This question is about enthalpy, entropy, free energy, and the industrial extraction of zinc.
Zinc can be extracted from its oxide ore, ZnO\text{ZnO}ZnO, using carbon. At high temperatures, the zinc is produced as a vapour. The following equilibrium is involved:
ZnO(s)+C(s)⇌Zn(g)+CO(g)ΔH=+349.0 kJ mol−1,ΔS=+284.0 J K−1 mol−1 \text{ZnO}(\text{s}) + \text{C}(\text{s}) \rightleftharpoons \text{Zn}(\text{g}) + \text{CO}(\text{g}) \quad \Delta H = +349.0\text{ kJ mol}^{-1}, \quad \Delta S = +284.0\text{ J K}^{-1}\text{ mol}^{-1} ZnO(s)+C(s)⇌Zn(g)+CO(g)ΔH=+349.0 kJ mol−1,ΔS=+284.0 J K−1 mol−1Explain why this equilibrium is classified as a heterogeneous equilibrium.
Write the expression for the equilibrium constant, KpK_pKp, for this reaction. Use parentheses and not square brackets.
The forward reaction is only feasible at high temperatures.
Another reaction involved in processing zinc is:
Zn(s)+H2O(g)⇌ZnO(s)+H2(g)ΔH=−105.5 kJ mol−1 \text{Zn}(\text{s}) + \text{H}_2\text{O}(\text{g}) \rightleftharpoons \text{ZnO}(\text{s}) + \text{H}_2(\text{g}) \quad \Delta H = -105.5\text{ kJ mol}^{-1} Zn(s)+H2O(g)⇌ZnO(s)+H2(g)ΔH=−105.5 kJ mol−1Using the standard enthalpy change of formation of H2O(g)=−241.8 kJ mol−1\text{H}_2\text{O}(\text{g}) = -241.8\text{ kJ mol}^{-1}H2O(g)=−241.8 kJ mol−1, calculate the standard enthalpy change of formation, ΔfH\Delta_f HΔfH, for ZnO(s)\text{ZnO}(\text{s})ZnO(s).
