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A reaction has rate constant $k=2.4 \times 10^{-4} \mathrm{~s}^{-1}$. Then, find the ratio of $t_{99.9}$ to $t_{50}$.
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Verified Answer
The correct answer is:
10
Unit of $k$ is $\mathrm{s}^{-1}$.
$\therefore$ It is first order reaction
$$
\begin{aligned}
t_{99.9 \%} & =\frac{2.303}{k} \log \frac{100}{0.1} \\
& =\frac{2.303}{k} \log 1000=\frac{2.303}{k} \log 10^3 \\
& =\frac{2.303}{k} \times 3 \log 10=\frac{2.303}{k} \times 3 \quad[\because \log 10=1] \\
t_{50 \%} & =\frac{0.693}{k}
\end{aligned}
$$
Ratio of $t_{99.9 \%}$ and $t_{50 \%}$ is $\frac{t_{99.9 \%}}{t_{50 \%}}=\frac{3 \times 2.303 / k}{0.693 / k}=10$.
$\therefore$ It is first order reaction
$$
\begin{aligned}
t_{99.9 \%} & =\frac{2.303}{k} \log \frac{100}{0.1} \\
& =\frac{2.303}{k} \log 1000=\frac{2.303}{k} \log 10^3 \\
& =\frac{2.303}{k} \times 3 \log 10=\frac{2.303}{k} \times 3 \quad[\because \log 10=1] \\
t_{50 \%} & =\frac{0.693}{k}
\end{aligned}
$$
Ratio of $t_{99.9 \%}$ and $t_{50 \%}$ is $\frac{t_{99.9 \%}}{t_{50 \%}}=\frac{3 \times 2.303 / k}{0.693 / k}=10$.
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