Isopropanol–Benzene VLE at 45 °C: 23-Chem-A1 Process Balances and Chemical Thermodynamics, December 2016, Q6

Question 6: Isopropanol–Benzene VLE at 45 °C (Part B — equal value)

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. The 45 °C ($T=318.15$ K) dataset. The third column, labelled “vapour pressure of isopropanol,” is in fact the total system pressure $P$ (kPa): at $x_1=1$ it reads $18.138=P_1^{\text{sat}}$ (pure IPA) and at $x_1=0$ it reads $29.829=P_2^{\text{sat}}$ (pure benzene). $R=8.314$ J·mol⁻¹K⁻¹.

Find. the $P$–$x$–$y$ (bubble/dew) diagram, the partial pressures $p_i=y_iP$, the activity coefficients via modified Raoult’s law, and the excess-Gibbs group $G^E/(x_1x_2RT)$.

Figure 3 — $P$–$x$–$y$ diagram at 45 °C. Bubble curve $P$ vs $x_1$ and dew curve $P$ vs $y_1$ meet at a pressure maximum ($P\approx36.45$ kPa near $x_1\approx0.30$): a minimum-boiling azeotrope from strong positive deviations.

Approach. Read the pure-component vapour pressures off the table endpoints; compute activity coefficients from modified Raoult’s law $\gamma_i=y_iP/(x_iP_i^{\text{sat}})$; then $G^E/RT=\sum x_i\ln\gamma_i$ and divide by $x_1x_2$.

(a) Bubble and dew curves. The bubble-point curve is the tabulated $P$ against liquid composition $x_1$; the dew-point curve is the same $P$ against vapour composition $y_1$ (Figure 3). Both start at $P_2^{\text{sat}}=29.83$ kPa ($x_1=0$) and end at $P_1^{\text{sat}}=18.14$ kPa ($x_1=1$), bulging above the Raoult straight line to a maximum of $\boxed{36.45\ \text{kPa}}$ — a minimum-boiling azeotrope.
(b) Partial pressures. By Dalton, $p_1=y_1P$ and $p_2=(1-y_1)P$. For example at $x_1=0.296$ ($y_1=0.2953$, $P=36.45$): $p_1=10.76$ kPa and $p_2=25.69$ kPa. The values at every data point are tabulated below and plotted in Figure 4. Both partial-pressure curves lie above their Raoult lines $x_iP_i^{\text{sat}}$, confirming positive deviations.
(c) Activity coefficients. Modified Raoult’s law $\gamma_i=\dfrac{y_iP}{x_iP_i^{\text{sat}}}$. At $x_1=0.296$: $$\gamma_1=\frac{0.2953(36.45)}{0.296(18.138)}=2.01\Rightarrow\ln\gamma_1=0.696,\qquad \gamma_2=\frac{0.7047(36.45)}{0.704(29.829)}=1.22\Rightarrow\ln\gamma_2=0.201.$$ The full set is in the table below and in Figure 5. Both $\gamma_i>1$ across the range (approaching the infinite-dilution values $\ln\gamma_1^\infty\approx2.0$ as $x_1\to0$ and $\ln\gamma_2^\infty\approx1.45$ as $x_1\to1$).
(d) Excess-Gibbs group. $G^E/RT=x_1\ln\gamma_1+x_2\ln\gamma_2$; dividing by $x_1x_2$ gives a mildly curved line whose end-points are the two Margules constants: $$\left.\frac{G^E}{x_1x_2RT}\right|_{x_1\to0}\approx A_{12}=2.0,\qquad \left.\frac{G^E}{x_1x_2RT}\right|_{x_1\to1}\approx A_{21}=1.45.$$ At $x_1=0.296$ the group is $\boxed{1.67}$; across the data it runs from 2.04 ($x_1=0.047$) to 1.45 ($x_1=0.966$), as tabulated and plotted in Figure 5. The positive, nearly-constant value (asymmetric two-suffix / van Laar behaviour) is the fingerprint of a positive-deviation, azeotrope-forming mixture.

Full data reduction (parts b–d). Every interior data point, computed with $p_1=y_1P$, $p_2=(1-y_1)P$, $\gamma_i=p_i/(x_iP_i^{\text{sat}})$ and $G^E/RT=x_1\ln\gamma_1+x_2\ln\gamma_2$. At the pure-component end-points ($x_1=0$ and $x_1=1$) the partial pressures are simply $p_2=29.829$ and $p_1=18.138$ kPa, and the activity coefficient of the absent component is undefined (0/0), so those rows are omitted from the $\ln\gamma$ columns.

$x_1$	$y_1$	$P$ (kPa)	$p_1$ (kPa)	$p_2$ (kPa)	$\ln\gamma_1$	$\ln\gamma_2$	$G^E/RT$	$G^E/(x_1x_2RT)$
0.0472	0.1467	33.633	4.93	28.70	1.751	0.010	0.0919	2.044
0.0980	0.2066	35.214	7.28	27.94	1.409	0.038	0.1721	1.947
0.2047	0.2663	36.271	9.66	26.61	0.956	0.115	0.2871	1.764
0.2960	0.2953	36.450	10.76	25.69	0.696	0.201	0.3477	1.669
0.3862	0.3211	36.292	11.65	24.64	0.509	0.297	0.3788	1.598
0.4753	0.3463	35.928	12.44	23.49	0.367	0.406	0.3873	1.553
0.5504	0.3962	35.319	13.99	21.33	0.338	0.464	0.3944	1.594
0.6198	0.3951	34.577	13.66	20.92	0.195	0.612	0.3535	1.500
0.7096	0.4378	33.023	14.46	18.57	0.116	0.762	0.3039	1.475
0.8073	0.5107	30.282	15.47	14.82	0.055	0.947	0.2266	1.456
0.9120	0.6658	25.235	16.80	8.43	0.016	1.167	0.1169	1.457
0.9655	0.8252	21.305	17.58	3.72	0.004	1.286	0.0481	1.446

Figure 4 — Part (b): partial pressures $p_1=y_1P$ and $p_2=y_2P$ (points) against the Raoult’s-law lines $x_iP_i^{\text{sat}}$ (straight lines). Both measured curves lie above their Raoult lines over the whole range — positive deviation.

Figure 5 — Parts (c) and (d): $\ln\gamma_1$, $\ln\gamma_2$ and $G^E/(x_1x_2RT)$ versus $x_1$. The $\ln\gamma$ curves cross near $x_1\approx0.47$; the $G^E/(x_1x_2RT)$ group falls gently from about 2.0 to 1.45.

Reading the plots: $\ln\gamma_1$ rises steeply as isopropanol becomes dilute (self-associated alcohol molecules are broken apart by the benzene), while $\ln\gamma_2$ climbs as benzene becomes dilute. The $G^E/(x_1x_2RT)$ group is nearly linear in $x_1$, so a two-parameter Margules model $G^E/(x_1x_2RT)=A_{21}x_1+A_{12}x_2$ with $A_{12}\approx2.0$ and $A_{21}\approx1.45$ represents the data well. One point is out of line: at $x_1=0.5504$ the printed $y_1=0.3962$ is higher than $y_1=0.3951$ at the richer $x_1=0.6198$, which is impossible for a smooth isotherm. It shows up as the small bump in the group (1.594) and the kink in $p_1$, so it is best treated as an experimental or transcription error, and the smooth curve should be drawn through the other points.

Quantity (at $x_1=0.296$ unless noted)	Result
$P_1^{\text{sat}}$ (IPA) / $P_2^{\text{sat}}$ (benzene)	18.14 / 29.83 kPa
Azeotrope pressure (max)	36.45 kPa near $x_1\approx0.30$
$p_1$ / $p_2$	10.76 / 25.69 kPa
$\ln\gamma_1$ / $\ln\gamma_2$	0.696 / 0.201
$G^E/(x_1x_2RT)$; end-points $A_{12},A_{21}$	1.67; 2.0, 1.45

Concept & Theory

Topic. Low-pressure VLE data reduction; activity coefficients and the excess Gibbs energy.

Key relations. Modified Raoult $y_iP=x_i\gamma_iP_i^{\text{sat}}$; $G^E/RT=\sum x_i\ln\gamma_i$; two-suffix/van Laar $G^E/(x_1x_2RT)$ with end-points $A_{12},A_{21}$.

Why this works. At these low pressures the vapour is essentially ideal, so all non-ideality sits in the liquid activity coefficients, obtained directly from the measured $x,y,P$. Plotting $G^E/(x_1x_2RT)$ linearises the composition dependence, letting the Margules/van Laar parameters be read as intercepts — a standard consistency and modelling check. Positive $\ln\gamma$ (activity above Raoult) drives the pressure maximum and hence the minimum-boiling azeotrope.

Common pitfall. Taking the third column literally as “IPA vapour pressure” instead of the total pressure — the endpoint values expose it as $P$. Also, near the azeotrope $\gamma$’s are sensitive to small $y$ errors, so the $G^E$ group scatters where $x_1x_2$ is small.

Source. Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics §§12.1–12.4 (gamma/phi VLE, excess properties, data reduction).

Question 6 of 6: Isopropanol–Benzene VLE at 45 °C

Question 6: Isopropanol–Benzene VLE at 45 °C (Part B — equal value)