Vapour–Liquid Equilibrium of Benzene/Acetonitrile: 23-Chem-A1 Process Balances and Chemical Thermodynamics, May 2014, Q7

Question 7: Vapour–Liquid Equilibrium of Benzene/Acetonitrile (Part D — 30%)

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. Modified Raoult’s law with one-parameter (symmetric two-suffix Margules) activity coefficients $\ln\gamma_1=x_2^2$, $\ln\gamma_2=x_1^2$. The system shows positive deviations ($\gamma_i>1$) and forms a pressure-maximum azeotrope.

Find. (a) $P$ and $y$ at the bubble point; (b) $x$ and $y$ at fixed $T,P$; (c) the dew-point $T$ and liquid composition.

P–x–y diagram at 20 °C. The bubble curve (solid) and dew curve (dashed) bow above the straight Raoult line and meet at a pressure-maximum azeotrope, reflecting the positive deviations. Part (a) fixes the liquid at x₁=0.20 and reads the bubble pressure and the equilibrium vapour y₁=0.33 along the tie line.

Approach

All three parts use modified Raoult’s law $y_iP = x_i\gamma_i P_i^{sat}$. A bubble-$P$ calculation is explicit; the fixed-$T,P$ case solves $\sum x_i\gamma_iP_i^{sat}=P$ for $x_1$ (two roots because of the azeotrope); the dew-$T$ case iterates on $T$ and $x$ until $\sum x_i =1$.

Part (a) — bubble pressure at 20 °C, $x_1=0.20$

Saturation pressures at 293.15 K. $$P_1^{sat} = 10^{\,4.01814-1203.835/(293.15-53.226)} = 0.1001\ \text{bar},\quad P_2^{sat} = 0.0933\ \text{bar}.$$
Activity coefficients. $\gamma_1=e^{x_2^2}=e^{0.64}=1.897$, $\gamma_2=e^{x_1^2}=e^{0.04}=1.041$.
Partial pressures and total. $$p_1 = x_1\gamma_1P_1^{sat} = 0.20(1.897)(0.1001)=0.03799\ \text{bar},\quad p_2 = 0.80(1.041)(0.0933)=0.07766\ \text{bar}.$$ $$P = p_1+p_2 = 0.1156\ \text{bar}.$$
Bubble pressure $P \approx 0.116$ bar.
Vapour composition. $y_1 = p_1/P = 0.0380/0.1156 = 0.328$.
$y_1 = 0.328,\ y_2 = 0.672$ (vapour is benzene-enriched: $y_1>x_1$).

Part (b) — boiling at 45 °C under 0.32 bar

Saturation pressures at 318.15 K. $P_1^{sat}=0.2979$ bar, $P_2^{sat}=0.2775$ bar. Both lie below 0.32 bar while the azeotrope pressure is higher, so the bubble equation $x_1\gamma_1P_1^{sat}+x_2\gamma_2P_2^{sat}=0.32$ has two liquid roots.
Benzene-lean root. Solving gives $x_1=0.103$ ($\gamma_1=e^{0.805}=2.24,\ \gamma_2=e^{0.0106}=1.011$); the vapour is $$y_1 = \frac{x_1\gamma_1P_1^{sat}}{P} = 0.214.$$
Root 1: $x_1=0.103,\ x_2=0.897$ → $y_1=0.214,\ y_2=0.786$.
Benzene-rich root. The second solution is $x_1=0.944$ ($\gamma_1=e^{0.0031}=1.003,\ \gamma_2=e^{0.891}=2.44$): $$y_1 = 0.882.$$
Root 2: $x_1=0.944,\ x_2=0.056$ → $y_1=0.882,\ y_2=0.118$.
Interpretation. Because the mixture is azeotropic, a given $T$ and $P$ between the pure pressures and the azeotrope can be satisfied on either side of the azeotrope; both compositions are physically valid boiling liquids.

Part (c) — dew point at 0.22 bar, $y_1=0.15$

Dew relation. At the dew point the first liquid appears with $x_i = y_iP/(\gamma_iP_i^{sat})$ and $\sum x_i = 1$; since $\gamma_i$ depend on the (unknown) $x$, iterate.
Iterate $T$ and $x$. Starting from $\gamma_i=1$ and updating, convergence gives a benzene-poor liquid $x_1\approx0.064$: $$\sum_i \frac{y_iP}{\gamma_iP_i^{sat}(T)} = 1 \quad\Rightarrow\quad T = 310.0\ \text{K}.$$
Dew point $T \approx 310$ K = 36.9 °C, with first-liquid $x_1\approx0.064$.

Part	Result
(a) bubble at 20 °C, $x_1=0.20$	$P=0.116$ bar, $y_1=0.328$
(b) 45 °C, 0.32 bar — lean root	$x_1=0.103,\ y_1=0.214$
(b) 45 °C, 0.32 bar — rich root	$x_1=0.944,\ y_1=0.882$
(c) dew at 0.22 bar, $y_1=0.15$	$T=36.9$ °C, $x_1=0.064$

Concept & Theory

Topic. Low-pressure VLE by modified Raoult’s law with a two-suffix Margules model.

Key relations. $y_iP = x_i\gamma_iP_i^{sat}$; $\ln\gamma_1=Ax_2^2,\ \ln\gamma_2=Ax_1^2$ with $A=1$; Antoine $P_i^{sat}(T)$; bubble-$P$: $P=\sum x_i\gamma_iP_i^{sat}$; dew-$T$: $\sum y_iP/(\gamma_iP_i^{sat})=1$.

Why this works. At these low pressures the vapour is ideal ($\phi_i\approx1$) so all non-ideality is in the liquid activity coefficients. Bubble-$P$ is explicit; the fixed-$T,P$ and dew-$T$ problems are implicit because $\gamma_i=\gamma_i(x)$, needing iteration. The equal-coefficient Margules form ($A=1$) makes $\gamma_1=\gamma_2$ at $x_1=0.5$, where the azeotrope sits.

Common pitfall. Missing the second root in part (b) — an azeotropic system gives two boiling compositions at one $T,P$. Also using Raoult’s law without $\gamma$ (would misplace the azeotrope and give the wrong $y$).

Source. Smith, Van Ness & Abbott, ch. 10 & 12 (VLE, activity-coefficient models, azeotropy).

Part	Result
(a) bubble at 20 °C, \(x_1=0.20\)	\(P=0.116\) bar, \(y_1=0.328\)
(b) 45 °C, 0.32 bar — lean root	\(x_1=0.103,\ y_1=0.214\)
(b) 45 °C, 0.32 bar — rich root	\(x_1=0.944,\ y_1=0.882\)
(c) dew at 0.22 bar, \(y_1=0.15\)	\(T=36.9\) °C, \(x_1=0.064\)

Question 7 of 8: Vapour–Liquid Equilibrium of Benzene/Acetonitrile

Question 7: Vapour–Liquid Equilibrium of Benzene/Acetonitrile (Part D — 30%)

Approach

Part (a) — bubble pressure at 20 °C, \(x_1=0.20\)

Part (b) — boiling at 45 °C under 0.32 bar

Part (c) — dew point at 0.22 bar, \(y_1=0.15\)