If a second order reaction involving a single reactant in `25%` complete in `20` minutes then what is its half life if initial concentration of the reactant is being made double at the same temperature.

Options

A. `50` min
B. `30` min
C. `60` min
D. `15` min

Answer

B

Explanation

`(1)/(C ) - (1)/(C_(0)) = kt` (for `2^(nd)` order reaction)
`(1)/((3)/(4)C_(0)) - (1)/(C_(0)) = k xx 20`
`(1)/((1)/(2)(2C_(0))) - (1)/(2C_(0)) = k xx t_(1//2)`
`rArr ((1)/(3C_(0)))/((1)/(2C_(0))) = (20)/(t_(1//2)) rArr (2)/(3) = (20)/(t_(1//2))`
`rArr t_(1//2) = 30` minutes