Reaction Mechanism and Rate Law反应机理与速率定律

Big Idea · Transformations (TRA)AP Science Practice 6 · ArgumentationAP link · Unit 5~45 min

Learning goals

  • Explain that the rate-determining step sets the overall rate, so when the first step is slow, the overall rate law is the rate law of the first step
  • Judge whether a mechanism is consistent with an experimental rate law; the steps must add up to the overall reaction, and the derived rate law must match the experiment with no intermediates in it
  • Recognize that several mechanisms may agree with the same rate law, so a rate law can support a mechanism but can't prove it

Before you start

  • Introduction to rate law (U5.2)
  • Elementary reactions (U5.4)
  • Introduction to reaction mechanisms (U5.7)

The slowest step is in charge

In a multistep reaction, the steps often have very different rates. However fast the others are, the whole reaction can go no faster than its slowest step, just as a production line can’t turn out products faster than its slowest station. The slowest step of a mechanism is the rate-determining step决速步 (or rate-limiting step).

The rate-determining step is an elementary step, so its rate law comes from its coefficients (U5.4). If the first step is the slow one, the overall rate equals that step’s rate:

overall rate law=rate law of the first step\text{overall rate law} = \text{rate law of the first step}

Fast steps after the rate-determining step have no effect on the overall rate.

An example: NO₂ + CO

OpenStax reports that below 225 °C, NO₂ + CO → NO + CO₂ has the experimental rate law rate = k[NO₂]². The proposed mechanism is

Step 1: NOX2+NOX2→NOX3+NO(slow)Step 2: NOX3+CO→NOX2+COX2(fast)\begin{aligned} &\text{Step 1: }\ce{NO2 + NO2 -> NO3 + NO}\quad\text{(slow)} \\ &\text{Step 2: }\ce{NO3 + CO -> NO2 + CO2}\quad\text{(fast)} \end{aligned}

Step 1 is the rate-determining step; it is bimolecular, with rate = k[NO₂]², which matches the experiment. CO appears only in step 2, the fast step, so the rate doesn’t depend on [CO]: that’s why the reaction is zero order in CO, as U5.2 described.

When is a mechanism consistent with a rate law?

To judge whether a mechanism agrees with experiment, check three things:

  1. The steps (multiplied where needed) add up to the overall reaction.
  2. The rate law written from the rate-determining step is the same as the experimental rate law.
  3. The rate law contains concentrations of reactants (or catalysts) only, never intermediates: an intermediate’s concentration is tiny and hard to measure, so it never shows up in an experimental rate law.

If the slow step isn’t the first step, its rate law usually contains an intermediate, which must be replaced using an earlier fast equilibrium; that’s the subject of U5.9.

Consistent isn’t the same as proven

Different mechanisms can lead to the same rate law. For example, the experimental rate law for 2NO + Cl₂ → 2NOCl in OpenStax is rate = k[NO]²[Cl₂]. It is consistent with a single termolecular step, and also with a two-step mechanism in which NOCl₂ forms quickly and then reacts slowly with NO. A rate law can rule out mechanisms that don’t match, but it can’t choose among those that do; that takes other evidence, such as detecting an intermediate.

For each case below, decide whether each of the four proposed mechanisms is consistent with the experimental rate law:

Try it: which mechanisms are consistent with the experimental rate law?

Overall reaction: NO₂ + CO → NO + CO₂
Experimental rate law: rate = k[NO₂]²

Mechanism A

  1. NO₂ + CO → NO + CO₂(slow)

Mechanism B

  1. NO₂ + NO₂ → NO₃ + NO(slow)
  2. NO₃ + CO → NO₂ + CO₂(fast)

Mechanism C

  1. NO₂ + NO₂ ⇌ NO₃ + NO(fast, reversible)
  2. NO₃ + CO → NO₂ + CO₂(slow)

Mechanism D

  1. NO₂ → NO + O(slow)
  2. O + CO → CO₂(fast)

The experimental rate law is from OpenStax (below 225 °C). B is the textbook’s proposed mechanism: the slow first step is two NO₂ molecules colliding. In C, the rate-determining step contains the intermediate NO₃; after substitution the rate law still has [CO] and [NO], which doesn’t match.

Example 1: A rate law from a slow first step

For the overall reaction 2A + B → A₂B, the proposed mechanism is A + B → AB (slow); AB + A → A₂B (fast). Write the rate law, and state how the rate changes when [A] doubles.

Step-by-step solution

  1. Check that the steps add up to the overall reaction.

    A+B+AB+A→AB+AX2B⇒2 A+B→AX2B\begin{aligned} &\ce{A + B + AB + A -> AB + A2B} \\ &\Rightarrow \ce{2A + B -> A2B} \end{aligned}

    After canceling AB, it matches the overall reaction.

  2. Find the rate-determining step. The overall rate is set by the slowest step: step 1, A + B → AB.

  3. Write its rate law. It is an elementary (bimolecular) step, so its coefficients are the orders:

    rate=k[A][B]\boxed{\text{rate} = k[\ce{A}][\ce{B}]}

  4. Check for intermediates. AB is an intermediate, but it doesn’t appear in this rate law, so nothing needs replacing; the rate law contains only reactants and can be compared with experiment.

  5. When [A] doubles. First order in A:

    21=22^1 = 2

    The rate doubles. Although A’s coefficient in the overall reaction is 2, the second A appears only in the fast step and doesn’t affect the rate.

Example 2: Finding step 1 from the rate law

The overall reaction 2X + Y → X₂Y has the experimental rate law rate = k[X]². The mechanism has two steps, and step 2 is fast: ? + Y → X₂Y. Work out the slow step 1.

Step-by-step solution

  1. Read the rate law. rate = k[X]²: the slow step involves two X particles and no Y, so Y is not in the slow step.

  2. Write the slow step 1. Two X particles collide, and the product must be able to combine with Y in step 2 to give X₂Y:

    X+X→XX2 (slow)\ce{X + X -> X2}\ \text{(slow)}

  3. Fill in the fast step 2. The ”?” is the product of step 1, X₂:

    XX2+Y→XX2Y (fast)\ce{X2 + Y -> X2Y}\ \text{(fast)}

    X₂ is an intermediate.

  4. Check the sum.

    2 X+XX2+Y→XX2+XX2Y⇒2 X+Y→XX2Y\begin{aligned} &\ce{2X + X2 + Y -> X2 + X2Y} \\ &\Rightarrow \ce{2X + Y -> X2Y} \end{aligned}

    This matches the overall reaction.

  5. Check the rate law. The slow step X + X → X₂ has rate = k[X]², matching experiment; Y appears only in the fast step, so the reaction is zero order in Y.

Practice: the first step is slow. What is the overall rate law?
  1. A + B → C (slow); C + A → D (fast)
  2. 2A → A₂ (slow); A₂ + B → A₂B (fast)
  3. A → C + D (slow); C + B → E (fast)
  4. B + B → B₂ (slow); B₂ + A → AB₂ (fast)
  5. A + B → AB (slow); AB + B → AB₂ (fast)
  6. A + A → A₂ (slow); A₂ + B → A₂B (fast); A₂B + B → A₂B₂ (fast)
  7. A → A* (slow); A* + B → AB (fast)
  8. B + A → C (slow); C → D + E (fast)

Common mistakes

Mistake 1: “The overall rate law comes from the last step (or from all the steps together).” The rate is set by the slowest step. Fast steps after the slow step don’t affect the rate law.

Mistake 2: “A rate law can include an intermediate’s concentration.” An intermediate’s concentration is tiny and hard to measure, so it never appears in an experimental rate law. If a derived rate law contains an intermediate, replace it first (U5.9).

Mistake 3: “If a mechanism matches the rate law, the mechanism is proven.” Several mechanisms can give the same rate law. A match only shows that the mechanism is possible; a mismatch rules it out.

Quick check

Mechanism: NO₂ + NO₂ → NO₃ + NO (slow); NO₃ + CO → NO₂ + CO₂ (fast). What rate law does it predict?

The reaction A + B → C has the rate law rate = k[A]. Which explanation is most reasonable?

Why does an intermediate’s concentration never appear in an experimental rate law?

For 2NO + O₂ → 2NO₂ (rate = k[NO]²[O₂]), two mechanisms both lead to this rate law. Which conclusion is correct?

The overall reaction A + 2B → AB₂ has the experimental rate law rate = k[A][B], and step 2 is AB + B → AB₂ (fast). What is the slow step 1 most likely to be?

For the I⁻-catalyzed decomposition of H₂O₂, the derived rate law is rate = k[H₂O₂][I⁻]. I⁻ is a catalyst, yet it appears in the rate law. Is that allowed?

A mechanism has a slow step followed by two fast steps. If one of the fast steps is replaced by an even faster step, how does the overall rate change?

AP-style free response

AP-style free response

The overall reaction 2A + B → A₂B has the experimental rate law rate = k[A]². This mechanism has been proposed:

Step 1: A+A→AX2(slow)Step 2: AX2+B→AX2B(fast)\begin{aligned} &\text{Step 1: }\ce{A + A -> A2}\quad\text{(slow)} \\ &\text{Step 2: }\ce{A2 + B -> A2B}\quad\text{(fast)} \end{aligned}

(a) Show that the steps add up to the overall reaction, and identify the intermediate.

(b) Derive the rate law from this mechanism, and state whether it agrees with the experiment.

(c) Predict how the rate changes when [B] doubles, and when [A] doubles.

(d) Another student proposes a single step, 2A + B → A₂B. Evaluate this mechanism using the experimental rate law.

(e) Small amounts of A₂ are detected in the reaction mixture. What does this finding mean for the mechanism in (a)?

Model answer and scoring

(a) A + A + A₂ + B → A₂ + A₂B; canceling A₂ gives 2A + B → A₂B. A₂ forms in step 1 and is used up in step 2, so it is the intermediate.

(b) Step 1 is the rate-determining step and is bimolecular: rate = k[A]², which agrees with the experiment.

(c) Doubling [B]: no change in rate (zero order in B, which appears only in the fast step). Doubling [A]: the rate quadruples.

(d) A single termolecular step would have the rate law k[A]²[B], which doesn’t match the experiment (zero order in B), so this mechanism is ruled out.

(e) Detecting the intermediate A₂ is evidence that supports the mechanism in (a), though it still doesn’t prove it.

Scoring: (a) correct sum and A₂ identified; (b) rate law from the slow step, compared with experiment; (c) both predictions; (d) writes the one-step rate law and compares it with experiment; (e) the evidence “supports” rather than “proves.”

Summary

  • The rate-determining step is the slowest step of a mechanism, and it sets the overall rate.
  • When the first step is slow, the overall rate law is the rate law of that elementary step; a reactant that appears only in fast steps is zero order.
  • A mechanism is consistent with experiment when its steps add up to the overall reaction, its derived rate law matches the experimental one, and no intermediate appears in the rate law.
  • A rate law can rule out a mechanism but can’t prove one; several mechanisms may agree with it.

AP tip: When evaluating a mechanism, write out the three checks in order: whether the steps add up to the overall reaction, the rate law from the rate-determining step, and whether it matches the experiment. To explain why a reactant is zero order, point out that it appears only in fast steps after the rate-determining step.

Practice this chapter: Unit 5 · Kinetics · 12 questionsPractice this topic: U5.8 · Reaction Mechanism and Rate Law · 5 questions

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