Enter ΔH in kJ/mol, the temperature in kelvin and ΔS in J/mol·K to get ΔG = ΔH − TΔS, and with it whether the process is spontaneous at that temperature.
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From the article Gibbs Free Energy ΔG = ΔH − TΔS — Spontaneity Explained.
For N₂(g) + 3H₂(g) → 2NH₃(g): ΔH° = −92.4 kJ mol⁻¹ and ΔS° = −198.3 J K⁻¹ mol⁻¹. Find ΔG° at 298 K.
First fix the units: ΔS° = −198.3 J K⁻¹ mol⁻¹ = −0.1983 kJ K⁻¹ mol⁻¹
TΔS = 298 × (−0.1983) = −59.09 kJ mol⁻¹
ΔG° = ΔH° − TΔS = (−92.4) − (−59.09) = −92.4 + 59.09 = −33.3 kJ mol⁻¹
Negative, so ammonia formation is spontaneous at 298 K. Watch the double negative in the subtraction — that is where careless working goes wrong.
Worked in full in Gibbs Free Energy ΔG = ΔH − TΔS — Spontaneity Explained.
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