CHM 3410 – Problem Set 5
Due date: Friday, October 16th
Do all of the following problems. Show your work.
"If things were simple word would have gotten around." - Jacques Derrida
1) If we assume Hvap is independent of temperature the following equation can be derived from the ClausiusClapeyron equation
ln(p2/p1) = - (Hvap/R) [ (1/T2) - (1/T1) ]
(1.1)
where p1 is the vapor pressure of the liquid at T 1 and p2 is the vapor pressure of the liquid at T 2.
a) Use equn 1.1 and the data given below to estimate the value for the vapor pressure of water (H 2O) at T =
50.0 C. Give your final answer in units of torr.
b) Equation 1.1 is derived assuming that the value for Hvap is independent of temperature. A better
equation may be obtained by including a term to account for the dependence of Hvap on temperature. Let
Hvap = a + b (T - T)
(1.2)
where a = Hvap, the enthalpy of vaporization at T, the normal boiling point of the liquid, and b = Cp,m = Cp,m(g) Cp,m() is the difference between the constant pressure molar heat capacities of the gas and liquid.
Using equn 1.2 and the Clausius-Clapeyron equation
d ln (p) = Hvap
dT
RT2
(1.3)
find an improved expression for ln(p2/p1) that takes into account the temperature dependence of Hvap.
c) Use your answer in b and the data below to estimate the value for the vapor pressure of water at T = 50.0
C. Give your final answer in units of torr.
d) Compare your estimated values for the vapor pressure of water found in part a and part c of this problem
to the experimental value at T = 50.0 C, pvap = 92.6 torr.
Hvap(H2O) = 40656. J/mol
T(H2O) = 100.00 C
Cp,m(H2O) = - 41.71 J/mol.K
2) "Synthetic air" is a mixture of nitrogen (N2) and oxygen (O2) with XN2 = 0.80. Find Gmix, Hmix, and Smix when
1.000 mol of synthetic air is produced by mixing an appropriate amount of pure nitrogen and pure oxygen at p =
1.000 atm and T = 25.0 C. Assume ideal behavior.
3) When ethylene bromide (EtBr) and propylene bromide (PrBr) are mixed they form, to a first approximation, an
ideal solution.
Consider a solution of EtBr and PrBr at T = 80.0 C. The vapor pressure above the solution is p = 139.
torr. The vapor pressures of the pure liquids at this temperature are p*(EtBr) = 172. torr and p*(PrBr) = 127. torr.
Find the mole fraction of EtBr in the solution and the partial prssure of EtBr in the vapor above the solution.
4) In class we derived the following expression for the free energy of mixing of two ideal gases initially at the same
temperature and pressure
Gmix = nRT [ XA lnXA + XB lnXB ]
(4.1)
The same procedure can be used to find Gmix when the two gases are originally at different pressures.
Consider the system below. 1.000 mol of argon (Ar) is initially confined to the left side of the container,
which has a volume V = 30.00 L, and 1.00 mol of helium (He) is initially confined to the right side of the container,
which has a volume V = 10.00 L. The barrier between the two sides of the container is removed and the gases are
allowed to mix until equilibrium is reached. Temperature remains constant at T = 300.0 K throughout the process.
Find Gmix. You may assume that argon and helium behave ideally. HINT: You cannot simply use equn 4.1 to
answer this question since it was derived assuming the initial pressures of the two gases were equal to each other.
5) A solution is formed by dissolving 66.83 g of vanadium tetrachloride (VCl 4, M = 192.75 g/mol), a nonvolatile
solute, in 1000.0 g of carbon tetrachloride (CCl4, M = 153.82 g/mol). The normal freezing point and freezing point
depression constant for carbon tetrachloride are Tfus = - 22.9 C and Kf = 30.0 kg.K/mol.
a) Based on the above data predict Tf, the depression of the freezing point for the above solution.
b) The observed depression of the freezing point for the above solution is Tf = 5.97 C, which is smaller
than that predicted in part a. This is due to the fact that vanadium tetrachloride dimerizes in carbon tetrachloride.
The reaction may be written as
2 VCl4(soln)  V2Cl8(soln)
Km = b(V2Cl8)/[b(VCl4)]2
(5.1)
where Km is the equilibrium constant for the reaction in terms of molality (b).
Using the information given above find Km for the dimerization reaction at the freezing point of the
solution.
Also do the following from Atkins:
Exercises
5.1 a Note that M(CH3COCH3) = 58.08 g/mol; M(CHCl3) = 119.38 g/mol.
Problems
4.8
4.10 Give the phase diagram for the pressure range from 0 to 1.5 atm and the temperature
range from - 40.0 C to 100.0 C.
5.4
EXTRA CREDIT - Is there evidence in the data in Problem 4.8 for temperature dependence in the value for
Hvap for nitric acid? Justify your answer - and if your answer is yes, find an expression for Hvap(T), the enthalpy
of vaporization as a function of T.
Solutions
1)
a) If we assume Hvap is independent of temperature, then
ln(p2/p1) = - (Hvap/R) [ (1/T2) - (1/T1) ]
Let
T1 = 100.0 C = 373.15 K
p1 = 760. torr
T2 = 50.0 C = 323.15 K
Hvap = 40656. J/mol
R = 8.3145 J/mol.K
Substitution into the above equation gives
ln(p2/p1) = - [(40656. J/mol)/(8.3145 J/mol.K)] [ (1/323.15 K) - (1/373.15 K) ] = - 2.0275
p2/p1 = exp(-2.0275) = 0.13166
p2 = (0.13166) p1 = (0.13166) (760 torr) = 100.1 torr
b) To take the temperature dependence of Hvap into account we say
Hvap = a + b (T - T) where T = 100.0 C = 373.15 K
Substitution into the Clausius-Clapeyron equatin gives
d ln (p) = Hvap = [ a + b (T - T) ]
rearrange
dT
RT2
RT2
d ln (p) = [ (a - bT) + bT ] dT
RT2
multiply both sides by dT and
integrate from p1, T1 to p2, T2
p1p2 d ln (p) = T1T2 [ (a - bT)/RT2 + b/RT ] dT
ln (p2/p1) = - [ (a - bT)/R] [ (1/T2) - (1/T1) ] + (b/R) ln(T2/T1)
c) Substituting the information from part a of the problem, and a = 40656. J/mol ; b = - 41.71 J/mol.K
ln(p2/p1) = - [ (40656. - (- 41.71) (373.15) ] J/mol [ (1/323.15 K) - (1/373.15 K) ]
(8.314 J/mol.K)
+ ( - 41.71 J/mol.K) ln(323.15/373/15)
( 8.3145 J/mol.K)
= - 2.8039 + 0.7217 = - 2.0822
p2/p1 = exp(- 2.0822) = 0.12465
p2 = (0.12465) p1 = (0.12465) (760 torr) = 94.7 torr
d) The experimental value for the vapor pressure at T = 50.0 C is p = 92.6 torr. The result in part c (2.3 %
higher) is a bit better than the result from part a (5.7 % higher). I'm not sure what is responsible for the remaining
difference, but I believe the biggest remaining error is in the assumption that the vapor phase behaves like an ideal
gas.
2) For ideal mixing
Gmix = nRT [XN2 lnXN2 + XO2 lnXO2] = (1.00 mol) (8.3145 J/mol.K) (298.15 K) [0.8 ln(0.8) + 0.2 ln(0.2)]
= - 1240. J
Smix = - nR [XN2 lnXN2 + XO2 lnXO2] = - (1.00 mol) (8.3145 J/mol.K) [0.8 ln(0.8) + 0.2 ln(0.2)] = 4.161
J/K
Hmix = 0.
3) Let
XE = mole fraction of ethylene bromide
XP = 1 - XE = mole fraction of propylene bromide
Then
XE pE* + XP pP* = XE pE* + (1 - XE) pP* = p
pP* + XE (pE* - pP*) = p
XE =
Then
(p - pP*) = (139. - 127.) = 0.267
(pE* - pP*)
(172. - 127.)
(and so XP = 0.733)
pE = XE pE* = (0.267) (172. torr) = 45.9 torr (and so pP = 93.1 torr)
4) We use the same procedure as used in class to find Gmix when the pressures were initially the same.
Gmix = Gf - Gi
Gi = nAr Ar,i + nHe He,i
Gf = nAr Ar,f + nHe He,f
where
j = j + RT ln pj
Let
pAr,i = initial pressure of Ar
pHe,i = initial pressure of He
Then, using Boyle's law
pAr,f = final pressure of argon = (30./40.) p Ar,i = (0.75) pAr,i
pHe,f = final pressure of helium = (10.40.) pHe,i = (0.25) pHe,i
So
Gi = nAr (Ar + RT ln pAr,i) + nAr (He + RT ln pHe,i)
Gf = nAr (Ar + RT ln 0.75 pAr,i) + nAr (He + RT ln 0.25 pHe,i)
= nAr (Ar + RT ln pAr,i) + nAr (He + RT ln pHe,i) + nArRT ln(0.75) + nHeRT ln(0.25)
Gmix = Gf - Gi
= nAr (Ar + RT ln pAr,i) + nAr (He + RT ln pHe,i) + nArRT ln(0.75) + nHeRT ln(0.25)
- [ nAr (Ar + RT ln pAr,i) + nAr (He + RT ln pHe,i) ]
= nArRT ln(0.75) + nHeRT ln(0.25)
= (1.00 mol) (8.3145 J/mol.K) (300.0 K) ln(0.75) + (1.00 mol) (8.3145 J/mol .K) (300.0 K)
ln(0.25)
= - 4175. J
Notice that we never had to find values for any of the pressures in the problem.
5) If all of the solute was in the orm of the monomer, VCl4, then
b = (66.83 g) (1. mol/192.75 g) = 0.3467 mol/kg
1.000 kg
The actual molality of solute particles can be found from the measured depression of the freezing point
Tf = Kf bsol
So
bsol = Tf =
Kf
5.97 K
(30.0 kg.K/mol)
= 0.1990 mol/kg
This lower value occurs because of the dimerization reaction. To find the value for K m we need the equilibrium
molalities of the monomer and dimer.
Compound
initial
change
equilibrium
VCl4
0.3467
- 2x
0.3467 - 2x
V2Cl8
0.0
+x
+x
Since bsol = b(VCl4) + b(V2Cl8)
0.1990 = (0.3467 - 2x) + x = 0.3467 - x
x = 0.3467 - 0.1990 = 0.1477 mol/kg
So
b(VCl4) = 0.3467 - 2 (0.1477) = 0.0513 mol/kg
b(V2Cl8) = 0.1477 mol/kg
Km =
b(V2Cl8) = (0.1477) = 56.1
[b(VCl4)]2
(0.0513)2
Exercise 5.1 a
Start by assuming 1.000 mol of the mixture. Let
XA = mole fraction acetone
XC = mole frac tion chloroform
m = XA MA + XC MC = (0.5307) (58.08 g/mol) + (0.4693) (119.38 g/mol) = 86.85 g/mol soln
So the total number of moles of solution is
n = 1000.0 g
1 mol = 11.514 mol soln
86.85 g
The volume occupied by 1.00 mol of solution is
Vm = (0.5307) (74.166 cm3/mol) + (0.4693) (80.235 cm3/mol) = 77.014 cm3/mol
So
V = n Vm = (11.514 mol) (77.014 cm3/mol) = 886.7 cm3
Problem 4.8
The data are tabulated below
T (C)
T (K)
0.0
20.0
273.15
293.15
40.0
50.0
313.15
323.15
70.0
80.0
90.0
100.0
p (kPa)
1.92
6.38
1/T (K-1)
ln p
0.0036610
0.0034112
0.65233
1.85317
17.7
27.7
0.0031934
0.0030945
2.87356
3.32143
343.15
353.15
62.3
89.3
0.0029142
0.0028317
4.13196
4.49200
363.15
373.15
124.9
170.9
0.0027537
0.0026799
4.82751
5.14108
The data are plotted below.
Plot of ln p vs 1/T
6
5
ln p
4
3
2
1
0
0.0026
0.0028
0.003
0.0032
(1/T) K-1
0.0034
0.0036
0.0038
The best fitting line is
ln p = 17.429 - (4570. K)/T
a) At the normal boiling point p = 1.000 atm = 101.325 kPa.
ln p = 17.429 - (4570. K)/T
ln p - 17.429 = (- 4570. K)/T
1/T =
and so T =
= ln (101.325) - 17.429 = 0.0028032 K-1
(- 4570. K)
ln p - 17.429
(- 4570. K)
1
0.0028032 K-1
= 356.7 K
b) Hvap = - m R = (4570. K) (8.3145 J/mol.K) = 38.00 J/mol
Problem 4.10
Start at the triple point
p = 36. torr = 0.047 atm
T = 5.50 C = 278.65 K
The solid-liquid boundary is, over the pressure range of the phase diagram, essentially a vertical line.
For the solid-gas and liquid-gas boundaries we use the general relationship
ln(p2/p1) = - (Hpt/R) [ (1/T2) - (1/T1) ]
So, for the solid-gas boundary
ln(p/0.047) = - 1275. [ (1/T2) - 0.003589 ]
and for the liquid-gas boundary
ln(p/0.047) = - 3705. [ (1/T2) - 0.003589 ]
Points for the phase diagram are tabulated below
solid-gas
triple point
liquid-gas
T(C)
p(atm)
- 40.0
- 30.0
- 20.0
- 10.0
0.0
5.5
10.0
20.0
40.0
60.0
80.0
100.0
0.019
0.024
0.030
0.036
0.043
0.047
0.058
0.091
0.203
0.414
0.777
1.363
The phase diagram is given below.
Problem 5.4
Let
W = water, nW = moles of water, VW = partial molar volume of water
C = copper II sulfate, nC = moles of copper II sulfate, VC = partial molar volume of copper II
sulfate
MW = 18.02 g/mol
MC = 159.61 g/mol
Then
V = nW VW + nC VC
dV = nW dVW + VW dnW + nC dVC + VC dnC
However, if we keep the mass (and therefore the number of moles) of water constant at 100.0 g, then dn W = 0.
From the Gibbs-Duhem equation
dVW = - (nC/nW) dVC
Substituting into our equation for V gives
dV = nW [ - (nC/nW) dVC ] + nC dVC + VC dnC = VC dnC
Solving for VC, we get
VC = dV/dnC
The data in the problem are tabulated below. Note we adjust the total mass of solution so that m(H2O) remains
constant at 100.0 g. This has a small effect on the mass of CuSO4, as indicated below.
m(soln) = 100.0 g [100.0/(100.0 - m(CuSO4))]
m(CuSO4) = m(soln) - 100.0 g
CuSO4 (g)
adjusted
mass soln (g)
adjusted
adjusted
mass CuSO4 (g) mol CuSO4
 (g/cm3)
V (cm3)
5.0
10.0
15.0
20.0
105.26
111.11
117.65
125.00
5.26
11.11
17.65
25.00
1.051
1.107
1.167
1.230
100.15
100.37
100.81
101.63
0.0330
0.0696
0.1106
0.1566
If we assume
VC = dV/dnC  V/nC
then we may use the above data to estimate the value for VC at each intermediate value of added CuSO4.
CuSO4 (g)
ave mass (g)
mol fraction
VC (cm3)
nC (mol)
VC (cm3/mol)
8.18
14.38
21.32
0.22
0.44
0.82
0.0366
0.0410
0.0460
6.0
10.7
17.8
0.0092
0.0160
0.0235
The data are plotted below
Plot of Vc vs Xc
Vc (ml/mol)
25
20
15
10
5
0
0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
Xc
The value for VC at XC = 0 (from extrapolation of the data) is VC = - 1.9 cm3/mol. This is close to the value VC = 1.4 cm3/mol for MgSO4 (Atkins, page 138).
EXTRA CREDIT
To look for evidence of curvature we can fit the data in problem 4.8 to a second order polynomial
ln p = a + b/T + c/T2
and see if the value of c (coefficient of the 1/T 2 term) is significantly different from zero.
Using the POLY program the best fitting second order polynomial to the data has
c = ( - 3.35  0.25) x 105 K2 (at 95% confidence limits)
Since c is significantly different from zero (at 95% confidence limits) there is evidence for curvature in the data.