$20 \%$ of a first order reaction was found to be completed at $10: 00 \mathrm{a} . \mathrm{m}$ at $11.30 \mathrm{a} . \mathrm{m}$. on the same day, $20 \%$ of the reaction was found to be remaining. The half life period in minutes of the reaction is

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Aicharya · Accepted Answer

For $t=90 \mathrm{~min}$ (from 10 a.m. to $11: 30$ a.m.) As $20 \%$ was already completed $	herefore a=80$ $$ \begin{aligned} & (a-x)=(100-80)=20 \ & \because \quad k=\frac{2.303}{t} \log \frac{a}{a-x} \ & k=\frac{2.303}{90} \log \frac{80}{20} \ & k=\frac{2.303}{90} \log 4=\frac{2.303 	imes 0.6020}{90} \ & k=0.015 \mathrm{~min} \end{aligned} $$ $\begin{aligned} & \because t_{1 / 2}=\frac{0.693}{k} \ & \Rightarrow t_{1 / 2}=\frac{0.693}{0.015}=46.2 \approx 45 \mathrm{~min}\end{aligned}$ ~

$20 \%$ of a first order reaction was found to be completed at $10: 00 \mathrm{a} . \mathrm{m}$ at $11.30…

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