Conisder the reaction mechanism:
`A_(2) overset(k_(eq))hArr 2A ("fast")` (where `A` is the intermediate.)
`A+B underset(k_(1))rarr P (slow)`
The rate law for the reaction is

Options

A. rate `=K[A_(2)][B]`
B. rate `=K[A_(2)]^(2)[B]`
C. rate `=K[A_(2)]^(1//2)[B]`
D. rate `=K[A_(2)][B]^(2)`

Answer

3

Explanation

From the slow step`:` rate `=k_(2)[A][B]………(i)`
From fast step `:K_(a)=([A]^(2))/([A]^(2))` or `[A]=K_(a)^(1//2)[A][B]^(1//2).......(ii)`
From `(i)` and `(ii)` rate `=k_(2)k_(a)^(1//2)[A]^(1//2)[B]=k[A_(2)]^(1//2)[B]`