Chemistry 360
Dr. Jean M. Standard
Problem Set 11 Solutions
1.
The vapor pressure of pure toluene is 400 torr and that of pure 1,2-dimethylbenzene is 150 torr at 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. Assume ideal solution behavior.
Assuming that the solution is ideal, it can be described by Raoult's Law,
Pi = x i Pi* ,
where Pi is the partial pressure of component i in the mixture, x i is the liquid phase mole fraction of
component i, and Pi* 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 = P1 + P2
€
= x1P1* + x 2 P2* .
Using the relation x 2 = 1 − x1 , the equation for total pressure (the bubble point line) becomes
€
€
P = x1P1* + (1 − x1) P2*
or
(
)
P = P1* − P2* x1 + P2* .
Solving this equation for the liquid phase mole fraction, we have
€
x1 =
P − P2*
.
P1* − P2*
Let us assign component 1 as toluene and component 2 as 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 torr − 150 torr
400 torr − 150 torr
= 0.92.
x toluene =
x toluene
Then, the liquid phase mole fraction of 1,2-dimethylbenzene is
€
x dimeth = 1 − x toluene
= 1 − 0.92
x dimeth = 0.08 .
€
2
1.) Continued
To determine the vapor phase mole fractions, the equation is
yi =
Pi
.
P
Knowing the liquid phase mole fractions, the partial pressures can be determined from Raoult's Law,
€
*
Ptoluene = x toluene Ptoluene
= ( 0.92)( 400 torr )
Ptoluene = 368 torr
and
€
*
Pdimeth = x dimeth Pdimeth
= ( 0.08)(150 torr )
Pdimeth = 12 torr .
Then, the vapor phase mole fractions are
€
Ptoluene
P
368 torr
=
380 torr
= 0.968
y toluene =
y toluene
and
€
Pdimeth
P
12 torr
=
380 torr
y dimeth =
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 = x i Pi* ,
where Pi is the partial pressure of component i in the mixture, x i is the liquid phase mole fraction of
component i, and Pi* 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 = P1 + P2
€
= x1P1* + x 2 P2* .
Using the relation x 2 = 1 − x1 , the equation for total pressure (the bubble point line) becomes
€
€
P = x1P1* + (1 − x1) P2*
or
(
)
P = P1* − P2* x1 + P2* .
Solving this equation for the liquid phase mole fraction, we have
€
x1 =
P − P2*
.
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
413.5torr − 221.6 torr
= 0.67.
xMeOH =
xMeOH
Then, the liquid phase mole fraction of ethanol is
€
xEtOH = 1 − xMeOH
= 1 − 0.67
xEtOH = 0.33.
€
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
1.0 atm − 0.264 atm
1.579 atm − 0.264 atm
= 0.560.
=
x propane
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
1.0 atm − 0.526 atm
2.947 atm − 0.526 atm
= 0.196.
=
x propane
(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 = x propane Ppropane
= ( 0.560)(1.579 atm)
Ppropane = 0.884 atm.
€
5
3
b.) Continued
Then, the mole fraction of propane in the vapor phase is determined from the definition,
y propane =
y propane
Ppropane
P
0.884 atm
=
1.0 atm
= 0.884 .
At –16.3˚C, the partial pressure of propane is
€
*
Ppropane = x propane Ppropane
= ( 0.196)( 2.947 atm)
Ppropane = 0.578 atm.
The vapor phase mole fraction of propane is determined from the equation,
€
y propane =
y propane
€
Ppropane
P
0.578 atm
=
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 = PA + PB
or P = x A PA* + x B PB* .
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
1.2 mol
1.2 mol + 2.3mol
= 0.343
xA =
xA
and
€
xB = 1 − xA
= 1 − 0.343
x B = 0.657 .
In the second case, the mole fractions are
€
1.2 mol
1.2 mol + 3.3mol
= 0.267
xA =
xA
and
€
xB = 1 − xA
= 1 − 0.267
x B = 0.733.
*
*
The total pressure, given by P = x A PA + x B PB , in the first case is
€
331torr = 0.343PA* + 0.657 PB* .
€
The total pressure in the second case is
€
€
347 torr = 0.267 PA* + 0.733PB* .
7
4.) Continued
*
There are two equations and two unknowns. One way to solve this is to solve the first equation for PA and
*
substitute into the second equation. So, solving the first equation for PA yields
0.343PA* = 331 − 0.657 PB*
or
PA*
€
= 965 −
€
1.915PB* .
Substituting this result into the second equation leads to the vapor pressure of pure B,
€
347 = 0.267 PA* + 0.733PB*
(
)
347 = ( 0.267) 965 − 1.915PB* + 0.733PB*
347 = 257.7 − 0.511PB* + 0.733PB*
89.3 = 0.222 PB*
PB* = 402 torr .
*
*
From the first equation, we had PA = 965 − 1.915PB . Substituting the vapor pressure of pure B allows us to
determine the vapor pressure
of pure A,
€
PA* = 965 − 1.915PB*
€
PA* = 965 − 1.915 ( 402 torr )
PA* = 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 pure n-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. For n-hexane, the
partial pressure is
€
*
Pnhexane = x nhexane Pnhexane
= ( 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 = Pnhexane + Pcyclohex
= 75.7 torr + 48.8 torr
P = 124.5 torr .
Then, the mole fraction of n-hexane in the vapor phase is
€
Pnhexane
P
75.7 torr
=
124.5torr
y nhexane =
y nhexane = 0.608.
The vapor phase mole fraction of cyclohexane is
€
y cyclohex =
y cyclohex
€
Pcyclohex
P
48.8 torr
=
124.5 torr
= 0.392 .
9
6.
6
The Henry's Law constant for carbon dioxide in water is 1.25×10 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
pi
.
KH
xi =
€
(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
€
xi =
30.0 torr
1.25× 10 6 torr
x i = 2.4 × 10−5 .
(b) For a partial pressure of 100 kPa
€ or 750 torr, the mole fraction of carbon dioxide dissolved in water is
xi =
750 torr
1.25× 10 6 torr
x i = 6.0 × 10−4 .
€
7.
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 2O
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 2O
10
7
a.) Continued
The moles of water in 1 L can be calculated using the density and molecular weight,
n H 2O =
=
DH 2O
MW H 2O
# 1000 mL &
1 g/mL
⋅%
(
18.015g/mol $ 1L '
n H 2O = 55.5mol/L .
From the definition of mole fraction, we can solve for the moles of air,
€
n air
n air + n H 2O
x air n H 2O
or n air =
.
1 − x air
x air =
Substituting, we can determine the moles of air,
€
n air =
x air n H 2O
1 − x air
(1.388 × 10 )(55.5 mol)
−5
=
1 − 1.388 × 10−5
n air = 7.70 × 10−4 mol.
Finally the molarity of air can be calculated,
€
[ air ]
n air
Lsolution
n air
≈
LH 2O
=
=
[ air ]
€
7.70 × 10−4 mol
1L
= 7.70 × 10−4 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,
KH =
€
pi
.
xi
The partial pressure of air is 1 atm. Substituting,
€
KH =
pi
xi
1atm
1.388 × 10 −5
= 72050 atm.
=
KH
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 H2O 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.