This question is about enthalpy, entropy, free energy, and the industrial extraction of tin.
Tin can be extracted from its oxide ore, SnO2\text{SnO}_2SnO2, using carbon. The following equilibrium is involved:
SnO2(s)+2C(s)⇌Sn(l)+2CO(g)ΔH=+358.5 kJ mol−1,ΔS=+392.4 J K−1 mol−1 \text{SnO}_2(\text{s}) + 2\text{C}(\text{s}) \rightleftharpoons \text{Sn}(\text{l}) + 2\text{CO}(\text{g}) \quad \Delta H = +358.5\text{ kJ mol}^{-1}, \quad \Delta S = +392.4\text{ J K}^{-1}\text{ mol}^{-1} SnO2(s)+2C(s)⇌Sn(l)+2CO(g)ΔH=+358.5 kJ mol−1,ΔS=+392.4 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 tin is:
Sn(s)+2H2O(g)⇌SnO2(s)+2H2(g)ΔH=−72.0 kJ mol−1 \text{Sn}(\text{s}) + 2\text{H}_2\text{O}(\text{g}) \rightleftharpoons \text{SnO}_2(\text{s}) + 2\text{H}_2(\text{g}) \quad \Delta H = -72.0\text{ kJ mol}^{-1} Sn(s)+2H2O(g)⇌SnO2(s)+2H2(g)ΔH=−72.0 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 SnO2(s)\text{SnO}_2(\text{s})SnO2(s).
