The half life for the virus inactivation if in the beginning $1.5 \%$ of the virus is inactivated per minute is (Given: The reaction is of first order)

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Aicharya · Accepted Answer

For the first order reaction for small finite change
$k_{1}=\frac{1}{[\mathrm{~A}]} \frac{\Delta[\mathrm{A}]}{\Delta t} \Rightarrow \frac{\Delta[A] /[A]}{\Delta t}=1.5 \% \mathrm{~min}^{-1}$
$=0.015 \mathrm{~min}^{-1}$
$t_{1 / 2}=\frac{0.693}{0015 \mathrm{~min}^{-1}}=46.2 \mathrm{~min} \approx 46 \mathrm{~min}$

The half life for the virus inactivation if in the beginning $1.5 \%$ of the virus is inactivated per minute…

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