The reaction: A 2B \rightarrow C 2D, is run three times. In the second run, the initial concentration of A is double that of the first run, and the initial rate of the reaction is double that of the first run. In the third run, the initial concentration of each reactant is double to 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

The reaction is first order with respect to reactant A and zero order with respect to reactant B.

Explanation:

A + 2B → C + 2D

The rate of a reaction is given as

Rate = k [A]ˣ [B]ʸ

where k = rate constant

x = order of the reaction with respect to reactant A.

y = order of the reaction with respect to reactant B.

For the first run

Initial Rate = k [A]ˣ [B]ʸ

For the second run, the concentration of A is doubled and the rate also doubles

New Rate = k [2A]ˣ [B]ʸ

New Rate = 2 × (Initial Rate) = 2 × k [A]ˣ [B]ʸ

k [2A]ˣ [B]ʸ = 2k [A]ˣ [B]ʸ

[2A]ˣ = 2 [A]ˣ

2ˣ [A]ˣ = 2 [A]ˣ

2ˣ = 2¹

x = 1

Hence, the reaction is of order 1 with respect to reactant A.

For the third run, the concentration of A and B are doubled and the rate of reaction is doubled.

New Rate = k [2A]ˣ [2B]ʸ

But we already know that x=1

New Rate = k [2A] [2B]ʸ

But New Rate = 2 × (Initial Rate) = 2 × k [A]ˣ [B]ʸ = k [A] [B]ʸ

k [2A] [2B]ʸ = 2k [A] [B]ʸ

[2B]ʸ = [B]ʸ

2ʸ [B]ʸ = [B]ʸ

2ʸ = 1

2ʸ = 2⁰

y = 0

Hence, the reaction is of order 0 with respect to reactant B.

Hope this Helps!!!