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The time required for a first order reaction to complete $90 \%$ is ' $t$ '. What is the time required to complete $99 \%$ of the same reaction?
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Verified Answer
The correct answer is:
$2t$
For the first order reaction,
$t=\frac{2303}{k} \log \frac{a}{a-x}$
Now putting the values
$\begin{aligned}
& t_{90 \%}=t=\frac{2303}{k} \log \left(\frac{100}{100-90}\right) \\
& t_{99 \%}=\frac{2303}{k} \log \left(\frac{100}{100-99}\right) \\
& \frac{t}{t_{99 \%}}=\frac{\frac{2303}{k} \log \left(\frac{100}{10}\right)}{\frac{2303}{k} \log \left(\frac{100}{1}\right)}=\frac{1}{2} \\
& t_{99 \%}=2 t
\end{aligned}$
$t=\frac{2303}{k} \log \frac{a}{a-x}$
Now putting the values
$\begin{aligned}
& t_{90 \%}=t=\frac{2303}{k} \log \left(\frac{100}{100-90}\right) \\
& t_{99 \%}=\frac{2303}{k} \log \left(\frac{100}{100-99}\right) \\
& \frac{t}{t_{99 \%}}=\frac{\frac{2303}{k} \log \left(\frac{100}{10}\right)}{\frac{2303}{k} \log \left(\frac{100}{1}\right)}=\frac{1}{2} \\
& t_{99 \%}=2 t
\end{aligned}$
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