Question

Calculate the vapor pressure at 80 C of a mixture of 35 mol % hexane and 65 mol % heptane. Calculate the mol fraction of each compound in the vapor in equilibrium with the mixture in (1). If a simple distillation were to be performed on this mixture( from question 1 ), what would be the composition of the initial distillate? Suppose the mixture was to be distilled using a very efficient fractional distillation column. what would the composition of the initial distillate be?

Answer 1

Given t 80 ⁰C 35 % hexane and 65 % heptane

Mole fraction of hexane = 35/(35 + 65)

                                              = 0.35 (X_hex)

Mole fraction of heptanes = 65/(35+65)

                                                 = 0.65 (X_hept)

P _total = p_hex * X_hex + p_hept * X_hept

           =(1062 * 0.35) + (462 * 0.65) (vapor pressure for hexane at 80C is 1062 and for heptanes is 462)

      = 371.7 + 300.3

     = 672

Hence total vapour pressure = 672

Partial pressure of gas is proportional to no :of moles of that gas

Hence mole fraction of each component is equal to ratio of its partial pressure to total pressure

X^’_hex = 371.7/672

          = 0.55

X^’_hept = 300.3/672

            = 0.446

So the mixture contains 55 % hexane and 45 % heptanes

I you perform a simple distillation the initial boiling temperature will be 80 ⁰C , hence the initial distillate consists of 55 % hexanes and 45 % heptanes.

In the case of fractional distillation we could get the pure hexane because even though the boiling temperature is 80 , the temperature at the top of the column will be 7o so only hexane distills out.