Unit 5 · Topic 5.8 Beta

Reaction Mechanism and Rate Law

The slowest step of a mechanism is the rate-determining step.

Practice 4: Model AnalysisPractice 6: Argumentation

Question set for this topic

Part 1 · Hook

Why this matters

At a stadium exit, thousands of fans walk quickly down wide corridors and then squeeze through a single turnstile. How fast the crowd empties has nothing to do with how fast people walk in the corridors; it depends only on the turnstile. In a reaction mechanism, the slowest step is that turnstile.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. In the mechanism (1) A + B → C; (2) C + D → E, what is C?

  1. An intermediate
  2. A reactant
  3. A product
  4. A transition state
Show the answer

C is made in step 1 and used in step 2.

  • Correct: An intermediate:
  • A reactant:
  • A product:
  • A transition state:

2. What is the rate law of the elementary step 2 X + Y → Z?

  1. rate = k[X]²[Y]
  2. rate = k[X][Y]
  3. rate = k[Z]
  4. It must be measured
Show the answer

An elementary step's coefficients are its orders.

  • Correct: rate = k[X]²[Y]:
  • rate = k[X][Y]:
  • rate = k[Z]:
  • It must be measured:

Part 4 · See it

See it first

A two-stage assembly line drawn as two pipes in a row. The first pipe is narrow and labeled step 1, slow; the second pipe is wide and labeled step 2, fast. Product leaves the end only as fast as material squeezes through the narrow first pipe, so the overall rate equals the rate of the slow step, and making the fast step faster changes nothing.
The overall rate equals the rate of the slow step. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Each later step needs a product of the step before itthe reaction cannot run faster than its slowest step
  2. When the first step is the slow onethe overall rate law is that step's rate law, written from its coefficients
  3. Reactants used only in later fast steps do not change the ratethey do not appear in the rate law (zero order)
  4. The predicted rate law is compared with the measured onea mismatch rules a mechanism out; a match makes it consistent with the data

Part 6 · Key ideas

Key ideas

  • The rate-determining step is the slowest step; the overall rate cannot be faster than it.
  • If the first step is slow, the overall rate law is the rate law of that step.
  • Reactants that enter only after the slow step do not appear in the rate law.
  • A mechanism is consistent with data if its predicted rate law matches the measured one. It is never proven.

Part 7 · Misconception

A common mistake

The wrong idea: The rate law can be written from the overall equation's coefficients.

What actually happens: Only an elementary step can be read that way. Write the rate law from the slow step of the mechanism and check it against the data.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Data table

Rate data and two proposed mechanisms

For NO₂(g) + CO(g) → NO(g) + CO₂(g) at a fixed temperature, a student measures initial rates:

Mechanism 1 (one step): NO₂ + CO → NO + CO₂

Mechanism 2: step 1 (slow): NO₂ + NO₂ → NO₃ + NO; step 2 (fast): NO₃ + CO → NO₂ + CO₂

Initial concentrations and initial rates
Trial[NO₂] (M)[CO] (M)Initial rate (M/s)
10.100.100.0050
20.200.100.0200
30.100.200.0050

1. What is the experimental rate law?

  1. rate = k[NO₂]²
  2. rate = k[NO₂][CO]
  3. rate = k[NO₂]
  4. rate = k[NO₂]²[CO]
Show the answer

Trials 1 → 2: [NO₂] doubles, rate × 4, so second order in NO₂. Trials 1 → 3: [CO] doubles, rate unchanged, so zero order in CO. rate = k[NO₂]².

  • Correct: rate = k[NO₂]²: Right: second order in NO₂, zero order in CO.
  • rate = k[NO₂][CO]: Doubling [CO] does not change the rate, so CO is not in the rate law.
  • rate = k[NO₂]: Doubling [NO₂] quadruples the rate, which is second order, not first.
  • rate = k[NO₂]²[CO]: CO is zero order, so it does not appear.

2. Which mechanism is consistent with the data, and why?

  1. Mechanism 2: its slow step gives k[NO₂]²
  2. Mechanism 1: it is the same as the overall equation
  3. Mechanism 1: a one-step path is simpler than two
  4. Both: they each add up to the overall equation
Show the answer

Mechanism 1 is a single elementary step, so it predicts rate = k[NO₂][CO], which does not match. In mechanism 2 the slow first step sets the rate: rate = k[NO₂]², which matches the data.

  • Correct: Mechanism 2: its slow step gives k[NO₂]²: Right: its predicted rate law matches the measured one.
  • Mechanism 1: it is the same as the overall equation: Summing to the overall equation is necessary, but its predicted rate law k[NO₂][CO] does not match the data.
  • Mechanism 1: a one-step path is simpler than two: Simplicity is not evidence; the predicted rate law must match.
  • Both: they each add up to the overall equation: Both pass the sum test, but only one predicts the measured rate law.

3. In mechanism 2, why does [CO] not appear in the rate law?

  1. CO reacts in the fast step, after the slow one
  2. CO is an intermediate that cancels out
  3. CO is a product of the slow step
  4. CO is present in a large excess in the trials
Show the answer

The overall rate is set by the slow step, NO₂ + NO₂ → NO₃ + NO. CO is used only in step 2, which waits on NO₃ from step 1, so speeding up step 2 by adding CO cannot speed up the overall reaction.

  • Correct: CO reacts in the fast step, after the slow one: Right: CO enters after the bottleneck.
  • CO is an intermediate that cancels out: CO is a reactant in the overall equation, not an intermediate.
  • CO is a product of the slow step: The slow step makes NO₃ and NO, not CO.
  • CO is present in a large excess in the trials: The trials change [CO]; it is not held in excess.

Model

A two-step mechanism for making NOBr

One proposed mechanism for 2 NO(g) + Br₂(g) → 2 NOBr(g):

  1. NO + Br₂ → NOBr₂ (slow)
  2. NOBr₂ + NO → 2 NOBr (fast)

4. What rate law does this mechanism predict?

  1. rate = k[NO][Br₂]
  2. rate = k[NO]²[Br₂]
  3. rate = k[NOBr₂][NO]
  4. rate = k[NO]²
Show the answer

The first step is slow, so it is the rate-determining step. Its rate law, written from its coefficients, is the overall rate law: rate = k[NO][Br₂].

  • Correct: rate = k[NO][Br₂]: Right: the slow first step sets the rate.
  • rate = k[NO]²[Br₂]: That is the rate law from the overall equation, which is not an elementary step.
  • rate = k[NOBr₂][NO]: That is the rate law of the fast step, and it includes an intermediate.
  • rate = k[NO]²: Br₂ is a reactant in the slow step, so it must appear.

5. Experiments show rate = k[NO]²[Br₂]. What does this tell you about the mechanism above?

  1. It is not consistent with the data
  2. It is consistent, since it sums to the overall equation
  3. It is consistent, since it has a slow step
  4. The data are wrong, since they match the overall coefficients
Show the answer

The mechanism predicts rate = k[NO][Br₂], first order in NO. The measured order in NO is 2, so the mechanism does not account for the data.

  • Correct: It is not consistent with the data: Right: predicted and measured rate laws disagree.
  • It is consistent, since it sums to the overall equation: Summing correctly is needed but not enough.
  • It is consistent, since it has a slow step: Having a slow step is not enough; its rate law must match the data.
  • The data are wrong, since they match the overall coefficients: A measured rate law can match the coefficients by chance; the data are the evidence.

6. In a mechanism, step 1 is slow and step 2 is fast. A chemist finds a way to make step 2 twice as fast. What happens to the overall rate?

  1. It halves
  2. It stays about the same
  3. It doubles
  4. It becomes four times as fast
Show the answer

Step 2 already uses its intermediate as fast as step 1 supplies it. The slow first step still limits the overall rate, so it barely changes.

  • It halves: Making a step faster cannot slow the reaction down.
  • Correct: It stays about the same: Right: the bottleneck is unchanged.
  • It doubles: Only speeding up the slow step would raise the overall rate.
  • It becomes four times as fast: Nothing about step 2 controls the overall rate here.

Part 9 · Summary

Summary

The slowest step of a mechanism is the rate-determining step. When the first step is slow, the overall rate law is that step's rate law, so reactants used only in later fast steps do not appear. A mechanism is consistent with experiment when the rate law it predicts matches the measured rate law.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections