CHM 3410 – Problem Set 5
Due date: Wednesday, October 13th
Do all of the following problems. Show your work.
1) Consider the system below
The left side contains 1.000 mol of N2 gas, while the right side contains 2.000 mol of O2 gas. Both sides have a
volume V = 10.00 L and are at a temperature T = 400.0 K.
The barrier between the two sides is removed and the two gases are allowed to mix until equilibrium is
achieved. Find Gmix, Smix, and Hmix for the process. (Hint: You may assume that N2 and O2 behave ideally, and
that T remains constant at 400. K. Note that you cannot simply use the equations derived in class for Gmix, Smix,
and Hmix, since those equations were derived for the case of equal initial pressures of gas. However, you can use
the same general method to obtain the appropriate equations for the change in the thermodynamic functions for
mixing).
2) There are two ways in which the equation for boiling point elevation is expressed
T = Kb XB
(2.1)
T = Kb bB
(2.2)
As shown in class, the value for Kb (the constant in equn 2.1) is given by the expression
Kb = RTb*2/Hvap
(2.3)
a) Using equn 2.1 and 2.3, find the value for Kb (the constant in equn 2.2). Note that your equation for K b
will involve MA (molecular mass of the solvent). Also note that in deriving your equation for K b you will need to
make the approximation that XB = nB/(nA + nB)  nB/nA.
b) Using the data in Table 2.3 of the appendix of Atkins find the value for K b for benzene (C6H6) and
carbon disulfide (CS2). Compare your calculated values to the experimental values given in Table 5.2 of the
appendix of Atkins.
3) Data on the partial pressures of iodoethane (I = CH 3CH2I) and ethyl acetate (A = CH3COOC2H5) in equilibrium
with a liquid solution of iodoethane and ethyl acetate, at T = 50. C, are given below.
XI
pI(kPa)
pA(kPa)
0.0000 0.0579 0.1095 0.1918 0.2353 0.3718 0.5478 0.6349 0.8253 0.9093 1.0000
0.00
3.73
7.03 11.70 14.05 20.72 28.44 31.88 39.58 43.00 47.12
37.38 35.48 33.64 30.85 29.44 25.05 19.23 16.39
8.88
5.09
0.00
a) Plot the above data. Plot pI, pA, and ptotal vs XI.
b) Find the numerical value for the Henry's law constant for ethyl acetate in iodoethane at T = 50. C.
c) Find the values for activity (a) and activity coefficient () for iodoethane and ethyl acetate for the data at
XI = 0.2353 and XI = 0.6349. In both cases use the solvent standard state for both iodoethane and ethyl acetate.
Exercises
5.3b
XB
PB(kPa)
At 310. K the partial vapor pressures of a substance B dissolved in a liquid A are as follows:
0.010
82.0
0.015
122.0
0.020
166.1
Show that this solution obeys Henry's law in this range of mole fractions and calculate the Henry's law constant at T
= 310. K.
5.6b
The addition of 5.000 g of a compound to 250.0 g of naphthalene lowered the freezing point of the
solvent by 0.780 K. Calculate the molar mass of the compound.
5.11b The mole fraction of N2 and O2 in air at sea level are approximately 0.78 and 0.21. Calculate the
molalities of the solution formed in an open flask of water at T = 25. C. (Hint: The appropriate Henry's law
constants are given in Table 5.1 of the appendix of Atkins. Assume a total pressure of 1.000 bar. Note that since the
solvent is water the molality and molarity of dissolved gas have approximately the same numerical value.)
Problems
5.15
The excess Gibbs free energy of solution of methylcyclohexane (MCH) and tetrahydrofuran (THF)
at 303.15 K was found to fit the expression
GE = RTx(1 - x) {0.4857 - 0.1077 (2x - 1) + 0.0191 (2x - 1)2 }
where x is the mole fraction of the methylcyclohexane. Calculate the Gibbs free energy of mixing when a mixture of
1.00 mol of MCH and 3.00 mol of THF is prepared. (Hint: The equation above is for GE. Recall that GE = Gmix Gmix,ideal. You are being asked for the value for Gmix in this problem.)
Solutions.
1) Following the procedure we used in class, we may say
Gmix = Gf – Gi
Assuming ideal gas behavior, we may say the chemical potential of the gas is given by the expression
A = A + RT ln(pA)
Let us call the initial pressures of N2 and O2 pN2 and pO2 (we have sufficient information to calculate the initial
pressures, but will not need to do so).
Then the partial pressures of N2 and O2 after mixing are (because Vf = 2 Vi)
pN2,f = (Vi/Vf) PN2 = pN2/2
pO2,f = (Vi/Vf) PO2 = pO2/2
So
Gf = nN2 [N2 + RT ln(pN2/2)] + nO2 [O2 + RT ln(pO2/2)]
Gi = nN2 [N2 + RT ln(pN2)] + nO2 [O2 + RT ln(pO2)]
Gmix = Gf – Gi = { nN2 [N2 + RT ln(pN2/2)] + nO2 [O2 + RT ln(pO2/2)] }
- { nN2 [N2 + RT ln(pN2)] + nO2 [O2 + RT ln(pO2)] }
= nN2 RT ln(1/2) + nO2 RT ln(1/2)
= - (nN2 + nO2) RT ln(2)
= - (1.000 mol + 2.000 mol) (8.314 J/mol.K) (400. K) ln(2) = - 6915. J
Since the gases are ideal, Hmix = 0.
Since G – H – TS, Gmix = Hmix - TSmix , and so
Smix = - Gmix/T = - (- 6915. J)/(400 K) = + 17.29 J/K.
2)
a) T = Kb XB
But XB = nB/(nA + nB)  nB/nA (when nB << nA)
If we assume 1000.0 g = 1.000 kg A, then
nB = bB (1.000 kg)
nA = 1000.0 g/MA
where bB = nB/kg A = molality of solute.
And so T = Kb XB = Kb bB (1.000 kg) = Kb MA (1.000 kg) bB
(1000.0 g/MA)
1000.0 g
= KbbB, where Kb = Kb MA 1.000 kg = (1.0000 kg) MARTb*2
1000.0 g
(1000.0 g) Hvap
b) The data are tabulated below
M (g/mol)
Tb* (K)
Hvap (kJ/mol) Kcalc (K.kg/mol) Kexp (K.kg/mol) % diff
C6H6
78.12
353.2
30.8
2.631
2.53
4.0
CS2
76.14
319.4
26.74
2.415
2.37
1.9
Compound
The agreement between the experimental and calculated values for Kb is quite good.
3)
a) The data are plotted below. Note that ptotal = pI + pA.
Plot of p vs XI
60
p (kPa)
50
40
30
20
10
0
0
0.2
0.4
0.6
0.8
1
XI
b) Normally you would find KB, the Henry’s law constant, by plotting pA vs XI and looking at the behavior
in the limit XI  1. However, for our data set even the smallest mole fraction of A is X A ~ 0.1, likely outside the
Henry’s law limiting behavior. So the best we can do is assume Henry’s law applies for this data point. Then
p A = K B XA
KB = pA/XA = (5.09 kPa)/(0.0917) = 55.5 kPa
c) For the solvent standard state (Raoult’s law)
aA = pA/pA* = A XA, and so A = aA/XA = pA/(XA pA*)
Based on the data pI* = 47.12 kPa, pA* = 37.38 kPa.
The activities and activity coefficients are tabulated below
XI
pI (kPa)
0.2353 14.05
0.6349 31.88
I
pA (kPa)
aI
29.44
16.39
0.2982 1.267
0.6766 1.066
aA
A
0.7876 1.030
0.4385 1.201
Exercise 5.3b
We may plot pB vs XB and look at the limiting slope by fitting the data to a low order polynomial. Such a
plot is given below.
Plot of pB vs Xb
200
pB (kPa)
160
120
80
40
0
0
0.005
0.01
0.015
XB
Since the data appears to closely fit a straight line we will use a linear fit to the data. Then
pB = - 0.48 kPa + (8267. kPa) XB
The Henry’s law constant is simply the slope of the line, so KB = 8267. kPa.
Exercise 5.6b
We may use freezing point depression
Tf = Kf bB , where for naphthalene Kf = 6.94 kg.K/mol
So
bB = Tf/Kf = (0.780 K)/(6.94 kg.K/mol) = 0.1124 mol/kg
And so nB = bB (kg naphthalene) = (0.1123 mol/kg) (0.2500 kg) = 0.02810 mol B
And so MB = mB/nB = (5.000 g)/(0.02810 mol) = 178. g/mol
0.02
Exercise 5.11b
Since the total pressure of air is ptotal = 1.000 bar, it follows from Dalton’s law that
pN2 = (0.78) (1.000 bar) = 0.78 bar = 78. kPa
KN2 = 1.56 x 105 kPa.kg/mol
pO2 = (0.21) (1.000 bar) = 0.21 bar = 21. kPa
KO2 = 7.92 x 104 kPa.kg/mol
From Henry’s law, pB = KB bB, and so bB = pB/KB. Since water is the solvent, we can assume molality and molarity
are the same (true to better than 1%)
So
[N2] = (78. kPa)/(1.56 x 105 kPa.kg/mol) = 0.0050 mol/kg = 0.00050 mol/L
[O2] = (21. kPa)/(7.92 x 104 kPa.kg/mol) = 0.000265 mol/kg = 0.000265 mol/L
Problem 5.15
Since GE = Gmix - Gmix,ideal, then Gmix = Gmix,ideal + GE (keep in mind this is per mole of solution, which
we will account for at the end). XMCH = 0.25 and XTHF = 0.75 (since 1.000 mol MCH and 3.000 mol THF)
Gmix,ideal = RT (XMCH ln XMCH + XTHF ln XTHF)
= (8.314 J/mol.K) (303.15 K) [ 0.25 ln (0.25) + 0.75 ln (0.750) ] = - 1417.3 J/mol
GE = RTx(1 - x) {0.4857 - 0.1077 (2x - 1) + 0.0191 (2x - 1)2 }
= (8.314 J/mol.K)(303.15 K) (0.25)(0.75) {0.4857 – 0.1077(- 0.50) + 0.0191(- 0.50)2} = + 257.2 J/mol
So on a per mole basis
Gmix,mol = - 1417.3 J/mol + 257.2 J/mol = - 1160.1 J/mol
Since there are 4.000 mol of solution formed
Gmix = (4.000 mol) (- 1160.1 J/mol) = - 4640. J