Elementary reactions — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: elementary reaction definition, reaction molecularity, rate law derivation for elementary steps, reaction intermediate identification, and construction of overall rate laws from multi-step elementary reaction sequences per AP CED guidelines.

You should already know: Rate law definition and how to calculate reaction orders from experimental data. Balancing chemical equations for net overall reactions. The definition of reaction rate and rate constants.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Elementary reactions?

An elementary reaction (or elementary step, when part of a larger mechanism) is a single-step reaction that occurs exactly as written, with no intermediate sub-steps between reactants and products. Unlike overall balanced reactions, which only describe the net result of multiple reaction events, elementary reactions represent individual collision events between reactant particles.

Per the AP Chemistry CED, this topic is a core building block of Unit 5: Kinetics, accounting for ~15-20% of the unit's exam weight, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections of the exam. The key distinguishing feature of elementary reactions is that the stoichiometric coefficient of each reactant directly equals its reaction order in the rate law, a rule that does not hold for overall reactions. This makes elementary reactions the foundation for all multi-step reaction mechanism problems on the AP exam.

2. Molecularity of Elementary Reactions

Molecularity is defined as the number of reactant particles that collide and react in a single elementary step. Because it counts discrete particles, molecularity can only be a small positive integer: 1 (unimolecular), 2 (bimolecular), or 3 (termolecular). Termolecular steps are very rare, because the probability of three particles colliding simultaneously with correct orientation and sufficient energy is extremely low; they only appear in a small number of fast radical reaction steps.

A common point of confusion is distinguishing molecularity from reaction order. Reaction order describes how the rate depends on concentration, and can be zero, fractional, or negative for overall reactions. Molecularity is only defined for elementary steps, and is always 1, 2, or 3. To find molecularity, you simply count the total number of reactant particles on the left-hand side of the balanced elementary step.

Worked Example

Identify the molecularity of each of the following elementary reactions:

$C_{4} H_{8} \to 2 C_{2} H_{4}$
$C l + C H_{4} \to H C l + C H_{3}$
$B r + B r + A r \to B r_{2} + A r$
Count the total number of reactant particles on the left side of the elementary equation.
Reaction 1 has only 1 reactant particle ( $C_{4} H_{8}$ ), so molecularity is 1, and the reaction is unimolecular.
Reaction 2 has two distinct reactant particles ( $C l$ and $C H_{4}$ ), so molecularity is 2, and the reaction is bimolecular.
Reaction 3 has three total reactant particles (two Br and one Ar), so molecularity is 3, and the reaction is termolecular.

Exam tip: AP MCQ often includes distractors with non-integer molecularity. If an option lists molecularity as 0, 1.5, or any non-integer, it is automatically incorrect — you can eliminate it immediately without further analysis.

3. Rate Laws for Elementary Reactions

For overall balanced reactions, reaction orders cannot be determined from the balanced equation — they must be measured experimentally. However, because an elementary reaction is a single collision event, the rate of the reaction is directly proportional to the concentration of each reacting particle raised to the power of its stoichiometric coefficient. This is because the probability of all required reactant particles colliding at the same time is proportional to the product of their individual concentrations.

For a general elementary reaction $a A + b B \to products$ , the rate law is: $rate = k [A]^{a} [B]^{b}$ where $k$ is the rate constant for the elementary step, $a$ is the reaction order with respect to A, and $b$ is the reaction order with respect to B. The overall order of the elementary step is simply the sum of $a$ and $b$ .

Worked Example

Write the rate law for the elementary reaction $2 H I \to H_{2} + I_{2}$ , and state the overall order of the reaction.

Confirm the reaction is explicitly stated to be elementary, so we can use stoichiometric coefficients as reaction orders.
The only reactant is HI, with a stoichiometric coefficient of 2.
Write the rate law directly from the coefficients: $rate = k [H I]^{2}$ .
Sum the reaction orders to get the overall order: 2 = overall order of the reaction.

Exam tip: In AP FRQ questions asking for a rate law for a forward elementary step, never include product concentrations. Only reactants appear in the rate law for forward elementary steps, which is what you will be asked for 99% of the time on the exam.

4. Elementary Steps in Multi-Step Reaction Mechanisms

Nearly all overall reactions are not single elementary steps — they proceed via a sequence of multiple elementary steps called a reaction mechanism. When you add all elementary steps in a mechanism together, you get the balanced overall reaction.

Two key types of species appear in mechanisms: reaction intermediates and catalysts. A reaction intermediate is produced in an early elementary step and consumed in a later step, so it never appears in the overall balanced reaction. A catalyst is consumed in an early step and regenerated in a later step, so it also does not appear in the overall reaction. For a mechanism to be valid, two conditions must hold: 1) the sum of elementary steps matches the experimental overall reaction, and 2) the rate law derived from the mechanism matches the experimentally determined rate law for the overall reaction.

Worked Example

A reaction mechanism for the conversion of ozone to oxygen is given below: Step 1 (fast): $C l + O_{3} \to C l O + O_{2}$ Step 2 (slow): $C l O + O \to C l + O_{2}$ Identify the reaction intermediate and the catalyst, and confirm the sum of elementary steps gives the overall reaction $O_{3} + O \to 2 O_{2}$ .

Track where each non-overall species is produced and consumed: Cl is consumed in step 1 and produced in step 2; ClO is produced in step 1 and consumed in step 2.
By definition, a catalyst is consumed first then produced: Cl is the catalyst. An intermediate is produced first then consumed: ClO is the reaction intermediate.
Add the two steps: Left side total: $C l + O_{3} + C l O + O$ , Right side total: $C l O + O_{2} + C l + O_{2}$ .
Cancel species that appear on both sides: Cl and ClO cancel, leaving $O_{3} + O \to 2 O_{2}$ , which matches the overall reaction.

Exam tip: Always double-check the order of production/consumption to avoid confusing intermediates and catalysts on FRQ questions — this is one of the most commonly missed points on mechanism problems.

5. Common Pitfalls (and how to avoid them)

Wrong move: Using stoichiometric coefficients from an overall reaction to write the rate law, the same way you do for an elementary reaction. Why: Students generalize the rule for elementary reactions to all reactions, forgetting that only elementary steps have orders matching coefficients. Correct move: Always confirm the reaction is explicitly labeled as elementary before using coefficients to get reaction orders; for overall reactions, only use experimentally derived orders.
Wrong move: Assigning a non-integer or zero molecularity to an elementary reaction. Why: Students mix up the definitions of molecularity (count of particles) and reaction order (can be any value). Correct move: Remember molecularity is only 1, 2, or 3 for elementary steps; eliminate any MCQ option with non-integer molecularity immediately.
Wrong move: Leaving reaction intermediates in the final overall rate law derived from a mechanism. Why: Students forget intermediates are not stable species and must be substituted out using equilibrium expressions for fast pre-steps. Correct move: Always substitute out intermediate concentrations using expressions from fast equilibrium steps before writing the final rate law.
Wrong move: Counting product particles to determine the molecularity of an elementary reaction. Why: Students count all particles in the equation instead of only reactants. Correct move: Only count the number of reactant particles on the left-hand side of the elementary step to find molecularity.
Wrong move: Labeling a catalyst as a reaction intermediate in a multi-step mechanism. Why: Students mix up the order of production and consumption for the two species types. Correct move: Follow the rule: catalysts = consumed first, produced later; intermediates = produced first, consumed later.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Which of the following correctly gives the rate law and molecularity for the elementary reaction $2 N O + O_{2} \to 2 N O_{2}$ ? A) Rate = $k [N O] [O_{2}]$ , unimolecular B) Rate = $k [N O]^{2} [O_{2}]$ , bimolecular C) Rate = $k [N O]^{2} [O_{2}]$ , termolecular D) Rate = $k [N O_{2}]^{2}$ , termolecular

Worked Solution: For elementary reactions, reaction orders equal the stoichiometric coefficients of reactants, so the rate law here must be $k [N O]^{2} [O_{2}]$ , eliminating options A and D. Molecularity counts the total number of reactant particles: 2 NO + 1 O2 = 3 particles, so molecularity is termolecular, eliminating option B. The correct answer is C.

Question 2 (Free Response)

The overall reaction $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ follows the three-step mechanism below: Step 1 (fast): $N_{2} O_{5} ⇌ N O_{2} + N O_{3}$ Step 2 (slow): $N O_{2} + N O_{3} \to N O + N O_{2} + O_{2}$ Step 3 (fast): $N O + N O_{3} \to 2 N O_{2}$ (a) Identify all reaction intermediates in this mechanism. (b) Write the rate law for the overall reaction consistent with this mechanism (use $k_{1}$ , $k_{- 1}$ , $k_{2}$ for the rate constants of step 1 forward, step 1 reverse, and step 2 respectively). (c) What is the overall order of the reaction based on this mechanism?

Worked Solution: (a) Track production and consumption: $N O_{3}$ is produced in step 1 and consumed in step 2 and 3; $N O$ is produced in step 2 and consumed in step 3. Both are produced early and consumed late, so $N O_{3}$ and $N O$ are the reaction intermediates. (b) The slow step (rate-determining step) is elementary, so its rate law is $rate = k_{2} [N O_{2}] [N O_{3}]$ . From the equilibrium in step 1, forward rate equals reverse rate: $k_{1} [N_{2} O_{5}] = k_{- 1} [N O_{2}] [N O_{3}]$ , so $[N O_{2}] [N O_{3}] = \frac{k _{1}}{k _{- 1}} [N_{2} O_{5}]$ . Substitute into the rate law: $rate = k_{2} \frac{k _{1}}{k _{- 1}} [N_{2} O_{5}] = k [N_{2} O_{5}]$ , where $k = \frac{k _{1} k _{2}}{k _{- 1}}$ . (c) The only reaction order is 1 for $N_{2} O_{5}$ , so the overall order of the reaction is 1.

Question 3 (Application / Real-World Style)

The radioactive decay of carbon-14, used for radiocarbon dating, is a unimolecular elementary first-order process with a rate constant $k = 1.21 \times 1 0^{- 4} year^{- 1}$ . A 10 g sample of ancient wood has an initial carbon-14 concentration of $1.65 \times 1 0^{- 10} mol/g$ . What is the initial rate of decay of carbon-14 in this sample, in moles per year?

Worked Solution: Since decay is an elementary unimolecular reaction, the rate law is directly derived from stoichiometry: $rate = k [^{14} C]$ . First find the total initial moles of C-14 in the 10 g sample: $10 g \times 1.65 \times 1 0^{- 10} mol/g = 1.65 \times 1 0^{- 9} mol$ . Substitute into the rate law: $rate = (1.21 \times 1 0^{- 4} year^{- 1}) (1.65 \times 1 0^{- 9} mol) = 2.00 \times 1 0^{- 13} mol/year$ . In context, this means the sample loses 2.00 × 10⁻¹³ moles of carbon-14 per year initially, which matches the expected slow decay of ancient carbon.

7. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
Elementary Step Rate Law	$rate = k \prod [X_{i}]^{n_{i}}$ , $n_{i}$ = stoichiometric coefficient of reactant $X_{i}$	Only valid for elementary reactions; does not apply to overall balanced reactions
Unimolecular Elementary Step	$A \to products$ , $rate = k [A]$	Molecularity = 1, overall order = 1
Bimolecular Elementary Step (A + B)	$A + B \to products$ , $rate = k [A] [B]$	Molecularity = 2, overall order = 2
Bimolecular Elementary Step (2A)	$2 A \to products$ , $rate = k [A]^{2}$	Molecularity = 2, overall order = 2
Termolecular Elementary Step	$a A + b B + c C \to products$ , $rate = k [A]^{a} [B]^{b} [C]^{c}$	Molecularity = 3, rare in mechanisms, overall order = 3
Molecularity	Count of reactant particles in an elementary step	Always 1, 2, or 3; never zero, negative, or fractional
Reaction Intermediate	Produced in early step, consumed in later step	Never included in final overall rate law
Catalyst	Consumed in early step, produced in later step	Not consumed in overall reaction; can appear in the rate law

8. What's Next

Mastering elementary reactions is the foundational prerequisite for working with full reaction mechanisms, the next core topic in AP Chemistry Unit 5 Kinetics. Without understanding how to write rate laws for elementary steps and identify intermediates, you cannot derive the overall rate law for a multi-step mechanism, which is a common high-weight FRQ question on the AP exam. Beyond kinetics, understanding elementary steps helps you interpret collision theory and activation energy, because each elementary step has its own activation energy and Arrhenius behavior. This topic also builds the foundation for understanding biological and industrial catalysis, where catalysts work by providing a new sequence of elementary steps with lower activation energy.

Follow-up topics: Reaction mechanisms Rate-determining steps Collision theory Catalysis

Elementary reactions — AP Chemistry Study Guide

1. What Is Elementary reactions?

2. Molecularity of Elementary Reactions

Worked Example

3. Rate Laws for Elementary Reactions

Worked Example

4. Elementary Steps in Multi-Step Reaction Mechanisms

Worked Example

5. Common Pitfalls (and how to avoid them)

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Question 2 (Free Response)

Question 3 (Application / Real-World Style)

7. Quick Reference Cheatsheet

8. What's Next

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