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Calculate the rate constant of the first order reaction if $80 \%$ of the reactant decomposes in 60 minutes.
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The correct answer is:
$2.68 \times 10^{-2}$ minute $^{-1}$
$\begin{aligned} \mathrm{k} & =\frac{2.303}{\mathrm{t}} \log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}} \\ & =\frac{2.303}{60} \log _{10} \frac{(100)}{(20)}=\frac{2.303}{60} \log _{10} \frac{10}{2} \\ & =\frac{2.303}{60}\left[\log _{10}(10)-\log _{10}(2)\right] \\ & =\frac{2.303}{60} \times 0.699=2.68 \times 10^{-2} \text { minute }^{-1}\end{aligned}$
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