1.4.4 (HL)—Equilibrium and Gibbs energy

At equilibrium for the stated reaction and temperature, Q = K and the current ΔG = 0. Substitution into ΔG = ΔG° + RT ln Q shows that ΔG° is generally not zero; it equals −RT ln K. Away from equilibrium, Q < K favours the forward direction and Q > K favours the reverse. Compare Q and K only for the same balanced reaction orientation and fixed temperature.

Compare Q with K to predict the immediate direction: Q < K gives ΔG < 0 for the forward reaction, while Q > K gives ΔG > 0 and favours the reverse. Keep ΔG for the current composition distinct from ΔG°, which describes standard-state reactants and products and fixes K at that temperature.

Worked $K$ and $Q$ example: for ammonia synthesis at 298 K, the local course book gives $\Delta G^\circ=-31.8\,\mathrm{kJ\,mol^{-1}}=-31800\,\mathrm{J\,mol^{-1}}$. From $\ln K=-\Delta G^\circ/(RT)$, $K=3.77\times10^5$, so products are favoured at equilibrium. If $Q=1.0\times10^6$, $\Delta G=-31800+(8.31)(298)\ln(10^6)=+2410\,\mathrm{J\,mol^{-1}}=+2.41\,\mathrm{kJ\,mol^{-1}}$; the forward reaction is then non-spontaneous because the current mixture has $Q>K$.