Differential Distillation - Benzene Toluene Separation

100 moles of Benzene (A) and Toluene mixture containing 50% (mole) of Benzene is subjected to a differential distillation at atmospheric pressure till the composition of benzene in the residue is 33%. Calculate the total moles of the mixture distilled. Average relative volatility may be assumed as 2.16

Calculations:

For system of constant relative volatility, the following equation is obtained after the simplification of Rayleigh's equation for differential distillation:

In the above equation, F is feed; W is residue; x_F is composition of feed; x_W is composition of residue; and a is relative volatility.

Given: x_F = 0.5; x_W = 0.33; F = 100; a = 2.16

Substituting these in the above equation,

ln (100/W) = (1/1.16) x ln {[(0.5(1 - 0.33)]/[(0.33(1 - 0.5)]} + ln [(1 - 0.33)/(1 - 0.5)]

ln (100/W) = 0.862 x ln (2.03) + ln (1.34)

ln (100/W) = 0.862 x 0.708 + 0.2927

ln (100/W) = 0.903

100/W = e^0.903

100/W = 2.467

W = 100/2.467 = 40.54 mole

Moles of mixture distilled = F - W = 100 - 40.54 = 59.46

The above can also be obtained by using the general form of Rayleigh's equation and by graphical method. This is given below.

The equilibrium curve relationship is given by

y* = ax/(1 + x(a - 1)) = 2.16x/(1 + 1.16x) à 1

For differential distillation, the following Rayleigh's equation is applicable:

 à 2

By using Equn.1 the following equilibrium data is obtained: (for the region of x = 0.33 to 0.5)

For the above data the following table of x Vs.1/(y* - x) is obtained:

Using the above data a graph of x Vs.1/(y* - x) is drawn between the limits of x = 0.33 to 0.5.

Area under the curve between the limits of x = 0.33 to 0.5, is = 0.904

i.e.,

ln (F/W) = 0.904

e^0.904 = F/W

2.4695 = F/W

Given F = 100 moles. Therefore,

W = 100 / 2.4695 = 40.5 mole.

Therefore, moles of mixture distilled = F - W = 100 - 40.5 = 59.5