In a first-order reaction $\mathrm{A} ightarrow \mathrm{B}$, if $\mathrm{k}$ is rate constant and initial…
- $\frac{\log 2}{\mathrm{k}}$
- $\frac{\log 2}{\mathrm{k} \sqrt{0.5}}$
- $\frac{\ln 2}{k}$
- $\frac{0.693}{0.5 \mathrm{k}}$
Solution
$k=\frac{2.303}{t} \log _{10} \frac{a}{a-x}$
when $\mathrm{t}=\mathrm{t}_{1 / 2}$
$k=\frac{2.303}{t_{1 / 2}} \log _{10} \frac{a}{a-a / 2}$
or $\mathrm{t}_{1 / 2}=\frac{2.303}{\mathrm{k}} \log _{10} 2=\frac{\ln 2}{\mathrm{k}}$
Asked in: JEE-TOPICTESTS-CHEMISTRY