The rate constant for a first order reaction is `"60 s"^(-1)`. How much time will it take to reduce the initial concentration of the reactant to its `(1)/(16)` th value ?

Explanation

If we carry out half - change four times, the concentration will reduce to `(1)/(16)`.
`t_(1//2)=(0.693)/(k)=(0.693)/("60 s"^(-1))=0.01155s`
Total time taken `=4xxt_(1//2)=4xx0.01155s=0.0462s`