Benzene
MW = 78.11 g/mole
Boiling Point = 80.0° C
ΔHvap = 32.0 kJ/mole
Toluene
MW = 92.14 g/mole
Boiling Point = 110.0° C
ΔHvap = 38.0 kJ/mole
At 300.0 K a solution is prepared with 10.0 g of benzene and 10.0 g of toluene. Determine the
number of grams of each chemical in the liquid and gas phases at a total pressure of 60.80 torr.
First calculate the vapor pressures of the pure materials at 300.0 K
using the Clausius Clapeyron Equation
ln(
ΔHvap 1 1
P2
)= −
( − )
P1
R T2 T1
P2 = ?
P1 = 760 torr
T2 = 300.00
T1 = 80.0°C = 353.15 K
P2 = 110.2 Torr for Benzene (PB°)
ΔH = 3.20 X 104 kJ/mole
T2 = 300.00
ΔH = 3.80 X 104 kJ/mole
T1 = 110.0°C = 383.15 K
P2 = 27.86 Torr for Toluene (PT°)
P2 = ?
P1 = 760 torr
A
B
C
D
E
F
G
1
x
P
x
P
TOTAL
y
y
2
Benzene
Benzene
Toluene
Toluene
PRESSURE
Benzene
Toluene
3
0.00
0.00
1.00
27.86
27.86
0.000
1.000
4
0.10
11.02
0.90
25.07
36.09
0.305
0.695
5
0.20
22.04
0.80
22.29
44.33
0.497
0.503
6
0.30
33.06
0.70
19.50
52.56
0.629
0.371
7
0.40
44.08
0.60
16.72
60.80
0.725
0.275
8
0.50
55.10
0.50
13.93
69.03
0.798
0.202
9
0.60
66.12
0.40
11.14
77.26
0.856
0.144
10
0.70
77.14
0.30
8.36
85.50
0.902
0.098
11
0.80
88.16
0.20
5.57
93.73
0.941
0.059
12
0.90
99.18
0.10
2.79
101.97
0.973
0.027
13
1.00
110.20
0.00
0.00
110.20
1.000
0.000
14
Column A is an arbitrary set of mole fractions for on of the components
Columns B & D are calculated using Raoult’s Law, Column E is simply the total of PT & PB, Columns F
& G are calculated from the definition of y: ( yT = PT/Ptotal). A plot of the x’s and the y’s versus
Ptotal yields the graph needed.
1
110
100
90
80
Y
70
60
50
40
30
20
.000
.100
.200
.300
.400
.500
.600
.700
.800
.900 1.000
X(Benzene)
This particular system is at a total mole fraction of benzene of 0.5412:
10.0 g /78.11 = 0.1280 moles B
10.0/92.14 = .1085 mole T
Total moles = 0.1280 + 0.1085 = 0.2366 & 0.1280/.2366 = 0.5412 = XB
Total pressure was given as 60.80 torr. Thus, the relevant portion of the graph is
2
liquid line
x's
0.5412
0.4000
y's
0.7250
gas line
Length of Gas Line = 0.5412 - 0.4000 = 0.1412
Length of Liquid Line = 0.7250 - 0.5412 = 0.1838
Total Line Length = 0.7250 - 0.4000 = 0.3250
To calculate moles of material in the liquid phase
Liquid Line moles of Liquid
=
Total Line
Total moles
nl
0.1838
=
0.3250 0.2366
nl = (0.1838/0.3250)*(0.2366) = 0.1338 moles
This can be used to calculate the number of moles of benzene in the liquid phase (since the mole fraction
of benzene in the liquid phase is 0.4000).
moles of B in liquid
= x B = 0.4000
total moles in liquid
nB (liquid )
= 0.4000
0.1338
nB (liquid ) = 0.400 * 0.1338 = 0.05352
&
the mass of benzene in the liquid phase is 4.18 grams
Doing a bunch of subtractions
Moles of toluene in the liquid phase = 0.1338 - 0.05352 = 0.0803 moles
Moles of toluene in the gas phase = 0.1085 - 0.0803 = 0.0282 moles
Moles of benzene in the gas phase = 0.1280 - 0.05352 = 0.0745 moles
Mass of toluene in the liquid phase = 0.0803 moles * 92.14 g/mole = 7.40 g
Mass of toluene in the gas phase = 0.0282 moles * 92.14 g/mole = 2.60 g
Mass of benzene in the gas phase = 0.0745 moles * 78.11 g/mole= 5.82 g
Check:
Mass of B (gas) + Mass of B (liq) = 5.82 + 4.18 = 10.0 g
Mass of T (gas) + Mass of T (liq) = 2.60 + 7.40 = 10.0 g
3
Additional Problem
The same solution is heated to 367.00 K at a total pressure of 1 atm. Determine the number of
grams of each chemical in the liquid and gas phases under these conditions
To construct a Temperature versus mole fraction diagram, the vapor pressures of the pure materials at the
various temperatures between the boiling points must be calculated from the Clausius Clapeyron Equation
(these are listed in columns B and C). Since the total pressure is 1 atm, the mole fractions of Benzene in
the vapor phase can be calculated (column D).
PT = yB PBo + yT PTo
PT = yB PBo + (1 − yB )PTo
1 = yB (PBo − PTo ) + PTo
1+ PTo
yB = o
PB − PTo
The mole fractions of Benzene in the liquid phase can be calculated from Raoult’s Law: PB = x B PBo
These values are in column E
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
A
Temperature
353.15
355.65
358.15
360.65
363.15
365.65
367.00
368.15
370.65
373.15
375.65
378.15
380.65
383.15
B
P(atm)
Benzene
1.0000
1.0796
1.1643
1.2544
1.3500
1.4514
1.5087
1.5590
1.6729
1.7934
1.9208
2.0554
2.1975
2.3474
C
P(atm)
Toluene
0.3630
0.3976
0.4349
0.4751
0.5184
0.5650
0.59161
0.6151
0.6688
0.7264
0.7881
0.8541
0.9246
1.0000
D
E
y (benzene) x (benzene)
1.0000
0.8833
0.7747
0.6736
0.5791
0.4907
0.4453
0.4078
0.3299
0.2564
0.1871
0.1215
0.0592
0.0000
1.0000
0.9536
0.9020
0.8449
0.7818
0.7122
0.6718
0.6357
0.5518
0.4599
0.3594
0.2496
0.1301
0.0000
4
380.0
Temperature (K)
375.0
370.0
365.0
360.0
355.0
.000
.100
.200
.300
.400
.500
.600
Mole Fraction of Benzene
.700
.800
.900 1.000
liquid line
0.5412
0.6718
0.4453
gas line
5
RELAVANT EQUATIONS ASSOCIATED WITH BINARY MIXTURES
Raoult’s Law
PA = xA P°A
PB = xB P°B
PTOTAL = P°B + (P°A - P°B) xA
yA = PA / PTOTAL
yB = PB / PTOTAL
yA = ( xA P°A ) / ( P°B + (P°A - P°B) xA )
A secondary but still useful relationship
PTOTAL = ( P°A P°B ) / ( P°A + (P°B - P°A) yA )
6