Question 5 of 6: Reactive Vapor–Liquid Equilibrium of n-Butane Isomerization

Question 5: Reactive Vapor–Liquid Equilibrium of n-Butane Isomerization (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. An isomerization reaching equilibrium in a two-phase system at 311 K. The reaction equilibrium constant is written with 1-atm standard states, the liquid is an ideal (Lewis–Randall) solution, and each component’s pure-liquid fugacity is $f_i^{L}=\phi_i^{sat}P_i^{sat}$ (Poynting ≈ 1).

Find. The liquid ($x_i$) and vapour ($y_i$) mole fractions of the two isomers at equilibrium.

Approach. At equilibrium each component has one fugacity common to both phases; write $K$ as the ratio of the isomers’ fugacities (= liquid fugacities via Lewis–Randall) to get the liquid composition, then recover the vapour composition from those same partial fugacities.

1. Reaction equilibrium in terms of liquid mole fractions. With 1-atm standard states, $K=f_{iso}/f_{n}$, and each phase fugacity equals the ideal-solution liquid value $f_i=x_i\phi_i^{sat}P_i^{sat}$: $$K=\frac{x_{iso}\,\phi_{iso}^{sat}P_{iso}^{sat}}{x_{n}\,\phi_{n}^{sat}P_{n}^{sat}}\;\Rightarrow\;\frac{x_{iso}}{x_{n}}=K\frac{\phi_{n}^{sat}P_{n}^{sat}}{\phi_{iso}^{sat}P_{iso}^{sat}}=2.24\frac{0.91(3.53)}{0.89(4.95)}=1.633.$$
2. Liquid composition. With $x_n+x_{iso}=1$: $$x_n=\frac{1}{1+1.633}=\boxed{0.380},\qquad x_{iso}=\boxed{0.620.}$$
3. Vapour composition. Treating the vapour mixture as ideal ($\phi_i^{V}\approx1$), each component’s partial fugacity equals its partial pressure $f_i=x_i\phi_i^{sat}P_i^{sat}=y_iP$. Evaluating $f_n=0.380(0.91)(3.53)=1.22$ atm and $f_{iso}=0.620(0.89)(4.95)=2.73$ atm gives $P=3.95$ atm and $$y_n=\frac{1.22}{3.95}=\boxed{0.309},\qquad y_{iso}=\frac{2.73}{3.95}=\boxed{0.691.}$$

The vapour is richer in iso-butane than the liquid ($y_{iso}=0.69$ vs $x_{iso}=0.62$) because iso-butane is the more volatile isomer (higher $P^{sat}$); the reaction equilibrium ($K=2.24>1$) already favours the branched isomer, and volatility amplifies it in the vapour.