The reaction A + B → 2 C has the rate law rate = k[A][B]³. By what factor does the rate of reaction increase when [A] remains constant but [B] is doubled?

The rate law is defined as the amount of reactant present in the initial concentration and used up during the course of the reaction. The rate depends on the following:-

According to the question, The expression of the rate is

To find the solution to the problem,

If [A] remains constant then that means [A2]= [A1]. If B doubled, then that mean [B2]= 2[B2]. To find what factor does the rate of reaction increases, we need to divide the first reaction rate with the second. In simple terms, the degree of the reactant will act as a coefficient to the product.

After solving the equation, the solution is as follows:-

Hence, The rate of reaction will be 8 times faster than the other

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8

Explanation:

To solve this question, we just need to put the new number into the equation. If [A] remain constant then that mean [A2]= [A1]. If B doubled, then that mean [B2]= 2[B2]. To find what factor does the rate of reaction increases, we need to divide the first reaction rate with the second. The calculation will be:

rate2/rate1= k[A2][B2]³     /     k[A1][B1]³

rate2/rate1= [A1][2B1]³      /     [A1][B1]³

rate2/rate1=  A1*8B1³       /     A1*B1³

rate2/rate1= 8/1= 8

The rate of reaction will be 8 times faster.

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