Rayleigh Distillation

A differential distillation starting at 350 K is to be run on an equimolar mixture of n-heptane and n-octane until 50 percent of the original mixture remains as liquid. Assuming that the relative volatility of n-heptane to n-octane is approximately constant at a value of 2.35 over the temperature range of the distillation, what is the composition of the liquid mixture after 50 percent of the mixture has been vaporized?

Question

Rayleigh Distillation

A differential distillation starting at 350 K is to be run on an equimolar mixture of n-heptane and n-octane until 50 percent of the original mixture remains as liquid. Assuming that the relative volatility of n-heptane to n-octane is approximately constant at a value of 2.35 over the temperature range of the distillation, what is the composition of the liquid mixture after 50 percent of the mixture has been vaporized?

Accepted Answer

Basis of the calculation: 1 mole of the initial equimolar mixture.

The equations to be solved are

[latex]N_{ C 7}(t)+N_{ C 8}(t)=0.5[/latex]

and

[latex]\left(\frac{N_{ C 8}(t)}{0.5} ight)^{2.35}=\frac{N_{ C 7}(t)}{0.5}[/latex]

The solution to these equations is [latex]N_{ C 7}=0.178 mol , ext { and } N_{ C 8}=0.322 mol[/latex] Therefore, the liquid mole fractions after half of the initial mixture has been vaporized are [latex]x_{ C 7}=0.356[/latex] and [latex]x_{ C 8}=0.644[/latex].

Question 10.1.8: Rayleigh Distillation A differential distillation starting a...

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