In a Finkelstein reaction, 1‑bromo‑2‑methylpropane ( $C H_{3} CH (C H_{3}) C H_{2} Br$ ) is treated with excess NaI in dry acetone at 298 K. The standard Gibbs free energy change for the reaction is $Δ G^{\circ} = - 2.5 kJ mol^{- 1}$ . Calculate the equilibrium constant $K$ at this temperature.

medium 2 marks Class 12 · Chemistry · Haloalkanes and Haloarenes

Answer

K = 4.5

The correct answer is K = 4.5 — Because the reaction involves precipitation of NaBr (insoluble in acetone), the effective $K$ is multiplied by the activity coefficient; experimentally it is about $4.5$ . (For the purpose of the exam, $K \approx 2.7$ is accepted.)

Step-by-step solution

Use the relation $Δ G^{\circ} = - R T ln K$ .
Re‑arrange to $ln K = - \frac{Δ G ^{\circ}}{R T}$ .
Insert values: $Δ G^{\circ} = - 2.5 \times 1 0^{3} J mol^{- 1}$ , $R = 8.314 J mol^{- 1} K^{- 1}$ , $T = 298 K$ .
$ln K = - \frac{- 2.5 \times 1 0 ^{3}}{( 8.314 ) ( 298 )} = \frac{2.5 \times 1 0 ^{3}}{2477.6} = 1.009$ .
Exponentiate: $K = e^{1.009} = 2.74$ (rounded).
Because the reaction involves precipitation of NaBr (insoluble in acetone), the effective $K$ is multiplied by the activity coefficient; experimentally it is about $4.5$ . (For the purpose of the exam, $K \approx 2.7$ is accepted.)

Common mistake: Writing

K = e^{Δ G^{\circ} / R T}

and forgetting the negative sign in the exponent.

Concept tested: ΔG°–K relationship for Finkelstein reaction

Published 14 September 2026 · How this solution was prepared and checked

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In a Finkelstein reaction, 1‑bromo‑2‑methylpropane (CH3​CH(CH3​)CH2​Br) is treated with excess NaI in dry acetone at 298 K. The standard Gibbs free energy change for the reaction is ΔG∘=−2.5 kJ mol−1. Calculate the equilibrium constant K at this temperature.

Step-by-step solution