CHM 3410 – Problem Set 6
Due date: Wednesday, October 14th
Do all of the following problems. Show your work.
"This paper contains many things that are new and interesting. Unfortunately, what is new is not interesting, and
what is interesting is not new." anonymous reviewer comments
1) Find F, the number of degrees of freedom, for the following systems.
a) A solution formed by mixing two miscible liquids, and in equilibrium with its vapor phase.
b) A solution formed by dissolving 0.100 mole of sodium chloride (NaCl) and 0.100 mol of sodium acetate
(NaCH3COO) in 1.00 L of water. The solution is in equilibrium with its vapor phase.
c) A mixture of nickel and iron metals forming a solid phase and a liquid phase (this is essentially what is
found in the core of the Earth).
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) Consider the system below. You have two gases initially at the same temperature and with equal initial volumes.
The barrier between the two gases is removed and the gases are allowed to mix. Find an expression and a numerical
value for G for this process. You may assume the gases are ideal for the conditions of the problem. (HINT: You
may use the same general procedure followed in class to find Gmix for mixing of ideal gases - however, you cannot
simply use the result obtained in class, since the initial pressures for the system shown below are not equal.)
4) 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 pressure and mole fraction of EtBr in the vapor above
the solution.
5) The vapor pressure of pure liquid water at T = 0.0 C is pw* = 4.579 torr. The equilibrium vapor pressure of
water above a solution of water and sucrose at the same temperature, with Xw = 0.918, is pw = 3.994 torr. Find aw
(the activity of the water) and w (the activity coefficient for water) in the above solution.
6) Data for solutions of carbon tetrachloride (C) and methanol (M) at T = 20.0 C are given below. Note that XC =
mole fraction of carbon tetrachloride in the liquid phase, Y C = mole fraction of carbon tetrachloride in the vapor
phase, and ptotal = total pressure of vapor above the solution.
XC
YC
ptotal (bar)
XC
YC
ptotal (bar)
1.00
0.90
0.80
0.70
0.60
0.50
1.00
0.65
0.55
0.51
0.49
0.48
0.128
0.180
0.203
0.210
0.213
0.214
0.40
0.30
0.20
0.10
0.00
0.47
0.46
0.45
0.43
0.00
0.214
0.213
0.212
0.207
0.120
a) What are pC* and pM*, the vapor pressures of pure carbon tetrachloride and pure methanol, respectively,
at T = 20.0 C?
b) Plot pC (vapor pressure of carbon tetrachloride), p M (vapor pressure of methanol) and ptotal vs XC. An
example of such a plot is given in Fig 5A.13 of Atkins. Also indicate in your plot dashed line showing what the plots
would look like for ideal solution behavior.
c) Find aC, C, aM, and M when XC = 0.60.
d) Find Gmix (free energy of mixing) and GE (excess free energy) when 0.60 moles of carbon tetrachloride
are mixed with 0.40 moles of methanol.
e) Use the data for XC = 0.7 and XC = 0.8 to test the validity of the Gubbs-Duhem equaltion. When applied
to partial molar Gibbs free energy (chemical potential) the equation says
dM = - (nC/nM) dC
Comment on how closely the experimental results agree with the result predicted from eq 6.1.
(6.1)
Solutions.
1) The phase rule says F = (C - P) + 2
a) There are two components and two phases, so F = (2 - 2) + 2 = 2
b) There are three components and two phases, so F = (3 - 2) + 2 = 3
c) Usually a mixture of two substances in the solid phase does not form a solution, and so one would expect
that there are two solid phases and one liquid phase present. Since there are two components, F = (2 - 3) + 2 = 1
However, since the problem was unclear as to how many distinct solid phases are present, I will also count
as correct the answer obtained when a single solid phase is assumed to exist, or F = (2 + 2) - 2 = 2
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)
G = Gf - Gi
Then
Gi = nAr (Ar + RT ln(pAr)) + nN2 (N2 + RT ln(pN2))
Let
pAr = initial pressure of Ar
pN2 = initial pressure of N2
Since the volume of each side of the container is the same, the temperature is constant, and the gases are ideal, then
when the barrier is removed the volume available for each gas doubles, and so the partial pressure of each gas will be
half of the initial value for pressure. Therefore
Gf = nAr (Ar + RT ln(pAr/2)) + nN2 (N2 + RT ln(pN2/2))
G = [ nAr (Ar + RT ln(pAr/2)) + nN2 (N2 + RT ln(pN2/2)) ]
- [nAr (Ar + RT ln(pAr)) + nN2 (N2 + RT ln(pN2))]
= nArRT ln(1/2) + nN2RT ln(1/2) = (nAr + nN2)RT ln(1/2)
= (3.00 mol)(8.3145 J/molK)(300.0 K) ln(1/2) = - 5187. J
4) 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* = 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)
YE = pE/p = (45.9 torr)/139. torr) = 0.330
(and so YP = 0.670)
5)
aW = pW/pW* = 3.994 torr = 0.8722
4.579 torr
Since aW = W XW
then
W = aW/XW = 0.8722 = 0.950
0.918
6)
a) pC* occurs when XC = 1.00, and pM* occurs when XC = 0.00 (which means XM = 1.00)
So
pC* = 0.128 bar and pM* = 0.120 bar
b) We can find the partial pressures using pC = YC ptotal, pM = ptotal - pC.
The values are tabulated below, and also plotted
XC
YC
ptotal (bar)
pC (bar)
pM (bar)
1.00
0.90
0.80
0.70
0.60
0.50
0.40
0.30
0.20
0.10
0.00
1.00
0.65
0.55
0.51
0.49
0.48
0.47
0.46
0.45
0.43
0.00
0.128
0.180
0.203
0.210
0.213
0.214
0.214
0.213
0.212
0.207
0.120
0.128
0.117
0.112
0.107
0.104
0.103
0.101
0.098
0.095
0.089
0.000
0.000
0.063
0.091
0.103
0.109
0.111
0.113
0.115
0.117
0.118
0.120
The straight lines correspond to the partial pressures predicted if the liquids both obeyed Raoult’s law. It is clear the
solution exhibits non-ideal behavior.
c) When XC = 0.60, aC = pC/pC* = (0.104 bar)/0.128 bar) = 0.812
Since
aC = CXC, then C = aC/XC = (0.812)/(0.60) = 1.35
Similarly,
aM = pM/pM* = (0.109 bar)/(0.120 bar) = 0.908
M = aM/XM = (0.908)/(0.400) = 2.27
When XC = 0.60, XM = 0.40, and so
d) Gmix – Gf – Gi
Gi = nC C* + nM M*
Gf = nC (C* + RT ln aC) + nM (M* + RT ln aM)
So
Gmix = [= nC (C* + RT ln aC) + nM (M* + RT ln aM) ] – [ nC C* + nM M* ]
= nCRT ln aC = nMRT ln aM = RT [ nC ln aC + nM ln aM ]
Using the information in the problem and the answer to part c, we get
Gmix = (8.3145 J/molK)(293.2 K) [ (0.600 mol)(ln(0.812)) + (0.400 mol)(ln(0.908)) ]
= - 398.7 J
For ideal mixing, Gmixi = RT [ nC ln XC + nM ln XM ]
= - 1641. J
The excess free energy of mixing, GE, is then
GE = Gmix - Gmixi = (- 398.7 J) – (- 1641. J) = 1242. J
e) The Gibbs-Duhem equation says
dM = - (nC/nM) dC
We will use the approximation
dM  M
dC  C
where  is the difference in chemical potential between XC = 0.70 and XC = 0.80
Note that (nC/nM) = (XC/XM). If we use the values half-way between XC = 0.70 and XC = 0.80
Then
(nC/nM) = (XC/XM) = (0.75/0.25) = 3.0
We can use the data to make the following table. Recall
So
XC
aC
aM
0.70
0.80
0.8359 0.8583
0.8750 0.7583
aC = pC/pC*, and aM = pM/pM*
C = C* + RT lnaC and M = M* + RT lnaM
C
M
C* - 0.179 RT
C* - 0.134 RT
M* - 0.153 RT
M* - 0.277 RT
C = ([ (- 0.134) – (- 0.179) ] RT = + 0.045 RT
M = ([ (- 0.153) – (- 0.277) ] RT = - 0.124 RT
MGD = - (3.0) (0.045 RT) = - 0.135 RT
The predicted value MGD = - 0.135 RT is close to the experimental value M = - 0.124 RT. The small difference
is due to roundoff error and the fact that setting d   is only approximate.