AP CHEMISTRY • KINETICS

Elementary Reactions

Understanding single-step molecular events whose rate laws can be written directly from their stoichiometry.

Historical Context & Motivation

The study of how fast chemical reactions proceed—chemical kinetics—gained rigor only when scientists realized that most overall reactions are composites of simpler, indivisible molecular events. Before this insight, chemists could measure macroscopic rates but struggled to explain why rate laws so often differed from what balanced equations predicted. The concept of elementary reactions resolved this puzzle by identifying the actual molecular collisions and rearrangements that constitute each step in a reaction mechanism.

1850s
Early Rate Studies
Ludwig Wilhelmy and others measured the rates of acid-catalyzed sucrose inversion, establishing that reaction speed depends on reactant concentration in quantifiable ways.
1884
van 't Hoff's "Études"
Jacobus van 't Hoff published Études de dynamique chimique, classifying reactions by their molecularity (unimolecular, bimolecular) and formalizing the idea that complex reactions proceed through discrete elementary steps.
1889
Arrhenius Equation
Svante Arrhenius proposed that an energy barrier must be overcome for an elementary step to occur, linking temperature dependence to the activation energy of each step.
1913
Bodenstein & Steady-State
Max Bodenstein applied the steady-state approximation to chains of elementary reactions in H₂ + Br₂, showing how complex rate laws emerge from sequences of simple steps.
1935
Transition State Theory
Eyring, Evans, and Polanyi developed transition state theory, providing a quantum-mechanical framework for the rate constants of individual elementary reactions.

The central question that drove this development remains the focus of modern kinetics: if a balanced equation does not directly reveal the rate law, what fundamental molecular events actually control the speed of a reaction? Elementary reactions are the answer—each one represents an irreducible molecular event whose rate law can be written by inspection.

Core Principles & Definitions

An elementary reaction (also called an elementary step) is a reaction that occurs in a single collision or molecular rearrangement event—it cannot be broken into simpler steps. Because the event happens exactly as written, the exponents in its rate law equal the stoichiometric coefficients of its reactants. This is the defining feature that distinguishes elementary reactions from overall reactions, whose rate laws must be determined experimentally.

1

Molecularity

The number of reactant particles involved in a single elementary step. Can be unimolecular (1), bimolecular (2), or rarely termolecular (3).
2

Rate Law from Stoichiometry

For an elementary step, exponents in the rate law equal the stoichiometric coefficients. For example, if A + 2B → products is elementary, then rate = k[A][B]².
3

Reaction Mechanism

A sequence of elementary reactions that, when summed, give the balanced overall equation. The mechanism must be consistent with the experimentally observed rate law.
4

Rate-Determining Step

The slowest elementary step in a multi-step mechanism. It acts as a kinetic bottleneck and largely dictates the overall rate law of the reaction.
5

Intermediates vs. Catalysts

An intermediate is produced in one elementary step and consumed in a later step. A catalyst participates in a step but is regenerated, appearing in the mechanism but not in the overall equation.
KEY TAKEAWAY
KEY TAKEAWAY

Visual Explanation — Molecularity

The three types of molecularity in elementary reactions. Unimolecular steps involve a single molecule decomposing or rearranging. Bimolecular steps involve two particles colliding. Termolecular steps require a simultaneous three-body collision and are exceedingly rare.

The diagram above illustrates the three categories of elementary reactions classified by molecularity. In a unimolecular step, a single energized molecule undergoes isomerization or bond breaking; the rate law is first order. In a bimolecular step, two species must collide with proper orientation and sufficient kinetic energy, yielding a second-order rate law. Termolecular events—requiring a simultaneous three-body collision with the right geometry and energy—are so improbable that nearly all proposed termolecular reactions are better explained as sequences of bimolecular steps. On the AP exam, you should assume that termolecular steps are essentially nonexistent in proposed mechanisms unless specifically stated otherwise.

Mathematical Framework — Rate Laws from Elementary Steps

The single most important mathematical principle for elementary reactions is that the rate law can be written directly from the balanced elementary step. This is not true for overall reactions. The distinction is critical and appears frequently on the AP Chemistry exam.

GENERAL ELEMENTARY RATE LAW
For aA + bB → products (elementary): rate = k[A]ᵃ[B]ᵇ
Here a and b are the stoichiometric coefficients from the elementary step (not the overall equation), and k is the rate constant specific to that step.
UNIMOLECULAR EXAMPLE
N₂O₄(g) → 2 NO₂(g) rate = k[N₂O₄]
One molecule of N₂O₄ decomposes; first-order rate law.
BIMOLECULAR EXAMPLE
NO(g) + O₃(g) → NO₂(g) + O₂(g) rate = k[NO][O₃]
Two different molecules collide; second-order overall (first order in each reactant).
BIMOLECULAR — SAME SPECIES
2 NO₂(g) → NO₃(g) + NO(g) rate = k[NO₂]²
Two identical molecules collide; rate law uses the coefficient as the exponent.
AP Exam Tip

When a mechanism has multiple elementary steps, the overall rate law is typically governed by the rate-determining step (RDS)—the slowest step in the sequence. The rate law for the overall reaction is the rate law of the RDS, provided that all species in that rate law are actual reactants (not intermediates). If the RDS rate law contains an intermediate, you must use a prior fast-equilibrium step to express the intermediate's concentration in terms of original reactants.

Reaction Mechanisms — Elementary Steps in Action

A reaction mechanism is a proposed sequence of elementary steps that accounts for the overall stoichiometry and the experimentally observed rate law. Consider the decomposition of ozone in the stratosphere, whose overall equation is 2 O₃(g) → 3 O₂(g). Experimentally, the rate law is rate = k[O₃]²[O₂]⁻¹. A one-step bimolecular collision of two O₃ molecules would predict rate = k[O₃]², which does not match the observed inverse dependence on O₂. The accepted mechanism involves two elementary steps with a reactive intermediate—atomic oxygen, O.

The two-step mechanism for ozone decomposition. Step 1 is a fast equilibrium producing atomic oxygen (the intermediate). Step 2, the rate-determining step, consumes O₃ and O. Substituting the equilibrium expression for [O] into the RDS rate law yields the experimentally observed rate law: rate = k[O₃]²[O₂]−1.

This example showcases several testable concepts. First, elementary steps must sum to give the overall balanced equation. Second, intermediates appear in the mechanism but cancel when steps are added—they never appear in the overall equation. Third, the rate law of the RDS initially contains an intermediate (O), which must be algebraically eliminated using the equilibrium expression from a preceding fast step.

Identifying Intermediates on the AP Exam

Worked Example — Deriving a Rate Law from a Mechanism

Consider the following proposed mechanism for the reaction 2 NO(g) + Br₂(g) → 2 NOBr(g):

  • Step 1 (slow): NO + Br₂ → NOBr₂
  • Step 2 (fast): NOBr₂ + NO → 2 NOBr

Determine the overall rate law predicted by this mechanism and identify any intermediates.

1
Step 1 — Verify the mechanism sums to the overall equationAdding Step 1 and Step 2: NO + Br₂ + NOBr₂ + NO → NOBr₂ + 2 NOBr. Cancel NOBr₂ (it appears on both sides): 2 NO + Br₂ → 2 NOBr. This matches the overall balanced equation.
✓ Mechanism is consistent with overall stoichiometry.
2
Step 2 — Identify intermediatesNOBr₂ is produced in Step 1 and consumed in Step 2. It does not appear in the overall equation.
NOBr₂ is the intermediate.
3
Step 3 — Write the rate law from the rate-determining stepStep 1 is the slow (rate-determining) step. It is bimolecular: NO + Br₂ → NOBr₂. Since it is elementary, the rate law follows directly from its stoichiometry.
rate = k₁[NO][Br₂]
4
Step 4 — Check for intermediates in the rate lawThe rate law contains only NO and Br₂—both are original reactants, not intermediates. No substitution is needed.
Overall rate law: rate = k[NO][Br₂] — second order overall (first order in each reactant).
Note on Reaction Orders

Elementary vs. Overall Reactions — Key Distinctions

The most common source of error on kinetics questions is confusing elementary reactions with overall reactions. The table below summarizes the critical differences.

Comparison of elementary and overall reactions
FeatureElementary ReactionOverall Reaction
DefinitionSingle molecular event; cannot be broken into simpler stepsNet transformation from reactants to products; may consist of multiple steps
Rate law sourceWritten directly from stoichiometric coefficientsMust be determined experimentally or derived from mechanism
MolecularityDefined (uni-, bi-, or termolecular)Not applicable — molecularity is undefined for overall reactions
IntermediatesMay produce or consume intermediatesIntermediates do not appear (they cancel)
Order vs. coefficientsAlways matchOften do not match
KEY TAKEAWAY
KEY TAKEAWAY

Connection to Advanced Theory — Transition States & Energy Profiles

Each elementary reaction passes through a transition state (also called an activated complex)—a high-energy, unstable configuration at the peak of the potential energy barrier separating reactants from products. Unlike intermediates, transition states cannot be isolated; they exist for approximately one vibrational period (around 10⁻¹³ s). In a multi-step mechanism, each elementary step has its own transition state and activation energy, and the energy profile shows multiple peaks separated by valleys where intermediates reside.

Elementary reactions in the context of advanced kinetic theory
ConceptElementary Reaction LevelAdvanced / College Extension
Activation energy (Eₐ)Each elementary step has its own Eₐ; the RDS has the largest Eₐ.Transition state theory relates k to Eₐ via k = (k_BT/h)e^(−ΔG‡/RT).
Potential energy diagramNumber of peaks = number of elementary steps.The shape and height of each barrier depend on the reaction coordinate and molecular geometry.
CatalysisA catalyst provides an alternative mechanism with a lower Eₐ for the RDS.Enzyme catalysis (Michaelis–Menten) models a two-step mechanism with an enzyme–substrate intermediate.
Steady-state approximationAt AP level: fast-equilibrium assumption used to eliminate intermediates.General steady-state: d[intermediate]/dt ≈ 0, applicable even when equilibrium is not established.

For the AP exam, you should be able to identify the number of elementary steps from a potential energy diagram (count the peaks), locate intermediates (valleys between peaks), and recognize the rate-determining step as the one with the highest activation energy barrier. In more advanced coursework, transition state theory and the steady-state approximation provide powerful tools for deriving rate laws from complex mechanisms without relying solely on the assumption of a single rate-determining step.

Practice Problems

1
Which of the following statements correctly distinguishes an elementary reaction from an overall reaction?
2
The elementary reaction 2 NO(g) + Cl₂(g) → 2 NOCl(g) has been proposed (hypothetically) as a single termolecular step. If this step is truly elementary, what is the rate law?
3
A reaction mechanism is proposed: Step 1 (fast equilibrium): A₂ ⇌ 2 A Step 2 (slow): A + B → AB What is the overall rate law predicted by this mechanism?
PROBLEM 4APPLIED
The overall reaction 2 NO₂(g) + F₂(g) → 2 NO₂F(g) is observed to follow the rate law: rate = k[NO₂][F₂]. (a) Explain why this rate law indicates the reaction is NOT a single elementary step. (b) Propose a two-step mechanism that is consistent with both the stoichiometry and the observed rate law. Clearly identify the slow step. (c) Identify any intermediates in your proposed mechanism. (d) A student claims that because the coefficient of NO₂ in the overall equation is 2, the reaction must be second order in NO₂. Evaluate this claim.
PROBLEM 5CRITICAL THINKING
The following mechanism is proposed for the reaction 2 A + B → C + D: Step 1 (fast equilibrium): A + B ⇌ X K₁ Step 2 (slow): X + A → C + D k₂ Experiment data at constant temperature: Experiment | [A]₀ (M) | [B]₀ (M) | Initial rate (M/s) 1 | 0.10 | 0.10 | 2.0 × 10⁻⁴ 2 | 0.20 | 0.10 | 8.0 × 10⁻⁴ 3 | 0.10 | 0.20 | 4.0 × 10⁻⁴ (a) Derive the predicted rate law from the proposed mechanism. (b) Using the experimental data, determine the order with respect to A and B. (c) Is the proposed mechanism consistent with the experimental data? Justify your answer. (d) Calculate the value of the composite rate constant k (= k₂K₁) including units.
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