The rate constant of a reaction with a virus is $3.3 \times 10^{-4} \mathrm{~s}^{-1}$. Time required for the…

The rate constant of a reaction with a virus is $3.3 \times 10^{-4} \mathrm{~s}^{-1}$. Time required for the virus to become $75 \%$ inactivated is
  1. $35 \mathrm{~min}$
  2. $70 \mathrm{~min}$
  3. $105 \mathrm{~min}$
  4. $17.5 \mathrm{~min}$

Solution

$t_{1 / 2}=\frac{0.693}{k}=2100 \mathrm{~s}=35 \mathrm{~min}$
$t_{75 \%}=2 t_{1 / 2}=2 \times 35=70 \mathrm{~min}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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