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Q. $99\%$ at a first order reaction was completed in $32$ min. When will $99.9\%$ of the reaction complete?

Chemical Kinetics

Solution:

$k\times32=ln\left(\frac{100}{1}\right) .....\left(i\right)$
$k_{2}\times t=ln\left(\frac{100}{0.1}\right) ...\left(ii\right)$
$eq\left(ii\right)\left(i\right)$
$\frac{t}{32}=\frac{3\, ln\, 10}{2\, ln\, 10}$
$t=48 min$