A reaction has rate constant $k=2.4 imes 10^{-4} \mathrm{~s}^{-1}$. Then, find the ratio of $t_{99.9}$ to $t_{50}$.

Question

A reaction has rate constant $k=2.4 	imes 10^{-4} \mathrm{~s}^{-1}$. Then, find the ratio of $t_{99.9}$ to $t_{50}$.

Aicharya · Accepted Answer

Unit of $k$ is $\mathrm{s}^{-1}$. $	herefore$ 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} 	imes 3 \log 10=\frac{2.303}{k} 	imes 3 \quad[\because \log 10=1] \ t_{50 \%} & =\frac{0.693}{k} \end{aligned} $ Ratio of $t_{99.9 \%}$ and $t_{50 \%}$ is $\frac{t_{99.9 \%}}{t_{50 \%}}=\frac{3 	imes 2.303 / k}{0.693 / k}=10$

A reaction has rate constant $k=2.4 \times 10^{-4} \mathrm{~s}^{-1}$. Then, find the ratio of $t_{99.9}$ to…

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