23-Chem-A1 Process Balances and Chemical Thermodynamics · May 2015
Question 5 of 7: Activity Coefficients and Excess Gibbs Energy of Acetone–Chloroform
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2015 — 04-Chem-A1 Process Balances and Chemical Thermodynamics. Three-hour, open-book exam; any non-communicating calculator permitted. Format: seven questions in three parts — answer one of Q1–Q2 (Part A, 15 marks), one of Q3–Q4 (Part B, 25 marks) and two of Q5–Q7 (Part C, 30 marks each); four questions totalling 100 marks constitute a complete paper. All seven are solved below for completeness. Property data (Cp coefficients, steam-table and thermochemical values) are stated explicitly in each Given block.
Reference texts: Felder, Rousseau & Bullard, Elementary Principles of Chemical Processes (4th ed., Wiley) — material & energy balances, humidity, phase equilibria and reactive systems; Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics (8th ed., McGraw-Hill) — excess Gibbs energy, VLE, activity-coefficient models and reaction equilibrium; supporting data from the NIST/ASME steam tables, Perry's Chemical Engineers' Handbook (9th ed.) and the NIST Chemistry WebBook.
Question 5: Activity Coefficients and Excess Gibbs Energy of Acetone–Chloroform (Part C — 30 marks)
Given. Azeotrope: $x_1=y_1=0.335$, $T=64.6$ °C, total $P=1$ atm; $P_1^{sat}=1.31$, $P_2^{sat}=0.98$ atm. Van Laar model $\ln\gamma_1=A_{12}\big[1+\tfrac{A_{12}x_1}{A_{21}x_2}\big]^{-2}$, $\ln\gamma_2=A_{21}\big[1+\tfrac{A_{21}x_2}{A_{12}x_1}\big]^{-2}$.
Find. (a) $\gamma_1,\gamma_2$ at the azeotrope; (b) $G^E$ and the sign of the mixing temperature change; (c) $y_1$ over the $x_1=0.12$ liquid; (d) the total pressure there.
Figure 5 — Acetone-chloroform T-xy at 1 atm: a maximum-boiling azeotrope at x=0.335, 64.6 C.
Approach. At an azeotrope $y_i=x_i$, so modified Raoult's law collapses to $\gamma_i=P/P_i^{sat}$. Those two coefficients fit the two van Laar parameters, which then predict $\gamma$ at any composition; partial pressures give the vapour composition and the total pressure.
(a) Activity coefficients. With $y_i=x_i$ and $P=1$ atm, $\gamma_i=P/P_i^{sat}$:
$$\gamma_1=\frac{1}{1.31}=\boxed{0.763},\qquad \gamma_2=\frac{1}{0.98}=1.020.$$
$\gamma_1<1$ signals the strong negative deviation of this maximum-boiling azeotrope.
(b) Excess Gibbs energy and mixing temperature. Using $G^E/RT=x_1\ln\gamma_1+x_2\ln\gamma_2$:
$$G^E=RT\big[0.335\ln0.763+0.665\ln1.020\big]=(8.314)(337.75)(-0.0770)=\boxed{-216\ \text{J/mol}}.$$
With $G^E=H^E-TS^E$ and enthalpic factors dominating, $H^E\approx G^E<0$: mixing is exothermic, so adiabatic mixing of the pure liquids gives a solution warmer than 64.6 °C.
(c) Fit van Laar, then evaluate at $x_1=0.12$. From the azeotrope, $A_{12}=\ln\gamma_1\big[1+\tfrac{x_2\ln\gamma_2}{x_1\ln\gamma_1}\big]^2=-0.196$ and $A_{21}=\ln\gamma_2\big[1+\tfrac{x_1\ln\gamma_1}{x_2\ln\gamma_2}\big]^2=0.664$. At $x_1=0.12$:
$$\gamma_1=0.809,\qquad \gamma_2=1.001.$$
Vapour composition from partial pressures $p_i=x_i\gamma_iP_i^{sat}$:
$$p_1=0.12(0.809)(1.31)=0.127,\ \ p_2=0.88(1.001)(0.98)=0.863\ \text{atm}\;\Rightarrow\;y_1=\frac{0.127}{0.990}=\boxed{0.128}.$$
(d) Total pressure. The sum of the partial pressures is the bubble pressure:
$$P=p_1+p_2=0.127+0.863=\boxed{0.990\ \text{atm}}.$$
The pressure lies between $P_2^{sat}=0.98$ atm (pure chloroform) and the 1 atm azeotrope, as expected for a liquid close to the chloroform end. With the printed data ($P_1^{sat}=1.31$ atm above and $P_2^{sat}=0.98$ atm below the azeotrope pressure) the azeotrope is not a true pressure minimum at 64.6 °C, so van Laar with parameters of opposite sign is only a fit to the data given, not a faithful model of the real negative-deviation system.
