Time required for $99.9 \%$ completion of a first order reaction is _______ times the time required for completion of $90 \%$ reaction.(nearest integer)

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{K}=\frac{1}{\mathrm{t}_{99.9 \%}} \ln \left(\frac{100}{0.1}ight)=\frac{1}{\mathrm{t}_{90 \%}} \ln \left(\frac{100}{10}ight) \ & \mathrm{t}_{99.9 \%}=\mathrm{t}_{90 \%} \frac{\ell \mathrm{n}\left(10^3ight)}{\ell \mathrm{n} 10} \ & \mathrm{t}_{99.9 \%}=\mathrm{t}_{90 \%} 	imes 3\end{aligned}$

Time required for \(99.9 \%\) completion of a first order reaction is _______ times the time required for…

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