`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. `0.75//k`
B. `0.69//k`
C. `029//k`
D. `0.10//k`

Answer

C

Explanation

`{:(,A , to , "Product"), ("Concentration intially" ,a, , 0),("Concentration after" , (a-x),,x),("time t" ,,,), ("Concentration after",,,), ("t_(1//4) , (a - (a)/(4)), , (a)/(4)):}`
For the first order kinetics ,
`K = (2.303)/(t) "log" ((a)/(a-x)) = (2.303)/(t_(1//4)) "log" (a)/((3a)/(4))`
`t_(1//4) = (2.303 log (4)/(3))/(k) = (0.29)/(k)`