The `t_(1//2)` of a first order reaction is 20 min . How long will it take for reactant concentration to drop from a to a/8?

Explanation

Let , after n number of half-life , the concentration reduces from a to a/8.
`therefore(a)/(2^(n))=(a)/(8) " or, "2^(n)=8 " or, " n=3`
Here half-life of the reaction is 20 min, Hence time required `=20xx3=60 "min"` .