This question is about enthalpy, entropy, free energy, and the industrial extraction of lead.
Lead can be extracted from its oxide ore, PbO\text{PbO}PbO, using carbon. The following equilibrium is involved:
PbO(s)+C(s)⇌Pb(l)+CO(g)ΔH=+112.5 kJ mol−1,ΔS=+215.3 J K−1 mol−1 \text{PbO}(\text{s}) + \text{C}(\text{s}) \rightleftharpoons \text{Pb}(\text{l}) + \text{CO}(\text{g}) \quad \Delta H = +112.5\text{ kJ mol}^{-1}, \quad \Delta S = +215.3\text{ J K}^{-1}\text{ mol}^{-1} PbO(s)+C(s)⇌Pb(l)+CO(g)ΔH=+112.5 kJ mol−1,ΔS=+215.3 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 lead is:
Pb(s)+H2O(g)⇌PbO(s)+H2(g)ΔH=−23.5 kJ mol−1 \text{Pb}(\text{s}) + \text{H}_2\text{O}(\text{g}) \rightleftharpoons \text{PbO}(\text{s}) + \text{H}_2(\text{g}) \quad \Delta H = -23.5\text{ kJ mol}^{-1} Pb(s)+H2O(g)⇌PbO(s)+H2(g)ΔH=−23.5 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 PbO(s)\text{PbO}(\text{s})PbO(s).
