The half -life of a first order reaction `ArarrB` is 10 min . What fraction of the reactant will be left behind after 1 h ?

Explanation

`t_(1//2)=(0.693)/(k) therefore k = (0.693)/(t_(1//2))=(0.693)/(10)=0.0693 "min"^(-1)`
`t=(2.303)/(k)log.([A]_(0))/([A]) " or, " 60 = (2.303)/(0.0693)log .([A]_(0))/([A]) " or, " ([A]_(0))/([A])=64`
or, `[A]=[A_(0)]xx(1)/(64)`