1. No category

CHEM 305
1.
Solutions for assignment #8
The molar volume of a certain solid is 0.142 L mol-1 at 1.00 atm at its normal melting temperature of 427.15
K. The molar volume of the liquid at this temperature and pressure is 0.1526 L mol-1. At a pressure of 11.8
atm the melting temperature increases to 429.26 K. Calculate the enthalpy and entropy of fusion of the
solid.
**Solution**
The Clapeyron equation describes the slope of the phase boundary on a P vs. T phase diagram in terms of
the entropy change and molar volume change between the two phases:
 S
dP
 fus
dT
fusV
 separate variables and integrate (assume that ΔfusS and ΔfusV are independent of T)
P2
 dP
fus S

fusV
P1
T2
 dT
T1

P 
fus S
fusV
T
Solve for ΔfusS:
P fusV
fus S 
T
ΔfusV can be calculated from the given molar volumes:
fusV  Vliquid Vsolid fusion is solid  liquid; therefore ΔfusV is Vliquid – Vsolid
 0.1526L mol1  0.142L mol1
 0.0106L mol1
ΔP and ΔT are:
P  P2  P1
 11.8 atm 1 atm
T T2 T1
 429.26K  427.15K
 10.8 atm
 2.11K
Substituting these into the expression for ΔfusS gives:
10.8 atm 0.0106L mol1
fus S 
2.11K
 0.054 L atmK 1 mol1


 101.316J L1 atm1
always convert your final answer to appropriate units!
 5.50 JK mol
To find the enthalpy, you might be tempted to use:
fus H Tfus fus S
1
1
Which Tfus would you use? It turns out that either would give an acceptable solution because ΔT is small.
However, a better, more rigorous approach would be to write the Clapeyron equation in terms of the
enthalpy, integrate it, then solve for ΔfusH:
 H
dP
 fus
dT T fusV

P2
 dP 
P1
fus H
fusV
T2
1
 dT
TT
1

fus H
T2
fusV T1
Solve for ΔfusH:
P 
ln
(continued on next page)
Copyright © 2014 by Eric B. Brauns
1
CHEM 305
fus H 
Solutions for assignment #8
P fusV
T 
ln  2 
 T1 
10.8 atm   0.0106L mol1 

ln 429.26K
427.15K
 23.233L atmmol1


 0.101kJ L1 atm1
 2.35kJ mol1
2. Calculate the freezing point of water at an elevation of 7000 feet. At this elevation, the atmospheric
pressure is 0.772 atm. The density of water at 273 K is 1 g/mL and the density of ice at 273 K is
0.92 g/mL. The enthalpy of fusion is 6.00 kJ mol-1.
**Solution**
We can use the integrated Clapeyron equation:
 H T
P  fus ln 2
fusV T1
Solve this for T2:
P fusV
lnT2  lnT1 
fus H

 P fusV 
  H 
 fus

T1 is 273 K and ΔP is:
P  P2  P1
T2 T1 exp 
 0.772 atm 1 atm
 0.228 atm
To determine ΔfusV, we need to use the densities to calculate molar volumes:
1
1
 0.92g mol 1000 mL 
 1g mol 1000 mL 


Vliquid  


Vliquid  


18g
1L
1L
 mL

 mL 18g

 0.018L mol1
 0.02L mol1
fusV  Vliquid Vsolid
 0.018L mol1  0.02L mol1
 0.002L mol1
Now we have all the quantities we need:
  0.228 atm   0.002L mol1 

1
1
101.316J
L
a
T2  273K  exp 

tm

6.00  103 J mol1


 273.002K
Pay attention to units! In the exponential, we
are dividing L atm by J—that doesn’t work.
3. Using a phase diagram for water (use any physical chemistry book), state which phase (solid, liquid, or gas)
of H2O has the lowest chemical potential:
a. 25°C at 1 atm
b. 25°C and 0.1 torr
c. 0°C and 500 atm
d. 100°C and 10 atm
e. 100°C and 0.1 atm
**Solution**
a. liquid; b. gas; c. liquid; d. liquid; e. gas.
Copyright © 2014 by Eric B. Brauns
2
CHEM 305
Solutions for assignment #8
4. ΔvapH of water is 40.67 kJ mol-1 at the normal boiling point. Many bacteria can survive at 100°C by forming
spores. However, most spores die at 120°C. Hence, autoclaves used to sterilize medical and laboratory
equipment are pressurized to raise the boiling point of water to 120°C.
a. At what pressure does water boil at 120°C?
b. What is the boiling point of water at the top of Pike’s Peak (elevation 14,100 ft), where the
atmospheric pressure is typically 446 torr?
**Solution**
Here, we are interested in the transition from a liquid (a condensed phase) to a vapor so we can use the
Clausius-Clapeyron equation:
d ln P vap H

2
dT

ln P2

ln P1
RT
d ln P 
vap H T2 1
R
T
T1
2
dT

ln
a.
vap H
P2

P1
R
1 1
  
T2 T1 
Solve for P2:
vap H  1 1 
ln P2  ln P1 
  
R T2 T1 

 vap H  1 1  
  
R T2 T1  

Just substitute the given values:
 40.67  103 J mol1  1
1 
P2   760 torr  exp  



1
1
 8.314 JK mol  393K 373K  
 1481torr
b. For this, we still use the Clausius-Clapeyron equation, but we solve for T2 (since we are given P2):
1
1
P2 
R
T2   
ln 
T1 vap H P1 
P2  P1 exp  
 1
8.314 JK mol1
446torr 
ln



1
3
760 torr 
 373K 40.67  10 J mol
 358.4K
1
5. What is the maximum number of phases that can coexist in a binary system?
**Solution**
The Gibbs phase rule is f  c  p  2 . For a binary system, c = 2, so the phase rule becomes f  4  p .
Since the minimum number of degrees of freedom is 0, the maximum number of phases is 4.
6. Water and phenol are partially miscible at 50°C. When these two liquids are mixed at 50°C and 1 atm, at
equilibrium one phase is 89% water by weight and the other is 37.5% water by weight. If 6.00 g of phenol
and 4.00 g of water are mixed at 50°C and 1 atm, find the mass of water and the mass of phenol in each
phase at equilibrium.
**Solution**
For this problem, let  be the phenol-rich phase and  be the water-rich phase. The weight fraction of
water in the entire mixture is 4 g/10 g = 0.4.
In this problem, the phase diagram is plotted vs. weight fraction rather than mole fraction as we have
done in class. This means that the lever rule can be written in terms of the masses of each phase rather
than mole fractions:
(continued on next page)
Copyright © 2014 by Eric B. Brauns
3
CHEM 305
Solutions for assignment #8
m L

m L
where
L  0.4  0.375
L  0.89  0.4
 0.025
 0.49
We can find the masses of each phase by writing the mass of one phase in terms of the other:
m  mtot  m
The lever rule becomes:
mtot  m L

m
L
solve for m

L
mtot
1 
m
L

mtot L  L

m
L

m 

mtot L
L  L
10 g   0.025 
0.49  0.025
 0.485 g
m is then:
m  mtot  m
 10 g  0.485 g
 9.515 g
Now we can calculate the masses of water and phenol in each phase. Start with the mass of water in
phase :
 
mHO  0.375  9.515 g
2
 3.57 g
The mass of phenol in this phase is found using mass conservation:
 
mphen
 9.515 g 3.57 g
 5.95 g
Do the same thing for phase :
mH O  0.89  0.485 g
2
 0.432g
 
mphen  0.485 g  0.432g
 0.053g
7. When water is shaken with benzene, one obtains two liquid phases in equilibrium: a saturated solution of a
small amount of benzene in water and a saturated solution of a small amount of water in benzene. Show
that the partial vapor pressure of benzene in equilibrium with a saturated solution of benzene in water is
equal to the partial pressure of benzene in equilibrium with a saturated solution of water in benzene at the
same temperature, provided that the vapors are assumed to be ideal (Hint: Think in terms of chemical
potentials.)
**Solution**
For this problem, let  and  be the two liquid phases. We start by writing the expression for the
chemical potential of benzene in each of the liquid phases:
(continued on next page)
Copyright © 2014 by Eric B. Brauns
4
CHEM 305
Solutions for assignment #8
 
 



ben
 ben
 RT ln xben
 
 

ben
 ben
 RT ln xben
At equilibrium, the chemical potentials are equal:
 
 
ben
 ben

 



 RT ln xben
 ben
 RT ln xben
ben
 

 

xben
 xben
 
Now we use Raoult’s law to write the mole fractions in terms of vapor pressures:
 
  
 
  
Pben  xben
Pben
Pben
 xben
Pben


 

xben

 
Pben

Pben
 
xben

These are equal:
 
 
xben
 xben
 
Pben

Pben

 
Pben

Pben

 
Pben

Pben

 
Pben  Pben
 
The partial vapor pressures are indeed equal.
8. Consider the following 2 component phase diagram. (a) By drawing lines on the figure, determine the
compositions of the two phases at T = T1 when a solution of A and B is 40% B. (b) If there are 10 moles of
solution, estimate the amounts of A-rich and B-rich components.
**Solution**
Let  be the A-rich phase and  be the B-rich phase.
a. The lines are drawn on the above graph. The mole fraction of B in phase  is ~0.2 and the mole
fraction of B in phase  is ~0.8.
b. For this problem, we use the lever rule:
N L

N L
where
(continued on next page)
Copyright © 2014 by Eric B. Brauns
5
CHEM 305
L  0.4  0.2
Solutions for assignment #8
L  0.8  0.4
 0.2
 0.4
The relative amounts of each phase is then
N 0.4

N 0.2
2
If there are 10 mol of solution, then according to mass conservation:
N  N  10 mol

solve for N
N  10 mol N
Substitute into the lever rule:
10 mol N 
2
N
Solve for N:
10 mol N  2N

10 mol  3N

10 mol
3
 3.33mol
N is then
N  10 mol 3.33mol
N 
 6.67 mol
Copyright © 2014 by Eric B. Brauns
6