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

Options

A. `76 min`
B. `66 min`
C. `56 min`
D. `46 min`

Answer

D

Explanation

For the first order reaction for small finite change
`k_(1) = (1)/([A])(Delta[Delta])/(Delta t) rArr (Delta[A]//[A])/(Delta t) = 1.5% min^(-1)`
`= 0.015 min^(-1)`
`t_(1//2) = (0.693)/(0.015 min^(-1)) = 46.2 min ~~ 46 min`