`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. 0.29/k
D. 0.10/k

Answer

C

Explanation

`{:(,A,to,"Product"),("initially",a,,0),("after time t",(a-x),,x),("after "t_(1//4),(a-(a)/(4)),,(a)/(4)):}`
For first-order kinetic
`k=(2.303)/(t)"log"((a)/(a-x))`
`k=(2.303)/(t_(1//4))"log"(a)/((3a)/(4))`
`t_(1//4)=(2.303"log"(4)/(3))/(k)`
`=(2.303(log4-log3))/(k)`
`=(2.303(0.6020-04771))/(k)`
`=(2.303xx0.1249)/(k)`
`=(0.2876)/(k)`
`=(0.29)/(k)`