Question 5 of 6: Ethanol–Benzene Azeotrope from a Single Tie Line

Question 5: Ethanol–Benzene Azeotrope from a Single Tie Line (Part B — 25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Non-ideal binary VLE at 45 °C. One measured tie line: liquid $x_{EtOH}=0.389$ ($x_{Bz}=0.611$), vapour $y_{EtOH}=0.434$ ($y_{Bz}=0.566$), $P=40.25$ kPa. Pure-component vapour pressures $P_1^{sat}=22.9$ kPa (ethanol, 1), $P_2^{sat}=29.6$ kPa (benzene, 2). Modified Raoult’s law $y_iP=x_i\gamma_iP_i^{sat}$ (low pressure, ideal vapour). The printed hint “few molecular interactions” (no association or specific chemical interactions) is read as licence for the two-parameter van Laar activity model, whose two constants are exactly fixed by one tie line. The data list prints “ethanol” twice; the second value, 29.6 kPa, must be benzene.

Find. Azeotrope composition and total pressure at 45 °C.

Approach. Back out the two activity coefficients from the single tie line via modified Raoult’s law, fit the van Laar constants $A,B$, then impose the azeotrope condition ($x_i=y_i$, equivalently $\gamma_1P_1^{sat}=\gamma_2P_2^{sat}$) and solve for the composition and pressure.

1. Activity coefficients from the tie line. $\gamma_i=\dfrac{y_iP}{x_iP_i^{sat}}$: $$\gamma_1=\frac{0.434(40.25)}{0.389(22.9)}=1.961,\qquad \gamma_2=\frac{0.566(40.25)}{0.611(29.6)}=1.260.$$ Both exceed 1 — positive deviations, the signature of a minimum-boiling azeotrope.
2. Fit the van Laar constants. Using $A=\ln\gamma_1\!\left(1+\dfrac{x_2\ln\gamma_2}{x_1\ln\gamma_1}\right)^2$ and the symmetric expression for $B$: $$A=1.594,\qquad B=1.885.$$ These reproduce the measured $\gamma$’s at $x_1=0.389$ exactly and define $\gamma_i(x)$ everywhere.
3. Impose the azeotrope condition. At an azeotrope the vapour and liquid compositions coincide, so modified Raoult’s law gives $P=\gamma_1P_1^{sat}=\gamma_2P_2^{sat}$, i.e. $$\ln\gamma_1-\ln\gamma_2=\ln\frac{P_2^{sat}}{P_1^{sat}}=\ln\frac{29.6}{22.9}=0.257.$$ Solving this with the van Laar $\gamma_i(x)$ (bisection) gives $$x_{EtOH}^{az}=\boxed{0.446\ (\text{44.6\% ethanol, 55.4\% benzene}).}$$
4. Azeotropic pressure. Evaluate either equal branch at $x_1=0.446$ ($\gamma_1=1.760$): $$P_{az}=\gamma_1P_1^{sat}=1.760(22.9)=\boxed{40.3\ \text{kPa.}}$$ The consistency check $\gamma_2P_2^{sat}=1.361(29.6)=40.3$ kPa confirms the result.

The measured operating point (38.9% ethanol, 40.25 kPa) sits essentially on top of the computed pressure maximum, which is exactly why the problem states an azeotrope exists at 45 °C: the given tie line is nearly the azeotrope itself, and the van Laar fit places the true maximum at 44.6% ethanol. As a model-sensitivity check, a two-parameter Margules fit to the same tie line gives essentially the same azeotrope (44.5% ethanol, 40.3 kPa), so the answer does not hinge on the choice of activity model.