In a Finkelstein reaction, 1‑bromobutane (${\mathrm{C}}_4{\mathrm{H}}_9\mathrm{Br}$) is treated with excess sodium iodide in dry acetone at 298 K. The standard Gibbs free energy change for the reaction ${\mathrm{C}}_4{\mathrm{H}}_9\mathrm{Br}+\mathrm{NaI}\to {\mathrm{C}}_4{\mathrm{H}}_9\mathrm{I}+\mathrm{NaBr}$ is +5.8\ \text{kJ mol^{-1}}. (a) Calculate the equilibrium constant $K$ at this temperature. (b) Explain why the reaction still gives a quantitative yield of ${\mathrm{C}}_4{\mathrm{H}}_9\mathrm{I}$ despite $K<1$.

Step-by-step solution

1. Use the relation $\Delta G^{\circ }=-RT\ln K$.
2. Convert $\Delta G^{\circ }$ to joules: +5.8\ \text{kJ mol^{-1}} = +5800\ \text{J mol^{-1}}.
3. Substitute R = 8.314\ \text{J mol^{-1}K^{-1}} and $T=298\text{K}$: -RT = -8.314\times298 = -2477\ \text{J mol^{-1}}.
4. Rearrange to $\ln K=-\Delta G^{\circ }/RT=-5800/2477=-2.34$.
5. Exponentiate: $K=e^{-2.34}\approx 0.09$.
6. Although $K<1$, NaBr is insoluble in acetone and precipitates out, removing it from the reaction mixture.
7. According to Le Chatelier’s principle, removal of a product shifts the equilibrium to the right, giving an essentially quantitative conversion to ${\mathrm{C}}_4{\mathrm{H}}_9\mathrm{I}$.

Concept tested: Equilibrium shift in Finkelstein reaction

Published 4 September 2026 · How this solution was prepared and checked