If the half-life of a first order reaction is T s , then show that after nTs `[1-(1/2)^(n)]` parts of the initial concentration of the reactant will take part in the reaction.

Explanation

If the initial concentration of the reactant in a first order reaction be a `"mol. L"^(-1)` , then the concentration left after 'n'
half -lives of the reaction is `(a)/(2^(n))=a((1)/(2))^(n)"mol.L"^(-1)` Therefore , the fraction of the initial concentration that has taken part in the reaction is `(a-a((1)/(2))^(n))/(a)=[1-((1)/(2))^(n)]`