If the rate constant for a first order reaction is $k$, then find the time required for completion of $80 \%$ of the reaction.

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Aicharya · Accepted Answer

First order rate constant is given as $k=\frac{2.303}{t} \log \frac{\left[A_0ight]}{\left[A_tight]}$ $80 \%$ completed, then $\left[A_0ight]=100$ $A_t=100-80=20$ $\begin{aligned} & k=\frac{2.303}{t} \log \frac{100}{20} \ & k=\frac{2.303}{t} \log \frac{10}{2} \ & k=\frac{2.303}{t}(\log 10-\log 2) \ & k=\frac{2.303}{t}(1-0.3) \ & k=\frac{2.303 	imes 0.7}{t} \Rightarrow t=\frac{1.6}{k} .\end{aligned}$

If the rate constant for a first order reaction is $k$, then find the time required for completion of $80…

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