Give the molecularity and rate law for the elementary step Cl + H_2 → HCl + H. [2 points]

College Board AP 2021 (style)-style practice: Section I (multiple choice). For the elementary step A + 2B → C, the rate law is (A) k[A][B] (B) k[A][B]^2 (C) k[A]^2[B] (D) it cannot be written without data. Justify your choice.

A 1-point conceptual MCQ. The answer is (B). For an elementary step, and only for an elementary step, the orders equal the stoichiometric coefficients: one A and two B give rate = k[A][B]^2. The step is termolecular. The trap is option (D): that caution applies to overall reactions, not to elementary steps.

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What is an elementary reaction, and why can its rate law be written directly from its molecularity?

Topic 5.4 Elementary Reactions: identify the molecularity of an elementary step and write its rate law directly from its stoichiometry, distinguishing elementary steps from overall reactions.

A focused answer to AP Chemistry Topic 5.4, covering elementary reactions, molecularity (unimolecular, bimolecular, termolecular), writing the rate law of an elementary step from its stoichiometry, and why this differs from overall reactions, with full worked examples.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

An elementary reaction is a single step that happens exactly as written in one molecular event, such as a collision or a bond breaking. Its molecularity is the number of reactant particles involved: unimolecular (one particle), bimolecular (two), or termolecular (three, which is rare because triple collisions are unlikely). For an elementary step, and only for an elementary step, the rate law can be written directly from the equation, with the orders equal to the stoichiometric coefficients. So an elementary step $\text{A} + \text{B} \rightarrow \text{products}$ has $\text{rate} = k[\text{A}][\text{B}]$ , and $2\text{A} \rightarrow \text{products}$ has $\text{rate} = k[\text{A}]^2$ . Overall reactions are sums of elementary steps and their orders must be found by experiment.

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What this topic is asking
Elementary reactions and molecularity
Writing the rate law of an elementary step
Why overall reactions are different
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What this topic is asking

The College Board (Topic 5.4) wants you to recognize an elementary reaction, state its molecularity, and write its rate law directly from its stoichiometry. The crucial distinction is that the orders of an elementary step equal its coefficients, whereas the orders of an overall reaction must be measured. This is the bridge between the experimental rate law and the molecular-level mechanism.

Elementary reactions and molecularity

A unimolecular step is one particle rearranging or falling apart, for example an excited or strained molecule decomposing. A bimolecular step is the common case: two particles collide and react. A termolecular step needs three particles to meet at once, which is so improbable that such steps are rare; most reactions avoid them by proceeding through a sequence of uni- and bimolecular steps.

Writing the rate law of an elementary step

This is why mechanisms matter. The rate of an elementary step is governed by collision frequency, which scales with the concentration of each colliding partner. Two A particles must both be present, so the rate goes as $[\text{A}]^2$ . The privilege of reading orders from coefficients belongs to elementary steps alone.

Why overall reactions are different

An overall reaction is the net result of a sequence of elementary steps (the mechanism, Topic 5.7). Its experimentally measured rate law reflects the slowest step and any prior equilibria, not the overall stoichiometry. That is why you cannot write the rate law of an overall reaction from its balanced equation: doing so would assume the whole reaction happens in one collision, which it almost never does. The match between coefficients and orders is the test of whether a step is truly elementary.

Rate laws for elementary steps

Write the rate law for each elementary step.

step 1 A unimolecular step

$\text{O}_3 \rightarrow \text{O}_2 + \text{O}$ . One particle reacts, so molecularity is 1 and $\text{rate} = k[\text{O}_3]$ .

step 2 A bimolecular step with two species

$\text{NO} + \text{O}_3 \rightarrow \text{NO}_2 + \text{O}_2$ . Two different particles collide, so $\text{rate} = k[\text{NO}][\text{O}_3]$ .

step 3 A bimolecular step with identical particles

$2\text{NO}_2 \rightarrow \text{NO}_3 + \text{NO}$ . Two $\text{NO}_2$ collide, so molecularity is 2 and $\text{rate} = k[\text{NO}_2]^2$ .

step 4 The general rule

In every case the orders equal the coefficients because each step is a single collision event written exactly as it happens.

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Q1. Give the molecularity and rate law for the elementary step $\text{Cl} + \text{H}_2 \rightarrow \text{HCl} + \text{H}$ . [2 points]

Cue. Bimolecular; $\text{rate} = k[\text{Cl}][\text{H}_2]$ .

Q2. Explain why termolecular elementary steps are rare. [2 points]

Cue. Three particles must collide simultaneously with the right energy and orientation, which is highly improbable, so reactions usually proceed by uni- and bimolecular steps.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (style)3 marksSection II (short FRQ). Consider the elementary step

2\text{NO}_2(g) \rightarrow \text{NO}_3(g) + \text{NO}(g)

. (a) State the molecularity of this step. (b) Write the rate law for this elementary step. (c) Justify why you can write this rate law from the equation here, but not for an overall reaction.

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A 3-point conceptual FRQ on elementary reactions.

(a) Molecularity (1 point): two molecules collide (two $\text{NO}_2$ ), so the step is bimolecular.
(b) Rate law (1 point): for an elementary step the orders equal the coefficients, so $\text{rate} = k[\text{NO}_2]^2$ .
(c) Justify (1 point): an elementary step happens in a single collision event exactly as written, so its rate is set by how often those molecules meet, which is captured by the coefficients; an overall reaction is the sum of several steps, and its rate is governed by the slowest step, so its orders must be found by experiment.

Markers reward the molecularity, the rate law from the coefficients, and the reasoning that only elementary steps reveal their rate law directly.

AP 2021 (style)1 marksSection I (multiple choice). For the elementary step

\text{A} + 2\text{B} \rightarrow \text{C}

, the rate law is (A)

k[\text{A}][\text{B}]

(B)

k[\text{A}][\text{B}]^2

(C)

k[\text{A}]^2[\text{B}]

(D) it cannot be written without data. Justify your choice.

Show worked answer →

A 1-point conceptual MCQ. The answer is (B).

For an elementary step, and only for an elementary step, the orders equal the stoichiometric coefficients: one A and two B give $\text{rate} = k[\text{A}][\text{B}]^2$ . The step is termolecular. The trap is option (D): that caution applies to overall reactions, not to elementary steps.

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Sources & how we know this

AP Chemistry Course and Exam Description — College Board (2020)