Chemistry 360 Dr. Jean M. Standard
Problem Set 11 Solutions 1.
The vapor pressure of pure pure toluene is 400 torr and and that of pure 1,2-dimethylbenzene 1,2-dimethylbenzene is 150 torr at 90˚C. 90˚C. Determine the composition of the liquid and vapor phases if toluene and 1,2-dimethylbenzene are mixed at 90˚C and a total pressure of 0.5 atm. atm. Assume ideal solution behavior. behavior. Assuming that the solution is ideal, it can be described by Raoult's Law, Pi
where
Pi is
* x i Pi ,
=
x i is
the partial pressure of component i in the mixture,
component i, and
*
Pi
the liquid phase mole fraction of
is the vapor pressure of pure component i. The total pressure pressure of the mixture mixture is given by
the sum of the partial pressures (which are determined from Raoult's Law), P
Using the relation
x
2
=
1"
x
1
=
P1
=
x1P1
P2
+
*
+
*
x 2 P2 .
, the equation for total pressure (the bubble point line) becomes
or
*
P
=
x 1P1
P
=
(
*
+
(1 *
P1
*
)
" x1 P2
)
" P2 x 1
+
*
P2 .
Solving this equation for the liquid phase mole fraction, we have *
P " P2
x1
=
*
P1
*
.
" P2
Let us assign component 1 as toluene toluene and component 2 as 1,2-dimethylbenzene. 1,2-dimethylbenzene. Then, the equation above becomes an equation for the liquid phase mole fraction of toluene, *
x toluene
=
P " Pdimeth *
*
.
Ptoluene " Pdimeth
Substituting and using the total pressure of 0.5 atm (= 380 torr), the liquid phase mole fraction of toluene is 380 380 torr torr " 150 torr torr
x
toluene
x
toluene
=
400 400 torr torr " 150 torr torr =
0.92.
Then, the liquid phase mole fraction of 1,2-dimethylbenzene 1,2-dimethylbenzene is x
dimeth
x
dimeth
=
1 "
=
1 " 0.92
=
x
0.08.
toluene
2
1.) Continued To determine the vapor phase mole fractions, the equation is y i
=
Pi
. P
Knowing the liquid phase mole fractions, the partial pressures can be determined from Raoult's Law, Ptoluene
*
=
x toluene Ptoluene
(
)(
)
=
0.92 400torr
Ptoluene
=
368 torr
Pdimeth
=
and
Pdimeth
*
x dimethPdimeth
(
)(
=
0.08 150torr
=
12torr .
)
Then, the vapor phase mole fractions are y toluene
Ptoluene =
P 368torr =
380torr
y toluene
=
y dimeth
=
0.968
and Pdimeth P 12 torr =
380torr
y dimeth
=
0.032.
Thus, we see that since toluene has the larger pure vapor pressure, the vapor phase is enriched in toluene.
3
2.
A solution of methanol and ethanol has a total vapor pressure of 350 torr at 50˚C. The vapor pressures of the pure components, methanol and ethanol, are 413.5 and 221.6 torr, respectively. Determine the composition of the liquid phase, assuming ideal solution behavior. Assuming that the solution is ideal, it can be described by Raoult's Law, Pi
where
Pi is
* x i Pi ,
=
x i is
the partial pressure of component i in the mixture,
component i, and
*
Pi
the liquid phase mole fraction of
is the vapor pressure of pure component i. The total pressure of the mixture is given by
the sum of the partial pressures (which are determined from Raoult's Law), P
Using the relation
x
2
=
1"
x
1
=
P1
=
x1P1
+
*
P2 +
*
x 2 P2 .
, the equation for total pressure (the bubble point line) becomes
or
*
P
=
x 1P1
P
=
(
*
P1
+
(1 *
)
*
" x1 P2
)
" P2 x 1
+
*
P2 .
Solving this equation for the liquid phase mole fraction, we have *
P " P2
x1
=
*
P1
*
.
" P2
Let us assign component 1 as methanol and component 2 as ethanol. Then, the equation above becomes an equation for the liquid phase mole fraction of methanol, *
x MeOH
P " PEtOH
=
*
*
.
PMeOH " PEtOH
Substituting and using the total pressure of 350 torr, the liquid phase mole fraction of methanol is 350 torr " 221.6 torr
x
=
MeOH
413.5torr " 221.6 torr
x
=
MeOH
0.67.
Then, the liquid phase mole fraction of ethanol is x
EtOH
x
EtOH
=
1 "
=
1 " 0.67
=
x
0.33.
MeOH
4
3.
The normal boiling points of propane and n-butane are –42.1˚C and –0.5˚C, respectively. The following vapor pressure data have been measured. T (˚C)
–31.2
–16.3
P*, propane (kPa)
160.0
298.6
P*, n-butane (kPa)
26.7
53.3
Assume that the substances form ideal solutions. (a) Calculate the liquid phase mole fraction of propane in the solutions in which liquid and vapor are in equilibrium at 1.0 atm pressure at –31.2˚C and –16.3˚C. For each of these temperatures, we can use an approach similar to that taken in Problems 1 and 2 to get the mole fraction of propane in the liquid phase. From Problem 1 (and the equation of the bubble point line), the liquid phase mole fraction is *
x1
P " P2 =
*
P1
.
*
" P2
Assigning propane as component 1 and n-butane as component 2, at –31.2˚C we have *
x propane
P " Pnbutane
=
*
*
Ppropane " Pnbutane
1.0atm =
x propane
=
"
1.579 atm
0.264 atm
"
0.264 atm
0.560.
Note that the pressures in the table above are given in kPa, so the conversion factor 1 atm = 101.32 kPa was used in the calculation of the mole fraction. At –16.3˚C, the liquid phase mole fraction of propane is *
x propane
P " Pnbutane
=
*
*
Ppropane " Pnbutane
1.0atm
"
0.526 atm
=
2.947atm x propane
=
"
0.526 atm
0.196.
(b) Calculate the vapor phase mole fraction of propane in each of the solutions. To obtain the vapor phase mole fraction of propane at –31.2˚C, the partial pressure is determined first from Raoult's Law, Ppropane
=
=
Ppropane
=
*
x propane Ppropane
(0.560)(1.579atm) 0.884 atm.
5
3
b.) Continued Then, the mole fraction of propane in the vapor phase is determined from the definition, y propane
=
Ppropane P 0.884 atm
=
1.0atm y propane
=
0.884 .
At –16.3˚C, the partial pressure of propane is Ppropane
=
Ppropane
*
x propane Ppropane
=
=
(0.196)( 2.947atm) 0.578 atm.
The vapor phase mole fraction of propane is determined from the equation, y propane
=
P 0.578 atm =
y propane
Ppropane
=
1.0 atm 0.578.
6
4.
A mixture of liquids A and B exhibits ideal behavior. At 84˚C, the total vapor pressure of a liquid solution containing 1.2 mol A and 2.3 mol B is 331 torr. Upon the addition of 1 more mole of B to the solution, the vapor pressure is 347 torr. Calculate the vapor pressures of pure A and pure B at 84˚C. The ideal solution can be described by Raoult's Law, *
Pi
x i Pi .
=
The total pressure of the mixture is given by the sum of partial pressures, P
=
P A
or P
=
x A P A
P B
+
*
+
*
x B P B .
We are given the total pressure for two different liquid phase compositions, and are asked to determine the pure vapor pressures. In the first case, the mole fractions are x A x A
1.2mol =
1.2mol
+
2.3mol
0.343
=
and x B
x B
=
1 " x A
=
1 " 0.343
=
0.657.
In the second case, the mole fractions are x A x A
1.2mol =
1.2mol
+
3.3mol
0.267
=
and x B
x B
The total pressure, given by
P
=
*
x A P A
+
1 " x A
=
1 " 0.267
=
*
x B P B ,
331torr
=
0.733.
in the first case is *
=
0.343 P A
=
0.267 P A
*
+
0.657 P B .
+
0.733 P B .
The total pressure in the second case is 347 torr
*
*
7
4.) Continued *
There are two equations and two unknowns. One way to solve this is to solve the first equation for P A and *
substitute into the second equation. So, solving the first equation for P A yields *
0.343 P A
*
=
331 " 0.657 P B
=
965 " 1.915 P B .
*
or
P A
*
Substituting this result into the second equation leads to the vapor pressure of pure B, *
*
347
=
0.267 P A
347
=
0.267
347
=
257.7 " 0.511 P B
89.3
=
0.222 P B
=
402 torr.
*
P B
+
0.733 P B
) ( 965 " 1.915 P B*)
(
*
+
+
*
0.733 P B *
0.733 P B
*
*
965 From the first equation, we had P A determine the vapor pressure of pure A, =
*
" 1.915 P B .
*
P A
*
=
965 " 1.915 P B
=
965 " 1.915 402 torr
*
P A
*
P A
Substituting the vapor pressure of pure B allows us to
=
(
195torr .
)
8
5.
Determine the mole fraction of each component in the vapor phase in equilibrium with a liquid phase in which there is a 1:1 molar ratio of n-hexane and cyclohexane. The vapor pressures of puren-hexane and cyclohexane are 151.4 torr and 97.6 torr, respectively. The partial pressures can be determined from Raoult's Law, *
Pi
x i Pi .
=
Since the molar ratio is 1:1 in the liquid phase, the liquid phase mole fractions are both 0.5. Forn-hexane, the partial pressure is Pnhexane
*
x nhexanePnhexane
=
( 0.5)(151.4 torr )
=
Pnhexane
75.7 torr .
=
For cyclohexane, the partial pressure is Pcyclohex
*
x cyclohex Pcyclohex
=
( 0.5)( 97.6 torr)
=
Pcyclohex
=
48.8 torr .
The total pressure is P
P
=
Pnhexane
=
75.7torr
=
124.5 torr .
Pcyclohex
+ +
48.8 torr
Then, the mole fraction of n-hexane in the vapor phase is y nhexane
Pnhexane =
P 75.7 torr =
124.5torr
y nhexane
=
0.608.
The vapor phase mole fraction of cyclohexane is y cyclohex
=
Pcyclohex P 48.8torr
=
124.5torr y cyclohex
=
0.392.
9
6.
The Henry's Law constant for carbon dioxide in water is 1.25 106 torr at 25˚C. Calculate the solubility of carbon dioxide in water at 25˚C when its partial pressure in air is (a) 4.0 kPa and (b) 100 kPa. Henry's Law is pi
=
K H x i . Solving for the mole fraction yields x i
pi
=
. K H
(a) For a partial pressure of 4.0 kPa or 30.0 torr (1 torr = 133 Pa), the mole fraction of carbon dioxide dissolved in water is 30.0 torr
x i
=
x i
=
1.25 " 10
6
#5
2.4 " 10
torr .
(b) For a partial pressure of 100 kPa or 750 torr, the mole fraction of carbon dioxide dissolved in water is
7.
750torr
x i
=
x i
=
1.25 " 10 6.0 " 10
6
#4
torr .
At 25˚C, the mole fraction of air dissolved in water is 1.388 10 –5. (a) Determine the molarity of the solution. The molarity is defined as moles of air divided by liters of solution,
[air ]
n air
.
=
Lsolution
Since the mole fraction is so small, we can assume that the liters of solution are equal to liters of water,
[air ]
n air =
.
L H O 2
In order to calculate the molarity, we need to determine the moles of air. The definition of mole fraction is x air
=
n air n air
+
n H O 2
.
10
7
a.) Continued The moles of water in 1 L can be calculated using the density and molecular weight,
n H O 2
D H O 2
=
MW H O 2
# 1000 mL & "% ( 18.015g/mol $ 1 L ' 1 g/mL
=
n H O 2
=
55.5mol/L.
From the definition of mole fraction, we can solve for the moles of air, x air
or
n air
=
=
n air n air
+
n H O 2
x air n H O 2
.
1 " x air
Substituting, we can determine the moles of air,
n air
x air n H O 2 =
1 " x air
(
1.388 # 10
"5
)(
)
55.5mol
=
"5
1 " 1.388 # 10 n air
=
"4
7.70 # 10
mol.
Finally the molarity of air can be calculated,
[ air ]
=
n air Lsolution
"
n air L H O 2 $4
7.70 # 10
mol
=
1L
[ air ]
=
$4
7.70 # 10
mol/L.
11
7.) Continued (b) Calculate the Henry's Law constant for air (in water). Compare your results with the literature values of 6.80 107 and 3.27 107 torr, respectively, for the Henry's Law constants of N2 and O2 dissolved in water at 25ºC. Henry's Law is pi
=
K H x i . Solving for the Henry's Law constant, K H
=
pi
. x i
The partial pressure of air is 1 atm. Substituting, K H
=
pi x i 1atm
=
1.388 " 10
K H
=
#5
72050 atm.
The literature values of the Henry's Law constants of N2 and O2 in water are 6.80!107 and 3.27!107 torr, respectively. Using 1 atm = 760 torr, the Henry's Law constants of N2 and O2 are 89500 and 43000 atm, respectively. The Henry's Law constant calculated for air lies in between these values as expected, since air is a mixture primarily of N2 and O2 (and it lies much closer to the value of N2 since air is about 78% N2).
(c) Would you expect the solubility of air to increase or decrease with an increase in temperature? The solubility of a gas in a liquid decreases as the temperature increases. This may be observed qualitatively for carbonated beverages. As the beverage warms up, you can see the bubbles of CO2 coming out of solution. For example, shown below are measured Henry's Law constants of CO2 in H 2O at various temperatures [from J. J. Carroll, J. D. Slupsky, and A. E. Mather, J. Phys. Chem. Ref. Data 1991, 20, 12011209]. Since the Henry's Law constant is inversely proportional to solubility, as the temperature goes up, the solubility of CO2 in H2O goes down.