For a given reaction of first order, it takes `20` minutes for the concentration to drop from `1.0 "mol liter"^(-1)` to `0.6 "mol litre"^(-1)`. The time required for the concentration to drop from `0.6 "mol litre"^(-1)` to `0.36 "mol litre"^(-1)` will be

Options

A. more than `20` minutes
B. less than `20` minutes
C. equal to `20` minutes
D. infinitly

Answer

C

Explanation

`(2.303)/(20) "log"(1.0)/(1.0 - 0.6) = (2.303)/(t)"log"(0.6)/(0.6 - 0.36)`
`:.` or `(1)/(20) log 2.5 = (1)/(t) log 2.5`
`:. t = 20` minutes