AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/5/5-4)
Unit 5 · Topic 5.4
5.4 Elementary Reactions
An elementary reaction happens in one step, in a single collision or a single particle breaking apart. Because of that, you can write its rate law straight from its coefficients. Steps that need three particles to collide at once are rare.
Key terms
- elementary reaction
- molecularity
- unimolecular
- bimolecular
Elementary steps
Most reactions happen as a series of simple steps. Each of those steps is an elementary reaction: it describes exactly which particles collide (or which single particle falls apart) at the molecular level.
Molecularity is the number of particles that take part in an elementary step.
| Molecularity | Particles in the step | Example | Rate law |
|---|---|---|---|
| Unimolecular | 1 | A → products | rate = k[A] |
| Bimolecular | 2 (different) | A + B → products | rate = k[A][B] |
| Bimolecular | 2 (same) | 2A → products | rate = k[A]² |
| Termolecular | 3 | A + B + C → products | rate = k[A][B][C] |
Why the coefficients work here
The rate of a collision depends on how often the right particles meet. If a step needs an A and a B to collide, doubling [A] doubles the number of collisions per second, and so does doubling [B]. That's why the rate is proportional to [A][B].
If a step needs two A particles to collide, doubling [A] gives twice as many A particles, each of which meets twice as many partners, so the rate goes up by 4. That's [A]².
This reasoning only works for elementary steps. For an overall reaction, the coefficients come from adding up several steps and say nothing about the collisions in any single one, so the rate law has to come from experiments.
Three-particle collisions are rare
For a termolecular step, three particles have to arrive at the same spot at the same instant, with the right energy and orientation. That's very unlikely compared with two particles colliding. So proposed mechanisms rarely include steps with three or more particles, and steps with four or more are not considered realistic at all.
When an overall rate law is third order, chemists usually explain it with a series of unimolecular and bimolecular steps instead, as in topic 5.9.
Reversible elementary steps
An elementary step can run in both directions. Each direction is its own elementary reaction with its own rate constant and its own rate law. For A + B ⇌ C, the forward rate is k₁[A][B] and the reverse rate is k₋₁[C].
When the forward and reverse rates become equal, the step is at equilibrium. You'll use exactly this idea in topic 5.9, and again in Unit 7, where the equilibrium constant turns out to be the ratio of the forward and reverse rate constants.
Elementary or overall?
A problem will tell you when an equation represents an elementary step, often by calling it a step in a mechanism. If it's just 'a reaction', don't assume it's elementary. Only an elementary step lets you write the rate law by inspection.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Rate laws for elementary steps
Write the rate law and state the molecularity of each elementary step: (a) O₃ → O₂ + O; (b) NO₂ + NO₂ → NO₃ + NO; (c) NO + O₃ → NO₂ + O₂.
Show the solutionHide the solution
- Step 1: (a) One particle breaks apart: unimolecular. rate = k[O₃].
- Step 2: (b) Two NO₂ collide: bimolecular. rate = k[NO₂]².
- Step 3: (c) One NO and one O₃ collide: bimolecular. rate = k[NO][O₃].
Answer: (a) unimolecular, k[O₃]; (b) bimolecular, k[NO₂]²; (c) bimolecular, k[NO][O₃]
- Example 2
Overall equation versus elementary step (classic trap)
The overall reaction 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g) has an experimentally measured rate law of rate = k[NO]²[H₂]. A student predicts rate = k[NO]²[H₂]² from the equation. Why is that wrong, and why couldn't this reaction happen in one step?
Show the solutionHide the solution
- Step 1: The overall equation isn't an elementary step, so its coefficients don't give the rate law. The measured rate law is first order in H₂, not second.
- Step 2: A single step with these coefficients would need four molecules (2 NO and 2 H₂) to collide at once, which is so unlikely that it isn't a realistic step.
- Step 3: So the reaction must happen through a mechanism of simpler steps, and the rate law has to come from experiment.
Answer: The coefficients of an overall reaction don't set its rate law; a four-particle collision is unrealistic, so the reaction occurs in several steps, and the measured rate law is k[NO]²[H₂].
Common mistakes
- Writing a rate law from the coefficients of an overall reaction.
- Treating 2A → products as first order. Two A particles colliding gives [A]².
- Proposing mechanisms with four-particle collisions.
On the exam
- Expect to write the rate law for a given elementary step, or to identify which step in a mechanism is termolecular or unlikely.
- If a question asks why a proposed one-step mechanism is unlikely, point to the number of particles that would have to collide at once.
Connected topics
Videos
Check yourself
4 questions on 5.4 Elementary Reactions. Pick an answer to see if you got it, and why.
The reaction NO(g) + O₃(g) → NO₂(g) + O₂(g) occurs in a single elementary step. If [NO] is doubled and [O₃] is tripled at constant temperature, the rate of the reaction increases by a factor of
Suppose the reaction 2NO(g) + Cl₂(g) → 2NOCl(g) occurred in a single elementary step. Which of the following gives its rate law and molecularity?
The reaction 2NO₂(g) + F₂(g) → 2NO₂F(g) has the experimentally determined rate law rate = k[NO₂][F₂]. A student claims the rate law should be rate = k[NO₂]²[F₂], based on the coefficients in the equation. Which of the following best responds to this claim?
Termolecular elementary steps are much less common than unimolecular and bimolecular steps. Which of the following best explains this?
0 of 4 answered