The dehydrogenation of ethane to produce ethene is represented by the following equation:
C2H6(g)⇌C2H4(g)+H2(g)ΔH=+137 kJ mol−1 \text{C}_2\text{H}_6(\text{g}) \rightleftharpoons \text{C}_2\text{H}_4(\text{g}) + \text{H}_2(\text{g}) \quad \Delta H = +137\text{ kJ mol}^{-1} C2H6(g)⇌C2H4(g)+H2(g)ΔH=+137 kJ mol−1Some standard entropies are given in the table below:
| Gas | S∘ / J K−1 mol−1S^\circ \ / \ \text{J K}^{-1}\text{ mol}^{-1}S∘ / J K−1 mol−1 |
|---|---|
| C2H6(g)\text{C}_2\text{H}_6(\text{g})C2H6(g) | 229 |
| C2H4(g)\text{C}_2\text{H}_4(\text{g})C2H4(g) | 219 |
| H2(g)\text{H}_2(\text{g})H2(g) | 131 |
Calculate the standard entropy change (ΔS∘\Delta S^\circΔS∘) for this reaction.
Calculate the Gibbs free-energy change (ΔG\Delta GΔG), in kJ mol−1\text{kJ mol}^{-1}kJ mol−1, for this reaction at 950 ∘C950\,^\circ\text{C}950∘C. (If you were unable to obtain an answer for Part 1, use a value of +110 J K−1 mol−1+110\text{ J K}^{-1}\text{ mol}^{-1}+110 J K−1 mol−1 for the entropy change. This is not the correct answer.)
The reaction is carried out at a higher temperature. Explain how this change in temperature affects the value of ΔG\Delta GΔG for the reaction.
