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Introduction to Reaction Mechanisms

AP Chemistry· difficulty 4/5

TS1 TS2 int reaction coordinate E

A two-step mechanism shows the second transition state higher than the first. What does this imply about rate-determining step and overall rate law if step 1 is fast equilibrium A+B⇌CA + B \rightleftharpoons C and step 2 is C+D→PC + D \rightarrow P?

  • A

    Rate = k[A][B]k[A][B]

  • B

    Rate = k[D]k[D]

  • C

    Rate = k[C][D]k[C][D]

  • D

    Rate = k[A][B][D]k[A][B][D]

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Explanation

Step 2 is rate-determining (higher TS). Rate = k2[C][D]k_2[C][D]. From fast equilibrium: [C]=K[A][B][C] = K[A][B], so Rate = k2K[A][B][D]k_2K[A][B][D] = k[A][B][D]k[A][B][D].

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