Equilibrium of a Gaseous Isomerization: 23-Chem-A1 Process Balances and Chemical Thermodynamics, December 2013, Q6

Question 6: Equilibrium of a Gaseous Isomerization (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. Gas-phase isomerization Isomer 1 → Isomer 2, $\Delta n_\text{gas}=0$, ideal mixture ($S^E=0$). $\Delta H^\circ_{298}=1.0$ kJ/mol; equal pure-isomer entropies $\Rightarrow \Delta S^\circ_{298}=0$; $C_{p1}=20$, $C_{p2}=25$ J/mol·K so $\Delta C_p = +5$ J/mol·K.

Find. (a) $\Delta H$ at 398 K; (b) equilibrium mole fractions at 298 K; (c) at 398 K.

Approach. Use Kirchhoff's law for the temperature shift of $\Delta H$, evaluate $\Delta G$ from $\Delta H$ and $\Delta S$ at each temperature (integrating the constant $\Delta C_p$), then convert $K$ to mole fractions via $K = y_2/y_1$ since $\Delta n_\text{gas}=0$.

Heat of reaction at 398 K (part a). With constant $\Delta C_p$, Kirchhoff's law gives $$\Delta H_{398} = \Delta H_{298} + \Delta C_p(398-298) = 1000 + 5(100) = \boxed{1500\ \text{J/mol}\;(1.5\ \text{kJ/mol})}.$$
Equilibrium constant at 298 K. Equal pure-component entropies make $\Delta S^\circ_{298}=0$, so $$\Delta G^\circ_{298} = \Delta H^\circ_{298} - T\Delta S^\circ_{298} = 1000\ \text{J/mol},\qquad K_{298} = e^{-\Delta G^\circ/RT} = e^{-1000/(8.314\cdot298)} = 0.668.$$
Mole fractions at 298 K (part b). For $\Delta n_\text{gas}=0$ in an ideal mixture the pressure cancels and $K = y_2/y_1$ with $y_1+y_2=1$: $$y_2 = \frac{K}{1+K} = \frac{0.668}{1.668} = 0.400,\qquad \boxed{y_1 = 0.600,\; y_2 = 0.400.}$$
Entropy and Gibbs energy at 398 K. March $\Delta S$ up with $\Delta C_p$: $\Delta S_{398} = 0 + \Delta C_p\ln(398/298) = 5\ln(1.3356) = 1.447$ J/mol·K, so $$\Delta G_{398} = \Delta H_{398} - 398\,\Delta S_{398} = 1500 - 398(1.447) = 924\ \text{J/mol},\qquad K_{398} = e^{-924/(8.314\cdot398)} = 0.756.$$
Mole fractions at 398 K (part c). Again $K=y_2/y_1$: $$y_2 = \frac{0.756}{1.756} = 0.431,\qquad \boxed{y_1 = 0.569,\; y_2 = 0.431.}$$ Heating drives the endothermic reaction toward the product, so $y_2$ rises from 0.400 to 0.431.

Quantity	Result
(a) $\Delta H$ at 398 K	1.5 kJ/mol
(b) 298 K equilibrium	$y_1 = 0.600$, $y_2 = 0.400$ ($K=0.668$)
(c) 398 K equilibrium	$y_1 = 0.569$, $y_2 = 0.431$ ($K=0.756$)

Concept & Theory

Topic. Chemical-reaction equilibrium for a gas-phase isomerization with a temperature-dependent $\Delta H$ and $\Delta S$ (Kirchhoff / van't Hoff).

Key relations. $\Delta H(T)=\Delta H_{298}+\Delta C_p(T-298)$; $\Delta S(T)=\Delta S_{298}+\Delta C_p\ln(T/298)$; $\Delta G = \Delta H - T\Delta S$; $K = e^{-\Delta G^\circ/RT}$; and for $\Delta n_\text{gas}=0$, $K=y_2/y_1$ (pressure-independent).

Why this works. Because one mole of gas becomes one mole of gas, the pressure and the standard-state choice drop out, and the equilibrium is purely a ratio of mole fractions. The equal-entropy statement zeroes $\Delta S_{298}$, so at 298 K equilibrium is governed by the small positive $\Delta H$ alone; heating then adds a favourable $T\Delta S$ term (the isomer with the larger $C_p$ acquires more entropy), pushing conversion up.

Common pitfall. Treating $\Delta H$ and $\Delta S$ as temperature-independent — with $\Delta C_p\ne0$ both drift, and although the drift is modest here it is what makes parts (a) and (c) differ from a naïve calculation.

Source. Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics §13.3–13.5 (standard property changes of reaction, temperature dependence of $K$).

Question 6 of 7: Equilibrium of a Gaseous Isomerization

Question 6: Equilibrium of a Gaseous Isomerization (equal value)