1. Tardigrade
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  3. Chemistry
  4. The time taken for 90% of a first order reaction to complete is approximately

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Q. The time taken for 90% of a first order reaction to complete is approximately

Chemical Kinetics

Solution:

$t 90 \%=\frac{2.303}{k} \log \frac{a}{a-0.9\,a} = \frac{2.303}{k}\log 10$
$\frac{2.303}{k}$
$t_{1/2}=\frac{2.303}{k}\log\frac{a}{a-a/2}=\frac{2.303}{k}\log 2$
$=\frac{2.303}{k}\times 0.3010$
$t 90 \% \: t_{1/2}=\frac{1}{0.3010}=3.3$
i,e.,$ t90\%=3.3 \times t_{1/2}$