An elementary step is a single molecular-level event in a reaction mechanism, representing one actual collision of particles. Unlike the overall reaction, an elementary step's rate law comes directly from its coefficients, and its molecularity counts how many particles collide in that step.
An elementary step is one piece of a reaction mechanism, a single collision event that actually happens at the molecular level. Most chemical equations you see are overall reactions, which are just the net result of several elementary steps added together. Each elementary step has its own rate constant and its own molecularity (unimolecular if one particle reacts, bimolecular if two collide, and rarely termolecular for three).
Here's the superpower that makes elementary steps special. For an elementary step, and ONLY for an elementary step, you can write the rate law straight from the coefficients in the equation. A step like NO₂ + NO₂ → NO₃ + NO has rate = k[NO₂]². You can never do that for an overall reaction, because the overall equation hides the mechanism. That's why the experimental rate law of the whole reaction is set by the molecularity of the slowest elementary step (EK 5.8.A.1), the rate-determining step.
Elementary steps live in Unit 5: Kinetics, specifically Topics 5.8 (Reaction Mechanism and Rate Law) and 5.9 (Steady-State Approximation). Learning objective AP Chem 5.8.A asks you to identify a reaction's rate law from a mechanism where the first step is rate limiting, and AP Chem 5.9.A covers the trickier case where it isn't, which forces you to use the pre-equilibrium approximation. Both skills depend on one idea, that each elementary step has a rate law you can read straight off its molecularity. This is also the bridge between the experimental side of kinetics (rate laws from data in Topics 5.2-5.5) and the molecular side (collisions and energy profiles in 5.6-5.7). A mechanism is only plausible if its steps add up to the overall equation AND its rate-determining step predicts the experimentally measured rate law.
Keep studying AP Chemistry Unit 5
Rate-Determining Step (Unit 5)
The rate-determining step is just the slowest elementary step in the mechanism. It acts like the slowest tollbooth on a highway, so the whole reaction's rate law matches that one step's molecularity, which is exactly what EK 5.8.A.1 says.
Intermediate (Unit 5)
Intermediates are species produced in one elementary step and consumed in a later one, like NO₃ in the classic NO₂ + CO mechanism. They cancel out when you add the steps, and they can't appear in a final rate law, which is why Topic 5.9 makes you substitute them out using the pre-equilibrium approximation.
Transition State (Unit 5)
Every elementary step has its own activation energy and its own transition state, so a two-step mechanism shows up as two humps on an energy profile diagram. The taller hump belongs to the slower, rate-determining step.
Rate Constant (Unit 5)
Each elementary step carries its own rate constant (k₁, k₂, k₋₁...). When the first step isn't rate limiting, the observed k for the overall reaction ends up being a combination of these step constants, which is the math behind the pre-equilibrium approach in Topic 5.9.
Multiple-choice questions love giving you a proposed mechanism with steps labeled slow and fast, then asking for the overall rate law. Your move is to write the rate law from the slow step's coefficients, then swap out any intermediate if one appears. A classic example is the mechanism NO₂ + NO₂ → NO₃ + NO (slow), NO₃ + CO → NO₂ + CO₂ (fast), which gives rate = k[NO₂]². Questions also run the logic backward, handing you an experimental rate law like rate = k[NO]²[O₂] and asking which mechanism is consistent with it. Vocabulary stems ask for the term describing the number of molecules in the rate-limiting step (that's molecularity). On FRQs, mechanisms show up alongside rate-law reasoning, like the 2022 short-response question on N₂O₅ decomposition with rate = k[N₂O₅]. You should be able to check that steps sum to the overall equation, identify intermediates, and justify why a mechanism matches (or fails to match) the experimental rate law.
The number one mistake in Unit 5 is writing a rate law from the coefficients of the overall balanced equation. That only works for elementary steps. An overall reaction like 2N₂O₅ → 4NO₂ + O₂ is the sum of several hidden steps, so its rate law (rate = k[N₂O₅], first order, not second) must come from experiment or from the rate-determining elementary step. If the question says 'elementary,' use the coefficients. If it doesn't, you need data or a mechanism.
An elementary step is a single molecular collision event in a mechanism, and the elementary steps must add up to give the overall balanced equation.
You can write a rate law directly from the coefficients of an elementary step, but never from the coefficients of an overall reaction.
Molecularity is the number of particles colliding in an elementary step, so a step with two reactant particles is bimolecular and second order.
When the first step is slow (rate limiting), the overall rate law is simply the rate law of that slow elementary step (EK 5.8.A.1).
When the first step is fast and reversible, intermediates sneak into the slow step's rate law, so you use the pre-equilibrium approximation to replace them with reactant concentrations (EK 5.9.A.1).
Each elementary step has its own rate constant, activation energy, and transition state, which is why multistep mechanisms show multiple humps on an energy diagram.
An elementary step is one single molecular event in a reaction mechanism, like two NO₂ molecules colliding to form NO₃ and NO. Each step has its own rate constant and molecularity, and the steps must add up to the overall reaction.
Only if it's an elementary step. For an overall reaction, the coefficients tell you nothing about the rate law. For example, 2N₂O₅ → 4NO₂ + O₂ has rate = k[N₂O₅], first order despite the coefficient of 2, because the rate law comes from the mechanism's slow step.
Every step in a mechanism is an elementary step, but only the slowest one is the rate-determining step. The rate-determining step is the bottleneck that controls the overall rate law, while fast steps mostly don't show up in it.
Molecularity counts how many reactant particles collide in that step. One particle is unimolecular, two is bimolecular, and three (termolecular) is rare because three-particle collisions are unlikely. It also tells you the step's order, so a bimolecular step is second order.
No. Intermediates are made in one elementary step and used up in another, so they can't be in the observed rate law. If the slow step contains an intermediate, you use the pre-equilibrium approximation from Topic 5.9 to rewrite its concentration in terms of reactants.