The initial concentration of both the reactants of a second order reaction are equal and `60%` of the reaction gets completed in `30s`. How much time will be taken in `20%` completion of the reaction?

Answer

5

Explanation

For second order:
`k_(2) = (1)/(t).(x)/(a(-x))` , [Let `a=1`]
`= (1)/(30 s) xx (0.6)/(1(1-0.6)) = (1)/(30) xx (0.6)/(0.4)`
Now for `20%` completion,
`k_(2) = (1)/(t).(x)/(a(1-x))`
(Since `k_(2)` is constant)
`(1)/(30) xx (0.6)/(0.6) = (1)/(t) xx (1)/(4)`
`t = (30)/(0.6) xx (0.4)/(4) = 5s`