`t_(1//4)` can be taken as the time taken for concentration of reactant to drop to `.^(3)//_(4)` of its initial value. If the rate constant for a first order reaction is `K`, then `t_(1//4)` can be written as:

Options

A. `0.10//k`
B. `0.29//k`
C. `0.69//k`
D. `0.75//k`

Answer

B

Explanation

`t_(1//4)=(2.303)/(k)"log"(a)/((3//4)a)=(2.303)/(k)"log"(4)/(3)`
`=(2.303xx0.125)/(k)=(0.29)/(k)`.