
362
STUDENT'S SO LUTIONS MANUAL
(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 (frl) (M) P fr + 2
411 , NAJL 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 molI) 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
www.elsolucionario.net
X
103 . It follows that
MOLECULAR INTERACTIONS
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 molI 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, dipolefree, 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 molI )
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 molI
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 highfrequency 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  ' 
&,
= 11 .58 D I.
p
www.elsolucionario.net
364
STUDENT'S SOLUTIONS MANUAL
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 molI x 273 K 1.00 atm X
4 3 I x 10 cm mol .
= 2.24
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 highfrequency contribution to Pm remains the same at 292.2 K then we have
N fL2
= Pm  Pm' = (57.57  5.66) cm 3 molI
_A_
9t:okT
Solving for fL we have
The factor
(
Therefore fL
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
(~I)
1(2
x (292 .2 K) 1/2 x (5.191 x 10 5 ) 1/2(m 3 / mo!) 1/2
= 11.58 D I.
The agreement is exact'
www.elsolucionario.net
MOLECULAR INTERACTIONS
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 LennardJones
potential is the usual LJ 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 LennardJones 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.
www.elsolucionario.net
366
STUDENT'S SOLUTIONS MANUAL
Solutions to theoretical problems P18.11
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 Ix
[(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
www.elsolucionario.net
=
MOLECULAR INTERACTIONS
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'
= rr2arcsin( 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
:::
~R2 + R2e v/ v', which is valid for all v.)
= ~mv2 , and so
= rr  2 arcsin (1 + 2 e~(E/E*)1/2 ) with E* =
~mv*2. This function is plotted as curve b
in Fig. 18.4.
www.elsolucionario.net
368
STUDENT'S SOLUTIONS MANUAL
Solutions to applications P18.19
(a) The energy of induceddipoleinduceddipole 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 JeyI) 4(1.0 x 10 9 m )6
= 6.6 x
?3
10  J
or l39 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 transNmethylacetamide is iL = (3.092 D) x (3 .336 X 10 30 C m D I) =
I 1.03
X
10 29 C m I
(semiempirical, 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
=
iL liL"zf (e)
4rreor
3
where/eel
= 1
3 cos
2
e
www.elsolucionario.net
MOLECULAR INTERACTIONS
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 molI) = 21 Jmol I = 2.1 x 10 2 kJmol 
l.
Thus, dipoledipo1e 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
www.elsolucionario.net
Materials 1: macromolecules and aggregates
19
Answers to discussion questions 019.1
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. Zaverage molar mass is defined through the formu la (eqn 19.4)
LN;M; Mz
=
2'
LN;M;
The Zaverage 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 stretchedout 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).
www.elsolucionario.net
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 nonpolar 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 waterinsoluble, fi lmforming 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. Selfassembled monolayers (SAMs) do not require assembl y by mechanical compression. SAMs form from charged materials that have adsorptiondesorption properties that promote selfassembly 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.
www.elsolucionario.net
372
STUDENT'S SOLUTIONS MANUAL Substrate
Figure 19.1(a)
selfassembly ~

F;oo",19.1(b)
Both methods yield wellorganized monolayers but LB films upon water provide better organizational control than is possible with spontaneous selfassembled films. However, not requiring mechanical compression, SAMs are much more versatile. The strong bonding of SAMs gives longlasting, stable films in contrast to the less stable van der Waals LB films.
Solutions to exercises E19.1(b)
The numberaverage molar mass is (eqn 19.1)
Mn =
~ LN;M; =
[3 x (62)
+ 2 ~ (78)] kg mol
I
I
= 68 kg mol  I
I
The massaverage 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 II  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 numberaverage molar mass, so
(m l / MI)MI (ml / Md
+ (m2/ M2) M2 + (m2/ M 2)
www.elsolucionario.net
MATERIALS 1: MACROMOLECULES AND AGGREGATES
373
(b) Lightscattering gives the massaverage molar mass, so
Mw = mIMI +m2 M 2 = (25) x (22) +(75) x (22/3) kgmol I =lllkg molII 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 sI[CRC Handbook] I 3 r = 4rr x (4.5 x 1O 9 m)3 x 1.00 x 10 kgm sI = 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
www.elsolucionario.net
374
STUDENT'S SOLUTIONS MANUAL
where we have assumed the data refer to aqueous solution at 298 K. (7.46 x 1O 13 s) x (8.314SJK 1 molI) x (298K)

~~~~~~~~~~ [I  (lOOOkgm 3 ) x (8.01 x 1O4 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)
1O6
~~~~
1O4
m s11
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 molI) 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 weightaverage 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) . .
www.elsolucionario.net
MATERIALS 1: MACROMOLECULES AND AGGREGATES
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 CC 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
Solutions to problems Solutions to numerical problems P19.1
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 leastsquares 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 minI ) 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
www.elsolucionario.net
376
STUDENT'S SOLUTIONS MANUAL
1.64
c ..... , ... ... , ... .
Eu
.E ~
1.62
1.60
o
40
20
60
I/ min
Figure 19.2
We see that the yintercept of a plot of the righthand side against c, extrapolated to c begin by constructing the following table using 1)0 = 0.985 g m I sI.
1)/1)0 c
I)
3
/ (dm gI)
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 leastsquares 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
www.elsolucionario.net
= 10.0716 dm3 gI
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
co (
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 yintercept 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 gI)
=
(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 MarkKuhnHouwinkSakurada equation [19.25] is
As the constant a may be nonintegral 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·
www.elsolucionario.net
378
STUDENT'S SOLUTIONS MANUAL
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 yintercept 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 solventdependent. 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 leastsquares 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 kgI and hence
(8 .314 J K  1 molI) x (298 K) _ (0.798 x 10 3 kg m 3 ) x (9.81 m s2) x (2 .043 x 10 3 m4 kg I)
www.elsolucionario.net
=
1
k II 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.
Solutions to theoretical problems P19.11
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(ybenzylLglutamate) is rodlike whereas polystyrene is a random coil (in butanol ).
P19.13
Call the constant of proportionality K , and evaluate it by requiring that L etMM = (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
www.elsolucionario.net
380
STUDENT'S SOLUTIONS MANUAL
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(MM)
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.
www.elsolucionario.net
MATERIALS 1: MACROMOLECULES AND AGGREGATES
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)dr
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
www.elsolucionario.net
382
STUDENT'S SOLUTIONS MANUAL
and
= (~) 1/2 x (3 Vs M) 1/ 3
R
5
g
4rrNA
( ~ ) 1/2 x
=
(3 vs/ em 3 gI) x em 3 gI 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 gI) 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 gI ,
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
www.elsolucionario.net
MATERIALS 1: MACROMOLECULES AND AGGREGATES
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
=1aT=O
.
Thus the internal energy is independent of the extension. Therefore
t
= aT =
T(~) aT = IT (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 (nonextended) state.
Solutions to applications P19.23
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.
www.elsolucionario.net
384
STUDENT'S SOLUTIONS MANUAL
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 molI) x (14.0 x
x (6.022 x
10
1023 molI) 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 7moldm3 1.07 x 107 g mol I = 9.35 x 1O 4 molm 3
and
nn°
no
For Hb, [J] = and P19.25
nn°
n0
= (28 m3 mol I) x (9.35 10 g dm 3
66.5 x 103 g mo]
X
10 mol m3) = 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 molI) = 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 ¢
= ReIo2, 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)
www.elsolucionario.net
MATERIALS 1: MACROMOLECULES AND AGGREGATES
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.
www.elsolucionario.net
386
P19.27
STUDENT'S SOLUTIONS MANUAL
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 SI) . 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 xintercept 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 bestfit straight line. The bestfit 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·
www.elsolucionario.net
MATERIALS 1: MACROMOLECULES AND AGGREGATES
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
0.8
f
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 Gc base pairs and at larger salt concentrations. Tm increases with the number of Gc 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 Gc pair. Low salt concentrations destabilize the double helix by inadequately contributing to the attractive forces between the solution and the sugarphosphate 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 molI = 104 g moli . 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 Ibutyl initiators) being incorporated into the polymer molecules . The most intense peak has a molar mass equal to that of n repeating groups plus
www.elsolucionario.net
388
STUDENT'S SOLUTIONS MANUAL
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 tbutyl groups, then
= 2M(tbutyl) = 2(4
M(terminal) and n
P19.35
=
x 12 + 9) g mol 1
M(peak)  M(Ag+)  M(terrninal) M(repeat)
=
=
114 g molI
25598  108  114 104
= 12441 .
The procedure is that described in Problem 19.7. The data are fitted to the MarkKuhnHouwinkSakurada equation. [1)]
= KM".,
[19.25].
The values obtained for the parameters are K
= 12.38
X
10 3 cm 3 gI 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
rrorbital interactions between heterocyclic rings.
www.elsolucionario.net
20
Materials 2: the solid state
Answers to discussion questions 020.1
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 closepacked (ccp) and hexagonal closepacked (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 bodycentered cubic (bcc) arrangement, which is not too much different from the closepacked 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 closepacked 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 Xrays 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.
www.elsolucionario.net
390
020.9
STUDENT'S SOLUTI ONS MANUAL
The FenniDirac 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 manyelectron 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 + /ldn for a onecomponent system. This may also be written dU = pdV + TdS + /ldN, and this /l is the chemical potential per particle that appears in the FD 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 FD 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 FD 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.
Solutions to exercises E20.1(b)
(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
www.elsolucionario.net
MATERIALS 2: THE SOLID STATE
E20.4(b)
391
The Bragg law is /1A
= 2d sine
Assuming the angle given is for a firstorder 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 facecentered 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
www.elsolucionario.net
392
E20.8(b)
STUDENT'S SOLUTIONS MANUAL
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 molI
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 facecentered cubic cell and would
be absent from a bodycentered 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 = sinI elOO
(~ (h 2 + k Z + P)I / Z)
= sin I (0.176(h 2 + k Z + p)I /Z)
= sinI (0.176(1 2 +0+0)1 /2) = 10.1° (checks)
ello = sinI (0.176(12 + 12 +O)I / Z) = 14.4° (checks)
ezoo = sin  I (0.176(2 2 + 0 + O)I /Z) = 20.6° (checks)
www.elsolucionario.net
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
A sin  \ 2d
e=
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 bodycentered 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.
www.elsolucionario.net
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 CH 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 22.5122.53 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 ratedetermining 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 .
Solutions to exercises E22.1(b)
E22.2(b)
v
= _ d[A] = _~ d[8] = d[C] = ~ d[D] = l.00moldm 3 s1 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 sI I
dt
Rate of consumption of A = 10.33 mol dm  3 S I I
www.elsolucionario.net
2 dt
dt
444
E22.3(b)
STUDENT'S SOLUTIONS MANUAL
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)
h
(I  0.100)
= 12 00 I
.
The halflife changes with concentration, so we know the reaction order is not 1. That the halflife increases with decreasing concentration indicates a reaction order < 1. Inspection of the data shows the halflife 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)
In (9.71 /7 .67) In 1  0.200
and
~ P2
= 55 .5 kPa = 1.92 28.9kPa
@] .
The rate law is 1 d[A] v =    =k[A]
2 dt
www.elsolucionario.net
THE RATES OF CHEMICAL REACTIO NS
445
The halflife 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 halflife 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 SI) 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 halflife 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=~
www.elsolucionario.net
446
STUDENT'S SOLUTIONS MANUAL
The halflife 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 secondorder constant are
dim ofv (dim of [A))2
amount x length 3 x time I
  
"'=
(amount x length 3 )2
= length 3 x
amount I x timeI
I
In molecule, m, s units, the units of k are m3 moleculeI s I
I
The dimensions of a thirdorder rate constant are dim of v (dim of [A])3
amount x length 3 x time I



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 secondorder 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 thirdorder rate constant in pressure units are dim of v (dim of p)3
press ure x timeI
=: ,:;3
(pressure)
= pressure
2.
x tIme
_I
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
www.elsolucionario.net
I
THE RATES OF CHEMICAL REACTIONS
447
Solving for [C] yields. after some rearranging [A]o[B]o{exp[kt([B]o  2[A]0)]  I} [C] = [B]'oexp=[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)
03 (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 sI 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 nthorder 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
ln Let tl /3 be the time at which [A]
d
(~ [A]O)III  [A]b II
[A]b  "[(Vn  I]
In
1n
= ~,~
so kt1 /3
an
= [A]0 / 3.
tl / 3 =
3" 1  I [A]In 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
www.elsolucionario.net
448
STUDENT'S SOLUTIONS MANUAL
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 sI
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 SI
= A exp (
Ea
(8.31451K I molI) 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 KI 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 297K31OK
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 sI
I 3 9.9 x 10 J mol) I (8.3145JKI mol ) x (297K)
I
E22.15(b) (a) Assuming that the ratedetermining step is the scission of a CH bond, the ratio of rate constants
for the tritiated versus protonated reactant should be [22.53 with hciJ
www.elsolucionario.net
= fiw = Ii(kj /1) 1/ 2]
THE RATES OF CHEMICAL REACTIONS
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 CI 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
www.elsolucionario.net
X
10 4 sI
)1
450
STUDENT'S SOLUTIONS MANUAL
Solutions to problems Solutions to numerical problems P22.1
A simple but practical approach is to make an initial guess at the order by observing whether the halflife of the reaction appears to depend on concentration. If it does not, the reaction is firstorder; if it does, it may be secondorder. Examination of the data shows that the first halflife 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 secondorder and confirm by plotting I/[Al against time. If the reaction is secondorder, 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 / secondorder /. 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)
www.elsolucionario.net
THE RATES OF CHEMICAL REACTIONS
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 halfLife is independent of concentration and that the reaction is therefore 1firstorder 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 leastsquares 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 twopoint method, that is,
R In (576/ 2.46)
41
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 molI I. P22.7
The data for this experiment do not extend much beyond one halflife. Therefore the halflife 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 threequarters lives will work. For a firstorder 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
www.elsolucionario.net
= k'
452
STUDENT'S SOLUTIONS MANUAL
T hus the threequarters life (or any given frac tional life) is also independent of concentration for a firstorder reaction. Examination of the data shows that the first threequarters 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 firstorder and the rate constant is approximately
k
0.288 0.288 ~ t3/4 80 min
= 
=
36 x 10 3 minI. .
A leastsquares fit of the data to the firstorde r integrated rate law [22.12b] gives the slightly more accurate result, k = 13.65 x 10 3 min  I I. The halflife 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 halflife. 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 halflife; therefore, we cannot see whether the halflife 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 halflife 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 firstorder but secondorder. Draw up the following table.
1
1
rims
[CIO]/(!Lmol dm 3)
( II[CIO])/(dm3!Lmoll)
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
www.elsolucionario.net
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 halflife depends on the initial concentration (eqn 22.16): '1 / 2
= __1  = k[ClO]o
I (2.37 x 10 7 dm 3 mol I sI )(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 firstorder in propene and thirdorder in HC!.
www.elsolucionario.net
= 3 when
454 P22.13
STUDENT'S SOLUTIONS MANUAL
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 thirdorder in HCI and firstorder 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)
www.elsolucionario.net
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.
Solutions to theoretical problems P22.21
A ;=" B d[A] dt
= k[A] + k'[B]
At all times, [Al Therefore, [B]
d~~]
+ [Bl
and
= [Alo
=
k' [Bl
+ k[Al .
+ [Blo·
= [Ala + [Blo 
= k [A ]
d[B] dt
[AJ.
+ k' {[Alo + [Blo 
[All = (k
+ k' )[Al + k'( [A]o + [B]o) .
To solve, one must integrate
f
(k
d[Al k' ([Alo
+ k' )[A] 
+ [Blo) = 
f
dt.
www.elsolucionario.net
456
STUDENT'S SOLUTIONS MANUAL
· . [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.
 2x) 2 x
(I
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
(Ao  2x)2 x (Ao  x)
ex (Ao  2x)2
+ _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. www.elsolucionario.net
THE RATES OF CHEMICAL REACTIONS
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) (Ao2x)2 Ao2x
10
+ (i/A6))
I
(i/Ao)
dx
Aox 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~lOr1  c~lor1
(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=
www.elsolucionario.net
[Alo  [Blo 2
457
458
STUDENT'S SOLUTIONS MANUAL
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
> 1, we conclude that benzoate ion is an I un competitive Iinhibitor of carboxypeptidase.
www.elsolucionario.net
482
STUDENT'S SOLUTIONS MANUAL
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 ( 10.010
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 zerocurrent potential for
the reduction of the metal ion is more negative than the reduction potential of the halfreaction 4H+ + 0 2 + 4e + 2H20, E" = 1.23 V. The zerocurrent 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 halfreactions 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)
www.elsolucionario.net
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)
= (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 molI)
I0.28 mg cm 2 d I I·
www.elsolucionario.net
.
539