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Maksim231197 [3]
3 years ago
6

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 co

nstant but [B] is doubled?
Chemistry
1 answer:
lawyer [7]3 years ago
7 0

Answer:

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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