CHM 3410 – Problem Set 6
Due date: Wednesday, October 20th (Note: The second hour exam will be on Friday, October 22 nd. It will cover
material from Chapter 4 (all), Chapter 5 (all except for 5.13), and Chapter 6 (6.1 to 6.4), plus handouts.)
Do at least 5 of the following problems. Your homework grade on this assignment will be based on your five highest
problems, for a maximum of 10 points. Show your work.
1) A phase diagram for a solution of A and B (at a fixed pressure of 1.00 atm) is given below. Based on the phase
diagram, answer the following questions.
a) Describe what happens when a closed system at an initial temperature T i = 60. C and an initial mole
fraction ZA = 0.400 is heated at a constant pressure p = 1.00 atm.
b) What is the normal boiling point for A and the normal boiling point for B?
c) Do A and B form an azeotrope? If so, give the location (temperature and mole fraction) for the
azeotropic mixture. Is the azeotrope a high boiling or a low boiling azeotrope?
d) Consider a system with 1.000 moles of substance, and a mole fraction of A in the system ZA = 0.400. At
T = 80.0 C, points a, b, and c are located at 0.284, 0.400, and 0.656, respectively. Find the total number of moles in
the liquid phase and the total number of moles in the vapor phase.
2) Consider the reaction of acetic acid (CH3COOH) in water
CH3COOH(aq) + H2O()  H3O+(aq) + CH3COO-(aq)
a) Give the expression for K, the equilibrium constant, for the above reaction in terms of activities. Give
the corresponding expression assuming ideal solute behavior.
b) Based on the data in Appendix 2.6 and 2.8 of Atkins find the numerical value for K for the above
reaction at T = 25. C.
c) Based on the data in Appendix 2.6 and 2.8 of Atkins find the numerical value for K for the above
reaction at T = 37. C (physiological temperature).
3) A phase diagram for a solution of A and B (at a fixed pressure of 1.00 atm) is given below. Based on the phase
diagram, answer the following questions.
a) Describe what happens when a closed system at an initial temperature T i = 80. C and an initial mole
fraction ZA = 0.400 is heated at a constant pressure p = 1.00 atm.
b) What is the normal boiling point for A and the normal boiling point for B?
c) Do A and B form an azeotrope? If so, give the location (temperature and mole fraction) for the
azeotropic mixture. Is the azeotrope a high boiling or a low boiling azeotrope.
d) Consider distillation of a mixture with a mole fraction of ZA = 0.400. Vapor is continuously removed by
condensation during the distillation. When does the mixture first boil? What is the initial value for the mole fraction
of A in the liquid phase and the vapor phase when the mixture first boils? How do the boiling temperature, the mole
fraction of A in the liquid phase, and the mole fraction of A in the vapor phase, change as the distillation proceeds?
Is it possible to obtain pure A by distillation from a mixture with an initial mole fraction Z A = 0.400? Why or why
not?
Exercises
5.15b Given that p*(H2O) = 0.02308 atm and p(H2O) = 0.02239 atm in a solution in which 0.122 kg of a
nonvolatile solute (M = 241. g/mol) is dissolved in 0.920 kg water at 293. K, calculate the activity and activity
coefficient of water in the solution.
5.17b By measuring the equilibrium between liquid and vapor phases of a solution at 30. C at 1.000
atm, it was found that XA = 0.220 when YA = 0.314. Calculate the activities and activity coefficients for both
components in this solution on the Raoult's law basis. The vapor pressures of the pure components at this
temperature are pA* = 73.0 kPa and pB* = 92.1 kPa. (Note: You are only being asked for the activities and activity
coefficients for A and B in the liquid phase).
6.10a The equilibrium constant of the reaction 2 C3H6(g)  C2H4(g) + C4H8(g) is found to fit the
expression ln K = A + (B/T) + (C/T 2) between 300. K and 600. K, with A = - 1.04, B = - 1088. K, and C = 1.51 x
105 K2. Calculate the standard reaction enthalpy and standard reaction entropy at 450. K.
6.19b For PbI2(s)  Pb2+(aq) + 2 I-(aq), K = 1.4 x 10-8 at 25. C and the standard Gibbs energy of
formation of PbI2(s) is - 173.64 kJ/mol. Calculate the standard Gibbs energy of formation of PbI 2(aq).
Problems
5.22
Consider the phase diagram in Figure 5.69, which represents a solid-liquid equilibrium. Label all
regions of the diagram according to the chemical species that exist in that region and their phases. Indicate the
number of species and phases present at the points labeled b, d, e, f, g, and k. (NOTE: I am having you skip the last
part of the problem involving the cooling curves.)
6.30
The dimerization of ClO in the Antarctic winter stratosphere is believed to play an important part
in that region's severe seasonal depletion of ozone. The following equilibrium constants are based on measurements
on the reaction
2 ClO(g)  (ClO)2(g)
T(K)
K
233.
4.13 x 108
248.
5.00 x 107
258.
1.45 x 107
268.
5.37 x 106
T(K)
K
280.
9.62 x 105
288.
4.28 x 105
295.
1.67 x 105
303.
6.02 x 104
273.
3.20 x 106
a) Derive the value for Hrxn and Srxn for this reaction.
b) Compute the standard enthalpy of formation and the standard molar entropy of (ClO) 2 given Hf(ClO) =
+ 101.8 kJ/mol and S(ClO) = 266.6 J/mol.K.
Solutions.
1)
a) The system initially forms a single liquid phase. When the temperature reaches 73 C, the liquid begins
to boil, with the vapor phase having a smaller mole fraction of A than the liquid phase. Boiling occurs between 73
C and 82 C, at which point all of the liquid has been converted into vapor. Above 82 C the system consists of a
single vapor phase.
b)
Normal boiling point for A = 90 C
Normal boiling point for B = 67 C
c) There is no azeotrope. The only liquids with a single boiling point temperature in this system are pure B
(XA = 0.00) and pure A (XA = 1.00), and pure liquids do not count as azeotropes.
d) We have two equations we can use here
n + ng = n = 1.000 mol
(ZA – YA) ng = (XA – ZA) n
where n = total number of moles in the liquid phase, ng = total number of moles in the gas phase, n = total number of
moles in the system. The second equation above is from the lever rule. Note that the mole fractions correspond to
the points given on the phase diagram, so YA = 0.284, ZA = 0.400, and XA = 0.656.
Substituting into the second equation gives
(0.400 – 0.284) ng = (0.656 – 0.400) n
0.116 ng = 0.256 n
ng = (0.256/0.116) n = 2.21 n
If we substitute this result into our first equation, we get
n + (2.21 n) = 3.21 n = 1.000 mol
n = 1.000 mol/3.21 = 0.312 mol
Notice that this answer is qualitatively consistent with the phase diagram. The “lever” to the liquid is longer than the
lever to the vapor, so the number of moles of liquid should be smaller than the number of moles of vapor.
2)
a) K = (aH3O+) (aCH3COO-) 
(aCH3COOH) (aH2O)
[H3O+] [CH3COO-]
[CH3COOH]
To get the second expression we have used aH2O  1 (since water is a solvent), and that the activities of the solutes
are equal to their molalities (for ideal behavior), which for water are approximately equal to molarity.
b) Since our textbook does not give a value for thermodynamic data for H 3O+, we need to rewrite the
reaction in terms of the Arrhenius acid/base theory.
CH3COOH(aq)  H+(aq) + CH3COO-(aq)
K = [H+] [CH3COO-]
[CH3COOH]
Notice that the expression for K is essentially the same (simply replacing H3O+ with H+), and so the numerical value
for K will be the same, but can now be calculated using the data in the Appendix of Atkins.
Grxn = [ (Gf(H+(aq)) + (Gf(CH3COO-(aq)) ] - Gf(CH3COOH(aq)
= [ 0.00 + (- 369.31) ] – [ (- 396.46) ] = 27.15 kJ/mol
Then ln K = - Grxn
RT
= -
(27150 kJ/mol)
(8.314 J/mol.K) (298. K)
= - 10.96
K = e-10.96 = 1.74 x 10-5
c) Since the temperature range is small (and since we do not have the heat capacity data to do a more exact
calculation) we can use the relationship
ln(K2/K1) = - (Hrxn/R) [ (1/T2) – (1/T1) ]
with T1 = 298. K K1 = 1.74 x 10-5 T2 = 310. K
Hrxn = [ (Hf(H+(aq)) + (Hf(CH3COO-(aq)) ] - Hf(CH3COOH(aq)
= [ 0.00 + (- 486.01) ] – [ (- 485.76) ] = - 0.25 kJ/mol
So
ln(K2/K1) = - [ (- 250. J/mol)/(8.314 J/mol.K) ] [ (1/310. K) – (1/298. K) ]
= - 0.004
And so K2/K1 = e-0.004 = 0.996
K2 = 0.996 K1 = 0.996 (1.74 x 10-5) = 1.73 x 10-5
The equilibrium constant is approximately constant over the temperature range of the problem. This is very unusual,
and not generally true.
3)
a) The system initially forms a single liquid phase. When the temperature reaches 89 C, the liquid begins
to boil, with the vapor phase having a smaller mole fraction of A that the liquid phase. Boiling occurs between 89 C
and 96 C, at which point all of the liquid has been converted into vapor. Above 96 C the system consists of a
single vapor phase.
b)
Normal boiling point for A = 94 C
Normal boiling point for B = 102 C
c) Yes, there is an azeotrope at mole ZA = 0.71, T = 83 C. It has a lower boiling point than the pure
liquids, and so is a low boiling azeotrope.
d) The mixture first boils at T = 89 C. Initially XA = 0.40, YA = 0.60. As distillation proceeds the mole
fraction of A in the system (ZA) decreases, since vapor enriched in A is being removed by the distillation. Both X A
and YA decrease as distillation proceeds. If we continue to remove vapor as we proceed (not what we would do in a
real distillation) then at the end the system will be pure B and boiling at T = 102 C (note that this is different
behavior from that observed in a closed system – distillation represents an open system, as vapor is being removed
from the system, changing the value of ZA for the system.
If the vapor is condensed and redistilled, the composition of the vapor in successive distillations will
converge on the composition of the azeotrope (ZA = 0.71), at which point no further purification can occur by
distillation. Therefore it is not possible to obtain pure A by distillation of a mixture with Z A = 0.40.
Exercise 5.15b
Let A = H2O, B = solute
Moles solute = nB = 122 g (1. mol/241. g) = 0.5062 mol
Moles water = nA = 920 g (1. mol/18.01 g) = 51.083 mol
XB = nB/(nA + nB) = (0.5062)/(51.083 + 0.5062) = 0.00981
And so XA = 0.99019
aA = pH2O/pH2O* = 0.02239/0.02308 = 0.9701
Since
aA = A XA, then A = aA/XA
So
A = 0.9701/0.99019 = 0.980
Exercise 5.17b
We first need the partial pressures of A and B in the vapor phase
pA = YA ptotal = 0.314 (101.325 kPa) = 31.82 kPa
pB = YB ptotal = (1 – YA) ptotal = (1 - 0.314) (101.325 kPa) = 69.51 kPa
We are using Raoult’s law to define ideal behavior for both components, and so
a = p/p*
So
 = a/X
aA = pA/pA* = 31.82/73.0 = 0.4359
A = aA/XA = 0.4359/0.220 = 1.981
aB = pB/pB* = 69.51/92.1 = 0.7547
B = aB/XB = 0.7547/0.780 = 0.968
Exercise 6.10a
Grxn = - RT ln K
At T = 450. K
ln K = - 1.04 + (-1088. K/450. K) + (1.51 x 10 5 K2)/(450. K)2 = - 2.712
So
Grxn = - (8.314 J/mol.K) (450. K) (- 2.712) = 10147. J/mol
Also
d lnK/d(1/T) = - Hrxn/R
and so
Hrxn = - R [ d lnK/d(1/T) ]
But d lnK/d(1/T) = d/d(1/T) [ A + B/T + C/T 2 ] = B + 2C/T = (- 1088. K) + 2 (1.51 x 105 K2)/450. K = - 416.9 K
And so Hrxn = - (8.314 J/mol.K) (- 416.9 K) = 3466. J/mol
Finally, since G = H – TS
Grxn = Hrxn – T Srxn
Srxn = (Hrxn - Grxn)/T = [ (3466. J/mol) – (10147. J/mol) ]/450. K = - 14.85 J/mol.K
NOTE: The problem in the book actually was for T = 400. K (I mistyped it in the problem set). At T = 400. K,
Grxn = 9370. J/mol, Hrxn = 2770. J/mol, Srxn = - 16.5 J/mol.K
Exercise 6.19b
Grxn = - RT ln K = - (8.314 J/mol.K) (298. K) ln(1.4 x 10-8) = 44.80 J/mol
But
Grxn = Gf(PbI2(aq)) - Gf(PbI2(s))
So
Gf(PbI2(aq)) = Grxn + Gf(PbI2(s))
= (44.80 kJ/mol) + - 173.64 kJ/mol) = - 128.84 kJ/mol
Problem 5.22
b) liquid and solid A
d) solid A and solid AB2
e) liquid, solid A and solid AB2
f) liquid and solid AB2
g) liquid, solid AB2 , and solid B
k) solid B and solid AB2
Note that I obtained the components by considering moving along a vertical line containing the point in question,
either beginning right above or right below the point.
Problem 6.30
a) The data are tabulated below
T (K)
K
1/T (K-1)
233.
248.
258.
4.13 x 108
5.00 x 107
1.45 x 107
0.004292
0.004032
0.003876
19.839
17.728
16.490
268.
273.
280.
5.37 x 106
3.20 x 106
9.62 x 105
0.003731
0.003663
0.003571
15.496
14.979
13.777
288.
295.
303
4.28 x 105
1.67 x 105
6.02 x 104
0.003472
0.003390
0.003300
12.967
12.026
11.005
ln K
The data are plotted below.
Based on the plot, I get
slope = 8785. K
intercept = - 17.61
And so at T = 298. K, K = 1.43 x 10 5
So
Grxn = - RT ln K = - (8.314 J/mol.K) (298. K) ln (1.43 x 105) = - 29.41 kJ/mol
Since (slope) = - Hrxn/R, it follows that
Hrxn = - R (slope) = - (8.314 J/mol.K) (8785 K) = - 73.04 kJ/mol
Finally (as shown in exercise 6.10a)
Srxn = (Hrxn - Grxn)/T = [ (- 73040 J/mol) – (- 29410 J/mol) ]/298. K = - 146.4 J/mol.K
b) We may find the standard enthalpy of formation and standard molar entropy as follows
Hrxn = Hf((ClO)2) - 2 Hf(ClO)
Hf((ClO)2) = Hrxn + 2 Hf(ClO) = (- 73.04) + 2 (101.8) = 130.6 J/mol
Srxn = S((ClO)2) - 2 S(ClO)
S((ClO)2) = Srxn + 2 S(ClO) = (- 146.4) + 2 (266.6) = 386.8 J/mol.K