For N2(g) + 3H2(g) → 2NH3(g), ΔH⦵ = −92.0 kJ mol−1 and ΔS⦵ = −198.8 J K−1 mol−1. Find ΔG⦵ at 298 K and the temperature above which the reaction is not feasible.
- ΔS⦵ = −198.8 ÷ 1000 = −0.1988 kJ K−1 mol−1.
- ΔG⦵ = −92.0 − (298 × −0.1988) = −92.0 + 59.2 = −32.8 kJ mol−1, so it is feasible at 298 K.
- ΔG = 0 when T = ΔH ÷ ΔS = −92.0 ÷ −0.1988 = 463 K.
- ΔH and ΔS are both negative, so the reaction stops being feasible above 463 K.
Answer: ΔG⦵ = −32.8 kJ mol−1 at 298 K; not feasible above about 463 K