Isomerization Equilibrium and a Dew Point: 23-Chem-A1 Process Balances and Chemical Thermodynamics, December 2017, Q5

Question 5: Isomerization Equilibrium and a Dew Point (Part B — equal value)

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Given. $T=400$ K; vapour pressures $P_A^{\text{sat}}=2$ atm, $P_B^{\text{sat}}=2.5$ atm; the gas is always at reaction equilibrium ($y_B/y_A$ constant, independent of $P$ for an equimolar isomerization); first liquid (dew point) appears at $P=2.2$ atm; ideal liquid solution (Raoult’s law, $\gamma_i=1$).

Find. (a) $K_g=a_B/a_A$ for the gas reaction; (b) $K_l=a_B/a_A$ for the liquid reaction.

Approach. At the dew point the vapour composition equals the overall (equilibrium) gas composition, and the incipient liquid satisfies Raoult’s law; the constraint $x_A+x_B=1$ fixes the vapour composition, hence both equilibrium constants.

Dew-point (Raoult) relations. The incipient liquid mole fractions are $x_A=y_AP/P_A^{\text{sat}}$ and $x_B=y_BP/P_B^{\text{sat}}$. Requiring $x_A+x_B=1$ at $P=2.2$ atm: $$y_A\frac{2.2}{2.0}+y_B\frac{2.2}{2.5}=1\;\Rightarrow\;1.10\,y_A+0.88\,y_B=1.$$
Solve for the vapour composition. With $y_A+y_B=1$: $1.10y_A+0.88(1-y_A)=1\Rightarrow0.22y_A=0.12$, so $$y_A=0.545,\qquad y_B=0.455.$$
Gas-phase equilibrium constant (part a). For A(g)↔B(g) with ideal gases, $K_g=(p_B/P^\circ)/(p_A/P^\circ)=y_B/y_A$: $$K_g=\frac{0.455}{0.545}=\boxed{0.833}.$$
Incipient-liquid composition. From Raoult at the dew point: $$x_A=y_A\frac{P}{P_A^{\text{sat}}}=0.545\frac{2.2}{2.0}=0.600,\qquad x_B=y_B\frac{P}{P_B^{\text{sat}}}=0.455\frac{2.2}{2.5}=0.400\;(x_A+x_B=1).$$
Liquid-phase equilibrium constant (part b). For A(l)↔B(l) with ideal solutions ($\gamma=1$), $K_l=x_B/x_A$: $$K_l=\frac{0.400}{0.600}=\boxed{0.667}.$$ Consistently, $K_l=K_g\,(P_A^{\text{sat}}/P_B^{\text{sat}})=0.833(2.0/2.5)=0.667$, the thermodynamic link between the gas- and liquid-phase constants through the pure-component vapour pressures.

Quantity	Result
Vapour composition at dew point	$y_A=0.545,\ y_B=0.455$
Incipient-liquid composition	$x_A=0.600,\ x_B=0.400$
(a) $K_g$ for A(g)↔B(g)	0.833
(b) $K_l$ for A(l)↔B(l)	0.667

Concept & Theory

Topic. Coupled reaction and phase equilibrium; dew point of a reacting binary via Raoult’s law.

Key relations. Equimolar-gas equilibrium $K_g=y_B/y_A$ (pressure-independent); Raoult $y_iP=x_iP_i^{\text{sat}}$; dew-point closure $\sum x_i=1$; phase-linked constants $K_l=K_g\,P_A^{\text{sat}}/P_B^{\text{sat}}$.

Why this works. Because the isomerization conserves moles, $K_g$ depends only on the composition ratio and is fixed throughout the compression. At the dew point the bulk vapour is still that equilibrium mixture, and the first drop of ideal liquid obeys Raoult’s law; imposing that the liquid fractions sum to one determines the compositions and both constants.

Common pitfall. Using $K_l=x_B/x_A$ with the vapour’s $y$ values, or forgetting that for an equimolar reaction the total-pressure terms cancel so $K_g$ is simply the mole-fraction ratio.

Source. Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics §13 (reaction equilibria) and §10 (VLE, Raoult’s law).

Question 5 of 6: Isomerization Equilibrium and a Dew Point

Question 5: Isomerization Equilibrium and a Dew Point (Part B — equal value)