Simultaneous Zn–C–O Equilibria — Partial Pressures in a Closed Reactor: 21-Mat-A1 Thermodynamics, Undated paper, Q3

Question 3: Simultaneous Zn–C–O Equilibria — Partial Pressures in a Closed Reactor (20%)

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. T = 897 °C = 1170 K; equimolar solid ZnO and C; the three formation ΔG°(T) expressions above; activities of pure solids = 1; the reactor starts with no other gas.

Find. Equilibrium P_Zn, P_CO, P_CO₂.

Approach. Combine the given formation reactions by Hess’s law to get K for each of the two stated equilibria, then close the system with an elemental mass balance — every oxygen atom in the gas phase came from a ZnO that released one Zn atom, so P_Zn = P_CO + 2P_CO₂ — giving three equations in three unknowns.

K for ZnO(s) + C(s) ↔ Zn(g) + CO(g). This equals [C+½O₂→CO] minus [Zn+½O₂→ZnO]: $$\Delta G^{\circ}_A=(-111700-87.65T)-(-460200+198T)=348{,}500-285.65T$$ $$\Delta G^{\circ}_A(1170\text{ K})=348{,}500-285.65(1170)=14{,}290\text{ J}$$ $$K_A=e^{-\Delta G^{\circ}_A/RT}=e^{-14290/(8.314)(1170)}=0.230=P_{Zn}P_{CO}$$
K for CO₂(g) + C(s) ↔ 2CO(g) (Boudouard reaction). This equals 2×[C+½O₂→CO] minus [C+O₂→CO₂]: $$\Delta G^{\circ}_B=2(-111700-87.65T)-(-394100-0.84T)=170{,}700-174.46T$$ $$\Delta G^{\circ}_B(1170\text{ K})=170{,}700-174.46(1170)=-33{,}418\text{ J}$$ $$K_B=e^{-\Delta G^{\circ}_B/RT}=e^{+33418/(8.314)(1170)}=31.05=\frac{P_{CO}^2}{P_{CO_2}}$$
Mass-balance closure. Every mole of Zn(g) that appears corresponds to one mole of ZnO decomposed, releasing one O atom into the gas phase as either CO (1 O) or CO₂ (2 O); no other oxygen source exists in the evacuated reactor, so $$P_{Zn}=P_{CO}+2P_{CO_2}$$ Substituting P_CO₂ = P_CO²/K_B and P_Zn = K_A/P_CO into this balance gives a single cubic in P_CO: $$\frac{2P_{CO}^3}{K_B}+P_{CO}^2-K_A=0$$
Solve the cubic numerically (bisection on P_CO>0): $$\boxed{P_{CO}=0.473\text{ atm}}\qquad P_{CO_2}=\frac{P_{CO}^2}{K_B}=\frac{(0.473)^2}{31.05}=0.0072\text{ atm}$$ $$\boxed{P_{Zn}=P_{CO}+2P_{CO_2}=0.473+2(0.0072)=0.487\text{ atm}}$$ Check: P_ZnP_CO = (0.487)(0.473) = 0.230 = K_A ✓

Species	Equilibrium partial pressure
P_Zn	0.487 atm
P_CO	0.473 atm
P_CO₂	0.0072 atm

Concept & Theory

Topic: Coupled gas-phase equilibria closed by an elemental mass balance.

Key relations: K = Π(activities)^ν, solids & pure condensed phases at unit activity; conservation of Zn and O atoms across the two simultaneous reactions.

Why this works: two equilibrium-constant expressions in three unknown partial pressures are under-determined; the reactor’s only oxygen source (ZnO) ties P_Zn to the total oxygen now carried by CO and CO₂, supplying the third, independent equation needed to solve the system exactly.

Common pitfall: assuming P_Zn = P_CO (true only if no CO₂ forms at all); ignoring the Boudouard reaction converts this into an unsolvable two-unknown, one-equation system.

Source: Gaskell, Introduction to the Thermodynamics of Materials, Ch. 9 & 13 (multi-reaction equilibria in pyrometallurgy, e.g. zinc smelting from ZnO–C mixtures).

Question 3 of 5: Simultaneous Zn–C–O Equilibria — Partial Pressures in a Closed Reactor

Question 3: Simultaneous Zn–C–O Equilibria — Partial Pressures in a Closed Reactor (20%)