`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

b) `t_(1//2)= (2.303)/k log 1/(3//4)`
`=2.303/k log 4//3`
`=(2.303 xx 0.1249)/(k) = 0.29/k`