Two reactants A and B separately shows two chemical reactions. Both reactions are made with same initial concentration of each reactant. Reactant A follows first order kinetics whereas reactant B follows second order kinetics. If both have same half lives, compare their rates (a) at the start of reaction (b) after the lapse of one half -life .

Two reactants A and B separately shows two chemical reactions. Both re

1.	Two reactants A and B separately shows two chemical reactions. Both reactions are made with same initial concentration of each reactant. Reactant A follows first order kinetics whereas reactant B follows second order kinetics. If both have same half lives, compare their rates (a) at the start of reaction (b) after the lapse of one half -life .
Answer» Solution :For `A :` rate `=k_(A)[A]^(1)`………..`(i)` and `(t_(1//2))_(A)=(0.693)/(k_(A))`…………`(ii)` For`B :` rate `=k_(B)[B]^(2)`………….`(iii)` and `(t_(1//2))=(1)/(ak_(B))` ………….`(iv)` where a is initial concentration `(a)` Initial rate of A, `r_(A)=k_(A)xxa` Initial rate of `B`, `r_(B)=k_(B)xxa^(2)` `:. (r_(A))/(r_(B))=(k_(A))/(k_(B))xx(1)/(a)` ..................`(v)` From EQS. `(ii)` and `(iv)` if `(t_(1//2))_(A)=(t_(1//2))_(B)`, then `(0.693)/(K_(A))=(1)/(k_(B)a)` or `(k_(A))/(k_(B))=0.693xxa`............`(vi)` `:.` From Eqs. `(v)` and `(vi)`, `(r_(A))/(r_(B))=(0.693a)/(a)=0.693` `(b)` After lapse of `1` half , the new rates are `R._(A)` and `r._(B)` `r._(A)=k_(A)xx(a)/(2)`, `r._(B)=k_(B)xx(a//2)^(2)` `:.(r._(A))/(r._(B))=(k_(A))/(k_(B))xx(2)/(a)` ................`(VII)` By Eqs. `(vi)` and `(vii)` `(r._(A))/(r._(B))=0.693xxaxx(2)/(a)=1.386`