Question

A solution of benzene and toluene is at equilibrium with vapour phase at 1 atm. Benzene and toluene form ideal solution. The vapour phase, which is at equilibrium with liquid (solution) is ideal gas. Activity of benzene in solution is 0.3.

a) Determine the temperature of the system at equilibrium.

b) Determine the vapour pressure of benzene at the system’s temperature.

Answer 1

a)

Given:

Benzene and toluene form ideal solution.

The vapor phase behaves ideally.

Therefore; Raoult's Law and Dalton's law is applicable.

Acitivity of benzene = Mole fraction of benzene (ideal solution)

Mole fraction of benzene; x = 0.3

Mole fraction of toluene; 1 - x = 0.7

From Raoult's Law;

Pb = x.Pb*

Pt = (1 - x).Pt*

where;

Pb and Pt are partial pressures of benzene and toluene

Pb* and Pt* are vapor pressures of benzene and toluene

From Dalton's law; the total pressure P

P = Pb + Pt

P = x.Pb* + (1 - x).Pt*

P = 1 atm

x = 0.3

Substituting these in the above equation

1 = 0.3Pb* + 0.7Pt*

Pb* and Pt* can be calculated by Antoine's equation.

Antoine Equation is given by:

log (P*) = A - B / (T + C)

P* = 10 ^ (A - B / (T + C))

Antoine Parameters for Benzene:

A = 4.726

B = 1660.652

C = -1.461

Antoinne Parameters for Toluene:

A = 4.078

B = 1343.943

C = - 55.743

Using these in the above equations:

Pb* = 10 ^ (4.726 - 1660.652 / (T - 1.461))

Pt* = 10 ^ (4.078 - 1343.943 / (T - 55.743))

Substituting these in the governing equation:

1 = 0.3(10 ^ (4.726 - 1660.652 / (T - 1.461)) ) + 0.7(10 ^ (4.078 - 1343.943 / (T - 55.743)))

which on solving gives:

T = 372 K

which is the temperature of the system at equilibrium.

b)

The vapor pressure of benzene at T = 372 K

Pb* = 10 ^ (4.726 - 1660.652 / (372 - 1.461))

Pb* = 1.9 atm

which is the required vapor pressure.