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