A first order reaction is $75 \%$ completed in 60 minutes, the time required for it's $50 \%$ completion is.
A first order reaction is $75 \%$ completed in 60 minutes, the time required for it's $50 \%$ completion is.
$120 \mathrm{~min}$
$60 \mathrm{~min}$
$40 \mathrm{~min}$
$30 \mathrm{~min}$
Solution
i. $[\mathrm{A}]_{0}=100, \quad[\mathrm{~A}]_{\mathrm{t}}=100-75=25, \mathrm{t}=60 \mathrm{~min}$
For first order reaction,
$\mathrm{k}=\frac{2.303}{\mathrm{t}} \log _{10} \frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]_{t}}$
$\therefore \mathrm{k}=\frac{2.303}{60 \mathrm{~min}} \log _{10} \frac{100}{25}$
$\mathrm{k}=\frac{2.303 \times 0.6020}{60 \mathrm{~min}}=0.0231 \mathrm{~min}^{-1}$
ii. The time required for $50 \%$ completion of reaction is,
$\mathrm{t}_{1 / 2}=\frac{0.693}{0.0231 \mathrm{~min}^{-1}}=30 \mathrm{~min}$