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 )

Solution :
For the first order reaction for small finite change
$${k_1} = \frac{1}{{\left[ A \right]}}\frac{{\Delta \left[ A \right]}}{{\Delta t}} \Rightarrow \frac{{\frac{{\Delta \left[ A \right]}}{{\left[ A \right]}}}}{{\Delta t}} = 1.5\% $$       \[{{\min }^{-1}}\]
\[=0.015\,{{\min }^{-1}}\]
\[{{t}_{\frac{1}{2}}}=\frac{0.693}{0.015\,{{\min }^{-1}}}=46.2\,\min \approx 46\,\min \]