The reaction A 2B<--> C 2D is run 3 times. In the second run, the initial concentration of A is double that in the first run, and the initial rate of the reaction is double that of the first run. In the third run, initial concentration of each reactant is double the respective concentrations in the first run, and the initial rate is double that of the first run. What is the order of the reaction with respect to each reactant, A and B respectively

The reaction is order 1 with respect A and order 0 with respect B.

Explanation:

Based on the reaction:

A + 2B → C + 2D

The law of reaction is:

K = [A]ᵃ [B]ᵇ

In the second run, the initial concentration of A is doubled and the initial rate was doubled too. That means the rate law with respect to A is:

K = [A]¹ [B]ᵇ

In the third run, A and B were doubled but the initial rate was just doubled. That means B concentration doesn't affect initial rate and rate law is:

K = [A]

Thus, the reaction is order 1 with respect A and order 0 with respect B.

- First-order with respect to A.

- Zeroth-order with respect to B

Explanation:

Hello,

In this case, given the information, we can conclude that since in the second experiment the concentration of A is doubled as well as the initial rate of reaction and the same happened in the third run even when the concentration of B was also doubled, we understand that no matter what happen to B, it is not involved in the rate expression as it does not affect the reaction rate, for that reason, this reaction is first-order with respect to A and zeroth-order with respect to B.

Best regards.

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