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If the rate constant for a first order reaction is $k$, the time $(t)$ required for the completion of $99 \%$ of the reaction is given by
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Verified Answer
The correct answer is:
$t=\frac{4.606}{k}$
Given, for a first order reaction, let initial concentration $=\left[R_{0}\right]$
After $99 \%$ completion, concentration
$\begin{aligned}
{[R] } &=R_{0}-0.99 R_{0} \\
&=0.1 R_{0}
\end{aligned}$
Using formula, $t=\frac{2303}{k} \log \frac{\left[R_{0}\right]}{[R]}$
$=\frac{2303}{k} \log \frac{100}{1}=\frac{4.606}{k}$
After $99 \%$ completion, concentration
$\begin{aligned}
{[R] } &=R_{0}-0.99 R_{0} \\
&=0.1 R_{0}
\end{aligned}$
Using formula, $t=\frac{2303}{k} \log \frac{\left[R_{0}\right]}{[R]}$
$=\frac{2303}{k} \log \frac{100}{1}=\frac{4.606}{k}$
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