Question 5 of 5: Oxide/Compound Stability — Reaction Quotient vs. Equilibrium Constant

Question 5: Oxide/Compound Stability — Reaction Quotient vs. Equilibrium Constant (15% — candidate’s choice of either alternative; both solved below)

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.

5a — ZrO₂ decomposition

Given. ΔG°(T) = 1,089,329 + 26.96T log₁₀T − 276.27T [J]; T = 2073 K; vacuum = 10⁻⁸ atm total pressure with the residual gas at air’s composition (≈21% O₂ by volume).

Find. K; P_O2,eq; whether the crucible can decompose under this vacuum.

Approach. Evaluate ΔG° at 2073 K, convert to K (with both solids at unit activity so K = P_O2), and compare that equilibrium P_O2 to the actual O₂ partial pressure supplied by the residual air in the vacuum.

1. ΔG° at 2073 K. $$\Delta G^{\circ}=1{,}089{,}329+26.96(2073)\log_{10}(2073)-276.27(2073)$$ $$\boxed{\Delta G^{\circ}=+701{,}980\text{ J}}$$
2. Equilibrium constant and P_O2. With a_ZrO2 = a_Zr = 1, K = P_O2: $$K=\exp\!\Big(\frac{-\Delta G^{\circ}}{RT}\Big)=\exp\!\Big(\frac{-701980}{(8.314)(2073)}\Big)$$ $$\boxed{K=P_{O_2,eq}\approx2.05\times10^{-18}\text{ atm}}$$
3. Actual O₂ present under the stated vacuum. A total pressure of 10⁻⁸ atm at air’s composition supplies $$P_{O_2,vac}=(10^{-8})(0.21)=2.1\times10^{-9}\text{ atm}$$ which is roughly nine orders of magnitude larger than the P_O2 the decomposition equilibrium would require.
4. Verdict. Because the residual O₂ in the vacuum (2.1×10⁻⁹ atm) vastly exceeds the equilibrium P_O2 for decomposition (2.05×10⁻¹⁸ atm), the system sits far on the ZrO₂-stable side of equilibrium. $$\boxed{\text{ZrO}_2\text{ crucible will NOT decompose under this vacuum}}$$

5b — FeO(s) + CO(g) ↔ Fe(s) + CO₂(g)

Given. T = 727 °C = 1000 K; formation ΔG°(T) for CO, CO₂, and FeO; furnace atmosphere 12% CO, 1.5% CO₂, 86.5% N₂ by volume.

Find. K at 1000 K; whether FeO forms on the iron sheet.

Approach. Build ΔG° for the target reaction by Hess’s law from the three formation reactions, convert to K, then compare the actual gas-composition reaction quotient Q = P_CO2/P_CO to K.

1. ΔG° for the target reaction = [C+O₂→CO₂] − [C+½O₂→CO] − [Fe+½O₂→FeO]: $$\Delta G^{\circ}=(-394100-0.84T)-(-111700-87.65T)-(-263700+64.35T)=-18{,}700+22.46T$$ $$\Delta G^{\circ}(1000\text{ K})=-18{,}700+22.46(1000)$$ $$\boxed{\Delta G^{\circ}=+3760\text{ J}}$$
2. Equilibrium constant. $$K=\exp\!\Big(\frac{-\Delta G^{\circ}}{RT}\Big)=\exp\!\Big(\frac{-3760}{(8.314)(1000)}\Big)$$ $$\boxed{K=\frac{P_{CO_2}}{P_{CO}}=0.636}$$
3. Reaction quotient of the actual furnace gas. $$Q=\frac{P_{CO_2}}{P_{CO}}=\frac{0.015}{0.12}=0.125$$
4. Verdict. Q (0.125) < K (0.636), so ΔG_rxn = RT ln(Q/K) < 0: the forward reaction FeO+CO→Fe+CO₂ is spontaneous, i.e. this gas mixture is reducing enough to strip oxygen from any FeO present. Equivalently, starting from bare metallic Fe, the reverse (oxidising) reaction is non-spontaneous. $$\boxed{\text{Pure FeO will NOT form; the iron sheet stays metallic}}$$