Question

The rate law of a reaction between the substance A and B is given by, rate=k[A]^n[B]^m. On doubling the concentration of A and making the volume of B half the ratio of new rate to the earlier rate of reaction will be?​

Answer 1

Answer:

4ⁿ

Explanation:

Given reaction is rate=k[A]ⁿ [B]ⁿ

Let the initial rate of reaction be k₁ and final rate be k₂

C=n/V⇒ if the volume(V) is halved concentration(c) doubles

k₂=[2A]ⁿ [2B]ⁿ = 4ⁿ [A]ⁿ [B]ⁿ

k₂ = 4ⁿ k₁

k₂/k₁ = 4ⁿ