Water-Gas Shift Reaction Equilibrium: 23-Chem-A1 Process Balances and Chemical Thermodynamics, December 2014, Q7

Question 7: Water-Gas Shift Reaction Equilibrium (Part C, 30 marks)

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. Charge 1 mol each of CO, H₂O, CO₂, H₂; $T=900$ K, $P=1.0$ bar. Formation data (298 K): $\Delta H^{\circ}_{r}=-41.2$ kJ/mol, $\Delta G^{\circ}_{r}=-28.6$ kJ/mol ($\Delta S^{\circ}_r=-42.1$ J/mol·K); $\Delta n_{\text{gas}}=0$.

Find. (a) % CO reacted at 900 K; (b) heat per mole CO to hold $T$; (c) temperature for 50% CO conversion.

Figure 6 — Equilibrium reactor charged with one mole of each species; the water-gas shift proceeds until CO+H₂O and CO₂+H₂ reach the equilibrium ratio set by K(T).

Approach. Because $\Delta n_{\text{gas}}=0$ the equilibrium constant depends only on mole numbers (pressure cancels); march $K$ from 298 K to 900 K with Kirchhoff’s corrections, solve for the extent, then use $\Delta H_r(T)$ for the heat duty and invert $K(T)$ for the 50%-conversion temperature.

Extent in terms of $K$. With $\xi$ mol of CO reacted, the amounts are $CO=1-\xi$, $H_2O=1-\xi$, $CO_2=1+\xi$, $H_2=1+\xi$. Since $\Delta n=0$, $$K = \frac{(1+\xi)^2}{(1-\xi)^2}\;\Rightarrow\;\xi = \frac{\sqrt{K}-1}{\sqrt{K}+1}.$$
Equilibrium constant at 900 K. Integrating $\Delta C_p$ (SVA polynomials) into $\Delta H_r(T)$ and $\Delta S_r(T)$ and forming $\Delta G_r(900)= \Delta H_r-T\Delta S_r = -6.35$ kJ/mol, $$K(900) = \exp\!\Big(\tfrac{-\Delta G_r}{RT}\Big) = 2.34.$$
(a) CO conversion. $\sqrt{2.34}=1.53$, so $$\xi = \frac{1.53-1}{1.53+1}=0.209 \;\Rightarrow\; \boxed{20.9\%\text{ of the CO has reacted}}.$$
(b) Heat to hold $T$. The reaction is exothermic; at 900 K $\Delta H_r(900)=-35.6$ kJ/mol. Releasing heat as $\xi=0.209$ mol reacts (per mole CO charged), $$Q = \xi\,\Delta H_r(900) = 0.209(-35.6) = \boxed{-7.5\ \text{kJ (removed) per mol CO charged}}.$$ (Equivalently 35.6 kJ removed per mole of CO that actually reacts.)
(c) Temperature for 50% conversion. $\xi=0.5$ needs $K = \big(\tfrac{1.5}{0.5}\big)^2 = 9$. Because the reaction is exothermic, $K$ rises as $T$ falls; solving $K(T)=9$, $$\boxed{T \approx 706\ \text{K} = 433\ ^{\circ}\text{C}}.$$

Quantity	Result
$K$ at 900 K	2.34
(a) CO reacted at 900 K	20.9%
(b) Heat removed to hold 900 K	7.5 kJ per mol CO charged
(c) Temperature for 50% CO conversion	≈ 706 K (433 °C)

Concept & Theory

Topic. Gas-phase reaction equilibrium with temperature-dependent $K$; heat of reaction and the effect of temperature on an exothermic equilibrium (Le Châtelier).

Key relations. $K = \prod (y_i)^{\nu_i}(P/P^{\circ})^{\Delta n}$ with $\Delta n=0$; $\ln K = -\Delta G_r^{\circ}/RT$; Kirchhoff $\Delta H_r(T)=\Delta H_{298}+\int\Delta C_p\,dT$, and $\Delta S_r(T)$ likewise.

Why this works. With no change in total moles, $K$ is a pure ratio of mole numbers and is pressure-independent, so a single quadratic gives the extent. The van’t Hoff/Kirchhoff march captures the modest fall in $K$ from $10^5$ (298 K) to ~2 (900 K), and inverting it locates the temperature that quadruples the equilibrium ratio for 50% conversion.

Common pitfall. Forgetting a species’ entropy/heat-capacity term when assembling $\Delta S_r$ or $\Delta C_p$, or assuming $K$ is temperature-independent (van’t Hoff with $\Delta H_r$ only) — adequate for a rough number but the Kirchhoff terms shift $K$ noticeably over a 600 K span.

Source. Smith, Van Ness, Abbott & Swihart, ICET §13.1–13.5 (chemical-reaction equilibria, temperature dependence of $K$).

Question 7 of 7: Water-Gas Shift Reaction Equilibrium

Question 7: Water-Gas Shift Reaction Equilibrium (Part C, 30 marks)