The order of a reaction is the exponent on a reactant's concentration in the rate law (rate = k[A]^m[B]^n), showing how strongly rate depends on that concentration. Orders come from experimental data, except for elementary reactions, where they match the stoichiometric coefficients of the colliding particles.
The order of a reaction tells you how sensitive the reaction rate is to each reactant's concentration. In a rate law like rate = k[A]^m[B]^n, the exponent m is the order with respect to A, n is the order with respect to B, and m + n is the overall order. First order in A means doubling [A] doubles the rate. Second order in A means doubling [A] quadruples the rate. Zero order means changing [A] does nothing to the rate at all.
Here's the rule that AP Chem hammers. For an overall reaction, you can only find orders from experimental data (like comparing trials in a table). The balanced equation's coefficients tell you nothing about order in that case. But for an elementary reaction, a single collision step, the rate law can be written straight from the stoichiometry of the particles that collide. That's the whole point of Topic 5.4. A step like A + A → products is second order in A because two A particles have to smash into each other for anything to happen.
This term lives in Unit 5 (Kinetics), specifically Topic 5.4 (Elementary Reactions), supporting learning objective 5.4.A, which asks you to represent an elementary reaction as a rate law using its stoichiometry. The essential knowledge here is that the rate law of an elementary step can be inferred from the particles in the collision, and that three-particle collisions are rare (so termolecular steps are unlikely). Reaction order is also the connective tissue of the entire kinetics unit. You need it to read a rate law, interpret the units of the rate constant, pick the right integrated rate law graph, and judge whether a proposed mechanism is consistent with experimental data.
Keep studying AP Chemistry Unit 5
Rate Law (Unit 5)
The order IS the exponent in the rate law. You can't write or interpret rate = k[A]^m without knowing what m means, so these two ideas are really one package.
Molecularity (Unit 5)
For an elementary reaction only, order and molecularity match. A bimolecular step is second order overall. This shortcut is exactly what LO 5.4.A tests, and it breaks down the moment the reaction is multi-step.
First-Order Reaction and Half-Life (Unit 5)
Once you know a reaction is first order, you unlock the constant half-life property and the linear ln[A] vs time plot. Order is the gatekeeper that tells you which integrated rate law and which graph to use.
Rate Constant (Unit 5)
The units of k depend on the overall order (s⁻¹ for first order, M⁻¹s⁻¹ for second order). On multiple choice, k's units are a sneaky way to hand you the order without saying it.
Reaction order shows up in two main flavors. First, data-table questions where you compare trials, see how the rate changes when one concentration doubles, and deduce the order for each reactant. Second, mechanism questions tied to Topic 5.4, where you write the rate law for an elementary step directly from its coefficients, or check whether a proposed slow step gives a rate law that matches the experimental one. You may also be asked to identify order from a graph (which plot is linear), from the units of k, or from half-life behavior. No released FRQ in recent years has used the exact phrase 'order of the reaction' as a standalone prompt, but determining and using reaction order is a staple of Unit 5 free-response parts about rate laws and mechanisms.
Molecularity counts the particles colliding in one elementary step (unimolecular, bimolecular). Order is the experimentally determined exponent in the rate law for the overall reaction. They're equal only when the reaction is a single elementary step. For a multi-step mechanism, the overall order is set by the slow step and the data, not by the balanced equation. If a question gives you an overall reaction and asks for the rate law, the answer is 'you need experimental data,' not 'copy the coefficients.'
The order with respect to a reactant is the exponent on its concentration in the rate law, and the overall order is the sum of all the exponents.
For an overall reaction, order must be determined from experimental data, not from the balanced equation's coefficients.
For an elementary reaction, the order does match the stoichiometric coefficients, because the rate law reflects which particles actually collide (LO 5.4.A).
Elementary steps requiring three or more particles to collide at once are rare, so termolecular steps in a proposed mechanism should make you suspicious.
Doubling a concentration doubles the rate for first order, quadruples it for second order, and changes nothing for zero order.
The order tells you which integrated rate law applies, which plot is linear, and what units the rate constant k must have.
It's the exponent on a reactant's concentration in the rate law, rate = k[A]^m[B]^n. The order shows how the rate responds when that concentration changes, and the overall order is m + n.
Only for an elementary reaction. For an overall reaction, no. You need experimental data, like comparing how the initial rate changes between trials when one concentration doubles. Assuming coefficients equal orders for an overall reaction is one of the most common wrong answers in Unit 5.
Molecularity counts colliding particles in a single elementary step, while order is the exponent in the rate law for the reaction as written. They match only for elementary reactions, which is exactly what Topic 5.4 covers.
Pick two trials where only one reactant's concentration changes, then compare the rate change to the concentration change. If [A] doubles and the rate doubles, it's first order in A; if the rate quadruples, second order; if the rate doesn't change, zero order.
The units of k must make the rate come out in M/s. A k in s⁻¹ means first order overall, M⁻¹s⁻¹ means second order, and M/s means zero order. It's a fast check that shows up in multiple choice.