The aqueous Finkelstein reaction C_4H_9Br + NaI → C_4H_9I + NaBr has a standard Gibbs free energy change of Δ G° = +4.0 kJ,mol^-1 at 298 K. Calculate the equilibrium constant K at this temperature.

The aqueous Finkelstein reaction $C_{4} H_{9} Br + NaI \to C_{4} H_{9} I + NaBr$ has a standard Gibbs free energy change of $Δ G^{\circ} = + 4.0 kJ, mo l^{- 1}$ at 298 K. Calculate the equilibrium constant $K$ at this temperature.

medium 3 marks Class 12 · Chemistry · Haloalkanes and Haloarenes

Answer

K = 0.20

The correct answer is $K = 0.20$ — 7. Thus $K a pp r o x 0.20$ .

1. Convert $D e l t a G^{c} i r c$ to joules: $4.0 kJ, mo l^{- 1} = 4000 J, mo l^{- 1}$ .
2. Compute $R T$ at 298 K: $R = 8.314 J, mo l^{- 1} K^{- 1}$ , so $R T = 8.314 \times 298 = 2477.3 J, mo l^{- 1}$ .
3. Use the relation $D e l t a G^{c} i r c = - R T l n K$ which gives $l n K = - D e l t a G^{c} i r c / (R T)$ .
4. Calculate $l n K = - 4000/2477.3 = - 1.614$ .
5. Exponentiate: $K = e^{- 1.614} = 0.199 a pp r o x 0.20$ .
6. Round to two significant figures as required.
7. Thus $K a pp r o x 0.20$ .

Common mistake: Using the wrong sign in the exponent or neglecting unit conversion

Concept tested: Equilibrium constant from ΔG°

Published 7 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked