This question is about thermodynamics.
Tungsten is extracted from its oxide via reduction with hydrogen gas at high temperature, according to the following equation:
WO3(s)+3H2(g)→W(s)+3H2O(g) \text{WO}_3(\text{s}) + 3\text{H}_2(\text{g}) \rightarrow \text{W}(\text{s}) + 3\text{H}_2\text{O}(\text{g}) WO3(s)+3H2(g)→W(s)+3H2O(g)The table below shows the relevant thermodynamic data for this reaction at standard pressure.
| Substance | WO3(s)\text{WO}_3(\text{s})WO3(s) | H2(g)\text{H}_2(\text{g})H2(g) | W(s)\text{W}(\text{s})W(s) | H2O(g)\text{H}_2\text{O}(\text{g})H2O(g) |
|---|---|---|---|---|
| ΔfH⊖ / kJ mol−1\Delta_f H^\ominus\ /\ \text{kJ mol}^{-1}ΔfH⊖ / kJ mol−1 | −843-843−843 | 000 | 000 | −242-242−242 |
| S⊖ / J K−1mol−1S^\ominus\ /\ \text{J K}^{-1}\text{mol}^{-1}S⊖ / J K−1mol−1 | 767676 | 131131131 | 333333 | 189189189 |
Explain why the standard entropy value for water vapour is significantly greater than that for solid tungsten.
State the temperature at which the standard entropy of solid tungsten is 0 J K−1mol−10\ \text{J K}^{-1}\text{mol}^{-1}0 J K−1mol−1.
Use the equation and the thermodynamic data provided to calculate the minimum temperature, in K\text{K}K, at which this reaction becomes feasible.
