Q4. ΔG° and ΔS° from an equilibrium constant
At 1200 K, the equilibrium constant for the reaction A₂ → 2A is found to be 3.3 × 10⁻³. Calculate ΔG° and ΔS° for this reaction at 1200 K. Given that ΔH° for the reaction is 216.65 kJ mol⁻¹.
Answer in brief
ΔG° = −RT ln K = +57.0 kJ mol⁻¹; ΔS° = (ΔH° − ΔG°)/T = +133.0 J K⁻¹ mol⁻¹.
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Coversstandard Gibbs energyequilibrium constantΔG° = −RT ln Kentropy changevan 't Hoff equation