The aqueous hydrolysis of tert-butyl bromide ( $(C H_{3})_{3} CBr$ ) at 30°C follows first‑order kinetics with a half‑life of $2.5 \times 1 0^{3} s$ . Calculate the rate constant $k$ in $s^{- 1}$ .

easy 2 marks Class 12 · Chemistry · Haloalkanes and Haloarenes

Answer

k = 2.77 \times 10^{-4}\ \text{s^{-1}}

The correct answer is $k = 2.77 \times 10^{-4}\ \text{s^{-1}}$ — Calculate $k = \frac{0.693}{2.5\times10^{3}} = 2.77\times10^{-4}\ \text{s^{-1}}$ .

Step-by-step solution

Write the first‑order half‑life expression: $t_{1/2} = \frac{l n 2}{k}$ .
Rearrange to solve for $k$ : $k = \frac{l n 2}{t _{1/2}}$ .
Substitute $ln 2 = 0.693$ and $t_{1/2} = 2.5 \times 1 0^{3} s$ .
Calculate $k = \frac{0.693}{2.5\times10^{3}} = 2.77\times10^{-4}\ \text{s^{-1}}$ .

Common mistake: Using

t_{1/2} = 1/ k

(zero‑order) instead of

ln 2/ k

Concept tested: First-order kinetics of SN1 hydrolysis

Published 5 September 2026 · How this solution was prepared and checked

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The aqueous hydrolysis of tert-butyl bromide ((CH3​)3​CBr) at 30°C follows first‑order kinetics with a half‑life of 2.5×103 s. Calculate the rate constant k in s−1.

Step-by-step solution

The aqueous hydrolysis of tert-butyl bromide ( $(C H_{3})_{3} CBr$ ) at 30°C follows first‑order kinetics with a half‑life of $2.5 \times 1 0^{3} s$ . Calculate the rate constant $k$ in $s^{- 1}$ .