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
- $35 \mathrm{~min}$
- $70 \mathrm{~min}$
- $105 \mathrm{~min}$
- $17.5 \mathrm{~min}$
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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