The first‑order hydrolysis of allyl bromide ($\mathrm{C}{\mathrm{H}}_2=\mathrm{CHC}{\mathrm{H}}_2\mathrm{Br}$) in 0.20 M NaOH at 25 °C has a half‑life of $4.2\times 10^2\text{s}$. (a) Calculate the rate constant $k$ in ${\mathrm{s}}^{-1}$. (b) If the initial concentration of allyl bromide is $0.15\text{M}$, find its concentration after $1.2\times 10^3\text{s}$.

Step-by-step solution

1. For a first‑order reaction, $t_{1/2}=\frac{\ln 2}{k}$.
2. Re‑arrange to obtain $k=\frac{\ln 2}{t_{1/2}}$.
3. Insert $t_{1/2}=4.2\times 10^2\text{s}$: k = \frac{0.693}{4.2\times10^{2}} = 1.65\times10^{-3}\ \text{s^{-1}}.
4. The integrated first‑order law is $[A{]}_t=[A{]}_0{e}^{-kt}$.
5. Insert $[A{]}_0=0.15\text{M}$, k = 1.65\times10^{-3}\ \text{s^{-1}} and $t=1.2\times 10^3\text{s}$:
6. $[A{]}_t=0.15e^{-(1.65\times 10^{-3})(1.2\times 10^3)}=0.15e^{-1.98}=0.15\times 0.300=4.5\times 10^{-2}\text{M}$.

Concept tested: First‑order kinetics and half‑life

Published 15 September 2026 · How this solution was prepared and checked