`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

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