The rate of reaction between two reactants A and B decreases by a factor of 4 if the concentration of reactant B is doubled. The order of this reaction with respect to reactant B is?Correct answer is '-2'. Can you explain this answer?

Question

Answer

Explanation:

Given, the rate of reaction between reactants A and B is affected by the concentration of reactant B. Let's assume the rate of reaction is given by the equation:

rate = k[A]^m[B]^n

Where k is the rate constant, m is the order of the reaction with respect to reactant A, and n is the order of the reaction with respect to reactant B.

We are told that if the concentration of reactant B is doubled, the rate of reaction decreases by a factor of 4. This means:

(rate when [B] = 2[B]_0) = (1/4)(rate when [B] = [B]_0)

Substituting these values into the rate equation, we get:

k[A]^m[2B]^n = (1/4)k[A]^m[B]^n

Simplifying, we get:

2^n = 1/4

n = -2

Therefore, the order of the reaction with respect to reactant B is -2.

Conclusion:

The order of the reaction with respect to reactant B is -2.

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