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In a first order reaction, the concentration of the reactant is reduced from $0.6 \mathrm{~mol} \mathrm{~L}^{-1}$ to $0.2 \mathrm{~mol} \mathrm{~L}^{-1}$ in $5 \mathrm{~min}$. What is the rate constant of the reaction? $(\log 3=0.4771)$
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Verified Answer
The correct answer is:
$0.219 \mathrm{~min}^{-1}$
Given,
$$
\begin{aligned}
t & =5 \text { minute } \\
A^{\circ} & =0.6 \mathrm{~mol}^{-1} \\
A & =0.2 \mathrm{~mol}^{-1} \\
k & =\frac{2.303}{t} \log \frac{A^{\circ}}{A} \\
& =\frac{2.303}{5} \log \frac{0.6}{0.2}=0.219 \mathrm{~min}^{-1}
\end{aligned}
$$
$$
\begin{aligned}
t & =5 \text { minute } \\
A^{\circ} & =0.6 \mathrm{~mol}^{-1} \\
A & =0.2 \mathrm{~mol}^{-1} \\
k & =\frac{2.303}{t} \log \frac{A^{\circ}}{A} \\
& =\frac{2.303}{5} \log \frac{0.6}{0.2}=0.219 \mathrm{~min}^{-1}
\end{aligned}
$$
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