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In a first order reaction, concentration of reactant is reduced to $(1 / 8)^{\text {th }}$ of concentration in 23.03 minutes. What is half-life period of reaction?
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Verified Answer
The correct answer is:
$7.7 \mathrm{~min}$
Initial concentration $\mathrm{a}_0=\mathrm{a}$ (let)
Final concentration $a_t=\frac{a}{8}$
$1^{\text {st }}$ order reaction
$$
\begin{aligned}
& \mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{a}_0}{\mathrm{a}_{\mathrm{t}}}=\frac{2.303}{23.03} \log \frac{\mathrm{a}}{\mathrm{a} / 8} \\
& \mathrm{k}=\frac{1}{10} \log 8=\frac{3}{10} \log 2=0.3 \log 2
\end{aligned}
$$
Half -life $\mathrm{t}_{1 / 2}=\frac{2.303 \log 2}{\mathrm{k}}$
$$
\begin{aligned}
& =\frac{2.303 \times \log }{0.3 \log 2} \\
& =7.7 \mathrm{~min}
\end{aligned}
$$
Final concentration $a_t=\frac{a}{8}$
$1^{\text {st }}$ order reaction
$$
\begin{aligned}
& \mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{a}_0}{\mathrm{a}_{\mathrm{t}}}=\frac{2.303}{23.03} \log \frac{\mathrm{a}}{\mathrm{a} / 8} \\
& \mathrm{k}=\frac{1}{10} \log 8=\frac{3}{10} \log 2=0.3 \log 2
\end{aligned}
$$
Half -life $\mathrm{t}_{1 / 2}=\frac{2.303 \log 2}{\mathrm{k}}$
$$
\begin{aligned}
& =\frac{2.303 \times \log }{0.3 \log 2} \\
& =7.7 \mathrm{~min}
\end{aligned}
$$
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