1. Tardigrade
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  3. Chemistry
  4. Ratio of time required for 99 % and 90 % completion for a first order reaction is

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Q. Ratio of time required for $99 \%$ and $90 \%$ completion for a first order reaction is

Solution:

$t_{99}=\frac{2.303}{k} \log \frac{100}{100-99}=\frac{2.303}{k} \times 2$
$t_{90}=\frac{2.303}{k} \log \frac{100}{100-90}=\frac{2.303}{k} \times 1$
$\frac{t_{99}}{t_{90}}=\frac{2}{1}$