Gibbs Free Energy And Equilibrium

Question

The standard Gibbs free energy change (ΔG°) for a reaction at 298 K is −5.70 kJ mol⁻¹. The equilibrium constant K for this reaction at the same temperature is approximately: (R = 8.314 J K⁻¹ mol⁻¹)

RankGuru · Accepted Answer

1. **Fundamental Relation**: $\Delta G° = -RT \ln K$.
2. **Rearrange**: $\ln K = \frac{-\Delta G°}{RT}$.
3. **Substitute**: $\ln K = \frac{-(-5700)}{8.314 	imes 298} = \frac{5700}{2477.6} = 2.301$.
4. **Solve for K**: $K = e^{2.301} \approx 10.0$.
5. **Alternative Approach**: Using $\Delta G° = -2.303 RT \log K$:
   $\log K = \frac{5700}{2.303 	imes 8.314 	imes 298} = \frac{5700}{5705.8} \approx 1.0$,
   so $K = 10^1 = 10$.
6. **Physical Meaning**: A moderately negative ΔG° gives K = 10, indicating products are moderately favoured at equilibrium — the reaction proceeds forward but not overwhelmingly so.

Gibbs Free Energy And Equilibrium

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