A reaction is of first order in reactant A and of second order in reactant B. How is rate of reaction affected when
(a) Concentration of B alone is increased to three times.
(b) The concentration of A as well as B is doubled.

Explanation

The reaction is first order in reactant A and second order in reactant k.
`therefore` rate=k[A]`[B]^2`
(a) Let [A]=a and [B}=b
`r_1=k[A][B]^2=k ab^2`
Concentrate of B is increased three times
[A]=a,[B]=3b
`r_2=k[A][B]^2=ka(3b)^2=9ab^2`
`(r_2)/(r_1)=(9(ab)^2)/(k ab^2)=9`
`r_2=9r_1`
i.e., rate of the reaction becomes nine times.
(b) [A]=a and [B]=b
`thereforer_1=k[A][B]^2=k ab^2`
Concentrations of both A and B are doubled
`therefore[A]=2a` and [B]=2b
`therefore r_2=k[A][B]^2=k(2a)(2b)^2=8kab^2`
`(r_2)/(r_1)=(8k" "ab^2)/(k" "ab^2)=8`
`r_2=8r_1`
i.e., rate of the reaction become, eight times