Question 5 of 5: Vapour–liquid equilibrium: Raoult's vs Henry's law

Question 5 — Vapour–liquid equilibrium: Raoult's vs Henry's law (10 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.

(a) Benzene–toluene — Raoult's law

Given. $x_B=x_T=0.5$ at $T=30\ \text{°C}$; Antoine constants as tabulated.

Find. System pressure $P$ and vapour composition $y_B,\,y_T$.

Why Raoult. Benzene and toluene are chemically very similar, adjacent aromatics that form a nearly ideal liquid solution over the whole composition range; both are present at high (50%) concentration. Raoult's law, $p_i=x_iP_i^{sat}$, is therefore the appropriate model.

1. Pure-component vapour pressures at 30 °C (Antoine). $$\log_{10}P_B^{sat}=6.906-\frac{1211}{30+220.8}=2.077\ \Rightarrow\ P_B^{sat}=119.5\ \text{mmHg},$$ $$\log_{10}P_T^{sat}=6.953-\frac{1343.9}{30+219.4}=1.565\ \Rightarrow\ P_T^{sat}=36.7\ \text{mmHg}.$$
2. System pressure by Raoult's law. The total pressure is the sum of partial pressures: $$P=x_BP_B^{sat}+x_TP_T^{sat}=0.5(119.5)+0.5(36.7)=\boxed{78.1\ \text{mmHg}}\;(0.103\ \text{atm}).$$
3. Vapour composition (Dalton). $$y_B=\frac{x_BP_B^{sat}}{P}=\frac{59.75}{78.1}=\boxed{0.765},\qquad y_T=1-y_B=0.235.$$ The vapour is enriched in the more volatile benzene ($y_B>x_B$), as expected.

(b) Ethane in water — Henry's law

Given. Gas at $P=20\ \text{atm}$ with $y_{\text{eth}}=0.010$; water at 20 °C; $H_{\text{eth}}=2.63\times10^{4}\ \text{atm/mole fraction}$.

Find. Mole fraction of ethane dissolved in the water, $x_{\text{eth}}$.

Why Henry. Ethane is a light, essentially non-condensable gas that is only sparingly soluble in water; it exists at very low mole fraction in the liquid, the regime where Henry's law $p_i=H_i x_i$ applies. (Raoult's law is invalid here — ethane has no meaningful liquid-phase vapour pressure at 20 °C, and water/ethane are grossly dissimilar.)

1. Partial pressure of ethane in the gas (Dalton). $$p_{\text{eth}}=y_{\text{eth}}P=0.010\times20=0.20\ \text{atm}.$$
2. Dissolved mole fraction from Henry's law. $$x_{\text{eth}}=\frac{p_{\text{eth}}}{H_{\text{eth}}}=\frac{0.20}{2.63\times10^{4}} =\boxed{7.6\times10^{-6}}.$$ The dissolved ethane is minute — consistent with the sparing solubility that justified Henry's law.

(The benzene and toluene Henry constants quoted in the exam are not used: at these high liquid concentrations Raoult's law, not Henry's law, is the correct model for part a.)