HW, Chapt 3 answers
Chapter 3, Background Thermodynamics.
1.
An organic liquid is equilibrated with pure water in a stoppered flask:
a) Assuming that the organic
liquid consists of pure toluene
(molecular weight 92.13, density
0.8669, aqueous solubility 515
aqueous
phase
mg/L at 25C), what is:
(i) the concentration of toluene in
the aqueous phase in mol/L and in mole fraction units;
(ii)
the aqueous activity coefficient and the activity of toluene in
the aqueous phase (neglect any dissolution of water in the organic
phase).
(iii)
the mole fraction based equilibrium constant K'12 for the
dissolution process and the concentration based equilibrium
constant K12;
(iv)
the partial molar excess free energy of solution of toluene in
the aqueous phase as well as in the organic phase at 25C;
(
organic
phase
(b)
Now assume that instead of consisting of pure toluene, the organic
phase consists of a 50:50 (volume: volume) mixture of benzene
(molecular weight 78.11, density 0.88765) and toluene. What will
be the concentration of toluene in the aqueous phase (in mol/L)
assuming that nonideal interactions are (to a first approximation)
independent of concentration?
2. Derive a relationship for a generic equilibrium const. ,Keq as a
function of free energy G
3. Derive an expression for Keq as a function of temperature
(a) Dissolution of pure toluene in water: let phase 1 represent
aqueous phase, phase 2 organic phase.
Answers:
a(i) Concentration of toluene in aqueous phase in mol/L and in
mole fraction units: Convert from mg/L to mol/L by converting
mg to g and dividing by molecular weight:
mg
1 mol
-3 g
515
 10
L
mg

92.13 g
 solubility of pure toluene = 5.59 x 10-3 mol/L
To convert from mol/L units to mol fraction units, multiply by
molar volume of mixture: Thus, x1 = (5.59 x 10-3 mol/L)(0.018
L/mol)
 x1 = 1.01 x 10-4 mole fraction (ii) Aqueous
activity coefficient and activity of toluene in aqueous phase. At
equilibrium, the chemical potential will be the same in each
phase:
1
1 =
x1
;
1 =
1
3
=
9.94
x
10
1.01 x 10 - 4
The activity of toluene in the aqueous phase is defined as:
activity = 1x1 from which it follows that
activity of toluene in aqueous phase = 1
(iii) Equilibrium constants for the dissolution process:
For the dissolution reaction:
toluenepure
tolueneaq
x1
K' 12 =
x2
Since in the pure toluene phase, x2 = 1, K'12 = x1 or
K'12 = 1.01 x 10-4
Alternatively, we can express a concentration based equilibrium
constant K12 as:
C1
K12 =
C2
To use this, we need to calculate the concentration of toluene in
the organic phase in mol/L units:
C2 = 0.8669
g
mL
1 mol
 103

= 9.41 mol / L
mL
L
92.13 g
5.59  10 -3 mol / L
K12 =
= 5.94  10 -4
9.41 mol / L
Note that K'12  K12.
(iv) Partial molar excess free energy of solution of toluene:
The partial molar excess free energy of solution is defined as
(SGI, p. 51)
partial molar excess free energy of solution = RT ln i
In the aqueous phase, i = 9.94 x 103, thus:
partial molar excess free energy of solution in aqueous phase =
(8.314)(298)[ln(9.94 x 103)] = 22.8 kJ/mol
In the organic phase, i = 1 (ln i = 0), and thus
partial molar excess free energy of solution in organic phase
=0
(b) Aqueous solubility of toluene if organic phase is 50:50
(vol:vol) toluene:benzene
Using the same reasoning as before, by equating the chemical
potential of the toluene in the aqueous phase with that in the
organic phase, we can show
1x1 = 2x2 (1 = aqueous phase)
We now need to determine x2, the mol fraction of the toluene in
the organic phase.
If we had 1 mL of toluene and 1 mL of benzene present, then the
total number of moles of organic liquid would be:
(
# moles = (1 mL )
0.8669 g
mL
)(
1 mol
92.13 g
)
(
+ (1 mL )
0.8876 g
mL
)(
= 2.077 x 10-2 moles in 2 mL, of which 9.410 x 10-3
moles consist of toluene and 1.136 x 10-2 moles consist of
benzene.
Thus, the mole fraction of toluene in the mixture is:
9.410 x 10 - 3
x2 =
= 0.453
2
2.077 x 10
x1 =
x2   2
(0.453)(1)
=
= 4.56 x 10 - 5 mole fraction
3
1
(9.94 x 10 )
To convert to mol/L units, divide by the molar volume of water:
4.56 x 10 - 5 moles / mol T OT
conc. in aq. phase =
= 2.53 x 10 - 3 mol / L
0.018 L / mol
1 mo
78.11