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Calculate the amount of reactant in percent that remains after 60 minutes involved in first order reaction. $\left(\mathrm{k}=0.02303\right.$ minute $\left.^{-1}\right)$
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The correct answer is:
$25 \%$
For a first order reaction,
$\begin{aligned}
& k=\frac{0.693}{t_{1 / 2}} \\
& t_{1 / 2}=\frac{0.693}{0.02303}=30 \mathrm{~min}
\end{aligned}$
Percent of reactant that remains after $t_{1 / 2}=50 \%$. $2 \times \mathrm{t}_{1 / 2}=60 \mathrm{~min}$
Therefore, percent of reactant that remains after $2 \mathrm{t}_{1 / 2}=25 \%$
$\begin{aligned}
& k=\frac{0.693}{t_{1 / 2}} \\
& t_{1 / 2}=\frac{0.693}{0.02303}=30 \mathrm{~min}
\end{aligned}$
Percent of reactant that remains after $t_{1 / 2}=50 \%$. $2 \times \mathrm{t}_{1 / 2}=60 \mathrm{~min}$
Therefore, percent of reactant that remains after $2 \mathrm{t}_{1 / 2}=25 \%$
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