On a multi-step energy profile, an intermediate appears at:

Answer: A local minimum (valley) between two TS peaks Intermediates are stable enough to occupy a valley (local minimum) between transition state peaks.

What is the role of NO?

Answer: Catalyst A catalyst is consumed in one step and regenerated in another, so it doesn't appear in the overall equation.

What is the predicted rate law?

Answer: rate = k[A]^2 Rate-determining step is step 1 (slow): rate = k[A]^2.

AP Chemistry · Topic 5.7

Introduction to Reaction Mechanisms Practice

Part of Kinetics.(TRA-5.A)

Practice questions

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Sample questions

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Sample 1difficulty 2/5
$A two-step energy profile shows two TS peaks separated by a valley where a labeled species sits.$
$On a multi-step energy profile, an intermediate appears at:$
- A
  $Only at the start$
- B
  $The highest peak$
- C
  $A local minimum (valley) between two TS peaks$
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- D
  $The lowest energy point overall$
Why
$Intermediates are stable enough to occupy a valley (local minimum) between transition state peaks.$
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Sample 2difficulty 3/5
$Two-step mechanism for ozone destruction. NO is consumed in step 1 and regenerated in step 2.$
$What is the role of NO?$
- A
  $Intermediate$
- B
  $Reactant$
- C
  $Catalyst$
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- D
  $Product$
Why
$A catalyst is consumed in one step and regenerated in another, so it doesn't appear in the overall equation.$
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Sample 3difficulty 3/5
$A two-step mechanism with the first step slow.$
$What is the predicted rate law?$
- A
  $rate = k[A][C]$
- B
  $rate = k[A]^2$
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- C
  $rate = k[A]^2[C]$
- D
  $rate = k[B][C]$
Why
$Rate-determining step is step 1 (slow): rate = k[A]^2.$
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Sample 4difficulty 4/5
$A + B \rightleftharpoons C$
- A
  $k[A][B]$
- B
  $k[D]$
- C
  $k[C][D]$
- D
  $k[A][B][D]$
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Why
$k_2[C][D]$
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Sample 5difficulty 4/5
$Step 1 (fast eq): A + B ⇌ C; Step 2 (slow): C + B \to D. The predicted overall rate law is$
- A
  $rate = k[C]$
- B
  $rate = k[A][B]$
- C
  $rate = k[A]²[B]$
- D
  $rate = k[A][B]²$
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Why
$[C] = K[A][B]; rate(slow) = k₂[C][B] = k₂K[A][B]² = k_obs[A][B]².$
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