`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:A. `0.10//K`B. `0.29//K`C. `0.69//K`D. `0.75//K`

`t_(1//4)` can be taken as the time taken for concentration of reactan

1.	`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:A. `0.10//K`B. `0.29//K`C. `0.69//K`D. `0.75//K`
Answer» Correct Answer - B `K = (2.303)/(t)"log"(a)/(a-x)` `= (2.303)/(t_(1//4))"log"(a)/(3a) = (2.303)/(t_(1//4))"log"(4)/(3)` `K = (2.303 xx 0.125)/(t_(1//4)) = (0.29)/(t_(1//4))`

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`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:A. `0.10//K`B. `0.29//K`C. `0.69//K`D. `0.75//K`

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