Example 4-8: Prediction of a heteroazeotrope with total liquid immiscibility

Describe the conditions at which the benzene(1) + water(2) binary mixture form three distinct phases at 1 atm (0.10135 MPa) pressure. Give the phase compositions.

Analysis:

When considering only the moderate temperature region (i.e. well below the water critical point), a number of simple approximations can be made:

• The hydrocarbons and water are entirely insoluble in each other in the liquid phase.
• The vapour phase is an ideal gas.
• The vapour pressures may be obtained using the Antoine equation:

where P^σ is expressed in torr and T in °C

The Gibbs phase rules states:

There are two components and three phases in the heteroazeotropic point, therefore only one degree of freedom. If pressure is fixed, the solution is defined.

Solution:

The diagram is constructed starting from the pure components. The boiling temperature of benzene is 353.15 K, that of water 373.15 K. At low temperature, two liquid phases are present, whose compositions are known:

for the hydrocarbon phase: x₁^HC=1, x₂^HC=0 and

for the aqueous phase: x₁^aq=0, x₂^aq=1.

When the vapour is in equilibrium with the aqueous phase, we have:

This is the water dew curve (y₁ vs T). Note that .

When the vapour is in equilibrium with the hydrocarbon phase, we have:

This is the organic dew curve (y₂ vs T). Note that .

When the vapour is in equilibrium with both aqueous and organic liquids, both equations (2) and (3) are true simultaneously. They can be summed to yield:

The total three phase pressure is equal to the sum of the two vapour pressures. Only one temperature satisfies this condition for P = 1 atm, that is T = 69.12 °C (342.27 K). At this temperature, the vapour phase composition can be calculated: y₂=0.3

Two dew curves are thus constructed. They meet at the three-phase point. The diagram is shown in figure 1.