The rate of a first order reaction is `0.69xx10^(-2) min^(-1)` and the initial concentration is 0.5 mol `L^(-1)` . The half life period is

Options

A. 3000 s
B. 0.33 s
C. 50 s
D. 100 s

Answer

A

Explanation

(A) Rate = k[C]
`0.69xx10^(-2) = k[0.5]`
or `k = (0.60xx10^(-2))/(0.5)`
`t_(1//2) = (0.693)/(k) = (0.693xx0.5)/(0.693xx10^(2))`
50 min = 3000 sec