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?

Question

Chemistry

Charisma

27 January, 03:14

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?

+5

Answers (1)

Know the Answer?

Johnny · Answer 1

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.