Half - life of a first order reaction is 15 min . Calculate rate constant of the reaction and the time required to complete 80% of the reaction. If the initial concentration of the reactant is doubled , how long will it take to reach 50% completion ? Give reason.

Explanation

`k=(0.693)/(t_(1//2))=(0.693)/(15)"min"^(-1) = 0.0462"min"^(-1)`
`t=(2.303)/(k)log.(a)/(a-x)=(2.303xx15)/(0.693)log.(a)/(a-0.8a)=34.84min`
Even if initial concentration of the reactant is doubled , still it will take 15min to complete 50% of the reaction