Reaction Equilibrium of ZnO(s) + CO(g) ⇌ Zn(g) + CO₂(g): 23-Chem-A1 Process Balances and Chemical Thermodynamics, May 2015, Q7

Question 7: Reaction Equilibrium of ZnO(s) + CO(g) ⇌ Zn(g) + CO₂(g) (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. $\Delta H^\circ_{298}$ and $\Delta G^\circ_{298}$ as tabulated; initial charge 1 CO, 1 CO₂, 1 Zn(g), 0.75 ZnO(s) at $T=1500$ K. Assumptions: ideal gases; solids at unit activity; $\Delta H^\circ$ and $\Delta S^\circ$ constant between 298 K and 1500 K (no $C_p$ data supplied); the tabulated Zn is taken as the reaction's gaseous product state.

Find. (a) the equilibrium pressure of the given mixture; (b) the pressure that just consumes all solid ZnO.

Approach. Build $\Delta G^\circ(1500)$ from $\Delta H^\circ_{298}$ and $\Delta S^\circ_{298}$, get $K$, then impose the equilibrium expression. Because $\Delta n_{gas}=+1$, pressure appears explicitly, so a specified gas composition fixes $P$.

Standard properties of reaction (298 K). $$\Delta G^\circ_{298}=(0-394.6)-(-318.4-137.2)=+61.0\ \text{kJ},\qquad \Delta H^\circ_{298}=(0-393.5)-(-348.0-110.5)=+65.0\ \text{kJ}.$$ $$\Delta S^\circ_{298}=\frac{\Delta H^\circ-\Delta G^\circ}{298.15}=\frac{4000}{298.15}=13.42\ \text{J/mol}\cdot\text{K}.$$
Equilibrium constant at 1500 K. With $\Delta H^\circ,\Delta S^\circ$ constant, $$\Delta G^\circ_{1500}=65000-1500(13.42)=44{,}876\ \text{J}\;\Rightarrow\;K=\exp\!\Big(\!-\frac{44{,}876}{(8.314)(1500)}\Big)=\boxed{0.0274}.$$
(a) Equilibrium pressure of the specified mixture. With solids at unit activity and $\Delta n_{gas}=+1$, $$K=\frac{y_{\text{Zn}}\,y_{\text{CO}_2}}{y_{\text{CO}}}\Big(\frac{P}{P^\circ}\Big).$$ The charge 1 CO : 1 CO₂ : 1 Zn gives $y_{\text{CO}}=y_{\text{CO}_2}=y_{\text{Zn}}=\tfrac13$, so $K=\tfrac13\,(P/P^\circ)$ and $$P=3K=\boxed{0.082\ \text{bar}}.$$
(b) Pressure to consume all the ZnO. Dissolving all 0.75 mol solid drives the reaction forward by $\varepsilon=0.75$: CO $=0.25$, CO₂ $=1.75$, Zn $=1.75$ (total gas 3.75). Because $\Delta n_{gas}>0$, the forward reaction is favoured by low pressure; equilibrium at that end-point gives the maximum allowable pressure: $$K=\frac{(1.75/3.75)(1.75/3.75)}{(0.25/3.75)}\Big(\frac{P}{P^\circ}\Big)=3.267\,(P/P^\circ)\;\Rightarrow\;P=\frac{K}{3.267}=\boxed{0.0084\ \text{bar}}.$$ Only at $P\lesssim0.0084$ bar can all the ZnO be reduced.

Quantity	Result
$\Delta G^\circ_{298}$ / $\Delta H^\circ_{298}$ / $\Delta S^\circ_{298}$	+61.0 kJ / +65.0 kJ / 13.42 J·K⁻¹
$K$ at 1500 K	0.0274
(a) Equilibrium pressure (given mixture)	≈ 0.082 bar
(b) Pressure to consume all ZnO	≈ 0.0084 bar

Concept & Theory

Topic. Heterogeneous reaction equilibrium with a gas-phase mole change; effect of pressure on conversion.

Key relations. $\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ$; $K=\exp(-\Delta G^\circ/RT)$; $K=\prod a_i^{\nu_i}$ with $a_{solid}=1$ and $a_{gas}=y_iP/P^\circ$; $\Delta n_{gas}=+1$.

Why this works. Solids drop out of $K$ (unit activity), so the equilibrium ties the gas mole fractions to the total pressure. When the gas composition is pinned (part a) the single unknown is $P$; when the endpoint composition is pinned by "all solid gone" (part b), $P$ is again determined. A positive $\Delta n_{gas}$ means lowering pressure pulls the reaction toward the gaseous products — the physical lever that lets you strip the ZnO.

Common pitfall. Leaving solid ZnO or Zn in the $K$ expression, or dropping the $P/P^\circ$ factor that carries the $\Delta n_{gas}=+1$ pressure dependence. Also note the stated Zn(0,0) datum forces the "product-state" assumption — flag it, as the exam invites.

Question 7 of 7: Reaction Equilibrium of ZnO(s) + CO(g) ⇌ Zn(g) + CO₂(g)

Question 7: Reaction Equilibrium of ZnO(s) + CO(g) ⇌ Zn(g) + CO₂(g) (Part C — 30 marks)