The half‑life of a first‑order reaction at 350 K is found to be 9 min. Calculate the rate constant $k$ (in s${}^{-1}$).

Step-by-step solution

1. Use the first‑order half‑life relation $t_{1/2}=\frac{ln2}{k}$.
2. Rearrange to find $k=\frac{ln2}{t_{1/2}}$.
3. Convert the half‑life to seconds: $9\text{min}=540\text{s}$.
4. Insert values: $k=\frac{0.693}{540},{\text{s}}^{-1}$.
5. Compute: $kapprox1.28\times 10^{-3},{\text{s}}^{-1}\approx 1.3\times 10^{-3},{\text{s}}^{-1}$.
6. Thus the rate constant is $1.3\times 10^{-3},{\text{s}}^{-1}$.

Concept tested: First‑order half‑life relation

Published 13 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked