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Student Solutions Manual to accompany
Physical Chemistry Fifth Edition
Ira N. Levine Chemistry Department Brooklyn College City University of New York Brooklyn, New York
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McGrawHill Higher Education $2 A Division of The McGrawHill Companies Student Solutions Manual to accompany
PHYSICAL CHEMISTRY, FIFTH EDITION IRA N. LEVINE Published by McGrawHill Higher Education, an imprint of The McGrawHill Companies, Inc., 1221 Avenue of the Americas, New York, NY 10020. Copyright © The McGrawHill Companies, Inc., 2002, 1995, 1988, 1983, 1978. All rights reserved. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of The McGrawHill Companies, Inc., including, but not limited to, network or other electronic storage or transmission, or broadcast for distance learning.
This book is printed on acidfree paper. 1234567890 BKM BKM 0321 ISBN 0072393602 www.mhhe.com
To the Student The purpose of this solutions manual is to help you learn physical chemistry. This purpose will be defeated if you use this manual to avoid working homework problems. You cannot learn how to play the guitar solely by reading books titled “How to Play the Guitar” or by watching other people play the guitar. Rather, most of your time is best spent actually practicing the guitar. Likewise, you won’t learn how to solve physical chemistry problems solely by reading the solutions in this manual.
Rather, most of your time is best spent actually working problems. Do not look up the solution to a problem until you have made a substantial effort to work the problem on your own. When you work a problem, you learn a lot more than when you only read the solution. You can learn a lot by working on a problem even if you don’t succeed in solving it. True learning requires active participation on your part. After you have looked up the solution to a problem you could not solve, close the solutions manual and work through the problem on your
own. Use the solutions manual as an incentive to work problems, not as a way to avoid working problems. Ira N. Levine
INLevine @brooklyn.cuny.edu
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Before using this manual, please read the preceding note To the Student.
Chapter  1.1
(a)
F.
(b)T.
(c)T.
(@)F.
(e)F. (A mixture of ice and liquid water has
one substance.)
(c) open; nonisolated.
1.2
(a)
Closed, nonisolated;
1.3
(a)
Three.
1.4
So that pressure or composition differences between systems A and B won't cause changes in the properties of A and B. Such changes can then result only from a temperature difference between A and B.
15
(a)
(b) Three; solid AgBr; solid AgCl, and the solution.
g
19.3
(b) open, nonisolated;
Ikg
(100 cm)’
cm? 1000g_
= 19300
(1m)’
troy oz 480 grains
lpound
kg/m’
e/
453.59¢g
m = pV= (19.3 g/cm?)(10° cm’) = 1.93 x 10’ g
(1.93 x 10’)($9.65/g) = $1.86 x 10° 1.6
(a)
ee
(b) Pe
(cha
(dale
1.7
(a)
32.0.
1.8
100 g of solution contains 12.0 g of HCI and 88.0 g of water.
(b) 32.0 amu.
(c) 32.0.
(12.0 g)(1 mol/36.46 g) = 0.329 mol;
4.885 mol.
LS
(a)
(d) 32.0 g/mol.
ny 9= (88.0 g)(1 mol/18.015 g) =
xXyc) = 0.329/(0.329 + 4.885) = 0.0631;
12.0 g/mol ee
ara 6.022 x10*°
atoms/mol
=
nyo =
1.99x10
293 ~ g/atom
Xy,9 =1 Xa = 0.9369.
1.10
18.0 g/mol
(b)
0.1480 g = nye(4.003 g/mol) + (0.01456 mol — nye)(20.18 g/mol). Nye = 0.00902 mol, ne = 9.00554 mol Xte = 0.00902/0.01456 = 0.619, mute = 0.0361 g
1.30
the The downward force of the atmosphere on the earth’s surface equals , weight W of the atmosphere, so P = W/A = mg/A and m = AP/g = 4nr’Ple 5
where ris the earth’s radius and P = 1 atm = 101325 N/m”. Thus oa 47(6.37 (6.37 10° ues 2 Nina) Z eens te X10° m) m)?(1. 9.807 m/s 1.31
(a)
Multiplication of both sides of the equation by 10° bar gives
P = 6.4 x 10° bar. (b) 1.32
460K.
(c) 1.2x 10° bar. (d) 312K.
Take one liter of gas. This volume has m = 1.185
g= my, +mMo,.
We have not = PV/RT =
(1.000 atm)(1000 cm?*)/(82.06 cm>atm/molK)(298.1 K) = 0.04087 mol.
Not = Ny, +No, =My,/Myn, +Mo,/Mo, = my, /(28.01 g/mol) + (1.185 g— my, /(32.00 g/mol) = 0.0487 mol. Solving, we get my, = 0.862 g, hence mo, = 0.323 g. Then ny, = 0.0308 mol and Bis 0.0101 mol; x9, = 0.0101/0.0409 = 0.247.
1°33
(a)>
Use of P=
P gives Py, = 0.78(1.00 atm) = 0.78 atm,
Po, =(0.21 atm,
(b) V=3000ft*.
Pa,=0.0093 atm,
Foo, = 0.0003 atm.
1 ft=12in. = 12 x 2.54 cm = 30.48 cm.
V = (3000 ft*®)(30.48 cm)*/ft® = 8.5 x 10’ cm®. mot = PV/RT = [(740/760) atm](8.5 x 10’ cm?)/[(82.06 cm°atm/molK)(293 K)] =
3.44x 10° mol.
ny, = Xn, Mor = 0.78(3.44 x 10° mol) = 2.63 x 10° mol.
my, = (2680 mol)(28.0 g/mol) = 75 kg.
Similarly, mo, = 23 kg,
Mar = 1.5 ke, Mco, = 4.5 x 1; g. We have P = mto/V=
(99.; kg)/(8.5 x 10’ cm*) = 0.00117 g/cm’. 1.34
f(x) is zero at the two points where fis a local minimum and where fis a
local maximum. maximum
135
f(x) is negative for the portion of the curve between the
and the minimum.
dy/dx= 2x +1 Atx
=) the slope 18 2¢1) aie
1.36
(a) 6x2e* — 6x3e**; (b) — 30xe™2" ; (c) I/x (not 1/2x); (d) 1/(1 — x); (e) I(x + 1) —x/(x + 1)? = W(x +1); (f) 2e 7/1 — &™); (g) 6 sin 3x cos 3x.
1.37
(a) (b)
y=2/(1—x) and dy/dx = 2(1 — x)’. d(xe*)/dx = 2xe* + 3x°e™; Vd? = 2e* + 6xe™* + 6xe™* + 9x76 = 2e* + 12xe* + 9x7e™*.
B(x
(c) Reminder.
1.38
(a)
dy = (10x —3 — 2/x’) dx. Work the problems before looking up their solutions.
x x
0.01 0:955m
0.1 0.794
0.00001 0.0001 0.001 ¢0.993.1495.0:999 Ise 0.99988
This indicates (but does not prove) that the limit is 1.
(by (xy
TOO 271?
OG 2720
1 Oy eel Oe nO oleae tote 2 1o27)
Hie: 2.1 L826
This suggests that the limit is e = 2.7182818....
1.39
(a)
Results on acalculator with 8digit display and 11 internal digits are: Ay/Ax = 277, 223.4, 218.88, 218.44, 218.398, 218.393, 218.4 for Ax = LOR elOgu Og al Ose 10°, 10°, 10°’, respectively. The best estimate is pac Sie\ey
(b)
dy/dx = 2xe* and at x = 2, dy/dx = 218.3926. A BASIC program for part (a) is
5 10 20 30 40 1.40
CX =0.1 FORN=1T07 X=2 CY = EXP((X + CX)*2)  EXP(X*2)_ R=CY/CX
(a) axcos(axy);
(b) —2byz sin (by’z);
(d) 0; (e) ae? /y(e™ + he
50 PRINT “DELTAX=";CX; “ RATIO=";R 60 CX =CX/10 70 NEXT N 80 END (c) (x“ly’)e,
1.41
(a) nR/P;
(b) 2P/nRT?.
1.42
Equation (1.30) gives dz = 2axy’ dx + 3ax°y? dy.
1.43
Partial differentiation of z = x°/y° gives
U5x OE eee ee
Oep ~3x40
dee a
az _9(_3x?)__ 15x! aan
hal
kl
Spaaniyie
ne
Sales
dz toes 7
y* ’ dydx
15a
dz
y*
dx dy
ody y?
P is a function of n, T, and V, so dP = (0P/dn)ry dn + (OP/0T)y,n aT +
1.44
(OP/0V)rn AV. Partial differentiation of P = nRT/V gives (dP/dn)r,v = RT/V = P/n (where PV = nRT was used), (OP/0T)y,n = nR/V = P/T, and (OP/0V)rn = _nRT/V’ = —P/V. Substitution into the above equation for dP gives the desired result. (Note that from P = nRT/V, we have
InP=Inn+InR+1nT—In V, from which dln P=dInn+dInT—d\n V follows at once.)
(b)
Approximating small changes by infinitesimal changes, we have
dn=An=0,
dt=AT=1.00K,
dV=AV=50cm’.
The original pressure is P = nRT/V = 0.8206 atm. Then AP = dP =
(0.8206 atm)[0 + (1.00 K)/(300 K) — (50 cm*)/(30000 cm*)] = 0.00137 atm.
(c)
The accurate final pressure is (1.0000 mol)(82.06 cm>atm/molK) x (301.00 K)/(30050 cm’) = 0.82197 atm. The accurate AP is 0.82197 atm — 0.8206 atm = 0.00137 atm.
1.45
1.000 bar = 750 torr = (750 torr)(1 atm/760 torr) = 0.987 atm.
Vm = V/n = (nRT/P)/n = RT/P = (82.06 cm*atm/molK)(293.1 K)/(0.987 atm) = 2.44 x 10* cm*/mol. 1.46
(a)
Division by n gives (P + a/V,~. (Vm — b) = RT.
(b)
The units of b are the same as those of Vm, namely, cm?/mol.
P and a/V,. have the same units, so the units of a are bar  cm®/mol’.
1.47
O = (1/Vn)(OVn/OT)p = (1/Vm)(c2 + 2¢3T— csP), where Vm is given by (1.40). K = — (1/Vn)(OVn/OP)r = —A/Vin)(—Ca — C57) = (c4 t+ CsT)/(C) + C2T + Gi — c4P —csPT).
1.48
(a)
p = m/V = (m/n)I(V/n) = M/Vm, 80 Vm = M/p = (18.0153 g/mol)/(0.98804 g/cm*) = 18.233 cm*/mol.
(b)
K = —(1/Vin)(OVn/OP)r and dVin/ Vm = —K AP at constant T. Integration gives In (Vin2/Vin1)= —K(P2 os Prat constant T.
K = (4.4 x 107! Pa ')(101325 Pa/1 atm) = 4.45 x 10° atm”! and In[Vm2/(18.233 cm?/mol)] = (4.46 x 10° atm™')(100 atm — 1 atm) = 0.0044, so Vm2/(18.233 cm*/mol) = e°° = 0.9956 and Vinz = 18.15 cm?/mol.
1.49
(a)
At constant P, the equation PVm = RT gives Vm = aT, where a = R/P is a positive constant. The isobars on a Vp, vs. T diagram are straight lines that start at the origin and have positive slopes. (As P increases, the slope decreases.)
(b)
For Vm constant, PVm = RT gives P = bT, where b = R/V» is constant. The
isochores on a P vs. T diagram are straight lines that start at the origin and have positive slopes.
Partial differentiation of
1.50
V=nRT(1 + aP)/P gives (OV/0T)p»=
nR(1 + aP)/P. The equation of state gives nR(1 + aP) = PV/T, so
(OV/OT)pn = V/T. Then @ = (1/V)(OV/0T)p.» = IT. Partial differentiation of V = nRT(1/P + a) gives (OV/0P)rn =
_nRTIP? = [PV/(1 + aP)\/P° =V/P(1 + aP), where the equation of state was used. Then k = —(1/V)(OV/0P)7 = 1/P(. + aP).
(b)
Solving the equation of state for P, we get P = nRT/(V— ank7T), partial differentiation gives (0P/07T)y = nR/(V — anRT) + an’R°TI(V — anRTY = PIT + aP’/T, where P = nRT/(V— anRT) was used. From (a), we have O/k
= P(1 + aP\/T = PIT + aP”/T, which agrees with Eq. (1.45). 1.51
For small AT, we have
Folate Wop VAL),
Since @ is an intensive property, we can take any quantity of water. For ! g, the equation V = m/p gives V = 1.002965 cm? at 25°C,  atm and
V = 1.003227 cm’ at 26°C, 1 atm. 1 1.003227 cm? 1.002965cm* _ 9 gqgn6 K' Hence a = 26°C 25°C 1.003 cm* Similarly, k = (1/V)(OV/0P)7 = —(/V)(AV/AP)r. At 25°C and 2 atm, we calculate V = 1.002916 cm’ for 1 g of water. 3 3 1 : 1.002916cm* —1.002965cm™ _ 4 94195 atm7!
Thus
k=—
1.003 cm”
2 atm —latm
_a_ ae 26x10*K ae Eq. (1.45) gives (22) ane K49 =1 kgm s?m?
(ec)
1 L= 10° cm = 10° (107 my = 10° m’
(Ameen 2.4
(ene
(f) kg.
kgm’ s~
=lkgm's~
ekeamt ce
ts
2
=mg Ax = (0.155 kg) (9.81 m/s )(10.0 m) = 15.2J mg dx (a) w=J? Fdxy=J;
(hime (c)
SAK  Ky Kpa hy 152 J. 4mv* = K and v= (2K/m)'” = [2(15.2 JO.155 kg)]'” = 14.0 m/s, since 
J= 1 kg m’/s*.
jee)
P = F/A = me/A = (0.102 kg)(9.81 m/s*)/(1.00 m*) = 1.00 N/m” = 1.00 Pa.
2.6
(ayer
Dad
(a)
(Dye
(chal
(yee
ey
oeDaal ge(2 ar,
area = length x height = (V2 — V\)P) = (5000 — 2000)cm°(0.230 atm) = 690 cm? atm. Wyey = —area = —(690 cm* atm)(8.314 J/82.06 cm” atm) =
69.9 J.
(b) 2.8
w.., =]7 P dV =P(V, V,)=ete.
w,, =]?
PdV =P(V, V,) = (275/760) atmx(875 — 385) cm" =
~177cm* atm (8.314 J/82.06 cm? atm)=—18.0J.
29
(a)
The area under the curve is the sum of the areas of a rectangle and a right triangle. The rectangle’s area is (V2 — V;)P2 = (2000 — 500)cm°(1.00 atm)
= 1500 cm’ atm.
The triangle’s area is Y2(base)(altitude) = “(V2 — V,)(P, — P2) =
(2000 — 500)cm*(3.00 — 1.00)atm = 1500 cm* atm.
Thus Wey = —3000 cm’ atm (8.314 J/82.06 cm*atm) = 304 J. (b)
Replacement of y and x with P and V in the straightline equation gives (P — Pi /(V— Vi) = (P2 — Pi)(V2— Vi). w= ; PdV= al Ae ee
a) Vers Vi) aaa
=
= P.(V; =V, IP, — PRVWVo nV MC2Vs ViV2)— (AV, V =P(V, —V, 4 (8 = 13 )\V5 V ) = as inia),
2.10
(a)
(b)
4
P
1
o>
2
E
1
A
2 »>
V
pagOI
V
Neglecting the dependence of specific heat on 7, we equate the heat gained by the water to that lost by the metal. The heat gained by the H20 is (24.0 g)(1.00 cal/g°C)(10.0°C) = 240 cal. Thus 240 cal = (45.0 g)cmetai(70.0 — 20.0)°C and Cmetai = 0.107 cal/g°C.
es.) iile mG) eka (Celie)
aly
2.12
(AlAl
213
Only (c).
2.14
cal 4.184J lday th tOnw (a) 220010: foal adil talaayes 07 Vee day
lIcal
24hr 3600s
(b) (6 x 10°)(107 J/s)(3600 s/hr)(24 hr/day)(365 days/yr) = 2 x fo 215
Since the process is cyclic, AU = 0. Hence g = w = 145 J.
2.16
(a) The bulk kinetic energy acquired by the 167ft fall is converted into internal energy, thereby warming the water by AT. The bulk kinetic energy equals the potentialenergy decrease mg Ah. The AU for a temperature increase of AT can be set equal to the heat g = mcp AT that would be needed to increase the temperature by AT, since the expansion work 1s negligible. Therefore mg Ah = mcp AT and
ap a SAN _ (9.80 m/s" (167 x12x2.54x10° m)_ Ical = 4.184] (1.00 cal/g°C)(10* g/kg) Cp
2s
(b) mg Ah = (2.55 x 10° cm*)(1.00 g/em*)(1 kg/10° g)(9.80 m/s”)(50.9 m) = Peg 10) Dali
We have 0 = AU, + AU2 = q, + Ww) + q2 + W2 = qi + q2 (since the wall is ngid);
therefore g2 = —q\.
2.18
This notation might seem to imply that g and w are state functions, which is
not so. There is no such thing as the change in heat for a system. There 1s only an amount of heat transfer for a process.
2319
Cool the water to some temperature below 25°C and then do enough stirring work to raise its T to 30°C.
2.20
V = Yk = ¥2(125 N/m)(0.100 m)° = 0.625 J = 0.149 cal. AY = 0.149 cal
(71;C)
+ m2C2)AT and AT =
(1.00 cal/g  °C)(112.g) + (20 g)(0.30 cal/g  °C)
AT = 0.00126 °C and the final temperature is 18.001°C Jaga
(a)
OO
OE
ett Che
Cbaeh
hue
Vinee
dE xy + dKpis. + mg dh + 0, so dE sys = —mg dh — dKpis.
(b)
dE xy = dq + dWimey = 0 + AWirey, SO AWirrey = dE sys. = —mg dh — dKpist. But
mg dh = (mg/A)A dh, where A is the piston’s area. Since mg/A = Pex, and A dh = dV, we have mg dh = Pex, dV and dWimey = —Pext dV — dK
pist
2.22
2
Eq. (2.33) gives Winey = — [°Pay AV — AK pise = —Pa [dV 0 = —P,,,(V. — V,) = —(0.500 bar)(4.00 dm*) = —2.00 dm? bar.
1 dm? = 1000 cm’ and 1 bar = 750 torr = (750/760) atm = 0.987 atm,
SO Wirev = 1974 cm? atm x (8.314 J/82.06 cm?atm) = —200 J. 2.23
(a) T; (b) F; (c) F.
2.24
All except force, mass, and pressure.
2.25
(a)
From the equation AH = qp.
(b)
It can mislead one into thinking that heat is a state function.
2.26
No. For example, in a cyclic process, AH is zero but g need not be zero, since q is not a state function.
2.27
AH = qp = 0 for the entire system. Since H is extensive,
H = H, + H2 and
AH = AH, + AH2 = qi + G2. Since AH = 0,7 q; + q2 = 0.
2.28
(a) T; (b) T.
2.29
(a)
Cpm= Cp/n and Cp = nCpm = (586 g/16.04 g mol!)(94.4 J/molK) = 3.45 kI/K.
(b)
(10.0 carat)(0.2 g/carat) = 2.00 g and Cp =nCpm=
(2.00 g/12.01 g mol7')(6.115 J/molK) = 1.018 J/K. cp = Cp/m = (1.018 J/K)/(2.00 g) = 0.509 J/gK.
2.30
vp = Vim=(mIVy! =p"! = (0.958 g/cm*)= 1.044 cm’/g.
2.31
(a) U;, (b) H.
2.32
uyr = (OT/OP)y,
AT = J; wyr aP, and AT ~ pyr AP for H constant. Thus AT =
(0.2 °C/bar)(—49 bar) = —10 °C. The final temperature is approximately 15°C.
17
2.33
Eq. (1.35) gives (QUm/OVm)r = (QU m/OP)H(OP/OVm)r. Partial differentiation ofP
= RT/Vm gives (OP/OVm)r = RT/V,, = —P°/RT. Hence, (QUm/OVm)r = —(QU p/OP)r P°URT.
2.34
(a)
(OUm/0Vm)r= (6.08 J/molatm)(1 atm)*/(82.06 cm°atm/molK)(301 K) =
(b)
2.46 x 107 J/cm’. Doubling P multiplies (0U,,,/0V,,)7 by 4 to give 9.84 x 10~ J/cm’.
(a)
Use of (1.34), (1.32), (2.64), and (2.53) gives
l= (= (= (=) = EE OP
(b)
),\0H
);\ OT
)p
(OH/0P),
ag (2.65) follows.
Partial differentiation of H = U + PV and use of (2.63), (1.35), (1.44), and (2.65) give
(=) (3) ‘i ) 4V (4 (=) “ret OPE
ony?
oP ye
OVE) COPD),
—Cp Wy, = Cy pw, Vk— PVk+ Vand the desired result follows.
Jashs)
(a)
py, = (OT/OV), =(QU/OV),/Cy.
(OU/OV), is intensive, since it is the
ratio of the changes in two extensive quantities. Cy is extensive. Therefore, doubling the size of the system at constant 7, P, and
composition will double Cy, will not affect (QU/dV), , and hence will cut u, inhalf. Therefore, 1, is neither intensive nor extensive, since it 1s not independent of the size of the system and 1s not equal to the sum of the p,’s of the parts of the system.
2.36
(ABE
2.37
(a)
aD) ar, (Cab
)elee(e) a
Since Tis constant, AU = 0 and AH = 0. (U and H of a perfect gas depend 2 2
on Tonly.) w= ff PdV = —nRT { V' dV =nRT In (V2/V,) =
—(5.00 mol) (8.314 J/molK)(300 K) In (1500/500) = 13.7 kJ. AU =q+w=0,sog =w = 13.7 kJ.
(b)
Since U and H are state functions, AU and AH are still zero. The work w 1S Zero.
T)=T, =300K. P>=nRT>/V> = (1.00 mol)(82.06 cm?atm/molK)(300 K)/(49200 cm’) =
2.38
0.500 atm. (An alternative solution uses P2V = P,V).)
(b)
Y= CemlCvm= (Cym+ R)/Cyvm = 2.5R/1.5R = 1.667. P, = nRT,/V, = 1.00 atm. For a reversible adiabatic process with Cy
constant, Eq. (2.77) gives P,V," = P2V2" and P2 = (V,/V>)‘P, =
(24.6 L/49.2 L)'! "(1.00 atm) = 0.315 atm.
T> = P2V>/nR = 189 K.
(c)
P/ atm 0.5
isotherm adiabat
(0)
V/L
0
2.39
(a)
24.6
49.2
q = O since the process is adiabatic.
Cy n= Com
R= 25Ris
I= Pi Vink =100 Ke
TT VilVoeei= (LOOK )(4,00)e t= 279 Ka dU = Cy dT and AU = Cy AT, since Cy is constant. Thus
AU = (0.0400 mol)2.5(8.314 J/molK)(119 K) = 98.9 J. AU=q+w=w=98.9J;_ AH = Cp AT = 138%]. (b)
From P)V," = P2V2", we get P\(nRT,/P\)' = P2(nRT2/P2)", which becomes
PIT) =P} 7," Hence (T1/T2)’ = (PoP)
and P2/P\ = (TWiT).
y = Cpm/Cym = 3.5R/2.5R = 1.40. P2/(1 atm) = (298 K/100 K)' 40 (2.98)>> = 0.0219, so Pz = 0.0219 atm = 2.22 kPa. 2.40
(a)
n=0.500 mol.
0.800 bar = 0.789 atm.
T> =P2V2/nR = 769.2 K.
7) = P)V\/nR = 384.6 K,
dU =CydT and AU = Cy AT
(0.500 mol)(1.5R)(384.6 K) = 2.40 kJ.
=nCy AT =
AH = Cp AT =
(0.500 mol)(2.5R)(384.6 K) =4.00 kJ. w=  Pdv =
~P(V> — V;) = (0.789 atm)(2000 cm>)(8.314 J/82.06 cm?atm) = 1.60kJ. gq =qp=AH =4.00kJ.
=
(b),
w=0 at constant VV.
Tp=PivVi/nR = 210, Keel
= 324k
AU= Cy ABR= 0.675 kI “AH =! Cp AT SailsKI
AU =9w = g;
SO. gd = 0.07 orn. 2.41 “ (a) Pe (Dy gleea(cjer.
2.42
(a) Process;
(b) system property;
(e) system property;
2.43
(d) PF; (e) EF:
C,,= dq ,/dT,,.
(f) system property;
2.44
(a)
(d) process;
(g) system property.
(a) dqpr > 0 and dT,, = 0, so Cp, = ©.
(c) dU =0 = dp, + dWpr; Hence C,, =.
(c) process,
ddpr = —dWpr;
(b) —c°, since dqp, < 0.
Wp, < 0 and dq,,>0.
dT,,=0.
(d) 0, since dqp, = 0.
Heat is required to melt the benzene, sog>0.
AH =q,>0. The
constantP work is w = —P AV; since the benzene expands on melting,
w O since the system contracts on melting.
(c)
gq =O for this adiabatic process. w is negative for an expansion. We have AU=q+w=w,so AU 0, since Cy > 0. From PV = nRT, it follows that AV > 0. Hence
AH = AU + A(PV) = AU + nR AT > 0.
20
w =—P AV 1/2" or 2" > 100. The minimum value of nis 7.
3.29
(a) q; (b) T.
3.30
In one minute of operation, w = (1000 x 10° J/s)(60 s) = 6 x 10'° J. Then e = 0.40 = wV/qu and qu = (6 x 10'° J)/0.40 = 15 x 10'° J in one minute. Hence qc 32
= 15x 10'°J6 x 10!° J=9 x 10'°J per minute. Use of g = mcp AT gives m = (9 x 10'” J)/(4.184 J/gK)(10 K) = 2 x 10” g. The density of water is 1 g/cm’, so 2 x 10° cm* = 2 million liters are used per minute.
3.31
(a)
q cannot be calculated, since g depends on the path and the path is not specified.
(b)
The pathdependent quantity w cannot be calculated.
(c)
dU = Or die=viGuadl
MAU —ancyedi=
na
bly dt =
na(T, — 7) + Yanb( Ty  i) = (4.00 mol)(25.0 J/molK)(500 — 300)K +
¥(4.00 mol)(0.0300 J/molK*)(500° — 300°)K? = 29.6 kJ. (d)
AH = AU + A(PV) = AU +nR AT = 29600 J + (4.00 mol)(8.314 J/molK)(200 K) = 36.3 kJ.
(e)
Brome 2.9) 7As
iF (Cy/T) dT + nR In (V2/V,). But Cy/T=
nCyw i = niall + 5), somsen
iF (a/T + b) dT + nR In (V2/V;) =
na n (T2/T) + nb(T2 — T\) + nR In [(nRT>/P2)/(nRT,/P)] = (4.00 mol)(25.0 J/molK) In (500/300) + (4.00 mol)(0.0300 J/molK’)(500 — 300)K + (4.00 mol)(8.314 J/molK) In [(500 K)(2.00 atm)/(300 K)(3.00 atm)] = 78.6 J/K.
BEEI
(a) Rev;
6yaKh)
(a)
Since U and S are extensive, the 10 g has the higher U and the higher S.
(b)
The vapor; the vapor.
(c)
The 40°C benzene; the 40°C benzene.
(d)
If the system at 300 and 310 K is adiabatically enclosed, it will spontaneously go to the state at 305 K. Since S increases in a spontaneous adiabatic process, the 305K system has the higher S. Since q = 0 and w is negligible for this process, the two systems have the same
(b) irrev;
(c) irrev;
(d) irrev;
i. (e)
Neither; the latm gas.
33
(e) irrev;
(f) irrev;
(g) rev.
3.34
PdV+VdP=d(PV),
VdV=d(V),
ddrev/T = dS, dqp = dH, and
dw,e/P = —dV, so b, c, d, f,h, andj = 0.
3555
(a)
3.36
The second law of thermodynamics is “hardly ever” violated.
3:37,
No. (A pilot plant using this method has been built.)
3.38
Suppose we could prepare a reservoir at Tc = 0 and could reduce the engine’s temperature to absolute zero. Then erey would become equal to 1. The Carnot
GP; k, Ot
(b)
Cp.m;
R
cycle would convert the heat gy completely into work. But this would violate the Kelvin—Planck statement of the second law. Hence we can’t achieve
absolute zero. (See also Sec. 5.11 in the text.)
3.39
(a)
AH = qr = 0.
(b)
Suppose the final state consisted of ice at 0°C, with no liquid present. A hypothetical path to attain this state is to warm the supercooled liquid from —10°C to 0°C and then to freeze all the liquid at 0°C. AH for warming the liquid is (1.00 cal/gK)(10.0 g)(10.0 K) = 100 cal and AH for freezing all the liquid is (79.7 cal/g)(10.0 g) = —797 cal. The overall AH is 697 cal, which is not 0. Hence the equilibrium state is not ice at 0°C. If the equilibrium state were all ice below 0°C, AH would be even more negative than —697 cal, and hence this is not the equilibrium state.
Therefore the equilibrium state must consist of ice and liquid water in equilibrium at 0°C. To satisfy the condition that AH = 0, the mass mice of ice produced must satisfy (79.7 cal/g)mice = —100 cal and mice = 1.25 g.
The mass of liquid remaining is 10.0 g — 1.25 g = 8.75 g.
(c)
A reversible path for the process is
liq. at 10°C —*> lig. at O°C —*9 mice + mig at O°C AS = AS, + AS, = (10.0 g)(1.00 cal/gK) In (273/263) — (79.7 cal/g)(1.25 g)/(273 K) = 0.373 cal/K — 0.365 cal/K = 0.008 cal/K, where (3.30) and (3.25) were used.
34
3.40
(a) J/K;
3.41
(a) False.
3.42
(a)
(b) J/molK;
(b) False.
(c) J; (d) Pa=N/m’*;
(c) True.
(d) False.
(e) no units;
(f) kg/mol.
(e) True.
False. (For example, a nonideal gas expanding into vacuum can undergo
a change in JT—the Joule experiment.)
(b)
rue:
(c)
False; S increases in an irreversible process in an isolated system.
(d)
False. For example, a Carnot cycle is reversible, has AV = 0 and w ¥ 0.
(e)
True, since S is a state function.
(f)
False, e.g., AS # 0 for (the irreversible) adiabatic expansion of an ideal gas into vacuum. False (unless T and P are constant).
False. (A counterexample is the melting of ice at 0°C.) False. (Reversibility and constant pressure are required.)
False. (A counterexample is the melting of ice at 0°C.) False. (Only the system has to return to its initial state.)
35
Chapter 4 te (e) be (ae
4.1
(a) ee(b) cE mca)
4.2
Processes a and b are reversible, isothermal, and isobaric and have AS = qplT.
(a)
AG = AH —T AS = qp — T(qgp!T) = 9. AA = AU T AS = qp + w  T(qp/T)
=w=J? PdV=PAV= —(1 atm)[(36.0 g)/(1.000 g/cm’) — (36.0 g)/(0.917 g/cm’)] =
(3.26 cm? atm)(8.314 J/82.06 cm*atm) = 0.330 J.
(b) (c)
As in (a), AG = 0 and AA = w =—P AV= —P(Vgas — Vig) = —PVeas = _nRT = —(0.50 mol)(8.314 J/molK)(353.2 K) = 1.47 kJ. Since the gas is perfect, the final T is 300K. AH = 0 and AU = 0, since U and H depend only on T for a perfect gas. From (3.29), AS = nR In (V2/V;) = (0.100 mol)(8.314 J/molK) In (6.00/2.00) = 0.913 J/K. AG=AHT AS =0  (300 K)(0.913 J/K) = —274 J. AS =274 J. Processes a and b are reversible and so have AA = AUT ASuniv = 0. For process c, there is no change in the surroundings, so
OS SOUT
NSS
D) BG oun (Clie @ pial:
4.3
(a) nye
4.4
Setting dU=0 in dU =T dS — P dV, we get
0=T dS, —PdV,,
so
(0S/O0V),, = (OS/0V), =PIT.
4.5
N=V,x=S,y=P, For dH = TdS + VdP = M dx + N dy, we have M =T, and (dM/dy), = (ON/0x), gives (OT/OP)s = (0V/0S)p. The equations in (4.45) are derived similarly from Eqs. (4.35) and (4.36).
4.6
(a)
Use of (QU/0V)7 = AT/x — P gives
(=) _ (3.0410 K~')(303.15 K) ee
ov
if
4.52x10~° atm!
36
—latm = 2040 atm
(b)
Use of tyr = ViaT— 1)/Cp gives pyr as 18.1cm?/mol
Oot nGlaKls
=
il
tLe tx
(1.987 cal)/(82.06 cm*atm) = 0.0221 K/atm 4.7
From (4.53), Cp.m— Cym = TVmoe/K. Vin = M/p = (119.4 g/mol)/(1.49 g/cm’) = 80.1 cm*/mol. TV m0C/« = (298 K)(80.1 cm?/mol)(0.00133 K7!)?/(9.8 x 107 atm7!) = (431 cm?atm/molK)(8.314 J/82.06 cm?atm) = 44 J/molK.
Then Cym = (116 — 44) J/molK = 72 J/molK. 4.8
Division of (4.30) by n gives (OH,,/0T)p = Cp.m. For (b), we use (4.48) divided by n; for (c), (4.47); for (d), (4.49)/n; for (e), (4.50)/n; for (f), (4.53)/n;
for (g), we use (0A/0V)7 = —P, which follows from (4.35). Substitution of numerical values gives: (a)
(0H,/0T)p = Cp.m = 40 cal/molK;
(b)
(OHm/OP)r = Vm — TV m& = 50 cm*/mol — (298 K)(50 cm?/mol)(10~> K7') = 35 cm’/mol = (35 cm?/mol)(1.99 cal/82 cm?atm) = 0.85 cal/atmmol: (QU/0V); = AT/k — P = (10° K')(298 K)/(10~ atm!) — 1 atm = 3000 atm = (3000 atm)(1.99 cal)/(82 cm?atm) = 70 cal/cm?;
(c) (d)
(OSp/0T)p = Cp m/T = (40 cal/molK)/(298 K) = 0.13 cal/molK?:
(e)
(0Sp/OP)7 = —AVm = —(10* K7')(50 cm*/mol)(1.99 cal)/(82 cm?atm) = —0.001> cal/molKatm;
(ff)
Coe Creel ncik= (40 cal/molK) — (298 K)(50 cm*/mol)(10° K~*)/(10~ atm”) = 40 cal/molK — (150 cm?atm/molK)(1.99 cal/82 cm?atm) = 40 cal/molK — 3.6 cal/molK = 36 cal/molK;
(g)
(dA/0V)7 = —P =—(1 atm)(1.99 cal/82 cm*atm) = 0.024 cal/em’, where (1.19) and (1.21) were used.
4.9
(a)
The Gibbs equation dU = T dS — P dV becomes dU; =T dS; — P dVr at
constant T. Division by dP7 gives dUr/dPr = T dSr/dP; — P dV;/dPr or (OQU/0P)7 = T(0S/0P)7 — P(OV/OP)z . But the Maxwell equation (4.45) gives (0S/0P)r = — (0V/0T)p, so (QU/OP)z = —~T(OV/0T)p— P(OV/OP)7 = TVa. + PV«. 37
4.10
(b)
From (1.35) we have (dU/dP)r = (QU/0V)r (OV/OP)r = _(QU/AV)7«V. Substitution of (4.47) gives the desired result.
(a)
Atconstant P, dUp = T dSp — P dVp. Division by dTp gives (QU/OT)p = T(OSIOT)p — P(OV/OT)p = Cp — PVa, where (4.31) and (4.39) were used.
(b) Cp = (OHIOT)p = [AU + PV)/OT]p = (QU/AT)p + POOVIAT)p, So (QU/0T)p = Cp P(OV/OT)p = Cp PVa
4.11
At constant T, we have dH7 = T dS; + V dPr. Division by dV7 gives (OH/0V)7 = T(0S/0V)7 + V(OP/OV)7 = T(OP/OT)y + V(OP/O0V)7 =
aT/K — 1/K, where (4.45), (1.42), (1.44), and (1.32) were used.
4.12
Liquids. Solids.
4.13
Differentiation followed by use of (4.51) gives
GT  al ar 4.14
1 (36) CS A
Ha TS
SR
=T dS —PdV becomes dU, =T dS, . Division by dT, and use of (4.29) gives Cy =(0U/0T)y =T(0S/0T),. At constant P, (4.34)
At constant V, dU
is dH p=T dS>p and division by dTp gives Cp = (0H /0T)p =T(0S/0OT)p. 4.15
From (2.63), ly = (QU/0V)7/Cy. Use of (4.47) gives the desired result.
Reminder:
4.16
Don’t look up the solution to a problem until you have made a serious effort to solve it.
Vin = RTIP + BRT. = (1/Vm)(OVm/OT)p = (A/V )(R + bPR)/P = RU + BPYPVm. K = —(/Vm)(OVm/OP)r = RT/P?Vm(QU/OV)7 = ATK — P = (1 + bP)P — P = bP”. ‘Cpm—Cv.m = TVm0/K = (R + bPRY/IR = R(1+bP). pyr = WaT — 1)/Cp= (V/Cp)[RT(1 + bP)/PVm — 1] = (V/Cp)\(PVm/PVm — 1) = 0.
38
4.17
Seon haan cmapnion iene . 1 2%)() of™) = {%) Ci eee cl at Ole) > DHE he 2
4.18
(a)
(b)
paw
VemiVgtet Vole) tbo 2K (QV/OT)p = aVo + 2bVo(T— 273 K) and (0° V/OT*) = 2bVo. (8Cp MOP) =T(0°Vm/OT)p = —2bVmoT, 80 (OCp.m/OP)r = ~2(0.78 x 1078 K)(298 K)(200.6 g/mol)/(13.595 g/cm?) = (6.85 x 10° cm? K™ mol"!)(1.987 cal)/(82.06 cm?atm) = ~1.66 x 10° cal/molKatm. ACpm = (0Cp.m/OP)7 AP = (1.66 x 10° cal/molKatm)(10* atm) = —0.02 cal/molK.
Cpm = 6.66 cal/molK — 0.02 cal/molK =
6.64 cal/molK.
Cp — Cy = TVo2/«. As in Prob. 1.47, & = (1/Vin)(c2 + 2c3T— csP) and K = (c4 + c5TV Vm. SO Cp — Cy = nT(co + 2¢3T— c5P)M(ca + €5T), since
4.19
ViV5. =:
(QU/OV)7 = ATI — P = T(c2 + 203T— csP)M(c4 + c5T) — P.
(0S/0P)7 = —QV = —n(c2 + 2c3T— csP).
Wye = (Vin CpmOT — 1) = (T/Cpm)(C2 + 2€3T — CsP) — VielCrm. (OS/0T)p = CplT. (OG/0P)r = V. (OVIOT)s = 1/(OTIOV)s = 1/(0P/0S)y = “OS/OP)y = ~(AS/OT)KOTIOP)y = —(Cy/IT)(OP/0T)y = —CvW/AT, where
4.20
(1.45) was used. Hence Os = —Cyk/TVa.
(b)
For a perfect gas, « = 1/T and k = 1/P [Eqs. (1.46) and (1.47)], so Os
_Cyl[PV = —CyinRT = —Cym/RT = V'(OVIOT)s, so dV/V =
—(Cym/RT) aT at constant S, and In (V2/V}) = (Cv,m/R) In (T)/T2) =
In (T,/T, )""® ;hence VoVi = (T,/T,)%"" or T2/T) = W/VEE
39
(c)
4.21.
(AV/OP)s = (OV/OT)s(AT/OP)s. From (a), (8V/0T)s = —Cyk/aT. From (4.44), (OTIOP)s = (QV/9S)p = [(AS/V)p} = ((AS/9T)p(OTIOV)p} = (Cp/T) \(OV/OT)p = (T/Cp)aV = ATV/Cp. So (OV/OP)s = (—CyK/aT)(ATV/Cp) = —CykV/Cp; Ks =V'(OV/OP)s = Cyk/Cp.
(dH/dP)r = V— T(OV/OT)p. For an ideal gas, V = nRT/P and (OV/0T)p = nR/P, so (OH/0P)7 = nRT/P — T(nR/P) = 0.
4.22
At25°C and 1 atm, Vm = RT/P = 24500 cm*/mol and Uintermoim = —@/Vm = ~(10° to 10”)(cm®atm/mol”)/(24500 cm*/mol) = (40 to —400)(cm?atm/mol)(1.99 cal/82 cm?atm) = —1 to10cal/mol, where (1.19) and (1.21) were used. At 25°C and 40 atm,
V., = RT/P = 610 cm?/mol and Uintermolm ~ —40 to 400 cal/mol. 4.23
(a)
AU vapm ~ —Uintermoim From Fig. 4.4, Uintermol.m = —35) kJ/mol at Vn = 107
cm’/mol.
(b)
AUvap.m = AHvapm— ACPVim)vap
VWmgas — Vmatiq = RT/P — 107 cm*/mol =
25200 cm?/mol. AU\ap.m = (6400 cal/mol)(4.184 J/cal) — (1 atm)(25200 cm?/mol)(8.314 J)/(82.06 cm>atm) = 24.2 kJ/mol.
4.24
(a)
a=(1.34~x 10° cm° atm/mol’)(8.314 J/82.06 cm*atm) = 1.36x10° Jem /mol’.” U= Usa t+ Onemolmie Ua (lo Rl alV, + const = (12.5 J/molK)T— (1.36 x YOAd cm?/mol)/Vm +const,
where U;;m 1s the molar molecular translational energy. (b)
For both the liquid and the gas, Utr.m = (3/2)RT = 1090 J/mol.
Vin,liq = M/p = (39.95 g/mol)/(1.38 g/cm’) = 28.9 cm*/mol.
Ey AMO ee Aa ~(1.36 x 10° J cm*/mol’)/(28.9 cm?/mol) = 4710 J/mol. Uintennolmgat = CVa gaspee OuOL, (c)
AU m.vap = Uintermol,m,gas a Uintermol.m,liq
4.7 kJ/mol.
4259
(a) ¢lee(D) el;
40
=
—19 J/mol
+ 4710
J/mol
=
4.26
Since T is constant, we have AA = AU — T AS and AG = AH —T AS.
From Prob. 2.49c, AH = 0 and AU = 0. From Prob. 3.13c, AS = 8.06 J/K. So AA = 0 — (400 K)(8.06 J/K) = —3220 J and AG = 3220 J.
4.27
Each process is isothermal, so AA = AU — T AS and AG = AH — T AS. (For g, w, AU and AH, see the answers to Prob. 2.44.)
(a)
AU=q+wand AS =q/T, so AA = w. Since w < 0, we have AA 0 and AA > 0.
(d)
AU=0O=AH. AA =T AS =T(q/T) =q =w p maha AC, /T. Integration gives AS°(72) — AS°(T;) = IF (NG, /Dae
5.41
(ayeeErom Eq, GO aAS a= AS ct lsgey (ACe/I) dl= ~173.01 J/molK + Ji000K (Aa/T + Ab + T Ac +T Ad) dT = ~173.01 J/molK + Aa In (1000/298.1) + Ab(1000 — 298.1)K + (1/2) Ac{(10007 — 298.17)K? + (1/3)Ad[(1000° — 298.1°)K* = 173.01 J/molK + (39.87 J/molK)(1.21003) + (0.11744 J/molK)(701.85) + 0.5(—9.8296 x 10°°)(911136)J/molK + (1/3)(2.8049 x 10°8)(9.7351 x 10°)J/molK = 174.51 J/molK.
(b) AStoo9 = ASSog + AC}998 In (1000/298.15) = —173.01 J/molK + (13.367 J/molK)(1.21016) = —189.19 J/molK. The approximation that AC;
is independent of T is a poor one.
5.42
Vin = RTIP + RTf(T) and (OVn/0T)p = R/P + Rf +RTF’. Substitution in (5.30) gives Smid(T, P) — Smre(T, P) = JO LRAT) + RIF ’(T)) dP = RPT) + Tf'(T)I.
5.43
A ¢Gy9g = Ay Hyg — T A ¢S3og  The formation reaction is C(graphite) + /202(g) + No(g) + 2H2(g) > CO(NH)2)2(c). Appendix data gives
for this reaction, AS° = >; viS },; = 104.60 J/molK — 5.740 J/molK —
V(205.138 J/molK) — 191.61 J/molK — 2(130.684 J/molK) = 456,69 J/molK = AyS3og urea»Then A ,G39g = 333.51 kJ/mol — (298.15 K)(0.45669 kJ/molK) = —197.35 kJ/mol.
5.44
(a)/
(AG? =AH* —T AS;= 1124.06 kJ/mol — (298.15 K)(0.39073 kJ/molK) = 1007.56 kJ/mol. AG; = 1036.04 kJ/mol — (298.15 K)(0.15290 kJ/molK) =
57
AG? =956.5 kJ/mol — (298.15 K)(0.0234 kJ/molK)
~990.45 kJ/mol. = —949.5 kJ/mol.
(b)
AG; (kJ/mol) = 2(237.129) + 2(300.194) — 2(33.56) — 3(0) = —1007.53. =990.41.
AG; /(kJ/mol) = 2(228.572) + 2(300.194) — 2(—33.56) —3(0) AG? /(kJ/mol) = 120.35 + 4(0) — 2(328.1) — 2(86.55) =
—949.6.
5.45
(a)
AG37) = AH 379 — T AS379 = ~1118.75 kJ/mol — (370 K)(0.037476 kJ/molK) = —980.09 kJ/mol.
(b) AG349 =1036.73 kJ/mol — (370 K)(0.15498 kJ/molK) = —979.39 kJ/mol.
5.46
(c)
AGiy) = 948.3 kJ/mol.
(a)
Gir = Hm20g — TSin29g = 0 — (298.15 K)(205.138 J/molK) = —61.16 kJ/mol.
(b)
Using the result of Prob. 5.22c, we get Gi193 = Hin29g — TSin208 =
—285830 J/mol — (298.15 K)(69.91 J/molK) = —306.67 kJ/mol. 5.47
LandoltBornstein data give 70.7 kJ/mol.
5.48
AHSo = 2(46.11 kJ/mol) — 0 — 3(0) = — 92.22 kJ/mol. AH
5000 =
AH
558 +);
Vi( Jel
ore

He
ont) =
=—9)
22
kJ/mol
ae 2(98.18
kJ/mol)
— 56.14 kJ/mol — 3(52.93 kJ/mol) = 110.79 kJ/mol.
5.49
The reaction is +N2(g) + 3 H2(g) — NH3(g). The Appendix gives A ¢H 49. = —46.11 kJ/mol. Use of (5.43) gives A ¢Gy99 = 46110 J/mol +
(2000 K)[—242.08 — 5 (223.74)  3 (161.94)](J/molK) = 179.29 kJ/mol.
5.50
AH 59g + T Li Vil Gur — Hn20g/TIi = AH 39 + T Di canny ellOe i>,
Wyle
crete
=
AH
563 aP AG;
=
58
AH
59, =
AG;.
AG?" —AGT™ = AH} —T AS? — (AH®™ —T AS2™) =
sypil
T(AS*" — AS3"™), since the difference between 1bar and 1atm enthalpies of a solid or liquid is negligible and H° of a gas is independent
of P. Since the effect of a slight change in P on S of a solid or liquid is
negligible, we have AS7* — AS#'™ = AS 7) — AS*™. = Measee VASE Pe Soery ) (0:1 094° T/moleR) cases Vi = (0.1094 J/molK)An,/mol and
AGS — AG#™ =T(0.1094 J/molK)An,/mol. (b)
Ha(g) + 102(g) > H20( 4). AyG353— AyG 398 = —(298 K)(0.1094 J/molK)(—1.5) = 48.9 J/mol.
a4
H
H

H

a
gaseous
»
H—C—C—O—H(g)
—
atoms
—
hemes
H
(a)

tH —
H
H 
—— Hq} €——_ OC


H
H
AH,/(kJ/mol) = 5(415) + 344 + 350 + 463 = 3232 AH,/(kJ/mol) = [6(415) + 2(350)] = —3190 AH = AH, + AH» = 42 kJ/mol
See)
(b)
AH/(kcal/mol) = [6(—3.83) + 2(—12.0)] — [5(—3.83) + (12.0) + (27.01)] = 11.17, so AH = 11.2 kcal/mol = 46.7 kJ/mol.
(c)
AH/(kJ/mol) = 2(41.8) — 99.6 (41.8 — 33.9 — 158.6) = 51.1 kJ/mol.
(a)
3C(graphite) + 4H2(g) + + 02(g) —>» CH30CH2CH3(g).
3C(g) + 8H(g) + O(g) ——
Nels = AH, +AH,.
Appendix data give
AH,/(kJ/mol) = 3(716.682) + 8(217.965) + 249.170 = 4142.94. Bond energies give AH , (kJ/mol) = [8(415) + 344 + 2(350)] = 4364. So A, H®/(kJ/mol)
= 4143 — 4364 =221.
59
(b)
A,A°/(kJ/mol) = 4.184[2.73 + 2(12.0) + 8(3.83)] =—217.2
(ec)
A,H°/(kJ/mol) =41.8 — 99.6 — 33.9 —41.8
=217.1
In the Benson—Buss bondcontnbution method, the carbonyl! group is treated as a unit and no explicit contnbution is made for the C—O bond. The effect of this bond is absorbed into the contributions for bonds to the carbonyl carbon. The bond contribution to S>, 59g of the F—CO bond is 31.6 cal/molK,
so the total bond contnbution 1s 2(31.6 cal/molK) = 63.2 cal/molK for COF»2(g). In addition, the quantity R In o must be subtracted to allow for molecular symmetry. For COF2, the symmetry number o is 2, since there are two indistinguishable onentations of the molecule (obtained by 180° rotation about the CO bond). The symmetry correction is —R In 2 = —1.38 cal/molK
and the predicted S* 493 is 61.8 cal/molK. AH 59g of vaporization refers to a change from liquid at 1 bar and 25°C to
vapor at  bar and 25°C. A path to accomplish this is the following 25°C isothermal path:
liq(1 bar) a liq(23.8 torr) = gas(23.8 torr) a gas(1 bar) AH® = Ay) + AHm2 + AHm3. As noted in Sec. 4.5, a modest change in pressure from  bar to 24 torr will have only a very slight effect on H and S of a liquid, so we can take AH, ;= 0. Also, since the vapor is assumed to behave ideally, its H depends only on T, and AH,, 3= 0. Thus AH® = AHm2 = 10.5
kcal/mol. For comparison, Appendix data give AH® = (241.818 + 285.830) kJ/mol = 10.519 kcal/mol. Next, AS° = AS}
+ ASm2 + ASm3. To a good
approximation, AS;,; = 0. Equations (3.25) and (3.29) and Boyle’s law give ASm2 + ASm3 = AHm2/T + R In (P\/P2) = (10500 cal/mol)/(298.1 K) +
(1.987 cal/molK) In (23.8/750) = 28.34 cal mol” K7' = AS°. The Appendix data give AS® = (188.825 — 69.91) J/molK = 28.42 cal/molK. Finally, AG® = AH® — T AS° = 10.5 kcal/mol — (298.1 K)(0.0283¢ keal/molK) = 2.03 kcal/mol. The Appendix gives AG° = 2.045 kcal/mol.
5.56
We use the 25°C path (where M is methanol) 2 3 4 M(“, 1 bar) >M(2£, 125 torr) >M(g, 125 torr) M(g,  bar) M(ideal gas,
60
1 bar). For this path, AH = Hy yy(g)— Him my = Ammgy—Aa — (A mya) = A;Hyg)—
Sp Hus), where H;, is the standardstate enthalpy of the
elements needed to form methanol. Since a moderate change in P has little effect on thermodynamic properties of a liquid, we have AH, = O (and AS, = 0). AH> = 37.9 kJ/mol.
AH; = 0 and AH, =0, where we neglected nonideality of
the gas. Then AH = AH, + AH? + AH; + AH4 = 37.9 kJ/mol and A; Hy,,)= —238.7 kJ/mol + 37.9 kJ/mol = —200.8 kJ/mol.
We have AS;
=0, AS2 = AH2/T
= (37900 J/mol)/(298 K) = 127. J/molK, AS3 = R In (V2/V;) = R In (P\/P2) = (8.314 J/molK) In (125/750) = —14.90 J/molK [where (3.29) was used], and AS, = 0, where the gas is approximated as ideal. Then AS = (127. — 14.9) J/molK = 112.2 J/molK. So Si, y(,)= 126.8 J/molK + 112.2 J/molK = 239 J/molK.
PAS
(a)
A, Hy. = (2n+2)boy +(#—Dbec
(b)
Breaking the formation reaction into two steps, we have
nC(graphite) + (n + 1)H2(g)
> nC(g) + (2n + 2)H(g) — CrH2n+2(8). So
A, H 29 = AH, + AH, = nA , H x9g{C(g)] +(2n+ 2)A , H y9g{H(g)] = (n—l)Dece — (2n + 2) Dey (c)
Equating the expressions in parts a and b, we have forn = 1:
Aboy = A ris lCig
+ 4A , H y9,[H(g)]—4D(CH) and
HyoglC(g)]+ A y H x9g(H(g)] —D(CH) . For n = 2, bey = 4A, 6bey + bcc = 2A , H y9g(C(g)] * 6A , H y9g[C(g)] — Dec — 6Dex Substitution of the result for bcy gives after cancellation:
bec = 0.5A , H y1C(g)]  Dec
(d) J; (e) m’/mol; (f) K.
5.58
(a) Pa: (b) J; (c) J/molK;
~ Pesky
If AH? is independent of T, then d AH°/dT= 0 and (5.18) gives AC*, =0. Then (5.37) gives AS; = AS;, .Q.E.D.
5.60
(a)
Equation (4.45) gives (AV/OT)p = (0S/0P)r. The 3rd law gives lim7_49 AS = () for an isothermal pressure change in an equilibrium system. Hence (0S/aP)r 3 0 as T > 0, and a = (1/V)(OV/0T)p — 0 as T— 0. 61
5.61
(b)
a=(1/V)(OV/0T)p =1/T [Eq. (1.46)], which goes to % as T > 0.
(a)
Nonzero.
(b)
Nonzero. ArH 5, refers to formation of the substance from elements in their stable forms at 298 K. The 298 K stable form of chlorine is Cl2(g). not Cl(g), and AH 5o. for /2Cl2(g) — Cl(g) 1s not zero.
(c)
Zero, since Cl2(g) is the stable form of an element.
(d)
Nonzero. (Entropies are zero at 0 K.)
(e)
Zero.
(f)
Zero, since 350 K formation of N2(g) from its stableform element(s) is a process in which nothing happens.
(g)
Zero.
(h)
Zero, since heatcapacities go to zero as T goes to zero, as shown by the Debye equation (5.31).
(i)
5.62
Nonzero.
The 25°C reaction (step c) is CHy(g)
+ 202(g) > CO2(g) + 2H2O( ¢ ) and has
AH = [393.509 + 2(—285.83) — (—74.81) — 2(0)] kJ = 890.4 kJ for burning 1 mole of CHy. We have AH, = AH, + AH, = 0 = 890 kJ + AH; and AH; = 890 kJ. From the Appendix, the heat needed to vaporize 2 mol of H2O is 2(—241.8 + 285.8) kJ = 88 kJ. In step b, the products are being heated from 25°C to the flame temperature 7. The heat required to do this is 88000 J + (1 mol)(54.3 J/molK)(T— 298 K) + (2 mol)(41.2 J/molK)(T— 298 K) + 2(3.76 mol)(32.7 J/molK)(T— 298 K) = 890000 J. So T — 298 K = 2096 K; T = 2400 K.
5.63
5.64
(a)
nC4Hj0(g), since larger molecules have larger entropies.
(b)
H2O(g), since gases have higher entropies than the corresponding liquids.
(c)
H2(g), which has larger molecules than H(g).
(d)
CyoHg(g).
(a)
AH" = apis positive, since heat is needed to vaporize the liquid. For liquid — gas, AS° > 0. 62
(b)
AH° is positive since energy is needed to break the bond. AS° is positive, since the number of moles of gas is increasing.
(c)
gis negative when the vapor condenses, so AH® is negative. For gas —> solid, AS° < 0.
(d)
(COOH))(s) + 202(g) 2CO2(g) + H2O( 2 ). We can expect the reaction to be exothermic, as is generally true for combustion reactions. Hence, AH° 0, since the number of moles of gases is increasing.
(e)
AS° Shh (bie ik
= (7771750)°/(548/750)°(442/750) = 3.41.
P pl
AG° =RT In K*, = (8.314 J/molK)(1000 K) In 3.41 = —10.2 kJ/mol.
(b)
Kp =(Pio,/Pso, Po, )P° = KeP° and Kp = Kp/P° = 3.41/(1 bar) = 3.41 bar’.
(GME
6.4
KSe Keo (RT CARP? oes = 3.41[(82.06 cm?atm/molK)(1000 K)(1 mol/1000 cm?)/(0.987 atm)? = 284.
A+B < 2C+3D. If 10.0 mmol of C is formed, then 5.0 mmol of A and 5.0 mmol of B must have reacted and (3/2)(10.0 mmol) = 15.0 mmol of D is formed. At equilibrium, na = 10.0 mmol, ng = 13.0 mmol, nc = 10.0 mmol, np
= 15.0 mmol; mo. = 48.0 mmol. Xp = 073125. Pe=226 tom,
x, = na/nio = 0.208,
xg =0.271,
Px=xaP =(0.208)(1085 tom) = 220 tor, Pp=339 torr,
xc =0.208,
Pe = 294 torn:
Kp= (Pc IP°)(Pp /P°)'/(P/P°)(Pp/P°) =
(226/750)*(339/750)°/(226/750)(294/750) = 0.0710.
AG®° =RT In Kp=
—(8.314 J/molK)(600 K) In 0.0710 = 13.2 kJ/mol.
6.5
Not = PV/RT = [(231.2/760) atm](1055 cm*)/(82.06 cm?atm/molK)(323.7 K) = 0.01208 mol. Let x mol of Br react to reach equilibrium. At equilibrium, 64
nxo/mol = 0.01031  2x, ng, = 0.00440 —x, nnopr=2x, Mot = 0.01471 — x= 0.01208; x= 0.00263. So mxo = 0.00505, ng,, = 0.00177, nnosr = 0.00526; xno = NNo/Mo = 0.418, Xp, = 0.1465, xNopr = 0.435. Pxo = xnoP = (0.418)(231.2 torr) = 96.6 torr, Pp, = 33.9 torr, Pros: = 100.6 torr,
P°=750torr.
K> = (PyopiP°)/(PNo/P°) (Pp, /P°) = 24.0.
AG? =RT In K* = (8.314 J/molK)(323.7 K) In 24.0 = 8.55 kJ/mol. 6.6
(Py/P°)'I (Py, /P°) = (0.12)"(720)(750) = 2.7 x 10°, which is less than K%. The system is not at equilibrium; Pyig) and hence nyig) Must increase to reach equilibrium.
6.7
Consider the reaction aA
_ Cre Gale
+ bB
paw ce UenIay pe kee
=cC + dD. Picea
Use of P; = x;P gives CP gee
=
K, CEP
ay a
2)(2  3)3  4)(4  5) = 2880.
6.8
abe jU+l) =(
6.9
(a) 120 (by SP areas
6.10
AG? = RT In Ky = —(8.314 J/molK)(298.1 K) In 0.144 = 4.80 kJ/mol. If
(fer
(do Fe (eye
ee) er, eh
AH? is assumed constant over the range 25°C to 35°C, then Eq. (6.39) applies and In(0.321/0.144) = [AH°/(8.314 J/molK)][(298.1 Kye 08 koe: AH° = 61.2 kJ/mol. AG° = AH® —T AS®° = 4800 J/mol = 61200 J/mol — (298.1 K)AS®° and AS®° = 189 J/molK.
6.11
We plot In K> versus 1/T. The data are InK,
1.406
T'/K'
0.002065
0.688
1.601
0.001873
0.00179,
65
Phfsie') 0.00174,
= 11399x + 22.085 Cease
2.5 N
In Kp
1.5
l OS 0 0.5 ] 1.5
{EYSCS TERYVc CE) CAG PAAAR I AAA Ce ARATE Ame
0.0017
PN
SV TS PT
0.0018
TS
LET
Gt
0.0019
LY)
SV
PEE
0.0020
1_J
0.0021
Tie The plot is very nearly linear with slope
[2.500 — (—1.500)]/(0.001718 — 0.002069)K' =11400 K; AH°/R = —11400 K; AH® = (11400 K)(8.314 J/molK) = 94.3 kJ/mol = 22.6 kcal/mol in
this range of T. Then AG.,,
=— RT In K, = (8.314 J/molK)(534 K) In 1.99
= —3.055 kJ/mol = —730 cal/mol. From AG® = AH® — T AS°, AS;,, = (AH® — AG°)/T = (94800 + 3055)(J/mol)/(534 K) = 183 J/molK = 43.7 cal/molK.
(b)
AH° = 22.6 kcal/mol = 94. kJ/mol, as in (a). (The near linearity of the
plot shows that AH° is essentially constant over this range of T.)
AG574 =—RT In K =10.7 kJ/mol = 2.55 kcal/mol.
A S574 = (AH 534 — AG55,)/T= 183 J/molK = 43.8 cal/molK. 6.12
Appendix data give AG 59, = 37.2 kJ/mol.
AG° =RT In K%. 37200 J/mol =
—(8.314 J/molK)(298.1 K) In Kp 99g and K > 599, = 3.0x 10°. AE Rie
celui Kee (Tein K p(T) =(AH*/R)1/T> —1/T)).
dln K%,/dT= In K apo. =
In (3.0 x 10°’)  [(87900 J/mol)/(8.134 J/molK)][(400 K)! — (298.1 K)"'} = ~5.99 and K% yop ~ 0.0025. 6.13
AG® = RT In Kp = 72400 J/mol = ~(8.314 J/molK)(600 K) In Kp and Kp = 4.9, x 107’ at 600 K. Integration of the van’t Hoff equation (6.36) with AH° assumed constant gives In(Kp/K, )= el TINUE) AUT ERE
In[26/(4.97 x 10’)] = [(217900 J/mol)/(8.314 J/molK)][1/T2 — 1/(600 K)] and T> = 1010 K. 66
AG° =RTInK § = RT[Ina+ bIn(T/K)—c/(T/K)]. Equation (6.36) gives AH° = RT7dInK/dT = RT?[b/T +(cK)/T*] = bRT+ (cRK). Then AS° =(AH°AG°)/T =bR+ R{Ina+bIn(T/K)} and AC? = dAH°/dT =bR. (b) At 300 K, AH? = (8.314 J/molK)[(—1.304)(300 K) + 7307 K] = 57.5 kJ/mol. At 600 K, AH? = 54.2 kJ/mol.
6.14
(a)
6.15
Substitution of AH°(T) = AH°(T;)
+ ACp(Ti)(T— T)) into (6.37) gives
In[K p(T2)/K p(1)] =R! [? AH (7, \ieo die
(A/T, /T,)+ dT = R'AH°(T,) R ACT)? WT T/T*) Re AG;@) (nG5/T)) Phi:
For No04(g) — 2NO2(g), AG3og = 4.73 kJ/mol, In K > 293 = 1.908, and K >19g 609 = = 0.148. AH io, = 57.2 kJ/mol; AC>..93 =2.88 J/molK. In K > 1.908 + (57200 J/mol)(8.314 J/molK) '(1/298.1 — 1/600)K7! + (8.314 J/molK) '(2.88 J/molK)[In(600/298) + 298/600 — 1] =
Dos 2andeK yy 6.16
aloe lee
(a)
Substitute (6.28) into (6.14).
(b)
In[K>(T2)/K »(T1)] = In K p (72) — In K
p(T) =
~AH°(T>)/RT2 + AS°(T2)/R + AH°(T WRT, — AS°(T)/R = AH°(T,)R (A/T, — 1/T2) if we take AH°(T2) = AH®(T)) and eC
i=
AS C1):
6.17
(a)
When 7) in the AH® expression in Example 5.6 is replaced by T and this expression is inserted into (6.37), the right side of (6.37) becomes
R'{? (That
T w+ pAb +3 AcT +1lAdT*)dT , where w =
AH; —T, Aa 17? Ab  17; Ac +T,* Ad . After integration, we have
) T,°)+ +4Ac(Ty LAD(T, T, R'[AaIn(T, /T,) + WT, Tz+') 1 Ad(T; T;)]= In[K°(T,)/Kp(7;)) (Eq. (1)]. Let T, = 298.15 K. We have
67
SUNS aon SINGop SIAR
eer 2(137.168) — 0] kJ/mol =
we ~514382 J/mol and In K > 59, = 207.50. Using data from Example 5.6, find w = —558488 J/mol for T; = 298.15 K. We can thus find Kol) using the preceding Eq. (1).
Substitution in Eq. (1) of part (a) gives In K (1000 K) = 207.50 + (1/8.3145)(48.25 — 1314.69 +41.21 — 14.93 + 2.28) = 47.01 and
(b)
K°(1000 K) =3 x 10”. 6.18
Equation (5.19) gives
=AH;, + f T JdT’ (2 +(2gc) —b)T’a+ AH; =AH; +J7. (2e(2ea\(T T,)+(f 5b\T” T7)+5(2g—c)(T° —T,’). Substitution in (6.37) and integration gives In KU) hie Gadlidhae
RAH? + (a2e)T + (1bf)T) + 5(c 2o\Te Ce Ty4)rt R™'(2e — a) In (TyT1) + fLb)(T2  THR + §(28 — Ty T; WR. 6.19
From (6.25), In K, = In K> — (An/mol) In T + const. Differentiation and use of (6.36) and (5.10) give
dinK; dinKp
dT 6.20
dT
seAn/mol _ AH® _ AnRT/mol _ AU®
je
Rie
RI
Ri
5
From (6.27), In K, = In K» — (An/mol) In (P/bar). Differentiation with respect
to T and use of (6.36) gives (0 In K,/0T)p = d In K p/dT = AH?°/RT”. Partial differentiation with respect to P gives (since Kp is independent of P): (0 In K,/OP)7 = —(An/mol)/P.
6.21
(a) T; (b) F.
6.22
Since P; = n,RT/V, the partial pressures are proportional to the moles. Let
zV/RT moles of CO react. The equilibrium partial pressures are then Pco =
342 torr—2,
Fo, = 351.4 torr—z,
Pooc, =z. Thus
68
439.5 torr = 342.0 torr —z + 351.4 torr — z +z and z = 253.9 torr.
Kp= (253.9/750)/(88.1/750)(97.5/750) = 22.2.
6.23
AG°/(kJ/mol) = —394.359 + (—879) — 2(619.2) =35. (35000 J/mol)/(8.314 J/molK)(298 K) In Kp = 14.1,
In K,
=—AG°/RT =
Kp =1.3 x 10°.
Let 2z moles of COF; react. The equilibrium amounts are n(COF>) =  — 2z,
n(CO2) =1+2z,
n(CF,4) =z. Because Kp is so large, we have 2z = 1 and z =
Vy, Hence, n(CO>) = 1.50 mol, n(CF4) = 0.50 mol, and
ye
tear reid WestieVed
1.3 x 10
6.24
ae)
_
n(COF2) =7 x 10~ mol
[n(GOE, )/rice li
In K> =AG°YRT= 1.258. K% = 0.284. (a)
Let z moles of A react. The equilibrium amounts are na = 1 z, 2c USC Oftias— \eives 2 e= 22, Np = oe, alld Mente Mat
0.284=
0.533 =
TET T EY ss lz
ees
2(1+ z)
P;
227 2a 1200 750
sae
(l1—z7)? \750
3339 Zz2 0333 1,
4
eos
Both times we took the positive square root, since z and  — z’ are positive.
na = 0.622 mol =ng,
nc = 0.756 mol = np.
(b) The equilibrium amounts are na = 1—z,
have
0.284 =
np= 2
Noe 2 Samp eWwe
[22/(3 + 22)]* (PIP*)* [((Q— zB + 2z)][2 zVGB+ 2z)]
0.00693, — 24/1 — 2)(2  2)(3 + 22) =0 z lies between 0 and 1, and trial and error gives the desired root as z = 0.530. Hence na = 0.470, ng = 1.470, nc = 1.060 = np.
6.25
Appendix AyG®° data give AG 39, = —2.928 kJ/mol. In Kp =AG°/RT= 1.18lfand K, =3.26. Kp = [P(HD)/P°]*/[P(H2)/P° ][P(D2)/P?] =
69
[P(HD)]°/P(H2)P(D2) = [n(HD)RT/V}"/[n(H2)RT/V) [n(D2)RT/V] =
(n(HD)]?/n(H>)n(D>). Let x mol of H2 react to reach equilibrium. The equilibrium amounts are then n(H2) = (0.300 — x)mol, n(D2) = (0.100 — x)mol, n(HD) = 2x mol. Substitution in the Kp expression gives after simplification: 0.227x? + 0.400x — 0.0300 = 0. The quadratic formula gives x = 0.072 and _1.83. Since we started with no HD, x must be positive. So x = 0.072, and n(H>) = (0.300 — x)mol = 0.228 mol, n(D2) = 0.028 mol, n(HD) = 0.144 mol.
6.26
(a) K%, = [P(NH3)/P°}/[P(N2)/P°}LP(H2)P°P = P(NH3)(P°)"/P(N2)[P(H2)) Use of P; = n,RT/V gives K‘> = [n(NH3)"/n(No)n(Ho) \(P°V/RT)
Now K°,(RT/P°V) = 36[(82.06 cm?atm/molK)(400 K)/(750/760)atm(2000 cm’)]° = 9960 mol. Let x mol of N2 react to reach equilibrium. The equilibrium amounts are (0.100 — x) mol of N2, (0.300 — 3x) mol of H2, and 2x mol of
NH3. So 9960 = (2x)/(0.100 — x)(0.300 — 3x)? = 4x7/3°(0.1 — x)(0.1 — x)
= 4x’/27(0.100 — x)’. Taking the square root of both sides, we get +99.8 = 0.385x/(0.100 — x)’. Since x must be positive (we started with no NH3),
the negative sign is rejected and 0.01 — 0.2x + x’ = 0.00386x. The quadratic formula gives x = 0.082 (the other root exceeds 0.1, which is impossible). Then n(N2) = 0.018 mol, n(H2) = 0.054 mol, and n(NH3) =
0.164 mol.
(b)
At equilibrium, n(N2)/mol = 6.200 — x, n(H2)/mol = 0.300 — 3x, and n(NH3)/mol = 0.100 + 2x. Then 9960 = (0.100 + 2x)7/(0.200 — x)(0.300 — 3x)° = W. To make W equal
9960, 3x must be close to 0.3. For an initial guess of x = 0.09, we find W = 26400, which is too big. For x = 0.08,
W = 2600. Repeated trial and
error (or use of the Solver in a spreadsheet) gives x = 0.0865. Then n(N2) = 0.113 mol, n(H2) = 0.0405 mol, n(NH3) = 0.273 mol.
6.27
The equilibrium amounts are ny, = 4.50 mol — E, ny, = 4.20 mol — 3&,
Nyy, = 1.00 mol + 2€. The requirement that each of these amounts be positive
gives €< 4.50 mol, & < 1.4 mol, § >0.5 mol. So 0.5 mol < € < 1.4 mol. Hence ny, lies between 4.50 mol — (—0.50 mol) and 4.50 mol — 1.4 mol; that
is, 3.1 mol < ny, Xpci, = 1  2x = 0.263.
6.32
(a)
Let No +3H2 ce 2NH; be called reaction I. Since reaction (a) has its coefficients equal to onehalf those of reaction I, we have K pa) = (Kies he = (36)? = 6)
(De Keer = Uke 0028, 6.33
For n > iso, AG j999 = —670 cal/mol and we find K > 1000 = 1.40 = Kpya, where
A = npentane, B = isopentane, C = neopentane. For n — neo,
AGjo99 = 4900 cal/mol and Kp jo99 = 9.085 = Kea. Eqs. (6.45) and (6.46) give Xa =Xn = W/(1 + 1.40 + 0.085) = 0.40,
xp = Xiso = 1.40/2.485 = 0.56,
XC = Xneo = 0.085/2.485 = 0.034.
6.34
AG 600 /(kJ/mol) = 1059.72 + 0 — 71.74 = 987.98. In K >6099 = (987980 J/molK)/(8.314 J/molK)(6000 K) = 19.804 and IKLei ORNs
6.35
AG° =RT In K>; AG
fe]
wrong
=RT In K, P, wrong
2500 J/mol = —RT In (Kp/K ;P,
wrong
*
AG° — AG‘ wrong =
). We get In (K'p/K p wrong) = —1.00 and
Kp wrong/Kp = 2.7. The error is a factor of 2.7.
72
6.36
(a)
L(g) e 2I(g). Let n* be the number of I, moles before dissociation, and let z moles of Ip dissociate to give 2z moles of I, leaving n* — z moles of
I,. The total number of moles at equilibrium is n* — z + 2z =n* + z. Then x; = 2v/(n* + z) and ci (n* — z)/(n* + z). We have n* = P*V/RT and n* + z= PV/RT, so z= PV/RT — n* = (P — P*)V/RT. Use of these
expressions for z, n* + z and n* gives x; = [2(P — P*)V/RT]/(PV/RT) = 2(P — P*)/P and x1, = [P*V/RT — (P — P*)V/RT]/(PV/RT) = (2P* — P)/P. (Dek
(P/P°)?/( xX), PIPoy= 2(R— P*)/P°}/{(2P* = P)/P?) =
4(P — P*)’/(2P* — P)P?°. (c)
Weuse the result of (b) to calculate the following Kp values:
K p973 = 4(0.0624 — 0.0576)7/[2(0.0576) — 0.0624]0.987 = 0.00177; K
P1073
OO
2s
Kp.1173
=
K
0.0493;
pi
=
Ou
25
We plot In K vs. 1/T. The data are InK,
6.34
OE
OO
+4.49
3.01
—1.757
Ee
TI
TS
One finds a straight line with slope [—1.19 — (6.73)]/(0.00075 — 0.00105)K” =18800 K. Eq. (6.40) gives AH? = (18800 K)(8.314 J/molK) = 156 kJ/mol = 37.4 kcal/mol.
y = 18850x + 13.054
>
7/.
te SF
cee
0.00075
eal a

pe PR NE
0.00085
tt tin Ns es (Ses Hl
0.00095
a
a
0.00105
ik
6.37
=11,(P/bar)"' /(P/atm)”* = K°/K*™ = (11,(P/bar)” VII, (P/a]tm)” II; (atm/bar)”' . For aA + bB Rese
ies dD,
]; (atm/bar)”' = (atrn/bar)*4
(1.01325)""™. (atm/bar)*”"™*! = (760 torr/750.062 torr)” = 73
S
6.38
At equilibnum, ny, /mol= 1 — x, ny, /mol = 3 — 3x, Nyy,/mol = 2x;
mol/mol = 4 = 2x. Py, = Xy, P= (My, /mor)P = [(1 — 2/4  2x))P, Py, = (BU x)/(42x)P, Py, = [2x/(4  2x)]P. Kp = a (Pap Po) (Pam pey( Py (P°), = (P°y' P(Aeon) Ox) (P°/P)*(16/27)(2  xyx7/(1  x)’ =K{.
(lx) =
x must be between 0 and 1. Taking the
positive square root of each side gives (P°/P)(4/27")(2 — x)x/(1  x) =4 4 a 2
or 4(2x — x2/(1 — x)? = (27K>)"? (P/P®) = s. Hence 8x — 4x? = s — 2sx +.x’s and (4 +s)x* — (25 + 8)x+s=O0or x’ — 2x + s/(s + 4) = 0. The quadratic formula
gives x= {2 +[4—4s/(s +4)]'7}/2 = 1 + [1 — ss + 4)]’”. Since x is less than 1, we take the minus sign: x = 1 — [1 s/(s + 4)]'”.
6.39
(a) No;
(b) no;
(e) no;
(c) no; (d) no;
(f) no;
(g) yes;
(h) no;
(i) yes.
6.40
(b).
6.41
(a)
See Fig. 6.11. results for n(CH4)/mol in order of increasing pressure are
0.00061, 0.053, 0.690, 1.355, 1.515, 1.62, 1.71. (b)
The NISTJANAF tables give at 1200 K:
AGA = 77.92 kJ/mol,
AG? =74.77 kJ/mol, K; = 2.46 x 10° and K> = 1.80 x 10°. One changes the values of T, P, K; and K> in the Fig. 6.9 spreadsheet. With the guessed values
of the equilibrium amounts taken as equal to the starting amounts, the Solver might not find a solution. The large values of K; and K2 indicate that one or both of the reactants CH, and H20 are present in small amounts at equilibrium.
The 0.01 bar 900 K equilibrium amounts in Fig. 6.10 have small amounts of the reactants, so it is a good idea to use the 0.01 bar 900 K equilibrium amounts as the initial guess for the 1200 K problem.
6.42
Suppose one equilibrium constant K; is very small with the value 1 x 10°. If the Solver changes the mole numbers and finds a calculated value of 8 x 10° for K,, then the calculated and true values of K, differ by 7 x 10; which is less than the Solver’s default precision of 1 x 10m and the Solver will declare
74
it has found a solution, even though the calculated K is 8 times the true K and at least some mole numbers must be greatly in error.
6.43
Let € additional moles of CO be formed when reaction 1 reaches equilibrium and let €; additional moles of CO2 be formed when reaction 2 reaches
equilibrium. The equilibrium amounts are then n(CHy) = 1 — & — &, n(H2O) = 1 — €; — 2&, n(CO2)=1+&, n(CO)=2+ &, n(H2) = 1 + 3&; + 4& (where the unit mol is omitted). The conditions that each of these amounts be positive give the inequalities: (I) 1>€, +€,; (lilacs mnLeon le (LVOaGy emmy) Scat ac. > =le Addition of (I) and (III) gives 2 > €, (VI). Addition of (I) and (IV) gives E, 0, so K% decreases as T decreases and the equilibrium shifts to the left.
(b)
As V decreases at constant T, the pressure increases and the equilibrium
shifts to the side with fewer moles of gas, i.e., it shifts to the left.
75
(c)
This removal decreases
P,,,_ , So to restore equilibrium, the reaction must 5
shift to the left.
(d)
Constant T and V addition of He does not affect the partial pressures of PCls, PCl3, or Clr, and there is no shift.
(e)
To keep P constant as He(g) is added, V must be increased. Since P; = n;RT/V, this volume increase will decrease each partial pressure P;
by the same percentage. Since the reaction has more moles of products than of reactants, the numerator of the reaction quotient Qp will decrease more than the denominator. Therefore the equilibrium will shift to the right to make Qp again equal to Kp.
6.47
From Prob. 6.19, d In K(/dT = AU°/RT * if AU® is positive, then K~ increases as T increases. Since c; = nj/V and V is held fixed, the mole numbers n; undergo
changes proportional to the changes in the concentrations, and the equilibrium shifts to the right if AU® > 0.
6.48
(a)
In QO, = In (T]i (x)” ] = Zi In)“ = Ev; In x; = Z; Vi In (1 Mor) = Ev; Inn; — (2; Vi) In (nj + n2 ++: +). We have O In Q,/dn; = vn; — AnI(nj + nz +  ~~) mol = (vj — xj An/mol)/nj.
(b)
Case I: suppose substance j is a reactant. Then v; is negative. To get a shift to the left, we want 0Q,/dn; > 0, so that addition ofjwill make Q, >
K,. We thus want v; — x; An/mol > 0 and v; > x; An/mol. Since v; is negative, An must be negative. Division of the inequality by the negative quantity An reverses its direction to give vj/(An/mol) < x;. The fact that An is negative means that the total moles of reactants exceeds the total moles of products, in agreement with condition (1). Case II: suppose
substancej is a product. Then v; is positive. We want 0Q,/dn; < 0, so that addition ofjwill make Q, < K, and shift the equilibrium to the nght. We
thus want v; — x,An/mol < 0 or v; < xjAn/mol. v; 1s positive, so An must be positive and division by An gives x; > vj/(An/mol). The fact that An is
positive means that the product total moles exceed the reactant total moles.
(c)
The left side of the reaction has the greater sum of the coefficients, so constant7andP addition of NH3 will never produce more NHs3. For N>, condition (2) is Xn, > (—1)/(2) = %, so for Xn, > 0.5, addition of N> at
76
constant T and P produces more N>. For Ho, condition (2) is Xy,> (—3)/(—2) = 1.5, which can never be satisfied.
6.49
(a)
K,= (1/5)7/(3/5)(1/5)° = 8.333. If x is the extent of reaction to reach the new equilibrium position, the new equilibrium amounts are n(N2) =
3.1 — x, n(H2) = 1 — 3x, n(NH3) =1 + 2x and
*
[(1+ 2x) (5.12x)]? [3.1—x)/(5.1—2x)][0 — 3x) /(5.1 2x)}°
The condition that the mole numbers be positive gives —0.5 < x < 0.333. Use of the Solver gives x = 0.0005438, so n(N2) = 3.1005438, riot =
5.1010876, x(N2) = 0.607820. (b)
K,= (0.4)? /(0.2)(0.4)° = 12.5. Ifx is the extent of reaction to reach the
new equilibrium position, the new equilibrium amounts are n(N>) = 14—x, n(H2) = 4 —3x, n(NH3) = 4 + 2x and
0
[(4 + 2x) /(202x)]’
~~ [02 — x) /(20 — 2x)][(4  3x) (20  2x)? The condition that the mole numbers be positive gives —2 < x < 1.333. Use of the Solver gives x = 0.12608, so n(N2) = 11.8739, not = 19.7478,
x(N2) = 0.6013.
6.50
The reaction rate at lower temperatures is too slow to make the reaction economically practical.
6.51
AG° =RTIn Kp =—2.303RT log Kp =
~2.303(8.314 J/molK)(500 K)[7.55 — (4830 K)/(500 K)] = 20.2 kJ/mol. din K3/dT =AH°RT. AH? = RT din Kp/dT =2.303RT d log Kp/dT= 2.303RT?(4830 K)/T? = 2.303(8.314 J/molK)(4830 K) = 92.5 kJ/mol. AS° = (AH? — AG°T = (72.3 kJ/mol)/(500 K) = 144.6 J/molK. Cp =(0H/OT)p. ACp = (0 AH°/0T)p = 0. 6.52
AG* refers to a change from pure, separated reactants, each in its standard
state, to pure, separated products, each in its standard state. AG® is not the change in G that occurs in the actual reaction mixture. The reacting system does not contain substances in their standard states. 77
6.53
K>, is a function of T only. Therefore only (d) will change Kp.
6.54
(a)
only and are AH° = ;v;H;>, ,.Enthalpies of ideal gases depend on T
(b)
unaffected by pressure changes or by mixing with other ideal gases; so the observed AH per mole of reaction does equal the reaction's AH”. Entropies of ideal gases depend strongly on P and the gases in the reaction mixture are not at  bar partial pressures. Therefore, the
entropies in the mixture do not equal the standardstate entropies and AS per mole of reaction differs substantially from AS®. (c)
6.55
Since AS, # AS®, it follows that AG, # AG®.
Let the superscripts 750 and 1000 denote standard states based on 750 torr and on 1000 torr, respectively. Kee
Ko ed =[(Pyo, POPU
(CA TE
code
oe 3d a
Sty
Cay
NEN:
(The gases are unaware of what choice of standard state has made, so the
equilibrium partialpressure ratio Pyo,/Py,o, 18 independent of the standard
state choice and cancels.) The Appendix AyG° data give A;G’7 = 4.73 kJ/mol and use of AyG*'° =RT In K3"° gives Kp” = 0.14. Then
K%'°° = (750/1000)0.14g = 0.11). 6.56
(a)
Since AH° 0, we have AG° < 0 at high T and K» > 1 at high T. Thus the highT limit has more trans isomer than cis isomer.
(c)
Even though the percentage of cis isomer continually increases as T increases, it is still possible to have more % cis trans than cis at high T. In the graph at the right, the % cis continually increases with
78
T but always remains below its asymptotic T = © limit of 43%. (d)
AG° = AH° —TAS°, where AS° and AH® are constant. In this equation,
the coefficient —AS° of T is negative, so AG® decreases as T increases. As noted in (a), K » decreases as T increases. We have AG°/T = — R In Kp, and since K > is decreasing as T increases, AG°/T is increasing as T increases, which is the opposite behavior as shown by AG*. (If this seems puzzling, note that AG° is negative.) (e)
Yes. For the reverse reaction, trans — cis, the results of (d) show that
AG? increases and K> increases as T increases. (The behavior of Kp is determined not by AG° but by AG°/T.) 6.57
0 = (d/dm) Dj (i — mx; — bY” = Di AHx)Oi— mx — b) = 2) xpi t+ 2m Dux, +2b xs
[b=Ouxoim pox

0 = (d/db) Xi (y; — mx — bY? = Li (i — mx; — b) = —Li yi + m Lixi+ nb, since ”,b=nb; thus [b= (Liyi—m Lx,)/n }. Equating the two bracketed equations for b and solving the resulting equation for m, we find m = (n Dixyi Lixi Di yd/(n plea — (>; x;)’]. Substituting this expression for m into b = (Yi yj —m L;x;))/n and using a common denominator for the terms on the right, we obtain the equation given in the text.
6.58
(a)
nj — Ni,0 = An; = vie, SO Nieq = Nio + vit.
ny, =3 mol —3§ (b)
nn, =
1 mol— Ge
My, = es
G=DinG,, (7, Pd = Li uw; (T, Pi) = Lini{w; (T) + RT In (P;/P°)), where the sum goes over N2, H2, and NH3, the n;’S are given in (a), and the P;’s
are: Py = [(3 mol — 3&)/(4 mol — 2&)]P,
Pyu, = [2E/(4 mol — 2&)]P,
Py, = (ny, /Mo)P = [(1 mol — &)/(4 mol  2§)]P. H= Lin, ; (T,Pi). Since H of an ideal gas is independent of P, H ehe= Heesand
H = YinjH*, ;, where the n,’s are given in (a). (c)
Results at € = 0, 0.2, 0.3, 0.4, 0.6, 0.8, 1.0 are: G/(kJ/mol) = 284.73,
288.45, 289.06, 289.20, 288.17, 286.07, 277.21; Hi(kJ/mol) = 23.55, 3.60, 6.37, 16.34, 36.29, 56.23, 76.18; TSI(kJ/mol) = 308.28, 292.05, 282.69, 272.86, 251.88, 228.83, 201.03. 79
10 HN =5.91
38 PN = ((1—X)/(42*X))*P
12 14 16 18 20 22 24 26 28 30 32 34 36
40 42 44 46 48 50 52 54 56 58 60 62
HH = 5.88 HA =38.09 G1 =97.46 G2 =66.99 G3 =144.37 T =500 R=8.314E3 PO=1 P=4 FOR X=0TO1 STEP 0.1 NN=1X NH =33*X NA =2*X
PH = ((33*X)/(4—2*X))*P PA = (2*X/(4—2*X))*P IF PA =0 THEN PA= 1E7 IF PN = 0 THEN PN = 1E7 IF PH =0 THEN PH = 1E7 GN = G1 + R*T*LOG(PN/PO) GH = G2 + R*T*LOG(PH/PO) GA = G3 + R*T*LOG(PA/PO) G=NN*GN + NH*GH + NA*GA H = NN*HN + NH*HH + NA*HA TS=HG PRINT “Xl=”";X;“G=”;G; “H=:H;“TS=";TS
64 NEXT X 66 END
6.59
(a)
Any reaction with An = 0; e.g., H2(g) + Clo(g)
— 2HCI(g).
(b)
From (6.36), to have K, independent of T requires that AH° = 0. Two mirrorimage species have the same energy, so AH®
dCHFCIBr(g)
is zero for
— /CHFCIBr(g). [Actually, because of the
nonconservation of parity, two mirrorimage molecules have an extremely tiny, experimentally undetectable energy difference. ]
6.60
(a)
The system discussed near the end of Sec. 6.6 with ny, = 3.00 mol,
ny, = 1.00 mol, nyy, = 1.00 mol has xy, = 3.00/5.00 = 0.600. If the reaction N2 + 3H2 «— 2NH; shifts slightly to the right with 0.100 mol of
Nz being consumed, the new amounts are ny, = 2.90 mol, m,, = 0.70 mol, nyy, = 1.20 mol, and xy, = 2.90/4.80 = 0.604. Thus x, has increased even though ny, has decreased.
(b)
Xj =NINo
ax; = dnjlnio — (nin ” ) dno. We have dn; = v; d& and dnyot =
d ND; yi ay dn; = pa Vj dE = (dj vj) dé = (An/mol) dé. SOx,
(Vino) AE — (xi/ntor)(An/mol) d& = n« [v; — x(An/mol)] dE. 80
=
6.61
Rodolpho is nght. The equations cited by Mimi show that (d/dT)(RT \n Kp) =AS°, so that the sign of AS° determines whether T In Ke
increases or decreases as T increases, but the sign of (d/dT)(T In Kp) can
differ from the sign of (d/dT)(In Kp)
6.62
(b).
6.63
(a) False.
(b) False; G is minimized only if the system is held at constant T and P. (c) True; see Eq. (6.4). (f) True; see Fig. 6.8. (i) False.
(d) False; see Sec. 6.6. (g) True.
(e) False; See Fig. 6.8.
(h) False; this is true only if An = 0.
() True.
(k) True; from (6.13) and (6.19), K, depends on P® but Kp does not.
(1) False.
Reminder:
(m) False;
(n) True.
Use the solutions manual as an incentive to work
problems, not as a way to avoid working problems.
8
Chapter 7 7.1
(a) F; (b) F.
thy?
(a)
p= 1,r=0,a=0;f=cp+2r—a= c= 2 (water and sucrose), The degrees of freedom are T, P, and sucrose mole fraction.
(b)meoseepa=l7=0,a=0;
ne
f=4 The degrees of freedom are 7, P, Xsucrose;
Xribose
(c)
c=3 (sucrose, ribose, water), p = 2 (solid sucrose, the solution); f=324+200=3(7,P, Xribose). Note that Xsucrose in the solution
saturated with sucrose is fixed. (d)
c= 3 (sucrose, ribose, water), p = 3 (solid sucrose, solid ribose, the solution); f=3—3+200=2 (T and P).
(e)
c=1 (water), p =2;f=1—2+200=1. The degree of freedom can be taken as either T or P. (Once T is fixed, P is fixed at the vapor pressure of water.)
(f)
c=2
[T and Xsuctose (or P
p= 2; f=22+200=2
(water, sucrose),
ANGE encrace):
Te)
(g)
c=2,p=3;f=23+200=1
(either T or P).
(h)
c= 2, p—3%f= 2 34 2 = (either Fore).
(a)
f=Cina—p +2 and p = Cina + 2 —f. The smallest possible fis zero, SO Pmax = Cd +2:
(b)
7.4
If
p= 10, then cing must be at least 8.
(q)erLO 2 Hi + Ole HPO, © Ht LPOseE Osea pi BPO 7a
apd
aha PO}
=
teeta
Bat
s
. The equilibrium conditions are: EPO, aaee
LoHO > aie
ee EO, ae
HPO; ;
My por: = Hye + Pypo2 > Uypoz = Myr + Hpo: Phe electroneutrality condition is Xi
=X a SX XPO;
+ 2X poz
+ 3x
= PO,
. There are 7
species (H2O, H*, OH’, H3PO4, Hx2PO;, HPO 4 , PO} ) and Eg. (7.10) gives Cing = 7
—4—1=2.
Hence f=2—1+253.
82
(b)
X,+ = X, and Ay = x, (assuming that no solid precipitates out of the solution). We shall neglect the ionization of water. (Its inclusion
would not change f.) The electroneutrality relation is Xe t+ Xea x
Br
tp
]
=
and is not an additional condition since it follows from the
two preceding equations. The species are K, Br, Na‘, Cl’, and H20. Equation (7.10) gives cing = 5 O— 2 =3. Hence
Ticks)
(a)
Cing =
300 =3 and
f=3—1+2=4.
f=31+2=4.
A reasonable choice is T, P, and
two of the mole fractions.
(b)
2NH3 < N2 + 3H2, $0 Hy, +34y, =2Myn, and r= 1. Hence cing= Ble
0 = 2and f=2 — l=
3
and
one mole fraction. (the other
two mole fractions are determined by the reaction equilibrium condition
and yy x, =1.
2Uyn, = Hn, + 3H, and xy, = 3xy, (since all the N2 and H2 come from decomposition of NH3). Hence cing = 311=1
andf=1—1+2=2.
[In (b) and (c), the catalyst was not counted in finding c. If the catalyst is considered to be a species whose amount can be varied, then c and Cina
are each increased by 1. If the catalyst is a solid, then p is increased by  f isunchanged; if the catalyst is in the gas phase, then fis increased and
by 1.] T and P.
(d)
We have r= 1. The mole fractions satisfy xy, +X, =1 and the condition that all the
N come from N> does not give an additional relation between
mole fractions’ fimc— p+2—ra=—214+2=)0=2
(e)
We have the reaction equilibrium CaCO3(s)
T and P.
ie CaO(s) + CO2(g) and the
phase equilibria CaCO3(s) = CaCO3(g) and CaO(s) = CaO(g). There is 1 reaction equilibrium condition. The phase equilibrium conditions have already been taken into account in deriving the phase rule; see Eqs. (7.3)(7.6). There are no relations between mole fractions. There are three species (CaCO3, CaO, COz) and Cina = 3 —1—0=2. Hence
fe 25382 'S) lerT.
83
7.6
Weshave the teactionsI) HON 2H’
1CNeand (2H. Ops Hf Ole
Let n, ,,.and n, ,,, denote the moles of H coming from reactions  and 2, respectively. Stoichiometry gives 1, 44+ = "Non and n, \+ = Noy > $0 Non
Ua
we As
H™
=n
1,H* “s do H*
= ip
Hg
and NaCl),p = 2,f=22+200=2
(T and P).
(a)
c=2 (HO
(b)
c=4(H,O, NaCl, Na’, Cl), p =2; r= 1, since we have the equilibnium
NaCl(s) a Na*(aq) + CI (aq); a = 1, since we have the electroneutrality
condition x Nav Gq) ae. Cla(aqy So f=42+211=2.
7.8
(a)
c=4,p=1,r=2 [HO < H*+OH and 2H,0 © (H)0)], anda= 1 (4
Cadi,
(b)
(a)
adie):
di
c=5,p=1,r=3 [the reaction in (a) and H.O + (HO);— (H20)s], (= NO
1.9
=)
ra=4 =) 427
= Xoy ef =O P+2—
1,
Here, the equilibrium conditions P® = p? = P’ =.. are eliminated. Instead of specifying 1 pressure, we must specify p pressures (where p 1s the number of phases). This increases fby p — 1, sof=cina—D+2+p—1 = Cig tel.
(b)
Here, the equilibrium conditions of Eqs. (7.3) to (7.6) are eliminated. There are c(p — 1) such conditions, so fis increased by c(p — 1) and f= cp+2ra+cp—c=cppt+2r——a=Cn—pt+t2+cp—c.
7.10
(aya
Gent
(a) Liquid;
7.12
(a)
Ub) To otc) Ea (d) bo (e) (b) gas;
(c) liquid;
ys (d) liquid;
ato)
(hh) ste)
(e) gas.
c=1,p=2;f=1—2+2—00= 1, as in Prob. 7.26.
(D)tec—ily pi ls feeee
(c) c=)
84
pee
eee
OO)
7.13
(a)
Treat the vapor as an ideal gas. The liquid’s volume is negligible
compared to the container’s volume. If the equilibrium vapor pressure is reached before all the liquid vaporizes, then the gas will be at 23.76 torr, and
(23.76/760) atm (10000 cm°) bE Nowe = 0.01278 mol BS RT (82.06 cm> atm/mok K)(298.15 K) Meas = (0.01278 mol)(18.015 g/mol) = 0.230 g AV
There are 0.230 g of vapor and 0.130 g of liquid.
(b)
With V = 20000 cm’, we would get 0.460 g of vapor if the equilibrium vapor pressure of 23.76 torr were reached before all the liquid vaporized. However, there is only 0.360 g of water present initially, so all the liquid vaporizes to give a system with only vapor present.
7.14
(a) Gas;
P/ atm 7:
i LOR
ae repepeticne
pemn ed
7.15
7.16
TA/
E
s
(b) solid;
(d) liquid;
(e) solid;
(f) liquid;
(g) gas.
(c) gas;
°
Isothermally and reversibly increase the pressure to the pressure at which ice melts at 10°C. Then reversibly freeze the water at constant T and P. Finally, reversibly and isothermally reduce the pressure on the ice to  atm.
(a)
Ar, which is larger and so has greater intermolecular attractions.
(b)
HO, due to hydrogen bonding.
(a)
saan
(b)
(c) C3Hs, which 1s larger.
2) cal/molK = 87 J/molK = AvapHn/(319.4 K), and Avapim =9.7 kcal/mol = 28 kJ/mol. AvapSm = 4.5R + R In 319.4 = 20.4 cal/molK = 85.4 J/molK = Naot (3194 K) and AvapHm = 6.52 kcal/mol = 27.3 kJ/mol.
85
7.18
7.19
The hydrogen bonding increases the degree of order in the liquid and hence decreases S of the liquid. Therefore, AS for the transition liquid — gas 1S increased.
(a)
The THErule gives Avapllmnbp.co (Sle) Kj (8.314 J/molK)(4.5 + In 81.7) = 74.0 J/molK and AyapHm,.nbp.co = 6.05
kJ/mol. Similarly we get 55.7 kJ/mol for anthracene, 168 kJ/mol for MgCl, and 295 kJ/mol for Cu.
(b)
Experimental values in kJ/mol (from Barin and Knacke, Reed et al., and the AIP handbook) are: CO—6.04, anthracene—S6.5, MgCl.—156, and
Cu—304.
7.20
Assuming ideal vapor and using (3.29) we have AS, = const.=
ASm1 + ASm.2 = AvapSimabp +R In (Vi,/Vimi) = AvapSm.nbp + Rin Vi, —R In (RTppp/1 atm) = AvapSmabp — R In (Tnbp/K) + R In Vi, — R In (R K/1 atm) = const.,
ke) const, Hk In Ve _ so AyapSmnbp = R In (Tipp
R In (R K/1 atm) = R In (Tnbp/K) + const’., which is the THE rule.
7.21
(a) T; (b) T; (c) F; (d) T; (©) F, @) F.
7.22
Assuming ideal vapor, neglecting Vinjiq iN Vm.gas — Vig, and neglecting the T
dependence of AyapHm, we use the integrated Clausius—Clapeyron equation (7.21). If state 1 is the normal boiling point 34.5°C and state 2 is 25.0°C, then In (P2/760 torr) = —[(6380 cal/mol)/(1.987 cal/molK)] x [(298.1 K)'  (307.65 K) '] = 0.332s = In (P2/torr) — In 760 and In (P2/torr) = 6.301; P2/torr = 545; P2.= 545 torr.
7.23
(a)
Integration of dP/dT = AH/(T AV) gives P2 — P, = (AH/AV) In (T2/T)) if
AH and AV are assumed constant. From Prob. 2.48, for one gram AH = 79.7 cal and AV = (1.000) 'cm? — (0.917)! cm? = 0.091 cm®. We have
Blan
79.7 cal
+ —0.091cm”
82.06cm* atm Wy 2215 K 1.987 cal
86
n———__— = 154 atm PASE
If AS and AV are assumed constant, then P2 — P; = (AH/T; AV)(T2 — T});
P, =latm+ (b)
82.06 cm? atm
79.7 cal
eee ee oe (273.15 K)(0.091cm3) 1.987 cal
OOK) ea aaaca ) (
With 272.15 K replaced by 263.15 K, we get P2 = 1350 atm if AH and AV are assumed constant or 1325 atm if AS and AV are assumed constant.
(c)
7.24
AH and AV change a lot for this large AP.
Use Eq. (7.24). For 1 g, AfusV = (1g)/(13.690 g/cm*) — (1 g)/(14.193 g/cm’) = 0.00259 cm?.
(a)
3
99 ii (Ty T,) =(234.25 Ky 202? 2.82 cal
eR,
82.06cm~
atm
K
T> = 234.25 K + 0.52 K = 234.77 K = 38.4°C
Tao
(b)
Replacing 99 atm by 499 atm in the above equation, we get AT’= 2.60 K and I> = 236.85 K =—36.3C.
(a)
Avapffmave = 40.7 kJ/mol and In(760 tort/P) =
—[(40700 J/mol)/8.314 J/molK][1/(351.5 K) — 1/(298.2 K)] = 2.49. We get P, = 63 torr, which is pretty close to the true value 59 torr. (b)
dP/dT = AHI(T AV) = AHI/TV gas. The true Voas 18 less than Vgas,ideal, SO the
true dP/dT is greater in magnitude than the dP/dT used in the calculation in (a). Therefore the change in P calculated allowing for nonideality will be greater in magnitude and the 25°C calculated vapor pressure will be less than 63 torr, likely bringing it closer to the true value 59 torr.
7.26
Since IT = const. X P, we have In (JT/K) = In const + In P. The slope of an
In(IT/K) vs. T' curve therefore equals the slope of an In P vs. 1/T curve, which equals —AsupHn/R, according to the Clausius—Clapeyron equation.
(b)
14)
2.18x 104 K = —AgupHp/(8.314 J/molK) and Asupm = 181 kJ/mol.
Equation (7.21) gives
760 torr
M3 J6torr
l
Vals
1.987 cal/molK \373.15K
AH m = 10.2 kcal/mol = 42.7 kJ/mol
87

298.15K
7.28
AH m = (539.4 cal/g)(18.015 g/mol) = 9717 cal/mol. (a)
Equation (7.21) gives In
1
1
9717 cal/mol
PB
1.987 cal/molK Ger eee z]
760torr
In(P/760 torr) = 0.667, P/760 torr = 1.950, P = 1480 torr (b)
These)
(a)
1 9717 cal/mol (1 , 446 torr ___ Toulon T60NOTun a1 .987 calimolo Katara WERT RON 10n Ke el =35 8)6 ne Saas A plot of In(P/torr) vs. 1/T has slope —AH,,)/R. —0:2934 1.2986 2.4214 In(P/torr)
Ke
24.204
25456
26.799
281d
10°/T
Okey An)
y = 7376.7x + 18.469
In (P/torr)
2.5
So
0.0024
ee
ee
ee
0.0025
ee
SS
0.0026
ES
ee
0.0027
eee
0.0028
0.0029
Tk The plot is linear with slope [0.76 — (—2.50)]/(0.00240 — 0.00284,)K"! = —7380 K = —AH,,)/(1.987 cal/molK) and AH, = 14.66 kcal/mol = 61.3 kJ/mol. (b)
—
Equation (7.21) gives
‘
B 1.845 torr
___
61300 J/mol
1
*
8.3145 J/moK \433.15K
In (P/1.845 torr) = 0.8240, P = 4.206 torr (Alternatively, the graph can be extrapolated.)
88
l 413.15K
(c)
Use of Eq. (7.21) gives
760torr
14600 cal/mol
1
1
"7.845torr
1.987 cal/mok K
IM
AIBNS.K
T6280 350 G (The true normal boiling point is 356.6°C. The error arises because AH, is not constant over the long temperature interval from 140 to 350°C.)
(d)
Setting up a spreadsheet like Fig. 7.8, one finds the regression analysis of the In P versus 1/T data gives intercept b = 18.4706 and slope —7377.53 (corresponding to A,,,H, = 61.34 kJ/mol), and these values give a sum of squares of residuals of P experimental versus P calculated of 7 x 10ue When the Solver is run to minimize this sum, we get 18.4326 and ~7362.4 as the intercept and slope, with the sum reduced to 6 x [On eis gives Avapffm = 61.2 kJ/mol. When this value is used in (b), we get P =
4.200 torr.
7.30
(a)
A plot of In(P/torr) vs. 1/T has slope AH,)/R.
In(P/torr) 10°/T
In (P/torr)
2.283 6.5295
3.544, 6.1293
4.6521 5.7753
5.6330 5.4600
Ke
6.5 6.0 a5) 5.0 4.5 4.0 335 3.0 2:0 2.0 0.0054
0.0056
0.0058
0.0060
0.0062 0.0064 0.0066 Ter?
The plot is linear with slope (5.82 — 2.08)/(0.00540 — 0.00660)K™' =
~3120 K = —AHp/(8.3145 J/molK) and AHn = 25.9 kJ/mol = 6.20 kcal/mol.
(b)
Equation (7.21) gives In(P/279.5 torr) = —[(25900 J/mol)/(8.314 J/molK)][(198 Ky = (838k el and Pe OL) tor 89
Ts
Trouton’s rule is AH,,/T= 10.5 R. At the normal boiling point, AV = Vin.gas = Vmitiq = Vingas = RI/P=RIp/Cpatm). Fora small change in P, we have dP/dT = AP/AT and the reciprocal of Eq. (7.18) becomes AT/AP = (AH m/T) 'AVm = (10.5R) RTp/(1 atm) = Tp/(10 +atm).
7.32
(a)
The 0°C path solid — liquid — gas shows that AsupH/m = Atusllm + Avapltm = 51.07 kJ/mol. (Pressure changes have no significance since P has little effect on H.)
(b)
The Clapeyron equation is dP/dT = AH,/(T AVm). For the solid—vapor line, AHm = Asupffm and AV = Vin.gas —
Vso. = RT/P — Mgo1/Psoi. =
(82.06 cm?atm/molK)(273 K)/[(4.585/760)atm] — (18.0 g/mol)/
(0.92 g/cm) = 3.716 x 10° cm?/mol — 20 cm*/mol = 3.716 x 10° cm?/mol, assuming ideal vapor. (dP/dT)so1gas
= AHm/(T AV) =
(51070 J/mol)/(273.16 K)(3.716 x 10° cm*/mol) = (5.031 x 10° J/cm?K)(82.06 cm?atm/8.3145 J) = 4.966 x 10~ atm/K = 0.3774 torr/K. For the liquid—vapor line,
AH m= AvapH{m = 45060 J/mol; AVm = 3.716 x 10° cm*/mol; (dP/dT)jigvap = 9.3330 torr/K. For the solid—liquid line,
AH = 6.01 kJ/mol, AVm = M/Ptiq — M/Psotia = (18.015 g/mol)(1.000 cm?/g — 1.0905 cm’*/g) = —1.630 cm*/mol, and
dP/dT = 1.012 x 10° torr/K. TERE
(a)
At 25°C, In (P/torr) = 18.3036 — 3816.44/(298.15 — 46.13) = 3.1602 and P = 23.58 torr. At 150°C, the Antoine equation gives In (P/torr) = 8.1810 ands? = 35/2 torr:
(b)
The Clausius—Clapeyron equation is d In P/dT= AH,,/RT”, with ideal vapor assumed and the liquid’s volume neglected. Differentiation of the Antoine equation gives at 100°C: d In P/dT= B/(T/K + CyK =
3816.44/(373.15 — 46.13)°K = 0.035687/K = AHm/RT? = AH p/(8.3145 J/molK)(373.15 K)? and AHnvap = 41.315 kJ/mol. 7.34
(a)
We make the approximations that the solid and liquid volumes are negligible compared to the vapor volume, that AH of vaporization and sublimation are constant, and that the vapor is ideal. Then Eqs. (7.21) and
(7.22) apply and show that In P varies linearly with 1/T. We plot In P vs.
90
1/T for the solid and join the two points by a straight line; we do the same
for the liquid. At the triple point, the solid and liquid vapor pressures are equal, so the intersection point of the two lines gives the triple point; this is found.to bePi=1.5.¢torra?
(b)
= 200 K.
Equation (7.21) gives In (10.0 torr/1.00 torr) =
(AgubHm/ 1.987 cal mol” K7')(1/177.0 K = 1/195.8 K) and AjupbHm = 8430 cal/mol. Also In (100.0 torr/33.4 torr) =
AvapHm/1.987 cal mol! K7')(1/209.6 K — 1/225.3 K) and AvapHm = 6550 cal/mol.
TESS)
Ags/m/(cal/mol) = 8430 — 6550 = 1880.
To use the ClausiusClapeyron equation to find the solid’s 1200°C vapor pressure, we need the enthalpy AjupH of sublimation of the solid and we need to know the solid’s vapor pressure at some particular temperature T~ We shall take T’as the triplepoint temperature Ty, since the solid and liquid vapor pressures are equal at 7, The triplepoint temperature differs only slightly from the normal melting point, so we take T;, = 1452°C = 1725 K. We use Eq. (7.21) to find the liquid’s vapor pressure at T,, = 1725 K; this equals the solid’s vapor pressure at 1725 K. Equation (7.21) gives for the liquid—vapor equilibrium: In (P2/P1) = —AHm(1/T2 — 1/T)) and 7
1.00
torr
0.100 torr
1
ap lm
Yo
1.987 cal/mok K
a
2078K
1
1879K
which gives AyapHm = 89.79 kcal/mol. Use of (7.21) for the liquid between 7;,
= 1725 K and 1606°C gives ee
A!
Lig
0100 torr ___89790.cal/mol
l
1725K
1.987 cal/molK\1879K
which gives P,, = 0.0117 torr. The paths solid — gas and solid — liquid — gas at the triple point give AjupH = AfusHT + Avapff. Thus AsupH = 4.25 kcal/mol + 89 79 kcal/mol = 94.04 kcal/mol. Now we use (7.21) for the solid:
pa
bene 0.0117 torr
li
pigunns4040‘eal/inol
1.987 cal/moK
(1473 K
wsinary
1725K
= 4.694
P=1.1 x 10~ torr 7.36
When the two forms are in equilibrium, their chemical potentials are equal. We thus want to make Gmai = Gmgr For each form, dGm = —Sm dT + Vm dP =
V,, dP at constant T. Solids are nearly incompressible, so we neglect the 91
change in Vm, with P. For each form, AGm = Vm AP or Gm(P2) = Gm(P1) + V., AP, where P; = 1 bar. Setting Gmai(P2) = Gm,gr(P2), we have Gmai(P1) + Vidi AP = Gmge(P1) + Vingr AP and
AP = [GmeP1) — Gmai(P1)V/(Vm.di —
Ver) = Ay G
gi/(Vinai — Vmgr), since Pi =
1 bar = P°. Using Vn = M/p, we get AP =
[(6070 J/mol)/(1.97 cm*/mol)](82.06 cm? atm)/(8.314 J) = 30400 atm and Pz = 30400 atm.
Dead
When gray (g) tin is more stable than white (w), we have Uy < Hy and Gm < Gmw. We have dGm = —Sm dT + Vm dP = —Sm GT at the constant P of 1 bar. Appendix data show that Sm > Sm, SO Gmw decreases faster than Gm» as T increases, and Gm, increases more rapidly than Gm.» as T decreases. At the temperature Teg with Gm. = Gm, the two forms are in equilibrium. Below Teg, we have Gm < Gmw and gray tin is more stables Let 7; = 25°C, 73 = Tq, _and
AT = Teg — 25°C. We have AGmn.g = Gm,g(Teq) — Gm,g(25°C) and AGmw = Gmw(Teq) — Gmw(25°C), 80 Gm,g(Teq) = Gm,w(Teq) becomes Gm,g(25°C) + AGm.g = Gmy(25°) + AGmw Or AGmg — AGmw = Gmw(25°C) — Gme(25°C) =
0 — 0.13 kJ/mol, where Appendix data were used. Since AGp = =? SredT at
constant P, we have AGm,g — AGmw = — Jf (Smg — Smw) aT = (Smw — SmgMT2 — T1) = (51.55 — 44.14)(J/molK)(T2 — 25°C), if we neglect the T dependence of Smg — Smw. Then —130 J/mol = (7.41 J/molK)(T— 25°C) and
Teg — 25°C = 17.5 K = 17.5°C, 80 Teg = 7 5es
7.38
In Example 7.7, Viner —
Vm,ai WaS assumed independent of pressure. At 25°C
and  atm, Vingr= 5.34 cm?/mol and Vina = oa! cm?/mol. Since graphite is more compressible than diamond, as P increases, Vm,gr Will decrease more
rapidly than Vin.gi and the difference Vin.pr—
Vmai Will decrease as P increases.
Since this volume difference occurs in the denominator of the expression for Pz —P,, P2— P; will be greater than calculated in Example 7.7 and P2 will be
greater than 15100 bar.
1h
Cp = (OH/0T)p = T(0S/0T)p. Since Cp > 0, both H and S increase as T increases.
A firstorder transition has AH > 0 and AS > 0, and so we see a sudden jump in H and in S at the transition temperature 7;,,. A secondorder transition has AH = 0 and AC? # 0, so there is no sudden jump in H but there is a sudden change
92
in the slope 0H/dT = Cp at T,,;. A lambda transition has AH = 0 and if Cp > © at T;; the H vs. T curve has a vertical (infiniteslope) inflection point at Tis.
Since (0S/0T)p = Cp/T, the Svs.T curves resemble the Hvs.T curves.
7.40
Lambda
Second order
First order
At T=0, w =Oand
o, =7/r=1.
If r=0, 6,=w/w = 1. This is a highly ordered state; with Cu and Zn atoms interchanged, each Cu is surrounded only by Zn atoms.
If r=w, then o, =0. In the upper (7 = 0) diagram, w = 0 and o, = 1. In the lower diagram, comparison with the upper diagram gives r= 16, w = 16, and 6, =
(16 — 16)/(16 + 16) = 0. At
T=0,n,p =n, ando,=21=
i
(f)
As T > ©, nyp = Yanp and Os = 2(%) 1=0.
(g)
In the upper (T = 0) figure, n,p = Np and o, = 1. In the lower figure:
np = 4(3) +2 + 4(3) +2 + 403) + 2+7=49, and ny = where 2424241424+143414342434+14+2+142+1=29, the counting was done by looking at the neighbors of the atoms in rows, 2, 4, 6, and 8. Hence o, = 2(29)/49 — 1 = 0.18.
(h)
From Sec. 7.5, both curves have infinite slope at 7: O¢
Os
Ty.
Th 93
7.41
At 0 K, the most stable arrangement in the crystal has the negative end of each molecule next to the positive end of the adjacent molecule. As T increases, the crystal becomes more disordered until there is a 5050 chance for the negative end of one molecule to be next to the positive end of the adjacent molecule.
7.42
(a)
At T=T, we have t=0. For T 0. As the positive
0.01285 goes to zero and Cp/(J/molK) goes quantity t goes to zero, t © =1 to B. Likewise, s is positive for T >7, and s ~ — 0 as the positive quantity s goes to 0. Hence Cp = 456 J/molK at 7) .
(b)
For T
(dP/dT)iigsgas The solid—vapor line has greater slope.
7.45
The pressure due to 10 cm of water is given by (1.9) as
P= pgh =
[1.00 (10° kg)/(10~ m*)](9.80 m/s)(0.10 m) = 980 N/m? = (980 J/m°)(82.06 x 10° m? atm)/(8.314 J) = 0.009, atm, where (2.7), (2.14), (1.19) and (1.20) were used. The total pressure 10 cm below the surface is
94
1.009; atm and Eq. (7.24) and data in the example that follows it give
0.009, atm =
THOT 7.46
15
79.7 cal 82.06cm? atm T, — 273.15 K 273.15K 1.987 cal —0.091cm*> KEE 3SC10FK:
Neglecting the change in water density with pressure and the change in g with depth, we have as the pressure at 3000 m: P= pgh=
(1.0 g/em?)(1 kg/10° g)(10° cm*/1 m*)(9.8 m/s*)(3000 m)= (2.9 x 107 Pa)(1 atm/1.01 x 10° Pa) =290 atm. P exceeds the 350°C vapor pressure of water and the stable phase is liquid water.
7.47
Let sc denote supercooled liquid water. All equations in this problem are for ~10°C. The phase equilibrium condition and Eq. (6.4) ZIVE Mice = solidi shi (he 950 torr/750 torr) and psc= vapor above sc= = Lvapor above ice=f,
torr/P;). But uw; + RT In (P;/750 torr). Subtraction gives Bice — Use= RT n (1.95
= Prob. 4.29b gives Hice — Msc = Gmiice — Gmsc = (2.76 cal/g)(18.0 g/mol)
_49.7 cal/mol. (This relation holds at 1 atm, but the effect of pressure on liquid e and solid thermodynamic properties is slight and can be ignored.) Therefor 49.7 cal/mol = RT In (1.95 tort/P;) = and (1.987 cal/molK)(263.1 K) In (1.95 torr/P;). We find 1.95 torr/P; = 0.9093
P, = 2.14 torr. (The experimental value is 2.15 torr.)
7.48
torr 0.01°C is the triplepoint temperature, so ice has a vapor pressure of 4.585 at 0.01°C.
7.49
(a)
above liquid A; The initial state (state 1) has partial pressures Pa and Pg
2) the liquid is subject to a pressure P = P, + Pg. The final state (state has partial pressures Pa + dPa and Pg + dPx above liquid A, which s in experiences a pressure P + dP, where dP = dP, + dPyx. Let the change be HEIOG and vapor A chemical potentials on going from state 1 to 2 2 ~ and du(A°*). The oe equilibrium conditions for states  and du(A ‘)
are [;(A ‘)= 1i(A®) and pofA’ ) = pp(A%)=pi(A‘) + ey Ve
dG A Ae) = u)(A%) + du(A*). So gies we Ha We have du(A ‘) = dT VVn(A ea wale iV‘(A ‘)dP, since T is constant. Also, BSA
95
(AS) = 1°(AS) + RT In (Pa/bar) and du(A%) = (RT/Pa) dP at constant T.
(b)
7.50
Hence Vm(A ‘) dP = (RT/Pa) dP = Vm(A*) dPa. Q.E.D. dPa/Px = [Vm(A ‘)/RT] dP. Integration with V,,(A “) assumed constant gives In (Pao/Pa.1) = [Vm(A_)/RT\(P2 P1). So In(Pa.2/23.76 torr) = (18.1 cm?/mol)(1 atm)/(82.06 cm?atm/molK)(298 K) = 0.000740 and Px,2/23.76 torr = 1.000740, Pa, = 23.78 torr.
Use the 25°C path: a
b
liq(1 bar) >
liq(23.766 torr) >
c
— vap(1 bar).
vap(23.766 torr)
We have dGm = — Sm dT + Vm dP = Vm dP at constant T. For the liquid, Ge Vm AP, and
82.06 cm? atm
== 1:8:J/mol
Step b is a constant7andP equilibrium process, so AG» = 0. For step c,
AG, = ? Vm dP = ; (RUIP) dP = RT In (Po/Pi) = (8.3145 J/molK)(298.15 K) In (750.062/23.766) = 8557.2 J/mol. AG, + AG, + AG; = 8555.4 J/mol. From the Appendix, AG/(J/mol) = =228572'+ 237129 = 8557. 7.51
(a)
Trouton’s rule is AH,/T; = 10.5R = c. Substitution into Eq. (7.22) gives
In (P/atm) = (c/R)(T;/T) + c/R = —(10.5R/R)T;,/T + 10.5R/R = 10.5(1 —T,/T).
Teo2
P/atm = exp{10.5[1 — (353.25 K)/T]J}.
(b)
P/atm = exp [10.5(1 — 353.25/298.15)] = 0.1436 and P = 109 torr.
(c)
In (620/760) = 10.5(1 — 353.2s5/T) and T= 346.5 K = 73.4°C.
(a)
Use of (7.21) gives at 0°C:
In (4.926/4.258) =
—[AH p/(1.987/cal/molK)][(274.15 K)"! — (272.15 K)"] and AHm.273 = 10.80 kcal/mol = 45.20 kJ/mol. Similarly, AHjn373 = 9.89 kcal/mol =
41.37 kJ/mol. Then ASm.273
= AH m.273/T = (10800 cal/mol)/(273.1 K) =
39.5 cal/molK = 165 J/molK; 111 J/molK.
AS)373
= AHm373/T = 26.5 cal/molK =
AG m= AH, —T AS, = 0, as it must be for constant7and
P equilibrium processes. The calculated 100°C AHpy is slightly in error, since the approximations used to derive Eq. (7.21) are less accurate the higher the vaporization temperature and the closer the temperature is to 96
the critical point; for example, H2O(g) is denser and hence less ideal at  atm and 373 K than at 5 torr and 0°C.
(b)
AH? and AS° are for the process liq — ideal gas at 1 bar. Let P be the 0°C vapor pressure. A convenient path is the following 0°C path:
2 3 4 liq(P°) > liq(P) > vap(P).— vap(P°) — ideal vap (P*)
Since moderate pressure changes have little effect on liquid thermodynamic properties, we can take AH; = 0, AS; = 0. Gas nonideality is slight at 1 bar, so we take AH4 = 0, AS4 = 0. Since the vapor is nearly ideal and H of an ideal gas depends on Tonly, we take AH; = 0. Thus AH3,, = AHm 2= 45.20 kJ/mol. Use of (3.25), (3.29), and Boyle’s law gives AS57, = ASm2 + ASm3 = AHm2/T+ R In (P/P*) =
(45200 J/mol)/(273 K) + (8.314 J/molK) In (4.58/750) = 123.2 J/molK.
7.53
(a)
dna=dn',—ad&,
dng=dn, —b dé, dng=edé&,
dng=fd§.Then
Y; p; dni = Ua(dn®, — a d&) + Up(dn; — b d&) + Wee d& + Upfdé = — bie + ee + flr) d&= pa dn’, +Up dnz, ua dn’, + tp dn, + (aa since the equilibrium condition for the reaction is Deval = —aua — bp + ele + flr = 0.
(b)
dG=S dT +VdP +>; pj dn; =—S dT + VdP + Us dn‘, + Up dna; at constant T, P, andn},dG = Ua dn, and Wa = (0G/dn ’,)r.P.n'y
7.54
(a) and (c). See Fig. 7. 1a.
7.55
(a) T; (b) T; (c) F; (@) F; ©) T; © F
97
&) 1. th) B®
F
Chapter 8 (b) Pa K!? m°/mol’ and m?/mol;
(c) m°/mol.
8.1
(a) Pa m°/mol’ and m*/mol;
8.2
As p > 0, Vm 2 &. In Eq. (8.2), b can be neglected compared with Vm to
give
P=RT/Vm—alV> = (1/Vin)(RT— a/Vm). AS Vm > 2, RT — a/Vm goes to
RT and P > RT/Vm. (Note that it’s harder to use (1.39) to get the limiting
behavior since both P and a/V2, go to 0.) In (8.4), the terms B/Vm, C/Vi,, . each go to 0, giving PVm = RT. When (8.3) is solved for P and b is neglected
compared with Vm, we get P = RT/Vm—a/V2,T'\” = (1/Vm)(RT— al/VinT '*) > REV 8.3
(a)
n=(28.8 g)/(30.07 g/mol) = 0.958 mol;
Vin = V/n = (999 cm*)/(0.958 mol) = 1043 cm*/mol. (82.06 cm? atm/moFK)(298. 1K) , pe 1043 cm?/mol
[ be 186 cm?/mol
a 1.06 x10* cm*/mol?
1043 cm?/mol
(1043 cm?/mol)?
=O
atm
The idealgas P is P = RT/Vm = 23.5 atm.
(b)
B' = B/RT= (186 cm*/mol)/(82.06 cm>atm/molK)(298.1 K) = 0.00760 atm”; C' = (C — B’YR’T’ = (1.06 x 10* cm*/mol? — 1867 cm*/mol*)/ (82.06 cm*atm/molK)?(298.1 K)* =4.01 x 107 atm~. Vin = [(82.06 cm?atm/molK)(298.1 K)/(16.0 atm)] x [1 — (0.00760 atm™')(16.0 atm) — (4.01 x 10° atm™)(16.0 atm)"] = 1327 cm*/mol. V=nVm = (0.958 mol)Vm = 1272 cm’. Also, Vavidealsl oy
8.4
Eq. (8.4) gives
cm?/mol and Viel
40> cm’,
P = RT(1/Vim + BIV> + GIVE. ++). Substitution into the nght
side of (8.5) gives PVm = RT[1 + B'RT(1/Vm + B/V2, + C/V>, ++) + OVA
OMCs + 2B/V> +)+]. We compare this expression for PVm with (8.4); equating the 1/V,, coefficients, we get B = B'RT. Equating the Aaa
98
coefficients, we get C = B'RTB + C'R°T’ = BVR’°T’ + C'R°T’ =
Ral (Bas G:
8.5
V_.=RT/P+B = (82.06 cm’atm/molK)(200 K)/(1 atm) — 47 cm*/mol = 16.36 L.
8.6
(a)
Bi =(B, + By/2 =387 cm’/mol. Let gas  be CHg. Then x; = 0.0300/0.1000 = 0.300 and x27 = 0.700. The equation at the end of Sec.
8.2 gives B = (0.300)°(42 cm*/mol) + 2(0.300)(0.700)(387 cm*/mol) + (0.700)°(732 cm3/mol) = 525 em*/mol. V/not = 10000 cm*/mol. P = [(82.06 cm?atm/molK)(298.1 K)/(10000 cm*/mol)] x [1 — (525 cm?/mol)/(1000 cm?/mol)] =r
(b)
s5 2 ath
With B,7 = —180 cm*/mol, we get B = 438 cm*/mol and P = 2.34 atm. Also, Pideat = RT/(V/niot) = 2.45 atm.
8.7
Vm = V/n = (200 cm*)/(2.483 mol) =
n = (74.8 g)/(30.07 g/mol) = 2.48 mol.
80.4 cm?/mol. (a)
(b)
p=
10.6K Perit K)(310.6
R _(@2.06cm RL
3_atm/molat me
Von
80.4 cm” /mol
Equations (8.18) and (8.2) give
2
K)’ (305.4 K)” ON
a Ber (BU0G eminatu mee
64(48.2 atm)
ee 50.105 cm® atm mol7
3) eraennls. ie (82.06 cm°atm/mok K)(305.4 K) eee Cr
8(48.2 atm)
_ (82.06 cm?atm/mol K)(310.6K) _ 5.50 10° cm°atm/mol? (80.4 cm*/mol)? 80.4 cm?/mol — 65.0cm?/mol P = 1655 atm — 851 atm = 804 atm
(c)
Equations (8.20), (8.21), and (8.3) give _ 0.4275(82.06 cm?atm/mol K)’ (305.4 K)"" rR
48.2 atm
9.73107 cm®atmK'/mol? 3 +e 0.08664(82.06 cm’  atm/molK)(305.4 K) wah. 48.2 atm
99
3/mol
(d)
P = RTI(Vm—b) — alVm(Vm + b)T '” = 720 atm — 548 atm = 172 atm Interpolation gives at 372°C: B =179 cm*/mol + (7.5/20)(22 cm’/mol) = —171 cm*/mol and C = 10119 em°/mol’. Bia (82206 cm?atm/molK)(310.6 K)/(80.4 cm?/mol)] x
[1 — (171 em?/mol)/(80.4 cm’/mol) + (10119 cm°/mol2)/(80.4 cm*/mol)’]
= 139 atm. Note: The observed pressure 1s 135 atm.
8.8
Mor = 0.2000 mol, V/ntor = 3500 cm*/mol. (a)
(b)
l) = P=RT/Vm = (82.06 cm*atm/molK)(313.1 K)/(3500 cm°/mo
7.34 atm. Let CH, be gas 1. For CoHy, a; = 27R°T? /64P. = 4.56 x 10° cm° atm/mol2, b, = RT,/8P, = 58.3 cm*/mol. For CO:, we find az = 3.61 x 10° cm® atm/mol” and b2 = 42.9 cm*/mol. Also, x; = 0.0786/0.200 = 0.393,
(c)
8.9
x) = 0.607. From the last paragraph of Sec. 8.2,
a = [(0.393)°(4.56 x 10°) + 2(0.393)(0.607)(4.56 x 10° x 3.61 x 102je7 (0.607)°(3.61 x 10°)}(cm®atm/mol*) = 3.97 x 10° cm°atm/mol’; b = 0.393(58.3 cm/mol) + 0.607(42.9 cm*/mol) = 49.0 cm*/mol. P = (82.06 cm?atm/molK)(313.1 K)/[(3500 — 49)cm*/mol] (3.97 x 10° cm®atm/mol’)/(3500 cm*/mol)” = 7.12 atm. Z=PVp/RT; P = ZRTIVm = 0.9689(82.06 cm*atm/molK) x (313.1 K)/(3500 cm*/mol) = 7.11 atm.
P =RT/Vm + RTB(TYV2, + RIC(TYV3,. (OP/Vm)r = 0 = RTIV:, — 2RTBIV?, —3RTCIV’,. (@°P/OV2, r= 0 = 2RTIV>, + ORTBIV), + 12RTCIV>,. RIAV. 8+ ORT BY met OR Le = On eC} 2RT.V2 .+ ORT,BV,,. + 12RT.C =0
(2)
Subtract twice equation (1) from equation (2) to get
B=—3C/V,,.. at T= T..
Substitution of this expression for B into equation (1) gives C(T;) =
V2 /3. Also, B(T.)
=—3C(TeVinc =— Vie. We have Z; = PeVnc /RT. =
Lee BCTV Vmet CIV 8.10
ea 1 a1
aan 3.
(a) From (8.18), b = RT,/8P. = (82.06 cm’atm/molK)(150.9 K)/8(48.3 atm) = 32.0cm*/mol; a= 27R°T//64P.= 100
27(82.06 cm?atm/molK)*(150.9 K)7/64(48.3 atm) = 1.34 x 10° cm® atm/mol*
(b)
Comparison of (8.9) with (8.4) gives as the van der Waals estimate:
B=b—ad/RT.
Byox = 32.0 cm*/mol — (1.34 x 10° cm®atm/mol’)/
(82.06 cm’atm/molK)(100 K) = 131 cm*/mol. Also, B20 k = 49.6cm*/mol, B3o0x =—22.4 cm*/mol, Bsoo x =—0.7 cm°/mol,
B,o00 x =15.7 cm*/mol. Agreement with experiment is fair. 8.11
We have AyapUm = @/Vm = ap/M and AvapHm, =~ ap/M + RT. For N> at its normal
boiling point, Avapf{m ~ (1.35 x 10° cm® atm/mol*)(0.805 g/cm*)/(28 g/mol) + RT = (3.9 x 10* cm? atm/mol)(1.987 cal/82.06 cm? atm) + (1.987 cal/molK)(77.4 K) = 1.1 kcal/mol. For HCI, Avapm =
(3.65 x 10° cm® atm/mol”)(1.193 g/cm*)/(36.5 g/mol) + RT=
2.89 kcal/mol + 0.37 kcal/mol = 3.3 kcal/mol. For H20,
Avapffm =
(5.46 x 10° cm® atm/mol’)(0.96 g/em*)/(18 g/mol) + RT =
7.8 kcal/mol. Agreement with experiment is fairly good.
8.12
Vin T in cell D2 is changed. The new graph shows we need to start at a smaller value, so the value in A9 is reduced until the graph starts at a substantially positive value for P. An initial Vm of 84 cm°/mol is found to be appropriate. 0 and The maximum in P is close to 12 atm, so the vapor pressure is between 12 atm; averaging these values, we enter an initial guess of 6 atm in C3. The values in columns A and B show that 6 atm corresponds to 2900 cm*/mol and to a value between 84 and 89 atm, so we use 2900 and 86 cm*/mol as the initial
guesses for V) and V. The Solver constraints for C3, E3, and G3 need to be appropriately modified. The Solver gives 3.01 atm, 6375 cm*/mol, and 86.0 cm?/mol as the vapor pressure, and molar volumes.
8.13
From Table 8.1 and Eqs. (8.20) and (8.21): = 29.68 cm*/mol, b = 0.08664(82.06 cm?atm/molK)(304.2 K)/(72.88 atm)
a = 0.42748(82.06 cm?atm/molK)(304.2 K)°5/(72.88 atm) = 6.375 x 10’
Fig. 8.6 K"? cm® atm/mol’. These values are entered into Cl and El of the shows no spreadsheet and the temperature is changed in D2. The new graph in A9. One finds minimum, indicating that we must start at a smaller Vm Value P is at 48 atm, so the 50 or 55 cm’/mol to be suitable. The local maximum in
the initial guess. vapor pressure is between 0 and 48 atm, and we use 24 atm as 101
The column A and B values show 24 atm to correspond to molar volumes of about 60 and 760 cm*/mol, so we use these as the initial Vm guesses. The
Solver constraints for C3, E3, and G3 need to be appropriately modified. The
Solver gives 38.7 atm, 57.2 cm?/mol, and 390 cm*/mol. 8.14
Table 8.1 and Eq. (8.18) give b = (82.06 cm?atm/molK)(369.8 K)/8(41.9 atm) = 90.5 cm?/mol,
a = 27(82.06 cm?atm/molK)*(369.8 K)7/64(41.9 atm) = 9.27 x 10° cm° atm/mol” When the van der Waals expression for P is substituted into
(8.24), we get PV? V,,) =] Ye(RT (Vn —b)—alV,)} dV,, , which gives ve Ve
(pe scoiaoad RT n Yel? +a
Ve m
m
m
eal! — Ven Ve
(8.25)vdW
m
The Fig. 8.6 spreadsheet is modified by changing the a (in an efficient way) the formulas in B9, B10,... to the expression for P, using the right side of Eq. (8.25)vdW and using the right side of the second equation of (8.2)
and b values, changing van der Waals (given above) in C4, with V,, replaced by
V. or V,”, in E4, and G4, respectively. The Solver constraints for C3, E3, and G3 need to be appropriately modified. The graph shows P is too high at 95 cm*/mol, and an appropriate starting Vm is 130 cm*/mol. The local maximum in P is at 22 atm, and 11 atm is a reasonable initial guess for the vapor pressure. The values in columns A and B show 11 atm corresponds to about 145 and 1885 cm*/mol, and these are reasonable initial guesses for Vm. The Solver gives 16.65 atm , 141.5 cm*/mol, and 1093 cm*/mol. 8.15
(a)
m=0.480 + 1.574(0.153) —0.176(0.153)° = 0.717.
2 2 J, 717], fe 0.42748(82. 8(82.06 cm? 2 atm/molK)’(369.8K)" 41.9 atm
_{298ISK 369.8K
v2))°
a = 1.08 x 10’ cm® atm/mol” (b)
We change the a value in C1 and delete the T'” factors in the
denominators of the formulas in cells C4, E4, G4, B9, B10,... (do this in
an efficient way). The Solver gives 9.47 atm, 97.6 cm*/mol, and 2144 cm°/mol.
102
8.16
(a)
Using wo = 0.153, we find k = 0.604. From T, = 369.8 K, P, = 41.9 atm,
and k = 0.604, we get b = 56.35 cm?/mol and a = 1.133 x 10’ atm cm°/mol’.
(b)
When the Peng—Robinson expression for P is substituted into (8.24), we
get P(V? Vi)= ie{RT(V,, —b) —al[Vn (Vp, +b) + D(V,, —b)}} dV, . Using the given integral with s = 2b and c = —b’, we get
sei m
ar Vn= fu(a +(1¥2)b Vi +d+2)b a Vie
m
bs
\Vo td +y2)b V.+0—V2)b
The Fig. 8.6 spreadsheet is modified by changing the a and b values, changing (in an efficient way) the formulas in B9, B10,... to the Peng—Robinson expression for P, using the nght side of Eq. (PR) (given above) in C4, and
using the right side of the Peng—Robinson equation with V,, replaced by Ve or V., in E4, and G4, respectively. An appropriate starting value is 85 cm?/mol in A9. The maximum in P is at 18 atm, and we take 9 atm as the
initial guess for the vapor pressure. This P corresponds to 86 and 2260 cm?/mol, which are the initial guesses for the molar volumes. The Solver gives
9.38 atm, 2143 and 86.1 cm*/mol. 8.17
(a)
For
P=RTMV,,—b)—a/[Vn Vin +b)T 7),
R feea s TieoP PT: Vib. DV AVeuEDLE aT Jy.
%

a —— Va Tae aie Neat
3a
eV (Velnb) Ie2 Then, using [foo + b)y"' dv = b  In[v/(v + b)], we get
A
oe ene
3a
Va( Ve
Pe b) les Gee we
ea ee Be ORD eV Ve
and addition of P AV gives the desired result.
(b)
Example 8.1 gave V,, =1823 cm*/mol and V,, = 100.3 cm°/mol. Then
PV” VL) = (10.85 atm)(1723 cm’/mol) = 1.869 x 10* cm? atm/mol =1894 J/mol. We have
Narr Behe
x10° cm® atm K'”/mol*) 1 1823(100 + 63) _ we 3(1.807 (69.7 cm /mol(298.15.K)) a) 9100.3018 23 + 63)
1.13 x 10° cm? atm/mol = 1.146 x 10° J/mol. Then A,,,H = 103
(11460 + 1894) J/mol = 13.3 kJ/mol. (The experimental value is 14.8 kJ/mol.)
8.18
Use of (8.20) and (8.21) gives a = 3.77 Xx 10° cm® atm K!/mol? and b = 92.4
cm?/mol. The Fig. 8.6 spreadsheet with a, b, and T changed can be used. One finds a minimum pressure of —14.5 atm at 210 cm’/mol.
8.19
= RT (Vm — RT¢ /8P.) (P+ 27R°T7/64P.V2 [P+ (27T2/64P, V2)(64P2V_ /9TZ Vm —Vnc!3) =(8PVinc/3T,)T
(P+3PIV2\(Vn —Vinc!3)=8PVineTy/3 (P. +3/V,\(V, —1/3)=8T,/3 8.20
(a)
P=RTlMVm—b) alTV>.. The criticalpoint conditions (OP/0Vm)r = 0
and (0*P/OV,.); =0 aS to
RT AV, —6)? =2a/T.Vz,, and REAV, 2sb) =3a/T Vane Cc
Division of the first equation by the secondeve Vee bo =2V ne
and
Vin. = 3b. Substitution in the first equation gives RT./4b7 = DneTh
TT. and
= (2/3)(2a/3bR)" * Substitution in the Berthelot equation at the critical point gives
“
~2 2/20)" _ 23/38 me NORGE RE 2b
3\3bR
9b?2\ 2a
DAelem Poise
Division of this P, equation by the preceding 7, expression gives P./T, = R/8b, so.
a3
Babe 21GP RR
ork Pay,me (Bye z. ase
Di Rigi 8Peal hetabove Peexpression gives
gPoR. 64 P. 2 oe1/2 Re pl/2 2) a 1/2 30375
TEN
(c)
Toe
em
Substitution of R = 8P.Vm,/3T, and the above expression for a and b into the Berthelot ee
peekOIE RET ei AMR, 64 P. TV?
followed by division by P.Vm ¢«gives
3 Aa SP.
104
OAS VSINTOR
ia
8.21
are
Ol
ShVewT a\l SPV. fe STEEP
Vg >
From (8.29), P, = 87,/3(V, +)  3/V,.
T, = TIT; = 310.6/305.4 = 1.017:
V, = VinlVine = V/NVm.e = (200 cm?)/(2.488 mol)(148 cm?/mol) = 0.543. Then P, = 8(1.017)/3(0.543 — 0.333) — 3/(0.543)° = 2.75 = P/P, and P = 2.75P.. = 1324 atm.
8.22
B = (82.06 cm*atm/molK)(150.9 K)(48.3 atm) '[0.597 — 0.462e° 70071509 KyT) = [153.1 — 118.429: 8”) cm?/mol. At T = 100, 200, 300, 500, and 1000 K, we get (in cm*/mol) —187.5, 47.7, 15.3, 6.3, 21.5. Agreement with experiment
is very good.
8.23
As noted in Sec. 8.8, at T and P°, H'! — Hm = JP [T0OVm/0T )p — Vl AP.
Differentiation of (8.5) gives (0VWdT)p = RP (1+ B'P+C'P+)+ RTP  (PB” + P°C’’ +), where the prime means differentiation with respect to T. Use of (8.5) gives T(0Vm/0T)p — Vm = REERN PB CLP Co).
SoH! SHa= ) RUB pee Cy)
de
Ri, [Pee
es 8(ps) Gee
sid 5, = JF [OVa/0T)p— RIP] dP = jf [RWB + CP ++ ) + RT(BY + PC’ +)] dP = R(B' + TB” )P° +
1R(C’ + TC")(P°)’+ . Finally, Gy — Gm = (Hy — Hm) — T(Sim— Sm) Use of the preceding results gives G = Gi==RIIBIP? + ME
8.24
(a)
ey,
Comparison of (8.9) with (8.4) gives the van der Waals estimates of the
virial coefficients as B = b — a/RT, C=b’,... . Use of (8.6) gives B' = b/RT—a/R°T’,... . Differentiation gives B’ =—b/RT* + 2a/R°T°, _.. . Substitution in the results of Prob. 8.23 gives H'4 — Hp =
RT?(b/RT* + 2a/R°T°)P° +   = (2a/RT— b)P° +    and gives Sees (aIRT Po + eee. 105
.
(b)
From (8.18), a = 27R°T2/64P, = 5.50 x 10° cm® atm/mol” and b = RT, /8P, = 65.0 cm*/mol. Then H'\ — Hm = (0.987 atm) x [2(5.50 x 10° cm® atm/mol’)/(82.06 cm?atm/molK)(298 K) 65.0 cm3/mol] = (379 cm?atm/mol)(1.987 cal/82.06 cm’ atm) = 9.2 cal/mol. Also ies ayng—A (yall) 10° cm® atm/mol*)/
(82.06 cm>atm/molK)(298 K)"](0.987 atm) = (0.74 cm? atm/molK)(1.987 cal/82.06 cm? atm) = 0.018 cal/molK. These are substantially smaller than the experimental values.
8.25
(a)
The Berthelot equation when solved for P and multiplied by Vp/RT is alRT *Vm. Use of (8.8) a eb GIR Vet Pave Rk Viyl VD) with x = b/Vm gives PVmn/RT= 1 + (b alRT°)/Vm + BV, tees, Comparison with (8.4) gives the Berthelot estimates of the vinal C= b’, ej Use 0ff(8:6) gives coefficients as B = b — alRT*, B' =DIRT  alR°T?, ... . Differentiation gives B* =—bIRT? + 3a/R°T*. = (3a/R? —b)P° +and
Use of Prob. 8.23 results gives He oH
SiS = (di Rigs) (b)
Poe
Substitution of the a and b expressions of Prob. 8.20a gives
H® — Hp = (81RT?/64P,T* — RT,/8P,)P° + +++ and S'\ — Sm=
(27RT? /32P,T*)P° +. (c)
Substitution in the results of (b) gives S — Sm = (27/32)(1.987 cal/molK)(305.4 K/298 K)*(0.987/48.2) = 0.037 cal/molK
and H'’ — Hm» = 15 cal/mol. Agreement with experiment is excellent.
(d)
gi — Sm = (27/32)(1.987 cal/molK)(430.8/298)°(1/77.8) = 0.065 cal/molK.
8.26
(a)
We use the following isothermal path (rg = real gas, ig = ideal gas): l
2
3
4
rg(Vin) > rg(Vm = 2) > ig(Vm = 09) > 1g(Vm) > 19(Vm.ia) From dAy = —Sm dT —P dVm, we get (dAn/OVm)r = —P. Then AA) =
aly bidVa = oer dV, oe AAn> = Osince AU —
and AS;
Oa
noted in Chap. 5); AAm3 =—J'" Pia dV{ =J" (RTIV,) dV’; AAma =
106
—[¥" (RTIV,) dVq =RT In (Vi8IVin) Amia(T, P)  An(T, P) = AAPIEE KAM (b)
AA, 3 NAM, S63 (RORT/V. ) dVl= REIN
S/V ay
Solving (8.3) for P and substituting in the result of (a), we get A‘ —An=
[% [RTIV. —b)— al Vo (Vi. + b)T'? — RTIV,,  Vg, — RT In (ViE/Vm) = [RT In(V.. — b)  (a/T"?)(—1/b) In(1 + b/ Vy, ) — RT In Vy, J =
RT In(V'2/Vm) = [RT In(1 — b/ Vp,) + (a/bT"*) INL + BIV,, UP = RT In(V"2/Vm) = RT In(1 = b/Vmn) + (a/bT'”) In + B/Vim) — RT In(V,/Vin)
(c) Si —Sm=@/0T), (Aim — Am) = _R In (1 — b/Vm) + (a/2bT *”) In(1 + b/Vm) +R In(ViC/Vin). U4 — Um = (A — Am) + T(S'S — Sm) = Gal2bT"*) In(1 + B/Vmn). 8.27
Differentiation gives dB/dT = (RT;/P.)(0.462)(0.7002T./T*)ee
=
0.3235(RT2/P.T) e970" _ From (8.6), B’ = B/RT and dB'/dT = (dBIdT)/RT _— B/RT?. Substitution of T, = 305.4 K, P, = 48.2 atm, and T= 298 K in the
equations for B and dB/dT gives B = 182 cm’/mol and dB/dT = 1.186 cm?/molK. Then B* = B/RT = 0.00744 atm”! and dB'/dT = (1.186 cm>/molK)/(82.06 cm?atm/molK)(298 K) +
(182 cm3/mol)/(82.06 cm?atm/molK)(298 K)° = 7.35 x 1 Onmatmnan Kae
Fron Probas235 i
teen
REce ecBdd ie
(1.987 cal/molK)(298 K)°(750/760)atm(7.35 x 10> atm! K~) = 13 cal/mol
and S4 — S,= RP°(B' + T dB'/dT) = (1.987 cal/molK)(0.987 atm)[0.00744 + 298(7.35 x 10>)Jatm™ =
0.028 cal/molK.
8.28
8.29
© xen a Cole OU) fo=0, f'@=0x), fC =1, fFOVas fF(ONE2) Oras ear be ef0) eo (eye. foc) = 14 (40) + (x — OY/2! 432 OV BI 4=Lextx tet: ) FON aS GIS, . fey ey eeepeneh As ipoee ix, (ey HOE Tea ae ras (eae ene /2iee eel ya a2 =O 2 een sae bon = Tye ee  @DV4+(x +(x YB 1/2 )(x—1 107
es
OES Ls FAO inl
iefan O)
hls
8.30
FO=IEa Ge ere Salextx2sex3i+:
8.31
(didx\(sin x) = cosx=1 — 3x7/3! + 5x'/5)—
8.32
x is in radians; 35° = 0.610865 rad. Substitution in (8.35) gives sin 35° = 0.5736.
8.33
The function 1/(x? + 4) has singularities at x = 2i and x = —21 (where + 4= 0). The distance between the origin (point a) and either of these singularities is 2,s0b=2.
8.34
(a)
5p 1"/n! =2.7166667,
= Dax 12! ene /Al
Deg l"/n! = 2.7182818,
es
Lj2o1"/n! =
2.7182818. The exact value is e = 2.7182818. (b)
For x= 10, we find 1477.6667, 12842.305, and 21991.482 for n =5, 10, and 20. The exact value is e!° = 22026.466. (For n = 30, one finds
22026.464.)
A BASIC program is 70
10 INPUT “X=";X 20 FORM=5TO20
80 NEXT N
30 S=1
90 PRINT "M=";M;" SUM=";S
40 NF=1
100 NEXT M
50 FORN=1TOM
110 END
60
8.35
STEPS
S=S + XAN/NF
NF =N*NF
From Sec. 8.3, fe= 1.67, =30(353 K)905.K une = 2.7 Vn = 2.7(78 g/mol)/(0.81 g/cm*) = 260 cm*/mol. From Sec. 8.4, Z, is usually between 0.25 and 0.30, so we take Z,. = P.Vm,/RTc = 0.275. Thus
P. =0.275
RT, Vine
(82.06 cm*atm/molK)(565 K)0.275 _ 260 cm*/mol
49 atm
(The experimental values are 562 K, 259 cm?/mol, 48 atm.)
108
8.36
Use the isothermal path: liq(I bar) = liq(23.766 torr) re vap(23.766 torr) _ ideal vap(23.766 torr) bs ideal vap(1 bar). For step (a), Prob. 7.50 gives
AGma = —1.8 J/mol. For steb (b), AGm,» = 0, since this is a reversible
constant7andP process. For step (c), Prob. 8.24a gives AGn,, = AHme— I ASmc = (alRT — b)P =
5.47x10° cm®atm/mol?__4, atm/molK)(298K) mol  760 3
ayes
mol
atm
eee =0.6 J/mol
82.06cm°
atm
For step (d), Prob. 7.50 gives AGmq@ = 8557.2 J/mol. The net AG,, 1s 8556.9 J/mol compared with 8555.4 J/mol in Prob. 7.50 and 8557 J/mol from data in
the Appendix.
8.37
(a)
Use of Eqs. (4.52) and (8.5) gives pyr = (1/Cp,m)[T(OVmn/0T)p — Vin] =
Cpl {T (RIP) + B'P + CP? ++) +(RTIP)(P dB'/dT + P* dC '/dT + Pa GaP eae fe I=(RTP)\UB (RT7/Cp m)(dB'/dT + P dC'/dT + P’ dD'/dT + ) lim wor = (RT?/Cpm)dB'/dT —>0
(b)
P= Equations (4.47) and (8.4) give (QU/0V)r = T(OP/0T)y — B RT ee Riel ae ::: ee B R T —
Velen
1+—+
ale
4+—
+ ——
Veve dh
Fe hoig ae
8.38
From (8.4), Vm — Vi? = (RT/P)(1 + B/Vn + C/V, +: ++) —RTIP = RT(B/PVm + C/PV2, ++). In the limit P > 0, we have PV, = RT and B, since Vm —>eeas P > 0. Vm  Vid + B+ C/Vm ++
8.39
(a)
As found in Prob. 8.24a, the van der Waals equation gives B' =
BIRT — alR2T2 and BY = b/RT? + 2a/R°T°. Substitution in the equation of Prob. 8.37a with all terms but the first neglected gives pyr = (2a/RT— b)/Cpm as the lowP estimate of wyr.
109
(b)
At the inversion temperature, the lowP
[jz is zero, So
(2a/RT— b)/Cp m= 0 and T;,p_50 = 2a/bR = 2(1.35 x 10° atm cm® mol™)/ (38.6 cm/mol)(82.06 cm?atm/molK) = 852 K, where N> a and b values in Sec. 8.4 were used. The lowP 298K estimate of yz 18 Wy7 = 2(1.35x10°
6
atmcm’
6
2
mol ~ ) ake
(82.06 cm> atm/moFK)(298 K) (10.3cm? eee
a
1
Ba
6.96 cal/moK
ee = (0.250 K/atm
82.06cm* atm where Cp is from the Appendix and (1.19) and (1.21) were used. Agreement with the experimental [17 1s good, but 7; is poorly predicted.
8.40
(a)
The larger size of a neon atom means that Ne has greater intermolecular
attractions than He, so Ne has a greater a, a greater T., and a greater Avapf1m. Ne atoms are larger so Ne has a greater b value.
(b)
C3Hsg has greater intermolecular attractions and larger size, and so has the greater a, T;, AyapHm, and b.
(c)
Due to hydrogen bonds, H20 has greater intermolecular attractions, and so has the greater a, T,, and AyapHm. The H2S molecule is larger than the H20 molecule, so H2S has the larger b.
8.41
From (8.18), b = RT,/8P,. = (82.06 cm>atm/molK)(190.6 K)/8(45.4 atm) =
43.1cm*/mol; a =27R°T 2/64P, = 2.27 x 10° cm® atm mol. Vmo =b + RTIP = 43.1 cm*/mol + (82.06 cm?atm/molK)(273 K)/(100 atm) = 267 cm?/mol. Vint = b+ RTMP + a/V>4)= (82.06 cm> /molK)(273 K) 100 atm + (2.27 x 10° cm® atm/mol?)/(267 cm?/mol)” Vn,1 = 213 cm*/mol. Vn2= b+ RTP + a/V2,,) = 192 cm3/mol; Vm3 = 182 43.1cm?/ mol +
cm?/mol. Successive calculations give (in cm?/mol): 176,3L 725 1702169. 168) 167, 166.6, 166, and 166. From Fig. 8.1a, for CHy at 100 atm and 0°C, Z=
0.785 = PVm/RT and Vm = 0.785RT/P = 0.785(82.06 cm’atm/molK)(273 K)/(100 atm) = 176 cm?/mol. A BASIC program is
110
8.42
10 DIM V(100)
50° FOR=s1
20 22 24 26 30 40
60 V(l+1) =B + R*TAP + A/V(I)A2) 70 PRINT"I="sI:" V=";V(1+1) 80 IF ABS(V(I+1) — V(I))/V(I) < 1E5 THEN STOP 90 NEXT  95 END
INPUT "A=";A INPUT "B=";B INPUT "P=";P INPUT "T=":T R = 82.06 V(1)=B+R*T/P
TO 99
T, = T/T, = (286 K)/(190.6 K) = 1.50 and P, = P/P. = (91 atm)/(45.4 atm) = 2.00. At these T, and P, values, Fig. 8.10 gives Z=0.83 = PV,/RT and V,;, =
ZRTIP. = 0.83(82.06 cm>atm/molK)(286 K)/(91 atm) = 214 cm*/mol. 8.43
(a)
Asin Prob. 7.33b, d In P/dT= (1/P)(dP/dT) = 0.035687/K and dP/dT= (0.035687/K)(1 atm) = 0.035687 atm/K at 100°C.
(b)
AV» = (82.058 cm>atm/molK)(373. 15 K)/(1 atm) — 452 cm*/mol —
19 cm*/mol= 30149 cm?/mol. Then AHm + (T AVm)(dP/dT) = (373.15 K)(30149 cm*/mol)(0.035687 atm/K) = (4.0148 x 10° cm?atm/mol)(8.3145 J/82.058 cm?atm) = 40.689 kJ/mol. 8.44
Cells for B and C are specified. One can set each of these to zero for the initial guess. The experimental P and V,, values are entered into columns A and B.
Formulas to calculate P from Eq. (8.4) are entered into column C. In column
D, the squares of the deviations between columnA and columnC pressures are calculated. The sum of the squares of the deviations is calculated 1n a cell, and the Solver is set up to minimize this sum by varying B and C. The result is B = 83.12 cm*/mol and C = 3330 cm®/mol’, with an excellent fit to the data. 8.45
(a) F;
(b) F.
Chapter 9 9.1
(a) mol/m*;
(b) mol/kg;
(c) no units.
9.2
Gis
9.3
(a)
n;=c;V= (0.800 mol/L)(0.145 L) = 0.116 mol.
(b)
(145 g) x 10.0% = 14.5 g;
(c)
m,=nwa =n,/(w — w;) = n/(w — n,M;), where w is the mass of the solution, w; is the mass of the HCl, n; is the number of moles of HCI, and
Cie
0.398 mol.
M,; is the HCI molar mass. Solving for n;, we get nj = mjw/(1 + mM;j) =
(4.85 mol/kg)(0.145 kg)/[1 + (4.85 mol/kg)(0.03646 kg/mol)] = 0.598 mol HCI. Alternatively, a solution with 1000 g of solvent has 4.85 mol
HCI, which is 176.83 g of HCl. The weight percent of HCI is [176.83/(1000 + 176.83)]100% = 15.03%. 15.03% of 145 g is 21.79 g of HCl, which is 0.598 mol HCl.
9.4
(a)
All quantities involved are intensive, so we can use any convenient
amount of solution. Let us take 1 dm? of solution. This amount of solution has 8.911 mol of CH3;OH, which is 285.5 g of CH3OH. Since the
solution is 30% CH3OH, the solution’s mass is (100/30)(285.5 g) = 951.8
g. Its density is (951.8 g)/(1 dm*) = 0.9518 g/cm’.
9.5
(b)
m;=n/wa = (8.911 mol)/(951.8 g — 285.5 g) = 0.01337 mol/g = 13.37 mol/kg.
(c)
PcH,on =™Mcy,on/V = (285.5 g/d dm*) wey 8)sl ep
Let us take 1000 g of solution. This contains (0.800 %)(1000 g) = 8.00 g of NH; and 1000 g — 8 g = 992 g of water. This is 0.4697 mol of NH3 and 55.06 mol of water. Then m; = nj/wa = (0.4697 mol)/(992 g) =
0.000474 mol/g = 0.474 mol/kg. Also, x; = 0.4697/(0.4697 + 55.06) = 0.00846.
9.6 The solution’s mass is m = pV = (1.2885 g/cm?)(1000 cm’) = 1288.5 g. The CsCl mass is (2.296 mol) x (168.358 g/mol) = 386.6 g. The solvent mass is
112
1288.5 g — 386.6 g = 901.9 g. The molality is (2.296 mol)/(0.9019 kg) = 2.546 mol/kg. Dad,
All quantities involved are intensive, so we can take any amount of solution. Take an amount of solution that contains one kg of water and hence contains 1.506 mol of KI. The solution’s mass is 1000 g + (1.506 mol)(166.00 g/mol) =
1250.0 g. The solution’s volume is V = m/p = (1250.0 g)/(1.1659 g/cm’) = 1072.1 cm*. The molarity is (1.506 mol)/(1.0721 L) = 1.405 mol/L.
9.8
cj = n;/V = nj/(w/p) = pn,/w, where w is the solution’s mass. Because the solution is very dilute, we have w = w; + Wa = Wa = NAMa, where A is the solvent. Then c; = pnj/naMa. Also, x; =n/(na + nj) = n/a, SO Cj = ox/Ma. For the molality, we have m; = nwa =nj/naMa = x;/M,. From c; = px/Ma and m; = x/Ma, we get c; = pmi.
De
As noted after Eq. (1.4), Ma = M, X 1 g/mol, so mp = np/naM a =
np/(naM,.a g/mol) = np/(naM, al0~> kg/mol) = (1000 np/naM,.a) mol/kg.
(cler: y(d)o0. (ey 1
9.10
(a)ee (Daron
9.11
V=nvV, + Ve=
Dab
aaa)
(alga)
Te
(0.500 mol)(18.63 cm?/mol) + (55.51 mol)(18.062 cm*/mol) = 1011.9 cm’. 9.12
Cp=nCp,
+ Bes. . An amount of 0.1000mol/kg solution that contains
1000 g of solvent has 0.1000 mol (which is 5.844 g) of NaCl and has an NaCl
weight percentage of [5.844/(5.844 + 1000)]100% = 0.581%. Taking 0.581% of 1000 g gives 5.81 g of NaCl in the 1000 g of solution, which is 0.0994 mol of NaCl. The percent water is 100 — 0.581 = 99.419%. The HO mass is 994.19 g, which is 55.186 mol H2O. So Cp = (55.186 mol)(17.992 cal/molK) + (0.0994 mol)(—17.00 cal/molK) = 991.22 cal/K.
3
V= nV, + n,Vq = 307.09 cm’ = [(72.061/18.0153) mol](16.488 cm*/mol) + [(192.252/32.0422) mol] Voy,on and Vou,ou = 40.19 em’/mol.
113
9.14
Take solutions that each contain 1000 g of H2O (constant nyo ). Let w be the solution mass. For the 12% solution: (1000 g)/w = 0.88000 and w = 1136.36 g;
this solution contains 136.36 g of CH3OH, which is 4.2557 mol of CH30H; the
solution has V = (1136.36 g)/(0.97942 g/cm’) = 1160.24 cm’. For the 13% solution: (1000 g)/w = 0.87000 and w = 1149.43 g; this solution has 149.43 g of CH3OH, which is 4.6636 mol of CH30H; the solution’s volume is V =
(1149.43 g)/(0.97799 g/cm’) = 1175.30 cm®. We have Voy.on =
(AV/AncH oH)7.Ping.g = (15.06 em’)/(0.4079 mol) = 36.92 em’/mol. To find Vis,0 » We now take solutions with 100 g of CH30H (constant noy.o4 )and do
the calculations the same way as for the constant ny,9 solutions. For the 12%
solution, we find nyo = 40.706 mol and V = 850.84 cm’. For the 13% solution, we find nyo = 37.148 mol and V = 786.54 cm’. Hence Vii,0 =
(AV/ Ano) Pincuson = (64.30 cm?)/(3.558 mol) = 18.07 cm*/mol.
915
(a)
Vof MgSO, at a given composition equals the slope of Fig. 9.3 at that composition. The slope is zero at the minimum, which occurs at Nyoso, = 0.07 moles in 1000 g of water, which is a molality of 0.07 mol/kg.
(b)
Infinite dilution corresponds to Nugso,
0. Drawing the tangent line to
the curve at n MgSO, = 0, one finds its slope P to be —3.s5 cm°/mol, which is
V of MgSOg. (c)
Drawing the tangent line to the curve at nyggo, = 9.05 mol, one finds its 3
—_
17
—_
17
1001.69, cm? = (55.509 mol) Viz, + (0.05 mol)(0.54 cm?/mol) and
Vino = 18.045 cm*/mol.
9.16
— waci= co
V (Na‘)+V (Cl),
Vino,=¥
(Kk )+ V_(NO;), and
Vyano,= V ~(Na*)+ V°(NO3). Then Vic, = V"(K*) + VV(Cl")=
Vino, + Vnaci — Vwano, = 38.0 + 16.6 — 27.8) cm’/mol = 26.8 cm*/mol. 9.17
Substitution of G =); nj; into
G = U + PV — TS gives the desired result.
114
—
9.18
A =(0A/0n, )F Pine where A is the Helmholtz energy of the solution, n; is the number of moles of 7 in the solution, 7 and P are temperature and pressure, and nz; Indicates that all mole numbers except 7 are held fixed.
9:19
H=U+
PV. Partial differentiation gives
(OH/8n; Jr. p.n,,,
9.20
(a)
=(OU/ON; Vr p.n,,, + PCOV/ON; Vr, p.n,,, »80 H, = U, + PV,.
G=Dinp= Din; + RT In (P/P°)). Taking P; = P° and P2 = P; in (4.65), we have for n; moles of pure gas i: G*¥(T,P,,n;) — G*(T, P°, ni)
=njRT J" P'dP = njRT In (P/P°). So G¥(T, Pi, ni) = G(T, P®, ni) + niRT In (PJP?) = n;G* ,(T, P°) + n:RT In (P/P°) = n,[p; + RT In (P/P*)), since G* ;=; for a pure substance. Hence the above sum for G of the mixture: becomes G =)
(b)
S= (O/0T)p,
7.G?(1,.P.1,).
G= Li [(OG*/0T)p,,J= Xi S¥(T, P,,n;), where (4.51)
was used.
(e)
Ha=G +78 =), Gt + PQ S* aia(G*
SS) = 2) (7. nj), where
P; is absent since H is independent of P for an ideal gas.
(d)
pp, Li H* = Li CQH*/OT),, = Li Ch, (T, ni). Cp= (OH/OT)p,, = (O/T) U=HPV=)>;
Dey (e)
a
= 2D; (A *— PV)= H* —PQi nRTIP) =D; (A *t nRT)
1).
Assuming the mixture is ideal, we have Cp = ae Ch (7, n,)=
CB.o, + Cé.co, ="0,CP.m.o, + "co, CP,m,co, = (0.100 mol)(29.355 J/molK) + (0.300 mol)(37.11 J/molK) = 14.07 J/K. 9.21
Taking d/0T of (9.32), we get (QA ni,G/OT)p,,= (OT)p,,Lini(G,  Gi) =
OT)pin,1= Li ni(S; + SA) = Limi SH; —LiniS; = Yi nil(OG,/OT)pn,OG, S* — § =—AnixS, where (9.30) was used.
9.22
n(H20) = 1.119 mol and n(C2HsOH) = 0.977 mol.
x(C2HsOH) =
0.977/2.087 = 0.468. For this composition, Fig. 9.9 gives V (H,O) =
115
16.8 cm’/mol, V(C,H,OH) =57.0cm*/mol. V= V,n, + V,n, = (16.8 cm?/mol)(1.11 mol) + (57.0 cm*/mol)(0.977 mol) = 74.3 cm’. 9.23
We draw the tangent line at Xethano! = 0.4. This line intersects the Xethanol =
0 axis at 1.1 cm’/mol = Vio — Viu,0 = Vu,o — 18.05 cm*/mol, and Vito = 16.95 cm?/mol. The tangent line intersects the Xethano! = 1 axis at 3 —l.15 cm’/mol
Vee
9.24
Sa = V,ethanol _
3cm
Vxm, ethanol =V.ethanol —~ 58.4 cm*/mol, and
mol.
The pure molar volumes are Vi,4,9 = M/p = (18.015 g/mol)/(0.99705 g/cm’)
= 18.07 cm*/mol; V* cy,oH = 40.71 cm*/mol. We plot AmixV/n vs. x4, (similar to Fig. 9.7). (a)
Drawing the tangent line at x9
=0, we find it intersects the x49 = 1
line at —3.6 cm?/mol. (With a reasonable choice of scale, the intersection
occurs off the paper and can be calculated by extrapolation.) Hence, at
X4,0 =9, Vio — Vi.u,0 =—3.6 cm*/mol = Vj,9 — 18.1 cm’/mol and Vit,0 = 14.5 cm*/mol. Of course, ie
Ve cujon = 40.7 cm?/mol at
Xy,0 =9.
Amix V/n
0.4
(b)
x(H,O)
0.6
The tangent line at Xy,0 = 0.4 intersects the x9 =0 line at 0.5 cm’/mol and intersects X49 = 1 at1.6 cm’/mol, so 116
VoH,OH 7 Vin.CH,OH = 0.5 cm’/mol and V,CH,0H = 40.2 cm?/mol; also,
Vii,0 = 16.5 cm*/mol. (c)
9.25
Vicu,on ~ Vm,cu,on = —!45 cm’/mol and V,CH,OH = 39.2; cm*/mol; Vio — V¥.n,0 =0.7 cm’/mol and Vy,9 = 17.4 cm*/mol.
(a)
pif =O + (3/2)cng\? + 2kng for naMa = 1 kg. Va = (OVidng)7
(b)
For mpg =  mol/kg, we have ng =  mol in a solution with naMag =  kg.
(c)
V =n,V, +ngVz, andn,Va =V —npV, =a + bnp + Chea + kt, = npp Qin = d= 5 Cee kn;,. Since na = (1 kg)/Ma, we get Vie ic
V, = 16.6253 cm*/mol + 1.5(1.7738 cm’/mol*”)(1.0000 mol)!" + 2(0.1194 cm?/mol’)(1.0000 mol) = 19.5248 cm*/mol.
(Ma/1000 g)(a— 1 cng” — kng) for nsMa = 1 kg. (d) (e)
(f)
mp = np/naMa = np/(1 kg) = np/kg and ng = mp kg. Substitution in the results of (a) and (c) gives the desired equations.
Vis,o = (18.0152 g/mol)[1002.96 cm* — 1(1.7738 cm?/mol*”) x (1 mol/kg)®” kg*? — (0.1194 cm*/mol’)(1 mol/kg)” kg”]/(1000 g) = 18.050 cm?/mol. In the infinitedilution limit, mp goes to zero and Vy in (d) becomes
V;° = b = 16.6253 cm’*/mol. 9.26
(a)
z= AmxV/n = (V— V*)/(na + np), So V— V* = (na + np)z and
V=(natnp)zt+ V* =(natnp)z+ naVan
(b)
+ nevi
n = V, for 4),,.7,p Taking (0/dn,4),,,.7,p Of the result of (a) gives (OV/O the left side and z + (na + np) (Oz/On,),,, + Vn. for the nght side. (Note that V* , is constant at fixed T and P and is independent of na.) Equating these expressions, we get Ve =r
(c)
(ddonny,, + Vana:
Equation (1.35) gives (dz/dn,),, = (dz/dx, )(0xg/dn,),,  Since Xp = np/(na + np), we have (0xp/0N 4) ng = —np/(na + np) = —xp/(na + Np) =
_xp/n. The result of (b) gives (02/dn4),, =(Va —Vik.4 — z)/n. Substitution of these expressions for dx,/dn, and dz/dn, into 117
dz/dn ,= (dz/dxp)(Oxp/dna) gives dz/dxp =(V* , — V4 +2)/xp. The ng subscript can be omitted from (dz/dx, ),,, because z = (V— V*)/n is an intensive quantity and so is a function of xg only and is independent of the size of the system and hence of npg. (d)
The plotted quantity AnixV/n is symmetric in A and B. Since the intercept at xp =.01S V, — V*,, the AB symmetry means that the intercept at va — OHS Ve cil Gaiaa Buti = Ouse —i8
9.27
The tangent to the AnixH/n versus Xj4,59, Curve at Xy,59, = 0.4 intersects X,50, = 9 at 13.9 kJ/mol = AA gig¢ 4,0 and intersects X4,,59, = 1 at16.4
kJ/mol = AA gre 1,50,  Drawing the tangent at x}, 59, = 0.333, we find AF site 4,80, = —95 kJ/mol and AH git y,80, = 226 kJ/mol.
9.28
NAH
SS eH:
(0 A nix
H/ON8g )r pn,
=
(OH/dn,
ee
ae (OH
*/Ong
)r pn,
.
But by definition, H, = (H/dng);.p,, Also, H* =nx H* , (T, P) + NgHy g(T,P), so (GH*/0ng)r p.n,= Hh.g Therefore (0A, ;,H/Ong)7 pn, = El — HX , = Adgitrs, where (9.38) was used.
9729
Subtraction of A;Hyxacys) from the apparent A, H° values give the following integral heats of solution per mole of NaCl vs. xnaci:
AFiintNacV/(kJ/mol)
1.874
2.347
3.016
3 hl
XNaC!
0.1
0.0625
0.03846
0.019608
As in the last example in Sec. 9.4, multiplication of AHintNaci by xNaci gives Amixf/n:
(AmixH/n)/(J/mol)
187.4
146.7
116.0
72.77
0
XNaCl
0.1
0.0625
0.03846
0.019608
0
We plot AmixH/n vs. xnacj and draw the tangent at xnac) = 0.05. The tangent line intersects Xnac) = O at 69 J/mol = Hino _ HyH,0 and intersects xnac; = 0.1 at
198 J/mol. Extrapolation gives the intersection at XNacr= 1 as
[69 + 10(198 — 69)]J/mol = 1.36 kJ/mol = Hyac} — H* naci
118
200.0
0
0.02
0.04
0.06
0.08
0.1
x(NaCl)
9.30
The NBS tables give A ;H 39. = —92.307 kJ/mol for HCl(g) and A ;H 50g = ~121.55 kJ/mol for HCI in 1 mol of H2O. Thus AHiint uci = —29.24 kJ/mol for a = solution with xc) = 0.500. From (9.37), Amixf = nuci AHintuci and Amix/nior xuci AHint uci = —14.62 kJ/mol at xuc1 = 0.5. Use of further data gives (values in
kJ/mol)
XHCI
0.500
0.400
0.333
0.286
0.250
0.200
AHin. uc)
29.24
40.36
48.65
53.17
—56.18
60.61
Nee Ano
14.62)
ol 615
16.22)
15 18
14.05
12.12
The We plot AmixH/mot VS. XHCI and draw the tangent line at xc) = 0.30.
intercepts at xyc) = 0 and xyc1 = 1 give AH git n,0 = —8.3 kJ/mol and AA aise acl
= 32
kJ/mol.
Amixf/n
119
9.31
(a) F; (b) T; (c) T; (d) F;
SREY
No. Models show that the AB intermolecular forces differ from the AA and BB intermolecular forces.
9753
(a) mle Dlg
9.34
The vapor is ideal and the pressure dependence of p of the liquid can be neglected.
9.35
peniest eo
C). He (d)eb;
a(e)
mOland Miet = 106) MOl
gat) a
ether= U4 12 ano
0 4 oe
Equation (9.44) gives AmixG = (1.987 cal/molK)(293.1 K) x
[(1.280 mol) In 0.5412 + (1.085 mol) In 0.4588] =—950 cal. Amixlt = 0)
AnixV= 0,
AmixG — Omit — 1DAmsS, SO Amino = (950 Caly@9s) Ky = 3.24
cal/K.
9.36
(a)
Men = (100.0 g)/(78.11 g/mol) = 1.280 mol. nyo = 1.085 mol. Xben = 1.280/(1.280 + 1.085) =0.5412. x4) =0.4588. Pren = 0.5412(74.7 torr)= 40.4 torr. Pio,=0.4588(22.3 torr)= 10.2 torr.
(b)
Prot= (40.4 + 10.2) torr = 50.6 torr. Use of P; = x; P gives
Xben = (40.4 torr)/(50.6 torr) = 0.798.
9.37
P=
Phex + Poct = Xhex, plaremit
le
Xs
Oe0e
nee) lace Xnex ere les Pe
(666 torr — 354 torr)/(1836 torr — 354 torr) := 0.2)
Phe = ne, hf = 021111836 torr = 287 torn ee (387 torr)/(666 torr) = 0.581;
9.38
ee
ICE
oe
= Poet) =
O21
0 780:
Xocty = 0.419.
Pi= x; P= x;P¥. 0.555(95.0 torr) =0.305P*,, and P*,= 173 torr. P= x; P¥ +x)P¥ = 95.0 torr= 0.305P*,, + 0. COBY fees 0.305(173 torr) +oe 695 Pe, and Pi», = 60.8 torr. We assumed an ideal solution, ideal vapor, and the pressure independence of LF.
120
9.39
(QypeRbea ie a= py PA SO x2 =n P#/Ps Also, P'= Pa + Po=
DP etal ex'sPEAS) 1x) P, so,x8 = Pa LOX ibe ct Dede, Piece P*)/ (oc, PETP® cl (b)
Yo),
If Bis benzene, then P3/P& = 74.7/22.3 = 3.35 and the equation in (a)
gives xp.y = 3.35 xy/(1 + 2.35 xp). We get vD
08
0
A
TONS
ORE
OT
ra
Xp
0
0.2
0.4
0.6
0.8
l
The plot resembles the upper curve in Fig. 9.18b. For toluene (t), we find
x? = 0.299 x!/(1 — 0.701 x; ) and
9.40
yen
0
0.070ae
016698
0:310=
*0:5457%
al
x
0
0.2
0.4
0.6
0.8
1
For an ideal solution, V= V* = n,V*, + n,V*, =n\Mj\/p} + n2M2/pz = m,/p* + m,/p% = (33.33 g)/(0.8790 g/cm’) + (33.33 g)/(0.8668 g/cm’) = 76.37 cm’. Then p = m/V = (66.66 g)/(76.37 cm’) = 0.8729 g/cm’.
9.41
(a) AmixCp = Cp— Cf = (QHIOT)p,,  (QH*/0T)>p,,, = (O/T)p,,,(H  H*) = (0 A, H/0T)p,,= 9, since AmixH = 0 for an ideal solution.
(b)
9.42
Cp= CB = Ch m1 + MC¥mo = [(100/78.11) mol](136 J/molK) + (100/92. 14)mol(156 J/molK) = 343 J/molK.
At xp = 0.50, the tangent is horizontal and intersects the xg = 0 and xp =  axes
at —0.415 kcal/mol = ta — WX = Up — YZ; the calculated result is WW; — W* = RT In 0.5 = (1.987 cal/molK)(298.1 K) In 0.5 =0.411 kcal/mol.
At xp = 0.25, the tangent line’s intercept at xg = 0 is 0.189 kcal/mol = Ua — Ux and that at xp =  is 0.815 kcal/mol = Lp — pZ; the calculated values are Ua — w* = RT In 0.75 =0.170 kcal/mol and pp — us = RT In 0.25 = —0.821 kcal/mol.
121
9.43
From (9.30), (9.31), (9.42), and (9.28): S; = —(0p,/0T)p,,, = —(OBF/OT) pn,Rinx;= S4;Rinx.
V, = —(0p,/0P);,, = (On#/0P); = Vinit
A, = wit TS, = pt + RT In x + TS5,, —RTInx)= Hj, (since we = Gy). 9.44
p= wo +RT In (P/P°) = wo + RT In (xP/P?) = wo + RT In (PIP?) + RT In x. For x; = 1, this equation becomes p* = ph; + RT In (P/P°), so Wj= W* + RTIn x;.
9.45
Substitution of Equation (9.42) into the phaseequilibrium condition py’ = uP
gives WX + RT In x,° = pt + RT In xP so xo = x? Similarly, x% = ae
9.46
(a)
The use of G=nju; + naa, for the solution gives G of the final state in the dilution process as G2 = njli2 + NA2WA2. For the initial state of the dilution process, G; = nil, + NAMA
+ (Ma2—Nai) UX.
AG of the
dilution process is G2 — G; and use of these expressions for G2 and G,
gives the desired result. (b)
Equating the right sides of (9.69) and (9.70) and using (9.71), we get
(9.72).
9.47
(a)
The solvent A is ethanol and Raoult’s law gives Pa = ah P=
Pen = 0.9900(172.76 torr)'= 171.03 torr. (177295
(b)
Ina
Poa =P
Po =
17 1103) torr = 6.92 tore:
gas mixture, P} = x;P, so x}, = 171.03/177.95 = 0.9611, and
Ken) = 0.92117 71.95 = 0.03880. (c)
Pi=Kjx! , 80 Ke = Poni x¢y) = (6.92 torr)/0.0100 = 692 torr.
(d) Pa = x4 PX = Pen = 0.9800(172.76 torr) = 169.30 torr; Pon = Ken x!,. = (692 torr)(0.0200) = 13.84 torr.
P = (169.30 + 13.84) torr = 183.14 torr.
Xeth = Pe/P = 169.30/183.14 = 0.9244 and xo, = Pon/P = 13.84/183.14 = 0.0756.
9.48
(a)
Pon = xg,P= 0.9794(438.59 torr) = 429.56 torr; Pen = (1 — 0.9794)(438.59 torr) = 9.03 torr. 122
(b)
Raoult’s law for the solvent chloroform gives Pon = x1,,P*,, and P*,, = (429.56 torr)/0.9900 = 433.90 torr.
9.49
(c)
Pet = Ken x), and Ken = (9.03 torr)/0.0100 = 903 torr.
(a)
At xcs, = 1 in Fig. 9.21b, P = 424 torr.
(b)
At x1,, =0.40 in Fig. 9.21a, the curves give P = 266 torr and Pen = Seon, SOM
9.50
= Eee — Oo
We have K; = P,/ x for very dilute solutions of 7. For CS» as solute, we draw
the line tangent to the P(CS2) curve at the point ibe = 0. Since P; = K;x;, the slope of this tangent line equals Kcg, ; the slope equals the intercept of the
tangent line with the ae = 1 line. We find Keg, = 1.25 x 10° torr in acetone as solvent. Drawing the tangent line to the P(ac) curve at the point te = 1 and
finding its intercept with Xces = 0, we get Kx. = 2.0 x 10° torr in the solvent CS2.
oSt
From (9.62), pw; — ee = RT \n (K;/P°) = (8.314 J/molK)(308 K) In (145/750) =——4 21 kKI/mol.
9.52
(a)
ny =8.14x 10° mol, Ny,9 = 555 mol; ip = 1.47 x 10°.
Ky, = Py, /xy, = (1.00 atm)/(1.47 x 10) = 6.82 x 10° atm. (b)
xy, = Py, /Ky, = (10.00 atm)/(68200 atm) = 1.47 x 10™. ny, = 8.14 x 10° mol and my, = 1.64 mg.
Eee)
xy, = Py,/Ky, = (0.780 x 760 torr)/(5.75 x 10’ torr) = 1.03 x 10°, Xo, = (0.210 x 760 torr)/(2.95 x 10’ torr) = 5.41 x 10%.
no, and ny, are
negligible compared to nyo = 5.55 mol, so ny, = (1.03 x 10°)(5.55 mol) =
5.72 x 10° mol and no, = (5.41 x 10°°)(5.55 mol) = 3.00 x 10” mol. my, = 1.60 mg and mo, = 0.960 mg. 123
9.54
(a)
/OP) »,But V,=(0n,/0P);, =(0/0P)r_, (Hy + RT Inx,) = (Op; (9.31) shows that (Op; /0P);
=.
.s0 Va=Ve.Since,, fonan
ideally dilute solution is a function of T and P only, its derivative
(Ou;/0P);,, is a function of T and P only, and (dy; /0P);,,, and V,° (which equals (On? /0P rn, ) are independent of concentration for concentrations in the ideally dilute range. Therefore V, (which equals
V,°) equals its limiting infinitedilution value V2 (b)
Using the intercept method of Fig. 9.7, we draw the tangent line to the curve at Xethano = 1. The tangent line intersects the Xethanoi = 0 vertical axis
at 4.7 cm*/mol = Vio —Vinn,o. since Vino = 18.0 cm*/mol at 20°C, we get Vis,0 = 13.3 cm’/mol in ethanol, in rough agreement with Fig.
on
9°55
Equations (9.30), (9.28), and (9.59) give: Si = —(Op;/0T) pn, =
—(0/0T) py, (fb? + RT In x) = S; —R Inx;. The infinitedilution limit of this equation gives Si = CS: + R In x;)”. We have i =Gi+TS;
57S,
= uw; + RT In x;
aye x; = H;. The infinitedilution limit of this equation gives
—oo
Hi; =e
9.56
Equations (9.17), (9.66) and (9.67) give AmixV=Diea n(V; — V*,) +
na(V, — V* 4) = Disa n(V,° — V* ,). Equations (9.68) and (9.66) give
AmixH=Disa n( H; — H*;)+na(H, — H*,)=Diean(H; H*,). ey
O= 2; VM; = 2, vj,(u; + RT In Kieg) = 2; Vib; + RT Z; In (Xieq )
PSNG Fe
RT In I]; (x;.4)"' =AG° + RT InK,, and AG° =RT InK,, where (1.68) and 1,eq (1.67) were used.
124
9.58
(a)
Bs = P,/K;, so as K; increases, ie decreases. The solubility of Oz in water decreases as T increases.
(b)
(H?° — AH; RT? = 0 In KJ/OT)p = (A In KJAT)p.
A In K; = In (3.52 X 10’) — In (2.95 x 10’) = 0.177. Hee —~ H;” =(8.314 J/molK)(298 K)’0.177/(10 K) = 13.1 kJ/mol.
(c)
The log of (9.62) gives RT In Kj/bar = py! — pj” = G; —G; and we have G, — G, = (8.314 J/molK)(298 K) In (44100 bar/1 bar) = 26.5 kJ/mol.
(d)
9.59
AS =(AH —AG)/T, so 5, — 5,” = [(13100 — 26500)/298] J/molK = 5 _133 J/molK.
The log of Eq. (9.62) reads In K; = In P® + (n>! — n°? RT. Partial + differentiation with respect to T gives (0 In Kj/0T)p = (et 24 ae WRT*
(ape! BT)p  dus’ /dTVRT = (Gi — Ge RIC # GSE Se DIRRE ov
CEP
oy
=e,
DEP
~,!
SOG
IP
sOR
h
2S MER? aoe eae
;
differentiation of In K; with respect to P gives (0 In K/0P)r = (0 uo! /OP)r=
Vi IRT= VeaiiRT (since 4," depends only on 7).
9.60
nbp,ben = Trouton’s rule is AvapHmanbp/Tp = 21 cal/molK. We find Ayapfm.
for the vapor 7420 cal/mol; AvapHm.nbp.to! = 8060 cal/mol. Equation (7.21) gives pressure of pure benzene at 120°C: In [P¥.,/(1 atm)] = and (7420 cal/mol)(1/353.2 K — 1/393.1 K)/(1.987 cal/molK) = 1.073
P*,, = 2.92 atm. Similarly, we find P%,, = 1.29 atm. Phen = Sed et
0.68(2.92 atm) = 1.99 atm.
Pro = 0.32(1.29 atm) = 0.41 atm.
P= 2.40 atm.
cy of xv, = Pren/P = (1.99 atm)/(2.40 atm) = 0.83. We assumed the accura
gases, an ideal Trouton’s rule, the T independence of AH of vaporization, ideal solution, and the pressure independence of p*.
9.61
png AmnG=G—G* =G—napt —ngpy. Then pa = (0G/0N4)7 + (0/0n 4), p.ng (NA ux +npyy + AmixG) = wx + RT In xa
125
naRT(OIn xa/Ona) Pe
B RT(O In xp/Ona) ny = MA + RT In xa +
naRT{I/na — 1/(na + ng)] + nBRT[1/(na + np)] = WX + RT In xa + RT — xaRT — xpRT = wx + RT In Xa, since xa + xp = 1. (The partial derivatives
of the logs are found as in Prob. 9.64b.)
9.62
(a) Use of (AG/OP)r= V gives AG, = & v* dP’ + J% V* dP’. We have AG> = 0 for this constant7andP equilibnum process. AG3 = AH3 —
T AS3 =—T AS3 = —naRT In (P¥/Pa) — npRT In (P3#/Pg), where (3.29) and Boyle’s law were used. AG4 = 0 (see the end of Sec. 6.1). AGs = 0 (constant7andP equilibrium process). AG¢ = IRs+P Vasp dP’.
(b)
They are small because G of a liquid varies only slowly with P.
(c)
With AG, and AGg¢ neglected, AG = AnixG = AG3 = naRT In (Pa/ PX) +
npRT In (P,/P%). (d)
9.63
Use (9.51).
Eq. (9.77) becomes AnixG/(ma + Ng) = RT[xa In a. P/ P=) +xp In Cee! Heep lh
At Xeth = 0.200, AmixG/(na + ng) = (8.314 J/molK)(318 K) x {0.200 In [(0.1552)454.53/172.76] + 0.800 In [(1 — 0.1552)454.53/433.54]} = —731 J/mol. At Xeth = 0.400, 0.600, 0.800, we get AmixG/mior
= —1034 J/mol,
—1554 J/mol, —997 J/mol, respectively. At xen = 0 and 1, AmixG/niot = 0.
9.64
(a) From (9.35), AmixS == (0 AppixGlOT) py,= 24 R In Xa — ng R In Xp (na + ng)xaxp(OW/0T)p. From (9.33), AH mix = AmixG + T AmixS = (na + Np)xaxplW— T(OW/0T)p]. From (9.34), AmixV = (0 AmixG/OP )7
=
(na + nNg)Xaxp(OW/OP)r.
(b)
AnixG = G G*, so G = G* + AnixG = napx(T, P) + ngps (T, P) + AmixG.
Then pa = (0G/On4)7 ping = WE + (0 AnixG/ON 4)z pn, » We have In x, = In na — In (na + np) and In xg = In ng — In (na + ng), SO 0 In xa/Ona = I/na — I/(ng + ng) and O In xp/dna = —1/(na + np). Also, xa xp(na + np) = nanp/(na + ng). Using these equations, we find La = ux + RT In xa +
naRT [1/nag — 1/(na + ng)] — ngRT/(na + ng) +
W(T, P)[np/(na + ng) — nanp/(na + npg)]= wk + RT In xy + W(T, P) x2.
126
Since AmixG is symmetric in A and B, by analogy to ta we have
Up = pe + RT In xg + WT, P) (c)
x4.
Equating a, in the solution to pa in the vapor above the solution, we get
ust RE In xa Wx3 = Ws oa; + RT In (Pa/P°). Also, for pure liquid A
in equilibrium with its vapor at T: WY = WA eas + RT In (P¥/P°). Subtraction of this equation from the preceding one gives RT In xa +
Wx? = RT In (Pa/ P¥), so In (Pal Pt) = In x4 + Wx3/RT; so Pal Pk = exp(In xa)exp(W x5 IRD), and Pa=xa Pr exp(W x; /RT). By symmetry,
Pp = Xp P§ exp(Wx4/RT). 9.65
AyapG° involves isothermal conversion of liquid at 1 bar to vapor at 1 bar. We use this isothermal path:
liq(P®) + lig(P,) > vap(P;) > vap(P?) AG, = 0, since moderate pressure changes have little effect on liquid thermodynamic properties. AG? = 0, since this is an equilibrium process at constant T and P. Assuming ideal vapor, we have AG3 = AH; — TWAS; =—T AS3
= —nRT In (P,/P°), where (3.29) and Boyle’s law were used. Therefore AyapG° = AGm.1 + AGm2 + AGm3 = —RT In (P;/P°) and P; = P° exp (—AyapG°/RT).
9.66
(a) UORHORVa, G(byeUnention Vim (c)s otic:
9.67
(a) All; (f) ideal;
9.68
9.69
(b) ideal;
(c) ideal and ideally dilute;
(d) all;
(e) ideal;
(g) ideal.
(a)
Neither; neither.
(b)
CCl, — Raoult’s law; CH30H — Henry’s law.
(c)
CH;0H — Raoult’s law; CCl4 — Henry’s law.
(d)
Both — Raoult’s law.
(a)
True, since the equilibrium condition at constant T and P is minimization of G. (b) False; see Fig. 9.5. (c) False. Ina liquid solution, intermolecular interactions are large. (d) T. (e) F. (f) T.
127
Chapter 10
10.1
(a) T; (b) T; (©) T; @)T; (e) F.
10.2.
(a) No.
(b) Yes.
(c) Yes. y; = aj/x; and a; in (10.3) depends on p;.
(d) Yes.
10.3.
From (10.4) and (10.7), uw; =y,* (7, P)+ RT a,;; > x;
Ina,,;. As x; 41, y,;; 1
and
1. As x; decreases (at constant 7 and P) from 1, Eq. (4.90) shows
that w; decreases. The equation p;
=—,;*+ RT Ina,, then shows that this
decrease in 1; means that a,,; decreases from its limiting value of  as x; decreases. Hence a,, can never be greater than 1. The definition (10.3) shows that a,; can never be negative. Hence 0 < a); < 1.
10.4
(a)
0G*/dn; = (0/dn,)(G — G*) = AG/An; — OG4/An; = wy; — w'? , where all derivatives are at constant T, P, njzi.
(b)
p= pw; + RT In yix;= (pw; + RT In x) + RT Iny; = tee + RTIn y;, so bi pie = RT In y; and use of the result of (a) gives the desired equation.
(c)
(0G*/dng),,. =(0/ Ong [ng + nc GE] =GE + (ng +nc)dGE/ On» (Eq. (a)]. We have (0G,,/Ong),,, = (OG;,/ OXp )(OXg /Ang),.. Also, x3 /Ong =(0/ Ong )[ng (ng tno) ']= (ng +nc)—ng (ng tne)? = (ng +c —N,)/@t,
+Nc) =Ne its + ne) =e (ng +Nc), So
(OG;,/ Ong), = (OG/OXp)xc M(ng +nc) [Eq. (b)]. Use of Eqs. (b) and (10.106) in Eq. (a) gives RT Iny,, =Gi + x¢(OGE/Oxg)>p. 10S
ee(ael
10.6
(a)
(bal
yii= x) P/x! PX.
Yicnt = (1 — 0.138)(304.2 torr)/(1 — 0.200)(295.1 torr) = 1.11.
128
Yieth = 0.138(304.2 torr)/0.200(102.8 torr) = 2.04. Acht = YichXchi = 1.11(0.800) = 0.889; ajeth = 2.04(0.200) = 0.408.
(b)
y= p; + RT Ina; = p* + RT In jj, So Wj — W* = RT In ay,. Meth — W*, = (8.3145 J/molK)(308.1 K) In 0.408 = —2300 J/mol. Mon — WA, = RT In 0.889 = 301 J/mol.
(c)
Egs. (9.32), (10.7), and (10.4) give AmixG = Di nj (Uj — w*).= > njRT In ay; = (8.3145 J/molK)(308.1 K)[(0.200 mol) In 0.408 + (0.800 mol) In 0.889] = —700 J. Alternatively, AmixG = (0.200 mol)(—2300 J/mol) + (0.800 mol)(—301 J/mol) = 701 J.
(d)
For an ideal solution, yj; = 1 and aj; = x;, So
AmixG = (8.3145 J/molK)(308.1 K) x [(0.200 mol) In 0.200 + (0.800) In 0.800] = —1280 J.
10.7
(a)
Cie Pips =o FIPS
di w= 0,020(0,02 kPa)/(19.92 kPa) = 0.176.
Viw = Aw/Xw = 0.176/0.300 = 0.586. At hp = (1 — 0.696)(5.03 kPa)/(2.35 kPa) = 0.651; Viho = 0.651/(1 — 0.300) = 0.930.
(b)
Since water is the solvent, Yi1,w = Yiw = 0.586. Example 10.2 in Sec. 10.3 gives Yiihp = Une / Kyp ) Yinp> where Knp = (Php/ wi )”. We plot hs P/ Sin
versus an and extrapolate to saa = 0. We find Knp = 0.62 kPa. (Use of a spreadsheet shows that a cubic or quartic polynomial gives an excellent fit to the data; the fitted equation has an intercept of 0.61 for the cubic
polynomial and 0.62 for the quartic.) So Yihp = (2.35 kPa/0.62 kPa)0.930 — ie ae
(c)
ail,hp = 3.5(0.700)
Aiw = Alw = 0.696.
=
We have AmixG = G — G* = Lin(pi we) = Dinkwi B) i)= >, n; RT \n ay;. An amount of solution with not = 1 mol has 0.300 mol of
water and 0.700 mol of H2O2, which is 5.405 g of water and 23.81 g of H5O>. The water wt. % is [5.405/(S.405 + 23.81)]100% = 18.50%; so 125 g of solution has 0.1850(125 g) = 23.1 g of water and 101.9 g of H2O2; thus ny = 1.284 mol and mp = 3.00 mol. Then AnixG = (8.314 J/molK)(333.1 K)[(1.284 mol) In 0.176 + (3.00 mol) In 0.651] = —9.75 kJ.
129
10.8
(a)
Let b=3.92. Equation (10.22) with B = Hg and A = Zn and state  being
pure Hg gives In Yup = —J? [xzo/(1 — X2zn)] d In (1 — bxzn) = b azn [x/(1 — x)(1 — bx)] dx = [b/(1 — b)][In(l — x) + b' In (1  bx)] 52 = [b/(b — 1)] In (1 — xzn) — [1/(6 — 1)] In C1  bxzn) = (3.92/2.92) In (1 — xz,) — (1/2.92) In (1 — 3.92xz,). (b)
10.9
Yuzn = 1 — 3.92(0.0400) = 0.843; aizn = Yu.znXzn = (0.843)(0.0400) = 0.0337. In Yue = (1/2.92) {3.92 In 0.960 — In [1 — (3.92)(0.0400)]}} = 0.00360; Yue = 1.0036. ayg = 1.0036(0.9600) = 0.9635.
From Eq. (10.12), G’ = RT Yj nj In y,;. So G'/n = RT Dj x; In yi = RT(Xac IN Yiac + Xchi IN Yichi). Walues of y; listed in Sec. 10.3 give at 35.2°C:
10.10
er
0
Grin
Ome 42 8) 9641
Xac
0.815
0.940
1
Guy
e698
24m
0
(a)
0.082
0.200
0.336
0.506
0.709
—~133.1.
+138.37—102:6 4 cal/mol
cal/mol
Inthe second example in Sec. 10.3, it was shown that ¥ , ;/Y,;=P7/K; for i # A. In the infinitedilution (x, —
1) limit, we have Yui —
1
[Eq. (10.10)] and y, ; > 7; ; in this limit, the boxed equation becomes
l/y;; =P#/K,
Equating the two expressions for P*/K, , we get
Vignes ae
(b)
From Fig. 10.3a, Yichi > 0.50 as xc
> 1. Therefore Yuen = 2.07,4) in
acetone.
10.11
(a)
Eq. (10.13) gives Fi,0 =4n,0Pi,o and Ayo = (23.34 torr)/(23.76 torr) =
9523) (b)
ay,9 = 22.75/23.76 = 0.9575. An amount of solution with  kg of solvent has 2.00 mol of sucrose and 55.51 mol of H2O, so XH.90 =
55.51/(55.51 + 2.00) = 0.9652. Then ay 5 = Yu,0%H,0 = 0.9575 = Yu,0 (0.9652) and yy,9 = 0.9920. an
130
10.12
All.
10.13
For an amount of solution containing  kg of water, ngycrose = 1.50 mol and
Ny,0 = Soe Pmolso x40 = So oiyo 7.01, —10.9737 and x4.1=10.0263: From
ae cl0.24)eVirie = YasueXA = 1.292/0.9737 =. 18327. Theniapac=Vitstexsuc = 19327,(010263) = 0.034984 lsondi suc =) Vmsuc iene, ) = 1.29201, 50y="1,94.
10.14
Equations (9.31) and (10.24) give V2, = (Ou, /OP); = (Op), ,/0P), = Vij = V,~, where an equation on page 284 in the text was used. From (9.30)
and (10.24), S” ==’, /oT)p =—(On;, /OT)p —R In Mam? =
Si; R In Mam® = (S;,+ R In x)” — R In Mam® = [S, + R In (x/Mam)]”, where an equation on p. 284 was used. We have x/Ma = nj/moMa; in the infinitedilution limit, no, = na and x/M, =nj/naMa = nj/wa = m;, where wa 1S
the solvent mass and m; is the solute molality. So Sj, = [S; + R In (mj/m°)]”.
Finally, (9.28), (10.24), the above relation between Sseand Seand the
relaliounly BiH D.284) vives
me ll etna
us, + RT In Mam® + T(S;,; —R In Mam’) = wi,;+ TS wa ie
10.15
pe ee
we, + RT In (yici/c°) = Wi; + RT In yusx. In the limit x4 > 1, Eqs. (10.29)
and (10.10) show that the activity coefficients go to 1; so this limit gives He ei
ae
ln (xic°/c;)_. We have x; = nj/mo = n/n and c; = nV =
n,/V* for xa near 1. Hence, (xic°/c))” = con Vx Ining = CVAAS Mee = Baa + RT In V* ,c’ . Substitution of this result in (10.29) gives yj = wi; + RT In VE co + RT In (ycici/c") = Wi, + RT In (VE yy, ;¢; ). Comparison ac) V4  We have with (10.23) shows that V* ,y.,¢; = Yusxi and Yo = (%)/V X; = Nj Nop = WAN WANot = WAM Not = PV {M;/No, (where w is mass) and Vi.4 =1V
fuNagsSO2i Vo.
c= Oamna/Nior = PamixXa. Then Yi = (4/VE AC)14 =
(pamixalci)(Ymi/xa) = (Pamilci)Y¥m,i. AlSO, YeiCi = PAMYm,i and (Y,ic¢;/c°)e° =
Pa(miym,/m°)M°, SO AciC° = Padmim® and a, = (Pam"/c")am,i
131
10.16
We use Eg. (10.22) written for Convention I and with A changed to B and B changed to C. Let state 1 be pure C, so ¥i,c.1 = 1 and xc = 1. Then (10:22)as
In vic2= —J? [xg /(—xg)] dIny,, [Eq. (a)] at constant T and P. We have In Yip = (W/RT)xe, so dIny,, =2(W/RT)x¢ dx¢ at constant T and P. Also, the integrand in Eq. (a) is xg/(1— xg) = (l xc )/X¢ . Hence Eq. (a) becomes
InYicg
=—2(WIRT) J? xc) die =20WIRT)(Xe —F x0) ? =
~2WIRT)(X¢.9 — 4x24 1+4) = 2(W/RT)[I= xp. $ (1 xg)" 4]= (W/RT)xp 5, and RT Iny;c = Wx;,, where the unneeded subscript 2 was dropped
10.17
Equation (10.106) gives RT In, = (0/dng)7 pp.[Wngnc (ng +Mc)1=
Wn l(ng +nc)—Wngncl(ng tc)? =Wine(ng +c) —Ngnc (ng +N)” = Wnél(ng +c)? = Wxe.
10.18
In the limit x,y; = 0 and x,, = 1, Fig. 10.3 gives ¥jchi = 0.50. Hence
In Vichi = (W/RT) x2, becomes In 0.50 = (W/RT) and W/RT = —0.693. For Xen =
0.494, the simplesolution model gives Iny,,. = (0.693)(0.494)* = 0.169 and
Yiac = 0.844. Also, Iny;.41 = (0.693)(0.506)" = 0.177 and Yicn = 0.837. The true values are 0.824 and 0.772.
10.19
(a)
From (10.12) and the definition of Gi in (10.32), we have Gi =
RT (Xe NY icht + Xhep INYi hep) and use of the data gives G£ i(J/mol) = 90.7, 197.99224,°1879 and 824tatien = Onl 0G90.5,.0.7, and 0%: (b)
We designate three cells for A, Az, and A3. A suitable initial guess is zero
for each of these, corresponding to an ideal solution. The xcn) values are put in column A, the Ge values in column B, the Redlich—Kister values
in column C, and the squares of the deviations of the column C values from the column B values in column D. We sum these squared deviations in a cell and use the Solver to minimize this sum by varying Aj, A>, and A3. The Solver gives A; = 0.334656, Az = 0.021853, and A; = 0.038433.
The fit is quite good, with deviations of typically 0.5 J/mol.
132
(c)
At xp =Xch = 0, the equations of Prob. 10.23 give In ¥j.ch) = A}
—A2 + A3 =
0.39494
Xchl = 0.4,
and Vichi = 1.48; also, In Yihep = 0 and Yithep =
Ati
we get In Yi,chi = 0.11188 and yych = 1.118; In Yihep = 0.06163 and Yihep =
1.064.
10.20
For the twoparameter fit, the spreadsheet of Prob. 10.19b can be used by setting A3 = 0 and omitting A3 from the By Changing Cells box of the Solver. One finds A; = 0.340293 and Az = 0.021853. The fit is much poorer than for
the 3parameter function, with the sum of squares of the deviations equal to 58 J*/mol? as compared with 1.44 J */mol” for the threeparameter function. For the fourparameter function, we modify the spreadsheet by adding a cell for Ag, modifying the formulas in column C, and including Aq in the By Changing Cells box. The sum of squares of deviations is reduced to 1.42 J*/mol* , which
is not a significant improvement over the 3parameter function.
10.21
From (10.12) and the definition of G& in (10.32), we have Gi = RT (Xen NY, cht + Xac NM Y1,ac) and use of the data gives G£ /(J/mol) = 179, 402, 557, 579, 429, 292, —103 at xcn = 0.918, 0.8, 0.664, 0.494, 0.291, 0.185, 0.060. The spreadsheet of Prob. 10.19 can be used with the data revised. We take B as chloroform. The twoparameter fit is poor with a sum of squares of deviations equal to 904 J*/mol? and deviations of 3 to 22 J/mol. The threeJ’/mol’parameter fit is fair, with a sum of squares of deviations equal to 258 J’/mol’ and deviations of  to 10 J/mol. The fourparameter fit is very good, with a sum of squares of deviations equal to 32 J’/mol’ and deviations of 1 to 3 J/mol. The four parameters are A; = —0.869763, Az = 0.22272, A3 = 0.10898,
Ag = 0.14581. The activity coefficients can be found from equations like those in Prob. 10.23, but since the fourparameter version of these equations was not given, this calculation is omitted.
10.22
Comparing GE = xgxcRTLA + Ap (%p — Xc) + Ag (Xp —Xc)” + Ay(XB — XC) +
.] with GE = xox,RT[Al + Ay (Xe — Xp) +Ag(%e — XB) +AU (Xe — XQ) + , we see that A’ = A,, Aj
=—A), A; = A;, Ay =—Ay, ete.
133
10.23
xg and xc are not independent. Substitution of xc = 1 — xg in (10.32) gives
G> = (xg — xp) RT[A, + A,(2x_ —1) + A; (2xg 1)’], 80 (GE /0xg) = (1—2x_)RT[A, + Ay (2xg —1) + A,(2x_ 1)7]+ (xp — xg )RT[2A, +4A,(2x, —1)]. Using xp = 1 — xc, we have Invyip = (RIG x0 (0G, /ox, =
oe arene, (2x)
Ae 218)
Xe (2x¢ —1[A, + A, (1 2x0) + 4,(1 2x¢)7]+ (1— xc) x@[2A, + 4A; (1 2x¢)] = (A, +3A, +5A3)x~ — (4A, +16A;)x0 +12A;xé . (The tedious algebra can be done by certain calculators or by symbolic algebra programs such as Maple V or Mathematica. If you don't have access to such resources, you are excused from doing the algebra.) Interchanging B and C in the final expression for In Y¥; 3 requires that the sign of Az be changed, as shown in Prob. 10.22, and this gives Iny,¢.
10.24
(a)
At infinite dilution of B, we have xc =  and the first equation of Prob.
10.23 with A3 = 0 becomes Iny;, = A, — A). With xg =  and A; = 0, the second equation of Prob. 10.23 becomes In y;~ = A; + A). Adding and subtracting these equations, we get the desired results. (b)
Let B be CCly. From the results of (a), A} = 0.5(1.129 + 1.140) = 1.134 and Az = 0.5(1.140 — 1.129) = 0.005. The equations of Prob. 10.23 then give
InYicca, = (A + SAy) Gc, =4A5Xsicl, = 0.1826 and Yjcc = 1.20;
Inyisic, = (4 BoAD cay +4A5Xoci, =, 0.407 and! ¥; ey 10.25
10.26
le 50.
(a) KCI>K*+CI; v,=1, v=1, u=1, z=1. (b) MgC > Mg**+2C l, r; v,=1, v_=2, z,.=2 z=1. (c) "MgSO, —"Mg*" + S027? vial, Val ye? 2 (d)
Ca3(PO4) 4 3Ca* 2 4+2PO7°; v,=3, V_=2) z,=2)> 7 =3)
(e)
Fora
V4 =n.
1:1 electrolyte, z, = _ = 1. KClis a 1:1 electrolyte. ye (Vo )t verve):
134
WP
Wie?
(a) y. =¥77"?.
(Cy.
2 (i=)
10:27)
Up = Vat
10283
(Vey =a)
(a) (b)
) w= Q)” Qn”.
ys)
eV See
(vs) =1' 1" and vz = I. (vs)? = 1)  2° and ve.= 1.587,
(cave = 7
(d)ui(vs),= 3. 2 andvs=2.951. 10.29
With
V,=V_,we
have
(ee
en
(v.)*
= (v,)"* (Va)
—
(v.)’* (v,)"* =(v,)?"* .From(v,)*"* =(v,)”*,we havevi =V,. 10.30 From Prob. 10.28(b), (v..)” =4 and (10.52) gives aygc), == 42 (m,/m°)’. 10.31
(a)
(b)
Equation (10.58) gives rN l (0.018015 kg/mol)2(4.800 mol/kg)
23.76 torr = 0.990 — 20.02 torr
From Eq. (10.56), aa = P,/P% = (20.02 torr)/(23.76 torr) = 0.843. One kg of water contains 55.51 mol of H2O. There is no significant ion pairing in the KCI solution, so x4 = 55.51/(55.51 + 4.80 + 4.80) = 0.853.
Equation (10.5) gives Ya = da/xa = 0.843/0.853 = 0.988.
10.32.
(c)
ay =0.843. xq = 55.51/(55.51 + 4.80) = 0.920. Ya = 0.843/0.920 = 0.916.
(a)
In¥n=am/m? + b(m/m°y + c(mim°?)° + d(m/m°y* +
[a (alm° + bm’/m° + cm? /Im® + dm?/m™) dm’ = 2am/m? +
(3/2)b(mim°y + (4/3)c(m/m?)? + (5/4)d(mim?°)’. (b)
Substitution of m/m° = 6.00 gives log Ym = 0.45415 and Ym = 2.85. One kg of H2O has 55.51 mol H20, and x(H20) = 55.51/(55.51 + 6.00) = 0.902;. Equation (10.24) gives Yi = Ymi/%a = 2700) 0202s= 3,16;
135
10.33
(a)
Atconstant T and P,0 =n, dua +n; duj = na(ORTMav dm; — RTMavm; do) + ni[(VRT/m;) dm; + VRTd In yi] and 0 = (n/m; — OnaAMa) dm; ~ n~»Mam; db + n; d In y;. We have m; = n/naMa.
Substitution of n4Ma = n,/m; into the preceding equation and division by n; gives d In y; = do + [(@ — 1)/m;] dm; at constant T and P. (b)
=
(ux —paVRTMavm, = RT In Yx,axa/RTMavm, = —In (Yx.aXa)/Mavmi.
At high dilution, y;,4 > 1 and In (yx,4xq) — In xa. Equation (8.36) gives for x, near 1: In x, =x, — 1. At high dilution, there is no ion pairing and
the electrolyte gives vn; moles of ions. Hence xa = 1 — vnj/nio = 1 —vnj/na. We have In x, = —vn,/na and @ > (vnj/na)/Mav(n/naMa) = 1.
(c)
Integration from the infinitedilution state (where y; = 1 and
@= 1) toa
solution with molality m gives In yi(m) = o(m) — 1 + HG —1)/m,] dm,.
10.34
Use of w, =m}'? and dw, = (1/2)m,''* dm, in the integral in (10.59) gives this
integral as J [( —1)/m,]2m!/? dw, =25*” (0 D/w,]dw,. 10.35
m(CI) = [0.0100 + 2(0.0050)]mol/(0.100 kg) = 0.200 mol/kg. m(K*) = 0.100 mol/kg. m(Mg?*) = 0.070 mol/kg. m(SO2)= 0.020 mol/kg. jk
Sai
1[0.200 + 0.100 + 27(0.070) + (—2)°0.020] mol/kg =
0.330 mol/kg.
10.36
For the electrolyte M, X, with stoichiometric molality m; (and no ion eee
y)
pairing), we have m, = v4m; and m_ = v_m,. Hence I, = 5 (z.V 4m; =f z°v_m;) =
zm,(v,z, 2
2 +v_z)= 42, lel (va + vm; = ay  from NaCIKCl, we get Ona.k and Wna.k.cl
10.43
Withee
Ven WEAVE Vie. Vict Vee 2 Vand Vai VietVe ==
as (7a) nrl—
becomes me = Ope les (lea a)
10.44
ys = OY+.
Substituting the m, and m_ expressions preceding (10.76) into (10.46) and using (1.68), we find
i= fb; + RT In {(yz)"(v.m{m°)"* [v_(1a)v,J (m,/m’ ) } = we? + RTIn {(y4.)° "+ (mlm? )"*"" (v,)"* (v_) [1@)v,/v_] J= po +
VRT In{y.a"*” (v,)"*” (v_)’" [1a@)v,/v_]" mm’ }=o + VRTIn{v, 7.0" [l—d— a)v,/v_]’"’m,/m’ }, which is (10.77) and (10.76).
10.45
From the m, and m_ equations preceding (10.76) and from m, =V4m, and
m= =Vv_m,;, we have m,/m> =a and m_/m™ = 1 — (1 —)v,/v_. Then
(m,/m? )**" (m_Im?)*y, =a" {1 (1 — a)v,/v_]’"’y.., which equals Yt eanclOw
10.46
7s
Use of the reactionequilibrium condition yp = by + LL, Eq. (10.74), and Eq. (10.38) gives G = nala + (V4n; — np), + (V_n; — Np) + nyp(y + PL) =
NaMa + (V4Py + V_LL)n; = Nala + nib. 138
10.47 (a) m(Pb**) = 0.100 mol/kg  0.43(0.100 mol/kg) = 0.057 mol/kg; m(NO3) = 0.200 mol/kg — 0.043 mol/kg = 0.157 mol/kg;
m(PbNO; ) = 0.043 mol/kg. In = 5[4(0.057) + 1(0.157) + 1(0.043)] mol/kg = 0.214m?°.
(b)
log y+ =—0.257 and y+ = 0.553. Equation (10.77) gives
yi. = (0.57)?[1 — 0.43(1/2)}?70.553 = 0.390. 10.48
As T increases, the water dielectric constant decreases, which increases the
interaction forces between ions, thereby increasing ion pairing.
10.49
For sucrose(aq), (298.1 K)A,S°
A ,G° =1551 kJ/mol = A, H° — TA,S° =2215 kJ/mol and A,S®° =—2227 J/molK.
A,S° = SS
°
elem ?
where the
entropy of the elements is S,,.,,elem = 125, [C(graphite)] + 11S, [H2(g)] +
5.5S5 [O2(g)] = 2635 J/molK, where Appendix data were used. Then ——
—2227 J/molK + 2635 J/molK = 408 J/molK.
AG3oq/(kI/mol) = 237.129 — 0 — (157.244) = 79.885. AH 59g /(kJ/mol) = 285.830 — 0 — (229.994) = 55.836. AS3o9q ((J/molK) = 69.91 — 0 — (10.75) = 80.66. (b) AG3og (kJ/mol) = 237.129 — 394.359 — 2(0) — (527.81) = 103.68. AH 59g /(kJ/mol) = 285.830 — 393.509 — 2(0) — (677.14) = 2.20. AS5og /(I/molK) = 69.90 + 213.74 — 2(0) — (56.9) = 340.5
10.50 (a)
10.51
(a)
Use of (10.91), (10.92), and (10.93) gives A ,G°/(kJ/mol) = 65.49 + 2(111.25) = 157.01.
A , H° kJ/mol) = 64.77 + 2(207.36) = 349.95. (b) 10.52
S°J/molK) = 99.6 + 2(146.4) = 193.2. AH°/(kJ/mol) = —240.12 — 167.159 — (411.153) = 3.87.
For this reaction, AG}o /(kJ/mol) = 1010.61 — (—261.905 — 744.53) = 4.18 (which corresponds to K° = 5.4.)
139
10.53
From (10.91) and (10.85), A ,G®° [HNO3(ai)] = A,G° [H*(ao)] + A ,G°[NO; (ao)] = A ,G°[NO; (ao)] . The NBS tables do not satisfy this
equality and so contain an error.
10.54
Since the two ways of writing the reaction refer to the same process, AS® must be the same for each. Hence S°(H*) + 5*(OH’)
= 5, (H350)=
5 (HO, ease (OH) — 25310). and) § (HO. (aq)Jims sl yak S? (H,O) =0 + 69.91 J/molK = 69.91 J/molK at 25°C. 10.55
)Bi,+V_we) =v, (OM)/0T) p+ p(v, S° =(0u°/0T) p = (0/0T v_(dp°/0T) p=v,5S,+Vv_S°. Subtraction of S;, from each side gives
A,S; =v, A,S2 +v_A,S°.Then A,H; = A,G; +T A,S; = vi(A,G, +T Ne Se eave Ne Ceetale A ,S*)=v1 A,Hi +v_A,H_.
10.56
(a)
In Eg. (10.81), Ym.isat and Misa are for P = P° = 1 bar. Since the solution is very dilute and nonionic, we can approximate Ym.isat aS 1 and (10.81)
gives A ,G59g[O2(aq)] = 0 — RT In 0.00115 = 16.8 kJ/mol.
(b) x; = P,/K; = (1 bar)/(30300 bar) = 3.30 x 10° = nj/mo = n;/ny,9 and
Nc,H, = "1,0 3.30 x 10) = (55.51 mol)(3.30 x 107) = 0.00183 mol in 1000 kg of water, so mc, = 0.00183 mol/kg. Then (10.81) gives
A ¢ G3og [C2H6(aq)] = 32.82 kJ/mol  RT In (0.00183) = 17.2 kJ/mol. 10.57
Use of Eq. (10.25) for 1; gives the following results. S; = —(ON,/0T) pn, =
—(pm.i/9T) p— (O/0T) p, [RTIn(Y,,,m/m® = Sp; — Rin (Y,,;m,/m*)— RT
In Ym/OT)p.»,
Vi = (OW/OP)p»,= (OW{/OP)p+
RT On Yn i/OP) py, = Ving + RTO, JOP).
Hy = bit TS; = Wry +
TS,,;— RT*(Olny,, 0D) p,, = H,,.mit — RE (OIny,,./00) p,.
140
10.58
From (10.91) with i= HCl, 131.23 kJ/mol = A ,G°[HCl(aq)] = A ,G° [H"(aq)] + A ,G° [Cl (aq)] = 0 + AG°[Cl(aq)] and A ,G° [Cl(aq)] = —131.23 kJ/mol. Similarly, (10.93) gives A ,H° [CI (aq)] = 167.16 kJ/mol.
Use of T A, S° [HCl(aq)] = A,H° [HCl(aq)]  A ,G° [HCl(aq)] gives A ,S° [HCl(aq)] = 120.5; J/molK = S°[HCI(aq)]  152, [H2(g)] 182 [Clo(g)] = S°[HCl(aq)] — 176.88 J/molK, so S° [HCl(aq)] = 56.36 J/molK = S°[H*(aq)] + S°(CI'(aq)] = S° [Cl(aq)]. 10.59
From (10.84), A ,G° [i(A)] = A G®° ((*) — VRT In (V2¥m.+.sat!Misa/M°), SO A Gog [KCl(aq)] = 409140 J/mol — 2RT In [1(0.588)4.82] = 414.30 kJ/mol. Equation (10.91) gives 414.30 kJ/mol = A ¢ Grog [K*(aq)] — 131.23 kJ/mol and
A , G39g[K*(aq)] = 283.07 kJ/mol. Equation (9.38) at 1 bar gives
AH SireKcyag) = 1722 kI/mol = H™ [KCl(aq)]  H;,[KCI(s)] = H° (KCl(aq)] — H°,[KCl(s)] = H° [K*(aq)] + H [Cl (aq) =H A[KCI(s)]. Subtraction and addition of the standard enthalpies of K, 5Ch, ande
onthe
right side of the last equation gives 17.22 kJ/mol = A ; H 59g [K*(aq)] + A ¢ A 9g (Cl (aq)] — A ¢ A x93 [KCI(s)] = A ; H p98 [K*(aq)] — 167.16 kJ/mol + 436.75 kJ/mol and AH ¥ 50g (K*(aq)] = 252.37 kJ/mol. To find S° [K*(aq)], we use the fact that AG, for dissolving KCI in water equals
A , Gy9g [KCl(aq)] — A ¢ Gyog [KCI(s)] = (283.07 — 131.23 + 409.14) kJ/mol = —5.16 kJ/mol. Since
AH 59, for the solution process is 17.22 kJ/mol, we have
AS° = (AH° — AG°\T = 75.06 J/molK for dissolving KCI in water. Then 75.06
ImolK = S°[KClK(aq)] — S°, [KCI(s)] = S°[K*(aq)] + S°[CI(aq)] 
S° [KCl(s)] = S°[K*(aq)] + 53.36 J/molK — 82.59 J/molK and S° [K*(aq)] = 104.3 J/molK.
10.60
As the charge increases, the ion binds more H2O molecules to itself, thereby increasing the degree of order in the solution and decreasing S of the solution
and S,.
141
10.61
(a)
From (10.104), In 62 = J7? (WVm/RT— 1/P) dP =
Jy? (/P + B' + C'P + D'P? +— 1/P) dP =  (BGP DLP aS) dP=B Pee P22
Py id to
Dropping the unnecessary subscript 2 on @ and P, we get the desired equation. (b)
Comparison of (8.9) with (8.4) gives B = b—a/RT,
C= b*,...; use of
(8.6) gives B' = (bRT —a)/R°T’, C’ = (2abRT —a°)/R°T",.. . and substitution into the result of (a) gives the desired result.
10.62 (a)
From Eq. (8.18), b = RT,/8P, = 42.9 cm*/mol and a = 27R°T2/64P. = 3.61 x 10° cm® atm mol. At 75°C, RT = 28570 cm?*atm/mol and Prob. 10.61b gives In @ =~[(2.38 x 10°)/(8.16 x 10* atm)](1 atm) — [(4.18 x 10')/(1.33 x 10'® atm’)](1 atm)? = 0.00292 and = 0.997 (compared to exper = 0.997). Replacing  atm by 25 atm in the preceding equation, we get In
= 0.0749 and 6 = 0.928 (compared to Qexper =
0.92). (b)
6; = OF (T, P) = 0* (75°C, 25 atm) = 0.928, where the result of (a) was
used.
10.63
fj = 0)P; = 0:x;P = 0.928(0.100)(25.0 atm) = 2.32 atm.
From (10.102) and (10.103), Gn = = n° + RT In (f/P°) = ° + RT In (OP/P°) = 2° + RT In (P/P°) + RT In @. For the corresponding ideal gas, @ = 1 and Gee
= pl = wo + RT In (P/P?). So Gm= Git + RT In 6 and In 6 = (Gm — G'9 RT. 10.64 (a)
(b)
10.65
(a)
Equation (4.65) gives AG = J> VdP = J? (nRT/P) dP = nRT In (P2/P;) = (1.000 mol)(8.314 J/molK)(273.15 K) In 1000 = 15.69 kJ = 3.75 kcal. AG=nAp=n[p° + RT In (f/P°) — W° — RT In (f/P°)] = nRT In (fo/f,) = NRT In ($2P2/o,P)) = (1.000 mol)(8.314 J/molK)(273.15 K) x In [1.84(1000)/0.9996(1.000)] = 17.1 kJ = 4.08 kcal. From (10.104), Ino equals the area under the V,,/RT— 1/P vs. P curve from 0 to 120 atm. The data are:
142
10ratrn
CVa/RES 1/Py AGse =4570
Platm
shud
10° atm (V,/RT— 1/P)
24.78 © —5.07e
(6
=
P/atm
20
40
Gemma
bae4102
100
120
80
0.0040 (V,/RT— 1/P) atm
45.42
60
pe E
0.0045 =
:
0.0050 0.0055 LE
fC
i)
SIE TELEMANN
0
20
40
+60.
80
=
"HOO  120 P/atm
Plotting the graph, cutting it out and weighing it, we find that the weight of the area between the —0.0040 horizontal line and the curve 1s 0.534
times the weight of the rectangular area shown in the figure. The rectangular area shown equals (0.0020 atm ')(120 atm) = 0.240, so the area between the —0.0040 line and the curve is (0.534)(0.240) = 0.128. Adding in the area (0.0040)(120) = 0.480 between the y = 0 axis and the
0.0040 line, we get an area of —0.608 as the value of In
=
jioam (y_/RT— 1/P) dP. So In $ = 0.608 and 6 = 0.544. Also, f= oP = 0.544 x 120 atm = 65.3 atm.
(b)
Problem 8.38 gives B = limp_49 (Vm — V0) = limp—o (Vm — RT/P). The data in part (a) give limp_s0 (Vn/RT— 1/P) = 0.00464 atm’. Hence B(—50°C) = (0.00464 atm !)(82.06 cm?atm/molK)(223 K) =
—85 cm*/mol.
10.66 (a)
B= [(82.06 cm*atm/molK)(126.2 K)/(33.5 atm)] x Meo
12,
4ge
143
cm fmol;
(b) B'=B/RT=5.7 x 10“ atm”. At  atm, Ing =B'P =
(5.7 x 107“ atm™)(1 atm) =—5.7 x 107 and $ = 0.99943. At 25 atm, In 6 =0.014; and @ = 0.9859.
10.67
(a)
Bi, = Ben/RT = (1040 cm*/mol)/RT = 0.04048 atm! =
5.326 x 10° torr. Bi, =—0.0569s atm! = 7.497 x 10™ torr”. In the mixture: In ben = In o%,(T, P) = BoP = (5.326 x 10> torr !)(301.84 torr) = 0.01608 and ocni = 0.9841; In Qcar = (7.497 x 10° torr ')(301.84 torr) = —0.02263 and car = 0.9776. In the pure vapors: In 4, = (—5.326 x 10~°)(360.51) = 0.01920 and o*, = 0.9810; In 6%,Car =0.01599; o%,,Car = 0.9841.
(b)
OFP* SOV ig = Or fi= ¥, x; f* and dix? P= 5x;
P/ x;O*P* . Then Yicn1
= (0.9841)(0.6456)(301.84 torr)/0.5242(0.9810)(360.51 torr) = 1.034 and Vicar = (0.9776)(0.3544)(301.84 torr)/0.4758(0.9841)(213.34 torr) = 1.047.
(c)
Here we take ; and * equal to  and ¥j,;= x? BE xiP*. We get Yichi =
1.031 and Yj.car = 1.054.
10.68
At constant T and P, we have du; = d(u* + RT In x;) = RT d n x; = (RT/x;) dx;. Therefore Ds n; di; =i]
di (nj/x;) dx; feel >; (niNtor/Ni) dx; =
RT not >; dx;. From >; x; = 1, it follows that ); dx;
=0. Hence );; n; du; = 0 at
constant T and P. This completes the proof.
10.69 There are 2Nq = 12 x 107° ions in 1000 cm? of solution. With uniform distribution, each ion is in the center of a cube of volume
(1000 cm*)/(12 x 107") = 8.3 x 10°” cm? and edge length (8.3 x 107? cm*)!? = 9x 10° cm=9 10.70
(a)
A, and this is the nearestneighbor distance.
Because AB attractions are weaker than AA and BB attractions, U and H of the solution will be higher than U and H of the corresponding ideal
solution. Hence Amx/? >0= A,
(b)
H".
Because AA and BB attractions are stronger than AB attractions and the molecules have similar sizes and shapes, A molecules in the solution
144
will tend to surround themselves preferentially with other A molecules (and similarly for B molecules). Hence the degree of order in the solution
is greater than in an ideal solution (where there is complete randomness in distribution of A and B molecules), and S of the solution is less than
Son
10.71
a
CARS AG".
band / Am asealAvens
(a)
An increase in z, increases the attractions between the positive ion and the negative atmosphere that surrounds it, thereby stabilizing the solution
and hence lowering Hl, and lowering Y,.
(b)
An increase in ionic diameter decreases the closest distance of approach between positive and negative ions, decreasing the attractions between them and hence increasing py, and increasing Y,.
(c)
An increase in ionic strength means an increase in the attractions between each cation and the anions in its atmosphere, which lowers w, and hence lowers Y+.
(d)
An increase in solvent dielectric constant decreases the interactions
between positive and negative ions, increasing [, and increasing Y;.
(e)
As T increases, the kinetic energy of random ionic motion increases,
which tends to distribute the ions more randomly and to thereby break up the ordered ionic atmospheres. This destabilizes the solution and increases pt, and y,.
10.72
(a)
Li (sin) = iv) and use of (10.51) gives wy; + VRT In (viyim/m?*) =
us” +RT In (P/P°), so In (Pi/P°) = (Me, — By? WRT+ In (veyimilm®)* and P; = P° exp [(W,; — Hi WRT\(vsyimlm°)’ = K(vsyimj/m°)’. For
HCl, v, = 1 =v., v=2, and vs = 1, so Pi = Kiym/m°y’.
(b)
K;= P° exp [(wi,; — Bi /RT).
we? = Wicugy: Pm = B°UH™(aq)] +
w°(Cl(aq)]. we; — Wy? = 4,G° [H"(aq)] + 4,G° [CI (aq)] A ,G° [HCl(g)] = [0 — 131.228  (95.299)] kJ/mol = 35.929 kJ/mol. K; = (1 bar) exp [(35929 J/mol)/RT] = 5.08 x 10°” bar.
145
P; = (5.08 x 107” bar)[0.80(0.100 mol/kg)/(1 mol/kg)]}* = 3.25 x 107’ bar = 0.00024 torr.
10.73
10.74
(a)
The following become  in the limit xy — I: Yiw, Yu.w, Yue, Ym.E
(b)
Only ye becomes .
(a)
True. When uy; = pH; in (10.3), then a; = 1.
(b)
False. For example, for Convention I, with x; < 1, yj might be such that (pesllbace
(c)
Miuemsee Lg.410.3).
(d)
messee Ege (105):
(e)
False.
(f)
False.
146
Chapter I  (a) T:
(b) T.
(a) F; (b) F; (c) F; (d) F. The H’ from H:20O ionization can make m(H*) exceed m if mis extremely small. (b). 11.4
(a)
m; = nwa = (4.603 g)(1 mol/46.026 g)/(0.5000 kg) = 0.2000 mol/kg. We neglect H” from H20. Let x = m(H") = m(HCOO). Then (11.15) gives Kz
= 1.80 x 107 mol/kg = yx*/(0.2000 mol/kg — x). With the initial approximation y+ = 1, we find x = 0.0059; mol/kg. Hence I, = 0.0059, mol/kg and the Davies equation gives y+ = 0.92. Use of this y+ in
the K, equation gives x = 0.00649 mol/kg. Then I, = 0.00649 mol/kg, y+ = 0.919, x = 0.0064, mol/kg.
(b)
m(KC]) = (0.1000 mol)/(0.500 kg) = 0.200 mol/kg. The KCI contribution
to I, is 0.200 mol/kg and the H” and HCOO contribution to J, is (from part a) about 0.006 mol/kg, so J, = 0.206 mol/kg. The Davies equation gives y+ = 0.745. Use of this y+ in the Kz equation gives x = 0.0079 mol/kg. This gives a revised I, of 0.2079 mol/kg, which gives y+ = 0.74s. We then obtain x = 0.0079 mol/kg.
(c)
If xis the moles of HCOOH that ionize per kilogram, then m(H”’) = x and m(HCOO) = x + 0.400 mol/kg, since the KHCOO
contributes 0.400
mol/kg of HCOO™. So 1.80 x 10~ mol/kg = v2 x(0.400 mol/kg + x)/(0.200 mol/kg — x). With the initial
approximation /,, = 0.400 mol/kg (due to the salt), the Davies equation gives Y: = 0.739. With x in the numerator and denominator neglected compared with 0.400 and 0.200 mol/kg, the Ka equation gives x = 1.69 x 10~* mol/kg. This x is small enough to neglect its contribution to I and to neglect it compared with 0.200 mol/kg.
(a)
A BASIC program 1s:
147
1OsG=i
75 XX=X
15 INPUT "KA";KA
80 GOTO 60
20 INPUT "M1";M1
85 PRINT "M=";M;" M(H+)=";X
25 INPUT "M2";M2
90 NEXT M
30 INPUT "DM";DM
95 STOP
35 PRINT "KA="KA
300 X=(—KA+SQR(KA*KA+4*KA*M*
40 FOR M=M1 TO M2 STEP DM
G*G))/(2*G*G)
45 GOSUB 300
310 RETURN
50 XX=X
400 =X
60 GOSUB 400
410 S=SQRiI)
65 GOSUB 300
420 LG =0.51*(S/(1+S)0.3*1)
70 IF ABS(XXX)/X bar.
11.30
Taking the activities of the solids as  and assuming ideal gases, we have
K° = P(CO2)/P(CO) = 1.15 = n(CO2)/n(CO). Since the initial n(CO2)/n(CO) value exceeds 1.15, the equilibrium position lies to the left. Let z moles of CO2 react to reach equilibrium. Then 1.15 = (5 — z)/(3 + z) and z= 0.72 mol. The equilibrium amounts are n(Fe304) = 2 2 MoOl. nO i= 312 nol
n(FeO) = 1.84 mol, n(COz2) = 4.28 mol.
11.31
K° = P(CO>)/P° = (183 torr)/(750 torr) = 0.244. (a)
(5.0 g)/(100.1 g/mol) = 0.050 mol CaCO3. We have Nco, =
(0.244 bar)(4000 cm?)/(83.14 cm?bar/molK)(1073 K) = 0.0109 mol. Hence the equilibrium composition is 0.039 mol of CaCO3, 0.0109 mol of CaO, and 0.0109 mol of CO.
155
(b)
The 0.0050 mol of CaCO; is not enough to give 0.0109 mol of CO2; therefore all the CaCO; dissociates to give 0 mol of CaCO3, 0.0050 mol of CaO, and 0.0050 mol of CO3.
11.32
At the pressures involved, we can assume idealgas behavior. Initially, n(CaO) = 0.00892 mol. K° = 0.244 = Peq(CO2)/(1 bar) and Peq(COz) = 0.244 bar = 183 torr.
(a)
The pressure of 125 torr is less than the equilibrium pressure 183 torr when all species are present, so no reaction occurs and the final amounts are n(CaCO3) = 0, n(CaO) = 0.00892 mol, and n(COz) =
(125/760) atm (4000 cm’)/RT = 0.00747 mol.
(b)
Peq(CO2) = 183 torr and neg(CO2) = (183/760) atm (4000 cm* RT = 0.0109 mol. The initial number of moles of COz is (235/760) atm (V/RT) = 0.0140 mol, so (0.0140 — 0.0109) mol = 0.0031 mol of CO; react.
Hence neg(CaCO3) = 0.0031 mol; neg(CaO) = (0.00892 — 0.0031) mol = 0.0058 mol.
(c)
The initial number of CO moles is n(CO>2) = (825/760) atm (4000 cm°)/ RT = 0.0493 mol. If the pressure P.g(COz) = 183 torr were reached, then
(0.0493 — 0.0109) mol = 0.0384 mol of CO would have reacted. But
only 0.00892 mol of CaO is present initially, so only 0.00892 mol of CO? react. The final amounts are thus n(CaO) = 0, n(CaCO3) = 0.00892 mol,
n(CO>) = (0.0493 — 0.00892) mol = 0.0404 mol.
L133
Eq. (11.4) holds: AG® =—RT In K° = 0 + 0 — 2(—95.299 kJ/mol) = 190.598
kd/MolaSO Keont= "alex 1Onee 11.34
The Lewis—Randall rule gives 9; = 6* (450°C, 300 bar). N2 + 3H2 pd 2NH3.
Let z mol of N> react to reach equilibrium. At equilibrium, ny, =1z,
My, = 332, and ny, =2z. [(3 — 32/4 —2z)P,
Pi=xP and Py =[(1z)/(422)]P,
P =
Pay, = [22/(4 — 2z)]P. The left side of (11.30) is
(4.6 x 10°°)[(0.91)7/(1.14)(1.09)*]"! = 8.2 x 10°. The right side is
(Pyu,/P)? 1(Py /P UP, IP’) = (PIP?) “(214
22)1G 232)6],—z) =
8.2 x 10°. Since P/P° = 300, we must solve 4z7(4 — 2z)7/3°(1 — z)* = 7.38. Taking the square root gives 2z(4 — 2z)/(1 — z)’ = 14.12, where the negative 156
sign in front of 14.12 is rejected, since z and 4 — 2z must be positive. Use of the quadratic formula to solve 18.1277 — 36.24z + 14.12 =0 gives z= 1.47 and
0.530. Since z must be less than  (to keep ny, > 0), we have z = 0.530 and
ny, = 0.470 mol, ny, = 1.41 mol, myy, = 106 mol.
11:35
N> + 3H — 2NH3. AGig = 2(6.49 kcal/mol) = 12.98 kcal/mol =—RT In K°. In K° = (12980 cal/mol)/(1.987 cal/molK)(700 K) = 9.332.
8.85 x 10°. The Lewis tule is 6; = *(T, P). For NH3, P, =
K°=
(500 atm)/(111.3 atm) = 4.49 and T, = (700 K)/(405.6 K) = 1.73. For No, P,; = 500/33.5 = 14.9 and T, = 700/126.2 = 5.55. For H2, P, = 500/(12.8 + 8) = 24.0 and T, = 700/(33.8 + 8) = 16.7. The Newton graphs of @ (the full reference is in
Sec. 10.11) give Oxy, = 0.86, On, = 1.26, and ,, = 1.16. The left side of Eq.
26)*]= 2.35 x 10~. Let 2w '(1.16) (11.30) equals (8.85 x 10°)/[(0.86)7(1. moles of NH3 decompose. The equilibrium amounts are n(NH3)/mol =  — 2w, n(N>)/mol = w, and n(H2)/mol = 3w.
mo/mol = 1 + 2w. Use OL B= 7,P cives
tor Ede (tie 30):
[((1—2w)/(1 + 2w)]? (507)? [w/(1 + 2w)][Bw/ (1+ 2w)]° (507)* 1631w* = (1 —2w)(1 + 2w)* = [(1 — 2w)(1 + 2w)]? = (1 — 4’). Let z= w. 2.35x10* =
Then 1631z? = (1 —4z)° = 1 — 8z + 162” and 1615z°+ 8z— 1 =0. We find z=
0.0225. Then w =z! = 0.150 and n(NH3) = 0.700 mol, n(N2) = 0.150 mol, and n(H>) = 0.450 mol.
11.36
In Example 11.1, we found Ip, = 1.96 x 10° mol/kg = m(H’). For this In, logio YHX) = 0.1(1.96 x 10*) and y(HX) = 1.00045, which when used in Eq. (11.14) makes no significant change in m(H"). In Example 11.2, J, = 0.200 mol/kg, so logi9 Y(HC2H302) = 0.0200 and y(HC2H302) = 1.047. Equation
(11.14) becomes
1.75x10~° mol/kg =
(0.746)? m(H,0* )(0.200 mol/kg) 1.047(0.100 mol/kg)
and m(H’) = 1.65 x 10° mol/kg, 5% larger than the answer in Example 11.2.
1137,
(a)
Assuming ideal vapor and taking the solid’s activities as 1, we have
K° = P(CO)/P°. Equation (11.32) gives [d In K°/0(1/T)]p = —AH°/R. We 157
plot In K° vs. 1/T. The data are:
In: Kom [OelG
BA8 5) 82237002 [Ogi 9.7
14 O932
0.046 8.57
Kk"!
y = 19988x+ 17.127 i eat:
0.00 0.50 l Bie at00. 1.50 2.00 2.50 3.00 3,5Q0
eee 8.89
OHHH
0.00085
tt
tt
0.00090 0.00095 0.00100
Cae

ee
Pes
ee
0.00110
The slope is [0.00 — (3.50)]/(0.000857 — 0.001031)K™' = ~2.01 x 10* K = AH°/R and AH® = 167 kJ/mol. At 1073 K, AG® = RT In K° = (8.314 J/molK)(1073 K) In (183/750) = 12.6 kJ/mol. Then AG° = AH® — T AS° and AS;5,3 = [(167000 — 12600) J/mol]/(1073 K) = 144 J/molK. (b)
11.38
From the extrapolated graph, we read In K° = 1.5 at 10°/T= 7.855, so K° = 4.5 and P(CO2) = 4.5P° = 4.5 bar.
Substitution of T/K = 298.15 into the equation gives log K°, = 13.998 and Ke
= 1.003 x 10°'*. We have AG° = RT In K° = —(8.3145 J/molK)(298.15 K) In (1.003 x 107!*) = 79.91 kJ/mol. (Cf. Prob. 10.50.) We have (0 In K°/0T)p = AH°/RT =
2.302585(0 log K°/0T)p. Differentiation of the K°, equation in the text (which is for P =  bar) gives (0 log K?./0T)p = (24746.26 K)/T° — 405.8639/2.302585T+ 0.48796/K — 0.000237 1(27/K”) = 0.03376 K~ at 25°C. Hence AH?® = 2.302585RT*(0.03376 K') = 57.45 kJ/mol at 25°C. AS593 = (AH® — AG°)/T = [(57450 — 79910) J/mol]/(298 K) =75.3 J/molK.
11.39 AV? =5.4 cm’/mol + 1.4 cm’/mol  (18.015 g/mol)/(0.997 g/cm?) = 22.1 cm*/mol. (0 In K°/OP)r = AV?/RT and J? dn K° = [ 1(AV°/RT) aP at 158
constant T. Neglecting the pressure dependence of AV°, we have In (K3/K; ) =
par/(100 x 107'*)] = ~(AV°/RT)(Pz — Pi) and In [Kop (22.1 cm?/mol)(83.14 cm?bar/molK) '(298 K) ‘(199 bar) = 0.178 = Ingk mate 32.236and Keay = U19,x 10m 11.40
(a)
As in Prob. 11.39, with the P dependence of AV° neglected, we have
—(AV°/RT)(P2 — P;) = 1n (K3/K, ) =1n 1.191 = 0.1748 and AV° = —0.1748(83.14 cm?bar/molK)(298.1 K)/(399 bar) = 10.9 cm*/mol.
(b)
'(298 K)'(P2 — P1) In 2 = 0.693 = (10.9 cm?/mol)(83.14 cm?bar/molK) and P> — P; = 1.58 kbar, so P2 = 1.58 kbar.
11.41
(a)
Since 1, = wx, we have H3 = H*,. Eq. (10.27) gives Hy; = 7. Hence AH° = v,H* , + Liza Vi; and substitution in (11.32) gives the desired result.
(b) (c)
K IK im=(Vxa%a)* Wien Voici) IV eae WADE GE ue yea (Ve iCi¥mitt:) =Miza Oa )’' =p, where b = Dien Vi
=
From (b), K,, =K,p4? and In K;, =In K? —bInp, + In (const), where const involves c° and m°. Hence (0In K,,/OT) p=(0In K7/0T) p—
b(0 In palOT)p = (0 In K2/0T) p+ boa = AH*/RT”, where we used the result of part (a) and used (0 In p/0T) p= (0/0T) p(Iinm—InV) =
~(1/V)(QV/0T) p=—0. Thus (0 In K2/0T)p = AH™/RT° — boy = AH IRT — Oa Died Vie
(d)
From (11.32) with the concentration scale used, AH?/RT? = AH™/RT° —
Oa Dea Vi = VaH® ,/RT + Liew (v;H7/RT? —v,0., ). Since AH; =
V,H* 4+ Lien V)Hi,, it follows that H2,= H —RT*a, fori# A. 11.42
(c).
11.43 AG” AG°= vai [2.,
+ v(H*)a[H* at a(H*) = 1077] 
py (H")] = VCH) {BHT at a) = 10']  1°(H")}. We ae Vikty + V(H" 159
have p; = W? + RTIn aj, so p[H" at a(H") = 107] = 1°(H*) + RT In 107 = °(H") — 16.118RT. AG” —AG° =16.118v(H‘)RT. 11.44
AG® = 2(4.83 kJ/mol) = 9.66 kJ/mol.
Q = (Pyy,/P°)?/(Py,/P° )(Py,/P’)” and
use of P; = x,P gives Q = [(1/7)3}°/{(2/7)3][(4/7)3]° = 0.0425. (AG/0E)rp = AG° + RT In Q = 9660 J/mol + (8.3145 J/molK)(500 K) In 0.0425 =
—3.47 kJ/mol. Since (0G/0€ )r.p < 0, the reaction proceeds to the right. 11.45
As found in Prob. 6.48, constant T and P addition of j will produce morej when x; > v,/A Vv where v; and A V7  have the same sign.
v; and A Vv  are
small integers and typical values of vj/A lv  are W2e 2/1 1/3, 2/5. etens ince the solution is dilute andj is a solute, we have xj O2(g), AG5og (kJ/mol) = 0 — 16.4 = 16.4, and the
equation in (a) gives In(K;m/P°) = AG°/RT = (16400 J/mol)/RT = 6.62: K;.,/P° = 747 and K;m = 747 bar. For CH4(aq) > CH,(g),
AG jog /(kJ/mol) = 50.72 + 34.33 =16.39; O61
11.47
(a)
0K
In(K;m/P°)
=AGYRT =
= 44ibar:
False. Intermolecular interactions between He and the gases in the
reaction may change the gases’ fugacity coefficients and so shift the equilibrium. (b)
True. See Fig. 4.9.
(Cymeralser See Pqr(lialo), (d)
False. AG° # AG in the reaction mixture.
(e)
(f)
False. The molalityscale standard state is a state in solutio n. False. See (e).
(g)
False.
160
Chapter 12 12.1
(a) T; (b) T.
12.2
(a) T; (b) F.
12.3
The contribution of sucrose to the vapor pressure can be neglected.
P =P, = a, P*. Since the solution is reasonably dilute, we can take aa = Xa and P=
x, Px. In 100 g of solution, there are (98.00 g)/(18.015 g/mol) = 5.440
mol of water and (2.00 g)/(342.3 g/mol) = 0.00584 mol of sucrose. Then x, = 0.99893 and P, = 0.99893(1074.6 torr) = 1073.4 torr.
12.4
Ves
(aye
(by,
(c) Fo (dele
(eel
eat
a(e)) &
Neu, = (0.226 g)(1 mol/72.15 g) = 0.003132 mol. mc,y,, = (0.003132 mol)/ (0.01645 kg) = 0.1904 mol/kg. kp= MyRT#7/AgsHma = (84.16 g/mol)(8.3145 J/molK)(279.62 K)*/(31.3 J/g)(84.16 g/mol) = 2.08 x 10* K g/mol = 20.8 K kg/mol. AT;=—kym = (20.8 K kg/mol) x (0.1904 mol/kg) = 3.96 K. Ty= 6.47°C — 3.96°C = 2.51°C. We assumed an ideally dilute solution and that only pure cyclohexane freezes out.
12.6
m = —ATy/ ky= (0.112 K)/(1.860 K kg/mol) = 0.0602 mol/kg = n/wa = n,/(0.0980 kg), so n; = 0.00590 mol of maltose. M; = w;/n; = (2.00 g)/(0.00590 mol) = 339 g/mol and M,,; = 339.
12.7
(a)
AT;= kpmp = —kpnp/Wa = —kywa/Mpwa and Mz = —kpw p/AT;wa.
(b)
For 100 g of solution, the 3% solution has wg = 3.00 g and wa =
97.00 g; Mg =(1.86 K kg/mol)(3.00 g)/(0.169 K)(97.0 g) = 0.3404 kg/mol = 340.4 g/mol. For the 6, 9, 12, and 15% solutions, we find Mg = 337.3, 334.5, 331.6, and 328.2 g/mol. Plotting Mg vs. wt. % maltose,
we get a nearly linear graph that extrapolates to Mg = 343 g/mol at 0 wt. os
161
B44
= 1,0033x + 343.43 ae Peer eae,
penne ene 
i
\
D
4
6
'
1
Mp
Om an]
(emt
AG
Wt. % maltose
12:8
1Bp = (UT* PAR) i ey pL Ade (1/T#)[1 — ATy/T# + (AT;/T#) +++ ].
12.9
(a) murea = [(1.00 g)/(60.06 g/mol)]/(0.200 kg) = 0.0833 mol/kg; kp=—ATjl Murea = (0.250 K)/(0.0833 mol/kg) = 3.00 K kg/mol. my = AT;/ky=(0.200 K)/(3.00 K kg/mol) = 0.0666; mol/kg = ny/(0.125 kg) and ny = 0.00833 mol. My = wy/ny = (1.50 g)/(0.00833 mol) = 180 g/mol. M,.y = 180. (bee
Anda
0, Rip ky = (200 g/mol)(1.987 cal/molK)(285 K)/
(3000 g K/mol) = 10.8 kcal/mol = 45.0 kJ/mol.
12.10
Let U denote CO(NH2)2.
AT;(A + Z) =kpm,,
AT;(A + U) = —kpmy; so 1.65
= AT;/(A + Z)/AT;(A + U) = m,/my = (nz/wa)(nu/wa) = nz/ny and nz = 1.65ny
= 1.65(0.679 g)/(60.05 g/mol) = 0.01866 mol. 29.0 g/mol;
Mz = (0.542 g)/(0.0186, mol) =
M,z= 29.0.
12.11 ky =M,RTEIA,,,Hp,4 = (18.0153 g/mol)(8.3145 J/molK) x (373.15 K)°/(40660 J/mol) = 512.9; K g/mol = 0.51295 K kg/mol. 12.12
We have 0.00313. mol of CjoHg and mg = (0.00313, mol)/(26.6 g) =
[el SaxulOwe mol/g. Hence k, = (0.455 K)/(1.18 x 107 mol/g) = 3860 K g/mol. Equation (12.19) gives AvapHm = (119.4 g/mol)(8.314 J/molK) x
(334.8 K)/(3860 K g/mol) = 28.8 kJ/mol = 6.89 kcal/mol. 162
12.13
(a)
We have Yaxa = aa and AT; = [R(T*)?/AtusE{m,a] In aa = (kp/Ma)(OM avi)
(b)
= —kpovm;.
In 96.0 g of water we have 4.00 g (or 0.02295 mol) of K,SO4. Hence
m; = (0.02295 mol)/(0.0960 kg) = 0.239 mol/kg. Then (0.950 K)/(1.86 K kg/mol)3(0.239 mol/kg) = 0.712. 12.14
= —AT;/kyvm; =
Equation (12.17) gives Dien Mm; = (2.37 K)/(14.1 K kg/mol) = 0.168 mol/kg; in
100 g of bromoform, we have 0.0168 total moles of solutes. Let P and P2 be phenol and its dimer. Then 2P z P>. The 2.58 g is 0.0274 mol of phenol. Let 2z mol of phenol react. At equilibrium, np/mol = 0.0274 — 2z, Np, =Z, and Mio = 0.0274 — z. Then 0.0168 = 0.0274 — z and z = 0.0106. So np = 0.0062 mol and np, = 0.0106 mol. The molalities are mp = (0.0062 mol)/(0.100 kg) = 0.062
mol/kg and mp, = 0.106 mol/kg. Thus Ky = (0.106 mol/kg)/(0.062 mol/kg)” =
27.6 kg/mol. 1215
Dien mj = AT;/ky = (0.70 K)/(5.1 K kg/mol) = 0.137 mol/kg = (Map + Man)/ (0.300 kg); SO Mnap + Nan = 0.0412 mol. We have 6.0 g = Mnap(128.2 g/mol) +
Nan(178.2 g/mol) = Mnap(128.2 g/mol) + (0.0412 mol — mpap)(178.2 g/mol). We get Mnap = 0.027 mol and nan = 0.014 mol.
12.16
(a)
In 100 g of solution, ng = (8.00 g)/(342.3 g/mol) = 0.02337 mol and mg = (0.02337 mol)/(0.092 kg) = 0.254 mol/kg. AT; =—(1.860 K kg/mol)(0.254 mol/kg) = 0.472s K.
(b)
In aq = (Afusflma/R)U/TF — W/Tp) = [(6007 J/mol)/(8.3145 J/molK)](1/273.15 — 1/272.665)/K = —0.00470 and a, = 0.9953;.
(c)
1ZA7
False.
12.18
(a)
100 g of solution has na = (92.00 g)/(18.0153 g/mol)
= 5.1068 mol of H>0, so xa = 5.1068/(5.1068 + 0.02337) = 0.99544 and VW, = Anixn = 0,99931/0.99544 = 0.99987. = (In aa)/Mam; = 0.00470/(0.018015 kg/mol)(0.254 mol/kg) = 1.027.
T=cRT= (0.282 mol/L)(1 L/10° cm’)(82.06 cm?atm/molK)(293.1 K) =O.) 5 alm.
163
(b)
12.19
Ve y,0 = (18.015 g/mol)/(0.998 g/cm*) = 18.05 cm*/mol. In ayo =—TIV* ,/RT ={(7.61 atm)(18.05 cm*/mol)}/ (82.06 cm?atm/molK)(293.1 K)] = 0.00571; ay,9 =0.99431. Xy,0 = 55.508/(55.508 + 0.300) = 0.99462. Yy, = 0.99431/0.99462 = 0.99969.
cg = IVRT = (6.1/760)atm/(82.06 cm°atm/molK)(273 K) = Boe x 10°’ mol/cm?. One cm’ of solution contains 3.58 x 10°’ mol of the
protein and contains 0.0200 g of protein; so Mg = (0.0200 g)/(3.53 x 107’ mol) = 56000 g/mol; M,.3 = 56000. This is approximate because the solution is not actually ideally dilute.
12.20
We have 5.55 moles of water and 0.00292 moles of sucrose. Equation (12.26) gives IT = [(82.06 cm>atm/molK)(298.1 K)/(18.07 cm’*/mol)] x (0.00292 mol)/(5.55 mol)] = 0.712 atm. We have II = pgh and h = II/pg =
x10’ erg a 0.712 atm oe ___ 8.134 =736cm (1.00 g/em*)(980.7 cm/s”) 82.06cm* atm 12.21
II = pgh. In the infinitedilution limit, p of the solution goes to p of water, and we Shall use p of water in the following calculations. For the first solution, II =
(0.996 g/cm*)(980.7 cm/s”)(2.18 cm) = 2129 erg/cm? and IT/pz = (2129 erg/cm*)/(0.00371 g/cm’) = 5.74 x 10° erg/g. The other solutions give Il/pg values of 6.29 x 10°, 7.18 x 107 and 8.10 x 10° erg/g. A plot of []/ps vs.
Pp is nearly linear and extrapolates to 4.54 x 10° erg/g at Op = 0. This intercept equals RT/Mg and Mg = (8.314 x 10’ erg/molK)(303.1 K)/(4.54 x 10° erg/g) =
5.55 x 10° g/mol; the number average molecular weight is 55500.
Il/Pg
8.50E+05 8. 00E+05
erg/g
7.50E+05
y = 31924x + 453459 I
7.00E+05 6.50E+05 6.00E+05 5.50E+05 5.00E+05 4.50E+05
 ! 
Ne
 !
OE 2  
eee 164
SPU
a ee ! aes 2
ee a
12.22
TI/pp ~ RT/My + A2RTPp + Az RTP3 = RT/Mp + ArRTPp + AZM ,RTP3/4 = [((RT/Mp)” + A>(MgRT)'”pp/2]° and taking the Square root of both sides gives the desired result.
12.23
(a)
Dies c; = IWRT = (7 atm)/(82.06 cm*atm/molK)(310 K) = 0.000275 mol/cm? = 0.275 mol/dm* = Orat
Hoe > Therefore Cg
=~
0.138 mol/dm? as compared with 0.15 mol/dm.. (b)
We shall approximate the concentration by the molalities. Then
Dies C; = [2(0.460) + 3(0.034) + 2(0.019) + 2(0.009)] mol/dm? = 1.078 mol/dm? = 0.001078 mol/cm*. Then I = (82.06 cm>atm/molK)(293 K)(0.001078 mol/cm*) = 26 atm. 12.24
We have V; = n;RT/P, so V; is proportional to n; and to x;. Then M, =
>; xM; and M,,, = (0.78)(28.01) + 0.21(32.00) + 0.01(39.95) = 29.0.
12.25
12.26
I = (RT/V* ,)In ag = (RT/VX 4(Mavi) = (ORT/V* ,MavniinaMa = oRTvnina Vi 4. (a)
a1 =Mar. Assuming ideally dilute solutions, we have px (Pi, T) +
RT In xa, = pe(Pit U1, 7) + RT In x2 and pi (P; + 11, TF)  pe(Pi, 1) = RT \n xa — RT In xa2 = —RTxp,) + RTxp.2, where (12.11) was used. From
(12.22), w*(Pi + TI, T) — wi (Pi, T) = Vi, U1, where the pressure
dependence of V* , is neglected. Hence Vi,,I] = RT(xp,2 — xB,1).
(b)
xp2=0.100/(0.100 + 55.51) = 0.00180 and xp, = 0.0200/(0.0200 + 55.51) = 0.000360. IT= (82.06 cm?atm/molK)(298.1 K)(0.00144)/(18.07 cm?/mol) = 1.95 atm.
sp A) At equilibrium, the equality sign holds in (4.12) and —P™ dV™ — P® av’ =
dA* + dA® = P* dV" +5; we dns  PP dV + Yi wh dn?, so Di wi dni* + gli dn® = 0. Let dn moles of substance jmove from phase « to phase B.
Then dn‘ = dn and dn‘ =—dn, we have FL
165
ne dn =O
and pi = ee
12.28 f=Ccina—p+2=2p+2=4pandp=4f. The minimum fis 0, so the maximum p is 4 in a binary system.
12.29
(a)
Yes. The horizontal tie line at 7; intersects the boundanes of the twophase region at ven = 0.23 (point L) and at xz = 0.68 (point Q).
(b)
No. The overall mole fraction is different for different points on the tie line.
12.30
As noted in Sec. 12.6, the upper curve is the Pvs. ie curve and the lower
curve is the Pvs.x, curve. We take B as acetone. Table 10.1 gives P and the ce and x, values that correspond to each P value. Plotting these points, we get a phase diagram with a minimum in P at Xacetone = 0.4.
12.31
The width between the vertical axes is 602
mm and xg = 0.3 lies at 18 mm.
We draw a vertical line at 18 mm, and then a horizontal line that goes from the liquid line at 18 mm to the vapor line; the intersection with the vapor line is at 452 mm (corresponding to xg = 0.75) and gives the result of the first distillation step; drawing a horizontal line from the liquid line at 454 mm, we find it intersects the vapor line at 58/2 mm, corresponding to xg = 0.97.
12.32
12.33
(a)
The width of the figure is 60 mm and x = 0.72 corresponds to 43.5 mm. Drawing a vertical line at 43 mm and then a horizontal line from the intersection of this vertical line with the liquid line, we find the horizontal line intersects the vapor line at 52’ mm, which corresponds to Xp = 2/2/00 =0:895.
(b)
in is given by point D as (36 mm)/(60 mm) = 0.60. Point G gives xp = (48% mm)/(60 mm) = 0.81.
The point giving the system’s state lies on the horizontal line at 7, and cannot lie to the left of point Q, since all points to the left of Q have some liquid present. Point Q corresponds to xp = (41 mm)/(60% mm) = 0.68, so xg = np/(ng + Nc) = 0.68; ng > 0.68np + 0.68nc; nc < 0.47n, = 0.47(2.0 0 mol) =
0.94 mol.
nc /(g atm) and v =
(67 cm*/g)(P/P*)/(1 + 0.38P/P*). With P/P* = 7.0, one gets v = 128 cm’/g. For the Freundlich isotherm, v = (72 cm?/g)(7)°7? alg cm*/g.
13.25
(a)
v=rIn(sP*)+rIn (P/P*), where P* =  atm. A plot of v vs. In (P/P*) is linear with slope r and intercept r In (sP*).
(b)
The data are
v/(cm*/g) In (P/P*)
101 “136 153 M62 "165. 166 1.253 2.303 2.815 3.246 3.512 3.669
As with the Freundlich isotherm in Prob. 13.24b, the points at high pressures show a strong curvature, indicating that the Temkin isotherm doesn’t apply at high P. (Note that it predicts v — % as P 4 .) We ignore the three highpressure points and draw a straight line through the remaining points. The slope is 33.4 cm*/g = r. The intercept 1s 59 cm*/g = (33.4 cm’/g) In (sP*) and s = 5.85 atm”.
vi(cm/g)
180 160 140 120 100 80 60 40
y = 33.30x + 59.29
ee Le el i Pra
SSS]
Ove
Se
SSS
O55
li
See
21
SSeS
5g)
eee
See
eee
Coes
ee
Soe
In (P/P*) 13.26
We have  + bP =cP/v, so a plot of P/v vs. P is linear.
13.27
(a)
The curve resembles Fig. 13.17b, indicating formation of more than a
monolayer. We have a type II isotherm and the BET equation is appropriate.
184
v/(cm’/g)
E \
0
1
Ua
ey
100
200
ao
0.0
t
i]
Af
ay es
300
400
1
A
eH
500
1
‘
600
700
P/torr
(b)
We plot P/v(P* — P) vs. P/P*, where (since 77 K is the normal boiling
point) P* = 760 torr. The data are
24969. 98 e318
10°[P/v(P* — P)\/(g/cm’)
0.07 winl 64am
10°P/P*
7 Agel
10°[P/v(P* — P\/(g/cm’) 10°P/P*
mele AS 6a S i 27 Ome Se) aeOSES
25
el One
293
atl
OnLO2L6 Oa
The low and mediumpressure points fit a straight line well, but the three points at high pressure deviate greatly from this line (the BET isotherm often works poorly at high pressures) and will be ignored. We draw a straight line of slope m and intercept b through the first six points. We find m=
exhaveuii=aol/O mon =! Omen SO) Donat iter by! = 0.842 cm’/g: also. c =
1.17 g/cm® and
ANC i= ek)Den
b=0.01, g/cm’. W
+0.0166 y = 1.1661x
(1 = mUmon) = 7 x 10'. OMe
;
+
LS.46
ee eae
eee
eon
2
1.4 [e2, 1 0.8 0.6 0.4 On2
0
0.4
0.2 185
0.6
0.8
P/P*
(c)
n=(l atm)(0.842 cm*V/RT= 3.76 x 10° moles of N> adsorbed per gram of sample. This is 2.26 x 10!” molecules, so the surface area of the
powder is (2.26 x 10!°)(16 x 107'° cm’) = 36000 cm? = 3.6 m’. 13.28
For 6 >  and 1/bP + 1 >> 1, so that
1/bP >> 0 and hence 1/bP >>  and bP /R)(A/T, 1/T,), where < AH, > is an average AH_, over the temperature range.
In (0.03/0.0007) =
[/(8.314 J/molK)][(873 K)  (773 K)'] and = —210 kJ/mol.
13.31
(b)
For @=0.10 and 500 to 600°C, = R[(873 K)' — (773 K)'}! x In (23/8) = —59 kJ/mol. For 600 to 700°C, = R((973 K)! — (873 K) ‘J! In (50/23) = —55 kJ/mol.
(a)
Multiplication of (13.40) by vc gives Pc/(P* — P) = (0/Vmon) X [1+ (c= I)P/P*] and )/Umon = PP*c/(P* = P)(P* + Pe — PY.
(b)
For P Cr gives (3) Cr’* + 3e7 > Cr. Hence AG; = AG; + AG;, which becomes —n3FF; =
n\ FE, — n2FF3; 80 &, = (mE, + nF; yng = [1(0.424 V) + 2(0.90 V)}/3 = —0.74 V.
14.32
£=£ —(RTInF) In [a(Zn**)/a(Cu**)]. The ZnSO, solution has Ip/m° = +[2°(0.002) + 27(0.002)] = 0.00800 and the Davies equation gives
log y(Zn**) = 0.163; y(Zn?*) = 0.68g. Also, In/m® = 0.00400 for CuSO,(aq); log y(Cu**) = 0.119; y(Cu?*) = 0.76. Use of ami = Yumi/m® gives a(Zn**) = 0.683(0.00200) = 0.00138 and a(Cu2*) = 0.00076. So £= 0.339 V — (0.762 V) — (RT/2F) In (0.00138/0.00076p) = 1.093 V. 14.33
(a) PtlAg AgCl(c) KCK(aq) HgsClo(c) Hg Pv. 194
(b) PtH»H2SO.(aq) Hg2SO.(c) Hg Pt’ (or we can use a Pb PbSO, halfcell).
14.34 Pt’H>HCl(aq) Clo Pt,
Pt HHCl(aq) AgCl(c) Ag Pr’, Pt Hz HCl(aq) Hg2Clo(c) Hg Pr’, Pt Ag AgCl(c) HCl(aq) Hg2Cla(c) Hg Pr’. 14.35
(a)
The halfreactions are Ag > Ag*(0.01m°) +e and Ag’(0.05m°)
+e >
Ag. The cell reaction is Ag*(0.05m°) > Ag*(0.01m°). We have & =
F,, —&, = 0. Then £= 0  (RT/F) In [y+,(0.01)/y+,.x(0.05)] = (0.02569 V){—1.6094 + In (y+.1/Y+.2)] = 0.04135 V — (0.02569 V) In (yx.1/Y+.r). We have Im, = 0.0100 mol/kg and In.r =
0.0500 mol/kg. The Davies equation for Ag” gives log y., = —0.0448 and Ys. = 0.902; also, log Y+,.2 = 0.08555 and y,.x = 0.821. Hence
f= 0.04135 V — (0.02569 V) In (0.902/0.821) = 0.0389 V.
14.36
(b)
£=0r— 6, >0, SO Or > Or.
(c)
The electrons flow from low to high potential and hence flow into the right terminal.
£=[(8.314 J/molK)(358.1 K)/2(96485 C/mol)] In (2521/666) = —0.0205 V, where Eq. (14.60) was used.
14.37
(a) T; (b) F.
14.38
(a)
[Ag+Cl — AgCl(s) +e] x2
Cl He Gl5(s)2en—>2Heme ) +2Hg > 2AgCl(s) 2Ag+Hg,Cl,(s (b)
(RT 2F) In [a(AgCl)a’(Hg)/a*(Ag)a(Hg2Cls)] =e posites (date activities of the solids can be taken as 1. Sof=/ =F, F, =
G=27
0.2680 V — 0.2222 V = 0.0458 V. = 0.0458, V.
(Cc)
=e
(d)
Since £=£/, we have dAOT = 0F/0T. So AS° = nF(0Z°/0T)p = 2(96485 C/mol)(0.000338 V/K) = 65.2 J/molK. AG® = Ske = 195
~2(96458 C/mol)(0.0458 V) =—8840 J/mol.
AH° = AG° + T AS® =
—8840 J/mol + (298.1 K)(65.2 J/molK) = 10.6 kJ/mol.
14.39 F =F, F, =0(0.01 V)=0.01 V. AG? =nFF° =RT In K° and In K° = nFPIRT = 2(96485 C/mol)(0.01 V)/(8.314 J/molK)(298.1 K) = 0.8. K° =2. 14.40
(a)
For the reaction 2Na* + H) > 2Na + 2H’, we have Fy, =fp —F) =
~2.714 V —0 =2.714 V. Then AG‘o, = —nFF = 2(96485 C/mol) x (2.714 V) = 5.237 x 10° J/mol = 2(0) + 2(0) 2 A Gog (Na*) — 2(0) and A ;G3og(Na*) = 2.619 x 10° J/mol = 261.9 kJ/mol. (b)
For the reaction 2Cl + 2H? > Cl, + Hp, we have £59. = 0 — (1.360 V) = —1.360 V. Then AGj, = —2(96485 C/mol)(1.360 V) = 2.624 x 10° J/mol = 0+ 0 2 A ¢Gy9g(Cl) — 2(0) and A Gyog (Cl) =
1.312 x 10° J/mol = 131.2 kJ/mol. (c)
For the reaction Cu** + H) > Cu + 2H*, £59, = 0.339 V 0 = 0.339 V and AG5og = 2(96485 C/mol)(0.339 V) = 6.54 x 10* J/mol = 0 + 2(0) — A Gyog (Cu’*)  0 and A ¢ Grog (Cu**) = 6.54 x 10* J/mol = 65.4 kJ/mol.
14.41
2e° + PblI(c) > Pb + 20(aq) Pb — Pb** (aq) + 2e° PbI,(c) — Pb** (aq) + 21 (aq) P= 0.365 V— (01126 V) = 0,239.V. eAGl == Ee — 2(96485 C/mol)(0.239 V) = 4.61 x 10° J/mol. In K},=AG°/RT =
~(46100 J/mol)/(8.314 J/molK)(298.1 K) = 18.6 and K°, =8 x 10°. 14.42 (a) £ = 1.360 V 1.078 V =0.282 V. AG® = 2(96485 C/mol)(0.282 V) = ~54400 J/mol. In K° =AG°/RT = (54400 J/mol)/(8.314 J/molK)(298.1 K) = 21.95 and K° = 3 x 10°.
196
(b)
£=0.282 V,n= 1, and AG? = 27200 J/mol.
In K° = 10.93 and K° =
6 x 10°.
(c)
The halfreactions are 2(Ag + Cl! > AgCl +e) and Cl; + 2e — 2CI. F = 1.360 V — 0.222 V = 1.138 V.
2.196 x 10° J/mol. (d)
In K° = 88.59 and K° = 3 x 10”.
This is the reverse of (c), so AG® = 2.196 x 10° J/mol and K° =
Gx10 (e)
AG®° =2(96485 C/mol)(1.138 V) =
jot= 350108 
The halfreactions are 2(Fe** Pe hee
2 eV P=20 440 N = 007 1aV=—1
e ) and Pew
2e => Fe,
AG a2 557 X 10° J/mol,
In K° =94.2, and K°=1x 10". 14.43
14.44
PIV = 0.23646 — 5.1144 x 10“(t/°C) —2.0628 x 10°((t/°C)> — 1.0808 x 10%(#/°C)’.
Cell 1 is Pt Fe?*(aq), Fe**(aq) ::Fe**(aq) Fe Pt’.
Cell 2 is Pt Fe2*(aq), Fe**(aq) ::Fe**(aq) Fe Pr’. Cell 3 is Fe Fe**(aq) ::Fe**(aq) Fe. All have the same cell reaction. For
example, Cell 3 has the halfreactions Fe’* + 2e” > Fe and Fe > Fe™* + 3e, and multiplication of these halfreactions by 3 and 2, respectively, gives the overall reaction as 3Fe** > Fe + 2Fe**. For Cell 1, =0.44 V0.77 V= 1.21V, nf =2(1.21 V)=2.42V, AG° =nF = 233 kJ/mol. For Cell IF = 0,04 V 30.77 V =—0.8L V5 nee = 3(0. 81) 2.43 V,, AG = _nFP = 234 kJ/mol. For Cell 3, & =0.44 V — (0.04 V) =0.40 V, nf = 6(0.40 V)=2.40 V;, AG® = 232 kJ/mol. AG° must be the same in all three ous cases, since all the cells have the same reaction. (Because of the spontane reaction 2Fe**(aq) + Fe(s) > 3Fe**(aq), an Fe** Fe electrode is not
reproducible.)
14.45
(a)
The halfreactions are Fe > Fe?* + 2e” and 2(Fe** + e° — Fe”*). The cell
reaction is Fe + 2Fe** > 3Fe”*. (b)
AG° =nFF = 2(96485 C/mol)[0.771 V — (0.440 V)] = ~2.33, x 10° J/mol. AS° = nF(0E°/0T)p = 2(96485 C/mol) x (0.00114 V/K) = 220 J/molK. AH® = AG? PAS es ~2.33, x 10° J/mol + (298.1 K)(220 J/molK) = —1.68 x 10° J/mol.
197
14.46
Equations (14.67) and (14.68) give at 10°C:
= 0.23643 V —
(4.8621 x 10 V/K)(10 K) — (3.4205 x 10° V/K*)(10 K)* + (5.869 x 10° V/K*)(10 K)° = 0.23123 V. Then AG? = —2(96485 C/mol)(0.23123 V) = 4.4621 x 10° J/mol. Equation (14.69) gives at
10°C: AS° = 2(96485 C/mol)[4.8621 x 10~ V/K + 2(3.4205 x 10° V/K’)(10 K) +3(5.869 x 10°” V/K*)(10 K)’] = ~106.69 J/molK. Then AH® = AG° + T AS° = 4.4621 x 10* J/mol + (283.15 K)(106.69 J/molK) = 7.4830 x 10* J/mol. From (14.66) and (14.67), AC; = 2FT[2c + 6d(T— To)] = 2(96485 C/mol) x (283.15 K)[2(—3.4205 x 10° V/K’) + 6(5.869 x 107° V/K?)(10 K)] = ~354.55 J/molK. 14.47
For. the cell Ag Ag:l Agl(c) Ag, we have the halfreactions
Ag — Ag’ +e
and Agl(c)
+e — Ag +T; the cell reaction is
Agl(c) > Ag* +I. Equation (14.63) gives & = RT In K,, /nF =
(8.314 J/molK)(298.1 K) In (8.2 x 10°'’)/(96485 C/mol) = 0.951 V =
Fe ods 14.48
Gio as Gr
0. OONV aHenceA t=. 0.152, V.
The halfreactions are Hy
> 2H* + 2e and 2(AgBr +e — Ag +Br). The cell
reaction is H2 + 2AgBr hae
yee
2Ag.
4 =0.073 V. Equation (14.46)
gives £=£ — (RT/2F) In {a(H‘)a(Br )*/Maddala cea he £ —(RTI/F) In [a(H*)a(Br)] = & = (RT/F) In (y+mj/m°)’. Hence 0.200 V = 0.073 V — [2(8.314 J/molK)(298.15 K)/(96485 C/mol)] In (0.100y+). We get
In (0.100y+) = 2.47 and ys = 0.84. 14.49
As in Prob. 14.40b, we find A,G°(Cl ) =~131.2 kJ/mol. Since u°(HCl) = .°(HCI) = n°(H") + p°(CI) [see Eq. (10.44)] and A ,G°(H") = 0, we have
A ,G°THCl(aq)] = 131.2 kJ/mol.
14.50
(a)
The halfreactions are Hy
> 2H* + 2e and 2[AgCl(c) +e > Ag+Ctr]J.
The cell reaction is Hy + 2AgCl(c)
+ 2H* + 2Cl + 2Ag. The solids’
activities can be taken as 1, and £= £ — (RT/2F) In {{a(H*)]*[a(Cl]*} =
& — (RT/F) In [a(H")a(Cl)]. Since we are writing H* rather than H30*,
198
we write the water ionization as H2O — H*+OH. Then Ki= a(H*)a(OH_)/a(H20) and a(H*) = K°.a(H,0)/a(OH_). So a(H")a(CI) =
)m(Cl )/y(OH )m(OH_ ). K°a(H,O)a(Cl )/a(OH ) = Kj,a(H,0)y(Cl Substitution in the above equation for £ gives the desired result.
(b)
As Im — 0, the y’s go to 1 and a(H20O) goes to 1. Hence
£— F —(RTI/F) In [K,m(Cl )/m(OH_
)J.
Soln K\, =(FIRT){& — £(RT/F) In [m(Cl )/m(OH )}* = [(96485 C/mol)/(8.314 J/molK)(298.15 K)](—0.8279 V) = —32.22s and
Ken 14.51
(a)
1.010105
The halfreactions are the same as in Prob. 14.50a and the cell reaction is
H> + 2AgCl(c) £
> 2H’ + 2Cl + 2Ag. As in Prob. 14.50a,
)]. For the ionization HX Lies xX ,we =F —(RT/F) \n [a(H')a(CYI
have K; = a(H")a(X )/a(HX) and a(H') = K,a(HX)/a(X). Hence
a(H*)a(Cl) = K; a(HX)a(Cl lax) = K, y(HX)m(HX)y(Cl )m(CY
yx
mC
ym”. Substitution in the above
equation for £ gives the desired result.
(b)
As Im — 0, the y’s — 1 and we get
InK? =(F/RT){& — &(RTI/F) in [(m(Cl )m(HX)/m(X )m°}}* = [(96485 C/mol)/(8.314 J/molK)(298.15 K)](—0.2814 V) =10.953 and
K? =1.75x 10°.
14.52
Sn+Pb2* © Sn2*+Pb. & =0.126
V+0.141
V=0.015 V. InK°=
nFEIRT= 2(96485 C/mol)(0.015 V)/(8.314 J/molK)(298 K) = 1.17 and ) = K° = 3.>. The solids’ activities can be taken as 1, so 3.5 = a(Sn**)/a(Pb** m(Sn**)/m(Pb**) = 2/(0.1 —z). We find z = 0.076. Hence m(Sn?*) = so 0.076 mol/kg and m(Pb**) = (0.024 mol/kg. The solution is reasonably dilute,
we expect the activity coefficients to be determined mainly by the ionic strength. Hence y(Pb™*) = y(Sn**) and the activity coefficients cancel in the expressions for K°.
199
14.53
4, =612 mV and 4; = 741 mV. Equation (14.73) gives pH(X) = 6.86 + [(612 — 741)10° V](96485 C/mol)/(8.314 J/molK)(298.1 K)2.3026 = 4.68.
14.54
In Eq. (14.78), we take ve = es , since the ionic strengths of the two solutions are equal. Then oP — o* = [(8.314 J/molK)(298 K)/(96485 C/mol)] x In (0.100/0.150) = 0.0104 V, where B is the NaNO3KNO; solution.
14.55 w= Sdand § = pd = (3.57 x 10° C m)(1.30 x 107° m) = 2.75 x 10°" C. Sle = (2.75 x 10°7° CV(1.60 x 10°'’ C) = 0.172.
14.56 (a)
>, O,x,
=(0.5e)(1.5 A) + (1.5e)(1.0 A) = 0.75e A =
~0.75(1.6 x 10!° C)(10'° m) = 1.2 x 10°? C m. The direction is in the negative x direction and the magnitude is 1.2 x 10° Cnn:
(b) x= D; Ox, =(e)(l A) = e A. py= X, Oy; IN,
=(e)(1 A) =
w= (2 +5)? =2'*¢ A=2.3 x10? Cm. The vector makes
an angle of 45° with the x and y axes and points from the negative charges’ region toward the positive charge.
(c) 14.57
; O,x; =(0.5e)(2.5 A) + (2e)(1.0 A) = 0.75e A.
Let a, b, and c be the x, y, and z coordinates of the new origin in the original coordinate system. If x; 1s the x coordinate of charge i in the original system,
then p, in the new coordinate system equals >’; Q;(x; —a) =; Q;x; 
ad; O; =X; O;x,; a0 =; Q,x;, whichis the same as in the original coordinate system. Similarly for py and p,. 14.58
(a)
Wro Wry = (1 — m/e (r, = ry ny (7,a0
(b)
= (mM — r2)(1) + 12)/ryr2(r7) + 12) =
)e
Forr>>d, we have r) = rz =r, and I/ry — Wr, = (7? —r2)/2r°. The law of cosines for triangle PAC gives ca = ee + d° — 2rd cos 0. (Because
r >> d, angle PAC is very nearly equal to angle PBC.) Then im a ie =
2rid cos 0  d’ = d(2r, cos 0 — d) = 2rd cos 0, since d
6.022 10”* mol!

—3(8.854107'? C?/Nm’)
and o = 1.25 x 10°? cm?C’/Nm? = 1.25 x 10°? C*m/N.
a/4n€p =
(1.25 x 107°? C?m/N)/41(8.854 x 1071? C?/Nm?) = 1.12 x 10°? m3 = 11.2 A?. 14.61
(a)
w=0 for CHy. Also, PV = (m/M)RT and M/p = RT/P. Equation (14.87)
gives
0.00094 atm/molK)(273.1K) (82.06 x 10% m? _= 3.001 1.00 atm 6.022107? mol”! 3(8.854x107!? C?/Nm’)
a = 3.10 x 107° C? N! m. Also, o/4it€9 = (3.10 x 10° C?m/NY 4n(8.854 x 107'? C?/Nm’) = 2.78 x 10°? m? = 2.78 A’. (b)
€+2=3.
M/p=RT/P.
w=0.
Equation (14.87) gives
€, — 1 = PN4O/RT€o = 4NPN4(O/4t€9)/RT = 4n(10.0 atm)(6.022 x 107° mol)(2.78 x 107° m’)/ (82.06 x 10° m?atm/molK)(373.1 K) = 0.0068; and €, = 1.00687. 14.62
Equation (14.87) applies. We plot Mp '(€, — 1)/(€, + 2) vs. 1/T. Noting that
M/p = RT/P for a gas, we have
201
10°Mp'(e, — 1)/(e, + 2) 10°/(T/K)
5.73 2602)
5.35 5.00; ) 2:380NN2049eew?
4.67 0664
4.31 LOIG
[10°Mp '(e, — 1)/(e, + 2)\/(m?/mol)
2077.7x + age 0.3535 Ue. slahmeiieeaialss y = Sacha
0, ee Se 5D) 5.0 4.8 4.6 4.4 4.2 0.0019)
SESS
50,0021,
Ssh
Sse
5.0.0023
eS Soave
0.00259"
sean
TEiKe
10.0027
where the units of the first line are m*/mol. The plot is reasonably linear with
slope 0.0208 m?K/mol = Nayt/9€ok. So p? = (0.0208 m?K/mol)9(8.854 x 107!? C?/Nm7)(1.381 x 10°77? J/K)/ (6.022 x 1073 mol!) = 3.80 x 10°? C? m? and u = 6.17 x 10°° C m. The intercept 1s 3.5 x 10° m*/mol = N,O/3€po, and
oe 3(8.854x10?C*/Nm7)(3.5x10°° m?/mol) 6.022107 mol! ot = 1.5 x 10° C*m/N [Using Eqs. (20.2) and (20.3), one finds that the dipole moment is 1.85 D.
Also, one find o/4m€ = 1.4 x 10°? m3 = 1.4 A?] 14.63
(a) Molecular,C m;
14.64
(a)
CSe, because of a greater a;
(b)
nCi9H22;
(c)
odichlorobenzene, because of a greater w.
(a)
Let f=(e, 1), +2). dffde, = Me, + 2) —(€,— 1)(e, [ep +2) = (e, — Dent 2)? = 3/(e, #2)" = 0.
14.65
(b) molecular,
202
C m? V7;
(c) macroscopic, no units.
+2) =
(b)
We have 1 < €, 0 for all €,, the minimum value offis
at the minimum value of €,; at €, = 1, fmin = 0. As €, 4 00, f approaches
its maximum possible value, which is 1.
14.66
ee
Then
oi 24 o* =
0.57440, 0.02569 V in P,P14100 ++ 0.0.0044P,,4649 0+ + 0.5P; 540
Let P(K"*) =
= (0.02569 V) In (48.4/682) = 0.068 V =68 mV
14.67
The halfreactions are Ag + Cl'(0.0100m°)
— AgCl+e and AgCl+e —
Ag + Cl (0.100m°). Equation (14.51) gives £= 4; + 0 (RT/F) In (0.100y_.r/0.0100y_7).
JIm.r = 0.100 mol/kg and the Davies equation
for Cl’ gives log y_.r = —0.1072 and y_z = 0.781. Similarly, y_., = 0.902. Then £
= 0.038 V — [(8.314 J/molK)(298.15 K)/(96485 C/mol)] In 8.66 = 0.093 V.
14.68
Im = 0.100 mol/kg. The Davies equation for H” gives log y, = —0.1072 and Y, = 0.781. Then a(H’) = (0.781)(0.100) = 0.0781 and —log a(H") = 1.107.
14.69
K°/K°, = {[a(Ag*)]°/a(Cu*)}/{ fam(Ag*)}/am(Cu’*)} = (0.997)7/0.997 = 0.997),
EF.
AG? =—RI In Kh =—nFF and
= (RI/nE) in K
so
=(RTnF\(In Kj, In K?)=(RT/nF) In K2/K>, =
—(RT/2F) n 0.997 = 0.000039 V = 0.039 mV, which is insignificant.
14.70
(a) C; (b) m;
(c) N/C=V/m;
(g) dimensionless;
14.71
(d) V; (e) V; (f) Cm;
(h) J/mol.
If the left and right electrodes are at different temperatures, the emf is nonzero.
203
Chapter I5 15. 1 (a)yF i
(bars
15.2
(a) T;
(b) F;
15:33)
(a)
ee
(c) T;
(d) T.
3nRT = 1.5(1.00 mol)(8.314 J/molK)(298 K) = 3720 J = 889 cal.
(b) 3720J. (c) (0.470 g)/(16.0 g/mol) = 0.0293 mol and 3nRT= 109 J.
15.4
(a)
(€,,) = 3kT = 1.5(1.381 x 10 J/K)(5S71.1 K) = 1.18 x 107°J.
G3) Sa 15.5
(€,,(T))/(€y(T;)) = (BKT2/2)/(3KT)/2) = T2/T; = (373.1 K)/(273.1 K) = 1.366.
15.6
Uims = (3RT/M)"”, SO Urms(Ne)/ Umms(He) = (Mrte/Mne)'”” = (4.0026/20.179)!? = 0.4454.
15.7
Ums =GRTy,/My,)” = (BRT, /Mo, )'” and Ty, = (My,/Mo, )To, = (2.016/32.00)(293.1 K) = 18.5 K.
15.8 0Ey = 3nRT= 3 PV = 1.5(1.00 atm)(90 x 10° cm*)(8.314 J)/(82.06 cm? atm) = 1.37 x 10’ J. The answer is the same for 40°C.
15.9
(a) Oto;
(b) coto : (ec) 0 (Fig. 15.5);
whose speed is in the range from vto v+ dv. 15.10
(a) T;
(b) F; (c) T;
(d) F.
204
(d) the number of molecules
15.11
(a)
=m/k=Nam/Nak = M/R. The interval is small enough to be considered
“infinitesimal” and Eq. (15.44) gives dN» =
4m.N (M/2MRT )* eM" 28T1)?dy = 0.0320 kg/mol ue 4n(6.02 x 107°  27(8.314 J/molK)(300 K) _ (0.0320 kg/mol)(500 m/s)? [som/s)* (0.001 m/s) =1.1«10!8 2(8.314 J/molK)(300 K) (b)
Considering the interval as infinitesimal and using (15.42) with x
replaced by z, we have dN,
= N(M/2nRT)'exp (—Mv2 eRI a0s—
(6.02 x 107*)[0.0320 kg/mol/2n(8.314 J/molK)(300 K)]'” x exp[—(0.0320 kg/mol)(150 m/s)*/2(8.314 J/molK)(300 K)] (0.001 m/s) = TAS 10. (c)
The fraction of molecules with x and z velocity components simultaneously in the ranges v, to v, + dv, and UV, to VD, + dU; 1s g(0,)2(V,) dv, dv, = (dN, /N )(dN,, /N) and the number of such
molecules is (dN,, /N)(dN, /N)N= dN, dN,_/N. From part (b), dN, 
= dN, =7.45x10'’ and dN, dN, /N = (7.45 x 10'7)/(6.022 x 10”) =
9.22 x 10''. 15.12 This probability equals dN/N = 4n(M/2nRT)*7 0 MY?87 9?dy= 4n[(0.0160 kg/mol)/27(8.314 J/molK)(300 K)}*” x exp[—(0.0160 kg/mol)(400 m/s)*/2(8.314 J/molK)(300 K)] x
(400 m/s)*(0.001 m/s) = 1.24 x 10°. 15.13
(a)
(b)
15.147
(a)
Fig. 15.6 gives G(v) = 0.0018; s/m at v= 500 m/s, so dNJN = G(v) dv = (0.0018; s/m)(0.0002 m/s) = 3.7 x 107’ and dN, = (6.02 x 107°)(3.7 x 10°’) = 2.2 x 10"” molecules. At 1000 K, Fig. 15.6 gives G(v) = 0.0008 s/m; dN,/N = (0.0008 s/m)(0.0002 m/s) = 1.6 x 10°77; dNy=1x 10"". Pr
ie G(v) dv= G(U;) re dv= G(v;)(U2 — V;) if G(v) is essentially
constant in the interval.
205
(b)
As in the example, M/2RT = 3.52 x 10° s*/m*. At v= 90.000 m/s, G(v) = exp[(90 m/s)°(3.52 x 10 s*/m’)] x "(3.52 x 10° s2/m*)*742(90 m/s)” = 1.17328 x 10~ s/m. Replacing 90 by 90.002, we find that at 90.002 m/s, G(v) = 1.17333 x 107 s/m, and
the change is 0.004%. 15.15
(a)
Equation (15.44) gives (dN,,, IN)I(dN ,,IN
y2eMHAT py? gM UAT — (42 /y?)exp[M (v7 —v3)/2RT ] = 5007 hs (0.0320 kg/mol)(15007 — 5007 )m7/s7 = 45000 2(8.314 J/molK)(298.1K) 15002.
(b)
From (15.44), (m/2nkT)’exp (mv{/2kT) 410; = (m/2nkT)*exp (—mv3/2kT) 4.05, and exp[m(v3 — v7;)/2kT] = 05/0;. Taking logs, we have m(v3 — v;)/2kT = 2 In (v2/v,) and T= M (v3 —v7)/[4R In (v2/01)] = (0.0320 kg/mol)(1500* — 5007)(m/s)*/ 4(8.314 J/molK) In 3.00 = 1750 K.
15.16
(fiw) + foOv)) = Tm [fidw) +.Aw)lgQv) dw = J" fidw)gQv) dw + es fw)g(w) dw = (fiw) + f,(w)). (b)
15217
(=
le = cf(w) dw = arhe fdw= c(f).
The fraction of molecules with x and y velocity components simultaneously in
the ranges V, to v, + dv, and Vy to Vy + dvy is g(Vx)g(Vy) dv, dvy = (m/2TtkT) ex
p
(—mv°/2kT) dv, dvy,5 where v” = v2x +02. y Molecules with speeds p
lying between v and v + dv have their velocity vectors lying within a thin annulus (ring) of inner radius v and outer radius v + dv. The area of this annulus is 1(v + dv)” — mv = 2mv dv, where the n(dv)’ term is negligible. The probability that v lies in this annulus is the following sum over the annulus:
Y (m/2nkT eT
dy, dv, = (ml 2nkT ye”(2a dv) =
(m/kT )e™”"*") dy, and this is the desired probability.
206
15.18
(a)
Since the integrand has the same value at —x as at +x, the areas on each side of the origin are equal, and the integral from —° to 0 equals the integral from 0 to °°.
(b)
For very small x, e ax
~ 1 and the graph resembles that of y = x. For
large x, the exponential factor dominates the factor x and the function
goes to 0. The graph resembles the v = 1 graph in Fig. 18.18. The area on the left side of the origin is negative and exactly cancels the area on the right side of the origin.
15.19
(a)
(b)
Let z = ax’. Then dz = 2ax dx and J* xe"® dx = (1/2a) J? &* dz= (/2a)e*" = (1/2a)(0 — 1) = 1/2a. (0/da) f= xe" dx = J" (/da) xe"™ dx = J® x2 dx = (0/0a)(1/2a) = —1/2a? and J xe°® dx = 1/2a”, Similarly, (8/0a) J" x3e°® dx =  ie we™ dx= (d/da)(1/2a*) = —1/a® and Ikexe™ dx= Ia’.
15.20
15.21
(a)
Vims = (3RTIM)'? = [3(8.314 J/molK)(500 K)/(0.04401 kg/mol)]'” = 532 m/s.
(b)
(v) = (8/31)! Urms = 490 m/s.
(c)
‘ie See
ae Beiaiatly,
dG(v)ldv = (m/2nkT)? [8mve "4? — Are? (mv/kT ee” *" ]. The equation dG(vy/dv = 0 gives 87Ump = 47unv;,,/kT and Ump = (2kT/m)'"" = (2RT/M)"”. (v = 0 and v = ~ also satisfy G’ = 0 but these are minima.)
15.22
(v2) = J® v’G(v) dv = 4n(m/2nkTY” JF pte MAT chy =
4n(m/2nkTy?(2410"7/292)(2kT/m)” = 3kT/m = 3RT/M, where integral 3 (with n = 2) of Table 15.1 was used.
207
15.23
(v3) = J? v'G(v) dv = 4m(m/2nkTY” Je vee” dv = 4n(m/2mkT)*?(2/2)(2kTim)y> = 2"?(kTimy? i”? = 2’?(RTIM)>"/"*. We have (v)(v7) = (8RT/M)'?(3RT/M) = 3(2°?)(RTIM) tt? # (v3).
15.24
(a)
(v,) = S7, vye(v,) dv, = (m/2nkT)"? J@,,ve" dv, = 0, where integral 4 of Table 15.1 was used. From Fig. 15.5, v, is as likely to be negative as positive, so (v,) =0.
(b)
Incalculating (v,), we average positive and negative values of v, and
get zero. In calculating (v~), all the (v2) values are nonnegative, and we must get (v2) to be positive.
15.25
(Cc)
Verms = (02) 7 = (kT/m)'? = (RT/M)'”, where (15.37) was used.
(d)
Zeto7 see. Figs 155:
(vt) = J. v4 g(v,) dv, = 2(m/2mkT)"” J” vte7”**" dv, .Integral 3 with n=2in Table 15.1 gives (v2) = 2(m/2nkT)'?(242'7/2°2)(2kT/m)>? = 3(kT/m)? = 3(RT/MY’.
15.26
Differentiation of the distribution function in Eq. (15.52) with respect to Ey
gives 2n(mkT) *?[5€,,/e S/T — (kT) eye €/*") = 0 and Exmp = ware (5/3. 15.27
(a) (b)
15.28
(2m)'71(u) = Jeeds = J¥[1 — 52/2 + (1/2!)54/4— (1/3!)s°/8 +J]ds= u—u'l6 + w/40 — u'/336 +. (2m)'"1(0.30) = 0.30 — (0.30)*/6 + (0.30)°/40 — (0.30)'/336 +  = 0.2956 and 1(0.30) = 0.118.
(a) The probability is J” G(v) dv = (m/2mkT*? JY e™” AT amy? dy = (Afr Vinp) Jo ew? re
dv = (4/n"? v3,)B,where Ump = (2kTIm)"?
2
and B = ips e ("¥m) »? dv. The integrationbyparts formula is J x ay
208
xy —
J y dx. Let x =v
SUC HeeR? Dace ph tien Bi
SU
SHU)
and dy = eo (emp
ly
as
ee Oe ee
Ao UR.) Ui e Wye!
2
a wD
2
“ vdv. Then y= =; mp e OE OO)
cm
2
~(V/ Vip)”
CU SUT = 20) Dro,
MIO? WN ee Dorp€° = 52/2 "Ump2 Wel as re= + 5Uinp is
oy (OT
h
ee Dd Dg) and. otG(0) a0.=
21(2"20' mp) — 20"(0/rmple (b)
The fraction with speed in the range from 0 to 4.243Ump is
212"? . 4.243) — 20 "(4.243)e ~(4243)" = 21(6.000) — (7.27 x 10°) =
2(0.4999999990) — (7.27 x 10°*) = 0.9999999253, since Fig. 15.10 gives 0.5 — 1(6) = 1.0 x 10°. The fraction with speeds exceeding 4.243Ump is
1 — 0.9999999253 = 7.5 x 10™. 15.29
(a)
The theory of relativity shows that speeds must be less than c, the speed of light in vacuum, so the correct range is 0 toc.
(b)
The number of molecules whose speed according to the Maxwell distribution lies between c and infinity is essentially zero at ordinary
temperatures.
15.30 From Eq. (15.58), 0.872 = (Mo,/M)' = [(32.0 g/mol)/M]'" and M = 42.1 g/mol. The only hydrocarbon with this molar mass is C3H6.
15.31
(a)
Equation (15.58) applies. We have dN/dt = Na(dn/dt) = Na(An/At) and
P = (An/At)(20MRT)'7/A ae (270(0.04496 kg/mol)(8.314 J/molK)(1690 K)]"” 7(0.001763/2)? m*
(10.510 (49.5x60) kg)/(0.04496 kg/mol) s 0 *m)°* atm 760 torr =0.0152 P = (2.03 N/m’an ) 82.06 (1 ae 8.314 J a
(b)
torr
Equation (15.67) applies. The order of magnitude of d is a couple of
angstroms. The order of magnitude of A is
209
(82 cm? atm/molK)(1690 K) ee 2" 1(2x 108 cm)? (0.015/760) atm (6x10? /mol) Since dnote = 0.2 cm, the condition A >> dhole iS met.
15.32
The CO; partial pressure is 0.00033 atm and Eq. (15.56) gives dNw/dt =
[1.0x (10°? m)}? (0.00033 atm)(6.02x107°/ mol) 8(8.314 J/molK)(298 i 4[82.06 (10°? m)? atm/molK](298 K)
7(0.0440 kg/mol)
=F 7x10 es”. The mass striking the leaf in 1 sec is [(7.7 x10'”)/(6.0 x 10” mol™')](44 g/mol) = 5.6 mg.
T5333
P = (0.010/760) atm (10.1325 Pa/atm) = 1.33 N/m?. As discussed in the (ext
we can take the evaporation rate as essentially equal to the rate at which molecules of vapor in equilibrium with the liquid strike the liquid. Equation (15.58) gives
adN___(1.33 N/m*)(6.02x10**/mol)(10 *m7) dt
[2m (0.3906 kg/mol)(8.3145 J/molK)(393 K)]!7_—
(8.9x10'’ molecules)
Lge 6.02x107* molecules
15.34
apie alee
ogUGe =0.58 mg  mole
I< (dN/dt) x Pr(i), where dN/dt is the effusion rate and Pr(i) is the probability of ionization. From Eq. (15.58), dN/dt « P/T'”. If x is the direction of motion of the molecular beam, the statements in the problem give Pr(i) I/(v,) «
1/T'””. Hence I « (P/T"?) x (A/T"”) and I~ PIT. 1535
(a) T; (b) F.
15.36
(a)
z(b) =(2 +2) collns./s =4 87,
(b)
Z,=[(2+242)s (1 x 10> cm’) =6 x 10° s7! cm. Nszo(byV= 3(4 s'W(1 x 10° cm) = 12 x 10° s"! cm™ # Zp. Noze(bYyV= 1(12 x 10° s' cm™) = 6 x 10° x"! cm™ = Zipp.
210
15.37
(a)
d = (v)(t) = (450 m/s)(4.0 x 107"? s) = 1.8 x 10°” m.
(b)
We have z(b) = 1/(5.0 x 107!° s) = 2 x 10” collisions per second and z;(c) = 1/(8.0 x 10°'° s) = 1.25 x 10” collisions per second. z,(b) is greater than
z(c). From Eqs. (15.60) and (15.61), these rates are influenced by the molecular diameters, the molecular concentrations, and the molecular
masses. Since no information is given about the concentrations (or the molecular masses), the question is defective and we cannot say which has
the larger diameter. From (15.66), Ay = (385 m/s)/(3.25 x 10’ s') =
Pep at ON mt 15.38
(a)
zo(b) = 2"°td 3(SRT/MM;)'P,Na/RT = 2'2(3.7 x 107! m)’[8(8.3145 J/molK)(298 K)/n(0.0280 kg/mol)]'” x (1.00 atm)(6.02 x 107°/mol)/(82.06 x 10° m?atm/molK)(298 K) = Ta lO ssa
(b)
Zo» = =Nozo(bV = 5 zo(b)(PNa/RT) = 5(7.1 x 10° s~')(1.00 atm) x (6.02 x 1073/mol)/(82.06 cm?atm/molK)(298 K) = 8.7 x 107° s' cm™.
(c)
1.0x 10° torr=1.3x 10° atm.
z,(b) is proportional to P and Zp, is
proportional to P*, so 2.(b)i= Ole lex 10° s')(1.3 x 10° atm/1 atm) =
9.3 5! and Zp = (8.7 x 1078s"! cem)(1.3 x 10°7/1)? = 1.5 x 10's"! cm. 15:39
Let
b= CO, andc=N>.
P,=0.97(4.7 torr)(1 atm/760 torr) = 0.0060 atm and
P,.= 0.00019 atm.
(a)
From (15.61), z(b) = 4107? (RT/Mp)' PpNa/RT =
4n"(4.6 x 107!° m)*[(8.3145 J/molK)(220 K)/(0.044 kg/mol)]'” x
(b)
(0.0060 atm)(6.02 x 107*/mol)/(82.06 x 10° m’atm/molK)(220 K) = 6.1x 10's". z.(b) = (810)!(ry + re) [RT(M  + M7')]'? PANa/RT =
(c)
(810)"/7(4.15 x 107!° m)°(RT)'"[(0.044 kg/mol)" + (0.028 kg/mol) '}'” x (0.0060 atm)Na/RT = 5.65 x 107s". z.(c) = 4r'2(3.7 x 10°'° m)[RT/(0.028 kg/mol)]'*(0.00019 atm)Na/RT= [oeetO. sss z(c) + 2(b) =o
(d)
10° s}+5.65;x10’s!'= 5.8) X Na ee
Zoe = Nozp(cV = Neze(b/V = 2(b)PeNa/RT =
(5.65 x 10” s')(0.00019 atm)Na/RT= 3.6 x 10° s' cm™,
211
15.40 (a) (b) (c) 15.41
A =RT/2'?nd’PN, = (82.06 cm’atm/molK)(300 K)/ 223.7 x 1078 cm)?(750/760) atm (6.02 x 107*/mol) = 6.8 x 10° cm. 2=(6.8 x 10° cm)(750/1) = 5.1 x 10° cm. 5.1 x 10° cm.
14100 ft = (14100 ft)(12 in./1 ft)(2.54 cm/I in.)(0.01 m/1 cm) = 4298 m. P = Poe M887 — (760 torr) exp[—(0.029 kg/mol)(9.81 m/s*)(4298 m)/ (8.3145 J/molK)(290 K)] = 458 torr.
15.42 Mg2/RT= (0.029 kg/mol)(9.81 m/s*)(30 x 12 x 0.0254 m)/ (82 14/molK)(298 K) = 000105 ho ope lene (1 atm)(1 — e °°!) = 0.00105 atm.
15.43 P/Py = e™"8"*" = 0.5 and z = (RT/Mg) In 0.5 = ~[(8.3145 J/molK)(250 K)/(0.029 kg/mol)(9.81 m/s”)] In 0.5 = 5060 m. 15.44
M, =0.97(44) + 0.03(28) = 43.5. P= Pre eat = (4.7 torr)exp[—(0.0435 kg/mol)(3.7 m/s”)(40000 m)/(8.314 J/molK)(180 K)] = 0.064 torr.
15.45
(a)
With T= To — az, (15.70) gives In(P’/P,) = —(Mg/R) {2 (Ty — az)!
(Mg/aR)\n(Ty — az) Be (Mg/aR) \n[(To — az’)/Tp] = In{(Ty  age)
SOR)
15.46
(a)
Te lena
eel ean ae (1h iz)
For Cy.m, the translation contribution is 3R/2, the rotational contribution is 3R/2, and the predicted vibrational contribution is 2[3(5) — 6]R/2 = OR, SO Cy.m is predicted to be 12R. Cpm=Cym+R and Cpm 1s predicted to
be 13R. (b)
No; the actual Cp is less than 13R at 400 K.
(c)
Cpm will reach 13R at high 7’s.
15.47 (a)
dz =
J*.g(w) dw=1= J®, Ae
dw = 2A(n!/2\(kTIc)"? and
A= (c/nkT)””, where integrals 1 and 2 in Table 15.1 were used.
212
(b)
(€,,.) =
1 2
15.48
ie Ewe(w) aw=A
ee cwe —cw2 /kT We
2Ac(2!10"7/271 N\(kTIc)*? _
T, where A = (c/mkT)'” and integrals 1 and 3 of Table 15.1 were used.
Lmy? = 5kT and Vims = (3kT/m)"" = [3(1.38 x 10°? J/K)(298 K)/(1.0 x 10°"? kg)]'” = 0.00035 m/s.
15.49
(v2) = 3RTIM and (v)* = 8RT/MM, so (v0?) # (v)°*.
15.50
The possible values of s are 1, 2, 3, 4, 5, 6 and the probability of each value is
15.51
(a)
1/6. Hence (s) =X, sp(s) = 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6) = 21/6 = 3.5. Also, (s?) =, s’p(s) = 1°(1/6) + 27(1/6) + 3°(1/6) + 4°(1/6) + 5(1/6) + 6°(1/6) = 91/6 = 15.17. (s)? = (3.5) = 12.25 # (s”).
(b)
oy = (vz) (v,)* =kTim—Oand o, = (kT/m)"”. (v,) =O and o = (kT/m)'”. The distribution function (15.42) is Vv
and the desired fraction is
(21) /?7'e"**/? Ones
oF
ce e
2 12 v,/26
dv;
=
2(2m)
z
W265
a.1
ie e
2 ane v,/26
dvx
=
2020) 207 fie * ods =2I(1) = 0.68, where we made the Sa)
2
substitution s = ,/o and used Eq. (15.51) and Fig. 15.10.
15.52
h(xy) = fx) + gy). We take (0/dx), of this equation. Let z = xy. The partial derivative of the left side is 0h(xy)/dx = (dh/dz)(0z/0x) = h'(z)y. The partial = y 'dfixldx. derivative of the right side is df(x)/dx = h’(z)y; we have h’(z) x 'dg(y)/dy. Hence Similarly, taking (d/dy), of h(xy) = fix) + g(y) gives K@=
x! de(ydy = y! dfvo/dx and y[dg(y)/dy] = x{dflx\/dx] = k. By the argument
dx and f(x) = that follows Eg. (15.33), k must be a constant. Hence df(x) = (k/x) k In x +a, where a is an integration constant.
15.53
the plane Molecules with v, = b have the tips of their velocity vectors lying in axes and is a v, = b. (This plane is parallel to the plane formed by the v, and v,
213
distance b from it.) The region corresponding to b < vx is larger than Ho, so hu, = ho, 
(e)
The H2 molecules are moving faster and so collide more often with the wall.
Substitution of 4; = w; (7) + RT In (P;/P°) into wp + Migz” = uP > Migz? gives
wh; (T) + RT In (P3/P°) + Migz” = 3 (T) + RT In (P?/P°) + Migz? or
RT In (P#/P?) = Mig(z’ — 2%). In (P2/P%) = —Mig(z — 2*)/RT and P® = P® exp [Mig(z — 2*)/RT]. 1557)
(a) See Fig. 15.7a.
15.58
A, zp(b), and Lhe
15.59
(a) Statistical mechanics.
15.60
(a) False.
(f) True.
(b) See Fig. 15.6.
(b) Yes.
(b) True. See Fig. 15.5.
(g) False.
(h)
(c) True.
(d) False.
True. See Eq. (15.13). (i) False.
214
(e) False.
Chapter 16 16.1
(a) T;
(b) T;
(c) F.
16.2
(a) Equation (16.1) gives g = K(AT/Ax)At = (0.80 J/Kcms)(24 cm’)(60 s)(50 K)/(200 cm) = 288 J. (b)
AS;oq = 0 since the rod’s state is not changed. The temperature of each
end of the rod differs only slightly from the temperature of the reservoir it is in contact with, so we can use dS = dg;ey/T = dq/T for each reservoir.
Theretore AS = AShorres + Aocoldres = (2008) (520s) + (288.J)/275 K) = 0.161 J/K.
16.3.
Cym=3R/2 for this monatomic gas, and Eq. (16.12) gives
Les (a9 ) 16\2
(8.314 J/molK)
(8.314 JmolK)(273 K) 
4)(6.02x1077/mol)(2.2 x107"° m)?  —_(0.00400 kg/mol)
k = 0.00142 J/Kcms. At 100°C, k = 0.166 J/Kms.
16.4
Cym+9R/4 = Cpm—R + 9R/4 = Cem + SR/4 = (35.309 J/molK) + (5/4)(8.3145 J/molK) = 45.7 J/molK. Then k = (5/16)(45.7 J/molK)[(8.314 J/molK)(298 K)/m(0.0160 kg/mol)]'"*/ [(6.02 x 107° mol')(4.1 x 107'° m)*] = 0.031 J/Ksm = 31x 10°3 K's! cm", where (16.12) was used.
16.5
p =M/Vm= (18.015 g/mol)/(18.1 cm’/mol) = 0.995 g/cm? = 995 kg/m® and k = 2.8(8.314 J/molK)[17.99/(17.72)(995 kg/m*)(4.46 x 107"? m?/N)]""7/ [(6.022 x 107 mol!)'2(18.1 cm*/mol)”*] = 6.05 x 10° JK! cm* ms" = 6.05 mJ K7! cm! $s".
(b) T;
(c) T;
(d) T.
16.6
(a) F;
16.7
The maximum (v,) for laminar flow makes Re = 2000, so (U,) max =
2000n/pd = 2000(0.0089 dyn s cm™)/(1.00 g/cm*)(1.00 cm) = 18 cm/s. 215
16.8
(a)
(32/760) atm (8.314 x 107 ergs)/(82.06 cm? atm) = 4.26. x 10° dyn/cm’. Equation (16.17) gives n = (mr°/8V)(AP/Ay\)t = [70.100 cm)*/8(148 cm?)][(4.266 x 10* dyn/cm7)/(24.0 cm)](120 s) = 0.0566 dyn cm” s = 5.66 cP.
(b)
In time t, a volume V = mr’ (d) flows through the pipe of crosssectional area Tir, where (d) is the average distance traveled by the fluid. We
have (d) = (v,) rand (v,) = (d)/t= (V/nr?)/t = Vitr’t = (7/8) AP / Ay , where Eq. (16.17) was used. So (v,) =
[(0.100 cm)?/8(0.0566 dyn cm~ s)](4.266 x 10* dyn/cm’)/(24.0 cm) =
39.3 cm/s. Then Re = p (v,) d/n = (1.35 g/cm*)(39.3 cm/s)(0.200 cm)/ (0.0566 dyn cm” s) = 187 < 2000, so the flow is laminar.
16.9
(a)
From (16.17), (P2 — Pi)/2 — y1) = (89/nr*)( Vit) = ~8[(0.04 dyn s cm~)/m(1.25 cm)*](5000 cm?)/(60 s) = 3.5 dyn/cem® = —35 Pa/m.
(b)
(v,) = Vit = (5000 cm*)/n(1.25 cm)"(60 s) = 17 cm/s.
(c)
Re = p(v,) din = (1.0 g/em*)(17 cm/s)(2.5 cm)/(0.04 dyn s/cm?) = 1100 < 2000, so the flow is laminar and use of Eq. (16.17) 1s justified. For a flow rate of 30 L/min, (Oy) and Re are 6 times as large and
Re = 6400 > 2000, so aortic flow is turbulent during vigorous activity.
16.10
P? — P? = (1.44  1.00) atm? = (0.44 atm’)(8.314 J)°/(82.06 x 10° m* atm)’ = 4.52 x 10° N/m’. Eq. (16.18) gives dn/dt = 1(0.000210 m)*(4.52 x 10° N*/m*)/ [16(0.0000192 N s/m*)(8.314 J/molK)(273 K)(2.20 m)] = 0.000018 mol/s, which is 0.00058 g/s.
16.11
Ne.u,, = N,0(Pc,,,tccH,, /PH,0%H,0) = (1.002 cP)(0.659 g/cm*)(67.3 s)/ (0.998 g/cm*)(136.5 s) = 0.326 cP.
16.12
(a)
The pressures at the left and right ends of C exert forces Pms” and —(P + dP)ms*, respectively, on C. The viscous force on C is given by
(16.13) as nA(dv,/ds) = n(2ns dy)(dv,/ds). So (P + dP)ns? + Ps? + n(21s dy)(dv,/ds) = 0 and dv,/ds = (s/2n)(dP/dy). Integration gives
216
Vy = (s?/4n)(dP/dy) + c, where c is a constant. Use of v, = 0 at s =r gives
c = (r’/4n)(dP/dy). Therefore v, = (1/4n)(r" — s°)(dP/dy). (b)
The volume of fluid in the shell that passes a given location in time dt
equals the volume of a cylindrical shell with length v,(s) dt and inner and outer radii s and s + ds; this volume is 1(s + ds)  Vy(s) dt — Ts Vy(S) d=
27sV,(s) ds dt, since (ds) is negligible compared with ds. Integration over all shells from s = 0 to r gives dV = [J§ 2msv,(s) ds] dt. Substitution
of dV = dmi/p and vy = (1/4n)(r° — s*)(dP/dy) gives dm/dt = (np/2n)(dPldy) J, (r’s — s*) ds = (r*p/8n)(dP/dy). Separating P and y and integrating from one end to the other, we have
= 4 dP = (dm/dt)(8r/nr'p) J>» dy, which becomes dm/dt = (tr'p/8n)(P; — P2)/(y2 — yi). We have dm/dt = d(pV)/dt = p dV/dt = pVIt, since the flow rate is constant with time. Hence (16.17) follows.
(c)
Substitution of p = PM/RT into (16.91), separation of P and y and
integration gives (RT/M)(dmldt)(8y/mr*) J}? dy = — Je PdP= 1 (p? — Py); use of (RT/M)(dm/dt) = RT d(m/M)/dt = RT dnidt gives
(16.18). 16.13
In time dt, the matter in the thin shell travels a distance Dy dt in the y direction.
The volume dV of matter in the thin shell that passes a fixed location in time dt
equals the length v, df times the shell’s crosssectional area d4; dV = vy dt dA. (The shell volume is the difference between volumes of cylinders of radii s + ds and s, and the volume of a cylinder equals its length times its crosssectional area. Hence the shell volume equals the shell length times the shell
that crosssectional area.) We have p = dm/dV, where dm is the thinshell mass
passes a given location in time dt, so dm = p dV = Pv, dt dA.
16.14
Equation (16.22) gives v = 2[(7.8 — 1.0) g cm (980.7 cm/s”)(0.050 cm)’/ 9(0.0089 dyn s cm’) = 420 cm/s. For glycerol, v =
2(7.8 — 1.25) 107 kg(10 m) *(9.807 m/s”)(0.00050 m)'/9(0.954 N s m™) = 0.0037 m/s = 0.37 cm/s.
16.15
' = From (16.25), @ = 5(MRT)'7/16nNan
0.1763(0.04401 kg/mol)!(8.314 J/molK)'7T'7/(6.022 x 10°/mol)n = 217
1.77 x 10°°° m? (7/K)!?/[y/(kg m7! s“')] = 1.77 x 1074 m? (T/K)'7/(n/P), where P stands for poise. At 0°C, d’ =1.77x 10% m’* (273)'7/(139 x 10°)
and d= 4.59 x 107° 16.16
m=4.59 A. At 490°C, d= 3.85 A; at 850°C, d= 3.69 A.
The diameters of H; and D> are the same and (16.25) shows that n is
proportional to M'”. So Np, /Nn, =(Mp, /Mu, )? and Np, =
(4.03/2.02)"(8.53 x 10° P) = 1.20 x 10° P. 16.17
(a)
M, =>) x)M; = 0.5(200 kg/mol) + 0.5(600 kg/mol) = 400 kg/mol.
My = (Xi xiM 7)/(i Mi) = [0.5(200 kg/mol)” + 0.5(600 kg/mol)”]/(400 kg/mol) = 500 kg/mol. (b)
16.18
Let us take 600 kg of each species. We then have 3.0 moles of the molecularweight 200000 species and 1.0 mole of the molecularweight 600000 species. Hence M,, = >; x,M; = 0.75(200 kg/mol) + 0.25(600 kg/mol) = 300 kg/mol. M, = [0.75(200 kg/mol)” + 0.25(600 kg/mol)*]/(300 kg/mol) = 400 kg/mol.
We plot (1, — 1)/pp vs. Pp. The points fit a straight line well, and extrapolation
to Pg = 0 gives [] = lim, _,o(n— 1)/pp = 0.147 dm*/g = 147 cm’/g. Then 147 cm’/g = (0.034 cm?/g)(Mp/M°)°® and Mg = 390000 g/mol. M,.p = 390000. 16.19
(a) sR (DSF
16.20
(a)
(Cc) Ty s(d)er (ey als a)
(Ax)ims = (2D1)'”, so t = (Ax)?_./2D = (1 cm?)/2(107! cm?/s) = 5 x 10% s=2~x 10" yr.
(b) ¢=(1cm’)/2(10” cm’/s) = 5 x 10° s = 2 x 107 yr. 16.21
(a)
(Ax)ms = (2Dt)' = [2(0.52 x 107 cm?/s)(60 s)]!? = 0.025 cm.
(b)
t= 3600s
(c)
t= 86400 s and (Ax)ms = 0.95 cm.
and (Ax)ms = 0.19 cm.
218
16.22
Let the origin be at the t = 0 location of the particle and let the particle be at
point (x, y, Z) at time ¢. Then (r?) = (x? + y? +27) = (x7) + (a) 4 za BY symmetry, Copa) = (2 acon ae Fons = (12) 1? = (6D1)"”,
16.23
3(x*) = 3(2Dr) = 6Dt and
From (16.35), k= 3[(AX)mms} mnr/tT. For t= 30s, k= 3(7.1 x 10“ cm)n x (0.011 dyns/em?)(2.1 x 107 cm)/(30 s)(290 K) = 1.26 x 107'° erg/K. Then Na = R/k = (8.314 x 10’ ergs/molK)/(1.26 x 107'® erg/K) = 6.6 x 10°*/mol. For t = 60s, we getk = 1.41 x 107'° erg/K and Na = 5.9 x 10” mol. For 90s, k= 1.07 x 107'° erg/K and Na = 7.8 x 10°? mol.
16.24
(Ax)= 4 (5.34+3.4190.440.5+3.10.23.5+414+0.31.0+2.6) um =—0.08 um. ((Ax)*) = (1/12)[(5.3)? + 3.4)? ++] um? = 6.28 pm’. (Ax)ims = 2.5 um.
16.25
(a)
Equation (16.42) and N/V = PN,/RT give Dj =
(3/8r02)[(8.314 J/molK)(273 K)/(0.0320 kg/mol)]'”” x (3.6 x 107! m)°[82.06 (10° m)’atm/molK](273 K)/ (1.00 atm)(6.02 x 10”? mol!) = 0.000016 m7/s = 0.16 cm//s.
16.26
(b)
Since Dj is proportional to 1/P, we have Dj = 0.016 cm’/s.
(a)
The volume of each cell is 8r> , and the molar volume is Vnj = 8r>N 1/3 pel
,
Hence’g=s5 (Ving Na)
(b)
16.27
Dj ~[(6.02 x 10°/mol)/(18.1 cm?/mol)]'°(1.38 x 107° ergs/K)(298 K)/ 27(0.00890 dyns/cm’) = 2.4 x 10° cm’/s.
The N> and HO molecules are not greatly different in size, so Eq. (16.38) is
appropriate. We have Dj, = (1.38 x 10°'° erg/K)(298 K)/
4n(0.0089 dyns/cm?)(1.85 x 10™ cm) = 2.0 x 10° cm’/s. 16.28
(a) Equations (16.25) and (16.42) give 6n/Sp = (61/32){v) A = (31/16)(v)A = Dy. 219
(b) Dj = 6nRT/SPM = 6(0.000297 dyns/cm*)(82.06 cm*atm/molK) x (273.1 K)/5(1.00 atm)(20.18 g/mol) = 0.40 cm’/s. 16.29
Since a hemoglobin molecule is much larger than a water molecule, we use Eq.
(16.37). We have Vm/Na = 47/3 and r) = (3Vin/4Nan)'? = [3(48000 cm?/mol)/4(6.02 x 107°/mol)m]'? = 2.67 x 10°’ cm. Then Dy = (1.38 x 107° erg/K)(298 K)/6n(0.0089 dyns/cm”)(2.67 x 107’ cm) = 9.2 x 10°’ cm’/s. 16.30
(a)
Let z=x*. Then d(x’)/dt = dz/dt = (dz/dx)(dx/dt) = 2x(dx/dt). Also, &(xdt? = (didt)[d()/dt] = (d/dt)[2x(dx/dt)] = 2(dx/dty’ + 2x(d’x/dt’). Substitution of these two equations into the equation in the problem transforms it to Eq. (16.33).
(b)
Averaging, we get 0 1f (d(x? )/dt) = 4m(d*(x*)/dt*)  (m(dx/dt)’) .
We have & = 1mv{ = 1m(dx/dt)’, so 2(e,) = (m(dx/dt)*). Also, €=€,+£,+€,and
(€) = (€,) + (€,) + (€,). By symmetry,
(€,) = (€,) = (€,); thus (€,) = (") = 3(96485 C/mol) x
[(0.03327 mol)/(1000 cm?*)](1.111 cm*)/(0.005594 A)(4406 s) = 0.434.
Then 7(CI) = 1 — 0.434 = 0.566.
16.45
The coulometer reaction is Ag’
+e = Ag. The 0.16024 g of Ag is 0.0014855
moles of Ag, so Q = (0.0014855 mol)(96485 C/mol) = 143.33 C. Letn= 0.0014855 mol. Then n moles of Cl enter the cathode compartment due to the reduction reaction AgCl + e = Ag + CI. The total number of moles of charge on the ions that cross the plane between the middle and cathode compartments is n; this charge is composed of tn moles of CI” leaving and t,n moles of K* Cl entering the cathode compartment. The net change in number of moles of the in the cathode compartment is thus n—tn= (1 t)n=tn, which is also change in number of moles of K* in this compartment. The final composition of KCI and of the cathode compartment 1s (0.0019404)(120.99 g) = 0.23477 g
120.99 g  0.235 g = 120.75s g of H20. The initial composition of this
223
compartment is 120.75; g of water plus x grams of KCI, where x/(120.755 + x)
= 0.0014941. We get x = 0.18069. The change in KC] mass in the cathode compartment is 0.05408 g, which is 0.0007254 moles. Thus, t.n = t,(0.0014855 mol) = 0.0007254 mol and t, = 0.4883. Then t = 1 —t, = 0.5117.
16.46
(a)
Using Eq. (16.84), we have A; (LiNO3) =A
> (Li?) +A
> (NO;) and the
following three equations: (mee (LiCl =A (Li +A, (Gly);
(2) Aj, (KNO3) =A; (K") +47, (NO3); GymA} (KGaA
a Ke ae (Ci)
We take (1) + (2) — (3) to get A>, (LiNO3) = A> (LiCl) + A> (KNO3) —
A® (KCI) = (90.9 + 114.5 105.0) cm?/Qmol = 100.4 Q7! cm? mol!.
(b)
(4) A; (HC) =A7, (H") +05 (CI); (5) Aj, (NaCl) = A> (Na*)
+A (CI);
(6) Aj, (NaC2H302) = A 7 (Na) + A= (C,H303). Taking (4) + (6) — (5), we get A; (H") + AZ,(C,H;,0;) = A* (HCI) + A, (NaC2H302) — A® (NaCl) = (426 + 91 — 126) cm?/Qmol =
391 cm’/Qmol.
16.47 From (16.85) for NaCl, t? = 0.463 = (1)A@ ,(96.9 cm? Q* mol!) and A, (Na*) = 44.9 cm7/Qmol. Then A* (CI) = (96.9 — 44.9) cm2/Qmol = 52.0 cm’/Qmol. We now use (16.84) for the other electrolytes. For NaNOs;,
106.4 cm’/Qmol = 44.9 cm’/Qmol + 4= (NO5) and A® (NO;) =
61.5 cm?/Qmol. For LiNO3, 100.2 cm2/Qmol = X~ (Li*) + 61.5 cm?/Qmol
and A (Li*) = 38.7 cm’/Qmol. For NaCNS, 107.0 cm?/Qmol = 44.9 cm?/Qmol + d.*, (CNS) and A* (CNS’) = 62.1 cm?/Qmol. For HCl,
192 cm*/Qmol = 4 (H*) + 52.0 cm?/Qmol and * (H*) = 140 cm’/Qmol.
For Ca(CNS)2, 244 cm’/Qmol = 4" (Ca*) + 2(62.1) cm?/Qmol and A (Ca**) = 120 cm?/Qmol.
224
16.48
(a)
upg =A™ p/zpF = (67.2 Q! cm’ mol ')/1(96485 C/mol) = Gog x10 cm Vv" Ss”. (b) vg =uR E= (6.96 x 107% cm’/Vs)(24 V/cm) = 0.017 cm/s. (c)
16.49
rp ~=zele/6nnus = 1(1.6022 x 10°" C)/ 6m(0.008904 x 107! Ns m™)[6.96 x 107 (10° m)*/Vs] = 1.37 x 107° m=1.37 A.
MST IN
Sia
ap
GS
(a)
A® =(73.5+71.4) cm’/Qmol = 144.9 cm*/Qmol.
(b)
A® =[2(73.5) + 159.6] cm?/Qmol = 306.6 cm*/Qmol.
A® =(106.1 + 159.6) cm?/Qmol = 265.7 cm’/Qmol. (d) A =[118.0 + 2(199.2)] cm?/Qmol = 516.4 cm*/Qmol.
(c)
16.50
From (16.85), i (Mg"*) = 1(106.1)/[1(106.1) + 2(71.4)] = 0.426 and (NO; )= 1 — 0.426 = 0.574.
16.51
From (16.85), f2 = v,A%,/AZ = 2A%,, /(259.8 Q7! cm’ mol"') = 0.386 and A® (Na*) = 50.1 Q7' cm? mol”!. From (16.84), AZ = 259.8 Q”' cm’ mol! =
2(50.1 Q7! cm? mol") + A%(SO2") and AS (SOZ) = 159.6 2"! cm? mol”. 16.52
We assume the electrodes are inert, so only Ag” and NO; carry the current in
the solution.
=t™ = v_A~_MviAR. t+V_An_) = 71.4/(62.1 + 71.4) =
0.535. If 10° mol of Ag’ is deposited according to Ag’ +e — Ag, then 105 Faradays of charge must have flowed through the circuit. The nitrate ions carry a fraction t_ of the current in the solution, so the nitrate ions carried L105, Faradays) of charge through the solution. Hence 10°t_ moles of NO,
crossed a plane in the solution, and this is (1.00 x 10~*)(0.535) mol = 0.535 mmol.
225
16.53
Because of its higher charge, Mg” is hydrated to a much greater extent than Na’ and so its radius is much greater than that of Na’. Therefore u in (16.70)
has similar values for these two ions.
16.54
16.55
(a)
AW(CI), u(Cl);
(b)
no (interionic forces differ in the two solutions).
x.
16.56 (a)
Use of(16.70) in (16.78) gives A™ , =zpFuy =ze eF/6mnrp. So InA@ =In(zzeF/6mrg) — In yn and d In A®/dT = (1/n) dn/dT.
(b)
(c)
dn/dT = AW/AT= (0.8705 — 0.9111)cP/(2 K) = 0.0203 cP/K and (1/n)(dn/dT) = (1/0.8904 cP)(0.0203 cP/K) = 0.023 K™, so din) id= 0.023) Kes Integration gives In (A* ,/A™ ,) = (0.023 K')(T7 — T;) and A™, = = A%1expl(0.023 K™')(T2  T;)] = (71.4 Q* cm? mole?” 89.9 QT cm? mol.
16.57 (a)
A® =(73.5 + 71.4) cm/Qmol = 144.9 cm?/Qmol. From (16.87), Am = 144.9 cm’/Qmol — [60.6 + 0.230(144.9)](cm?/Qmol)(0.00200)!” = 140.7 cm’/Qmol. K = A,c = (140.7 cm?/Qmol)(0.00200 mol)/ (1000 cm*) = 0.000281 Q7! cm.
(b)
Neglecting the conductivity of the water, we have
R=pl/A= ¢/KA=
(10.0 cm)/(0.000281 Q7' cm™')(1.00 cm?) = 35500 Q. 16.58
K,.=[H'][OH ], since y+ = 1 at the extremely low J,, value in this solution.
Equation (16.90) with c, = 0 in the denominator gives the initial estimate c, ~
(5.47 x 10° Q7 cm'V/[(350 + 199) cm?/Qmol] = 9.96 x 107!! mol/em? =
9.96 x 10° mol/dm*. We have S = 60.6 cm’/Qmol + 0.230(549 cm?/Qmol) = 186.9 cm’/Qmol. Using the initial estimate of c, in the denominator of Eq.
(16.90), we have c, = (5.47 x 10° Q7! cm"!)/
[549 cm*/Qmol — (186.9 cm?/Qmol)(9.96 x 10°°)"”] = 9.96 x 107! mol/em?
226
= 9.96 x 10° mol/dm’, as before. Hence K; = (9.96 x 10° mol/dm’)* = 9.9, x 107° mol?/dm*. 16.59
(a)
Since z, =z_, Eq. (16.90) applies. We have A, +A
=
(118.0 + 159.6) cm?/Qmol = 277.6 cm7/Qmol and
S =8[a+ bm, i, Cy
+A%_)] = 996 cm/Qmol. Therefore 221A
Ome@arcems.
~ 277.6 cm?/Qmol — (996 cm?/Qmol)[c, /(10~> mol/em?)]'/?
With c, = 0 in the denominator, we get the initial estimate c, =
7.96 x 10° mol/em>. With c, = 7.96 x 10° mol/cm’, we get the improved estimate c, = 1.17 x 10° mol/cm?. Further repetitions yield the successive estimates 1.30 x 10° mol/cm’, 1.35:x107 mol/cm’, 1.36 x 10° mol/cm’, and 1.37 x 10> mol/cm’. Hence c, =
1.37 x 10°? mol/dm’. The solution is dilute, so we can take the molality
as 0.0137 mol/kg. The Davies equation with J, = 0.0548 mol/kg then gives log y. =—0.353 and y+ = 0.443. We can neglect the difference between molalityscale and concentrationscale activity coefficients in this dilute solution, so the concentrationscale Ksp 18 Ksp =
(0.443)°(0.0137 mol/dm*)? = 3.7 x 10° mol’/dm’*.
(b)
No.
Ksp= a [Ca**][SO re ]. We found the ionic concentrations and Ys.
The existence of the additional equilibrium Ca** + SO,” = CaSO,(aq) does not invalidate our work.
16.60
We have A”, +A%,_ = (350 + 40.8) cm7/Qmol = 391 cm*/Qmol and S = [60.6 + 0.230(391)] cm?/Qmol = 150.5 cm’/Qmol. Therefore 4.95x10° Q!' cm”!
CS
+
OO
TT
391cm?/Qmol  (150 4.cm?/Qmol)[c, /(10* mol/em*)}"”
With c, = 0 in the denominator, we get the initial estimate c, =
1.27 x 1077 mol/cm’. Recalculation with this c, value in the denominator gives
c, = 1.27 x 1077 mol/cem? = 1.27 x 10“ mol/dm’, which is the H30*
concentration. For this dilute solution, we can take the ionic molalities as 1.27 x 107 mol/kg. The Davies equation then gives Y1 = 0.987. Neglecting the slight differences between concentrationscale and molalityscale y’s, we have
794}
K. = (0.987)(0.000127 mol/dm?)7/[{(0.001028 — 0.000127) mol/dm*] = 1.74 x 10° mol/dm?. 16.61
We have A*,, +A*_ = (106 + 160) cm*/Qmol = 266 cm*/Qmol and S = 8[60.6 + 0.23(266)] cm?/Qmol = 974 cm*/Qmol. Then
3
6.156x10° Q' cm!
*
266 cm?/Qmol — (974 cm?/Qmol)[c, /(10°3 mol/em?*)}'”
With c, = 0 in the denominator, we get the initial estimate c, = 2.31 x 107’ mol/cm*. With this value of c,, we get the improved estimate c, =
2.45 x 10°’ mol/cm®. Another repetition yields the final result c, = Dida ax 10°’ mol/cm? = 2.46 x 10~* mol/dm*. For this dilute solution, we can take the ion molalities as 2.46 x 10~ mol/kg, so J, = 9.8 x ir mol/kg. The
Davies equation gives logy+ = —0.061 and y: = 0.87. We can assume the molalityscale and concentrationscale activity coefficients to be equal in this
dilute solution. We have [Mg”*] = [SO2 ]= 2.46 x 10~* mol/dm? and [MgSO,(aq)] = 2.50 x 10“ mol/dm? — 2.45; x 10“ mol/dm? = 4.5 x 10% mol/dm*. Hence, K, = (4.5 x 10° mol/dm?)/(0.87)°(2.46 x 107+ mol/dm*)” =
100 dm*/mol. 16.62
Omitting the S term and using Eq. (16.58), we have for (16.90):
CoN
16.63
eh
Nycl (Nae een
enema
cae =e (ee PA),
The Onsager equation predicts ac!” dependence of A,, in dilute solutions, so we plot A, vs. c’”. The points give a rather good straightline fit; the intercept
atc =\00seA —270:8icm /Qamal: 275 A p/(Q7 cm’/mol)
= 14.095x + 270.77
at
265 260
22 250 245
IN
O02
SV
hes
0.459
Ne
0,Gian 0) Cres lems
ee
ie
oe [c/(mmol/L)]"”
228
ed
16.64
10 A=60.6:B=0.23
60 FOR J=1 TO 1000
15 INPUT “ZPLUS”;ZP
70 M=CP
20 INPUT “LAM+IN/(CM2/OM—
80 CP=K/(LS*(CP/.001)10.5) 90 IF ABS(CPM)/CP m = 380000 A = 0.0038 cm, where u” was taken from p. 513 of the
text. In 10 s, (AX)mms = (10/1)'7(38 wm) = 120 pm. 16.66
(a) Increases;
(b) increases;
(Cc) increases;
(d) increases;
(e) increases;
(f) increases.
16.67
Thermal conduction: g (units J), k (unitsJ Kom oe as T (units K), where the orderis W, L, B. Viscous flow: py (kg m/s), 1 (N s/m? ), Vy (m/s). Diffusion: n; (V). (mol), Djx (m 1s), Cj (mol/m’). Electrical conduction: Q (C), K(Q"! m’), o
229
Chapter 7 17.1
(a) F; (b) F; (c) T; (d) F; (e) T; (f) F; (g) T; (h) F.
17.2
(a) s'; (b) Lmol's7!; (c) L’ mol™ s™.
17.3
(a).
17.4
0.002 mol L7! s7!.
Ulds)
(a)
na/V=Pa/RT = (0.10 atm)/(82.06 cm?atm/molK)(298 K) =
(b)
r=[I/(2)] d[N2Os]/dt = 7.1 x 10° mol/dm*s and d[N2Os]/dt =
4.09 x 10™° mol/cm? = 4.09 x 10° mol/dm*. r= k[N2Os] = (1.73 x 10° s“!)(4.09 x 10°? mol/dm’*) = 7.1 x 10° mol/dm’*s. J=rV=(7.1 x 10° mol/dm*s)(12.0 dm*) = 8.5 x 107’ mol/s.
1.4 x 10°’ mol/dm’s. (c) (d)
(1.4 x 1077 mol/dm?s)(6.02 x 107/mol)(1 s)(12 dm*) = 1.0 x 10"*. N20; — 2NO2 + 40>. Here, J= —dny9,/dt as compared to J = — = dny o,/dt for 2N20s > 4NO? + O2. So J = 2(8.5 x 10°’ mol/s) = 17x 107 mol/s.
r=J/V=
14x 10° mol/dm?*s.
r= k[N2Os] and
k = r/[N2Os]; since r is doubled, so is k, and k = 3.46 x ie ee [In (c),
the approximation dn/dt = An/At was used. This approximation is justifiable, as follows: We have (0.0041 mol/L)(12 L) = 0.049 mol, which is 3 x 10°” molecules present; therefore a change of  x 10°
molecules will not significantly change the rate dn/dt over a period of 1 s.]
17.6
From (17.2), J = (1/v;)(dn,/dt), where the stoichiometric coefficient v; is
negative for reactants. From (4.97), dn; = v; d&, so J = (1/v) v,(d&/dt) = dE/dt.
Ait
Cy =Ng/V =(Ng/Na)MV/N,) =nghVIN,)=[BIN,. r=—(1/b)d[B/dt. ro = (I/b)dC,/dt = (/b)N ,d[B\/dt = N,r. So rc/r=Na. For simplicity,
230
let the rate law be r = k[B]”. We have r =kCy = k[B]" Nj. Hence re/r=
(kc /k)Nj . Also rc/r = Na. Hence (kc/k)NA =Naand k. =kNI. 17.8
(a)
PaV=naRT and Pa = [AJRT. The equation (1/a)dP,4/dt = kpP.
becomes —(RT/a)d[A]/dt = kp[A]"(RT)" and r = —(1/a)d{A]/dt = (RT)" 'kp[A]". Comparison with r = k[A]" gives kp = K(RT)'™.
(b) Leo
Yes.
r, = k,[A] and rz = k2[B]. Since [B] might be much greater than [A], it is possible for rz to exceed r; even though k, exceeds ko.
17.10
(a)
Since the first step produces one C molecule while the second step consumes two C’s, the first step must occur twice for each occurrence of the second step. The second step produces one F and the third step consumes one F, so the second and third steps have the same stoichiometric number. The stoichiometric numbers are thus 2 for the first step, 1 for the second step, and 1 for the third step. Adding twice the first step to the second and third steps, we get
2A +2B+2C+F+B—2C+2B+F+2A is 3B > 2B +G.
(b)
17.11
+ Gand the overall reaction
Species A is consumed in the first step, regenerated in the last step, and does not appear in the overall reaction. Hence A is a catalyst. B is a reactant. C and F are reaction intermediates. B and G are products.
From Eq. (17.5), the units of k are (dm?/mol)"* st. Comparison with the value
of k shows thatn —1=1,n =2.
17.12
r, = k,[A]. Since [A] decreases exponentially with time (Fig. 17.3), so does rj. r2 = k2[B].
[B] increases from 0 to a maximum and then decreases to zero (Fig.
17.3) and so does rp.
L713
(a) k,[B], [BJ]; (b) k:[B], Ai[B] —ko[E], k2[E]; (c) k,[B] + k[E], k,[B] — K.\[E] — ko[E], ko[E]; (d) —(ki + k3)[B] + Ki[E] + ks[F], k,[B] — k.i[E], k3[B] — ks[F]. 231
17.14
(a) F;
17.15
(a)
(b) F;
(c) T (assuming Vis constant).
[AV/[A]o = e7*4', so 0.65 = e 4955) and In 0.65 = —ka(325 s). We get ka = 0.00133 s!. From Eq. (17.11), k = kala = ka/2 = 0.00066 s™. (b) t = —(1/ka)In([A]/[AJo)= [1/(0.00133 s”')] In (0.30 or 0.10) = O0DiS Olas:
17.16
(a)
73 x 10°s')= Kati = 0.693. From (17.11), ka = ak = 2(1.
(b)
10° s71(24% 36005) _ 46% [A] = [Aloe “*’ = (0.010 mol/dm’) 73
10° s"') = 2.00 x 10° s. 3.46 x 10s. So typ = 0.693/ka = 0.693/(3.46 x
5.0 x 10“ mol/dm’.
17.17 d{AVdt=kslAT,
J?LAT? d[A} =—J? ka dt, 1[AT*]i =ka(o— 11),
W/[A]3 — WA]? = 2ka(t2 —t1) or L/LA]* — 1/[A]9 = 2kat, where we took t, = 0 and
17.18
(a)
=
Eqs. (17.22) and (17.19) with A = NO2, B = F2, a= 2, and b=  apply.
[A]o = (2.00 mol)/(400 dm?) = 0.00500 mol/dm? and [B]po = 0.00750 mol/dm?. (a[B]o — b[A]o)kt = [2(0.00750) — 0.00500](mol/dm*)(38 dm?/mols)(10.0 s) = 3.80. Using [B]/[B]o = 1  ba'[A]o/[B]o + ba '[A]/[B]o, we have for Eq.
(17.22): 3.80 = In {(0.667 + [66.7 dm*/mol][A])/(200 dm*/mol)[A]} and e°®° = 0.00333/(dm*/mol)[A] + 0.333. We find [A] = 7.5 x 10° mol/dm> = [NO}]. Then na = (400 dm*)(7.5 x 10° mol/dm’*) = 0.0300 mol NO. Let 2z moles of NO; react in 10 s. Then 2 — 2z = 0.0300 and z = 0.985.
Then ng, = 3—z= 2.015 moles and nyo,p = 2z = 1.97 mol.
(b)
r = k[NO>][F2] = (38 dm*/mols)(0.00500 mol/dm*)(0.00750 mol/dm*) = 0.00142 mol/dm*s initially. After 10 s, part (a) gives [NO>] = 7.5 x 10>
mol/dm? and [F2] = (2.01; mol)/(400 dm*) = 5.04 x 107° mol/dm’. So r= (38 dm*/mols)(7.5 x 10° mol/dm*)(5.04 x 107 mol/dm’*) = 1.44 x 107 mol/dm?s.
232
17.19
(a)
As noted in Section 1.8, (d/dx) J S(x) dx = fix). Using this identity, we
have dy/dx = e"(dwldx)(J efx) dx +c) + eM Te"MAR X)]. Since w(x)
= J g(x) dx, we have dw/dx = g(x) and dy/dx = e"g(J efdx+c) +f, The night side of the differential equation is f+ gy = f+ ge"(J e"fdx +c), which is the same as dy/dx. Hence, y is the correct solution. (b)
In 19539) iweitake y =(B], x =7,/f{x) = RiAlpes , (x) = —k>. Then
w=] ky dt=IKot and y=[B] = e "(Je k[A]oe™' dt+c) = eo" {ki LA]oe2  1" Nk — ki) +c} = ki[Aloe ky — ky) + ce". To
evaluate c, we use the fact that [B] = 0 at r= 0; so0 = ki [A]o/(k2 — ki) +
and ¢ = —ky[A]o/(k2 — ki). Hence [B] = {kilA]o/(kz  ki) }(e™" — e"),
17.20
No. A kinetically reversible reaction is one where a significant amount of back reaction occurs. A thermodynamically reversible process must go through equilibrium states only, so a kinetically reversible reaction is a thermodynamically irreversible process.
17.21
bromiEqn@d: ld) sr kK{A]? integrates to [A] = [A]o/(1 + akt[A]o), and
r = k[A]° becomes r= k[A]¢/(1 + akt{A]o)”. 17.22
r= k{A]® =k =—d[A]/dt, which
(A)
integrates to [A] =—kt+c.
At t=10, [A]o=.0 +.c. So [A}=
[Alo
[A]o — kt. The graph is linear.
ez3
For n = 1, the integrated rate law is (17.30), for which [A] reaches 0 only at t = oo, For n # 1, the integrated rate law is (17.28). Setting [A] = 0 in (17.28), we see there are two cases to consider. If n < 1, then the left side of (17.28) is zero
when [A] = 0, and the time 7” required for [A] to reach 0 satisfies 0 = 1 +
[A]g(n— Lkat* and f° = 1/[A]§
‘(1 —n), which is positive, since n < 1.
Hence [A] reaches 0 in a finite time when n < . Forn> 1, the left side of
(17.28) becomes infinite when [A] equals 0 and an infinite amount of time is required for [A] to reach 0.
233
17.24
a! d{A\/dt = k{A]’[B]. Use of (17.19) gives J?TAT? ([Blo — ba'[A]o + ba [A] [A] =  \? ak dt. A table of integrals gives J [1/x*(p + sx)] dx =—l/px + (s/p’) In [(p + sx)/x]. We have x=[A],
p=[Blo ba'[A]o, and s = ba™', so
Blo —ba™'[A]p LORIN i
ba"!
 (Bly —ba [A] LAI!
[A]
([B]) —ba '[A]y)°
,
1 =—ak(t, —t,)
ae
a
soma
a[Bl)—b[Alp [Alo [A] [B]od[A]o
Me
[AVILA]
where we used (17.19) and took t; = 0 and f2 = 1.
17:25
17.26
(a)
(1.5)°(3) = 6.75.
(b)
(3[A]o)" = 27([A]o)” and n = 3.
In equation (17.28), let [A]/[A]o = a. Then Ce ale (Al ee (n — 1)Katg and ty == (a! ~"—1)/[A] ae (n — 1)k,y. Taking the log of each side, we get the desired result for n # 1. Forn = 1, we set [A]/[A]o = & in (17.30) to get
Q=e “kala and In a= —Katg.
17.27
For n # 1, Eq. (17.51) gives log 6 = log ((a! ~*_ 1)/(n — 1)]. For
a = 0.05 and n
= 0, log @ = log[(0.05 — 1)/(1)] = 0.0223; for n = 1/2, 3/2, 2, and 3, one finds log 6= 0.1911, 0.8416, 1.2788, and 2.2999, respectively. For n = 1, we have log o = log (In &) = log (In 0.05) = 0.4765.
17.28
(a)
We use the fractionallife method, plotting log ta vs. log [A}o. The data are log ta
Did
2525
2.954
31057
log [A]o —2.090
—2.191
—2.509
—2.726
The slope of the straight line through these points 1s 0.44 =  —n, so n= 1.44 = 1.5. The order is 3/2.
234
BQ. goneneene ge enn g ee YEH OMM2IA+ 1.8495 log (to/s)
B05
poe
Zui) 2.95 2.90 23 2,80) ore
Tae hae To die
eee
Ce See ee eee es f74 Pd Eje2 denote det oe ied=3Ee ed el aia een Mined Po  alate)
SOLGf IM eyMEPeiay SOR
Oe).
aay) a ae)
log {[A]o/(mol/L) } (b)
Putting [A]/[A]o = & = 0.69 and n = 1.5 in Eq. (17.28), we get ka = 0.408/1,[A] a . Substitution of the four pairs of tf and [A]o values gives
10°ka/(dm**/mol'*s) as 7.67, 7.64, 8.14, and 8.25. Averaging, we get ka = 7.9 x 10° dm*/mol’”s. 17.29
(a)
Both plots give pretty good fits to a straight line. The log [A] vs. t plot (corresponding to n = 1) gives a better fit, but considering the inaccuracy of kinetics data, one could not absolutely rule out n = 2 from the plot. n=1 Log({A]/c®)
SU
Rete 0.99989
Nes
1/(LA]/c?)
[
140
0
5
ORS
eiSige Ole eS t/min
(b)
ie.
Saas
Lege
0
aaa
5
I
Oa
R? = 0.997
SE A
ae
ee?
Oe)
t/min
A good fit is obtained for n = 1 and for n = 3/2; the fit for n = 2 is not
good. To decide between n =  and n = 3/2, one would need data at later times.
235
17.30
B = S20  and A = C3H7Br.
Equation (17.22) with a =  and b= 1 applies. Let
Then a[B]o — b[A]o = 57.1 mmol/dm’. We use [A] — [A]o = [B]  [B]o to calculate the [A] concentrations. A plot of In {({B]/[B]o)/[A]/[A]o)} =
In z vs. t is linear with slope (a[B]o — b[A]o)k. The data are
A t/s
0
0.104
0.189
0.474
= 1.035
1110
2010
5052
ie
The slope is 9.29 x 10° s“!, so k = (9.29 x 10° s“')/(0.0571 mol/dm’) = 1.61 x 10°? dm? mol! st. y = 9.21E05x + 2.89E03 1.2
ee poe
Le n
ana
asa
See
see
ay Sal) lS
alin eile Tie iee a
(lim thick sts Valle eI TR fi!
0
eS
2000
SS
Fe
4000
0
6000
a

IE es Ne
8000
ea
10000
ale
eel
12000
t/s
17.31
(a)
The reaction has the form 2A — B. We have P = Pa + Pg = (ca + Cp)RT. Let 2z mol/dm? of A react to form z mol/dm? of B. Then ca = Cao — 22 and cg = z= (cao — Ca). Hence P = (ca + Y2Ca0 — Y2Ca)RT =
Y2 (Cx +Cao)RT. Also, Po = Ca oRT, SO Ca = (2P — Po)/RT and & = Ca/cao = 2P/Po — 1, where Po = 632 torr. We calculate the & values and plot o
vs. log t; comparison with the Powellplot master curves shows the order is 2. Alternatively, we can use the fractionallife method: The calculated values of 100c,/(mol/dm*) are 1.0925 >a G 84560457403.
0.8929, 0.7349, 0.6252, and 0.4780. We plot ca vs. t and take a = 0.75.
From the graph, we find the times needed for
236
ca/(mol/L)
Gitte
Oo ON
a
a
natin ale oh Sled hark ratant
gla
0.004
estes gy
0
ae
Pee grease eg
2000
mig
4000
ecto
6000
et
pe
8000
a
10000 12000 t/s
100ca/(mol/dm*) to undergo the following changes: 1.6 to 1.2, 1.4 to 1.05, 1.2 to 0.90, 1.0 to 0.75, 0.8 to 0.6. Values of log fo.75 are
log to.75
3.13805
ate 04
log [100ca/(mol/dm’)]
0.204
0.146
log(to75/s)
0.079
0
3.467 —0.097
= 1.1008x + 3.3693
350
S710
eee jae oon.
FY
1 PS a ET
E(TOF0'05"
OE
0:00
0.059
a
0.108
OE
0:15
40200)
log[100c,4/(mol/L)]
where Ca is the initial SOM
(b)
A concentration. The plot has slope —1.1 = 1 —n,
25
Equation (17.16) applies. We plot I/ca = 1/[A] vs. t. The data are
(mol/dm*)/ca
t/s
59.11
64.29
69.66
74.34
85.15
0
367
731
1038
1751
237
ete.
(mol/dm*)/c,
0
2000
4000
6000
8000
10000
12000
t/s
The slope is 0.0141; dm’ mol s' = ka = ak = 2k, and k = 0.0071 dm? mol"! s1. 17.32
(a)
The first two columns of data show that tripling [A]o at constant [B]o and [C]o triples ro, so
@ = 1. The first and third columns show that tripling
[B]o at constant [A]o and [C]o multiplies the rate by 9, so B = 2. The
second and fourth columns show that y = 0. (b)
ro=k[A],[B] ro and use of the first column of data gives
k = (0.0060 c? s'/(0.20 c°)(0.30 c°)’ = 0.33 dm® mol™ s™. (c)
For the rate law (1) in Eq. (17.6), [HBr] is initially zero and initialrate
data would yield the erroneous result r = k[H2][Brz] ue
17233
In the first run, we have [A]o >> [B]o, so [A]o 1s essentially constant. The B concentrations are 0.400, 0.200, 0.100, and 0.050 mmol/dm? at 0, 120, 240,
and 360 s. The halflife is thus constant at 120 s; hence the order with respect to B is 1. In the second run [B]o 1s constant and the A concentrations are 0.400, 0.200, 0.100, and 0.050 mmol/dm’ at 0, 69, 208, and 485 ks. The halflives are
69, 139, and 277 ks. The halflife is doubled when the A concentration is cut in half, so t1/2 1s proportional to 1/[A] and [see Eq. (17.29)] the order with respect
to A is 2. Hence r = k[A]’[B]. On the first run, r = —d[B]/dt = k{A]?[B] = kp[B]. The reaction is pseudo first order and Eq. (17.14) gives [B] = [Blo eke.
we have kg = —f! In ([B]/[B]o) = (120 s)! In % = 0.00578 s™! = k(0.400 mol/dm?*)’ and k = 0.036 dm°/mol?s.
238
17.34
(a)
We plot a=[A]/[A]o vs. log t; comparison with the Powellplot master
curves shows the order is 3/2. (Alternatively, the fractionallife method can be used.)
(b)
From (17.28) with n = 1.5, a plot of ([A]o/[A])'” vs. tis linear with slope 1[A]o°k,. The data are (AISA) t/s 0
lems 100
1206) 200
294% 300
1399 400
1612) 600
201 1000
The slope is 0.00101 s' = 0.5(0.600 mol/dm’*)'7k, and ka = 0.0026, s! dm*? mol” = ak =k.
CN
y = 0.001013x + 0.997930
ae
oie oes oe a
DO ai
te
nad '
1.8 £ pone manne rern5y fo  ro= =~
yp eee lt
co eae ,
Le ee ee Fe a
: Soe ante 8 Se
Oe ee ieee
1.0 0
a
200
LA
400
cy
ee
600
800
eee
el
1000
1200
t/s
17.35
For the first run, [A]o >> [B]o, so [A] is essentially constant. A Powell plot of
a = [B]/[B]o vs. log t shows the order with respect to B is 2. (Alternatively, the fractionallife method can be used.) For the second run, [A] is also essentially constant. For runs  and 2, we have r; = —d[B]/dt = k[A] at [B]° =j,[B]’ and
ry = K{A] 02 (B} = jalB]’. Hence jy/j2 = kLA] 9, /KLA] 9.2 = ((A]o,1/[A]o.2)%. The pseudo order is 2; a plot of 1/[B] vs. tis linear with slopej.For run 1, such a
plot has slope 0.0119 dm? mol s“'; for run 2, the slope is 0.0067, dm? mol”! s7!. Then (0.800/0.600)* = 0.0119/0.0067; = 1.773 and o = (log 1.773)/(log 1.333) = 1.99 = 2. Also, j) = k[A] 9, andk = (0.0119 dm? mol! s~!)/(0.800 mol/dm?)” = 0.0186 dm” mol? s™'. (Data from the second run give 0.0186 dm’ mol? s”'.)
239
Run 1
1/ {[B]/(mol/L)}
y = 0.00671x + 502.61523
1200 ee
200
ee ee
0
25000
50000
Run 2
1/{[B]/(mol/L)}
y =0.01192x + 502.53457
eee
bees
Pepa eS reel
500
75000 100000
ee2 a ee
0
25000
50000
(a)
:
ee!
100000
t/s
t/s
17.36
75000
3
The regression line has slope 1.289 x 107 L mmo!
a
es
s”
and ka = 0.1285
Lmo!! s”.
(b)
The Solver gives 1.289 x 10“ L mmol"! s”.
(c)
The 1/{A] vs. t regression line has slope ka = 1.3647 x 107 L mmot!!
min”! = (1.3647 x 107)(1000/60) L mol s"' = 2.2745 x 10° L mol! s™. Also (Sec. 17.3) k = ka/3 = 7.58 x 107 L mol! s”'. For a leastsquares fit
of the [A] vs. t data, the Solver gives ka = 1.2901 x 107 L mmol! min"!
= 2.150 x10~ L mol! s7!, and k = ka/3 = 7.17 x 10° L mol s"' 1s 4, (a) T;
17.38
(b) T.
For an elementary reaction, k;/k, = K. We have —RT In Kp = AG®° = [2(51.31) — 97.89] kJ/mol = 4730 J/mol. Then In K, =1.908 and K;, =
0.148. From Eq. (6.25), K% = K® RT/(bardm*/mol) and K° = 0.148(bardm*/mol)/(0.08314 dm?bar/molK)(298.1 K) = 0.00597; so K, =
0.00597 mol/dm?. Then ky = ky/K; = (4.8 x 10* s')/(0.00597 mol/dm*) = 8.0 x 10° dm? mol"! s™!.
17239
From (17.4), r= —d[A]/dt = ¥2 d[C]/dt, so d[A]/dt = —r and d[C]/dt = 2r. From (17.5) for this elementary reaction, r = k[A][B]. Hence d[A]/dt = —k[A][B] and a[(C]/dt = 2k[A][B]. 240
17.40
(a) T;
17.41
The charge is not balanced.
17.42
k, and k> have different dimensions (units) and can’t be compared with each
(b) F.
other.
17.43
The rds composition is given by rule 1 as HyNO2Br. One possibility is Hebe
wes
(rapid equilib.), HBr + HNO2 — ONBr + H20
(slow rds),
ONBr + Ce6HsNH2 > CeHsN3 +H20+Br.
17.44
(a)
For runs  and 3, the CIO” initial concentration is cut in half while the other initial concentrations are kept constant; since the rate is cut in half,
we conclude that the order with respect to CIO is 1. Similarly, runs 2 and 3 tell us that the order with respect to I is 1. For runs 3 and 4, the OH initial concentration is multiplied by 4 and the rate is quadrupled; so the order with respect to OH is —1. Thus r= k{IJ[ClO )/[OH J. For run 1, 0.00048 c°/s = k(0.00200 c°)(0.00400 c°)/(1.000 c°) and k = 60 can
(b)
We assume a mechanism with a ratedetermining step (rds). From rules 1b and 2 in Sec. 17.6, the total composition of the rds reactants is fe Olt
OH
oro
ClO
> =; ana inespecics Olims a product
in an equilibrium step that precedes the rds and OH’ does not appear in the rds. To keep 2x — 1 nonnegative, x must be  or greater. The simplest assumption is x = 1. This gives the rds composition as CIIOH .A plausible mechanism that meets the preceding two requirements and that
gives the proper stoichiometry is OCI + H2O — OH +HOCI (rapid equilib.); HOC] +I — HOI+ CT (rds); HOI+ OH — OT + H20 (rapid).
17.45
The ratedeterminingstep reactants’ composition is NO2CI. One plausible > NO; + Cl (slow), Cl + NO2Cl + NO) + Cl (rapid). NO); + Cl (slow), Cl + Cl > Cl (rapid), with Another possibility is NOzCl
mechanism is NO,CI
stoichiometric numbers 2 and 1, respectively, for steps 1 and 2.
241
17.46
The ratedeterminingsteps reactants’ composition is CrT1°*. One possibility is
Ons Tl* S Gr Tl slow), Il Grams Gr + ble (rapid)sAnother possibility is Cr°* + TP* 2 CrTI* (equilib.), CrTP°* 3 Cr°* + TI** (slow), erapid), Teer One 17.47
The ratedeterminingstep reactants’ composition is NO2F. A possible mechanism is NO> + F2 > NO2F + F (slow),
F + NOz
— NOszF (rapid).
Another possibility is NO2 + F2 = NOsE (equilib.), NO2F2
— NO2F + F
(slow), F + NO2 — NO%F (rapid).
17.48
XeF,
17.49
The ratedeterminingstep reactants have composition N2Os. A likely mechanism is the slow step N20; NO» + NO; followed by a series of rapid
+ NO
> XeF; + NOF.
steps that yield the correct stoichiometry. One of the many possibilities for this series of rapid steps is NO2 + ClO
NO;3Cl
+O,
O+0O—
+ NOC] + OCI,
OC] + NO3 —
Og. (The stoichiometric number of all steps but the last
is 2.)
17.50
The ratedeterminingstep reactants have overall composition HeTl**, and Hg* is a product in an equilibrium that precedes the ratedetermining step. Two
other mechanisms besides that given in Example 17.1 are: Hg3* + TI*
HgTI** + Hg” (rapid equilib.), HgTI°* > Hg** + TI* (slow); and Hg3* + TI**
— Hg* + TI* + Hg” (rapid equilib.), Hg* + TI’* > Hg”* + TI* (slow). 17.51
The reverse reaction would then proceed by the onestep mechanism N2 + 3H2 occur.
17.52
> 2NH3. But a tetramolecular elementary step is far too unlikely to
(a)
d[O2}/dt = 2k,[O][O3] + k\[O3][M] — k1[O2][O][M]. d[O3]/dt = —k,[O3][M] + k1[O2][O][M] — k2[O][03].
(b)
d[O}/dt = 0 = k\[O3][M] — k,[02][O][M] — k2[O][O3], so k,[O3][M] — k_,[O2][O][M] = k2{O][O3]. Substitution into the expressions
242
for d[O2)/dt and d[O3]/dt in (a) gives d[O2]/dt = 3k,[O][O3] and
d[O3]/dt = —2k2[O][O3].
(c)
From (b) we get [O] = k;[O3][M]/(A_1[O2][M] + k2[03]). We have
r =—Y2 d{O3]/dt = k2[O][O3] = kik2[O3]°[MJ/(K1[O2][M] + k2[Os]) = kik2[O3]}"/(k1[O2] + k2[O3])/[M]). Also, r = (1/3)d[O2)/dt = k2[O][Os] = etc:
(d)
If step 1 is in equilibrium, then k\/k_; = [O2][O][M]/[O3][M] and [O] =
k,[O3]/k_,;[O2]. As noted in the problem statement, r= + d(O)/dt =
ks[O][O3] = kiks[O3]°/k1[02].
17.53
(e)
If k2[O3]/[M] > k2[O3]), the second term in the denominator of the steadystate expression can be neglected, thereby giving the ratedeterminingstep expression.
(a)
d{NO]/dt = 0 = k,[NO2][NO3] — k[NO][NO3] and k.[NO][NO3] =
ky[NO2][NO3]. d[NO3]/dt = 0 = ka[N20s] — ka[NO2][NOs] kp[NO2][NO3] — k[NO][NO3] = ka[N2Os] — kKa[NO2][NO3] ky[NO2][NO3] — k,[NO2][NO3] and [NO3] = ka[N20s]/(ka + 2k»)[NOz]. Then r= $4[N20s)/dt = 3 (—ka[N2Os] + k_a[NO2][NO3]) =
5KalN20s] = 5 kakalN2Os)/ (kK_q + 2kp) = [kakp/(ka + 2k») [N205] = k[N2Os] and k = kgkp/(k_a + 2k»).
(b)
If step b is the ratedetermining step and step a is in equilibrium, then Kal kq = [NO2][NO3]/[N20s]. The rate of the reaction equals the rate of
the ratedetermining step b (since the stoichiometric number of step b is 1), so r= kp[NO2][NO3] = (kpka/k_a)[N20s].
(c)
If k_4 >> 2k,, then the steadystate rate law of part (a) reduces to the rate
determiningstep rate law of (b). (Of course, this is a necessary condition for the validity of the ratedeterminingstep approximation. )
(d)
From this problem, we see that k of the N»O5 decomposition is a function of ka, ka, and kp. Hence, it is clear that the mechanism of the reaction in
Prob. 17.49 starts off with steps a and b of the mechanism in Eq. (17.8). After step b, we need steps that give the correct stoichiometry. A possible mechanism is N2Os5 i NO; + NO; (rapid equilib.), NOz
+ NO3 >
NO + O> + NO; (slow, rate determining), NOz + ClxO — NO CI + OCI, OC] + NO > NO,Cl, NO2Cl + 02 — NO3Cl + O, O + O > Oj. (The
stoichiometric number of all steps but the last is 2.) 243
17.54
For (17.60), r= % d[NOz]/dt = k2[N202][O2]; the initial equilibnum gives
[N,O,}/[NO}? = K,.1 = ky/k; and [N02] = (ki/k.1)[NO]’, so r= (koky/k_;)[NO]*[O2]. For (17.61), r= % d[NOo}/dt = k2[NO3][NO] and k)/kK., = [NO3]/[NO][O9], so r= (koki/k1)[NO][O2]. For (17.62), r = k[NO}°[O2]. 17.55
The exactnecultaml.00 siisfAl= 1A loan = Gl.00 mol leven Gala amines — 0.860708 mol/L and is 0.637628 mol/L at 3.00 s. For this reaction g([{A]) in
(17.64) is g([A]) = —k[A]" = K[A] and [A]; = [A]o — k[A]o At. The first number in column A (the time column) is 0 and the next entry in column A is the first entry plus Ar. The first number in column B is  and the next entry contains the Euler formula. The results for 1 s and 3 s are 0.85873 and 0.63325 mol/L for At = 0.2 s and 0.85973 and 0.63546 for Ar = 0.1 s.
17.56
The exact value at 1.00 s is [Eq. (17.17)]
[A] =
(1.00 mol/L)/[1 + (0.15 L/mols)(1.00 s)(1.00 mol/L)] = 0.869565 and is 0.689655 at 3.00 s. If the Prob. 17.55 spreadsheet is properly set up, one need only change n from  to 2 to get the Euler results for this problem, which turn out to be 0.86627 and 0.68421 mol/L for At = 0.2 s and 0.86795 and 0.68697 mol/L for At = 0.1 s. 17.57
We have g[A] = A[A] in the modifiedEuler formulas. The top entries in the r and [A], columns are 0 and  respectively, and the top entry in the [A]n41/2 column is [A]1/2 = [A]o — k[A]o At/2. The n = 1 modifiedEuler results at 1 and
3 s are 0.860728 and 0.637672 mol/L for At = 0.2 s and 0.860713 and 0.637639 mol/L for At = 0.1 s. Comparison with the exact values in Prob. 17.55 shows the modifiedEuler results are quite good and much better than the Euler results.
17.59
(a) F; (b) F; (c) F.
17.60
k= Ae™*e!®T
k(T>)/k(T) = exp [(Eg/R\(T)! T 5'))
k720 = (0.0012 dm*/mols) exp er
alK 7 K ) 0.018 dm? mol! $7!
244
17.61
k= Ae *«'®T andink=In A — E,/RT. We plot In k vs. 1/T. The data are
i @eOay
ie
ay
a
KOE
Mast)
RSC
SOR
ee
kee.
The slope is 19500 K! = E,/(8.314 J/molK) and E, = 162 kJ/mol. The
intercept at 1/T =0 is 25.9 = In(A c° s);
A=7 x 10'° dm’ mo! s'. (The
intercept is calculated from the slope and one point on the graph.
y = 19458x + 24.943 Inte. Ss)
. bee
0.0014
0.0015
a
0.0016
0.0017 1/(T/K)
17.62
k= Ae **!®T  ky [kp = exp [(Eg/R)(WUT) — W/T2)], In (hy, /kp,)= (E,/R)(1/T; — 1/T>) = In (0.000030/0.0012) = [E,/(1.987 cal/molK)][1/(700 K) — 1/(629 K)] and Ey = 45.5 kcal/mol. ie
keealRT —
(0.0012 dm? mol"! s”') exp[(45500 cal/mol)/(1.987 cal/molK)(700 K)] = [5< 10!! dm? mol! s™.
17.63
k=
Ae &*'®"
As T ©, E,/RT goes to 0 and k goes to A. As T 4 &, the
collision rate goes to infinity and the fraction of collisions having at least the activation energy goes to 1, so the collisiontheory picture leads one to expect the rate to increase without limit as T — 9, rather than approaching an upper limit as predicted by the Arrhenius equation. 245
17.64
(a)
Ink=InAE,/RT. A plot of In N vs. 1/T (where N is the chirping rate)
has slope —E,/R. We have
IiVN ie Siisaemeed
83, ~ 4/60.
107T
03408
G5dme
44cm
ke.
The plot is linear with slope —6.48 x 10° K"! = E,/(1.987 cal/molK) and Exes) eto kcal/mol.
y = 6476.7x + 26.906 Sh In N
alan lee sie Tal lima ie 4.4
et
0.00335
0.00340
0.00345
0.00350
1/(T/K)
(b)
At 14.0°C, 10°/T = 3.4825 K”' and we read the value In N = 4.35 from the graph. Thus N = 77 per minute. The rule gives the Fahrenheit temperature as 40 + '4(77) = 59°F. Actually, 14.0°C is 57°F, so the
crickets are in error by 2°F. 17.65
(a)
A =2.05 x 10'* s! and E, = 24.65 kcal/mol.
(b)
k(0°C) = (2.05 x 10!? s!) x exp [(—24650 cal/mol)/(1.987 cal/molK)(273.15 K)] = 3.87 x 10°’ sl.
(c)
The units of k and the rate law in Eq. (17.6) show that the order is 1. Equations (17.15) and (17.11) give katy2 = 0.693 = akty/2 = 2kty2 and
t/2 = 0.693/2k. From the given k(T) expression, we find k(—50°C) =
1.47 x 107"! s“' and k(50°C) = 4.36 x 10~* s"!. We find 11/.(50°C) = 2.36 x 10'° s; ty2(0°C) = 8.95 x 10° s; and t4/2(50°C) = 795 s.
246
17.66
This fraction of collisions equals e*"’ = exp(—Nae/RT) =
exp[—(80000 J/mol)/(8.3145 J/molK)T] = exp[9622/(T/K)]. (a) exp(9622/300) =e?" =1.2x 10°"; (b) exp(9622/310) =e?! = 3.3 x 10°"; (c)
17.67
Cpe e887
alinti(Each 10°C increase nearly triples the fraction.)
From (17.43), the observed rate constant is k = kj + kz =
Aye @
+ A,e °**" From Eq. (17.68), Ea =RT d In dT =
RT°(d/dT) In (ky + ky)=RT? (ky + 2) \(dk\/dT + dk/dT)=
[RT M(ky + ko) [(Ea/RT)A 6 08” + (Eq o/RT)Are 88] = (Eaiky + Eg 2k2)i (ki ko). 17.68
AH 593 = (393.509 + 90.25 + 110.525 — 33.18) kJ/mol = —225.91 kJ/mol. Since An, = 0, we have AH® = AU®, and use of Eay— Ea» = AU*® gives
116 kJ/mol — E,, = —226 kJ/mol. So Eq», = 342 kJ/mol.
17.69
The equilibrium condition for step 1 gives k,/k_; = [C][D]/[A][B]. We have
= d[G]/dt = k[C]’ = ko(i [A] [BV/K1[D])°=A kok, [A] [B]/[D]’, so k = manke vand Ae ©eteen Ae 0) Ae —]
aa
a2 Be Ae eae Then
= 2E,) + Eg2 —2Ea,1 = (240 + 196 — 192) kJ/mol= 244 kJ/mol.
17.70
a7.71
= (Ea RXWT, WT2)
yy (ko/ky) =
(a)
klk, = AeFa! RT  Ag Fa! RN
(b)
(E,/R)(A/T, — 1/T>) and In 6.50 = [Eq/(8.314 J/molK)] x [(300.0 K)! — (310.0 K)'] and E, = 145 kJ/mol = 34.6 kcal/mol. ko/ky = exp [(19000 J)/(8.314 J/molK)}[(300 Ky '— (310 K)'] = 1.28.
(a)
The linear regression slope is —12459 K"! =E,/R, and E, = 24.76 kcal/mol. The intercept is 30.81, = In(A/s) and A = 2.42 x [Opec lhe
absolute percent errors range from  to 5% and the sum of the squares of
the deviations is 1.9 x 10’ s™. (b)
Since the sum of the squares of the deviations is so small, it is helpful to minimize 10!° times this sum, rather than the sum itself. One finds E, =
25424 cal/mol and A = 6.55 x 10'* s"! as the optimum values with a sum 247
of squares of deviations equal to 8.91 x 10°'°. The absolute percent error
is 0.1% at the highest T and about 10% at the three lowest T’s. 17.72
(a)
ForH2 +h — 2HI, AG‘4o) =—23.6 kJ/mol = —RT In Kp and Kp = 57.7 =
K,, since An = 0. Then KS = k/ky = 0.064/0.0012 = 53 = (58)'”, so Sie
(b) 17.73
K's = K, = ky/ky = 0.0025/0.000030 = 83.
kikp— Kon (a)
kr/kp = K. Let
r= kpZ be the rate law of the reverse reaction. At
equilibrium, the forward and reverse rates are equal, so kpZeq =
KIBO. jeg SO.se leatealey and Zen BLO) sen SOuge eal kig leq ii Ke BT SO ge i BiO. leis slew Duetetor k,Z = ky[Br ][SO7 ]° [H*V[SO3"]°. (b)
k/ky = K\?. From part (a), ke/kp = Zeq/[BrO 5JeqlSO 3” JeqlHJeq =
Ki? = [Br] ig,[SO4 Jeg IBrO31.¢ [SO ]2q, and k,Z = ol Bilwe (SO al:
17.74
1He(BrOs iSO:
aio
kplkp = K's and In ky — In ky = (1/s) In K,. Differentiation of this equation with
respect to T and use of Eq. (17.68) and the result of Prob. 6.19 gives
Eq/RT — Eqy/RT’ = (1/s)AU°IRT’ and Ey — Ea» = AU°s. IWATE)
We reverse the designations of forward and back reactions and use primes to denote the newly designated constants; i.e., ky =ikyand:k, = kealso. kK = 1/K,. Since the mechanism of the reverse reaction consists of the reverse of the mechanism of the forward reaction and (as noted on p. 546 of the text) the
reversereaction’s ratedetermining step is the reverse of that for the forward
reaction, we have s’ = s. Substitution into k/k, = K).” gives k,/ky = (1K; Nee rye and k,/k,
+ vis’ =(K.)°.Q.E.D.
248
17.76
Consider two chemical reactions
aA+bB > cC+dB
(), pP+qQ > rR+wW
(I)
Let us form a third reaction that is the sum of m times the first reaction and n times the second reaction:
(Hl) maA + mbB + npP + ngQ mcC +mdB+nrR+nwW The concentrationscale equilibrium constants of these reactions are
Ki = (C} (DIAN IBY’, Ku = [RYEW) "(PYLON
Km = (C)"(DI" [RI (W) "A BY PP"1Q)” (where all concentrations are equilibrium concentrations) and we see that Kw = K{' Kj,. A similar result holds for a reaction formed by multiplying more than two reactions by integers and adding them. Since the elementary reactions of a mechanism multiplied by their stoichiometric numbers and then added
yield the overall reaction, we have K, = K;!K3? K," = II; (K;)" , where K, is the equilibrium constant of the overall reaction and Kj, K2,, Km are
the equilibrium constants of the elementary reactions. But we know that for an elementary reaction K; = kj/k_; [Eq. (17.53)], so K = pike
1729
Step (b) has stoichiometric number s = 1, so k¢/ky = K i . Let r, = kpZ be the rate law of the reverse reaction. We know that the forward reaction has rate law rp= kf N2Os]. At equilibrium, k{N2Oseq = koZeq and Zeg/[N2Osleq = kp/ko =
K, = [NOo] 4,[O2] eq /IN2Os] 2q» 80 Po = keZ = ko[NO2]}"[O2}/[N205]. 17.78
1/kuni = kK1/kik2 + 1/[M]k1. Since PoV = NokT, we have [M] = Po/RT and VWkuni = K\/kik2 + RT/Pok\. A plot of 1/kuni vs. 1/Po is linear with slope RT/k, and intercept k_\/k\k2. The data are
Iki 10°/P)
10440 9.09
9620 4.74
9010 1.316
9260 2.58
s tom
1
The slope is 1.83 x 10° s torr = 240 s atm = (0.08206 dm’atm/molK) x
(743.1 K)/ky and k, = 0.253 dm? mol! s'. The intercept is RTTO ES A Hl Kinip2 SO Kuni pee = 1140 x 107 serand kay/k3 = (8770 s)(0.253 dm? mol"! s“') = 2220 dm*/mol.
249
ne ee
__y= 182844x+8772.2
W(Kyni/S”') 10000 9500 9000 Read liames 0e laghy Sahl [pa lieatgeeliont’
9599
Cees
0.000
0.002
tt
0.004
0.006
0.008
0.010
1/(P)/torr)
17.79
The B and C molecules are smaller than the A molecules and so undergo collisions with A less often than A molecules do.
17.80
d{Br2]/dt = —k\{[Bro][M] + k,[Br]?[M] — k3[H] [Bro]. d{Br]/dt = 2k,[Br2][M] — 2k_,[Br]?[M] — k2[Br][H2] + k2[HBr][H]+4[H] [Bro].
17.81
(a)
Step (1) is the initiation step, (2) and (3) are the propagation steps, and
(4) is the termination step.
(b)
The chainpropagating steps (2) and (3) occur many times for each occurrence of step (1) or step (4). The overall reaction is therefore the
sum of steps (2) and (3), namely, CH3CHO (c)
— CH, + CO.
We use the steadystate approximation for the intermediates CH; and
CH3CO: d[CH3]/dt = 0 = ki[CH3CO] — k.[CH3][CH3CHO] + k3[CH3CO] — 2k4[CH3]?. d[CH3CO}/dt = 0 = k2[CH3][CH3;CHO] — k3[CH3CO]. Addition of these two equations gives 0 = ki[CH3;CHO] —
2k4[CH3]° and [CH3] = (k;/2k4)![CH3;CHO]"”. Then r = d[CHy]/dt = ko[CH3][CH3CHO] = k2(ki/2k4)"”” [CH3CHO]*”. (Alternatively, we can find [CH3CO] and use r = d[CO]/dt.)
17.82
The reverse of step I has a nearzero E,, so E, of step I equals AU 7; From the
Appendix, AH; = 104 kcal/mol and AU; = 103 5 kcal/mol at 300 K. This 250
103+ kcal/mol E, is far higher than the 45 kcal/mol E, of Br;
+M
2Br + M [see the discussion after Eq. (17.95)], so the dissociation of Hz by M
can be neglected. Step II is the reverse of step 3 in (17.88). Appendix data give AH, = AU; =41 ; kcal/mol = E52
angie, 3 = Ea y=
421 kcal/mol (since E,.3 =  kcal/mol). Ea, is far higher than the 18 kcal/mol E, of Br + H,
~ HBr + H, so Br reacts preferentially with H2
rather than with HBr, and reaction II can be neglected. The atom combination
> HBr + M has E, 1 = O. Reaction 3 in (17.88), namely, H + Br, > HBr + Br, has E,3 =  kcal/mol, not much different from Eq; thus the rate constants k3 and ky are of the same order of magnitude. We have r3 = k3{H][Br2] and ny = kin{H][Br][M]. The concentration of the reactant Bry is
reaction
H + Br+ M
high and is of the same order of magnitude as the M concentration. The very low concentration of the reactive intermediate Br makes ry = (1.6 x 10'! dm*?/mol"”?s)(1.9 x 107° mol/dm*)'” x exp {[(18000 — 40600) cal/mol]}/(1.987 cal/molK)(298 K)} = 3 x 10° dm? mol! s"!. Then kp = Ae
17.85
Ps2
(a)
[Root] = (fki/k) 7)" = (0.008 mol/dm*)'”7(0.5 x 5 x 10° s7')'” /(2 x 10’ dm*/mols)"”* = 1.0 x 10°’ mol/dm’. (DP) = kp[M\/(fkik,)' [1] = (3 x 10° dm*/mols)(2 mol/dm°)/ (0.5 x 1000 dm?/mol:s”)'7(0.008 mol/dm?)'” = 3000. —d[M]/dt = kp (fkilky) [MI] = kp[M] [Rio] = (3000 dm*/mols)(2 mol/dm°) x (1.0 x 10°’ mol/dm?) = 0.0006 mol/dm?s. d[Po]/dt = ki[Rtor]” = (2 x 10’ dm?/mols)(1.0 x 107 mol/dm?)” = 2 x 107’ mol/dm*s.
(b)
When termination is by disproportionation, two polymer molecules (instead of one) are formed whenever two radicals combine. This doubles
d[PioJ/dt and hence cuts (DP) in half. Thus d[Pro/dt = 4 x 10° mol/dm?s and (DP) = 1500. The other quantities are unchanged.
17.86
The initiation reaction 2M — 2R contributes a term 2k,[M]° to —d[M]/dt and (17.97) is modified to —d[M]/dt = 2k,[M]° + k,[M][Rio?]. To apply the steadystate condition d[Rtot:]/dt = 0, we note that the initiation step has (d[Rtot:]/dt); =
2fki[M]°. So 0 = (d[RioV/dt); + (d[Riov Vdt), = 2fki[M}  2k,[Rro]” and [Riot] = (fki/k,)'""[M]. Substitution in the above —d[M]/dt equation gives —d[MJ/dt = [2k; + kp(fkilk,)?)[M]’. Use of the above [Rio] expression in (17.105) gives
d[Prol/dt = ki{Rro]” = fki{MI]°. Equation (17.104) becomes (DP) =
—(d{MV/dt)/d[Pros\/dt = [2ki + kp(fki/ ki)! Vp. 17.87
d[AJ\/dt = k{A] + kG
Let [A]eg and [C]eg be the equilibrium concentrations under the new conditions, and let x = [A]eg — [A]. Then dx/dt = —d[A]/dt. Since 2 moles of C are formed when  mole of A reacts, we have [C]eg — [C] = 2x.
Then —dx/dt = —ky({A]eq — x) + kx((C] 2, + 4x1Ceq + 4x°) =
252
—kp[Aleq + Kol] a + xkp(kplkp + 4[Cleq + 4x). At equilibrium, d[A]/dt = 0 and the first equation in this paragraph gives —k{Aleq + ko[C] Be = 0. Since the perturbation is small, [C] is close to [C]eq and we can neglect 4x in comparison
with 4[C]eg. We then have dx/dt = —xkp(k/kp + 4[Cleq) = ~t~'x, where t =
(ky + 4kp[C]eq) .Integration gives x = xoe* or [A] — [Aleq = ([A]o  Nee
17.88
(a) (b)
CH3;CH2CH3, CH3CH3, CH;CH»CH2CH;, and N2. CH3CH2CH;3 and N> (cage effect).
17.89
x Equation (17.112) gives kp = 27(6.02 x 107°/mol)(4 x 10° cm) Sap! (8.4 x TOs cm//s) = 1.3 x 10'? cm?/mols = 1.3 x 10'° dm? mol!
17.90
(a)
). Then From Eggs. (17.114) and (17.1 15), In kp = In T— In n + In (const.
Eq. (17.68) gives Eq = RT? d\n kp/dT= RT’ [l/T — (1/n)dn/aT] =
RT — (RT’/n)dn/aT. (b)
.023 K') Eq = (1.987 cal/molK)(298 K) — (1.987 cal/molK)(298 K)*(0 = 4.7 kcal/mol = 19 kJ/mol.
(Cc)
ar
ro
25
20.3
12.5
6.45
WiSIo
0.800
0.400
0.200
0.050
(bd) Ie
17.91)
(aie
17.92
The data give
dm’s/mmol dm*/mmol
y = 38.98x + 4.7321
1/{ro/(mmol/Ls)] 40 35 30 phs)
0
0.2
0.6
0.4 253
0.8
1
1/{[S]o/(mmolV/L) }
The slope is 39 s = Ky/k2[E]o and the intercept is 4.7 dm?s/mmol = 1/k2[E]o.
So ky = (2.8 x 10° mol/dm?)"'(4700 dm?s/mol)"! = 7.6 x 10° s' and Ky = (39 s)/(4.7 dm?s/mmol) = 8.3 x 107° mol/dm’.
17.93
Since [P] =~ 0 and since [ES] is assumed small, it follows that [S] = [S]o. From (17.122) with [S] = [S]o = Ku = (kK: + k2)/k; and [P] = 0, we have [ES}/[E]o =
(ky + ko/(kKy + ko + ky + ky) = %. From (17.125), ro = ka[E]oKm/2Kmu = k2[E]lo/2 = ro max/2, SINCE Tomax = Kal[E]o:
17.94
We plot log t1/2 vs. log Po. The data are log ti2
0.88;
0.56,
0.239
log Po
2.423
2.114
M .763
log(t;/min)
Ow
y = 0.9858x  1.5106
(CR
Lo
clFES RY) ES
LS
EES)
2
RAY SS
Ne
CE
VyWY a ER
eee
HU
VY
ees ae
US TW
ee
(SUE
0

ee)
log(Po/torr)
The slope is 0.99 = 1 —n andn=0.
17.95
The overall reaction is N2 + 3H2
2NH3. To produce this overall reaction
from the steps listed, we must multiply steps a, b, c, d, e and fby 1, 3, 2, 2, 2, and 2, respectively, and then add them. These are the stoichiometric numbers of steps a to f. The fact that the ratedetermining step probably has stoichiometric number s =  indicates that the ratedetermining step is probably step (a), Nz + 2*
17.96
— 2N*.
The rate of the bimolecular desorption reaction 2A(ads) > A2(g) is proportional to (n,/A) (na/A) and so is proportional to 64 ee
254
kgs, . The
rate rag; of the adsorption reaction A2(g) — 2A(ads) is proportional to the
A»surface collision rate, which is proportional to P. Since two adjacent vacant sites are required for A2 to be adsorbed dissociatively, the adsorption rate is also proportional to (1 — @,)(1 — 8,), the square of the fraction of vacant sites.
Hence, rags = Ka(1 — 9,)°P. SO rads = Faes and ka(1 — 9,)°P = kg, ork? (1 O4)P'7/k'!? = O,. Solving for 04, we get 0, = (kalka)?P'7/[1 + (kalka)'P'"), which is (13.38) with b = ka/ka. 17.97
(a)
We assume that each CO molecule occupies one adsorption site. (This isn’t always true.) The number of sites per cm’ is (2.3 x 10°’ mol) x
(6.02 x 107°/mol) = 1.33 x 10". (b)
The total number of sites is (1.38 x 10!°/cm7)(5.00 cm’) = 6.9 x 10'°. The number of occupied sites is (9.2 x 10'° mol)(6.02 x 10°7/mol) =
5.5 x 10'*. So 0 = (5.5 x 10'*)/(6.9 x 10'*) = 0.080. The adsorption rate
per unit area is r; = (5.5 x 10'*)/t(5.00 cm’) = (1.1 x 102 cm~)t! and the equation in Sec. 17.18 that defines s becomes s =
(1.1 x 10!“/em?)(2nMRT)'7/tPNa. We have tP = (0.43 x 10° torr  s) x (1 atm/760 torr)(101325 Pa/1 atm) = 5.7 x 10° Pasands=
(1.1 x 10'4 10¢ m™)[2m(0.028 kg/mol)(8.314 J/molK)(300 K)}""7/ (5.7 x 10° s N/m’)(6.02 x 10°°) = 0.67. Since 8 is close to 0, this is approximately So.
17.98 (a) (b)
/RT
ty. = 0.693/kges = 0.693 08"
MAges =
5.3.%10 s:
17.99 d=(2D1)'? =(2Dye
1)? =
dis 100"” = 10 times d at 1 s, which is 3.7 x 10° cm.
17.100 (a)
With Egaas = 0, Eq. (17.71) gives Eades = —AU jag = ATjas AH ays/RT IRT Ages
/Ag., = 0.693¢
tp = 0.693/kges = 0.693e"**"'
0 60fe0e
J/mol)/(8.314 J/molK)(300 S08
255
sr) oo
107’ S.
=
(b)
180s.
(c) 4.6x 10!’ s.
17.101 Let the halfreaction be M* + ne” = M* or its reverse (where z and z’ are the charges on the species M). Let B in Eq. (17.128) be the species e . Then
val=nand r,='val7'dnp/dt = (1/n) dn, /dt, where n_ (not to be
confused with n, the number of electrons in the halfreaction) is the number of
moles of electrons. Since the Faraday constant is the absolute value of the
charge per mol of electrons, we have
Q =Fn_ ,so dn__ /dt = F'dQ/dt = I/F and r, = (/AnF)I =jinF. 17.102 N = (0.000420 g)(1 mol/347 g)(6.022 x 107*/mol) = 7.29 x 10!” atoms of *°U. Then A = A/N = (9.88 x 10° s"')/(7.29 x 10'”) = 1.355 x 10° 71. ty. = 0.693/A = 0.693/(1.355 x 10°! s') = 5.11 x 10'* s = 1.62 x 10° yr. 17.103 (a)
(20.0 g)(1 mol/63.0 g)0.00200(6.022 x 107°/mol) = 3.82 x 107° atoms of 7H. Then A = AN = 0.693N/t/2 = 0.693(3.82 x 10°°)/ (12.4 x 365.25 x 24 x 60 x 60 s) = 6.77 x 10!' 571.
(b) A= Ape =Apen
= (6.77 x10! /s)270 9936 20124) — 4.79 x 10!! s1.
17.104 The rates of the two decay modes are (dN/dt),
= —A,N and (dN/dt)> = A2N.
Since dNior = (dN); + (dN)2, the total decay rate is the sum of the contributions
of the two modes: (dN/dt) = (dN/dt), + (dN/dt)2 = —(A, + A2)N = AN. Then Ati = 0.6931 and ty2 = 0.6931/A = 0.693 1/(A; + Ad).
17.105 (a)
(1.00 g)(1 mol/12.01 g)(6.022 x 107*/mol) = 5.01 x 107” atoms in one gram of carbon. Then N = A/A = At)/2/0.693 =
(12.5 min™')(5730 x 365.25 x 24 x 60 min)/0.693 = 5.43 x 10!° atoms of C. So, % '4C = 100(5.43 x 10!%)/(5.01 x 1077) = 1.08 x 107!° %. (b)
A =
Age
CAre
eee
(c)
0.0296 min! (g Cy)". 7.0 = 12.5e°° 079%.
sat [12.5 min”!
(g Gye
ce
ee)
In (7.0/12.5) = 0.693/(5730 yr) and
t = 4800 yr.
256
sm
17.106 Let A = 2°U and B = 25U. We have Na = Naoe **" and Ng = Ngoe **’. We want Nao = Ngo at t= 0. So Na/Np = exp[(Ap — Aa)t] = exp[0.6931(1/ti/2.8 — Wth/2,)]. So In (99.28/0.72) =
0.693z[1/(7.0 x 10° yr) — 1/(4.51 x 10° yr)] and 1 = 5.9 x 10” yr. (In 2)(—1)t/t 17.107 A = Age™ and A = 0.693/ty/2 = (In 2)/ti2, 80 A = Age
= Ay(s) 17.108 (a)
t/ty)2
1/2
=
Ao2
(I)t/t
1/2
"”.
Let the steps be called I, II, Ill, IV. Step II requires two *He nuclei, but only one >He is produced in step II. Hence the stoichiometric number of step II is 2; this requires that the stoichiometric number of step I be 2. The stoichiometric number of step IV is also 2, so as to get rid of the two positrons formed when step I is multiplied by 2. Multiplication of steps I, Il, IW, and IV by 2, 2, 1, and 2 gives as the overall reaction
4'H+2°e — jHet+ 2v+6y. (b)
AUm is 2.6 x 10'? J/mol. There are (3.9 x 107° J)/(2.6 x 10° J/mol) = 1.5 x 10'* moles of “He produced each second.
(c)
The number of neutrinos produced each second is AG
ayes 10'* mol)(6.02 x 107°/mol) = 1.8 x 10°°. The area of a sphere of
radius 1.5 x 10° km is 4n(1.5 x 10!! m)? = 2.8 x 10” m’. One cm? =
10~ m? and the number of neutrinos hitting a square centimeter of the
Beir ni
second isGksecl0 )1¢10s, ms )/(2 SO)
el Om
(which is a lot of neutrinos).
17.109 (a) 10;
17.110 (a)
(b) 5;
(c) 10;
(d) 6.
For '3CH4, the answers are 10;5; 10; 7.
= For this elementary reaction, r = kmax{ B][C] = —d[B]/dt = —(d/dt)(np/V)
—(d/dt)(Np/NaV) = —(1/V)(dNp/dt)/Na = Zac/Na and kmax = Zpc/Na[B][C]. At each collision, one B molecule disappears, and Zgc is the collision rate per unit volume, so Zgc = —(1/V)(dNp/dt).
(b)
= We have [B] = Np/VN and [C] = Nc/VNa, so (15.62) gives Kmax
= ZaclNa{B][C] = Na(rs + rc) (8RTm)'(M5 +My (6.02 x 1073/mol)(4 x 107° m)?[8(8.314 J/molK)(300 K)m]"* x
257,
[(0.030 kg/mol)! + (0.050 kg/mol)']'” = 1.8 x 108 m* s! mol"! = 1.8 x 10'! dm? mol"! $7. J = (1/v;) dnj/dt (e) T. From (17.2) and (4.97), = dé/dt. (f) F. See Eq. (17.77). (g) T. (h) F;e.g., see (17.60). (i) $i (j) T. (k) F. (1) F(r increases with tin an explosion). (m) F. (n) F.
17.111 (a) T.
(b) T.
(c) T.
(d) T.
258
Chapter 18 ay"
18.1
(a)
ve
dR/idv =0=  C2 So 3 = (AVmax/KT) eT
enan —1)
=xe/(e  1),where x =
hVmax/kT. Then 3e* — 3 = xe‘ and multiplication by e“ gives x + 3e™ = 3.
(b)
For x = 0, 1, 2, 3, the function x + 3e” equals 3, 2.104, 2.406, and 3.149. So the nonzero root lies between 2 and 3. We have (3 — 2.406)/
(3.149 — 2.406) = 0.80, so interpolation gives x = 2.80. For x = 2.80, 2.81, 2.82, 2.83, we find x + 3e~* = 2.98243, 2.99061, 2.99882, 3.00704. The root lies between 2.82 and 2.83, and interpolation gives x = 2.8214. The
Excel Solver gives 2.82143937... . (Use Options to change the Precision
(c)
for Omme) At 300 K, Vmax = kTx/h = (1.3807 x 10°*° J/K)(300 K)(2.821)/ (6.626 x 104 Js) =1.76 x 10’? s’. At 3000 K, Vmax is 10 times as large, namely, 1.76 x 10'* s'. From Fig. 21.2, these frequencies lie in the infrared.
(d)
T = Vmaxxk= (6.626 x 107 J s)(3.5 x 10" s1)/2.821(1.38 x 107 J/K) = 6000 K.
18.2
(e)
Vmax = kTx/h = (1.38 x 107? J/K)(306 K)2.821/(6.626 x 10°" Js) = 1.80 x 10'° s'. Infrared.
(a)
The total emission per unit time and per unit area is Jo Rv) dv =
(2nhic?) J° [vi(e"*? — 1)] dv. Let z = Av/KT; then dz = (h/kT) dv. We
(b)
dz — have J® R(v) dv = (2mh/c*)(kTIh)* Jy ize = 8/15 = 2K TSC. (2mh/c2)(kTIh) The emission rate is (21K T*/15c7h?)(4nr’)= 9 8° (1.38110 —23 ULSY(5800 K)* Oligo Te ays Saou 15(2.998 x10° m/s)? (6.626 x10 **Js)’ (similar to the value given in Prob. 72LGS),
(c)
In 1 year, AE = (3.95 x 10° J/s)(365.25 x 24 x 60 x 60 s) = 1.25 x MO Ip SOAR = (1.2< 10 412.998 x 10° m/s) = 14% 10 kp.
259
18.3
(a)
hv=@+
i mv’ and hvin, = @. For K we have Vin, = ®/h =
(2.2 eV)(1.60 x 107!? J/eV)/(6.626 x 1074 J s) = 5.3 x 10'* s! and Aine = C/Vine = (3.0 x 10!° cm/s)/(5.3 x 10'4 5!) = 5.7 x 10° cm. For Ni we find Vin, = 1.2 x 10'° s7! and Ag = 2.5 x 10° cm. (b)
Kyes; Nino.
(c)
Lv’ = hv — © = (6.63 x 10°" J s)(3.00 x 10° m/s)/(4.00 x 1077 m) —
(2.2 eV)(1.60 x 107? eV) = 14x 10
J=0.9 eV.
18.4
E = hy = hcld = (6.626 x 1074 J s)(2.998 x 10° m/s)/(700 x 10” m) = 210 oo)
18.5
Ephoton = hv = held = (6.626 x10 J s)(2.998 x 10° m/s)/(590 x 10? m) = 3.37 x 10°!° J. Then 100 J/s = N(3.37 x 107"? J) and N = 2.97 x 10° photons/s.
18.6
hv=O+ Lmy* = ® + Kmax and Kmax = hv —®.
[0 aKa eresoes A le 256 10°\4v/s"!
91593,
el. OS
OS
GISi213. 18 7.408!
= 8490
where we used v = c/X. The slope is 6.53 x ioe ergS=h. @®can be found from the graph as the value of hv at Kmax = 0 or as the negative of the intercept at v = 0. We find ® = 2.8; x 10°” erg = 1.8 eV. y = 0.652x  2.840
eo
10'*Kmay/ergs
VE ES
4
5
a
6
a
iy
Se
a
8
a
ee
ae
9
10°'4v/s"!
260
10
18.7
18.8
(a) (b)
A=hA/mv= (6.626 x 10° J s\/(1.67 x 107’ kg)(6.0 x 10° m/s) = 6.6 x 10°? m. 2A=(6.626 x 10°* J s)/(0.050 kg)(1.20 m/s) = 1.1 x 10° m.
(a)
sinO=A/w anddA =h/m», so sin 9 = h/mvw =
6.26x10°** J s
):
(9.11x1077! kg)(6.0x10° m/s)(2400x10'° m) i 5.05 x 10 and 0 = 0.0289° = 5.05 x 10~ rad. (b)
Let z be the width of the central maximum. Figure 18.4 gives tan 0 = PE
(c)
18.9
DE
52/(40 cm) and z = 2(40 cm) tan 0.0289° = 0.040 cm.
Ap, = 2h/w = 2(6.63 x 10°" J s)/(2400 x 107° m) = 5.5 x 10°’ kg m/s.
AxAp, > hand Ap, > A/Ax = (6.6 x 104 J s)/(1 x 107"? m) = ~
Gx
ie
kgm s!. We have Ap, = A(mv,) = m Av, and Av, = Ap,/m 2
(6.6x 10g kgm sW(9.1 x 105) ke} = 9x 10° m/s, which is very large.
18.10
The time ¢ and the 9 spatial coordinates x,, yj), Z),X2» V2» Zz, 3, V3, 23 Of the three particles.
ren(c) L, (dye
13:11
(a) Fb)
18.12
z? =z2* and z = (zz*)!”.
18.13
ie) ie (ty: 7.
(a)
Bl2 l= 2
(by
B20
(c)
cos 6 +7sin 0  = [(cos 8 + i sin 8)(cos 8 —7 sin 9)!" =
(d)
(cos’0 + sin?@)! = 1'*=1. [23274  ua (ex tae Cer uO ee a (Oey? ay
1B=2nGa24So0'+4
Pe ise
Let Isis19and rsjg19denote the left side and right side of (18.10), respectively. We have Is}g10= rSig.10 10 see if c'¥ is a solution of (18.10), we replace ‘¥ in
(18.10) by c¥ and see if (18.10) is satisfied. With c'¥ as the proposed solution, the left side of (18.10) becomes (—h /i)(d/dt)(c¥) = c(h /i)(O'¥/dt) = c 1818.10. 261
The night side becomes
~(h?/2m)[0(CPVOx? ++
J— + (h7/2m,)[0?(CP)lox7 +++  1+ Ve =
CTS} 19. Since V is a solution of (18.10), we have Isig10= rSig.10. Then c ISig.10 = C FSig19 and so c® satisfies (18.10).
18.14
f° [4 (1ANe(Oyn(o) dr] dO do = J? [4g@)A(O)LS, Ar) dr] dO do = Ji,f(r) dr x J? [J e(@)h(o) dO] do = J‘ fir) dr J4(8) dO J? h(o) do
18.15
The result of Prob. 18.14 with r, 8, @ replaced by x, y, z, gives Wale Iga (2 fc?"
eat
e728 1" (2imc2)"/2 e729 1° (2 Ic? )"2 @2e° 10° dx dy dz =
(2 Imc?)!!2 fe, ee gilts dx (2inc?)'!? J~., eV le? dy. (2/mc?)!? ha popes AES
Use of integrals 1 (with n = 0) and 2 in Table 15.1 gives (2Imc2)'"? J, ernie Pe NO OLE Be ice kee 2(2htc? yn
11 2\(C 12)" 7 = . By symmetry the y and z integrals also equal 1
and V is normalized. 18.16
The 9 spatial coordinates x,, ¥,, Z}, X95 Yo» Za» 3» Y3> 23 Of the three particles.
18.17
(a) I; (b) F; (c) T, (d) T.
18.18
The timedependent Schrodinger equation is more general, since the timeindependent equation applies only to stationary states.
18.19
Let f= fi + if) and g = g; + ig2, wheref) 1s the real part of f,andfo is the
coefficient of the imaginary part of f.Then (fg)* = [(fi + i/2)(g1 + ig2)]* =
[igi —frg2 + i(fogi + fig2))* =figi —frg2 — igi + fig2). Also, f*g* = (fi + if2)*(e1 + ig2)* = (fi — tf2)(g1 — 182) =figi —fog2 — Wf2e1 + fig2) = (fe)*. 18.20
Let Isig.24 and rs;g.24 denote the left and right side of (18.24). To see if ky is a
solution of (18.24) we replace y in (18.24) by kw and see if (18.24) is satisfied.
With ky as the proposed solution, the left side of (18.24) becomes
(h* /2m,)[0? (ky)/Ox) +J— + Vik) = 262
k[ (—h*/2m, \(0°w/dx; +:)+Vy)]
= k sig.24. The right side becomes
Eky = k(Ew) = k rsig.24. Equation (18.24) is Isig.24 = rsig.24 and multiplication
by k gives k sig24 = k rsjg.24, so ky is a solution of (18.24).
18.21 (a) Yes. The integral " e?” dx with a >0 is finite, as shown by integral 1 with n = 0 and integral 2 in Table 15.1.
(b) No. The integral of the square of this function is infinite.
(c) No. J®, x? dx =), x? dx +JE x? dx =—x' 2, x7' [P= 0(0) (d)
No. For x TA
p
+n° = 3,6, 9, 11, 12, 14, so there are 6 energy levels in the
given range.
(ears
(f)Prite)o Ge
18.33
(ajerra(b) Fs (c) F; (d)akay
18.34
(a) Yes;
18.35
Since the wave function is an eigenfunction of the Hamiltonian (energy) operator, we must get the eigenvalue 25h7/8ma’.
18.36
(a)
(b) no;
(c) yes;  (d) no.
AB (x) — BAf (x) = (d/dx”)Lxflx)]  x[(d"/dx")flx)] = (d/dx) (xf (x) + fx) — xf (20) = xf (x) + £0) + £0) — fC) = FO).
(b)
(A+ B)(e* +cos 2x) = (d2/dx? + e* +008 2x] = (Pidx)(e* + cos 2x) +x(e* +cos 2x) =d’(e* idx + d*(cos 2x)/dx + 2
Xe it x COS 2y= Je
18.37
2
2
+.4x7e)
2
2
ta
—4icos 2x + xe,
tt COS 2X
(a)
(
ip and(_i+)* are nonlinear; the others are linear.
(b)
A (f+g)=[(hA?/2m,)V? +V\(f+2)= [(h?/2m,)V? — (f+ 9) + Vf + g) = (h?/2m, Vi f(h?/2m, )V? g —+++ Vf+ Ve =[(h?/2m)V? —+ Vif+ [(h?/2m,)V?  + Vg = Hf + He , where the definition of the sum of operators and the fact
that (0°/dx? \(f+g) = 0°fldx; + 0°g/dx; were used. Similarly, one finds
H (cf) = cHf .So H is linear.
266
18.38
(a) When
B operates on g(x), it turns g into another function, which we shall
call f(x). When A operates on f(x), we get another function, so ABg(x) is a function.
18.39
18.40
(b) Operator.
(c) Function.
(d) Operator.
(a)
p. = [(h/i)(d/dx)} = (h7/i)(07/Ox’).
(b)
pt = [(A/i)(d/dz)}* = f4 04/02").
(e) Function.
(d*/dx*)(sin 3x) = (d/dx)(3 cos 3x) = —9(sin 3x), so sin 3x is an eigenfunction of
d°/dx? with eigenvalue —9.
(d*/dx’)(6 cos 4x) = 96 cos 4x = 16(6 cos 4x), so
6 cos 4x is an eigenfunction of d’/dx* with eigenvalue —16. (d’/dx’)(5x°) = 30x, which does not equal a constant times 5x°; so 5x° is not an eigenfunction of
Pid.
(@ldx?)x' = 2x? # (const.)x'. (d’/dx*)(Be™) = 75e™* = 25(3e™),
and the eigenvalue is 25.
18.41
(a)
(d*/dx’) In 2x = —1/x? # (const.)In 2x.
(p,)=J. w*p,w deaJ®, y *(h/i)(0/dx)w dx= (Ali) 9. w *(w/dx) dx + (hli) Jo w*(Ow/dx) dx + (h/i) J? w *(dw/dx) dx = (h/i) i? w*(dw/dx) dx = (h/i) (2/a)(nt/a) ip sin (nttx/a) cos (nttx/a) dx =
(2nnh/ia? (a/nt) > sin? (nmtx/a) 6 =O, since sin nt =O for 1,2,3,...
(b)
—
. (We used the fact that y* = 0 outside the box.)
(x) = [2 w*xy dx = J8 (2/a)'” sin (nmxla)x(2/a)"”
sin (nttx/a) dx =
(2/a) ikex sin? (nmtx/a) dx. A table of integrals gives J x sin’ cx dx = dg — (x/4c) sin 2cx  (1/8c’) COS ZEX SOY
=
(2/a)[ +.x°— (ax/4nm) sin (2nx/a)  (a?/8n?1) cos (2nnx/a)} : ~ (2/a)(4.a? — a?/8n°1 + a’/8n71) = a/2, since sin 2nt = 0 and cos 2nt = 1 sal 2s {OWE
(c)
ok
(x*) = iF wey dx = (2/a) Ik x’ sin? (nmtx/a) dx. A table of integrals gives  x sin? cx dx = x°/6 — (x7/4c  1/8c°) sin 2cx — (x/4c°) cos 2cx, so
(x?) = (2/a)7/6 — (ax’/4nt — a?/8n°n°) sin (2nmx/a) (a*x/4n°n’) cos (2nttx/a)] 3 = (2/a)(a°/6 — a /An?t) = a?/3 — a7/2n*n’.
267
18.42 The timeindependent Schrodinger equation Hy = Ey for (18.64) is Ey and (Ay + How ++ Hw) = Ew (1). w= ) (H, + A, ++ +H Taking y = fi(qi)f2(q2)    f(r), we have Aw = Ay titan flq2) : f(g) = (fa = ay) Afi, since H, involves only q;. Equation (1) becomes (fp   Af
file des Division by fiz * + A fe a f= + (fifse Po bafets gives (I/f;) Hifi + (/f) Hof t:+:+ (Uf) H,f=E (2). By the same kind of argument used after Eq. (18.39), each term on the left side of equation (2) must be a constant. Calling these constants E), E2,...,E,, we have (1/fi) Af S12) or Af = Efi, etc., and equation (2) gives FE, +E.+::+£,=E.
18.43 y = (2/a) sin (mymxj/a) sin (nymxx/a). E=n?h?/8mia? + n3h'/8moa°. (b) vee, quantum number.
18.44
(a) nu, frequency;
18.45
Viight = (Eupper — Elower/h = [(Dupper We 5 )AVose — (Viower + 5 )AVosc\/h =
(Dupper — Viower)Vose = (8 — 7)Vose = Vosc = 6.0 x 10, cue
18.46
Squaring the curves in Fig. 18.18, we get the following curves (note the unequal peak heights): y?
18.47
(a)
From Fig. 18.18, Wo is a maximum at x = 0; likewise, Wo is
at x = 0 and this is the most probable value of x.
268
amaximum
(b)
dy? /dx = 0 = (402/m)'2(2xe" — 2ox°e
+), sox= t1/a!”? =
+(h /2tvm)"?. (x = 0 is a minimum.)
18.48
adyoldx*
=
(o/n)!*(d?/dx)e
(wn)'*(—oe pur
ax2/2
=
(o/n)'*(d/dx)(axe
ax*/2
)=
2a axe ox?/2 )= (02x? — a) Wo =
(L6n4v?m?x/h? — 42°’vm/h) yo. Equation (18.73) gives k = 4r’v'm, so Lkx*yo = 2n’v*mx Wo. Then —(h 7/2m)(d’yo/dx’) + Lk Yo = (21° v?mx" + Lhv)yo+ 2m vmxWo= LhvYo = EoWo.
18.49
Fyn de = (403/m)!? Jere dx = 2(408/m)'? JF
Pe" dx =
2(403/n)!/?(270/7/2302) = 1, where we used integrals 1 and 3 (with n = 1) in Table 15.1. 18.50
(a)
(x) = J° w*xy dx = (O/T)
1/2 ie co
Me
d= 0, where integral 4 (with n
=) in Table 15.1 was used. This result is obvious from Fig. 18.18.
(b)
(x2) = [Pee de = (ovm)!? [ee
dx = 2(a/n)!? JP Pe dx =
2(ov/n)? (2107/23) = 1/2 = h/8n°vm, where we used integrals 1 and 3
in Table 15.1.
()
(p,) = Pow* bv de= (on)? [ee2(Al@dNe*7dx= (o/n)?( A 1i)(—a) J”, xe “ax" dy = 0) (from integral 4).
18.51
18.52
=K+V=1m(dddty + Lk? = Lim{(kim)'"A cos [(k/m)"7t + b]}* + kA? sin? [(k/m)'7t + b] = 1 kA*{cos" ((kim)71 + b] + sin? [((k/m)'"t + b]}
(a)
by
(b)
mad?xidt? = m(a’/dt){A sin [(k/m)!71 + b]} = —mA(K/m) sin [(k/m)'"t + B] =—k{A sin [(k/m)'t + b]} = —kx
(a) (b)
v = (1/2n)(k/m)!? and k = 41°v°m = 417°(2.4 s')°(0.045 kg) = 10.2 N/m. E= 1kA* =0.5(10.2 N/m)(0.04 m) = 0.0082 J=(v+ 1)hv,sov+ + = (0.0082 J)/(6.626 x 10°" J s)(2.4s"') =5.2 x 10° =v. 269
18.53
(a)
The Hamiltonian is the sum of three onedimensional harmonicoscillator
Hamiltonians, one for each coordinate; the separationofvariables theorem [Egs. (18.65) and (18.66)] gives the energy as the sum of three
onedimensionalharmonicoscillator energies: E=E,+ E, +E, = (vx + 2)hVy + (Vy + Ya)hvy + (v; + Y2)hv:., where
= OF EO Dy
Dy
U7 =O) ee eee ani On eters,
Vv, = (1/20)\(k/m)!2, vy = /2n)(kim)'?, v = (1/20)(k/m)'”, where m is the particle’s mass. (b)
18.54
The lowest level has v, = Vy = Vv; = 0 and E = ShWvx + Vy + V:).
H = (1/2u)(uv? + pos + uv?) + V4 (1/2M\(M'vy + M’v? +M’v;)=
V+luwoi tv; +v2)+ 1M(vy +0} +7). Using Eq. (18.77), we have v, = dxldt = dx,/dt — dx\/dt = Vx2 — Vx. Similarly, Vy = Vy2 — Vy,1 and VD: = Vz2 — Vz. Since X = (mx; + m2x2)/M, we have vx = dX/dt = [m(dx,/dt) + m2(dx2/dt)\/M = (my0x,1 + MyVx,2)/(m + m2); similar equations
hold for vy and vz. SoH = V+ 5[m,mz/(m, + m2)] X 9) 2 2 ergy BP oe (my +m )(m, + m2) ~ X (On SP
v2, + (m2 Wee,
2mymyW,1Vx2
MM2Vx,1Vx,2
V+ $(m +m)
=
+MZV_2 FMV,
9
+7) = aa
2 2 [(m, + m2z)mvy, + (mm, + m2)mqv., +°°° J =
V+ lomv%,2 + mv? 4 +md>, +mV>\, +mv2, +mW=,)= 2 2 V+ imyv; 2 + imy 3 =Vtm? v7 /2m +m35v5/2m2 = p;2 /2m, + p5/2m + V. (The dots indicate similar terms in y and z.)
18.55
(a) (b)
(c)
w= mymp/(m, + m2) = [(12.0 g/mol)/Na][(16.0 g/mol)/Na]/ [(28.0 g/mol)/Na] = (6.86 g/mol)/Na = 1.14 x 10> g. [= pd? =(1.14 x 10° kg)(1.13 x 10°'° m)’ = 1.45 x 10“ kg m’. Ep = JJ + 1)h7/21. hh? 121 = (6.626 x 10°" J s)?/807(1.45 x 10° kg m?) = 3.83 x 10°? J. For J=0, 1, 2, 3, we have Exot = 0, 7.66 x 10° J, 23.0 x 10°" J, 46.0 x 10°*° J, respectively. The levels are (2J + 1)fold degenerate, so the degeneracies are 1, 3, 5, 7.
(d)
ForJ=Oto 1, AE=7.66 x 107° J0=7.66x 10° J=hv= (6.626 x 104 J s)v and v = 1.16 x 10'' s"'. For J= 1 to 2, AE= (23.0 — 7.66)10° J = Av and v = 2.32 x 10's. 270
18.56
(a)
Let Nx(a—x) be the normalized function. So Ie [Nx(a — x)]*Nx(a — x) dx
=1andN = 1/[J%x*(a—x)? dx]'”. From Example 18.8, Jox°(a —x) dx = a°/30, so N = (30/a”)'”.
(Dem) = Gla) [xia xx ae de = 30/0") In (Gx ax ex) ax — (30/a°)(a'/5 — a'/3 + a'/7) = 30a7/105 = 2a°/7 = 0.2857a’. The true value is found by setting n = 1 in Prob. 18.41c to give (x?) =a7(1/3  1/2r’) = 0.2827a*. The error is 1.1%.
18.57
The value of k that minimizes the variational integral W satisfies OW/dk = 0.
We have 0W/dk = 0 = (h2/ma’)[(8k + L)/(2k — 1)  2(4K + K)/(2k  1)] = (h*/ma?)(8k° — 8k — 1)/(2k — 1)* and 8k* — 8k — 1 = 0. The solutions are k = 1.112372 and —0.112372. The negative value of k makes } = e at x = 0 and so
is rejected. For k = 1.112372, W= (h7/ma’)(4k° + k)/(2k — 1) = 4.94949 h*/ma
2
= 4.94949h7/47ma* = 0.125372h?/ma’ compared with the true value h7/8ma* = 0.125h*/ma*. The percent error is only 0.30%.
18.58
(a)
I o*o dx = [6x°(a—x)°x"(a — x) dx=
§2 (ax! — 4a°x? + 6a’x° — 4ax’ + x°) dx = a°/5 — 2a°/3 + 6a°/7 — 12 + a’/9 = a°/630. We have H 6 = (h7/2m)(d’/dx*)(x’a’ — 2ax? + x*) = ~(h7/2m)(2a’ — 12ax + 12x’). So J? o* H 6 dx = _(h7/m) lfx(a —x)"(a’ — 6ax + 6x’) dx = —(h?/m) Ik (ax
Sax +19ax, — 18ax° + 6x°) dx =
~(h2/m)(a’/3 — 2a’ + 19a’/5 — 3a’ + 6a7) = h°a"/105m = h’a'/420n'm. Then  o* Ho dx/J o*o dx = (h’a"/420n’m)(630/a’) = (3/20°)(h7/ma’) =
0.152h7/ma? ~ Egs. The true Egs is h?/8ma” = 0.125h"/ma’. The error is 22%. (b)
18.59
Itis discontinuous at x = a, since o = 0 outside the box.
H = H°+ 4H’, where H ° is the particleinabox Hamiltonian operator and
H’ =kx for0D™,c,W,,) where w, = 2'’* sin(nnx) . From Eq. (18.98), * iF (x* — x*)sin(ntx) dx. A table of integrals (or use of
the website integrals.wolfram.com or a calculator that can do symbolic integration) gives Jx? sin kxdx =k? (2—k?x”)coskx + 2k °xsin kx and
Jx? sin kxdx = k3 (6x —k?x?)coskx +k“ (3k?x° —6)sin kx. We get C, = —2"? (nn) 3[4(1)” + 2], where sinnm =0 and cosmm =(1)" were used.
So G = D™,(2) (nn) °[4(1)" + 2] sin nx. We set up a spreadsheet with x values going from 0   in steps of 0.02 in column A, the values of F at these
273
the points in column B, and the values of the first, second,..., fifth terms in series G in columns C, D, E, F, and G. In column H we sum the first three
terms of the series and in column I we sum the first 5 terms of the series. The
data in columns B, H, and I are graphed versus x on the same plot. The fiveterm sum gives a more accurate representation of F’ than the threeterm function. For example, some values are
 ee Se mnie a le  F [ies [00153 oos44 [loos [o.1as [0.126  0072 x
18.67
18.68
(a)
Since y’ dx isaprobability and probabilities are dimensionless, lw?
(b)
has units of length’ and w has units of length”. The SI units of y are m!”* for a oneparticle onedimensional system. lw? dx dy dz is dimensionless and y has units of length >”.
(c)
lw? dx, dy, dz, dx2 dyz dz2 is dimensionless and y has units of length”.
The blackbody function (18.2) depends on the combinations of constants
hic? and h/k. In 1900, c was known reasonably accurately, so by fitting the observed blackbody curves Planck obtained values for both h and k. Use of R= Nak then gave Na. Use of F = Nae then gave e.
18.69 (a) (b) (c) 18.70
E=n7h’/8ma’ and doubling a multiplies E by “4. E=J(J+1)A7/21 = JJ + I)h?/2ud’ and doubling d multiplies E by 4. E= hv =h(1/2n)(k/m)'” and doubling m multiplies E by 1/2.
(a) T. (b) T. The future state is predicted by integrating the timedependent Schrodinger equation. (c) T. (d) T. (e) F. “Sum” must be replaced by “product” to make the statement true.
274
(f) T.
(g) F.
(h) F.
(i) F.
Chapter 19 19°
(a) bab)
19.2.
The equation shows that 4m€ has the same units as Q1Q)/rV, which are C?7/(m J=C?N"! m™”’, since lJ=1N m.
19.3
c* = I/po€o, so 1/49 = ploc2/4n = (40. x 107 N sIC*)c7/41 = 10°C? N s? C?,
19.4
=
els (c)iT:
Q1Q2/4m€Eor.
(a)
V=(1.602 x 10°” C)*/4m(8.854 x 107!” C?/Nm’)(3.0 x 107! m) = Gal X10 J = (7.7 x 100.3) (MeVil.602 x10 2.1) = 4. grey
(b)
Let the electrons be numbered  and 2, let the proton be p and let e denote the proton charge. Then V = V;> + Vip + Vrp = (1/411€9) x
[(e)7/(3.0 x 107'° m) — e7/(4.0 x 107! m) — €/(5.0 x 107! m)] = [(1.602 x 107° C)7/4n(8.854 x 1071? C2/Nm?)](1.167 x 10° m7!) = E21 x 10 = (2, Fx 10° Te V/N602 < 10) Wyse ev. 19.5
Q/m decreases at high v due to the relativistic increase of mass with speed.
19.6
V=4nr°/3, 80 VauelVatom = Foy¢/P io, = (107? cm)3/(10° em)? = 1 x 1072.
19.7
Vm = M/p = (197 g/mol)/(19.3 g/cm’) = 10.2 cm*/mol. Vatom = (10.2 cm?/mol)/(6.02 x 107*/mol) = 1.69 x 102 cm? = ¢3 and €=2.57x 10° cm=2.6A.
19.8
(a) T; (b) T; (c) F, wis #0 atr=0.
(i) F. 19.9
(a) 8; (b) r
(Cc) o.
275
(d) T; (e) F; (f) F; (g) T:; (h) F:
19.10
(a) T; (b) T; () F
19.11
(a)”
“OFT; 2h
(b)
5, 4, 3, 2,1, 0, 1, 2, 3, 4, 5.
(a)
The only n=  state is lso, so the degeneracy is  (..e., nondegenerate), if
19.12
spin is not considered.
(b)
The states 2s, 2p;, 2po, 2p_; have the same energy, so the degeneracy 1s 4.
(c)
The states 3s, 3p1, 3po, 3p1, 3d2, 3d), 3do, 3d_, 3d_2 have the same
energy and the degeneracy is 9. (The general formula is n’.)
19.13
These are hydrogenlike species, so £ = —(Z7In*)(13.60 eV). The ionization energy IE is —E for the ground state, n = 1.
19.14
(a)
JIBS 2°(13.60 eV) = 54.4 eV and the ionization potential IP is 54.4 V.
(b)
IE= 37(13.60 eV) = 122.4 eV and IP = 122.4 V.
AE =(13.60 eV)(1.602 x 107! J/1 eV)(1/3” — 1/2”) = 3.026 x 10°”
J=hv
and v = (3.026 x 107! J)/(6.626 x 10°°4 J s) = 4.567 x 10'*/s. d= clv = (2.9979 x 10° m/s)/(4.567 x 10'*/s) = 6.564 x 10°’ m = 656.4 nm. ONS
For a hydrogen atom, ft = m.m,/(me + mp) =
(9.1095x1L0*! kg)(1.67265x10~’ kg) =9.1045x107*! kg (9.1095x10*!kg +1.67265x107’ kg) We have a = 4m€oh 7/e* = 4m1(8.854 x 10°!” C*/Nm’)(6.6262 x 107°" J  s)”/ 4n°(9.1045 x 10°! kg)(1.6022 x 10°” Cy’ = 5.295 x 107"! m = 0.5295 A. 19.16
This is a hydrogenlike species, so its energy levels are given by Eq. (19.14) as
E =—(Z’In’)[e?/(4m€o)2a] = (Z’/n’)[we*/2(4m€0)" h 7]. Let m, be the electron mass; the positron has mass m,. SO positronium = 7 :/(me + m,) = m,/2, as
compared with pw = m, for an H atom. Since E 1s proportional to pt, each positronium energy level is half the corresponding Hatom energy. The positronium ionization potential 1s thus *(13.6 V)=6.8 V.
276
19.17
Using Table 19.1 and Eq. (19.24), (r) = Jw*ry dt=1'(Z/ay x
Jo" I SG etre ""r* sin © dr dO do = (Z’Ima’) Jp re?” dr J" sin 0 dO J5™ do. A table of integrals gives J ce dz= e(2 /b —327/b* + 62/b* — 6/b*), sO
JS re?" dr = 6/(2Z/a)* = 3a*/8Z', since e“”" vanishes at r =o. (This result also follows from the definite integral JF z"e~’* dz = n!/b"*' found in
most tables.) Then (r) = (Z*/ma*)(3a*/8Z*)(2)(2m) = 3a/2Z. 19.18
e = 1 + ib + (i)°/2! + (i)°/3! + (10)*/4! + (iQY/5! ++ = 1 + id — 07/2! — ig°/3! + 0°/4! + i0°/5! 4 . Also, cos 6+ isin d=(1 — 07/2! + 94/4! —)+i(0 67/314 0/5!)= 1 + id — 67/2! — i9°/3! + 0°/41.4 9/5! + =e".
19.19
2p, = 2"'7(2p, + 2p1) = 2? (1/80?)(Z/ayrre "“r sin O (e® + e*). We have e° +e =cos O +i sin d + cos (d) + isin (0) = cos 6 +7 sind + cos oi sin @ = 2 cos 9, so 2p, = (217/23 n!”)(Zlayrre "rr sin 8 cos = n'(Z/2a)?xe "4, where Eq. (19.7) was used. Also, 2py = (2p; — 2p:liv2 = (27)77i)(1/820"”)(Zlayr?e @" r sin 0 (e’® — e"*). We have ee" Ove
19.20
(a) (b)
=cos 6 + isin b— (cos d —i sin o) = 2i sin , so 2Py = izlayice
a
rsin @ sino = mo (Zi2aye
vest
a,+ib; = az + ib2. We must have a; = az and b, = bp. (2ny ei" = (2m) tein + 27) (2m) Vein e2nmi 55  = o2tuni cos (27m) + i sin (27m), where (19.21) was used. Equating the real parts
and the imaginary parts of this last equation [as shown in part (a)], we have cos (27m) =  and sin (2mm) = 0. The cosine function equals  only
for angles of 0, +27, +4, +67, ... and the sine vanishes at these angles. Therefore 21071 = 0, +20, t4n,...
19.21
andm=0, 11,172.78.
Wo). has the form bze~”’; b and d are constants. Along the z axis, x = 0 = y and
ra(e+y +7)? = (27) =z. Thus W>,, = bze“""! along the z axis. Near z = 0, e4*! = 1 and Wop. = bz (a straight line through the origin). For large values
of z, the exponential causes y to fall to zero. Also, W>,_ Pelelce  along the z axis. lw? is parabolic near the origin and is positive for negative values of z. The graphs are: 277
2 Ilr
19.22
We have 0.9 = (1/32na°) J (2  rlay’r’e dr JF sin 0 dO J," db = e98!4(_p? /2@? — ra — 1 —r3,/8a"] + 1, where a table of integrals was used. Let v = r,/a. We must solve e(4v" +80+8+ v') = 0.8. Trial and error (or a spreadsheet Solver) gives v = 9.125 and r2; = 9.125a = 4.83 A.
19.23
de 2"V0x = (de ""“/ar)\(Orlax). We have r = (x? + y’ + 2’)'” and dr/dx = Gs ag BayObra die Se de" sox = (Zxlraye™". Pe Ol0x" =
(d/dx)(Zxe?""Ira) = Ze*"/ra — x{(0/dr)(Ze""/ra)\(Or/dx) = Ze" Ira + (PLP ayer" + (Zelarye 2". Similar equations hold for 0’/dy"
and 07/02 of e~”". Then (07/dx* + 07/dy" + O°/0z)e~" = Bo7em Gira aye DL ira jen + (Ze + y+ 2varje" = =376 "Ira + (Lila jew it (Zirayee haw Ze6 “Iran Z janes = ~2ue*Ze~""/4nepr hh? + (Z’/a*)e“", where (19.14) was used. The Hamiltonian
operator is (19.5) and H wy;=(h7/2p)(0°/dx° + 0°/dy" + 7/dz*)[t (Layee) (Ze7/Aneor)[1 (Zia?"") = (Ze*/4neor) mt 7(Z/ay7e "" — nt (Zlay?( h7/2p)[—2we?’Ze —"/4nerh? + (Zla)e"") = (h 27a
19.24
ee (Zlaye
oh A ~[(Z267/2(4nep)a][e2(Z/a)ye2e2""] = EW.
(V) = (Ze*/4neyr) =—(Ze7/4neo) J wt. r'wi; dt=
~(Z'e*l4repa’n) Jon [EIS ere?" r° sin 0 dr dO db = ~(Z'e7/4neoa’n) JF re?" dr J5 sin © dO J2" do. Using either the definite integral JF r"e”” dr =n'/b""' or the indefinite integral J re?” dr =
—e’'(rlb + 1/b’), we get JF re?“"" dr = a /4Z°. Then (V) = —(Z'e7/Aneoa’n)(a’/4Z’)2(2n) = Z’e”/4neoa for the ground state. 19.25
(a)
Wehave: (r) = J WS, T Wop, dt=
(2°/32na°) J5" Ji S5 re?" cos @ r re""* cos 0 r’ sin 8 dr dO do =
278
(2°/32na?) Je re" dr §2" do Jp cos’ @ sin @ d@. A table of definite integrals gives JF r"e”” dr= n\/b"*' for b > 0 and na lr re~ dr=5!a°/Z. Let
positive integer. So
t= cos 0; then dt = sin 0 dO and
J” cos’®@ sin 8 dO = J;! 1 dt = 2/3. So (r) = (Z/32ma’)(120a°/Z°)(2/3)(21) = SalZ. (b)
The 2p and 2p, orbitals have the same shape and the same size and differ only in spatial orientation. Since r does not depend on spatial orientation, (r) must be the same for the 2p, and 2p, states.
GS
Oy Saat
eee
lh Pe" dr Ji" cos’ dbx
if sin’ d@. The r integral was found in (a). A table of integrals gives
J cos’ do = to+ + sin 2 and J sin°@ d0== cos 0  i cos 8 sin’0 and we find (r),, = (Z/32na°)(120a°/Z°)(1)(4/3) = SalZ.
19.26
Equations ee =) and (19.19) give the groundstate Hatom radial distribution
re". The maximum is found by setting the function as Ryie r= 4(Z/ay derivative equal to zero: 0 = 4(Z/a) ‘(Ore ma (22Zr’la)e°“"") and r = a/Z. (The root
19.27
r=0 is a minimum.)
Letc = 2.00 A. To find the desired probability, we integrate y*w dt over the volume of a sphere of radius c. The angles go over their full ranges and r goes from 0 to c. Table 19.1 and Eq. (19.24) EN the probability as (1/ma*) x
jel ale en 7 sin
drdo dy =(1/na®) JSr°e?”* dr Jf sin 0 dO J5" do. The
radial integral has the same form as the radial nee in Prob. 19.22 except that r2, is replaced by c. So J6 re?"
dr=
oe
S (+ ac’ + lac+ ta)t+ 4a.
1 The 0 and 9 integrals are given in Prob. 19.22, and the akira probability iS
_%!4(22}a? + 2cla + 1) + 1. We have cla = (2.00 A)/(0.5295 A) = 3.777, and the probability is 1 — &7°7[2(3.777)° + 23.777) + 1] = 0.981.
19.28
[2"@? do = J3" @*@ do = (1/2) J" ein gin® dy = (1/2r0) J2" do= 1.
19290
Ra) el
Dye
rc) al
279
19.30
Let 0 be the angle between the positive z axis and an angularmomentum
vector. For m = +1 in Fig. 19.10, cos @ = L/L =
/./2h = 1/V2 =0.7071
and 0 = 45°. Form = 0, 0 = 90°. For m =1, 8 = 180° — 45° = 135°.
19.31
(a)
(b)
L=rp sin B, where f is the angle between r and p. For circular motion, the velocity vector v is perpendicular to the radius, and so is p =mv; thus B = 90° and sin B = 1. Since p = mv, we get L = mur. The L vector is perpendicular to both r and p and r and p lie in the plane of the circular motion. Hence L is perpendicular to the plane of the circular orbit.
19.32
19.33
(a)
From Sec. 19.4, L = [J+ )]'"h =0, since / = 0 for the Is state.
(b)
From Sec. 18.3, Bohr had L = mur = nh/2n = h/2r for the ground state. The Bohr theory had the wrong value of L].
+p)?A = (12)? = V2h = V2(6.626x10™ Js)/2n = IL] = TAS log On eis:
19.34
[S= [s(s + 1]? = [0.5(1.5)]'7(6.626 x 10°" J s)/2n = 9.13 x 10° Js.
19735
Let 0 be the angle between the z axis and a spin vector. For m, = +i, we have
cos 0 = }nl.{3/4h = 1/V3 = 0.57735 and 0 = 54.7°. For m, = — O'="180?
19.36
19.37
1 7 >
547 = 12553":
(a)
Electronic orbital angular momentum; L = [/(/ + Oa
(b)
z component of electronic orbital angular momentum; L; = mh.
(c)
Electronic spin angular momentum; S = [s(s + 1)]'7h.
(d)
z component of electronic spin angular momentum; S, = mh.
(a)
For s = 3/2, Eq. (19.29) gives m, = 3/2, 1/2, 1/2, and —3/2. The possible z components of the spin are m,h . The length of the spin vector is Sis
ln = sV15h . The possible orientations are 280
(b)
cos@= 1.5/5 15h = 0.7746 and 6 = 39.2°.
19.38
(a) SF: e(b) GF; (oc) 2; (d).T:
19.39
(a)
Neither, since f(2)g(1) # tf(1)g(2).
(b)
Symmetric, since g(2)g(1) = g(1)g(2).
(c)
Antisymmetric, since f{2)g(1) — g(2)fU) = —[f(1)2(2)  gC)f{2)].
(d)
Symmetric.
19.40
(e) Antisymmetric.
The true groundstate energy of He is 79.0 eV (Sec. 19.6). The variational
value —86.7 eV is less than the true E,,; this violates the variation theorem (18.86), so there must be an error in the calculation.
19.41
There is one electron, soS =s =
anid 2Sue lee
SO
(a) 2S 2c(b). ¢Bew (0) 97D: 19.42
28 +1=4,so S = 3/2. The code letter F means =s
19.43
(a)
(b) (c) 19.44
(a)
(b)
Total electronic orbital angular momentum; L = [L(L + Di
ne
Total electronic spin angular momentum, [S =[SO+ 17h. = Msh. z component of total electronic spin angular momentum; S.
= 2, so\LI= IL Dla G te D means Foeteeends , *P 3/7, *Py).
(c)
L=3.and S =2, so the levels are 2 Fed aay Mie Ey yey
(d)
L=2 and S = 1s; the levels are Dh, tase Dyk
282
19.49
Let the electrons be numbered 1, 2, and 3. The nuclear charge is 3e. As was done with He in Eq. (19.32), we use the electron mass m, in the Hamiltonian.
Then H =(h7/2m,)V? (h7/2m,)V3 = (h7/2m.)V 5— 3e7/4neor 3e7/Ameor2 — 3e7/A4n€or3 + e/AnEor\> + e’/Aneori3 + e/4n€orr, where r; is the
distance between electron  and the nucleus.
19.50
Let fand g denote the n =  and n = 2 spatial functions, i.e.,
f= (2/a)'” sin (mx/a) and g = (2/a)'” sin (2nx/a). With interelectronic repulsion ignored, the spatial wave function is a product of oneelectron spatial functions. Analogous to Eqs. (19.42) and (19.43), we form the linear
combinations 27"7[f{1)g(2) + A2)g(1)] and 27 [f(1)g(2) — fl2)g(1)] that don’t distinguish between the electrons. To satisfy the Pauli principle, the symmetric spatial function must be combined with the antisymmetric twoelectron spin function (19.38) and the antisymmetric spatial function must be combined with
one of the symmetric spin functions. The approximate wave functions are therefore
27(R1)g(2) + fl) 12" Lax) B)  B(L)a(2))
Ja 1)a(2) 2" [A 1)g(2) — fl2)g(1) 27" TA1)g(2) — fl2)g(D)IBO)B(2) 7” + BU) a(2)] [ae1)B(2) 211) 9(2) — A2)g(L))]2
The first wave function has S = 0. The second, third and fourth have S =  and have the same energy as one another (since they have the same spatial factor). According to Hund’s rule, the S =  functions lie lower.
19.51
The groundstate configuration 1s 1s?2s°. To make the approximate wave function antisymmetric, we use a Slater determinant. Analogous to Eq. (19.51), we have 2s5(1)Bd) 2s5(1)a(1) Is(DBC) Is()ad)
_
fls(2)a(2)
1s(2)B(2)
2s(2)0(2)
— 2s(2)B(2)
Mee
1s(3)a(3)
1s(3)B(3)
25(3)a(3)
25(3)B(3)
1s(4)a(4)
1s(4)B(4)
25(4)a(4)
25(4)B(4)
where N is a normalization constant (equal to 1/24 ).
19.52
H (1s), Li (152s), B (1s?2s?2p), C (1s2s°2p*), N (1s72s?2p), O (1s72s2p'), S #0 and F (1572s?2p”) all have one or more unpaired electrons and so have
283
and have paramagnetic ground states. He (157), Be (1572s°), and Ne (1572s?2p°)
have all electrons paired, have S = 0 and L = 0 and do not have paramagnetic ground states. (Ne has two 2p electrons with m = +1, two 2p electrons with m = 0, and two 2p electrons with m = —1, and so has total orbital angularmomentum quantum number L = 0.)
19.53
We want the energy needed for ;gAr'’* > Ar'**. The ion Ar'’* has one electron and so is a hydrogenlike species. From Eq. (19.18) with n = 1, the ionization
potential is (18)°(13.6 V) = 4406 V. 19.54
€ = —(Z2,.eff In*)(13.6 eV). (a)
In Li, the first ionization potential is for removal of a 2s electron,
so 5.4 eV = (Z>,/2”)(13.6 eV) and Zet, = 1.26. (b)
93eV= (Ze /2°)(13.66 eV) and Zerr = 1.65. (The increase over Li is due
to the poor screening of one 2s electron by the other.)
19.55
(a)
If s = 3/2, the m, values are 3/2, 1/2, 1/2, 3/2. For s = 3/2, the electrons
are still fermions and the Pauli exclusion principle still holds. The four values of m, mean that 4 electrons (instead of 2) can go in each orbital. The Ls, 2s, and 2p subshells would therefore hold 4, 4, and 12 electrons
(double their capacities for s = ¥2). The groundstate configurations are
15°, 1s*2s*2p, and 1s‘25'2p”. (b)
Fors = 1, the electrons would be bosons and there would be no
restriction on the number of electrons in a spinorbital. The groundstate configurations would therefore be Is*, 1s°, and Is'’.
19.56
(a)
The outer electron in K is further from the nucleus, so Na has the higher ionization potential.
(b)
The ineffective screening of one 4s electron by the other makes Zegr
greater in Ca than in K, so Ca has the higher ionization potential. (C))
Ch
(d)
Kr.
284
1957,
For Z = 10, the figure gives ./e/€,,
= 8, 2.2, and 1.¢. Since €4 = —13.6 eV, we
get €),=—870 eV, &; =66 eV, and €2, = —35 eV. Substitution in € =
—(Z24 In*)(13.6 eV) gives Zetts = 8, Zettas = 44, Zett.2p ~ 3.2. 19.58
(a) T;
19159
Nitrogen, with 3 unpaired electrons.
19.60
Ionization energy data in Sec. 19.8 show that AE = 5.1 eV for Na > Na’ +e,
(b) T.
so E(Na* +e) > E(Na) and E(Na’ + 2e) > E(Na+e ). Electron affinity data give AE =0.5 eV forNa+e — Na, so E(Na )< E(Na+ e ). The lowest
energy (most stable) system is Na ; the highestenergy system 1s Na’ +2e.
19.61
(a) Sr;
(b) F;
(ce) K;
(d) C;
(e) CI.
Cl and Ar are isoelectronic and the
higher Z in Ar means a smaller size.
19.62
d= es bigx, where gx = Nyexp(—Cir/ao). Replacement of Z by & in the Is
orbital in Table 19.1 gives Ny = 1 '(Cy/ao)””. So o = bigi + b2g2 ++» + bsgs = rao) 0.7681 ¥(1.417/a)*exp(1.417 r/ao) + 0.23310'7(2.377/ao)”exp(2.377 + 0.041107 '7(4.396/a9)*exp(—4.396r/ao) — rao) + 0.0020 "”(7.943/a9)”exp(7.943rlao). 0.0102 (6.527/ap)>exp(—6.527
19.63
e=ee ROO
1s(2)a(2)
eee

25(2)a(2)
2771 s(1)a(1)25(2)an(2) — 15(2)0(2)25(1)0(1)] = 15(1)2s(2) — 1s(2)2s(1)}o(1)@(2), which is the S = 1, Ms = 1 function in 21 Fig. 19.13. Replacement of & by f in the preceding equations shows that D4 equals the S = 1, Ms =—1 function. r=eae
Is(a(l) 2s(DBC) 1s(2)0(2) 2s(2)B(2)
211 s(1)2s(2)a(1)B(2) — 28(1)18(2)B(1)(2)]. Interchange of o and B in D> gives D3 = 2/7[1s(1)2s(2)B(1)a(2) — 2s(1)1s(2)ox(1)B(2)]. We have [ax 1)B(2) + B(L)aX(2)]  2s(1)1s(2)2"" x 22(D, + D3) = 27!?{ 1s(1)25(2)27 285
[B(1)a(2) + a(1)B(2)]}, which is the S = 1, M, = 0 function. Similarly,
ue)
— D3) is found to be the S = 0, Ms = 0 function.
19.64 [27 J" [2 P sin @dr dO do = J2" do J% sin @ dO Jé r° dr = 2n(2)(a'/3)= 4 na’.3 3
19.65
19.66
(a)
Eis proportional to a7a so Ey > ENa :
(b)
The ionization energy of K”.
(c)
The energylevel spacing for these oneelectron species is proportional to Z°, so V is proportional to Z’. Thus Viet > Vu and Ay >A 
(d)
These quantities are equal.
(a)
Particle in a box; rigid rotor.
(b)
Harmonic oscillator.
(c)
Hydrogenlike atom.
19.67 (a)
dt= nr’ = 210.0010 Ay’ = 4.19 x 10” AY. lw? = (/na*)e?"? = [2(0.5295 Ay =2.14A~.
? dt=9.0x 10”.
(b) fw)? = (/na%)e2 9 = 0.324. dt=4.19x 10° A’. (c)
lw? dt=1.4x 10°. w? =1.35 x 10% and y? dt=5.7 x 107".
19.68 From (19.25) and (19.19), this probability is Pr=R1,(r)°r° dr = (4a )e?"r
19.69
dr.
(a) (b) (c) (d)
Pr =4(0.5295 A) 2620 109295)(g 100 A)*(0.001 A) = 0.000185. 4(0.5295 A)*exp[2(0.500/0.5295)](0.500 A)?(0.001 A) = 0.00102. 0.00062. 4.2 x 10°’. (See also Fig. 19.8.)
(a)
dt=dxand«o with nz — n, odd are forn = 1 3 2, n=2—3,andn=3
— 4. We get 6.82 x 10s Hz, 1.14 x 10!° Hzaand
1.59 x 10'° Hz. 21.12
v=
(Eupper = Eiewern)/h = [ANupper(upper + 4) — Attiower(Miower + 4)\/h =
a[(Niower + 3)(Niower + 3 + 4) — Mower(Mower + 4)]/h = (ONtower + 21)a/h, where Tike ees:
21.13
3 De
Toe
Using u and @ for upper and lower, we have
v=h'A[K,(K, + 3)—K (K, + 3)) = An UK, + (K, +4)K (K, +3)] = 305
2Ah"(K, +2), where K, = 1, 2,3,... Then 60 GHz = 2Ah”'(4); A=7.5h GHz: v=(15 GHz)(K, +2). 135 GHz = (15 GHz)(K, +2) and Keats 21.14
T = 10%. For A = 0.1, T= 10°! = 0.79 and 21% is absorbed. ForA = 1, T = 0.10 and 90% is absorbed. For A = 10, T= 10°'° and 99.99999999% is absorbed.
21.15
aan © COnIV = PIREand T= B/ino =a ele (a) Pp, ¢/RT = (10° dm*/molcm)[{(10/760) atm](1.0 cm)/ (0.08206 dm?3atm/molK)(298 K) = 5.35. T= 10°°* = 4.2 x 10°. (b) T=10>°%=16x10™.
21.16
T = Dl/h.o = 108". (a) ecg’ = (150 dm*/molcm)(10~ mol/dm’)(1.0 cm) = 0.15 and T=10, °=0.71, (b) T=10'° =0.033.
21.17
A = log (Ip/I) = log (T') = log (1/0.083) = 1.08. c = (0.080 g/cm3)(1 mol/14600 g)(10° cm*/1 dm*) = 0.0055 mol/dm*. T= I/Ip = 10! and € = (1/cl) log T = (0.0055 mol/dm*) (0.010 cm)! log 0.083 = 2.0 x 10 dm? mol” cm".
21.18
A2/A, = [log (h.o/b)2\/[og Ch.o/D)1) = €2¢B.2!2/€.¢B 1h = Cp.2/Ccp.1 = 2. So log (h.o/I))2 = 2 log (1/0.60) = 0.444 and Kh o/I, = 2.78. We have
T = 1/2.78 = 0.36 and 36% of the light is transmitted.
21.19
Let the subscripts 3 and 4 denote Fe(CN) re and Fe(CN) ar respectively. Use
of A = (€3c3 + €4C4)l and c3 + c4 = 1.00 x 10°° mol/dm’ gives 0.701 = [(800 dm?/molcm)c3 + (320 dm*/molcm)(0.00100 mol/dm? — c3)](1.00 cm). We find c3 = 7.94 x 10“ mol/dm’. Then c3/c3.9 = 0.000794/0.00100 = 0.794. The % reacted is 20.6%.
306
21.20
(a) T; (b) T; (c) F; (d) F; () T; (f) T; (g) F; (hb) T.
21:21
(a) F:
“(b) T;°
(c) F.
21.22 Division of Eq. (21.25) by he gives Do/he = D,/he  19, +45 ,x. (a)
Do/he = [79890 — +(2359) + =(14)] cm! = 78714 cm; Do = (78714 cm™')(100 cm/m)(6.6261 x 10~** J s)(2.9979 x 10° m/s) = 1.5636 x 10°'® J = 9.759 eV.
(b)
Do/he = [90544  +(2170) + =(13)] cm”! = 89462 cm”; Do = 1.7771 x 107'8 J = 11.092 eV.
21.23
(a)
D, is the depth of the electronic energy curve E, and k, equals E, (R,). In the Born—Oppenheimer approximation, E,(R) is found by solving the electronic Schrodinger equation (20.7) in which the nuclei are fixed; the nuclear masses do not occur in (20.7) and (20.6). Hence, E,(R) is the
same for 7H*°Cl and 'H™Cl, which have the same nuclear charges. From Eq. (21.25), Do differs from D, by the zeropoint vibrational energy. The vibrational frequency v, equals (1/2m)(k/)"” * The reduced mass u differs
substantially for 7H*°Cl and 'H*°Cl, so their v,’s differ and their Do’s differ.
(b)
For 'H*°Cl, Do/hc = D/hc  V, + V,x, =
[37240 — +(2990.9) + 7(52:8)] em" =35758 em); Do = 7.103; x 107? J = 4.4334 eV. From Eqs. (21.23) and (18.79):
VepcvVeuc! = (MucYMpc)7; Hci = 1.00782(34.969)(g/mol)/35.977Na = (0.97959 g/mol)/Na; Mpc = (1.9044 g/mol)/Na. So V, pc =
(2990.9 cm™!)(0.97959/1.9044)'* = 2145.1 cm”. Also, De.pci = Dene. For “HCl we then have Dolhe = [37240 — 5(2145) + oa)Lcms = 36181 cm™! and Dp = 7.187 x 10°!” J = 4.486 eV (where we neglected the change 1n Vx).
21.24 VYhv, = YekA*. But v, = (1/2m)(k/p)'’, sok = 417pv2; “hv, = 2n’uv? A’ and A = (h/4ruv,)!””. For H°°Cl, p = (1.0)(35.0)g/(36.0)(6.02 x 107°) = 307
1.6 x 104 g and v, = (2.998 x 10!° cm/s)(2991 cm”) = 9.0 x 10" s', so A =[(6.63 x 1074 J s\/4n2(1.6 x 1077 kg)(9.0 x 10'/s)]'? = 1.1 x 107" m=
0.11 A. For '*Ny, p= 14(14)g/28Na = 1.16 x 10 g, v.=7.1 x 10° s™ and
A=0.045 A. 21.25
Use of (21.31) gives A = c/v = c/2(J + 1)B, so 2.00 cm = c/2(3)B and B = c/(12.0 cm). We then have A = c/2(J + 1)c(12.0 cm) = (6.0 cm)/(J + 1).
hoes? = (6.0 cm)/7 =.0.86 cm.
21.26
From Eqs. (21.31), (21.33), and (21.15): Vyos+1 = 2V + 1)Bo = 2Bo = 2h/8tIp= h/4r? wR? and Ro = (h/pv)'"/21. (a)
Use of the table of atomic masses gives Hh = mym)/(m, + m2) = 
(b)
(1.0078250)(78.91834)g/(79.926165)(6.02214 x 10°°) = 1.652431 x 104 ¢ for 'H’°Br and p = 1.652945 x 10 g for 'H®'Br. For 'H’’Br, Ro = [(6.62607 x 107° J s)/(1.652431 x 1077 kg)(500.7216 x 107/s)]'"/2n = 1.424257 A. For 'H®'Br, we get Ro = 1.424257 A. ForJ=1 to 2, v = 2( + 1)Bo = 4Bo = 2Vs041 = 2(500.7216 GHz) = 1001.4432 GHz with centrifugal distortion neglected.
(c)
For the J=0
to  transition, Vps/Vusr = 2Bo,pp/2Bo.HBr = LHBR/Mpsr, Since
Ro is essentially unchanged on isotopic substitution.
Upp; = (2.014102)(78.91834)g/(80.93244)(6.02214 x 10°*) = 3.261264 x 10°" g. UnBr/Mpsr = 1.652431/3.261264 = 0.506684 and Vpsr = 0.506684(500.7216 GHz) = 253.708 GHz. 21.27
From Eq. (21.31), Vsss41 = 2BoV + 1), so Bo?’ K*’Cl) = (22410 MHz)/2(3) = 3735 MHz. (a)
v= 2(3735 MHz)! = 7470 MHz.
(b)
The reasoning in Prob. 21.23a shows that R, for the two isotopic species is the same; further, Ro should differ only very slightly for the two species (see, for example, Prob. 21.26). Equations (21.33) and (21.15)
then give Bo?’ K*°C1)/Bo(?K*’Cl) = In??K* C/o? KCl) = w??K7Cl/uCPK Cl) = 18.9693/18.4292 = 1.02931, where the reduced masses were found from mamp/(ma + mp). Then BoC KRG) =
308
1.02931(3735 MHz) = 3844+MHz and vj9..; = 2(3844+ MHz) =
7689 MHz for *°K*°Cl. 21.28
The equation preceding (21.37) is ¥ = V origin + BU’ + 1) BS’ + 1). For P. branch lines, J =J” —1 and/(V +1)J 4g =e a
21.29
cae TG
ne 1) = a) Ie so
Vp =v
J
+ 1)=
origin — 2B ae
Eq. (21.39) multiplied out gives Vp = V origin + 2B, J” + 1) Gv’ (J? + 3S” +2) + &, 0" + I") — &, (J + 1). With centrifugal distortion neglected, Eq. (21.26) gives Vp = (Eysiy — Ey
Ve +2 a, (v’ o
)— Vixt0 +o) ys" cs Due
. 2\s
Gi. (WH EF #1) = 910
eB Poe
he =
2)
Vv,(0”  ‘) ‘2 Voxe(0" +
0) = V_2(0° £0 =
Dios BJS’
+ 1) fs
A=) +
B,(2J” +2)+ &, 0°77 +I") — & v(I + 3” + 2)  &, Ul” + 1), which Eq. (21.34) shows to be the same as the above multipliedout form of Eq. (21.39), verifying (21.39). When multiplied out (21.40) is
Vp =Voigin — 2B” + GU"  I) + GW" +I) + &, J". Then Fp = (Ey. — Eyyhe
and use of (21.26) leads to the multipliedout form of
(21.40).
21.30
ve = (1/2m)(ke/)!? and ke = 41°v 2p = 41°? V2p = 400°(2.998 x 10'° cm/s)” x (1580 cm™)°(15.99491) g/2(6.0221 x 107) = 1.176 x 10° dyn/cm = 11.76 mdyn/A = 1176 N/m, where we used LL = mama/(ma + ma) = ma/2.
21.31
(a)
From Eq. (21.34), Vorigin(O > LI = V, —2V,%e and Vorigin (0 — 2) = DUNE GY fxHence BV one, (0 => 1) Vote tO > aaa Vee
3(2886.0 cm”) — 5668.0 cm™ = 2990.0 cm'. Then V, xe = LV, = Vorigin (0 > 1} = 5(2990.0  2886.0)em™ = 52.0 cm”. (b)
From (21.34), Voigin (0 > 6) = 6(2990.0 cm) — 42(52.0 cm) = 15756 cm .
309
21.32
(a)
The distance of a line from the band origin is the change in rotational energy for the line's transition. The selection rule is AJ = +1 and the spacings between rotational levels increase as J increases. Hence the two lines closest to the band origin involve the J = 0 and 1 levels. The 2139 cm’ line is lower in frequency and energy than the band origin and so must be the J = 1 — 0 line, where the rotational energy is decreasing.
The 2147 cm”! line has J = 0 > 1. The 2151 cm" line has J = 1 > 2. The 2135.5 cm” line has J = 2 > 1.
(b)
Let a, b, c, d be the four given wavenumbers in order of increasing
se
epumber. From (21.40) and (21.39), we have
Be
(2B,
20)
en40 er
4B,
p= Vor (2B, — 26, )1&, =Voigin — 2B, + &, erat (2B 20g NaC d= Vonen
ea ee a 30.
(2B, —20,)246, = Vorigin + 4B, — 80,
Three spreadsheet cells are designated for Voisin» B,, and &,. The initial guess for Vici, Can be taken as the average of b and c, namely 2143.25 cm’'.The initial guesses for B, and @, can be taken as zero. The four formulas for a, b, c, and d are entered into four cells and the squares of the deviations of the a, b, c, and d formula values from the observed
values are calculated, and summed. The Solver is set up to minimize the
sum of the squares of the deviations by varying V origin
?
B, yagd.
subject to the constraints that B, and @, be positive. An excellent fit to
the observed lines is found with V,,,.;, ongin = 2143.2695 cm"!, B, = 1.93023
m',and @, = 0.016411. (c)
From (21.16) and (21.15), B, =h/8n* UR? and R, = (h/8n7pB,)'”?. 1 = [(12)(15.994915)/27.9949 15]9/(6.02214 x 107°) = 1.138500 x 10 g.
B, = B,c = (1.93023 cm ')(2.997925 x 10'° cm/s) = 5.78668 x 10!°s7!,
R. = [(6.62607 x10 J s)/8107(1.138500 x 10° kg)( 5.78668 x 10!° sy] = 1.12863 x 107'® m = 1.12863 A, which is smaller than Ro in Example Ba:
310
21.33
With centrifugal distortion neglected, the v = 0 vibrationrotation levels are
given by Eq. (21.26) as Eviprot = 5 Ve — Fhvexe + hB IJ + 1)  shad foi For
03 Evirot(O) =
5 Ve =
Nee
We
have Evibrot\) a E\ibrot(O) =
h(Be 5Oe)J(J + 1) =(B,  5&, acJ(J +1). (B, — 4G, helkT= [10.594 — (0.31)] cm” (100 cm/m)(6.6261 x 10°" J s)(2.9979 x 10° m/s)/ CE S50 77% 1Oe2 J/K)(300 K) = 0.050064. The degeneracy of each level is
2J + 1, so the Boltzmann distribution law gives the relative populations as
NjINo = (2J + Ve"? = (2 + 1) exp [(B,  + &, \acJJ + L/KT] = (Oy ll \camaeneds ae Werrind al N,/No
21.34
0 1
1 2 3 2.) 14 yr3a7 03 ae) 328395
4 5 hp3150d SeeUY
6 1.588
B,—rotational constant; o©,—vibrationrotation interaction; D—centrifugal distortion; v.x.—anharmonicity.
21.35
(a)
From Fig. 20.17 O2 has 4 more bonding electrons than antibonding
electrons, O53 has 5 net bonding electrons, and O; has 3 net bonding
electrons. Therefore O 5 has the strongest bond and the largest k, and O 5 has the smallest ke.
(b)
Use of Fig. 20.14 shows that N> has 6 net bonding electrons and N35 has 5 net bonding electrons, so N2 has the stronger bond, the larger k, and the larger Vv, since V, = (1/2)(ke/p)”?.
(c)
N>zhas a triple bond and O; a double bond. The N2 bond is stronger and Nz has the larger ke.
(d)
Ex = J(J + 1)A7/21, where
I= wR2. An Na atom is heavier and larger
than an Li atom, so Naz has the larger pt and the larger R,. So Exot jai 1S greater for Lip.
21.36
(a) T;
21.37
(a)
(b) T;
(c) T;
(d) F.
A C2 axis and two symmetry planes.
311
(b)
A C3 axis (through the CCI bond) and three symmetry planes (each one containing C, Cl, and one F).
(c)
The molecule is square planar. A C4 axis perpendicular to the molecular plane; an Sq axis and a C) axis, each coincident with the C4 axis; a
symmetry plane coincident with the molecular plane; four symmetry planes perpendicular to the molecular plane; four C2 axes in the molecular plane (two pass through pairs of opposite F’s and two lie between the F’s); a center of symmetry.
(d)
The structure is trigonal bipyramidal. A C3 axis; an S3 axis coincident with the C3 axis; a (horizontal) plane of symmetry containing the
equatorial Cl’s; three planes of symmetry that each contain the two axial Cl’s; three C2 axes, each lying in the horizontal symmetry plane.
21.38
(e)
The VSEPR method shows the structure is a squarebased pyramid. A C4 axis and four symmetry planes.
(f)
A C? axis perpendicular to the molecular plane; a center of symmetry; two C2 axes in the molecular plane; three symmetry planes—one coincident with the molecular plane and two perpendicular to it. See ProbsZ1.3 9c.
(g)
A C.. axis through the nuclei and an infinite number of symmetry planes that each contain the C... axis.
(h)
A C.. axis (which is also an S.. axis), an infinite number of symmetry planes through this axis, a symmetry plane perpendicular to this axis, a center of symmetry, an infinite number of C2 axes perpendicular to the molecular axis.
(a)
The symmetry elements (Prob. 21.37a) are a C2 axis and two symmetry planes, which we call 6, and 6». The symmetry operations are E ‘ eo :
(obey Oy,0:
(b) 21.39
(a)
Moves a nucleus at x, y, z to —x, —y, z.
(b)
From x, y, Z, (0.x) Va—c:
312
(c)
The S, rotation about the z axis consists of a C > rotation about z followed by reflection in the xy plane. From the answers to (a) and (b) this moves a point at x, y, z to —x, —y, z. We see that Ss =].
21.40
(a)
The three principal axes intersect at the B nucleus (which is the center of mass). One principal axis is perpendicular to the molecular plane (and coincides with the C3 axis). The other two principal axes lie in the molecular plane; one of these can be taken to coincide with a BF bond,
and the other is perpendicular to this one. (As in XeF4, the orientation of the principal axes 1s not unique.)
(b)
The three principal axes intersect at the center of mass, which lies on the C2 axis. One principal axis coincides with the C2 axis. The second lies in the molecular plane and is perpendicular to the C2 axis. The third is perpendicular to the molecular plane.
(c)
The three principal axes intersect at the C nucleus. One principal axis coincides with the molecular axis (which is a C.. axis); the other two can
be taken as any two axes through the C that are perpendicular to the molecular axis and perpendicular to each other.
21.41
21.42
(a)
SF, has more than one noncoincident C4 axis and is a spherical top.
(b)
IF; (which is a squarebased pyramid) has one C4 axis and is a symmetric top.
(c)
One C2 axis. Asymmetric top.
(d)
One C3 axis. Symmetric top.
(e)
One Cz axis. Symmetric top.
(f)
One C.. axis. Symmetric top.
(g)
One C; axis. Symmetric top.
(h)
Symmetric.
The principal axes intersect at the center of mass, which is the B nucleus. One principal axis is the C; axis. For this axis, J. = 2 mir; =
3(18.998 amu)(1.313 A)” = 98.3 amu A’. The other two principal axes lie in the molecular plane and we can take one of them to coincide with a BF bond.
313
Hence, In = Xj mir2,, = 2(18.998 amu){(1.313 A)(sin 60°)]° = 49.1 amu A®. With one C3 axis, BF; is a symmetric top, so J, = J, = 49.1 amu A’. I, can also
be calculated by taking distances from the F atoms to an inplane line through B and perpendicular to a BF bond: J, =
(19.0 amu)(1.313 A)? + 2[(cos 60°)(1.313 A)]?(19.0 amu) = 49.1 amu A’. 21.43
One principal axis of this symmetric top is the C3 axis through the axial bonds. For this axis, J = 3(34.97 amu)(2.02 iP = 428 amu A’. Another principal axis
can be taken to coincide with an equatorial P—Cl bond, and for this axis J =
2(34.97 amu)(2.12 A)’ + 2(34.97 amu)[(2.02 A)(sin 60°)]* = 528 amu A’. Since this is a symmetric top, the moments of inertia about the principal axes that are perpendicular to the C3 axis are equal, and the third principal moment
is 528 amu A’. 21.44
The molecule is a symmetric top with J, 4 I, = I.. From Eq. (21.45), Ero/h =
BU(J +1) +(A—B)K’ =[BU(J +1)+(AB)K’Ic. (a)me
FOr, Oth — 0 and Eas
On bore
el ke — = On Per or,y
—leand
K =0, Eph = 2Bc = 2(0.05081 cm')(2.9979 x 10!° cm/s) = 3.046 x 10° s'. For J=1 and K =1, E,./h = (2B +(A B)]c = (B+ A)c = (0.2418 cm')c = 7.249 x 10° 71. (b)
From Eq. (21.47), v = 2B(J + 1) = 2Bc (J+ 1) = 2Bc, 4Bc,... = 3.046 x 10’ s!, 6.093 x 10’ s!,... = 3046 MHz, 6093 MHz.
21.45
(a)
(b)
(ce)
Let the molecule lie on the positive half of the z axis with the origin at the
oxygen nucleus. Then Zcom = [12(1.160 A) + 31.972071(2.720 A)]/ 59.966986 = 1.682 A. Io =D; mir? = [(15.994915)(1.682)° + 12(1.682 — 1.160)" + 31.972071(2.720 — 1.682)7](g A)*/(6.02214 x 1074) = 1.3777 x 10° g A? = 1.3777 x 10° kg m*. Vyon1 = 2(J + 1)Bo. Bo = h/817Ip = (6.62608 x 10°*4 J s)/ 87°(1.3777 x 10 kg m’) = 6.091, x 10’ s7!. vo_5; = 2By = 12.18 GHz: V1 52 = 4Bo = 24.36 GHz; V2_,3 = 6Bo = 36.55 GHz.
314
21.46
Bo = Boc = (0.39021 cm™!)(2.99792 x 10!° cm/s) = 1.16982 x 10's"). Ip = h/81By= (6.62607 x 10** J s)/817(1.16982 x 10'° s"!) = 7.1738 x 10° kg m?. The center of mass is at the carbon atom and the principal axes pass through this atom. If dis the CO bond length, then Jp = >; mir? = mod’ + mod’ =
2mod’, so d = (Ig/2mo)'”” = [(7.1738 x 10° kg m’)(6.02214 x 107°)/2(15.994915 x 107 )kg]!” = 1.162 x 107° m= 1.62 A. 21.47
21.48
(a)
From the VSEPR method, SO; is nonlinear; 3V— 6 =9 —6 =3.
(b)
Linear. 3V5=7.
(c) 3V6=9.
We look for sets of integers 7, 7, k such that 36577 + 15957 + 3756k is slightly greater than 7252. Systematic trial and error (best done by first setting j = 0 and looking for i and k values that fit, then setting 7= 1 and looking for i and k,
then settingj= 2, etc.) gives the v;v3v4 possibilities for the 7252 cm! band as (calculated frequencies in parentheses)
21.49
(a)
5 i hv; = she eS
200 (7314),
101
(7413),
002
(7512).
+hc(1340 + 667 + 667 + 2349) cm” =
4.99 x 10°°° J=0.311 eV.
21.50
21.51
21.52
(b)
+he(3657 + 1595 + 3756) cm! = 8.95 x 10°? J = 0.558 eV.
(a)
Inactive, since the dipole moment remains unchanged.
(b)
Active, since the dipole moment changes.
Ve
(L/2I) ak)
Ue
(a)
The C=C bond is stronger and has the greater & and the greater v.
(b)
C—Hhas
(c)
Bending vibrations are generally lowerfrequency than stretching, so C—H stretching has the greater v.
the smaller yp and the greater v.
Vea) C= (1/2n)(k/)" "Ic. Isotopic substitution does not affect the electrons
and hence doesn’t affect k. We have pt = mym2/(m, + m2) = mym/m2 = m,, 315
where m = my (Or mp) and mz is the mass of the rest of the molecule, and we
used m7 >> m). Therefore cp = 2Ucu and Vep = Voq/2"”” = (2900 cm ')/2"" = 2050 cm”!.
21.53 v =(1/2n)(k/p)” and k = 4r°v2p = 407 9 2C7. cn = 12(1) g/13(6.02 x 10°°) = 1.53 x 1074 g, and uco = 12(16) 2/28(6.02 x 10”) = 1.14 x 10° g. So kcu = 410°(3000 cm™!)*(100 cm/m)?(3.00 x 10° m/s)?(1.53 x 107°’ kg) = 489 N/m. Also, kco = 41°(1750 cm™')*(100 cm/m)*(3.00 x 10° m/s)* x (1.14 x 10°°° kg) = 1240 N/m. 21.54
21.55
(a) T;
(b) T;
(Lele
(2 er.
(a)
(c) F;
(d) T (provided it is not a spherical top);
(e) T.
The rotational levels of a linear molecule are E,., = BhJ(J + 1)
[Eq. (21.45) with K = 0] and the pure rotational Raman selection rule is
AJ = +2. So Vo — Vscar = AE/h = t(Bh/h)[(VJ + 2)(VJ + 3) —J(J + 1] = +(4J + 6)B, where J=0, 1, 2,... . The spacing between adjacent lines is [4(/ + 1) + 6]B (4/ + 6)B = 4B. (b)
4B =7.99 cm! and B = 1.998 cm! = 199.8 m!. We have B = Bia=
h/8n°Ic = h/8r?cuR? and R = (h/8r°cu B )". Also, 1 = mym2/(m, + mo) = m;/2m, = mj/2 and R =
2(6.626x107*4 J s)(6.022x1023 mol!)
1/2
81° (2.998x10* m/s)(0.01401 kg/mol)(199.8 m7!) 1.098 x 10°'° m = 1.098 A (as in Table 21.1). (c)
The lowest J is 0 and Vo — Vscat = [4(0) + 6]B = 6B.
(d)
vo =c/Ag Vo — Vscat +0.00359 5.547, x
= (2.9979 x 10° m/s)/(540.8 x 10° m) = 5.543; x 10!4 57. = +6B = +6 Bc = +6(199.8 m™!)(2.998 x 10° m/s) = x 10'* s7'. Vecat = 5.5435 x 10'4 5! + 0.00359 x 10'4 57! = 10'* s! and 5.5395 x 10'4 s7!. Then Ascat = C/Vscat = 540.45 nm
and 541.15 nm.
21.56
(a) F; (b) T;
(c) F.
316
21.57
For the Balmer series, n, = 2 in Eq. (21.50); for the series limit, ng = ce and 1/A = R/4 = (109678 cm !)/4. Then A = 3.647 x 10° cm = 364.7 nm.
21.58
For the Paschen series, n, = 3 in Eq. (21.50); the first three lines have ng =
4,5, and 6. So 1/A = R(1/9 — 1/16), R(1/9 — 1/25), R(1/9 — 1/36). We get A =
1.8756 x 10~* cm, 1.2822 x 107 cm, 1.0941 x 107 cm. 21.59
Li** is a hydrogenlike atom and Eq. (19.14) gives the energy levels. The Li nucleus is substantially heavier than the H nucleus, so we can take pt equal to the electron mass. So E = —9[21’me*/(4m€9)’n°h’] and 1/A = v/c = AE /he =
9[21?me*/(4m€o)*ch*(1/n? — 1/n2) = 9(109736 cm ')(1/1 — 1/4) = 7.4072 x 10° cm”. Then A = 1.3500 x 10° cm. 21.60
Using the notation of Fig. 21.38 and Eq. (21.52), we have Do of the B state as
Dj, = hVcont — hVoo = (6.6261 x 10 J s)(2.998 x 10° m/s) x [1750.5 x 107?° m)! — (2026.0 x 107"? m)"] = 1.543 x 10°” J = 0.963 eV. Do of the ground state is given by (21.52) as
Dy, = (6.6261 x 10** J s)(2.9979 x 10° m/s)(1750.5 x 10°'° my! x
(1 eV)/(1.60218 x 107'? J)  1.970 eV = 5.11 eV. 21.61
Ny\/Nj9 = 0.181 = G/L) and Ae/kT = In(0.181/3) = 2.81. So T= 0.356Ae/k. From (21.15), (21.33), and (21.32), Ae = 1(2)h7/210 = h’/4rI= 2 Byhe = 2(B,  ; &, )hc = 2(1.931 — 0.009) cm (1 cm/0.01 m) x
(6.626 x 10° J s)(2.9979 x 10° m/s) = 7.636 x 10° J. T= 0.356 Ae/k = 0.356(7.63¢ X 10°? J)/(1.3807 x 10°°° J/K) = 1.97 K. 21.62, (a) 1
e(D) elec) cP se(d)ul pe(e)ah a(0) F) (2) ste
ee Ueh ss.() a0;
(kK) Pee:
21.63 F =BQvsin 0 = (1.5 T)(1.60 x 10°? C)(3.0 x 10° m/s) sin 6 = (7.2 x 10°? N) sin 8. (a) F=0. (b) F=(7.2x 10° N) sin 45° =5.1 x LOpaea (c) 7.2x10°3N. (d) 0.
317
21.64
From the paragraph before (21.61), m = Qur/2 =
21.65
(a)
(2.0 x 107! €)(2.0 x 10° m/s)(25 x 107! m)/2 = 5.0 x 10° J/T.
e/2m, = (1.602176 x 10"! C)/2(1.672622 x 10° kg) = 4.78942 x 10’ Hz/T, where (21.54) was used.
21.66
21.67
(b)
y/2n = (4.78942 x 10’ Hz/T)5.58569/2n = 42.5775 x 10° Hz/T.
(a)
Equations (21.65) and (21.63) give E = —yh BM, =
(b)
15000 G = 1.50 T; energies are the same as in (a).
(a)
From (21.67) and (21.63), v = yB/27 =
(b)
27.32 MHz.
~1.792(4.7894 x 10’ s'/T)1.792(6.626 x 10°" J s)(2m) '(1.50 T)M7 = ~(1.35g x 10°26 J)M,. Since I = 3/2, M, = 3/2, 1/2, 1/2, and 3/2. The levels are —2.04 x 10° J, 0.679 x 107° J, 0.679 x 10° J, 2.04 x 10°° J.
(4.7894 x 10’ s'/T)1.792(1.50 T)/2m = 2.049 x 10’/s = 20.49 MHz.
21.68
From (21.67), V,/Vg =Yal/¥g  and v(°C) = (10.708/42.577)(600 MHz) = 151 MHz.
21.69
(a)
Equations (21.67) and (21.63) give B = 2nv/y =
2n(60 x 10° s!)/[5.5857(4.7894 x 10’ s'/T)] = 1.41 T.
21.70
(b)
(300/60)(1.41 T) = 7.05 T.
(a)
M, = +2 and —'2. The energy separation is given by (21.65) as AE =
y 2B = (4.7894 x 10’ s"'/T)5.5857(6.626 x 10°" J s)(2)'(1.41 T) = 3.98 x 10° J. The levels are nondegenerate and the population ratio is
eoMEAT = exp [(3.98 x 10°°° JV/(1.381 x 10°? J/K)(298 K)] = exp (0.00000967) = 0.9999903.
(b)
An increase in B increases the separation between energy levels, thereby
producing a greater population difference between the initial and final states. Hence the absorption intensity increases.
318
21.71
From (21.72), vi — vj = 10°(60 x 10° Hz)(1.0) = 60 Hz.
21.72
From the tables in the text, 6 is 2 to 3 for the CH; protons and is 9 to 11 for the CHO proton; J is 1 to 3 Hz. The CH; doublet has three times the total intensity of the CHO quartet. Thus (all splittings are about 2 Hz) (a) i
21.73
eae Hz
(a)
300 Hz
ee
(b) 00 Hz12 cee 1500 Hz a=
From (21.70) and (21.68), increasing 5; means decreasing 6; and
increasing V;, SO V, increases to the left. (b) To the left.
(c) To the nght.
(d) From (21.70) and (21.69), increasing 5; means decreasing 6; and
decreasing By;,S0 By; increases to the nght.
21.74
(a)
One singlet peak.
(b)
One proton NMR peak that is split into a doublet by the F.
(c)
One singlet.
(d)
The four methylene protons give a quartet of relative intensity 2; the six methyl protons give a triplet of relative intensity 3.
(e)
The (CH3)2 protons give a doublet of relative intensity 6; the CH proton
gives a septet of relative intensity 1.
(f)
Two equalintensity singlet peaks. The CH groups are not equivalent
and don’t split each other.
(g)
Three quartet peaks of equal intensity. In each of the three quartets, the 4 lines have equal intensity but only two of the 3 spacings are equal.
(h)
The CH; protons give a triplet of relative intensity 3, the CHO proton gives a triplet of relative intensity 1; the CH2 protons give an octet of relative intensity 2. 319
21.75
(a)
(b)
nts: one for This molecule has two different proton—'’F coupling consta ent coupling the 'H and '°F nuclei that are cis to each other, and a differ constant for 'H and !*F nuclei that are trans to each other. r to the In this molecule, the F atoms lie in a plane that is perpendicula coupling plane containing the CH? group, and there is only one HF
constant. 21.76
All peaks except in (b) are singlets.
(a)
One peak.
(b)
One peak that is split into a doublet by the Ee
(c)
One peak.
(d) (e)
Two peaks of equal intensity. Two peaks with 2:1 intensity ratio.
(f)
Three equalintensity peaks.
(g)
Two peaks of equal intensity.
(h)
Three equalintensity peaks.
(Db), 59 (c)r3
21.77
4 kinds of carbons); (a) 4 (there are
21.78
In Fig. 21.44, 100 Hz corresponds to a length of 28 mm and J corresponds to a
length of 22 mm, so J = (2 /28)(100 Hz) = 8 Hz.
21.79
(a) (c)
Unchanged; see Eq. (21.70) and the following paragraph. Each v and the difference between the v’s is multiplied by 10; see Eqs.
(21.68) and (21.72). (e) Unchanged, as noted after (21.73).
21.80 From Eq. (21.69), 6; = 1 — 27Vspec/¥iBo,i. $0 8; = 10°(Grer — 51) =
10°(—21Vspec!'¥i)(U/Bo,ret — 1/Bo.i) = 10°Bo(Bo,et — Bo,i)/Boset Bo,i. Since 0; rotational levels is
smaller for D2 and the fraction in J = 0 is greater for Ho.
21.129 (a)
joule, newton, watt, pascal, hertz, coulomb, volt, tesla, ohm, kelvin, ampere, siemens;
(b)
21.130 (a)
(b)
poise, debye, gauss, angstrom, svedberg, dalton, torr, bohr, hartree.
aL, F. It cannot change its rotational state by absorption or emission of radiation, but can change its rotational state during collisions.
(c)
F. Counterexamples are CHy and BF3.
(d)
F. The energy might be transferred to another molecule in a collision. 330
(e)
F. A counterexample is H2O.
(f)
ly
(g) (h) (i)
F. This formula is only for linear and sphericaltop molecules. 18.
(j) (k)
le Ei This question is too silly to answer.
331
Chapter 22 22.1
(a) T, V, and mole numbers. (b)
he V, I
UC oor
(c) Over states.
E ,/kT
22.2
Z=;e
"
is dimensionless.
22.3.
The Helmholtz energy A is extensive, so Az is 25/10 = 2.5 times A. Hence Ad/A, = 2.5 = (kT In Z2)/(AT In Z;) = (In Z2)/(n Z;).
22.4
G=A+PV=KT InZ + VET(0 In Z/OV)7.y, =KTV[(V™ In ZV] y, 
22.5
Let a subscript o denote the partition function and thermodynamic properties before the constant b is added to the levels. Then
IE J rss pee ah
en ale yi obit = eZ, and In Z = —b/kT + In Zo.
(a)
P=kT(0 In Z/V);y, =kTO In ZJOV)7y, = Po:
(b)
U=kT'(0 In ZT);y, =kT [b/kT’ + 0 In ZJOT) yy, ] = b + Uo.
(c) S=U/T+k\n Z= DIT+ UST kb/kT + k ln Zo = UST + k In Zo = So. (d) A=kT In Z=KkT(b/kT + In Z,.) = b — kT In Z, = b + Ao. 22.6
From p; = exp(BE,)/Z [Eq. (22.15)], we have In pj = BE; — In Z.
So Y
p; In pj = B Dj pjEj + In Z Dj pj = (kT) 'U + In Z = S/k, where (22.3),
(22.33), and 2; pj = 1 were used. Erk
©
ie
:
.
221
ee
As dimensionless:
22.8
From the equation before (22.53), the number of available translational states
is roughly 60V(mkT/h?)*” = [60(10 cm*)(1 m?/10° cm*)/(6.6 x 1074 J s)*] x [(0.020 kg/mol)(1.38 x 10°” J/K)(300 K)/(6.0 x 10°°/mol)]}*? = 3 x 10°”.
332
22.9
(a)
Each dot with n,, n,, and n, being positive integers corresponds to a stationary state with quantum numbers nx, ny, n,. The states with Er S Emax Satisfy Eq. (22.52). The distance of a dot from the origin is r= (n* +n ; sete :)"”? so the the positive square root of (22.52) is
The region defined by this inequality is a sphere of
PaASmV hea
radius rmax = (8mV"?h Emax)” and the number of dots in 1/8th of this sphere equals the number of quantum states with energy €; S Emax} We take 1/8th of the sphere because n,, ny, and n, must each be positive,
which is true only in 1/8th of the sphere. (b)
Fig. 24.8 shows the 4 cubes at the same altitude that share a dot, and 4
more cubes above these 4 also share this dot. The number of translational
states with €, < max is then +(Sar;
) = (n/6)[(8mV"Fh?  3kT)'?PF=
(1/6)(24mVe nh?RT)”.
22.10
(a) T; (b) F.
22.11
(a) (b)
We get 3628810. 9.332621569 x 10'°’, where we used 100!” = (107)! = 1077!
(c)
Using (LO.
eee
=
101024
and py
= (ene yl? =
(3.72007597601 x 10~4)!'° = 507595.88975 x 10°", we get 4.0238726006 x 10°°°’, 22.12
In (300!) =In 1 +1In2+1n3++1n 300. A BASIC program is 10 S=0
40 NEXT 
20 FOR !=1 170300
50 PRINT S
30 S=S+LOG(I)
60 END
One finds In (300!) = 1414.905850. Also, N In N — N = 300 In 300 — 300 = 1411.134742.
22.13
From (22.51) and (22.56), In Z = Nc In zc — In Nc! + Np In zg — In Ng! +: Ne In zc —In Nc! + Ng In zp — Np In Ng + Ng +: and (0 In ZIONB) TV Neo =
In zg — In Ng — 1 + 1 = In (zp/Ng). Substitution in Eq. (22.41) gives pp =
—RT n (zp/Np). 333
22.14
(a) F; (b) F; (c) T; (d) T.
22.15
€, =nzh’/8ma’ and Ae, = (h’/8ma’)[(n, + 1)? — n2] = (h7/8ma*)(2n, + 1). From €, = n2h’/8ma* = kT we get ny = a(8mkT)'"/h = [(0.02 m)/(6.6 x 107*4 J s)][8(0.028 kg/6.0 x 107°)(1.38 x 10773 J/K)(273 K)}"” and n, = 1.1 x 10°. Then Ag,J/kT = (h?/8ma’)(2n, + IWkT = 1.8 x 10°”.
22.16
At room T, kT = (1.38 x 107? J/K)(298 K)(1 eV/1.6022 x 107'? J) = 0.026 eV; =. kT/he = (1.38 x 10°? J/K)(298 K)/(6.63 x 10** J s)(3 x 10° cm/s) = 207 cm"; RT= (1.987 cal/molK)(298 K) = 0.59 kcal/mol = 2.5 kJ/mol.
22.17
With anharmonicity neglected, (N,)/(No) =e plain AT joa tol JkT —hv,/kT
Cae eh:
(a)
Avo/kT= (6.626 x 107*4 J s)(6.98 x 10!'7/s)/(1.381 x 10°? J/K)(298 K) =
11.24 and (N,)/(Ny) =e7''** =.0.000013. (b) (N,)/(Ng) =e "7! = 0.044. (c) Ye 7 =046. 22.18
Use Eq. (22.76). Let x = AV/KT = hcV/kT = (6.6261 x 10°" J s) x (2.9979 x 10® m/s)(2329.8 cm™!)(100 cm/m)/(1.38065 x 10°*° J/K)T = Gs5 20K) Tawa eee! ch cnt) / (lee (a) x = (3352 K)/(298.15 K) = 11.243. zyip = W/(1 —e 79) = 1.000013. No = (6.0221 x 107°)1/1.000013 = 6.0220 x 107°. N, = Nae ''743/1.000013 = 7.888 x 10'°. (b) x = (3352 K)/(1073.15 K) = 3.1235. zyip = 1.0460. No = Na(1)/1.046 = BS TpalO aeN= Naer 4 a)104.65 932 xl Oaee (c) x=1.0241. zp = 1.5604. No =3.859x 107. N, = 1.386 x 107°.
22.19
With centrifugal distortion neglected, the rotational levels are BohJ(J + 1). The
degeneracy of each level is 2J + 1, and (N,)/(No) =3¢770"/*7 fe® =
3 ?BohlkT =O Boh/k = 2(5.96 x 10!°/s)(6.626 x 104 J s)/(1.381 x 10°23 J/K) = 5.72 K. 334
(a) (b)
(N,)/(No) aaa a2
= 3076.72 K)/(200 K) _ 2.92.
K)/(600 K) o
2.97.
(Cc)
22.20
(a)
The vibrational levels are nondegenerate and (N 1)/(No) ayes
e'", We have (N>)/(No) = e2""T = (Ni/No)’. We find (0.528)? = 0.279, as is observed. Therefore there is an equilibrium distribution in these levels.
(b)
In (N{/No) = AV/KT and T = —hv/k In (N,/No) =
(c)
(6.626 x 10° J s)(6.39 x 10!” s)/(1.381 x 10°73 J/K)(In 0.528) = 480 K. N3/No = e°""" = (Ni/No)’ = (0.340) = 0.0393. This is an approximation since anharmonicity has been neglected.
22.21
With the +1 neglected, (22.77) becomes (N,), =e ®/*” /eM". Use of the Prob. 22.13 result gives —w/RT = In(z/N), so eY*" = el) = 2/N. Hence
(N,)., =e °"" Niz, which is (22.69). 22.22
In Zpp =—BUN/Ng + D, (41) In [1 + e PH €) ), Where Eqs. (1.67) and (1.68) were used. Since  >> eP(#/"a®)  we can use Eq. (8.36) to expand the log and
we need include only the first term in the sum; so In Zhe ~
—BUNIN, + >, (£1 [te P/N =")  = BuN/N, +X, eH NAEOT
Ror
/kT kT 46, (N,) ). (N,)
=N= ne. 8 el
N(x, e*'*" ) and p/NakT= In N
tun sock eh Bie
In(X, e ©"/*" ). Hence In Zin =~NInN+
Nind,e *!’ +N=NInz—In N! and Zpp ~z'/N! for (N,) 0.42 (the maximum on the above graph) for T = 300 K and Cy decreases for temperatures above 300 K.
Equations (22.76) and (22.90) give (N,)/N =e 8? fein =e zyip = UW evi = d  eeuey.
22.29
(a) (d)
Wein = 1 —e7'° = 0.999955. 1—e'=0.632. (e) 0.095.
(a)
(N,)/N =e oweulKt Feri
(b) 1—e° =0.950.
Ce ues
eee
(c)
1—e? = 0.865.
ye
e POvn/T (7 _ @Ovn/T) where (22.75), (22.90), and (22.88) were used.
(b) From (22.75) and (22.76) with
v=0: (No)/N = Uzviy and zyiy = N/(No).
A similar formula holds for Zrot. (c)
Ovm=Avi/k = hVclk = (hclk)(2143 cm’) = 3084 K, as in the first
example in Sec. 22.6. At 25°C: Oyi,/T= 10.344;
(Ny)
(G022 6107 i= 2 ns 41 )i=.6,022
1058.
(Ne View em(leten es) = 1.94 210%, At 1000°C: @yit/T= 2.422. (Ny) = 5.49 x 10°. (N,) = 4.87 x 107. (d) For ©yi, = 3352 K and v = 1, we have (N,)/N = €©? OT] — ¢C352 KT),
We find TIK 0 (N,)/N 0
2000 4000 6000 8000 10000 12000 15000 0.152. 0.245 0.245 0.225 0.204 0.184 0.160 337
N> is not a harmonic oscillator, so anharmonicity and the finite number of vibrational levels make the highT values inaccurate.
(N,N
O03"
7
O22 Osa O
1
0

4000
=
8000
12000
16000
T/K
2230 na Na IN\al = 0 =(didine =) (v + 1)(OQy/Te PPT
latesme aye (Ge Een
vib /T
at
= 0 and v = (v + Le?” , so Qvi/T =
—In[v/(v + 1)] and T = Oyjp/In[(o + 1/0] = Ovin/InC. + 1/v). For Oyii/T= In[v/(o
iy 1)), we
get (N,) ING
e In{v/(v+ 1)] as ee
UinluKos)) ae
exp{In[v/(v + 1)]°} —exp{In[v/(v + L]’*!} = v'/(v + LP  v7 (SP ae ee Sains be 22.31
+ Y=
Yes. Despite the fact that €, > €,in Eq. (22.72), N, might exceed N, if the degeneracy g, exceeds g,. (For example, see Prob. 22.19.)
22,325 (A) ez» cn(levels\ SinCale bewkekl Gav Gahpeek sme Meme (b)
= 3.055
Equation (22.71) gives (N(e,)) = Ng,e°9/*" Jz,
Sov N)) = (G02 sql Ossie is Qac e 1 oo aa. (N>) = (6.02)x 10ay3a uy” 9/3.93p ye met OF (N;) = (6:02 x 10ssbe@Be pane 12095. =[ieTlel Og: (c)
AsT—3 ~~, we havee o"*
5 ¢°%=1 andz514+34+5=9.
So (N,) — (6.02 x 10/9 = 0.669 x 10°’, (N,) — (6.02 x 1077)3/9 = 2.01 x 10°, and (N3) — (6.02 x 107°)5/9 = 3.35 x 10°. 22.33 (a) N!=123NandInN!=In1+In2+In3++mnN=>D%, Inx. 338
(b)
For large N, the main contributions to the sum come from the later terms
(where x is reasonably close to N); the later terms don’t vary greatly as we go from x to x + 1. (For example, In 50 = 3.912 and In 51 = 3.932.) So
Eq. (22.79) can be used.
22.34
(c)
Dea Inx~ JY Inxdv=(xInxx* =NInNN+1=NINN EN.
(a)
Let 0 = Ojo. Then Oz = DF (20 + Le M*"", a = 0, and f(/) = (2/ + le so0
J corresponds to n,
Dy So f(O) = 1. Differentiation gives
flee O(2i I)iTleat? 1)" ony ONT f(D) = [602 + DIT + 0°27 + 1)}/PJe8077 FED =I T 12072) Te 0 Os LITlen a fa) ——120/ T1205 20 17s f°) = (602) + DOT’ — 20027 + 18+ (27 + 1P04/TYe 88 tT
Te [12005/ Teal SOQ 411)07/T 2 3027
10
Te OF)
at
e IT ¢Q) = 1200/7? — 1806°/T + 300°/T* — 65/T°. Noting that J* (27 + 1)e°Y "+" ay = J" e"" dw = TIO, we have G2 WO 1422 0/1/12 + Gl20/T + 120711 6 1 20 (12007/T° — 1800°/T* + 3004/T* — 6°/T°)/30240 ++ and
Zrot = (T/6@ro)(1 + Oro 3T + 7, /15T? + 40? /315T? +). rot
(b)
O,o/T = 85.3/273.15 = 0.3123. Equation (22.85) gives Zor = 1/2(0.3123) = 1.601. Equation (22.86) gives Zror =
1.601[1 + 0.3123/3 + (0.3123)7/15 + 4(0.3123)7/315 ++
]=1.779 and
the error is 10%.
(c)
22.35)
Oro/T= 2.862/273.15 = 0.01048. Then Z,o: = 1/2(0.01048) = 47.72 and Zot = 47.72[1 + 0.01048/3 + (0.01048)7/15 + 4(0.01048)°7/315] = 47.72(1.0035) and the error is 0.35%.
Urop = NRT*(d In Zo)/dT) = nRT?(d/dT){In T — In 6©,o:) = nRT.
Uviy = MRT? d In zyp/dT = —nRT*[d In (1 = dT] = RT" (v/kT?)/ (1
~ eT) = (nRhv/k)i(e"™"" — 1). Uy = nRT (d/dT) In ge1o = 0.
339
22.36
Hence, As T > ©, the Taylor series for e* shows that eovr/T _» 1 4 Oyi/T.
22.37
Si: = UlT +Nk In 2 — Nk(In N  1) =
/ TY = nR. TY (Ovi/ Cyviy 9 NR@viv
3nR + NK{3 In (2nmkih’) + > In T+ In V]—Nk InN + nR = snk +
nR In [(2nmk/h?y?T??(NKTIP)(AI/N)] = >nR +nR In [(2nm)7(kT)Y/hP. Srot =
T + NK In Zot = NR + nk In [T/OQ,zot]. Urol
nRQwiT(eo* =1)—nRin( —e 9"). 0+nR
Svib = Uvir/T+ Nk In Zvib =
Sa=Uesl T+ NE In 21 =
In 8e1,0 = nR \n £el0
22.38
Siem@/R = 2.5 + 2.5 In (T/K) — In (P/bar) + 1.5 In M, + In{(2n x 10° kg/mol)*” (1.38065 x 10° J/KY"?K” x (6.02214 x 107/mol)*7(6.62607 x 10° Js) *(10° N/m*)'] = 2.5 + 2.5 In (T/K) — In(P/bar) + 1.5 In M, — 3.65171 = 2.5 In (T/K) — In(P/bar) + 1.5 In M, — 1.1517.
22.59
For these gases of closedshell monatomic molecules, S = Sy.
Equation (22.108) gives Sj,793 (He) = (8.3145 J/molK)(1.5 In 4.0026 + 2.5 In 298.15 —In 1 — 1.1517) = 293 (Ar) = 154.85 126.15 J/molK. Similarly, S*,59g(Ne) = 146.33 J/molK, Sj,
JmolK, $2,593 (Kr) = 164.09 J/molK. 22.40
For H, geio = 2 and Eqs. (22.107) and (22.108) give S= late beim els (8.3145 J/molK)(1.5 In 1.0079 + 2.5 In 298.15 — In 1 — 1.1517 + In 2)=
114.72 J/molK.
Cp 203 = Cy m29g + R = 20.79 J/molK.
22.41
Cy 29g = (3/2)R = 12.47 JimolK.
22.42
Equations (22.94) and (22.95) give Uj, 593 — Umo = Utem = (3/2)RT =
1.5(8.3145 J/molK)(298.15 K) = 3718 J/mol for each gas.
340
22.43 S=(UUp/T+kInZ; InZ=[S—(UUp\/TVk= [154.8 J/K — (3718 J)/(298.1 K)](1.3807 x 107? J/K)! = 1.03 x 10” =
2.303 log Z. So log Z = 4.48 x 10" and Z = 10**8*!" for 1 mole. Z=*/N! and In Z=NInz—InN!=NlInzNInN+N.
Solnz=(InZ+NInNN)/N
= (1.03 x 10°°)/(6.02 x 107) + In (6.02 x 1078) — 1 = 70.9 and z=6 x 10°”. 22.44
(a)
V5 = VP 2 V Xa23586'em OC 4ercms) — 2530.0'enn Be=iBs 5 &, = 1.998 cm™ — 10.017 cm”) = 1.9895 em”. hclk = (6.62607 x 107*4 J s)(2.99792 x 10° m/s)/(1.38065 x 10°*° J/K) = 0.0143877 m K = 1.43877 cm K. Ovi» = Vo he/k = (2330.0 cm ')(1.43877 cm K) = 3352.3 K.
(b)
(c)
Orot = (1.9895 cm™')(1.43877 cm K) = 2.862 K. 2 = (2MMKTINa)?V/h?. V=nRTIP = 12310 cm>. 2, = (0.012310 m*) x [27(0.0280 kg/mol)(1.3807 x 10°? J/K)(300 K)/(6.022 x 107*/mol)]*7/ (6.626 x 10°47 Js)? = 1.78 x 10°”. Zrot = T/O@ro = (300 K)/2(2.862 K) = 52.4. eee ee (lee le) OOU0 tare gale V = nRTIP = 1.026 x 10° cm? and zy, = 3.57 X 10°73 Zot = 437; pa
22.45
(a)
hte
WAIsh
ores yan
Ovip = Vy hclk = 5696 K and Ovit/T = 19.104. Orr = By hc/k = 29.58 K. Equations (22.105)—(22.108) give at P =  bar:
S°..m, tr = (8.3145 J/molK)(1.5 In 20.0063 + 2.5 In 298.15 — 1.1517) =
146.22 J/molK. S* sot = (8.3145 I/molK){ 1 + In[(298.15 K)/1(29.58 K)]} = 27.53 J/molK.
5°.iy= (8.3145 J/molK)[19.104/(e171 — 1) = Ind = e*)] = 8 x 107 J/molK.
SO
(b)
S°,, = 0 (since all electrons are paired).
ee Saito mmet om)
mee  ae
aOR
Equations (22.100)—(22.102) give: Cy um = 15(8.3145 J/molK) =
12.47 molK.
C2com = 8.3145 S/molK. Cy vin =
1? %/(e!? 1?— 1)° = 1.5 x 10° J/molK. (8.3145 J/molK)(19.104)°e So Cy m293 = 20.79 J/molK.
341
(c)s 22.46
29410 S/mol ks
(Cp = Cet
p= m’/2m = m/2 = +(126.90 g)/(6.022 x 10 yea 0530 108 cee he (1.0536 x 10°? g)(2.67 x 10% em) =7.51x 10° gem’. On = h?/810Ik= 0.0536 K. Qyp = hvo/k = 306.9 K and ©,i,/T = 0.6138. (a)
From Egs. (22.94)(22.98) at 500 K:
Uj. = (3/2)RT=
1.5(8.3145 J/molK)(500 K) = 6236 J/mol. U},.,o¢ = RT =4157 J/mol. U*,sy= (8.3145 J/molK)(306.9 K)/(e?*'** — 1) = 3011 S/mol. U?,4=0. U%,509 — Uino = 13404 J/mol. OO)
(c)
Pea
ee Ons
13404 J/mol + (8.3145 J/molK)(500 K) = 17561 J/mol. From (22.105)(22.108): S°,,, = R(L.5 In 253.8 + 2.5 In 500 — 1.1517) = 188.65 J/molK. S°,so = R{1 + In[500/2(0.0536)}} = 78.55 J/molK. S? ap = R(0.6138/(e°°'** — 1) — In(1 — €°*!**)] = 12.50 J/molK. S220,
(dq)
Uae
SOS)
hy SV
Oia,
Giisoo Umo =4mso0 — TS Ey a
any
17561 J/mol — (500 K)(279.70 J/molK) = —122.29 kJ/mol. 22.47
From Fig. 21.13 the purerotational lines are at 2Bo, 4Bo, 6Bo, . . . , so the
separation is 2B, = 20.9 cm’! and By = 10.45.cm™’. Also, Vp = 2885 em So
Orot = Byhclk = 15.04 K; vip = Vy hc/k = 4151 K and Oyip/T = 13.92. We shall assume the spectral data are for the predominant species HCl and shall
calculate S° 593forH Cl. At 25°C: S*. = R(1.5 In 36.0 + 2.5 In 298.1 — In 1 — 1.1517) = 153.5 J/molK. S° rot = R{1 + In[298.15/1(15.04)]} = 33.1 J/molK. See R[13.92/(€i ae 1) = Intl ee 2) = 0.0001 1/moleKeens S0.5 jos = 1 80,0,)/mol Ke
342
y= 0)
22.48
(a)
For this relatively light diatomic molecule, only translation and rotation
contribute to Cy and Cy,,,= (3/2)R + R = 2.5R = 20.79 J/molK.
Cpom= Cy m +R =3.5R = 29.10 J/molK. The true value is 29.13 J/mol
K. (b)
All gases of relatively light diatomic molecules have Cp,,= 3.5R = 29.1 J/molK.
22.49
€.,,/k = (0.0149 eV)(1.6022 x 107! J/eV)/(1.38065 x 1077? J/K) = 172.9 K. Zeal = elo He Baines Sin ward + Ze 2 TAL 30 Kozy ai 2iei2e aise a= 2.006. At 150 K, z1 = 2.63. At 300 K, z,) = 3.12.
22.50
Using the z.; expression found in Prob. 22.49, we have Ue, = NkT°(d Ineze al) =
(nRT"/z1)(dzeVdT) = (nRT7/z1)(2(173 K)T 761? 87) = nRB46 Kye 72, and Ueim = (2875 J/molje 1? 6"y[2 + 26177 8) = (1438 Jmol/e"? 7 + 1). Cveim = dela = (1438 J/mol)[(173 KVP Jen (eB OF + 17 = (2.49 x MO
K/mol)T7e"'? ST Kens we
ie We find (note the maximum in
Cv cim)
Cvetm/(J/molK)0.86 T/K 30
2.52 3.47 3.42 2.67 2.02 1.30 0.88 0.64 45 60 90 120 150° 200 5250, 300
C\e.n/(J/molK)
4x ntxe txt (x4+0 2951. (a), saxselex (b)
—»). Sos= +°)=1.Wd
The Taylor series for 1/(1 — x) about x = 0 is given by Eq. (8.8) as x + store x Weel te ee 343
22.52
(a)
Zvib = ss 2 =eanay hal = » Seeseres
= Bol tds aE ie, BORAT —
eVAKTI( — e""1) where Eq. (22.89) was used.
(b)
Uyy = nRT(0 In 2viv/OT)y =
In vip = —SAV/AT—In (1  eT
nRT[L hvlkT? + &(avikTYL — =
=Niv +
nRhv/k(e""" — 1), which differs from Eq. (22.97) by M(hv/2); this agrees with the Prob. 22.5b result.
22.53
From Eq. (22.100), Cvirm = 3R/2. The translational levels are so closely spaced
that this result holds at all temperatures not extremely close to 0 K. From p. 846, O,o1 is typically of order of magnitude  K (except for hydrides). Figure 22.11 shows that Cy,;otm = R for T above 1.30,. From p. 846, Ovi is typically 10° K. Figure 22.10 shows that Cy,vib.m is negligible below 0.10,» and increases to R as T increases to  or 2 times ©yjp. So:
CHeya/ Cym
2.5R
1.5R
TIK 1
22.54
10
100
1000
Zot = [(298 K)/1(2.77 K)][1 + (1/3)(2.77/298) +   +] = 107.5. Fron et (a)
Le */2] = J(J + 1)k©yot, where (22.83) was used.
Use of Eq. (22.71) gives
[eyo (NZ) VN = Zi
Dye (QJ + LexplJU + @ro/T) =
Zin Jy (2S + expl(J? + N@ro/T) dd = 25), 17*7 exp[(@,o/T)w] dw = Zo (—T/OQyo:)[EXp(—306@yo/T) — 1] = (107.9) '(298/2.77)[1 — 2002 77/298) —
0.939, which is 94%; we took w = J’ + J and dw = 2J + 1. The sum has 17 terms, so 17 was used as the upper limit of the integral.
344
(b)
The required sum is (107.9)!
D'6, (20 + Le ~ +2728). Direct
evaluation gives 0.932 = 93.2%. The individual percentages are 0.927, 2.73, 4.38, 5.80, 6.93, 7.71, 8.15, 8.26, 8.07, 7.63, 7.00, 6.25, 5.44, 4.61, 3.82, 3.09, and 2.44 for J/=0, 1, 2,..., respectively.
A BASIC program
iS 10 S=:0
70
20 A=2.77/298.1
80 NEXT J
30 FORJ=0TO
16
S=S+B
90 PRINT "SUM=";S
40 B= (2*J+1)*EXP(A*(J*J+J))/107.9
95 END
50° P = 100*B SOPRRINT d=40, POP=—e
22.55
v =(k/p)'/2m. N> and F; have roughly the same p values, but N> has a triple bond and F; has a single bond, so ky, >> kp, (cf. Table 21.1). The high ky,
makes vy, and ©,;,, high, so the vibrational contribution to Seeneeis negligible at room temperature. The light mass of H makes Uy > Vp, and S Vin m298,HE is the heaviest molecule and has the largest moment of inertia and the smallest ©,o:. N> is the lightest molecule and also has the shortest bond
(b)
length (since it has a triple bond), so N> has the smallest / and the largest One v= (1/2n)(k/p)'”. The large p: of Br2 makes this singly bonded species have the smallest v and hence the smallest Oyj». The high & and small pt of N==N makes vy, and ©,,,n, the largest.
(c)
The low value of ©, for Brz2 makes more of its excited rotational levels
populated and makes Z,,; p,, largest.
(d)
The small v,,, and O,i,.5,, make excited vibrational levels populated and Z,i,p,, 18 largest.
(e)
All these species have essentially the same Cy,rot.m at room temperature
since they all have essentially attained the highT limit Cyrot.m = R at room temperature.
(f) 22.60
Bry, for which excited vibrational levels are accessible at room 7.
Zr is very roughly equal to the number of translational energy levels that have significant populations at temperature T. The particleinabox translational levels have €, proportional to 1/m (Chap. 18), so an increase in molecular mass m lowers the translational levels and allows more of them to be populated at T, thereby increasing Z,, and increasing St.
22.61
(a)
At 20 K, Zot = T/O@yor= (20 K)/(29.577 K) = 0.676. At 30 K, Zrot = 1.014. At 40 K, Zot = 1.352.
(b)
Zrot = 0.6762[1 + 1/3(0.6762) + 1/15(0.6762)° + 4/315(0.6762)°] = 1.136 at 20 K. At 30 K, Zoe = 1.426. At 40 K, Zor = 1.742.
346
(c)
A BASIC program is 1
Ont =1293577.
60 IF A< 1E12 THEN 90
fq ooreyeasiid)
70
S=S+A
SUMINEU Teta rsT
80 NEXT J
40 FOR J =0 TO 10000
SOIRRINT ZROT=2S
50 A= (2*J + 1)*EXP(J*(J + 1)*TH/T)
95 GOTO 20 99 END
he exact values are 7.4; =1.1 565 at 20 Koo — 14I at 30 Kee 1.7439 at 40 K.
22.62 S=U/T+kInZ=kT@ In Z/0T)yy, +k In (“/N!) = kT(0 In (2“/NV/OT) yy, +NkInz—kIn N! = NKT(Q In /OT)y y, + Nk In zk In N! = (NKTIz)(0Z/0T) yy, +NkInzDse°""" /z—kIn N!. Use of dv/0T= (O/0T) Xe 88/" = (UT?) &, ee 8"
(Nk/z) Inz
gives S = (NizT) Ds ee 8"
+
se!" —k InN! =NKY, [(e/kT + In z\(e *8""" /z)] — k In NI.
From (22.69), x; =e ~€s/KT 7 and In x; =—€,/kT —1n z, so
S =Nk ¥, x5 In x;—k In N!. The other condition is (N,) &, all terms after the second go to zero and U,. > n(RT— RO,o/3).
22.64
(a)
We have the original orientation plus the following: 120° and 240° rotations about the C3 axis give 2 indistinguishable orientations; rotation
about a C> axis followed by rotations about the C3 axis give 3 more indistinguishable orientations. So o = 6. )
%
(c)
1 (the same as for a heteronuclear diatomic).
347
(d)
120° and 240° rotations about each of the four C3 axes produce 2(4) = 8
indistinguishable orientations; 180° rotations about each of the three C2 axes (these coincide with the S4 axes—see Fig. 21.22) produce 3 more
indistinguishable orientations. Adding in the original orientation, we get eel”. (e)
2 (the same as for a homonuclear diatomic).
(f) There are three C2 axes and o = 4.
(Cay er 22.65
S*_, = R(L.5 In 34.08 + 2.5 In 298.15 — 1.1517) = 152.87 J/molK. Ay =hI8r7clao, By = h/8n°clyo, Cy = etc. Equation (22.109) becomes Zrot =
(!7/6)(kTIncy?(AI Ay By Cy)”. Using o = 2, we get Zr = 125.9. So $2, = RU.5 + In 125.9) = 52.68 I/molK. @vin,i/T=(2615 em™ Jhc/kT = 12.62. Ovip2/T =5.709. Ovin.3/T= 12.68. Sip m= RU12.62/(e'7* — 1) In (1 — e712) + 5.709/(e7? — 1) — In (1 — >) + 12.68/(e'7 — 1) 
In (1 —e7!?*)] = 0.186 J/molK. S%,, =0. Adding, we get So, = 205.74 J/molK. 22.66
Siem = R15 In 44.01 + 2.5 In 298.15 — In 1 — 1.1517) = 156.05 J/molK. O.S
By hc/k = 0.561 K. For this linear molecule, Eq. (22.105) gives
Srom =R{1 + In [(298.15 K)/2(0.561 K)]} = 54.73 J/molK.
Ovin./T=
(1340 cm! )hc/kT = 6.466; Qyin.2/T = (667 cm )hc/kT = 3.219;
Ovin3/T =
3.219;
Oyin.4/T= 11.34. Equations (22.118) and (22.106) give S\,
R[6466/(e OS 1) Sn Cle)
eS 21D Ve
wes) oe
=
nl er) +
11.34/(e!!34 — 1) — In (1 — e!!™4)] = 0.3616R = 3.01 J/molK. Si,_ =0.
S°,x93= (156.05 + 54.73 + 3.01 + 0) J/molK = 213.79 J/molK. 22.67
Cy tnm = (3/2)R.
Cy vipsm — ¥ at high T for each vibrational mode s.
Cyrtm — Kat high T for linear molecules; Cy om nonlinear molecules.
348
— (3/2)R at high T for
(a)
Linear. 3(3)  5 = 4 normal modes.
54.04 J/molK.
(b)
Cy,, > RU.5+4+1)=6.5R=
Cp,, =Cy , +R = 62.36 J/molK.
Nonlinear. 3(3)  6 =3 normal modes.
Cy,, > R(1.5+3 + 1.5) =6R=
49.89 JimolK. C% _—> 58.20 J/molK. (c)
12 normal modes.
Gre)
22.68
35.03)
Cy , > R(1.5 + 12 + 1.5) = 15R = 124.72 J/molK.
/moles
Since z, is the same as for a gas of diatomic molecules, U, is given by (22.95). For a gas of linear molecules, Z,o 1s the same as for a gas of diatomics, so
(22.96) gives Zo. Use of Eq. (22.109) gives for a gas of nonlinear molecules:
Urot = NRT d\n ZjodT = nRT d(= In T + const)/dT =
=
3nRT. Equation (22.110) gives Uvin = SrRinidiabimi le (iesen *nRT(didT) Yin (1 =e ©: mR Ou es 1) 22.69
(a)
yi= =nRT Valet

COVE
= etat
)) =
AtOK, there is no translational, rotational, or vibrational energy (above
the zeropoint energy) for the H2 molecules; AU 4 is determined by the
change in electronic energy and AU 9 = (4.4780 eV)Na = (4.4780 eV)
Rip The translational contribution to Um of 2H(g) is
2( >RT) = 3RT. Hence AU9, — AUg = 3RT— 5 RT = 5 RT. Then AH 59g °
°
5
1
°
= AU 59g + AngRT/mol = AUg + Asie 4+ RT =AU> + =RT = 432.06
kJ/mol + 1.5(0.0083145 kJ/molK)(298.15 K) = 435.78 kJ/mol. The Appendix gives 2(217.965 kJ/mol) = 435.93 kJ/mol.
22.70 (a) Hoy —Hijg = Usp +RTUnny =U inte — Uno + RT =GI2)RT+ RT = 2.5(8.3145 J/molK)(298.15 K) = 6.1974 kJ/mol. (b) 2.5RT = 20.786 kJ/mol. 349
(C)i
Gp
hao
har oO San
ge aT
io ean rae YC
take H;,, —H},9 from (a) or (b) and use (22.108) for S 7, [Alternatively, (22.123) can be used.] Gj,59g — Hi, = 6.1974 kJ/mol — (298.15 K)(0.0083145 kJ/molK)[1.5 In 39.948 + 2.5 In 298.15 — 1.1517] = —39.970 kJ/mol. (d)
22.71
Gi jo00 — Amo = 20786 kJ/mol — 180.00 kJ/mol = —159.22 kJ/mol.
T _ From (22.123) with 7,9 =U ;,o, we have z = Ne —(GmrHmo)/R
(6.022 x 107*)exp[(257.7 J/molK)/(8.314 J/molK)] = 1.739 x 10°”. 22.72
A—Up=kT In Zand In Z=—(A — UpWkT.
AUp = U Up TS=
H — Ho) TS, since there is negligible difference between H — Ho and U — Up
for a solid. So A — Up = [523 J/mol — (298.15 K)(2.377 J/molK)](1 mol) =
~185.4J. 22.73
In Z= (185.7 JKT =4.51 x 10”.
Substitution of (22.124) into )); vu; =0 gives
0= ¥, VUmo, — RTE; v,In(z,/N,;) = AUS — RT In], (z,/N,)”']. So AUCIRT = In{T]; [(z,/VN.,)/(N,/VN, 1" } = In{T, [(z,/VN..)/c, J}. Then eXpAUG/RI)Y=l1(GZIVNS)/ cc) Nl, GivN) “Wiklete) lsc Ket
P(e
= exp AU Rd oleae VN)
22.74 S=kIn Wand AS =k In (Wena Winitiat), SO Wrinad Winitiat = Co. AS mix= —R(1.00 mol In 0.5 + 1.00 mol In 0.5) = 11.53 J/K. So Wena Winitiat = exp[(1 1.53 J/K)/(1.381 x 107? J/K)] = exp(8.35 x 107°). Let
10” = e83*! Taking base10 logs gives y = 8.35 x 107 log e = 3.63 x 10, SO Wena Winitial = 102
22.75
ee
S=kIn Wand In W= S/k = (191.61 J/K)/(1.3807 x 10°73 J/K) = 1.388 x 1029, where the Appendix was used.
350
22.76
Stom = SNan =kIn
Wrom —kIln
Wan
=k In [eo
Wnan] —kln
Wan
=
kine") +k In Wyan — k In Wyan = K(10"2) = (1.4 x 107 I/K)10" = 107" I/K, which is utterly negligible.
22.77
(a) Yes;
22.78
(a)
Ne.
(b)
Each has 2 electrons. The single +2charged nucleus in He holds the 2 electrons more tightly than the two +1charged nuclei in H2, so H2 is more polarizable.
(c)
Ch.
(d)
Each has 18 electrons. H2S is more polarizable.
(e)
CrHe.
22.79
(b) no;
(c) no.
[In the first printing, Eq. (22.133) has an error; the 47t€, should be squared. ]
Var’ = —2[(1.60 D)(3.336 x 10°°° C m/D)]*/ {3(1.38 x 10°? J/K)(298 K)[4n(8.854 x 107!” C?/Nm7)]?} =106 x 10°” J m®. Vaigr® = 2[(1.60 D)(3.336 x 10°°° C m/D)]*(6.48 x 10°? m*)/ [4(8.854 x 107!? C?/Nm7] = 33 x 10°” J m°. Vaisp?” = [3(11.3 eV)7/2(22.6 eV)](1.602 x 107"? J/eV)(6.48 x 10° m*)’ = =5/0 10; Im. 22.80
22.81
F =—dVidr = 4e(120'/r? — 60°/r’) = (24¢/r)[2(a/r)"” — (G/r)°). € = 0.013 eV =2.1 x 107! Jando =3.8 A. 10> (Selo mileG.e/8), — (4.8/8) = =a (ane eto 10> NO (Cc) 30. < 107 N, Ne (byl
(a) (b) (ce) (d)
lOn
v=Alr'? Bir’. 0 =A/o'? — Bio®, so Bo’ = A and v = Bo'/r'? — Bir’. dv/dr =12Bo°/r'? + 6Bir’ = 0. So r8,, = 26° and rmin = 2'°0. € = (ce) — V(rmin) = 0 — [BO*(2!a)!?  Bi(2'*a)*] = B/40° and B = 4o°e. Substitution of 6 = rmin2”” in (22.136) gives v =
Ae(r!2 /4r'? 76 /2r°) = e[(rmin/7)'? — 2(rmin/ 7].
22.82
22.83
(b)
Tmin = 2! = 2'°(3.5 A) = 3.9 A. min = 2!°(8.6 A) = 9.65 A.
(a)
As noted on p. 868 of the text, for a nonpolar molecule € = 1.3k7) =
(a)
1.3(1.38 x 10°23 J/K)(27.1 K) = 4.86 x 10°” J.
22.84
(b)
From Sec. 8.3, T, = 1.67», 80 € = 1.3k(Te/1.6) = 3.42 x 107" J.
(a)
We approximate the intermolecular potential by the LennardJones
potential. F(r) = dV/dr = 4e (126'7/r?’ + 60°/r’) = 24e(20'4/r'? ~ o°/r’) = c/r'? — d/r'. The F (r) curve has the same general
appearance as the v(r) curve, being 0 at r = 9, negative (i.e., attractive) for large r, positive for small r, and o at r= 0.
(b)
0 = 24e(20'7/b"3 — o/b’), so 20° = b® and b = 2"o. [Cf, Prob. 22.81b; F(r) = 0 at the minimum in v.]
22.85
22.86
(a)
Xe (higher M);
22.87
(a)
v/e = 4[(6/2'%o)'* — (6/2'%o)"] = 4(~ — +) = =I (this is the minimum),
(b)
4[(o/1.50)!? — (o/1.50)°] = 0.320;
(c)
0.062;
(a)
For an ideal gas, /= 0; for linear molecules Eq. (22.145) becomes
22.88
(b) C2H5OH (H bonding);
(d) —0.016;
(c) H20.
(e) —0.005.
Zeon = (40) J§ sin 0, dO,  [5 sin O@v dOy J5” do,  J2" doy x Jo dx J dy: [6 dz  + S5 dxy J8 dyy JS dey=(4m) %2%(20)% (abc) = 352
V™. For an ideal gas of nonlinear molecules, (47) is replaced by
(81°) = (2n) “(411)” and we have the additional integral factors Ro aaee
(b)
Jo" dx = (2m), so we still get Zoon = VY.
P=kT(0 In V*/OV)rN = NKT(O In V/OV) = NKTIV.
22.89 B=2nNal? (eT — 1) dr =2nNalJ§ (CO — Dr drt [5 (€  Vr ar] = —2nNa(J¢r° dr + 0] = 2mNad’/3 = 4Nal =n(d/2)"]. 22.90
At 100 K, log(kT/e) = log[(1.381 x 10° J/K)(100 K)/(1.31 x 107" J)} = 0.0229. For this log(kT/e), Fig. 22.22 gives Big Na]—9 4nd
B= 5.4
(3.74 x 108 cm)*(6.022 x 1077/mol) = 170 cm*/mol. At 300 K, log(kT/e) =
0.500; Fig. 22.22 gives Bio’Na =0.1 and B= 3 cm*/mol. At 500 K,
log(kT/e) = 0.722; B/o*N, = 0.5 and B = 16 cm*/mol.
22.91 (a)
B=2mNalo(e — lr dr—2nNa Jo (Ee — Vr? dr 2nNa Jz (e  Dr? ar = 2nN,0°7/3 — 2nNa(e™" — 1)(a? — 0/3 + 0 = 20Na(O? — a )e""/3 + 2nNaa’/3.
(b)
Substitution of numerical values gives B(100 K) =163 cm?/mol,
B(300 K) = —4 cm?/mol, B(500 K) = 17 cm*/mol. (c)
22.92
From (8.5) and (8.6), Z= 1+ B'P=1+BP/RT = 1 — (163 cm°/mol)(3.0 atm)/(82.06 cm?>atm/molK)(100 K) = 0.94.
U=kT°(A In Z/OT)yn = kT In Ziel )v.n + kT In Zeon/OT)v.n = Via + kT°(N)[(1/Vm)dB/dT+ (1/2V 7 )dClaT ++ = Uig — nRT’[(1/Vm)dBlaT+ (1/2V>)dClaT ++). S=U/T+klnZ= UsalT —nRT [(1/Vm)dBIdT + (1/2V2 )dClaT ++) + k An Zig + KN In V— NE[(/Vm)B + (1/2V2,)C ++] Nk In V= Sig —nR{(A/Vm)(B + T dBIdT) + (1/2V2, (C + T dCldT) +   «J, since Sia = ) = UsalT +k in Zig. G=A+ PV =KT In Z + KTV In Zeon/OVrN _kT In Zig— KIN In V + KTN(B/Vm + C/2V5,+ +>) +kIN In V+ kTVN(A/V + B/VVm + C/VV2, ++) = Gia + NRT(2B/Vin + 3C/2V2, +), since Gig =
Aia + (PV)ia = KT In Zig + NAT.
353
22.93
A=kT
In Zand Afalse = —kT In Zfatse, SO Afalse — A = kT In (Ztatse/Z) =
(1.38 x 1073 J/K)(298 K) In 1019"? = (4.11 x 107?! J)2.30 log 101° = (9.5 x 10°77! J(£10'5) = F 10° J, which is negligible.
)IN= 6 22.94 (N,)IN=e §""" Iz, s0 (Ng, 22.95
22.96
z= Iz = 1/154.1 = 0.00649.
(a)
Since A is proportional to n and A = 4T In Z, we see that In Z is proportional to n. From (22.81), Zr is proportional to V, which is proportional to n; sO Z is proportional to n. Since Zr is proportional to n, z iS proportional to n.
(b)
Zot, Zvib» aNd Ze) are independent of n.
(c)
None.
(d)
Zrot, Zvibs and Zel.
Every thermodynamic system in equilibrium.
354
Chapter 23 23.1
Az=keFa!®l = ke Fw!RT ol? — Nand? e'?(4RT/1Mg)'”, where Eqs. (17.69), (23.5), and (23.4) were used.
23.2
(a)
Equation (23.6) gives A =
(6.0 x 107°/mol)n(3.4 x 107'° m)*[8(8.314 J/molK)(500 K)/r]'”” x [mol/(0.030 kg) + mol/(0.048 kg)]'"(2.72)'" = 3 x 108m? s7! mol! =3 x 10'* cm? mol! st.
(b)
8x 10!! cm? mol! s™ = p(3 x 10'* mol! s“') and p = 0.003.
23.3
391 nm.
23.4
(V,c) = IF Vreg(Vre) Bre = Womyl2MKT)"” Jy Vie EXP(MrcV eekly dvs = " = (2kT Item)". (2KTIm,)/2. 2(myel 2UKT)
23:5
The energy per molecule is (9.6 kcal/mol)/Na =
[(9600 cal/mol)/(6.02 x 107°/mol)](4.184 J/1 cal)(10’ erg/1 J) = 6.7 x 10°" erg.
23.6
(a)
p1H{H, d=0.930 A (Example 23.2). Let the origin be at D and let the molecular axis be the x axis. The center of mass 1s at Xem = > miX/ Mor
= [2.014(0) + 1.008d + 1.008(2d)]/4.030 = 0.698 A. The principal axes pass through the center of mass (which is on the molecular axis and 0.698 A away from D); one principal axis is the molecular axis; the other two are perpendicular to this axis.
I, pn, = Li mir, =Nq' [(2.014)(0.698)° + 1.008(0.930 — 0.698)" + 1.008(1.860 — 0.698)?](g/mol) A? = 2.397N, (g/mol) A’. (b) For Hp, fy, = PR? = (mj,/2my)R° = ‘mpR® = N; *(1.008)(0.741)°(g/mol) A? = 0.2767N (g/mol) A*. Then Iy.pn,/ly, = 2397/0.2767 = 8.66. Since 04, =2 and py, =I, the rotationalpartitionfunction ratio is 2(8.00) =e
355
23.7
(a)
Zvinn, =[lexph¥c/kT)]; hVc/kT =(6.626 x 10> sis) (4400 cm™!)(2.998 x 10!° cm/s)/(1.3807 x 10°°° J/K)(450 K) = 14.07; Zyipn, =(1e7'*°”)' = 1.000. The activated complex has 3 ordinary vibrations, with wave numbers
1764, 870, and 870 cm”! and
zi is the product of factors for each of these vibrations. We have
AV guclkT = h(1764 cm” )c/kT = 5.640, AV penac/kT = h(870 cm” )c/kT = 2.78). Then z?,=(ler ss )) (lee) = 1.141 bom (22.81) we find: (2° /VV[(zu.p/V)(Z un, /V)) = (pn, fp my, P(r /2nkTY” = [4.030/(2.014)2.016]°(mol/10~ kg)*”(6.022 x 10°°/mol)*” x (6.626 x 10° J s)>[27(1.3807 x 10°73 J/K)(450 K)]}°? = 5.51 x 107! m?=5.51 x 10° cm’. NakTh ‘exp (—Ae§/kT) = (6.022 x 107°/mol)(1.381 x 107? J/K)(450 K)(6.626 x 10°" Js)! x exp{(—5.79 x 10°? J)/[(450 K)(1.381 x 10°? J/K)]} = 5.07 x 10°*/mols. Then k, = (5.07 x 10°7/mols)(5.51 x 10°? em®)(17.3)(1.141)(1) = iy
(b)
5.5 x 10’ cm? mol! $7.
23.8
Replacing 450 K by 600 K in Prob. 23.7b, we find NakTh ‘exp (Ne
6.94 x 10°*/mols. Replacing 450 K by 600 K in Prob. 23.7b, we find t
(25 /V@eD/V)@ wn, (V)] =3.58 x 10 m°, Zinn, = 1.000, zii, = 1.323. So k, = (6.94 x 10°*/mols)(3.58 X 10° cm*)(17.3)(1.323)(1) = 5.7 x 10!° cm?/mols. The tunneling correction 1s 7.5/5.7 = 1.3, which is less
than that at 450 K; tunneling becomes less important as T increases.
23.9
VY =vie=(1/2nc)(Kp)'”.
Uh,
kis the same for H2 and D2 so Vp,/Vu, =
ye ( (covalent solid);
(d) MgO (greater ionic charges). For H20(£) > H20(g), Ec.208 = AH 39, = 44.0 kJ/mol.
Ecou = —(1.6022 x 107!? C)?(1.74756)(6.0221 x 10°°/mol)/ 4n(8.8542 x 107}? C?/Nm7)(2.798 x 107'° m) = 867.8 kJ/mol. 24.7
n= 1+72n(8.854 x 107!? C?/Nm*)(2.798 x 107'° m)*/ (3.7 x 107! (cm?/dyn)(m2/10* cm?)(dyn/10~ N)}(1.602 x 107! C)°1.74756 = 8.4.
24.8
(a)
From (24.11) with Ro at O K approximated by Ro at 25°C, we have
n= 1+72n(8.854 x 107'? C?/Nm7)(3.299 x 10°'° m)* (10° Pa)/ (5.5 x 10°)(1.602 x 10°’? C)71.74756 = 10.6. From (24.9), E. = (1.602 x 107!? C)?1.74756(6.022 x 107°/mol)(1 — 1/10.6)/ 4n(8.854 x 107!? C?/Nm*)(3.299 x 107'° m) = 666 kJ/mol. 362
(b)
From Fig. 24.17, the nearestneighbor distance Ro is onehalf the length d
of the diagonal of the unit cell. As in Fig. 15.1 and Eq. (15.1),
d’ =a’ +a’ +a’ and d = 3"a, where a is the edge length. So Ro =
¥2(3")a = ¥2(3")(4.123 x 107! m) = 3.571 x 107° m. Fig 24.17 shows
that there is one CsCl ion pair per unit cell, so the molar volume is
Vino = Nag’ = Na(2Ro/3"”)’ = 8R4,.Na/3>” and Eq. (24.10) becomes n = 1 + 32me9(3'”)RG/Ke°M= 1 + 32n(8.854 x 10°? C?/Nm2)(3") x (3.571 x 107"? m)*(10° Pa)/(6 x 10°°)(1.602 x 107!? C)?1.762675 = 10.9. Eq. (24.9) gives Ex = (1.602 x 107'? C)*1.762675(6.022 x 107/mol) x (1 — 1/10.2)/4m(8.854 x 107'* C*/Nm’*)(3.571 x 107'° m) = 620 kJ/mol. In both (a) and (b), the theoretical value is less than the experimental
value due to neglect of the dispersion energy.
24.9
Vin = 2NaR’, SO (OVm/OR)r.p = 6NaR?. Taking P as 1 atm = 101325 N/m? and putting R equal to the NaCl equilibrium value 2.80 A, we have
P(0V m/OR)r p = (101325 N/m”)6(6.02 x 107*/mol)(2.80 x 107!° m)* = 2.9 x 10'° N/mol. Taking Um = Ep, we have (0U,/OR)r.p = (E,/OR)rp =
(e7/4m€) MNAR 9 (—Ro/R° + R2/R"*') = (e7/4€o) MNa(RO'/R"*! — 1/R?). At the equilibrium value R = Ro, QU/OR is zero, but for a value R = 2.81 A that
deviates by 0.01 A from Ro, we have, using n = 8.4 and
e*/Amteg = (1.6 x 107!” C)7/40(8.85 x 10°12 C? N! m”) = 2.3 x 10°78 N m’, 0U,/OR = (2.3 x 10°78 N m’)(1.75)(6.02 x 107°*/mol) x (2. 8054107,.m)) #8 loelOaeiny ce t/2:81 4109 mA — 8.0 x 10'° N/mol, and P(0Vp/9R) is negligible compared with 0U,/OR. 24.10
(a)
(0’Up/OV2,)r = T(0°P/OVm OT) — (OP/AVm)r. AS T — 0, the first term on the right side of this equation goes to zero and (0°U,»/0V 5,)r>
~(OP/OVm)r = 1/KVm, Where K = —V_! (OVn/0P)r was used. (b)
Use of Vee = c!?R and Vet — eR A in (24.8) gives the desired equation
for —E,; differentiation of this equation gives 201k jd
=
(e7/4ne.MNaRo ((1)(3)VinoVm "SHS — DV oVm I pan, = (€/4MEOMNAR9Vino X Setting Vm = Vmo. We get J°EplOV (1 —n)/9. Substitution of this result into the OK equations
363
 =O Un/OVinnny = 1/KVmo gives (e7/4me)MNaRo! Xx PE /OV>pony Vz2)(n— 1/9 = 1/KVmo; solving for n, we get (24.10). 24.11
A given positive ion has two negative ions at a distance R, two positive ions at 2R, two negative ions at 3R, two positive at 4R, etc. The potential energy of interaction between one positive ion and all the other ions is
(e2/4me€o)(2/R + 2/2R — 2/3R + 2/4R — +  +) = —(e*/4rt€o MIR, where M =
2(1 — 1/2 + 1/3  1/4 + 1/5 ). By symmetry, the potential energy of interaction between a given negative ion and all the other ions is MR by Ng and division by 2 (to —(e7/4m€9M4/R. Multiplication of ~2(e7/4mte9 avoid counting each interionic interaction twice) gives Ecoui =
—(e7/4m&) MNA/R, as in Eq. (24.4). Equation (8.36) with x = 2 gives In 2 =
fea
24.12
4 =2 In 2 =i38622" so 4e 1/3 —1/
(a)
—E, = (6.022 x 107*/mol)(118 K)(1.381 x 10° J/molK) x [24.264(3.50 A/3.75 A)!? — 28.908(3.50 A/3.75 A)°] and E, = 8.3 kJ/mol.
(b)
At equilibrium, 0E,/0R = 0E,/0R = 0 =
Nat{—12(24.264)o'7/R" + 6(28.908)6°/R’}; so RS = [12(24.264)/6(28.908)]o° and Ro/o = 1.09. The experimental value is Ro/o = (3.75 AV/(3.50 A) = 1.07. 24.13
E, = 24(6.022 x 107°/mol)(0.0101 eV)(1.602 x 10°” J/eV) x [(3.50/3.75)!* — (3.50/3.75)°] = 5240 J/mol = 1.25 kcal/mol.
24.14
(a)
Each of the 8 points at the corners is shared with a total of 8 unit cells.
The point within the unit cell is not shared. So each unit cell has 8/8 + 1 = 2 lattice points.
24.15
(b)
8/8 +2/2=2
(a)
The unit cell has 8/8 + 6/2 = 4 lattice points. There is one basis group at each lattice point, so each unit cell has 4 basis groups.
(b)
Each unit cell has 8/8 =  lattice point and therefore has  basis group.
364
24.16
An orthorhombic lattice has 90° angles and Eq. (24.12) gives
Z = pabcNa/M = (2.93 g/cm’)(4.94 x 10° cm)(7.94 x 108 cm) x (5.72 x 10° cm)(6.022 x 107*/mol)/(100.09 g/mol) = 3.96 = 4. There are 4 formula units and hence 4 Ca”* ions per unit cell. 24.17
A tetragonal lattice has 90° angles and has a = b. Equation (24.12) gives
p = MZ/abcNg = (79.899 g/mol)2/(4.594 x 10° cm)?(2.959 x 107° cm) x
(6.022 x 107*/mol) = 4.249 g/cm’. 24.18
The c intercepts are all at o0, so the Miller / index is 0 in each case. With origin at the leftmost dot in the third row from the bottom, the leftmost p2 plane intercepts the a axis at  and the b axis at —’. The reciprocals of these intercepts give the Miller indices as (120). With origin at the sixth dot in the bottom row, the s3 surface intersects the a axis at  and is parallel to the b axis (intercept at oo). The reciprocals give (100) as the Miller indices. [The p>
planes can also be called (i200)
24.19 (11)
(10) e
ie 24.20
o_o
(02) _@
(12)

e
o_o
e
(11) e
ry
—e—_e—_e—
@
e
A bodycentered lattice has 8/8 + 1 = 2 lattice points per unit cell. There is one basis group per lattice point, so there are 2 basis groups per unit cell. The basis therefore has 16/2 = 8 molecules of COCI.
24.21
For a spherical atom of radius r inscribed in a cubic unit cell of edge length 2r, the atom’s volume is snr and the unit cell’s volume is (2r)° = 87°. The
percentage of occupied space 1s (£17°/8r°) 100% = (1/6)100% = 52.4% and there is 47.6% empty space.
365
24.22
The shaded atom touches the six atoms that have a single dot; all seven of these atoms lie in the (111) plane. The shaded atom also touches the three atoms with two dots [which lie below the (111) plane of the shaded atom] and
touches the three atoms with three dots [which lie above the (111) plane of the shaded atom]. (The two atoms drawn with broken circles lie on the back faces
of the unit cells.)
24.23
A cubic unit cell has right angles and has a = b =c, so Eq. (24.12) gives
0 = MZ/abcNa = (58.10 g/mol) 4/a*(6.022 x 10”*/mol) = 2.48 g/cm’. We get a = 5.38 x 10° cm = 5.38 A. As is clear from Fig. 24.16b, the nearestneighbor distance is Yaa = 2.69 A. 24.24
The CsCl space lattice is simple cubic with Z = 1; Eq. (24.12) gives a =
MZ/ONa = (212.8 g/mol) 1/(4.44 g/cm*)(6.022 x 107*/mol) = 7.96 x 10°? cm’, so a = 4.30 A. From Fig. 24.17a, the nearestneighbor distance is half the length of the diagonal of the cubic unit cell, which is 113 a =3.72 A.
24.25
From Fig. 24.18 and the associated discussion, the lattice is facecentered cubic with a = b = c; there are 8 F ions and 8/8 + 6/2 = 4 Ca’* ions per unit
cell, so Z = 4. Equation (24.12) gives a= MZ/pPNa = (78.08 g/mol)4/
(3.18 g/cm?)(6.022 x 107*/mol) = 1.63 x 10°** cm? and a = 5.46 A. 24.26
The lattice is facecentered cubic with 90° angles and a = b = c. Equation
(24.12) gives a = (MZ/pNa)'” = [(12.011 g/mol)8/(3.51 g/cm*)(6.022 x 1077/mol)]'? = 3.569 A. Nearest
366
neighbor atoms are at points 0 0 0 and