Excess Gibbs Energy and Vapour–Liquid Equilibrium of a Binary Mixture: 23-Chem-A1 Process Balances and Chemical Thermodynamics, December 2013, Q5

Question 5: Excess Gibbs Energy and Vapour–Liquid Equilibrium of a Binary Mixture (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. Binary liquid at 298 K, 1.0 bar total (air head). $P_A^{sat}=0.8$ bar, $P_B^{sat}=0.5$ bar; for the 50/50 mixture $H^E = -10$ kJ/mol and $S^E = -1$ J/mol·K; symmetric (two-suffix) model $G^E = C\,x_A x_B$.

Find. (a) $G^E$ at 50/50; (b) total $\Delta S_\text{mix}$ for 1 mol A + 1 mol B; (c) sign of the deviation from Raoult's law; (d) $\gamma_A,\gamma_B$ at 50/50; (e) the gas-phase composition above the 50/50 liquid.

Approach. Assemble $G^E = H^E - TS^E$ (the enthalpy of mixing is the excess enthalpy for a liquid), get the total mixing entropy from ideal + excess parts, then use the symmetric model to extract activity coefficients and finally the partial pressures.

Excess Gibbs energy (part a). By definition $G^E = H^E - T S^E$, with $H^E$ equal to the enthalpy of mixing: $$G^E = -10{,}000 - (298)(-1) = \boxed{-9702\ \text{J/mol} \;(-9.70\ \text{kJ/mol})}.$$
Total entropy of mixing (part b). For one mole of mixture $\Delta S_\text{mix} = -R\sum x_i\ln x_i + S^E$. The ideal part at 50/50 is $-R\ln0.5 = 5.763$ J/mol·K; adding $S^E=-1$ and scaling to the two moles actually mixed, $$\Delta S_\text{total} = 2\big[5.763 + (-1)\big] = \boxed{9.53\ \text{J/K}}.$$
Direction of deviation (part c). $G^E<0$, so both activity coefficients are below unity and every partial pressure $p_i = x_i\gamma_i P_i^{sat}$ falls below its Raoult's-law value: the equilibrium vapour pressure is lower than that of an ideal solution — a negative deviation. Physically, the strongly exothermic mixing ($H^E=-10$ kJ/mol) signals A–B attractions stronger than the A–A and B–B interactions, so molecules are held more tightly in the liquid and their escaping tendency (fugacity) drops.
Fit the model constant and the activity coefficients (part d). At 50/50, $G^E = C(0.5)(0.5)=0.25\,C$, so $C = -9702/0.25 = -38{,}808$ J/mol. The symmetric model gives $\ln\gamma_A = (C/RT)x_B^2$ and $\ln\gamma_B=(C/RT)x_A^2$; at $x_A=x_B=0.5$, $$\ln\gamma_A = \ln\gamma_B = \frac{-38{,}808}{(8.314)(298)}(0.25) = -3.916 \;\Rightarrow\; \boxed{\gamma_A=\gamma_B \approx 0.020}.$$ (Equivalently $\ln\gamma = G^E/RT$ at the symmetric midpoint — a useful check.)
Gas-phase composition (part e). With air insoluble, each organic partial pressure is $p_i = x_i\gamma_i P_i^{sat}$: $$p_A = 0.5(0.0199)(0.8) = 0.00798\ \text{bar},\qquad p_B = 0.5(0.0199)(0.5) = 0.00499\ \text{bar}.$$ Air makes up the balance of the 1.0 bar: $p_\text{air}=1-0.00798-0.00499 = 0.9870$ bar. Dividing by 1.0 bar, $$\boxed{y_\text{air}=98.7\%,\quad y_A=0.80\%,\quad y_B=0.50\%.}$$

Quantity	Result
(a) $G^E$ (50/50)	−9.70 kJ/mol
(b) $\Delta S$ for 1 mol A + 1 mol B	+9.53 J/K
(c) Deviation from Raoult	Negative — vapour pressure lower than ideal
(d) $\gamma_A = \gamma_B$ at 50/50	≈ 0.020
(e) Gas phase	98.7 % air, 0.80 % A, 0.50 % B

Check

The very small $\gamma\approx0.02$ follows directly from the unusually large $|H^E|=10$ kJ/mol supplied in the problem; it is the mathematically consistent result of the symmetric model, and signals extremely strong negative deviation. If a milder $H^E$ were intended the activity coefficients would rise toward unity, but the numbers as stated give $\gamma\approx0.02$.

Concept & Theory

Topic. Excess properties, activity coefficients from a two-suffix (symmetric Margules) model, and modified-Raoult VLE.

Key relations. $G^E = H^E - TS^E$; ideal mixing entropy $\Delta S^{id}=-R\sum x_i\ln x_i$; symmetric model $\ln\gamma_A=(C/RT)x_B^2$, $\ln\gamma_B=(C/RT)x_A^2$; modified Raoult $p_i = x_i\gamma_i P_i^{sat}$.

Why this works. The single-parameter form $G^E=Cx_Ax_B$ is the simplest model that vanishes for the pure components and is symmetric in the two species; differentiating $nG^E$ with respect to $n_i$ produces the partial molar excess Gibbs energy, i.e. $RT\ln\gamma_i$. The sign of $C$ (set by $G^E$) therefore fixes the sign of the deviation from ideality.

Common pitfall. Equating the enthalpy of mixing with the excess Gibbs energy (it is the excess enthalpy), or forgetting to include the ideal $-R\sum x\ln x$ term in the total mixing entropy — the excess entropy is only the deviation from ideal.

Source. Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics §11.6–12.1 (excess properties, activity coefficients, modified Raoult's law).

Question 5 of 7: Excess Gibbs Energy and Vapour–Liquid Equilibrium of a Binary Mixture

Question 5: Excess Gibbs Energy and Vapour–Liquid Equilibrium of a Binary Mixture (equal value)