FQ II Resolvidos

Prévia do material em texto

5 Simple mixtures 
Answers to discussion questions 
OS.1 At equilibrium, the chemical potentials of any component in both the liquid and vapor phases must be 
equal. This is justified by the requirement that, for systems at equilibrium under constant temperature and 
pressure conditions, with no additional work, 6.G = 0 [see Section 3.5(e) and the answer to Discussion 
question 3.3] . Here 6.G = 11-;(v) - 11-; ( 1) , for all components, i , of the solution; hence their chemical 
potentials must be equal in the liquid and vapor phases. 
OS.3 All of the colligative properties are a function of the concentration of the solute, which implies that the 
concentration can be determined by a measurement of these properties. See eqns 5.33, 5.34, 5.36, 5.37, 
and 5.40. Knowing the mass of the solute in solution then allows for a calculation of its molar mass. For 
example, the mole fraction of the solute is related to its mass as follows: 
The only unknown in this expression is MB which is easily solved for. See Example 5.4 for the details 
of how molar mass is determined from osmotic pressure. 
OS.S A regular solution has excess entropy of zero, but an excess enthalpy that is non-zero and dependent on 
composition, perhaps in the manner of eqn 5.30. We can think of a regular solution as one in which the 
different molecules of the solution are distributed randomly, as in an ideal solution, but have different 
energies of interaction with each other. 
ES.1(b) 
Solutions to exercises 
Total volume V = nA VA + nB VB = n(XA VA + XB VB) 
Total mass m = nAMA + nBMB 
n = 1.000kg(103 g/kg) = 4.6701 mol 
(0.3713) x (241.1 g/mol) + (l - 0.3713) x (I98.2g/mol) 
E5.2(b) 
E5.3(b) 
E5.4(b) 
92 STUDENT'S SOLUTIONS MANUAL 
v = n(XA VA + XB VB) 
= (4.6701 mol) x [(0.3713) x (188.2 cm3 mol-I) + (1 - 0.3713) x (176.14 cm3 mol-I)] 
= 1843.5 cm3 1 
Let A denote water and B ethanol. The total volume of the solution is V = nA VA + nB VB 
We know VB; we need to determine nA and nB in order to solve for VA. 
Assume we have 100 cm3 of solution; then the mass is 
m = pV = (0.9687 g cm- 3) x (100 cm3) = 96.87 g 
of which (0.20) x (96.87 g) = 19.374 g is ethanol and (0.80) x (96.87 g) = 77.496 g is water. 
77.496 g 
nA = = 4.30 mol H20 
18.02 g mol- 1 
19.374 g -
nB = I = 0.4205 mol ethanol 
46.07 g mol-
V - nB VB 100 cm3 - (0.4205 mol) x (52.2 cm3 mol-I) 
--- = VA = ---------=--------
nA 4.30 mol 
= 18.15 cm3 
Check that PB/XB = a constant (KB) 
0.010 0.015 0.020 
(PB /XB) / kPa 8.2 x 103 8.1 x 103 8.3 x 103 
KB = pix, average value is 18.2 x 103 kPa I 
In Exercise 5.3(b), the Henry's law constant was determined for concentrations expressed in mole 
fractions. Thus the concentration in molality must be converted to mole fraction . 
1000 g -
meA) = 1000 g, corresponding to n(A) = I = 13.50 mol nCB) = 0.25 mol 
74.1 g mol-
Therefore, 
0.25 mol -
XB = = 0.0182 
0.25 mol + 13.50 mol 
using KB = 8.2 x 103 kPa [Exercise 5.3(b)] 
P = 0.0182 x 8.2 x 103 kPa = 11.5 x 102 kPa I 
E5.5(b) 
E5.6(b) 
E5.7(b) 
E5.8(b) 
We assume that the solvent, 2-propanol, is ideal and obeys Raoult 's law. 
49.62 
xA(solvent) = p jp* = -- = 0.9924 
50.00 
MA(C3H80) = 60.096 g mol- I 
250g -
nA = I = 4.1600 mol 
60.096 g mol-
nA nA 
XA = nA +nB =-
nA + nB XA 
nB = nA C~ -1) 
= 4.1600 mol (0.9~24 - I) = 3.186 X 10-2 mol 
MB = 8.69 g 2 = 273 g mol-I = 1270 g mol-I 1 
3.186 x 10- mol . . 
Kr = 6.94 for naphthalene 
mass ofB 
MB=---
nB 
nB = mass of naphthalene · bB 
t3.T (mass of B) x Kr 
bB = - so MB = .,------::c--:-:--:---:----:--= 
Kr (mass of naphthalene) x t3.T 
(5.00 g) x (6.94 K kg mol-I) 1 _ I 1 
MB = = 178 g mol (0.250 kg) x (0.780 K ) 
nB nB 
t3.T = KrbB and bB = -----
mass of water V p 
P = 103 kg m-3 (density of solution ~ density of water) 
nv 
nB = RT 
n I 
t3.T = Kr -- Kr = 1.86 K mol- kg 
RTp 
SIMPLE MIXTURES 93 
(1.86 K kg mol-I) x (99 x 103 Pa) 2 
t3.T = = 7.7 x 10- K (8.314 JK- I mol-I) x (288K) x (l03kgm-3) 
Tr = 1-0.077 °C I 
t3.miXG = pV(! In! +! In!) = -pVln2 
= -(100 x 103 Pa) x (250cm3) C;6 ::3) In 2 
= -17.3 Pa m3 = 1-17.3 J 1 
t3. . S = -t3. mixG = 17.3 J = 16.34 X 10-2 J K- I 1 
mix T 273 K . . 
94 STUDENT'S SOLUTIONS MANUAL 
ES.9(b) t.mixG = nRT LXJ InxJ [5 .18] 
J 
'" -t.mix G t.mixS = -nR ~xJ InxJ [5.19] = ---
J T 
n = 1.00 mol + 1.00 mol = 2.00 mol 
x (Hex) = x (Hep) = 0.500 
Therefore, 
t.mix G = (2.00 mol) x (8.314 J K- 1 mol-I) x (298 K) x (0.500 In 0.500 + 0.500 In 0.500) 
= 1-3.43 kJ 1 
+3.43 kJ 1 - I 1 t.mixS = 298 K = +11 .5 J K 
t.mixH for an ideal solution is zero as it is for a solution of perfect gases [7.20]. It can be demonstrated 
from 
ES.10(b) Benzene and ethyl benzene form nearly ideal solutions, so 
To find maximum t.mixS, differentiate with respect to XA and find value of XA at which the derivative 
is zero. 
Note that XB = 1 - XA so 
dln x 
use -- =-: 
dx x 
d XA 
-(t.mixS) = -nR(lnxA + I -!n(l - XA) - I) = -nRln--
dx I -XA 
I 
= 0 whenxA =-
2 
Thus the maximum entropy of mixing is attained by mixing equal molar amounts of two components. 
nB mB / MB mE ME 106.169 
- = I = -- x - = - = --- = 1.3591 
nE mE / ME mB MB 78.115 
mB = 1 0.73581 
mE 
ES.11 (b) With concentrations expressed in molalities, Henry 's law [5.26] becomes PB = bBK. 
Solving for b, the molality, we have bB = PB / K = xPlolal/K and Plolal = Palm 
E5.12(b) 
SIMPLE MIXTURES 95 
For N2 , K = 1.56 x 105 kPa kg mol-I [Table 5.1] 
0.78 x IOl.3kPa 10 Ik - I I b = = .51 mmo g 
1.56 x 105 kPa kg mol-I 
For 0 2, K = 7.92 x 104 kPa kg mol - I [Table 5.1] 
0.2 1 x 101.3 kPa I I I b = = 0.27 mmol kg-
7.92 x 104 kPa kg mol-I 
bB = PB = 2.0 x 101.3 kPa = 0.067 mol kg- I 
K 3.0 1 x 103 kPa kg mol - I 
The molality will be about 0.067 mol kg - I and, since molalities and molar concentrations for dilute 
aqueous solutions are approximately equal, the molar concentration is about 10.067 mol dm- 3 1 
E5.13(b) The procedure here is identical to Exercise 5. 13(a) . 
/::;.rusH (I I) In XB = -- X - - - [5 .39; B, the solute, is lead] 
R T* T 
(
5.2X 103 J mol - I ) (I 1) 
= 8.314 J K- I mol-I x 600 K - 553 K 
= -0.0886, implying that XB = 0.92 
n(Pb) XBn(Bi) 
XB = , implying that n(Pb) = ---
n(Pb) + n(Bi ) I - XB 
. . 1000 g 
For I kg of bismuth, n(BI) = I = 4.785 mol 
208.98 g mol -
Hence, the amount of lead that dissolves in I kg of bismuth is 
(0.92) x (4.785 mol) ~ 
n(Pb) = I _ 0.92 = 55 mol , or ~ 
COMMENT. It is highly unlikely that a solution of 11 kg of lead and 1 kg of bismuth could in any sense be 
considered ideal. The assumptions upon which eqn 5.39 is based are not likely to apply. The answer above 
must then be considered an order of magnitude result only. 
E5.14(b) Proceed as in Exercise 5. 14(a) . The data are plotted in Figure 5.1, and the slope of the line is 
1.78 cm/(mg cm-3) = 1.78 cm/(g dm- 3) = 1.78 x 10- 2 m4 kg-I . 
96 STUDENT'S SOLUTIONS MANUAL 
Figure 5.1 
Therefore, 
(8.314 J K- I mol-I) x (293.15 K) 1 -I 1 M = = 14.0 kg mol (1.000 x 103 kgm-3) x (9.81 m s-2) x (1.78 x 10-2 m4 kg-I) 
ES.1S(b) Let A = water and B = solute. 
PA 0.02239 atm I I 
aA = - [5.43] = = 0.9701 P~ 0.02308 atm 
0.920 kg 
nA = = 51.05 mol 
0.01802 kg mol-I 
and nB = 0.122 kg = 0.506 mol 
0.241 kg mol-I 
5105 0.9701 ~ XA = . = 0.990 and YA = -- = 0.980 
51.05 + 0.506 0.990 
ES.16(b) B = Benzene /LB(I) = /LB(I) + RT InxB [5.25, ideal solution] 
RTlnxB = (8.314 J K- I mol-I) x (353.3 K) x (In 0.30) = 1-3536 J mol-II 
Thus, its chemical potential is lowered by this amount. 
PB = aBPB [5.43] = YBXBPB = (0.93) x (0.30) x (760 Torr) = 1212 Torr 1 
Question. What is the lowering of the chemical potential in the nonideal solutionwith Y = 0.93? 
ES.17(b) YA = PA PA = 0.314 
PA + PB 101.3 kPa 
PA = (101.3 kPa) x (0.314) = 31.8 kPa 
PB = 101.3 kPa - 31.8 kPa = 69.5 kPa 
ES.18(b) 
ES.19(b) 
SIMPLE MIXTURES 97 
PA 31.8kPa ~ 
aA=-= =~ p~ 73.0 kPa 
PB 69.5 kPa ~ 
aB=-= =~ Ps 92.1 kPa 
YA = aA = 0.436 = [ill 
XA 0.220 
aB 0.755 ~ YB=-=--=~ 
XB 0.780 
I = ~ "2:,;(bdb") z1 [5 .71] 
and for an MpXq salt, b+ / b" = pb/ b£>, b_ / b" = qb/b" , so 
I = ~(Pz~ + qz~)b/b" 
1= I (K3[Fe(CN)6D + I (KCI ) + I (NaBr) = ~ (3 + 32) b (K3 [F:~CN)6D + b(~:I) + b(::Br) 
= (6) x (0.040) + (0.030) + (0.050) = 1 0.320 1 
Question. Can you establish that the statement in the comment following the solution to Exercise 5. 18(a) 
holds for the solution of this exercise? 
b 
1= I (KN03) = b" (KN03) = 0.110 
Therefore, the ionic strengths of the added salts must be 0.890. 
(a) 
(b) 
b 
I (KN03) = b'" so b(KN03) = 0.890 mol kg- I 
and (0.890 mol kg- I) x (0.500 kg) = 0.445 mol KN03 
So (0.445 mol) x (10 1.11 g mol- I) = 145.0 g KN03 1 must be added . 
I b b 
I (Ba(N03h) = 2 (22 + 2 x 12) b" = 3 b" = 0.890 
0.890 " - I b = -3-b = 0.2967 mol kg-
and (0.2967 mol kg- I) x (0.500 kg) = o. 1484 mol Ba(N03h 
So (0. 1484 mol) x (261.32 g mol-I) = 138.8 g Ba(N03h 1 
ES.20(b) Since the solutions are dilute, use the Debye-Hiickel limiting law 
log Y± = - lz+z_IAlI /2 
I I 
1= 2 L z1(bdb£» = 2 { I x (0.020) + 1 x (0.020) + 4 x (0.035) + 2 x (0.035)} 
I 
= 0.125 
logy± = -I x 1 x 0.509 x (0.125)1 /2 = -0.17996 
(For NaCI) y± = 10-0. 17996 = 1 0.6611 
ES.21(b) 
98 STUDENT'S SOLUTIONS MANUAL 
Alz z_IIl/2 
The extended Debye-Htickellaw is log y ± = - ----'----'-+-~-
1 + BII/2 
Solving for B 
( I AIZ+Z_I) (1 0.509) B = - [Ifi + log y± = - (b/bG )I /2 + log y± 
Draw up the following table 
b/(mol kg-I) 5.0 x 10-3 10.0 x 10-3 50.0 x 10-3 
y± 0.927 0.902 0.816 
B 1.32 1.36 1.29 
Solutions to problems 
Solutions to numerical problems 
PS.1 PA = YAP and PB = YBP (Dalton's law). Hence, draw up the following table. 
PAikPa 0 1.399 3.566 5.044 6.996 7.940 9.211 10.105 
XA 0 0.0898 0.2476 0.3577 0.5194 0.6036 0.7188 0.8019 
YA 0 0.0410 0.1154 0.1762 0.2772 0.3393 0.4450 0.5435 
PBikPa 0 4.209 8.487 11.487 15.462 18.243 23.582 27.334 
XB 0 0.0895 0.1981 0.2812 0.3964 0.4806 0.6423 0.7524 
YB 0 0.2716 0.4565 0.5550 0.6607 0.7228 0.8238 0.8846 
The data are plotted in Fig. 5.2. 
11.287 12.295 
0.9105 
0.7284 
32.722 36.066 
0.9102 
0.9590 
We can assume, at the lowest concentrations of both A and B, that Henry's law will hold. The Henry's 
law constants are then given by 
KA = PA = 115.58 kPa I from the point at XA = 0.0898. 
XA 
KB = PB = 147.03 kPa I from the point at XB = 0.0895. 
XB 
PS.3 Vsalt = (av) mol- I [Problem 5.2] 
ab H20 
= 69.38(b - 0.070)cm3 mol- I with b == b/(mol kg-I). 
Figure 5.2 
Therefore, at b = 0.050 mol kg-I , Vsalt = 1-1.4 cm3 mol-I I. 
The total volume at this molality is 
v = (1001.21) + (34.69) x (0.02)2 cm3 = lool.22cm3 . 
Hence, as in Problem 5.2, 
SIMPLE MIXTURES 99 
(1001.22 cm3 ) - (0.050 mol) x (-1.4cm3mol- l ) 1 2 _I 1 
V(H20) = = 18.04cm mol . 55.49 mol 
Question. What meaning can be ascribed to a negative partial molar volume? 
PS.S Let E denote ethanol and W denote water; then 
v = nE VE + nw Vw [5.3]. 
For a 50 per cent mixture by mass, mE = mw , implying that 
V 
which solves to nE = --""77--;-;--
MEVW' 
VE+--
Mw 
nE Furthermore, XE = ---
nE +nw M · I+~ 
Mw 
Since ME = 46.07 g mol-I and Mw = 18.02 gmol- I, 
XE = 0.2811 , XW = 1 - XE = 0.7189. 
ME 
- = 2.557. Therefore 
Mw 
PS.7 
100 STUDENT'S SOLUTIONS MANUAL 
At this composition 
Vw = l7 .5cm3 mol- I [Fig.5. l ofthetextl. 
lOOcm3 
Therefore, /lE = 1 1 = 0.993 mol , (56.0cm3 mol - ) + (2.557) x (l7 .5cm3 mol - ) 
/lw = (2.557) x (0.993 mol) = 2.54 mol. 
The fact that these amounts correspond to a mixture containing 50 per cent by mass of both components 
is easily checked as follows : 
mE = /l EM E = (0.993 mol) x (46.07 g mol-I ) = 45.7 g ethanol, 
mw = /lWMW = (2.54 mol) x (I8 .02gmol- l ) = 45.7 g water. 
At 200 e the densities of ethanol and water are, 
PE = 0.789gcm - 3, Pw = 0.997gcm-3. Hence, 
VE = mE = 45.7g =!57.9 cm3! ofethanol, 
PE 0.789 gcm- 3 
mw 45.7g 1 31 Vw = - = 3 = 45.8cmofwater. 
Pw 0.997 gcm-
The change in volume upon adding a small amount of ethanol can be approximated by 
where we have assumed that both VE and Vw are constant over this small range of /lE. Hence 
3 (l.oocm3) x (0.789 g cm-3»)! ! L'l V "" (56.0 cm mol-I) x 1 = +0.96 cm3 . (46.07 g mol - ) 
L'lT 0.0703 K _I 
bB = - = 1 = 0.0378 mol kg . 
Kf 1.86 K / (mol kg- ) 
Since the solution molality is nominally 0.0096 mol kg-I in Th(N03)4, each formula unit supplies 
0.0378 ~ . . . . . 
0.0096 ",, ~. (More careful data, as descnbed m the ongmal reference gIves \J "" 5 to 6.) 
PS.9 The data are plotted in Figure 5.3. The regions where the vapor pressure curves show approximate straight 
lines are denoted R for Raoul! and H for Henry. A and B denote acetic acid and benzene respectively. 
As in Problem 5.8, we need to form YA = PA * and YB = PB * for the Raoult 's law activity coefficients 
XAPA XBPB 
and YB = PB for the activity coefficient of benzene on a Henry's law basis, with K determined by 
xBK 
extrapolation. We use p* A = 7.3 lcPa, Ps = 35.2 lcPa, and K*B = 80.0 kPa to draw up the following 
table. 
SIMPLE MIXTURES 101 
0.2 0.4 0.6 0.8 1.0 
XA Figure 5.3 
XA 0 0.2 0.4 0 .6 0.8 1.0 
PA/kPa 0 2.7 4.0 5.1 6.7 7.3 
PB / kPa 35.2 30.4 25.3 20.0 12.4 0 
aA(R) 0 0 .36 0.55 0.69 0.91 l.oo[PA/p~] 
aB(R) 1.00 0.86 0.72 0.57 0.35 O[PB/PS] 
YA(R) 1.82 1.36 1.15 1.14 l.oo[PA /XAP~] 
YB(R) 1.00 1.08 1.20 1.42 1.76 
-[PB/XBPS] 
aB(H) 0.44 0.38 0.32 0.25 0.16 O[PB/KB] 
YB(H) 0.44 0.48 0 .53 0.63 0.78 l.oo[PB /xsKBl 
GEjs defined as [Section 5.4] 
GE = 6mixG(actual) - 6mixG(ideal) = nRT(xA In aA + XB In aB) - nRT(xA InxA + XB InxB) 
and, with a = yx, 
GE = nRT(xA In YA + XA In YB). 
For n = I, we can draw up the following table from the information above and RT = 2.69 kJ mol - I. 
XA In YA 0 
XB In YB 0 
GE /(kJ mol-I ) 0 
0.2 
0. 12 
0 .06 
0.48 
0.4 
0.12 
0.11 
0.62 
0.6 
0.08 
0.14 
0.59 
0.8 
0.10 
0.11 
0.56 
1.0 
o 
o 
o 
102 STUDENT'S SOLUTIONS MANUAL 
PS.11 (a) The volume of an ideal mixture is 
PS.13 
so the volume of a real mixture is 
v = Videal + VE. 
We have an expression for excess molar volume in terms of mole fractions. To compute partial molar 
volumes, we need an expression for the actual excess volume as a function of moles. 
nln2 ( al(nl-n2») 
so V = nlVrn.1 +n2Vrn,2 + --- ao + . 
nl +n2 nl +n2 
The partial molar volume of propionic acid is 
_ ( av ) _ aon~ al (3nl - n2)n~ 
VI - - - Vrn I + 2 + , 
ani P,T,1I2 ' (nl + n2) (nl + n2)3 
1 VI = Vrn,1 + aoxi + al (3xl - X2)X~ I· 
That of oxane is 
(b) We need the molar volumes of the pure liquids, 
MI 74.08gmol- 1 3-1 
Vrn I = - = 3 = 76.23cm mol 
PI 0.97174gcm-
86.13 gmol- I 3-1 
and Vrn2 = 3 = 99.69cm mol 
, 0.86398 g cm-
In an equimolar mixture, the partial molar volume of propionic acid is 
VI = 76.23 + (-2.4697) x (0.500)2 + (0.0608) x [3(0.5) - 0.5] x (0.5)2cm3mol-1 
= 1 75.63cm3 mol-II 
and that of oxane is 
V2 = 99.69 + (-2.4697) x (0.500)2 + (0.0608) x [0.5 - 3(0.5)] x (0.5)2 cm3 mol-I 
= 1 99.06 cm3 mol-I I. 
Henry's law constant is the slope of a plot of PB versus XB in the limit of zero XB (Fig. 5.4). The partial 
pressures of C02 are almost but not quite equal to the total pressures reported. 
.... 
OJ 
.D 
---
80 
60 
--;::; 40 
o 
U 
'i:( 
20 
o 
0.0 0.1 0.2 0.3 
Figure 5.4 
Linear regression of the low-pressure pointsgives KH = 1371 bar I. 
The activity of a solute is 
PB 
aB = - =XBYB 
KH 
so the activity coefficient is 
SIMPLE MIXTURES 103 
where the last equality applies Dalton's law of partial pressures to the vapor phase. A spreadsheet applied 
this equation to the above data to yield 
pfbar Ycyc Xcyc Ye02 
10.0 0.0267 0.9741 1.01 
20.0 0.0149 0.9464 0.99 
30.0 0.0112 0.9204 1.00 
40.0 0.00947 0.892 0.99 
60.0 0.00835 0.836 0.98 
80.0 0.00921 0.773 0 .94 
PS.1S C E = RTx(l - x){0.4857 - 0.1077(2x - I) + 0.0191(2x - 1)2} 
with x = 0.25 gives CE = 0.1021RT. Therefore, since 
~mix C(actual) = ~mix C(ideal) + nCE, 
~mixC = nRT(XA In xA + XB InXB) + nCE = nRT(0.25 In 0.25 + 0.75 In 0.75) + nCE 
= -0.562nRT + 0.1021nRT = -0.460nRT. 
Since n = 4 mol and RT = (8 .314 J K - I mol - I) x (303 .15 K) = 2.52 kJ mol-I , 
~mix C = (-0.460) x (4 mol) x (2.52 kJ mol - I) = 1-4.6 kJ I. 
PS.17 
104 STUDENT'S SOLUTIONS MANUAL 
Solutions to theoretical problems 
= gRTxA (\ - XA) + (1 - xA)gRT(l - 2xA) 
= gRT(1 - XA)2 = gRTx~. 
Therefore, 1 /LA = /L~ + RTlnxA + gRTx~ I· 
PS.19 nAdVA + nBdVB = 0 [Example 5.1]. 
PS.21 
nA Hence - dVA = -dVB . 
nB 
Therefore, by integration, 
jVA(XA ) XA dVA Therefore, VB(XA , XB) = VB (0, 1) - --. 
VA (0) 1 -XA 
We should now plot XA / (\ - XA) against VA and estimate the integral. For the present purpose we 
integrate up to VA (0.5 , 0.5) = 74.06 cm3 mol-I [Fig. 5.5], and use the data to construct the following 
table. 
74.11 
0.60 
1.50 
73.96 
0.40 
0.67 
73.50 
0.20 
0.25 
72.74 
o 
o 
The points are plotted in Fig. 5.5, and the area required is 0.30. Hence, 
V (CHCI3; 0.5 , 0.5) = 80.66 cm3 mol-I - 0.30 cm3 mol-I 
= 180.36cm3 mol-I I. 
InaA 
¢=--. 
r 
I 1 
Therefore, d¢ = - - d In aA + 2" In aA dr, 
r r 
1 
d InaA = - InaAdr - rd¢. 
r 
(a) 
(b) 
'< 
'i' 
~ 
>< 
1.5 
1.0 
0.5 
o 
72 
SIMPLE MIXTURES 105 
f 
I 
~ 
1;( 
--0 
~ ~ 0: ~ 
73 75 
Figure 5.5 
From the Gibbs-Duhem equation, XA dJ1.A = Xs dJ1.s = 0, which implies that (since J1. = J1.G + 
RT In a, dJ1.A = RTd In aA , dJ1.s = RTd In as ) 
XA d InaA 
dlnas = - - dlnaA = ---
Xs r 
I 1 
= -"2 In aA dr = dl/> [from(b)] = -I/>dr = dr = dl/> [from(a)] 
r r 
= I/>dlnr +dl/>. 
Subtract din r from both sides, to obtain 
as (I/> - I) 
din - = (I/> - I)d Inr+dl/> = --dr+dl/>. 
r r 
. . . (as) (YSXS) Then, by mtegratlOn and notmg that In - = In -- = In (YS)r=O = In I = 0, 
r r=O r r=O 
PS.23 A(s) ;=' A(I) . 
J1.~(s) = J1.~(I) + RTlnaA 
and ~fusG = J1.~ (I) - J1.~ (s) = - RT In aA· 
-~fusG 
Hence, In aA = ~. 
din aA I d (~fusG) ~fusH 
-- = --- -- = -- [Gibbs-Helmholtzeqn] . 
dT R dT T RT2 
For ~T = T( - T, d~T = - dT and 
106 STUDENT'S SOLUTIONS MANUAL 
Therefore, 
dlnaA -MA -MAdt:.T 
-- = -- and dlnaA = ----
dt:.T Kf Kf 
According to the Gibbs-Duhem equation 
which implies that 
nB 
and hence that d In aA = --d In aBo 
nA 
din aB nAMA I 
Hence --- = -- = --
'dt:.T nBKf bBKf 
We know from the Gibbs-Duhem equation that 
XA d In aA + XB d In aB = 0 
f XB and hence that J d In aA = - - d In aBo XA 
Therefore In aA = - f xB d In aB. 
XA 
The osmotic coefficient was defined in Problem 5.21 as 
I XA ¢ = -- In aA = -- In aA. 
r XB 
Therefore, 
XA f XB I lob 1 lob 1 lob 1 lob ¢=- -dlnaB=- bdlnaB=- bdlnyb=- bdlnb+- bdlny 
xB xA bob 0 bob 0 
1 lob 
=1+- bdlny. 
b 0 
From the Debye-Hiickellimiting law, 
In y = -A'b I /2 [A' = 2.303AJ. 
1 
Hence, dIn y = -2A' b-1 /2db and so 
I ( I ') {b 1/2 1 (AI) x _2b3/2 --11- _IAII /21. ¢ = 1+ b -2A 10 b db = I - 2 b 3 3 
COMMENT. For the depression of the freezing point in a 1, 1-electrolyte 
-D.fusH (1 1) and hence -rl/> = --- - - - . 
R T T* 
_ D.fusHxA (~ _ 2.) _ D.fusH xA (T* - T) "" D.fuSH xA D. T 
Therefore, I/> - R T T* - R IT* R T*2 Xs Xs Xs 
D.fusHD. T 
"" vRbsT*2MA 
MRT*2 
where v = 2. Therefore, since Kf = --, 
D.fusH 
~T 1/>---- 2bsKf . 
Solutions to applications 
PS.2S In thi s case it is convenient to rewrite the Henry 's law expression as 
(1) At PN2 = 0.78 x 4.0 atm = 3.1 atm, 
SIMPLE MIXTURES 107 
mass ofN2 = 3.1 atm x lOOgH20 x 0.18 ~gN2/ (g H20atm) = IS6~gN2 1. 
(2) At PN2 = 0.78 atm, mass of N2 = 114 ~g N21. 
(3) In fatty tissue the increase in N2 concentration from 1 atm to 4 atm is 
PS.27 (a) i = I only, N, = 4, K, = 1.0 X 107 dm3 mol - ' , 
v 4 x IOdm3 ~mol- ' 
[AJ I + IOdm3 ~mol-' x [A( 
The plot is shown in Fig. S.6(a) . 
(b) i = I; N, = 4, N2 = 2; K, = 1.0 X 105 dm3 mol-' = 0.10 dm3 ~mol-', 
K2 = 2.0 X 106 dm3 mol - ' = 2.0 dm3 ~mol- '. 
v 4 x 0.IOdm3~mol-' 2 x 2.0dm3 ~mol-' 
[AJ 1+ 0.IOdm3 ~mol-' x [AJ + 1+ 2.0dm3 ~mol-' x [A( 
The plot is shown in Fig. S.6(b). 
PS.29 By the van ' t Hoff equation [S.40], 
cRT fl=[BJRT= -. 
M 
108 STUDENT'S SOLUTIONS MANUAL 
40.0 r---------------, 
30.0 
<' 
"f 20.0 
10.0 
0.0 L.....L......L....l.-.L-.1-.L....I-..L.......I-.L....I-..L.....L-.L.-'-.L-.1--'--'--' 
0.0 0.2 0.4 0.6 0.8 1.0 
[Aj/(dm3 I1mol-l) Figure 5.6(a) 
5.-----------------------------, 
4 
3 
2 
2 8 10 
Figure 5.6(b) 
Division by the standard acceleration of free fall, g, gives 
n c(R/g)T 
g M 
(a) This expression may be written in the form 
cR' T 
n ' =--M ' 
which has the same form as the van't Hoff equation, but the unit of osmotic pressure (n ') is now 
force/area 
length/ time2 
(mass length)/(area time2) mass 
length/time2 area 
This ratio can be specified in g em - 2. Likewise, the constant of proportionality (R') would have the 
units of R/g, 
energyK- 1 mol-I 
length/ time2 
(mass length2/ time2) K- 1 mol-I ~---=-----'----;;---- = mass length K-1mol- l . 
length/ time2 
This result may be specified in I g cm K- I mol-I I. 
, R 
R=-
g 
8.3 1447 J K- I mol-I 
9.8066Sms-2 
I I (103 0) x (102
m
Cm) = 0.847840 kg m K- mol- kg '" 
I R' = 84784.0 g cm K- I mol-I I. 
In the following we will drop the primes giving 
cRT []=-
M 
and use the [] units of g cm- 2 and the R units g cm K- I mol-I. 
SIMPLE MIXTURES 109 
(b) By extrapolating the low concentration plot of [] Ie versus e (Fig. S.7(a» to e = 0 we find the 
intercept 230 g cm- 2 I g cm- 3 . In this limit the van' t Hoff equation is valid so 
500 
450 
;::-- 400 
I § 
00 
....... 
I' 350 E 
u 
~ 
....... 
~ 
§:300 
250 
RT RT 
- = intercept or M = ---
M intercept 
(84784.0gcmK- 1 mol- I) x (298.ISK) 
M- ---~-----~----~-------
- (230gcm- 2)/(gcm-3) , 
1M = 1.1 x 105 g mol- I I. 
Polyisobutylene in chlorobenzene 
at low concentrations 
Intercept: 230 gcm- 2 Ig cm- J 
• 
200 ~--~--~--~--~--~--~--~--~ 
0.000 0.010 0.020 
c/(gcm - J ) 
0.030 0.040 
Figure 5.7(a) 
110 STUDENT'S SOLUTIONS MANUAL 
(c) The plot of n/c versus c for the full concentration range (Fig. 5.7(b» is very nonlinear. We may 
conclude that the solvent is good. This may be due to the nonpolar nature of both solvent and 
solute. 
(d) n/c = (RT /M)(l + B'c + C'c2). 
Since RT / M has been determined in part (b) by extrapolation to c = 0, it is best to determine the 
second and third virial coefficients with the linear regression fit 
(n /c)/ (RT/M) - 1 = B' + C'c, 
c 
R = 0.9791. 
B' = 21.4cm3 g-', 
C' = 21lcm6 g- 2, 
standard deviation = 2.4cm3 g-'. 
standard deviation = 15 cm6 g-2. 
(e) Using 114 for g and neglecting terms beyond the second power, we may write 
1 
(I + "2 B'c) . 
CH 3 
o-CI tCH' - ~ t 
CH3 
Polyisobutylene in chlorobenzene 
7~ ,------------------------------, 
• 
6~ 
5~ 
a 
u 
~ 4~ 
• §: 
3~ 
2~ 
I~ 
o 
0.00 0.050 0.100 0.150 0.200 0.250 0.300 
c/(gcm- 3) Figure 5.7(b) 
SIMPLE MIXTURES 111 
We cansolve for 8 '; then g(8')2 = C', 
RT 1M has been determined above as 230 g cm-2/g cm-3. We may analytically solve for 8 ' from 
one of the data points, say, IT Ic = 430 gcm-2 Igcm - 3 at c = 0.033 g cm-3. 
(
430 g cm- 2 Ig cm-3) 1/2 1 , -3 
-23-0--'g'--c-m---=2'--1 g=-c-m----=-3 - I = 2. 8 x (0.033 g cm ). 
, 2 x (1.367 - I) - 3-1 
8 = = 22.2 cm g 
0.033gcm- 3 
Better values of 8 ' and C' can be obtained by plotting ( ~) 1/2 / (~) 1/2 against c. This plot 
is shown in Fig. S.7(c). The slope is 14.03 cm3 g- I. 8' = 2 x slope = 128.0 cm3 g- I I. C' is then 
1196 cm6 g-21. The intercept of this plot should theoretically be 1.00, but it is in fact 0.916 with a 
standard deviation of 0.066. The overall consistency of the values of the parameters confirms that g 
is roughly 114 as assumed. 
6.0 
5.0 
N 4.0 ~ 
r---... 
~ I ~ 
'---""' 
--........... 
3.0 
N 
r---... 
t::1 <.> 
'---""' 2.0 
1.0 
0.0 
0.00 0.05 0.10 0.15 
c/(gcm- J ) 
0.20 0.25 0.30 
Figure S.7(c) 
Phase diagrams 
Answers to discussion questions 
06.1 Phase: a state of matter that is uniform throughout, not only in chemical composition but also in 
physical state. 
Constituent: any chemical species present in the system. 
Component: a chemically independent constituent of the system. It is best understood in relation to the 
phrase 'number of components' which is the minimum number of independent species necessary to 
define the composition of all the phases present in the system. 
Degree of freedom (or variance): the number of intensive variables that can be changed without disturbing 
the number of phases in equilibrium. 
06.3 See Figs. 6.I(a) and (b). 
Liquid 
Solid 
p 
Critical point 
Triple point 
T Figure 6.1(a) 
p = constant 
Liquid A and B 
t 
T 
B 
,/ 
T* B 
Eutectic 1 
Solid B and solid AB 
06.5 See Fig. 6.2. 
E6.1(b) 
o 
A 
Liquid A 
and B 
SolidAB 
Solid A 
SolidAB 
Liquid A 
and B 
SolidAB 
Solid B 
SolidAB 
0.5 
Solutions to exercises 
Eutectic 2 
Solid AB and solid A 
0.50 
xA ~ 
1.0 
B Figure 6.2 
PHASE DIAGRAMS 113 
A 
Figure 6.1(b) 
E6.2(b) 
114 STUDENT'S SOLUTIONS MANUAL 
19kPa - 18kPa ~ 
XA = 20kPa _ 18kPa = ~ A is 1, 2-dimethylbenzene 
XAP~ (0.5) x (20 kPa) - ~ 
YA = = = 0.526 "'" L.22J 
Ps + (p~ - Ps) XA 18 kPa + (20 kPa - 18 kPa)0.5 
YB = 1 - 0.526 = 0.474 "'" 0.5 
PA = YAP = 0.612p = XAP~ = xA(68.8 kPa) 
PB = YBP = (I - YA)P = 0.388p = XBPS = ( I - XA) X 82.1 kPa 
YAP 
YBP 
XAP~ 0.612 
and 
XBPS 0.388 
(0.388) X (68 .8)XA = (0.612) x (82. 1) - (0.612)(82. I)xA 
26.694xA = 50.245 - 50.245xA 
50.245 ~ ~ 
XA = =~ XB = 1-0.653 =~ 
26.694 + 50.245 
P = XAP~ + XBPs = (0.653) x (68.8 kPa) + (0.347) x (82.1 kPa) = 173.4 kPa 1 
E6.3{b) (a) If Raoult 's law holds, the solution is ideal. 
PA = XAP~ = (0.4217) x (110.1 kPa) = 46.43 kPa 
PB = XBPs = (I - 0.4217) x (94.93 kPa) = 54.90 kPa 
P = PA + PB = (46.43 + 54.90) kPa = 101.33 kPa = 1.000 atm 
Therefore, Raoult 's law correctly predicts the pressure of the boiling liquid andl the solution is ideal I. 
(b) PA 46.43 kPa I I VA = - = = 0.4582 
. P \0 1.33 kPa 
YB = I - YA = 1.000 - 0.4582 = 1 0.5418 1 
E6.4{b) Let B = benzene and T = toluene. Since the solution is equimolar l B = ZT = 0.500 
(a) Initially XB = ZB and XT = ZT ; thus 
(b) 
P = XBPs + XTPT [6.3) = (0.500) x (9.9 kPa) + (0.500) x (2.9 kPa) 
= 4.95 kPa + 1.45 kPa = 16.4 kPa 1 
PB 4.95kPa ~ ~ 
YB = - [6.4) = = L222.J YT = I - 0.77 = ~ 
P 6.4kPa 
(c) Near the end of the distillation 
YB = ZB = 0.500 and YT = ZT = 0.500 
E6.5(b) 
E6.6(b) 
PHASE DIAGRAMS 115 
Equation 6.5 may be sol ved for XA [A = benzene = B here] 
YBPT ___ C_O._500_)_x_C_2_.9_k_Pa_) ___ = 0.23 
(9.9 kPa) + (2.9 - 9.9) kPa x (0.500) XB = * ( * *) PB + PT - PB YB 
XT = 1 - 0.23 = 0.77 
This result for the special case of ZB = ZT = 0.500 could have been obtained directly by realizing that 
Y B Cinitial) = XT (final) ; YT Cinitial) = XB (final) 
p (final) = XBPS + XTPT = (0.23) x (9.9 kPa) + (0.77) x (2.9 kPa) = 14.5 kPa 1 
Thus in the course of the distillation the vapor pressure fell from 6.4 kPa to 4 .5 kPa 
See the phase diagram in Figure 6.3. 
(a) YA =@1D 
(b) XA = 1 0.671 YA = 1 0.9251 
155 
150 
145 
140 
or e 
135 
130 a 
b 
125 
120 
0 0.2 0.4 0.6 0.8 1.0 
XA Figure 6.3 
AI3+, H+ , AICl), AI (OHh, OH- , Cl-, H20 giving seven species. There are also three equilibria 
AICI3 + 3H20 ;==> Al (OHh + 3HCI 
AICl) ;==> AI3+ + 3CI-
H20 ;==> H+ +OH-
and one condition of electrical neutrality 
Hence, the number of independent components is 
C = 7 - (3 + I ) = [IJ 
116 STUDENT'S SOLUTIONS MANUAL 
E6.7(b) NH4CI(s) ;=' NH3(g) + HCI(g) 
E6.8(b) 
E6.9(b) 
(3) For this system 1 C = I 1 [Example 6.1] and 1 P = 21 (s and g). 
(b) If ammonia is added before heating, 1 C = 21 (because NH4CI, NH3 are now independent) and 
1 P = 21 (s and g). 
(3) Still 1 C = 21 (Na2 S04, H20) , but now there is no solid phase present, so 1 P = 21 (liquid solution, 
vapor). 
(b) The variance is F = 2 - 2 + 2 = 0. We are free to change any two of the three variables, amount 
of dissolved salt, pressure, or temperature, but not the third. If we change the amount of dissolved 
salt and the pressure, the temperature is fixed by the equilibrium condition between the two phases. 
See Figure 6.4. 
+ 10 
- 10 
- 30 
-50 
- 70 
- 90 
Figure 6.4 
E6.10(b) See Figure 6.5. The phase diagram should be labeled as in figure 6.5. (a) Solid Ag with dissolved 
Sn begins to precipitate at QI , and the sample solidifies completely at Q2. (b) Solid Ag with dissolved 
Sn begins to precipitate at bl, and the liquid becomes richer in Sn. The peritectic reaction occurs atb2, and 
(a) 
b 
(b) 
a 
800 Liquid 
L + Ag solid 
contaminated 
U with Sn 460°C 
~ Bi b2 <c Ag)Sr + Ag 
L+ Sn L + Ag)Sn cdntaminated 
I 
solid solid b) wit Sn 
Be 
200 Sn + Ag)Sn solids 
Sn Ag)Sn Ag Time Figure 6.S 
PHASE DIAGRAMS 117 
as cooling continues Ag3Sn is precipitated and the liquid becomes richer in Sn. At b3 the system has its 
eutectic composition (e) and freezes without further change. 
E6.11 (b) See Figure 6.6. The feature denoting incongruent melting is circled. Arrows on the tie line indicate the 
decomposition products. There are two eutectics: one at XB = 1 0.531, T = 1 T21; another at XB = 1 0.821, 
T=@]. 
o 
A 
0.33 0.67 
Xs B Figure 6.6 
E6.12(b) The cooling curves corresponding to the phase diagram in Figure 6.7(a) are shown in Figure 6.7(b). 
Note the breaks (abrupt change in slope) at temperatures corresponding to points G"b" and b2. Also 
note the eutectic halts at G2 and b3. 
(a) 
o 
A 
0.33 0.67 
Xs 
I 
B 
(b) 
--- --- ----+-------"------"----,. 
Figure 6.7 
118 STUDENT'S SOLUTIONS MANUAL 
E6.13(b) Rough estimates based on Figure 6.41 of the text are 
(a) XB ~ 1 0.75 1 (b) XAB2 ~ @]] (c) XAB2 ~ 1 0.61 
E6.14(b) The phase diagram is shown in Figure 6.8. The given data points are circled. The lines are schematic 
at best. 
1000 Liquid 
900 
800 
700 Solid 
o 0.2 0.4 0.6 0.8 
Figure 6.8 
A solid solution with x(ZrF4) = 0.24 appears at 855 °e . The solid solution continues to form, and 
its ZrF4 content increases until it reaches x(ZrF4) = 0.40 and 820 °C. At that temperature, the entire 
sample is solid. 
E6.15(b) The phase diagram for this system (Figure 6.9) is very similar to that for the system methyl ethyl ether 
and diborane of Exercise 6.9(a). The regions of the diagram contain analogous substances. The solid 
compound begins to crystallize at 120 K. The liquid becomes progressively richer in diborane until the 
liquid composition reaches 0.90 at 104 K. Atthat point the liquid disappears as heat is removed. Below 
104 K the system is a mixture of solid compound and solid diborane. 
Figure 6.9 
PHASE DIAGRAMS 119 
E6.16(b) Refer to the phase diagram in the solution to Exercise 6.14(a). The cooling curves are sketched in 
Figure 6.10. 
(a) (b) (c) (d) (e) 
95 
93 
91 
:.: h' 89 
87 
85 
83 Figure 6.10 1 --+ 
ES.17(b) (a) When XA falls to 0.47, a second liquid phase appears. The amount of new phase increases as XA falls 
and the amount of original phase decreases until, at XA = 0.314, only one liquid remains . 
(b) The mixture has a single liquid phase at all compositions. 
The phase diagram is sketched in Figure 6.11 . 
54 
52 
50 
;;.; 48 
-.... 
"" 46 
44 
42 
40 
38 
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 
XA Figure 6.11 
Solutions to problems 
Solutions to numerical problems 
PS.1 (a) The data, including that for pure chlorobenzene, are plotted in Fig. 6.12. 
(b) The smooth curve through the x, T data crosses x = 0.300 at 1391.0 K I, the boiling point of the 
mixture. 
(c) We need not interpolate data, for 393.94 K is a temperature for which we have experimental data. 
The mole fraction of I-butanol in the liquid phase is 0.1700 and in the vapor phase 0.3691. According 
P6.3 
120 STUDENT'S SOLUTIONS MANUAL 
400 
TIK 395 
390 
385 
0.0 0.2 0.4 
xor y 
0.6 
ex 
O y 
x (0.3, 393.94 K) 
0.8 
Figure 6.12 
to the lever rule, the proportions of the two phases are in an inverse ratio of the distances their mole 
fractions are from the composition point in question. That is, 
~ = ~ = 0.3691 - 0.300 = 1 0.5321. 
I1vap l 0.300 - 0.1700 
PA = aAP~ = YAXAP~ [5 .45]. 
PA YAP YA=--=-- · XAP~ XAP~ 
Sample calculation at 80 K: 
O.II(lookPa)( 760TOrr) 
Y0 2 (80 K) = 0.34(225 Torr) 10 1.325 kPa ' 
Y0 2 (80 K) = 1.079. 
Summary: 
T / K 77.3 78 80 82 84 86 88 90.2 
Y02 0.877 1.079 1.039 0.995 0.993 0.990 0.987 
To within the experimental uncertainties the solution appears to be ideal (y = I) . The low value at 78 K 
may be caused by nonideality ; however, the larger relative uncertainty in y(02 ) is probably the origin 
of the low value. 
A temperature--composition diagram is shown in Fig. 6.13(a). The near ideality of this solution is, 
however, best shown in the pressure--composition diagram of Fig. 6.13(b). The liquid line is essentially 
a straight line as predicted for an ideal solution. 
P6.S A compound with 1 probable formula A3B exists I. It melts incongruently at 700°C, undergoing the 
peritectic reaction 
A3B(S) ~ A(s) + (A + B, I) . 
PHASE DIAGRAMS 121 
92 
X8 
:.: ;::; 84 
SO 
76 
0 20 40 60 80 100 
x(02) or y(02) Figure 6.13(a) 
700 
S()() 
300 
100 
o 20 40 60 80 100 
Figure 6.13(b) 
The proportions of A and B in the product are dependent upon the overall composition and the 
temperature. A eutectic exists at 400°C and XB ~ 0.83. See Fig. 6.14. 
P6.7 The information has been used to construct the phase diagram in Fig. 6.15(a). In MgCu2 the mass 
24.3 ~ 48.6 r::;:;l percentage of Mg is (100) x =~, and in Mg2Cu it is (100) x =~. The 24.3 + 127 48.6 + 63.5 
initial point is Q I , corresponding to a liquid single-phase system. At Q2 (at 720°C) MgCu2 begins 
to come out of solution and the liquid becomes richer in Mg, moving toward e2. At Q3 there is solid 
MgCu2 + liquid of composition e2 (33 per cent by mass of Mg). This solution freezes without further 
change. The cooling curve will resemble that shown in Fig. 6.15(b). 
P6.9 (a) I Eutectic: 40.2 at% Si at 1268°C I, 1 Eutectic: 69.4 at% Si at 1030°C I. 
Congruent melting compounds: Ca2 Si mp = 13 14°C 
CaSi mp = 1324°C 
122 STUDENT'S SOLUTIONS MANUAL 
£-! 
<i> 
1300 
LiquidA& B 
1100 
900 
Solid A 
700 
Solid A LiquidA& B 
500 SolidA3B SolidA3B 
300 
Liquid A3B & Solid B 
0 0.20 0.40 0.60 0.80 1.00 
XB Figure 6.14 
(a) (b) 
1200 
800 
400 
Mg Time Figure 6.15 
1 Incongruent melting compound: CaSi2 mp = 1040°C 1 melts into CaSi(s) and liquid (68 
at% Si). 
(b) At 1000°C the 1 phases at equilibrium will be Ca(s) and liquid (13 at% Si) I. The lever rule gives the 
relative amounts: 
nCa iJiq 0.2 - 0 ~ 
---- -~ 
nbq - ICa - 0.2 - 0.13 - . . 
(c) When an 80 at% Si melt it cooled in a manner that maintains equilibrium, Si(s) begins to appear at 
about 1250°C. Further cooling causes more Si(s) to freeze out of the melt so that the melt becomes 
more concentrated in Ca. There is a 69.4 at% Si eutectic at 1030°C. Just before the eutectic is 
reached, the lever rule says that the relative amounts of the Si(s) and liquid (69.4% Si) phases are: 
nSi lliq 0.80 - 0.694 . .. 
--::------::----::-:c- = 0.53 = relative amounts at T slightly higher than 1030°C . 
nliq lSi 1.0 - 0.80 
Just before 1030°C, the Si(s) is 34.6 mol% of the total heterogeneous mixture, the eutectic liquid 
is 65.4 mol %. 
PHASE DIAGRAMS 123 
At the eutectic temperature a third phase appears-CaSi2 (s). As the melt cools at this temperature, 
both SiCs) and CaSi2(s) freeze out of the melt while the concentration of the melt remains constant. 
At a temperature slightly below 1030°C, all the melt will have frozen to SiCs) and CaSh(s) with the 
relative amounts: 
0.80 - 0.667 
1.0 - 0.80 
= 1 0.665 = relative amounts of T slightly higher than 1030°C I. 
Just under 1030°C, the SiCs) is 39.9 mol% of the total heterogeneous mixture; the CaSi2(s) is 
60.1 mol%. 
A graph of mol% SiCs) and mol% CaSi2 (s) vs. mol% eutectic liquid is a convenient way to show 
relative amounts of the three phases as the eutectic liquid freezes. See Fig. 6.16. Equations for the 
graph are derived with the law of conservation of mass. For the silicon mass, 
where n = total number of moles. 
WSi = Si fraction in eutectic liquid = 0.694 
XSi = Si fraction in SiCs) = 1.000 
YSi = Si fraction in CaSi2(s) = 0.667 
ZSi = Si fraction in melt = 0.800 
Freezing of eutectic melt at 1030°C 
0.7 r------.------,------,------.------,------.------. 
- mol fraction CaSi1 
.. ...... mol fraction Si 
0.6 
0.5 
0.4 
0.3 
0.2 
0.1 
0.1 0.2 0.3 0.4 
mol fraction liq 
Freezing proceeds toward left 
Figure 6.16 
0.5 0.6 0.7 
124 STUDENT'S SOLUTIONS MANUAL 
This equation may be rewritten in mole fractions of each phase by dividing by n: 
ZSi = (mol fraction liq)WSi + (mol fraction Si)XSi + (mol fraction CaSi2)Ysi. 
Since, (mol fraction Iiq) + (mol fraction Si) + (mol fraction CaSi2) = I 
or (mol fraction CaSi2) = I - (mol fraction liq + mol fraction Si), we may write: 
ZSi = (mol fraction Iiq)WSi + (mol fraction Si)xsi 
+ [I - (mol fraction liq + mol fraction Si)]YSi. 
Solving for mol fraction Si: 
I f . S· (ZSi - YSi) - (WSi - YSi)(mol fraction Iiq) mo ractIon 1 := , 
XSi - YSi 
mol fraction CaSi2 := I - (mol fraction Iiq + mol fraction Si). 
These two eqns are used to prepare plots of the mol fraction of Si and mol fraction of CaSi2 against 
the mol fraction of the melt in the range 0-0.65. 
Solutions to theoretical problems 
P6.11 The general condition of equilibrium in an isolated system is dS = O. Hence, if ct and f3 constitute an 
isolated system, which are in thermal contact with each other 
P6.13 
dS = dSa + dS,B = O. (a) 
Entropy is an additive property and may be expressed in terms of U and V. 
S = S(U, V). 
The implication of this problem is that energy in the form of heat may be transferred from one phase to 
another, but that the phases are mechanically rigid, and hence their volumes are constant. Thus, dV = 0, 
and 
( aSa ) (as,B ) I 1 dS = -a - dUa + - dU,B = -dUa + -dU,B [3.45] . Ua v aUf3 v Ta Tf3 
1 I r---~ 
But, dUa = -dUf3 ; therefore Ta = Tf3 orl Ta = Tf3I· 
Solutions to applications 
(i) Below a denaturant concentrationof 0.1 only the native and unfolded forms are stable. 
(ii) At denaturant concentration of 0.15 only the native form is stable below a temperature of about 
0.70. At temperature 0.70 the native and molten-globule forms are at equilibrium. Heating above 
0.70 causes all native forms to become molten-globules. At temperature 0.90, equilibrium between 
P6.1S 
P6.17 
P6.19 
PHASE DIAGRAMS 125 
molten-globule and unfolded protein is observed and above this temperature only the unfolded form 
is stable. 
C = I; hence, F = C - P + 2 = 3 - P. 
Since the tube is sealed there will always be some gaseous compound in equilibrium with the condensed 
phases. Thus when liquid begins to form upon melting, P = 3 (s, I, and g) and F = 0, corresponding 
to a definite melting temperature. At the transition to a normal liquid, P = 3 (t, l', and g) as well, so 
again F = O. 
To examine the process of zone levelling with the phase diagram below, Fig. 6.17, consider a solid on 
the isopleth through a I and heat the sample without coming to overall equilibrium. If the temperature 
rises to a2 , a liquid of composition b2 forms and the remaining solid is at a; . Heating that solid down an 
isopleth passing through a~ forms a liquid of composition b3 and leaves the solid at a~. This sequence 
of heater passes shows that in a pass the impurities at the end of a sample are reduced while being 
transferred to the liquid phase which moves with the heater down the length of the sample. With enough 
passes the dopant, which is initially at the end of the sample, is distributed evenly throughout. 
A B 
liquid 
solid 
a: 
o 
Composition, xB Figure 6.17 
The data are plotted in Fig. 6.18. 
(b) From the tie line at 2200°C, the liquid composition is y(MgO) = ~ and the solid x(MgO) = 
10.351· 
The proportions of the two phases are given by the lever rule, 
~ = n(liq) = 0.05 = I 0.41. 
12 n(sol) 0.12 
(c) Solidification begins at point c, corresponding to I 2640°C I. 
126 STUDENT'S SOLUTIONS MANUAL 
10 
8 
';' 6 
c.. 
o 0.3 
~ .............. , 
x(MgO) 
:::::: ~ 4 .... ... .. , .... .. ..• ... 
2 ......... ·····T 
o 
0.2 0.4 0.6 
x or y 
v 
P6.21 (a) The data are plotted in Fig. 6.19. 
2640 
1.0 
0.8 
Figure 6.18 
e x 
O y 
x (0.500, 6.02 MPa) 
1.0 
Figure 6.19 
(b) We need not interpolate data, for 6.02 MPa is a pressure for which we have experimental data. The 
mole fraction of C02 in the liquid phase is 0.4541 and in the vapor phase 0.9980. The proportions of 
the two phases are in an inverse ratio of the distance their mole fractions are from the composition 
point in question, according to the lever rule 
~ = ~ = 0.9980 - 0.5000 = 110.851. 
n vap I 0.5000 - 0.4541 
Chemical equilibrium 
Answers to discussion questions 
07.1 The position of equilibrium is always determined by the condition that the reaction quotient, Q must 
equal the equilibrium constant, K. If the mixing in of an additional amount of reactant or product 
destroys that equality, then the reacting system will shift in such a way as to restore the equality. That 
implies that some of the added reactant or product must be removed by the reacting system and the 
amounts of other components will also be affected. These adjustments restore the concentrations to their 
(new) equilibrium values. 
07.3 (1) Response to change in pressure. The equilibrium constant is independent of pressiIre, but the indi-
vidual partial pressures can change as the total pressure changes. This will happen when there is 
a difference, f:l.ng , between the sums of the number of moles of gases on the product and reactant 
sides of the chemical equation. The requirement of an unchanged equilibrium constant implies that 
the side with the smaller number of moles of gas be favored as pressure increases. 
(2) Response to change in temperature. Equation 7.23a shows that K decreases with increasing tem-
perature when the reaction is exothermic; thus the reaction shifts to the left, the opposite occurs in 
endothermic reactions. See Section 7.4 (a) for a more detailed discussion. 
07.5 (a) Consider the metals M and Z, which, for the sake of simplifying discussion, form I: I oxides hav-
ing the formulas MO and ZOo Z will spontaneously reduce MO provided that the ZO line upon 
the Ellingham diagram lies above the MO line (this statement assumes that the vertical f:l.rG axis 
decreases upward). In this case the standard Gibbs energy for the reaction MO(s) + Z(s) -+ M(s) + 
ZO(s) will be negative. Figure 7.10 of the text indicates that Fe will reduce PbO, CuO, and Ag20 . 
(b) Using f:l.rG"(ZnO) = -318 kJmol- i at 25°C (Table 2.7) and a slope that is common for all the 
oxides, we may add the approximate line for ZnO in the Ellingham diagram as shown in Fig.7 .!. 
The ZnO curve passes under the reaction (iii) curve at about 1300°C so that is the estimate of the 
lowest temperature at which zinc oxide can be reduced to the metal by carbon. See Fig. 7.1. 
07.7 Electrode combinations that produce identical cell compartments with differing concentrations only 
(electrolyte concentration cells) have a cell potential dependence upon the liquid junction potential 
and the concentration difference. If the cell has identical compartments with either gaseous or amal-
gam electrodes (electrode concentration cell), the cell potential will depend upon the gas pressure 
differences or the amalgam concentration differences but will not have a liquid junction potential. 
Other electrode combinations produce cells for which the cell potential depends upon the half-reaction 
reduction potentials. 
07.9 The pH of an aqueous solution can in principle be measured with any electrode having an emf that is 
sensitive to H+ (aq) concentration (activity). In principle, the hydrogen gas electrode is the simplest 
E7.1(b) 
128 STUDENT'S SOLUTIONS MANUAL 
,. 
"0 
E 
:;:: 
"" . 
"" "l 
-500 
--400 
- 300 
-200 
- 100 
0 
+100 
o 500 1500 
Temperature, OI OC 
2000 2500 
Figure 7.1 
and most fundamental. A cell is constructed with the hydrogen electrode being the right-hand electrode 
and any reference electrode with known potential as the left-hand electrode. A common choice is the 
saturated calomel electrode. The pH can then be obtained by measuring the emf (zero-current potential 
difference), E, of the cell . The hydrogen gas electrode is not convenient to use, so in practice glass 
electrodes are used because of ease of handling. 
Solutions to exercises 
N204(g) ~ 2N02(g) 
Amount at equilibrium (I - a)n 2an 
I -a 2a 
Mole fraction - - --I +a I+a 
(I - a)P laP 
Partial pressure 
I+a l+a 
Assuming that the gases are perfect, aJ = p~ 
p 
K = (PNOzlp f»2 = 4a2p 
(PN20 4 I pf» (I - ( 2)pf> 
4a2 
Forp=pf> K= ---
, 1 -a2 
CHEMICAL EQUILIBRIUM 129 
(a) 1 ~r G = 0 1 at equilibrium 
(b) 
(c) 
0'=0.201 K = 4(0.201)22 = 1 0.16841 1 
1-0.201 
~rG~ = -RTInK = -(8.314JK- I mol-I) x (298K) x In(0.16841) 
E7.2(b) (a) Br2(g) .= 2Br(g) a = 0.24 
E7.3(b) 
Amount at equilibrium (I - a)n 
Mole fraction 
I-a 
--
1+0' 
( I - a)P 
Partial pressure 
1+0' 
Assuming both gases are perfect aJ = PJ 
P~ 
K = (PBrlp~)2 = 4a2p 
PBr2 /p~ (I - 0' 2)p~ 
4(0.24)2 _ ~ 
= 1 _ (0.24)2 = 0.2445 = ~ 
2an 
20' 
--
1+0' 
2aP 
1+0' 
(b) ~rG~ = -RTInK = -(8.314JK- I mol-I) x (1600K) x In(0.2445) 
1 = 19kJmol- 1 1 
(c) lnK(2273K) = In K(l600 K) _ ~rH~ (_1 ___ 1_) 
R 2273 K 1600 K 
_ (ll2 x 103 mOl - I) 
= In (0.2445) - I x (-1.851 x 10-4) 
8.314JK-1 mol -
= 1.084 
K(2273 K) = el.084 = 1 2.96 1 
v(CHCI3) = 1, v(HCI) = 3, V(CH4) = -I , V(CI2) = -3 
(a) ~rG~ = ~fG~(CHCb , I) + 3~fG~(HC1, g) - ~fG~(CH4, g) 
= (-73.66kJmol- l ) + (3) x (-95.30kJmol- l ) - (-50.72 kJ mol - I) 
= 1-308.84kJmol-1 1 
~rG~ -(-308.84 x 103 Jmol - I) -
InK = ---[7.8] = = 124.584 
RT (8.3145JK- I mol-I) x (298.15K) 
K = 1 I.3 x 1054 1 
E7.4(b) 
130 STUDENT'S SOLUTIONS MANUAL 
(b) 6.,H" = 6.fH" (CHCI), l) + 36.fH"(HCI, g) - 6.rH"(CH4, g) 
= (- 134.47kJmol- l) + (3) x (-92.31 kJmol- l) - (-74.81 kJmol- l) 
= -336.59 kJ mol - I 
InK(50 0C) = InK(25 0C) _ 6.,H" (_1 _ _ _ 1_) [7.25] 
R 323.2 K 298.2 K 
= 124.584 _ (-336.59 x 1~3 J mOil-I) x (-2.594 x 10-4 K- I) = 114.083 
8.3145JK- mol -
K(50 °C) = 13.5 x 1049 1 
6.,G" (50 °C) = -RT In K(50 °C) [7.17] = -(8.3145JK- 1 mol- I) x (323.15K) x (114.083) 
= 1-306.52kJmol- 1 1 
Draw up the following table. 
A + B ~ C + 
Initial amounts/mol 2.00 1.00 0 
Stated change/mol +0.79 
Implied change/mol -7.09 -7.09 +7.09 
Equilibrium amounts/mol 1.21 0.21 0.79 
Mole fractions 0.1782 0.0302 0.1162 
(a) Mole fractions are given in the table. 
(b) Kx = nx~J , 
J 
(0.1163) x (0.6745)2 ~ 
Kx = =~ (0.1782) x (0.0309) 
(c) PI = XIP. Assuming the gases are perfect, QJ = PI/p" , so 
K = (Pe /le, ) x (Po / p,,)2 = Kx (.£) = Kx when P = 1.00 bar 
(PA / P"' ) x (PB / p") p" 
(d) 
K=Kx=~ 
6.,G"' = -RTlnK = -(8.314JK-1 mol-I) x (298K) x In(9.609) 
= 1-5.6kJmol- 1 1 
2D 
3.00 
+1.58 
4.58 
0.6742 
Total 
6.00 
6.79 
0.9999 
CHEMICAL EQUILIBRIUM 131 
E7.5(b) At 1120 K, /).rGf7 = +22 x 103 J mol- I 
E7.6(b) 
E7.7(b) 
/).rGf7 (22 x 103 Jmol- I) -
InK(1l20K) = -- = - = -2.363 
RT (8.314JK- 1 mol-I) x (l120K) 
K = e-2.363 = 9.41 x 10-2 
InK2 = InKI _ /).r
H f7 (~ _ ~) 
R T2 TI 
Solve for T2 at In K2 = 0 (K2 = I) 
I RlnKI I (8.314JK- 1 mol-I) x (-2.363) I - 4 
- = -- + - = + -- = 7.36 x 10-
T2 /).rH f7 TI (125 x 103Jmol- l ) 1120K 
T2 =11.4 xlO3KI 
d(lnK) -/).rHf7 
Use--=---
d(l I T) R 
We have InK = -2.04 - 1176K (~ ) + 2.1 x 107 K3 (~ Y 
T = 450K so 
_ /).r
H f7 
= -1l76K + (2.1 x 107 K3) x 3 (_1_)2 = -865K 
R 450K 
/).rH f7 = +(865K) x (8.314Jmol- 1 K- I) =1 7. 19lkJ mol-I 1 
/).rGf7 = -RT In K 
{ 
1176K 2.1 x 107 K3} 
= -(8.314JK- 1 mol-I) x (450K) x -2.04 - 450K + (450K)3 
= 16.55 kJ mol-I 
f7 /).rH f7 - /).rG f7 
/).rS = ---T---
=1-21JK- l mol- 1 1 
7.191 kJmol- 1 - 16.55kJmol- 1 
----------- = -20.79JK-1 mol-I 
450K 
U(s) + ~H2(g) ~ UH3(S), /).rG f7 = -RTIn K 
At this low pressure, hydrogen is nearly a perfect gas, a(H2) = (Plpf7). The activities of the solids are 1. 
E7.8(b) 
E7.9(b) 
132 STUDENT'S SOLUTIONS MANUAL 
Hence, InK = In - = --In!!..... ( 
P )-3/2 3 
pe 2 pe 
e 3 P b.fG = -RTln-2 pB" 
( 3) ( 139Pa ) = - X (8 .314JK-
1 mol - I) x (500 K) x In 5 
2 1.00 x 10 Pa 
= 1-41.0kJmol- 1 I 
Kx = n x/J [analogous to 7.16] 
J 
The relation of Kx to K is established in Illustration 7 .5 
n(PJ)VJ[ . PJ] Kx = J pe 7.16wlthaJ = pe 
Therefore, Kx = K (pi pe ) -v, Kx ex p-v [K and pe are constants] 
v = 1 + I - 1 - I = 0, thus I KA2 bar) = KAl bar) I 
5.0g 
Initial moles N2 = I = 0.2380 mol N2 
28.01 gmol -
2.0g - 2 
Initial moles 0 2 = I = 6.250 x 10- mol 0 2 
32.00gmol-
N2 02 NO 
Initial amount/mol 0.2380 0.0625 0 
Change/mol - z - z +2z 
Equilibrium amount/mol 0.2380 - z 0.0625 - z 2z 
Mole fractions 
0.2380 - z 0.0625 - z 2z 
--
0.300 0.300 0.300 
K = Kx (; ) v [v = ~ VJ = 0] , then 
(2zI0.300)2 
K = Kx = -,--------,'----;-:--:-:-,---,----.,-
(
0.2380 - z) x (0.0625 - z) 
0.300 0.300 
4z
2 
= 1.69 x 10- 3 
(0.2380 - z)(0.0625 - z) 
Total 
0.300 
o 
0.300 
( I ) 
E7.10(b) 
E7.11(b) 
CHEMICAL EQUILIBRIUM 133 
4z2 = 1.69 x 10-3{0.01488 - 0.300Sz + Z2 ) 
= 2.514 X 10- 5 - (S.078 X 1O- 4 )z + (1.69 X 1O- 3)Z2 
4.00 - 1.69 X 10- 3 = 4.00 so 
4z2 + (S.078 X 1O- 4)z - 2.S14 X 10- 5 = 0 
-S.078 X 10-4 ± {(S.078 X 10-4 )2 - 4 X (4) X (-2.SI4 X 10- 5»)1 /2 
z = 8 
= ~(-S .078 X 10- 4 ± 2.006 X 10- 2) 
z > 0 [Z < 0 is physically impossible] so 
z = 2.444 X 10- 3 
2z 2(2.444 X 10- 3) 1 - 2 1 
XNO = -- = = 1.6 X 10 
0.300 0.300 
In K ' = 6.JH f> (~ _ ~) 
K R T T' 
T = 31OK, T' = 32SK ; 
f> Rln (~) 
so 6.JH = (~ _ ~) 
T T' 
K' 
let- = K 
K 
(8.3 14JK- 1 mol- I) _ 
Now 6.JHf> = « 1/ 3IOK) _ (1/ 32SK)) X InK = SS.84kJmol- l lnK 
(a) K = 2 6.JHf> = (SS.84 kJ mol- I) X (In 2) = 139 kJ mol- II 
6. r H f> = (SS.84kJmol- l ) X (In!) =1-39kJmol-1 1 
I (b) K = -
2 
(a) 
p = p(NH3 ) + p (HCI) = 2p(NH3) [P(NH3) = p(HCI)] 
K = n a;i [7 .16]; 
J 
PJ 
a(gases) = G; 
p 
K = (P(NH3» ) X (P(HCI») = p(NH3)2 = ~ X (!!...-) 2 
pf> pf> pf>2 4 pf> 
At 427 °C (700 K) , K = ~ X (~: ~:) 2 = 19.241 
I (1IISkPa) 2 ~ At4S9 °C (732K) , K = 4' X lookPa =~ 
134 STUDENT'S SOLUTIONS MANUAL 
(b) I'::..,G" = -RTlnK [7.S] = (-S.3141 K-1mol - l ) x (7ooK) x (In 9.24) 
=1-12.9kJmo1- 11 (at 427° C) 
(C)I'::..H"::::; R1n (K'jK ) [7 .25] 
, (l j T - 1j T') 
(S.3141 K- 1mol- l ) x In (31.0Sj9.24) I - I I 
::::; =. +161 kJmol . (11700 K) - ( 1/732 K) 
E7.12(b) The reaction is 
E7.13(b) 
For the purposes of this exercise we may assume that the required temperature is that temperature at 
which K = 1, which corresponds to a pressure of 1 bar for the gaseous products. For K = 1, In K = 0, 
and I'::..,G" = O. 
Therefore, the decomposition temperature (when K = I) is 
I'::.. H " 
T=-'-I'::..,S& 
CUS04 . 5H20 (s) ;=' CUS04 (s) + 5H20 (g) 
I'::..,H& = [(-771.36) + (5) x (-241.S2) - (-2279.7)] kJ mo1- 1 = +299.2 kJmo1- 1 
I'::..,S" = [(109) + (5) x (1SS.S3) - (300.4)] JK- 1 mol- I = 752.8JK- 1 mol- I 
299.2 x 1Q3Jmol - 1 ~ 
Therefore, T = 1 1 = ~ 
752.SJK- mol-
Question. What would the decomposition temperature be for decomposition defined as the state at which 
K = I j2? 
PbI2 (S);=' PbI2(aq) Ks = 1.4 X 10-8 
I'::..,G" =-RTlnKs=-(S.314JK- lmol - l) x (29S.15K) x In (1.4 x 10-8) 
= 44.S3 kJ mol - I 
I'::..fG" (PbI2 , aq) = I'::..,G& I'::.. + I'::..fG& (PbI2, s) 
= 44.S3kJmol- 1 - 173.64kJmol- 1 
=1-12S.S kJmol- 1 1 
CHEMICAL EQUILIBRIUM 135 
E7.14(b) The cell notation specifies the right and left electrodes. Note that for proper cancellation we must equalize 
the number of electrons in half-reactions being combined. 
For the calculation of the standard ernfs of the cells we have used E~ = t; - Et, with standard electrode 
potentials from Table 7.2. 
(a) R: AgzCr04(S) + 2e- -+ 2Ag(s) + CrO~-(aq) +0.45 V 
L: Clz(g) + 2e- -+ 2Cl-(aq) 
Overall (R - L): AgzCr04(S) + 2Cl-(aq) -+ 2Ag(s) + CrO~-(aq) + (CIZg) 
(b) R : Sn4+(aq) + 2e- -+ Sn2+(aq) 
L: 2Fe3+(aq) + 2e- -+ 2Fe2+(aq) 
Overall (R - L) : Sn4+(aq) + 2Fe2+(aq) -+ Sn2+(aq) + 2Fe3+(aq) 
(c) R : MnOz(s) + 4H+(aq) + 2e- -+ Mn2+(aq) + 2Fe3+(aq) 
L: Cu2+(aq) + 2e- -+ Cu(s) 
Overall (R - L) : Cu(s) + MnOz(s) + 4H+(aq) -+ Cu2+(aq) + Mn2+(aq) 
+2HzO(l) 
+1.36 V 
-0.91 V 
+0.15 V 
+0.77 V 
-0.62 V 
+1.23 V 
+0.34 V 
+0.89V 
COMMENT. Those cells for which E~ > 0 may operate as spontaneous galvanic cells under standard 
conditions. Those for which E~ < 0 may operate as nonspontaneous electrolytic cells. Recall that E~ informs 
us of the spontaneity of a cell under standard conditions only. For other conditions we require E. 
E7.1S(b) The conditions (concentrations, etc.) under which these reactions occur are not given. For the purposes 
of this exercise we assume standard conditions. The specification of the right and left electrodes is 
determined by the direction of the reaction as written. As always, in combining half-reactions to form 
an overall cell reaction we must write half-reactions with equal number of electrons to ensure proper 
cancellation. We first identify the half-reactions, and then set up the corresponding cell . 
(a) R: 2HzO(I) + 2e- -+ 20H-(aq) + Hz(g) 
L: 2Na+ (aq) + 2e- -+ 2Na(s) 
and the cell is 
or more simply 
I Na(s)INaOH(aq)IHz(g)IPt I 
(b)R: Iz(s) + 2e- -+ 21-(aq) 
L: 2H+(aq) + 2e- -+ Hz(g) 
and the cell is 
or more simply 
I PtIHz(g)IHI(aq)llz(s)1 Pt I 
- 0.83 V 
-2.71 V 
I +1.88V I 
+0.54 V 
o 
I +0.54 V I 
136 STUDENT'S SOLUTIONS MANUAL 
(c) R: 2H+(aq) + 2e- ~ H2(g) 
L : 2H20(I ) +2e-~H2(g)+20H-(aq) 
and the cell is 
or more simply 
I PtIH2(g) IH20(l)IH2(g) IPt I 
O.OOV 
0.083 V 
10.083 V I 
COMMENT. All of these cells have Ee. > 0, corresponding to a spontaneous cell reaction under standard 
conditions. If Ee. had turned out to be negative, the spontaneous reaction would have been the reverse of 
the one given, with the right and left electrodes of the cell also reversed. 
E7.16(b) (a) e. RT E=E --lnQ v=2 
vF 
Q n VJ 2 2 = aJ = aH+aCl - [all other activities = 1] 
J 
= a~a:' = (y+b+)2 x (y_b_)2 [b == : e. here and below] 
RT () 2RT Hence, E = E e. - 2F In y1b4 = E e. - Fin (y±b) 
(b) 6.rG = -vFE [7.27] = -(2) x (9.6485 x 104 C mol-I) x (0.4658 V) = 1-89.89 kJ mol-II 
(c) logy± = -IZ+Z_WI /2[5.69] = - (0.509) x (0.010)1 /2 [I = b for HCI(aq)] = -0.0509 
y± = 0.889 
MT () E e. = E + Fin (y±b) = (0.4658 V) + (2) x 25 .693 x 10-3 V x In (0.889 
= I +0.223 V I 
The value compares favorably to that given in Table 7.2. 
vFEe. 
E7.17(b) In each case In K ="RT"" [7.30] 
(a) Sn(s) + CuS04(aq) ;=: Cu(s) + SnS04(aq) 
R : 
L: 
Cu2+ (aq) + 2e- ~ Cu(s) + 0.34 V } 
Sn2+ (aq) + 2e- ~ Sn(s) _ 0.14 V + 0.48 V 
In K = (2) x (0.48 V) = +37.4, 
25.693 mV K = 11.7 x 10
16 1 
x 0.010) 
E7.18(b) 
P7.1 
(a) 
R: 
L: 
Cu2+ (aq) + e- -+ Cu(aq) + 0.16 V } _ 0.36 V 
Cu+ (aq) + e- -+ Cu(s) + 0.52 V 
-0.36V -
CHEMICAL EQUILIBRIUM 137 
InK = = -14.0, 25.693 mV 
R: 2Bi3+(aq) + 6e- -+ 2Bi(s) 
L : Bi2S3(S) + 6e- -+ 2Bi(s) + 3S2-(aq) 
Overall (R - L): 2Bi3+(aq) + 3S2-(aq) -+ Bi2S3(S) v = 6 
vFE<7 6(0.96 V) -
In K = -- = = 224 
RT (25.693 x 10-3 V) 
K = e224 
K = aBi2S3(S) _ M5 _ e224 
a2 a3 - [Bi3+]2 [S2- ]3 -Bi3+ (aq) S2- (aq) 
In the above equation the activity of the solid equals I and, since the solution is extremely dilute, 
the activity coefficients of dissolved ions also equals 1. Substituting [S2-] = 1.5[Bi3+] and solving 
for [Bi3+] gives [Bi3+] = 2.7 x 10-20 M. BhS3 has a solubility equal I to 1.4 x 10-20 M.I 
(b) The solubility equilibrium is written as the reverse of the cell reaction. Therefore, 
Ks = K - I = l/e224 = 15.2 x 10-98 1. 
Solutions to problems 
Solutions to numerical problems 
(a) !:ire? = -RTlnK = -(8.314JK- I mol- I) x (298K) x (lnO.I64) = 4.48 x 103 Jmol- I 
= I +4.48kJmol- 1 I· 
(b) Draw up the following equilibrium table. 
h Br2 IBr 
Amounts (l-a)n 2an 
Mole fractions 
(1 - a) 2a 
(1 + a) (1 + a) 
Partial pressure 
(l-a)p 2ap 
(1 + a) (l + a) 
P7.3 
P7.5 
138 STUDENT'S SOLUTIONS MANUAL 
Withp = O.l64atm, 
2 I 
a =-5 ' a = 0.447. 
2a (2) x (0.447) I I 
PIEr = -- X P = X (0.164atm) = 0.101 atm. 
1 + a I + 0.447 
(c) The equilibrium table needs to be modified as follows. 
P = Plz + PBr2 + PIEr, 
PIEr = XIErP, Plz = XI2P 
. (1 - a)n 
wIth XBrz = [n = amount of Br2 introduced into container] (1 + a)n + nl2 
2an 
and XlEr = -----
(1 +a)n+nI2 
K is constructed as above [7.16], but with these modified partial pressures. In order to complete 
the calculation additional data are required, namely, the amount of Br2 introduced, n, and the 
equilibrium vapor pressure of 12(S). nl2 can be calculated from a knowledge of the volume of the 
container at equilibrium which is most easily determined by successive approximations since PI2 is 
small. 
Question . What is the partial pressure of IBr(g) if 0.0100 mol of Br2 (g) is introduced into the 
container? The partial pressure of 12(S) at 25°C is 0.305 Torr. 
3 _ . 
U(s) + "2H2(g) ~ UH3(S), K = (P l pB) 3/2 [ExerCIse 7.7(b)] . 
B 2 d In K 2 d (3 B) 6.fH = RT -- [7.23a] = RT - --lnpl p 
dT dT 2 
= -~RT2 d lnp 
2 dT 
= -~RT2 (1.464 X 104K _ 5.65) 
2 T2 T 
= -~R(l.464 x 104K - 5.65T) 
= 1-(2.196 x 104 K -S.4ST)R I. 
or 6. r C;: = -- = IS.4S R I. ( a6. f HB) aT p 
K=~. pB 
CHEMICAL EQUILIBRIUM 139 
Since t':;.rc& and In K are related as above, the dependence of t':;.rc& on temperature can be determined 
from the dependence of In K on temperature 
t':;.rc& = -RTlnK = -RTln.E..-
p" 
I I (17.1kPa) " 
= -(8.314J K- mol- ) x (400 K) x In [p = Ibar] 
100.OkPa 
= +13.5kJmol- 1 at400K. 
t':;.rc&(T) t':;.rc&(T' ) W ( I I ) 
T - T' = t':;.r T - T' [7.25]. 
Therefore, taking T' = 400 K, 
t':;.rc&(T) = (_T_) x (13 .5 kJ mol-I) + (78 kJ mol-I) x (I _ _ T_) 
~K ~K 
I (13.5-78)kJmOI- I ) (T) 
= (78 kJ mol- ) + 400 x K . 
That is, t':;.rC&(T) / (kJ mol-I) = 178 - 0.161 x (T /K) I. 
P7.7 The equilibrium we need to consider is A2(g) <=' 2A(g). A = acetic acid. It is convenient to express the 
equilibrium constant in terms of a, the degree of dissociation of the dimer, which is the predominant 
species at low temperatures. 
A A2 Total 
At equilibrium 2an (1 + a)n (1 + a)n 
2a I-a 
Mole fraction -- --
I+a I+a 
Partial pressure 
2ap C ~:)p l+a p 
The equilibrium constant for the dissociation is 
We also know that 
pV m 
pV = ntotalRT = (1 + a)nRT, implying that a = -- - I and n = -
nRT M 
In the first experiment, 
pVM (101.9 kPa) x (21.45 x 1O- 3dm3) x (120.1 g mol- I) 
a = -- - I = - 1 = 0.392. 
mRT (0.0519 g) x (8.314 kPa dm3 K- I mol- I) x (437 K) 
140 STUDENT'S SOLUTIONS MANUAL 
H K (4) X (0.392)2 X (764.3/750.1) ~ 
ence, = 1 _ (0.392)2 = ~. 
In the second experiment, 
pVM (101.9 kPa) x (21.45 x 1O-3dm3) x (120.1 g mol- I) 
Ct = -- - I = - 1 = 0.764. 
mRT (0.038 g) x (8.314 kPa dm3 K- I mol-I) x (471 K) 
(4) x (0.764)2 x C64.3 ) 
Hence, K = 750.1 = §]. 
I - (0.764)2 
The enthalpy of dissociation is 
K' ( 5.71 ) Rln- Rln --
,r& K · 0.740 - I 
I1rfl = (~ _ ~) [7.25, ExerCISe 7.1O(a)] = (_1 ___ 1_) = +103 kJ mol . 
T T' 437K 471K 
The enthalpy of dimerization is the negative of this value, or 1-103 kJ mol- II (i.e. per mole of dimer). 
P7.9 The equilibrium I2(g) .= 2I(g) is described by the equilibrium constant 
4 2 (p) 
x(I)2 P Ct pB 
K = --2 x ~ = 2 [Problem 7.7]. 
x(l2) p I - Ct 
nRT 
If pO = V ' then p = (1 + a )pO, implying that 
We therefore draw up the following table. 
937 K 1073 K 1173 K 
p/atm 0.06244 0.07500 0.09181 
104nl 2.4709 2.4555 2.4366 
pO/atm 0.05757 0.06309 0.06844 
Ct 0.08459 0.1888 0.3415 
K 11.800 x 10-3 1 11.109 x 10-2 1 14.848 x 10-2 1 
2 (dinK) I 2 (-3.027 - (-6.320») I1Er = RT x ~ = (8.314 J K-Imol- ) x (1073 K ) x 200 K 
= 1 +158 kJ mol-I I. 
CHEMICAL EQUILIBRIUM 141 
P7.11 The reaction is 
The equilibrium constant is 
Let h be the uncertainty in !::"rFr so that the high value is h + the low value. The K based on the low 
value is 
( -M{~ ) (!::"rSS) (-!::"r~gh) ( h ) (!::"rSS) KlowH = exp RT ow exp ~ = exp RT exp RT exp ~ 
= exp (:T ) KhighH. 
KlowH ( h ) 
So Khi ghH == exp RT . 
KlowH ( (289 - 243) kJ mol- I ) 1 81 (a) At 298 K, -- = exp = 1.2 x 10 . 
KhighH (8.3145 x 10- 3 kJ K- I mol-I) x (298 K) 
b At 700 K, -- = ex = 2.7 x 10 . KlowH ( (289 - 243) kJ mol-
I ) 1 31 
( ) KhighH P (8.3145 x 10-3 kJ K- I mol-I) x (700 K) 
P7.13 (a) I=~{(;t zi +(;t zq[5 .711=4(:t». 
For CUS04, I = (4) x (1.0 x 10-3) = 14.0 x 10-3 1. 
For ZnS04, I = (4) x (3.0 x 10- 3) = 11.2 x 10- 2 1. 
(b) logy± = -lz+z_IAII /2. 
logy±(CUS04) = -(4) x (0.509) x (4.0 x 10-3)1 / 2 = -0.1288, 
y±(CUS04) = 1 0.74 1. 
logY±(ZnS04) = -(4) x (0.509) x (1.2 x 10-2)1 / 2 = -0.2230, 
y±(ZnS04) = 1 0.60 I· 
(c) The reaction in the Daniell cell is 
Cu2+ (aq) + SO~- (aq) + Zn(s) -+ Cu(s) + Zn2+ (aq) + SO~- (aq). 
142 STUDENT'S SOLUTIONS MANUAL 
(d) 
(e) 
a(Zn2+)a(SO~- , R) 
Hence, Q = -----::-------=---a(Cu2+)a(SO~-, L)where the designations Rand L refer to the right and left sides of the equation for the cell reaction 
and all b are assumed to be unitless, that is, b/bf7 . 
b+(Zn2+) = b_(SO~-, R) = b(ZnS04). 
b+(Cu2+) = b_(SO~- , L) = b(CUS04)' 
Therefore, 
rG 6. r G" -(-212.7 x 103 J mol-I) I I 
c. = ---[7.28] = = +1.102 V . 
vF (2) x (9.6485 x 104 C mol-I) 
25.693 X 10-3 (25.693 x 10-3 ) -E=r=- v VlnQ=(1.102V)- 2 V In(5.92) 
= (1.102V) - (0.023V) = 1 +1.079 V I· 
P7.15 The electrode half-reactions and their potentials are 
R: Q(aq) + 2H+(aq) + 2e- --+ QH2(aq) 
L: Hg2Cl2 (s) + 2e- --+ 2Hg(1) + 2CI- (aq) 
Overall (R - L): Q(aq) + 2H+(aq) --+ QH2(aq) + Hg2Cl2(s), 
.. a(QH2) Q(reactlOn quotient) = 2 2 
a(Q)a (H+)a (Cl 
r 
0.6994V 
0.2676V 
0.4318 V 
Since quinhydrone is an equimolecular complex of Q and QH2, m(Q) = m(QH2) and, since their 
activity coefficients are assumed to be I or to be equal, we have a(QH2) ~ a(Q) . Thus 
25.7 mV 
E = r - InQ [Illustration 7.10]. 
v 
\!(~ - E) (2) x (0.4318 - 0.190) V 8-
In Q = = = 18. 2 25.7 mV 25.7 x 10-3 V ' 
- 8 Q = 1.49 x 10 . 
CHEMICAL EQUILIBRIUM 143 
For HCI(aq), b+ = b_ = b and, if the activity coefficients are assumed equal, a 2 (H+) = a 2 (CI-); 
hence 
1 
Q = a 2 (H+)a2 (CI ) - a4 (H+)' 
( 1)1/4 ( 1 )1/4 Thus a(H+) = - = = 9 X 10-3 , Q 1.49 X 108 ' 
pH = - log a(H+) = [IQ). 
RT 
E = r - -ina(H+)a(CI-) [Section 7.8]. 
F 
a(H+) = y+b+ = y+b; a(CI-) = y_b_ = y_b [b = ; here and belOW] . 
a(H+)a(Cn = y+y_b2 = ylb2 . 
2RT 2RT 
E = r - -In b - -In y±. F F (a) 
Converting from natural logarithms to common logarithms (base 10) in order to introduce the Debye-
HUckel expression, we obtain 
,.-.& (2.303) x 2RT (2.303) x 2RT 
E = c. - F log b - F log y± 
= r - (0.1183 V) 10gb - (0.11 83 V) log y± 
= r - (0.1183 V) 10gb - (0.1183 V) [-l z+z_ IAII /2] 
= r - (0.1183 V) 10gb + (0.1183 V) x A X b l / 2 [I = b]. 
Rearranging, 
E + (0.1183 V) 10gb = r + constant x b l / 2 . 
Therefore, plot E + (0.1183 V) log b against b 1 / 2, and the intercept at b = 0 is £'7 IV . Draw up the 
following table. 
bl(mmol kg- I) 1.6077 3.0769 5.0403 7.6938 10.9474 
(; y/2 0.04010 0.05547 0.07100 0.08771 0.1046 
EIV + (0.11 83) log b 0.27029 0.27109 0.27186 0.27260 0.27337 
The points are plotted in Fig. 7.2. The intercept is at 0.26840, so £'7 = +0.26840 V. A least-squares 
best fit gives £'7 = I +0.26843 V I and a coefficient of determination equal to 0.99895. 
144 STUDENT'S SOLUTIONS MANUAL 
0.274 
¢ 0.272 
-c 
........ 
-c 
0;; 
..2 
r') 
~ 
-0 0.270 
+ 
> 
~ 
0.268 
o 0.02 0.04 0.06 0.08 0.10 
(b/ b'7 )1/2 Figure 7.2 
For the activity coefficients we obtain from equation (a) 
r-E b 0.26843-E/ V b 
In y± = 2RT I F -In be = 0.05139 -In be 
and we draw up the following table. 
1.6077 3.0769 5.0403 7.6938 10.9474 
In y± 
y± 
-0.3465 -0.05038 -0.6542 -0.07993 -0.09500 
0.9659 0.9509 0.9367 0.9232 0.9094 
P7.19 The cells described in the problem are back-to-back pairs of cells each of the type 
Ag (s) IAgX (s) IMX (bl) IMxHg (s) . 
R: M+ (bl) + e- ~ M xHg (s) (Reduction ofM+ and formation of amalgam) 
L: AgX (s) + e- ---+ Ag (s) + X- (bJ) 
R-L: 
Hg 
Ag (s) + M+ (bl ) + X- (b)) --+ MxHg (s) + AgX (s) , v = 1. 
RT 
E=r- -lnQ. 
F 
CHEMICAL EQUILIBR IUM 145 
For a pair of such cells back to back, 
Ag (5) IAgX (5) IMX (bl) IMxHg (s) IMX (b2) IAgX (5) lAg (5) , 
Rt RT 
ER = r - FIn QR, EL = r - Fin QL, 
-RT QL RT (a (M+) a (X-))L 
E = - - In - = - In -7---7-----i----+------;~ 
F QR F (a (M+) a (X- ))R 
(Note that the unknown quantity a (MxHg) drops out of the expression for E.) 
With L = (1) and R = (2) we have 
2RT bl 2RT Y±(1) 
E=- In -+ -In--. 
F b2 F y±(2) 
Take b2 = 0.09141 mol kg-I (the reference value), and write b = :~ . 
E=- In--- +ln --- . 2RT (b y± ) 
F 0.09141 y± (ref) 
For b = 0.09141, the extended Debye-Hlickel law gives 
(-1.461) X (0.09141) 1/2 -
logy±(ref) = 1/2 + (0.20) x (0.09141) = -0.2735, (I) + (1.70) x (0.09141) 
Y±(ref) = 0.5328. 
( b y± ) Then E = (0.05139 V) x In 0.09141 + In 0.5328 ' 
E b 
In y± = 0.05139 V - In (0.09141) x (0.05328) 
We then draw up the following table. 
bl (mol/kg- I) 0.0555 0.09141 0.1652 
EIV -0.0220 0.0000 0.0263 
y 0.572 10.5331 0.492 
0.2171 
0.0379 
0.469 
1.040 
0.1156 
0.444 
1.350 
0.1336 
0.486 
A more precise procedure is described in the original references for the temperature dependence of 
r (Ag, AgCl, CI-) ; see Problem 7.20. 
146 STUDENT'S SOLUTIONS MANUAL 
P7.21 (a) From (aG) = V [3.50] , 
ap T 
we obtain -- = flr V. ( aflrG) ap T 
Substituting flrG = -vFE [7.27] yields 
(b) The plot (Fig. 7.3) of E against p appears to fit a straight line very closely. A linear regression 
> 8 
~ 
analysis yields 
Slope 'I =-2-.4-8-0-x-I-0---3-m-V-a-tm-------'1 ~ standard deviation = 3 x 10- 6 mV attn-I . 
Intercept= 8.5583 mY, standard deviation = 2.8 x 10-3 mY. 
R = 0.99999701 (an extremely good fit). 
From flrV 
( -2.666 x 10-6 m3 mol-I) 
1 x 9.6485 X 104 C mol-I . 
Since J = VC = Pam3, 
Therefore 
Pam3 
C=--
V 
or 
V 
Pa 
- X = 2.80 x 10- Vatm-(aE) (2.666 x 10-6 ) V 1.01325 x 105 Pa 6 I a? T,n - 9.6485 X 104 Pa atm 
= 12.80 x 10-3 mV atm- I I. 
This compares closely to the result from the potential measurements. 
13 
12 
II 
10 
9 
8 
o 500 1000 1500 
p/atm Figure 7.3 
P7.23 
CHEMICAL EQUILIBRIUM 147 
(c) A fit to a second-order polynomial of the form 
E = a + bp + cp2 
yields 
a = 8.5592 mY, standard deviation = 0.0039 mV 
b = 2.835 x 10-3 mVatm- l , standard deviation = 0.012 x 10-3 mVatm- 1 
c = 3.02 x 10- 9 m V atm -2, standard deviation = 7.89 x 10-9 m V atm- I 
R = 0.999 997 II. 
This regression coefficient is only marginally better than that for the linear fit, but the uncertainty in 
the quadratic term is > 200 per cent. 
(aE) = b + 2cp. ap T 
The slope changes from (aE) = b = 2.835 x 10- 3 mVatm- 1 
ap min 
to (aE) = b + 2c(l 500 atm) = 2.836 x 10- 3 mVatm- l . 
ap max 
We conclude that the linear fit and constancy of (~:) are very good. 
(d) We can obtain an order of magnitude value for the isothermal compressibility from the value of c. 
a
2
E = _~ (aLlrV) = 2c. 
ap2 vF ap T 
2vcF 
V 
(
82.058 cm3 atm) 2(1 ) x (3 .02 x 10-12 Vatm- 2) x (9.6485 x 104 Cmol- I ) x 
8.3145J 
(KT )cell = -----------------(--:--,----:-----:-----:-)-----'------'---
(lcm3 jO.996g) x 18.016g 
I mol 
= 13.2 x 10-7 atm- I 1 standard deviation "'" 200 per cent 
where we have assumed the density of the cell to be approximately that of water at 30°C. 
COMMENT. It is evident from these calculations that the effect of pressure on the potentials of cells involving 
only liquids and solids is not important ; for this reaction the change is only::>:: 3 x 10- 6 V atm- 1 . The effective 
isothermal compressibility of the cell is of the order of magnitude typical of solids rather than liquids; other 
than that, little significance can be attached to the calculated numerical value. 
We need to obtain Ll rH f7 for the reaction 
~H2 (g) + Uup+ (aq) ---+ Uup (s) + H+ (aq) . 
We draw up the thermodynamic cycle shown in Fig. 7.4. 
P7.25 
148 STUDENT'S SOLUTIONS MANUAL 
Ej(H) 13.6eV -11.3eV 
H(g) + Uup+(g) t. hydHB- (H+) 
1 x 4.5eV H+(aq ) + Uup+(g) 
1H2 + Uup+(g) -5.52eV 
3.22eV H+(aq )+ Uup(g) 
A 
1H2 + Uup+(aq) -1.5eV 
x W(aq) + Uup(s) 
Figure 7.4 B 
Data are obtained from Tables 10.3, lOA, IIA, 2.7, and 2.7b. The conversion factor between eV and 
kJmol- 1 is 
1 eV = 96A85 kJ mol-I 
The distance from A to 8 in the cycle is given by 
6. rW = x = (3.22 eV) + G) x (4.5 eV) + (13.6 eV) - (11.3 eV) - (5.52eV) - (1.5 eV) 
= 0.75eV. 
6. r s"" = s"" (Uup, s) + s"" (H+ , aq) - !s"" (H2,g) - s"" (Uup+ , aq) 
= (0.69) + (0) - G) x (1.354) - (1.34) meV K- 1 = -1.33 meV K- 1• 
6. rcr = 6. r W - T6. r s"" = (0.75eV) + (298.15K) x (1.33 meVK- 1) = +1.l5 eV 
which corresponds to I + III kJ mol-I I. 
The electrode potential is therefore -~~cr , with v = 1, or 1-1.15 V I. 
Solutions to theoretical problems 
We draw up the following table using the stoichiometry A + 38 ~ 2C and 6. nJ = vJ~. 
A 8 C Total 
Initial amount Imol 3 0 4 
Change,6.nJ / mol -~ -3~ +2~ 
Equilibrium amount Imol I - ~ 3(1 -~) 2~ 2(2 - ~) 
Mole fraction 
I - ~ 3(1 - ~) ~ 
2(2 -~) 2(2 - ~) 2-~ 
P7.27 
Since K is independent of the pressure 
27 
a2 = - K , a constant. 
16 
Therefore (2 - ~)~ = a (; ) x (1- ~)2, 
which solves to ~ = I - ( 
I )1/2 
1+ apjpB 
CHEMICAL EQUILIBRIUM 149 
We choose the root with the negative sign because ~ lies between 0 and 1. The variation of ~ with p is 
shown in Fig. 7.5. 
0.1 \0 100 \000 
ap/p-&- Figure 7.S 
log y± = _AII /2 = _AC I / 2 In y± = -2.303AC I / 2 
- 2.303AC I / 2 y± =e 2 -4606AC
I/2 y± = e . 
Ks = S'(S' + C) x e-4.606ACI /2 
2 Ks We solve S' + S' C - 2 = 0 
y± 
P7.29 
150 STUDENT'S SOLUTIONS MANUAL 
( )
1/ 2 
, 1 2 4Ks I Ks 
to get S = - C + - - - C "'" --
2 Yf 2 CYf 
. ? 4606ACI/2 Therefore, since Y:f = e- . 
K e- 4.606AC I/2 S,""' _ 5 ___ _ 
C 
Solutions to applications 
6.C = 6.c? + RTlnQ [7 .11] . 
In equation 7.11 molar solution concentrations are used with 1 M standard states. The standard state (G) 
pH equals zero in contrast to the biological standard state (EEl) of pH 7. For the ATP hydrolysis 
we can calculate the standard state free energy given the biological standard free energy of - 31 kJ mol - I 
(impact 17.2). 
6.CEB = 6.c? + RTln(l0-7M/ IM) 
6.G" = 6.CEB - RT In(10- 7M/ I M) = -31 k1 mol - I - (8.3141 mol-I K- I )(310 K) In(10- 7) 
= -31 kJmol- 1 + 41.5 kJ mol - I = +11 kJ mol- I. 
This calculation shows that under standard conditions the hydrolysis of ATP is not spontaneous! It is 
endergonic . 
The calculation of the ATP hydrolysis free energy with the cell conditions pH = 7, [ATP] = [ADP] = 
[Pi] = 1.0 x 10 -6 M, is interesting. 
6.C = 6.G" + RT In Q = 6.G" + RT In{[ADP][P;- ][H+]/ [ATP](1 M)2 } 
= +11 k1 mol - I + (8.314 J mol - IK- I)(310 K) In(l0-6 x 10- 7) = +11 k1 mol - I -77 k1 mol-I 
= -66 k1 mol - I. 
The concentration conditions in biological cells make the hydrolysis of ATP spontaneous and very 
exergonic. A maximum of 66 kJ of work is available to drive coupled chemical reactions when a mole 
of ATP is hydrolyzed. 
P7.31 Yes, a bacterium can evolve to utilize the ethanol/nitrate pair to exergonically release the free energy 
needed for ATP synthesis. The ethanol reductant may yield any of the following products. 
ethanol ethanal elhanoic acid 
The nitrate oxidant may receive electrons to yield any of the following products. 
NO) --+ NO; --+ N2 --+ NH3 . 
nitrate nitrite dinitrogen ammonia 
CHEMICAL EQUILIBRIUM 151 
Oxidation of two ethanol molecules to carbon dioxide and water can transfer 8 electrons to nitrate during 
the formation of ammonia. The half-reactions and net reaction are: 
2[CH)CH20H(I) -+ 2C02(g) + H20(l) + 4H+(aq) + 4e- l 
NO)"(aq) + 9H+ (aq) + 8e- -+ NH) (aq) + 3H20(1 ) 
6.r~ = -2331 .29 kJ for the reaction as written (a Table 2.5 and 2.7 calculation). Of course, enzymes 
must evolve that couple thi s exergonic redox reaction to the production of ATP, which would then be 
available for carbohydrate, protein, lipid, and nucleic acid synthesis. 
P7.33 The half-reactions involved are: 
R: cytox + e -+ cytred t:;yt 
L: Dox + e -+ Dred ~ 
The overall cell reaction is : 
R - L = cytox + Dred ;=: cytred + Dox E" = t:;yt - ~ 
(a) The Nernst equation for the cell reaction is 
At equilibrium, E = 0; therefore 
( 
[Dox leq ) ([cyt] ) Therefore a plot of In --- against In -[ ] ox is linear with a slope of one and an intercept 
[Dredleq cyt red 
of :T (~t -~ ) . 
(b) Draw up the following table. 
I ([Dox leq ) 
- 5.882 -4.776 -3.661 -3.002 - 2.593 -1.436 -0.6274 
n [Dred leq 
( [cytox leq ) 
-4.547 - 3.772 -2.415 
In [cytredleq - 1.625 - 1.094 -0.2120 -0.3293 
152 STUDENT'S SOLUTIONS MANUAL 
Th ( 
[Doxleq ) . ( [cytox]eq ) . 
e plot ofln -[D] agamst In [ IS shown in Fig. 7.6. The intercept is -1 .2124. Hence 
red eq cyt,ed]eq 
RT ~t = F X (-1.2124) + 0.237 V 
=0.0257V X (-1.2124) +0.237 V 
= 1 + 0.206 V I· 
0 
-\ 
g- - 2 
1 
B 
-3 
-~ 
0 
B -4 
:5 
-5 
-6 
-5 -4 -3 -2 - I o 
Figure 7.6 
P7.35 A reaction proceeds spontaneously if its reaction Gibbs function is negative. 
6. r G = 6. r(7 + RT In Q 
Note that under the given conditions, RT = 1.58 kJ mol-I. 
(i) 6.r G/(kJ mol-I) = 6. r(7(I) - RTlnpH20 = -23.6 - 1.58 In 1.3 x 10-7 = +1.5. 
(ii) 6. r G(kJ mol-I) = 6. r (7(2) - RTln(pH2oPHN03) 
= -57 .2 - 1.58ln [(1 .3 x 10- 7) x (4. 1 x lO- IO)J = +2.0. 
(iii) 6.r G/(kJ mol - I) = 6. r (7(3) - RT In(p~20 PHN03) 
= -85.6 - 1.581n[(1.3 x 10-7)2 X (4.1 X 10- 10) ] = -1.3 . 
(iv) 6. r G/(kJ mol-I ) = 6. r (7(4) - RT In(p~20PHN03) 
= -85.6 - 1.58In[(1.3 x 10-7)3 X (4. 1 X 10- 10)] = -3.5 . 
CHEMICAL EQUILIBRIUM 153 
So both the dihydrate and trihydrate form spontaneously from the vapor. Does one convert spontaneously 
into the other? Consider the reaction 
which may be considered as reaction (iv) - reaction (iii). Therefore t;.rG for this reaction is 
We conclude that the dihydrate converts spontaneously to the 1 trihydrate I, the most stable solid (at least 
of the four we considered). 
PART 2 Structure 
8 Quantum theory: introduction and principles 
Answers to discussion questions 
08.1 At the end of the nineteenth century and the beginning of the twentieth, there were many experimental 
results on the properties of matter and radiation that could not be explained on the basis of established 
physical principles and theories. Here we list only some of the most significant. 
(1) The energy density distribution of blackbody radiation as a function of wavelength. 
(2) The heat capacities of monatomic solids such as copper metal. 
(3) The absorption and emission spectra of atoms and molecules, especially the line spectra of atoms. 
(4) The frequency dependence of the kinetic energy of emitted electrons in the photoelectric effect. 
(5) The diffraction of electrons by crystals in a manner similar to that observed for X-rays. 
08.3 The heat capacities of monatomic solids are primarily a result of the energy acquired by vibrations ofthe 
atoms about their equilibrium positions. If this energy can be acquired continuously, we expect that the 
equipartition of energy principle shou ld apply. This principle states that, for each direction of motion 
and for each kind of energy (potential and kinetic), the associated energy should be ~ kT. Hence, for 
three directions and both kinds of motion , a total of 3 kT , which gives a heat capacity of 3 k per atom, 
or 3 R per mole, independent of temperature. But the experiments show a temperature dependence. The 
heat capacity falls steeply below 3 R at low temperatures. Einstein showed that, by allowing the energy 
of the atomic oscillators to be quantized according to Planck 's formula , rather than continuous, thi s 
temperature dependence cou ld be explained. The physical reason is that at low temperatures only a few 
atomic oscillators have enough energy to populate the higher quantized levels; at higher temperatures 
more of them can acquire the energy to become act ive. 
08.5 If the wavefunction describing the linear momentum of a particle is precisely known, the particle has 
a definite state of linear momentum, but then, according to the uncertainty principle, the position of 
the particle is completely unknown as demonstrated in the derivation