Use the given data at 500 K to calculate ΔG°for the reaction
2H2S(g) + 3O2(g) → 2H2O(g) + 2SO2(g)
| Substance | H2S(g) | O2(g) | H2O(g) | SO2(g) |
| ΔH°f(kJ/mol) | -21 | 0 | -242 | -296.8 |
| S°(J/K·mol) | 206 | 205 | 189 | 248 |
Solution:
The change in free energy (ΔG°) is calculated as,
ΔG° = ΔH° - T ΔS°
For the given reaction,
2H2S + 3O2 = 2H2O + 2SO2
ΔH° is calculated as,
ΔH° = (2 x ΔH°SO2 + 2 x ΔH°H2O) - (2 x ΔH°H2S - 3 x ΔH°O2)
= (2 mol x -296.8 kJ mol^-1 + 2 mol x -242 kJ mol^-1) - ( 2 mol x -21 kJ mol^-1 + 3 x 0 kJ mol-1) = -1035.6 kJ
ΔS° is calculated as,
ΔS° = (2 mol x S°SO2 + 2 mol x S°H2O) - ( 2 mol x S°H2S + 3 mol x S°O2)
= ( 2 mol x 248 JK-1 mol^-1 + 2 x 189 JK-1 mol^-1) - ( 2 x 206 JK-1 mol^-1 + 3 mol x 205 JK-1 mol^-1)
= -153 J K^-1= - 0.153 kJ K^-1
Therefore,
ΔG° = ΔH° - T ΔS°
= -1035.6 kJ - ( 500 K x - 0.153 kJ K-1)
= -1035.6 kJ + 76.5 kJ
= -959.1 kJ
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