Elementary Reactions - AP Chemistry

SECTION 1

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.

SECTION 2

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.

Molecularity

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

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.

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.

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.

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.

SECTION 3

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.

SECTION 4

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.

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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.

SECTION 5

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.

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Identifying Intermediates on the AP Exam

SECTION 6

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.

Worked example

1
Step 1 — Verify the mechanism sums to the overall equation

Adding 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 intermediates

NOBr₂ 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 step

Step 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 law

The 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).

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Note on Reaction Orders

SECTION 7

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.

Feature	Elementary Reaction	Overall Reaction
Definition	Single molecular event; cannot be broken into simpler steps	Net transformation from reactants to products; may consist of multiple steps
Rate law source	Written directly from stoichiometric coefficients	Must be determined experimentally or derived from mechanism
Molecularity	Defined (uni-, bi-, or termolecular)	Not applicable — molecularity is undefined for overall reactions
Intermediates	May produce or consume intermediates	Intermediates do not appear (they cancel)
Order vs. coefficients	Always match	Often do not match

Comparison of elementary and overall reactions

SECTION 8

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.

Concept	Elementary Reaction Level	Advanced / 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 diagram	Number of peaks = number of elementary steps.	The shape and height of each barrier depend on the reaction coordinate and molecular geometry.
Catalysis	A 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 approximation	At AP level: fast-equilibrium assumption used to eliminate intermediates.	General steady-state: d[intermediate]/dt ≈ 0, applicable even when equilibrium is not established.

Elementary reactions in the context of advanced kinetic theory

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.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Which of the following statements correctly distinguishes an elementary reaction from an overall reaction?

A. An elementary reaction's rate law must be determined experimentally, whereas an overall reaction's rate law can be written from its stoichiometry.
B. An elementary reaction occurs in a single step, and its rate law exponents equal the stoichiometric coefficients of its reactants.
C. An elementary reaction always involves exactly two reactant molecules colliding.
D. An elementary reaction has no activation energy because it occurs in a single step.

Show answer (B)

B is correct. An elementary reaction is a single molecular event whose rate law can be written directly from its stoichiometry—this is the defining characteristic. A is wrong because it reverses the relationship: it is the overall reaction whose rate law must be determined experimentally, not the elementary reaction. C is wrong because elementary reactions can be unimolecular (one molecule) or termolecular (three molecules), not only bimolecular. D is wrong because every reaction—elementary or otherwise—requires some activation energy to proceed.

Question 2: BASIC CALCULATION

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?

A. rate = k[NO][Cl₂]
B. rate = k[NO]²[Cl₂]
C. rate = k[NOCl]²
D. rate = k[NO]²[Cl₂]²

Show answer (B)

B is correct. For an elementary reaction, the exponents equal the stoichiometric coefficients: the coefficient of NO is 2 and the coefficient of Cl₂ is 1, so rate = k[NO]²[Cl₂]. A is wrong because it uses an exponent of 1 for NO instead of 2. C is wrong because products never appear in the rate law of a forward reaction. D is wrong because Cl₂ has a coefficient of 1, not 2, in the elementary step.

Question 3: INTERMEDIATE

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?

A. rate = k[A₂][B]
B. rate = k[A][B]
C. rate = k[A₂]^(1/2)[B]
D. rate = k[A₂]²[B]

Show answer (C)

C is correct. The RDS (Step 2) gives rate = k₂[A][B]. A is an intermediate, so we eliminate it using the Step 1 equilibrium: K₁ = [A]²/[A₂], which gives [A] =

(K₁[A₂])^{1/2}

. Substituting: rate = k₂

(K₁[A₂])^{1/2}

[B] = k[A₂]^(1/2)[B]. A is wrong because it treats A₂ as first order without performing the equilibrium substitution. B is wrong because A is an intermediate and must not appear in the final rate law. D is wrong because squaring A₂ has no basis in either the RDS rate law or the equilibrium expression.

Question 4: APPLIED

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.

Show answer

(a) If the overall reaction 2 NO₂ + F₂ → 2 NO₂F were a single elementary step, its rate law would be rate = k[NO₂]²[F₂] (exponents matching coefficients). The observed rate law is rate = k[NO₂][F₂] — first order in NO₂, not second order. This discrepancy proves the reaction proceeds through multiple elementary steps. (b) Step 1 (slow): NO₂ + F₂ → NO₂F + F. Step 2 (fast): NO₂ + F → NO₂F. The rate-determining Step 1 is elementary and bimolecular, so rate = k₁[NO₂][F₂], which matches the observed rate law. The two steps sum to 2 NO₂ + F₂ → 2 NO₂F. (c) The fluorine atom (F) is an intermediate: it is produced in Step 1 and consumed in Step 2. (d) The student's claim is incorrect. Stoichiometric coefficients determine rate-law exponents only for elementary reactions. For overall reactions, rate-law exponents must come from experiment. In this case, NO₂ is first order despite its coefficient of 2.

Question 5: CRITICAL 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.

Show answer

(a) The rate-determining step (Step 2) gives rate = k₂[X][A]. Since X is an intermediate, use the fast equilibrium from Step 1: K₁ = [X]/([A][B]), so [X] = K₁[A][B]. Substituting: rate = k₂K₁[A][B][A] = k₂K₁[A]²[B]. Letting k = k₂K₁, the predicted rate law is rate = k[A]²[B]. (b) Comparing Experiments 1 and 2: [A] doubles (0.10 → 0.20) while [B] is held constant, and the rate increases by a factor of 4. Since 2ⁿ = 4, n = 2 (second order in A). Comparing Experiments 1 and 3: [B] doubles while [A] is held constant, and the rate doubles. Since 2ᵐ = 2, m = 1 (first order in B). (c) The mechanism predicts rate = k[A]²[B], which is second order in A and first order in B, matching the experimentally determined orders exactly. The mechanism is consistent with the data. (d) Substituting Experiment 1 data: 2.0 × 10⁻⁴ M/s = k(0.10 M)²(0.10 M) = k(1.0 × 10⁻³ M³). Solving: k = 2.0 × 10⁻⁴ / 1.0 × 10⁻³ = 0.20 M⁻²s⁻¹. Verification with Exp 2: 0.20 × (0.20)²(0.10) = 0.20 × 4.0 × 10⁻³ = 8.0 × 10⁻⁴ M/s ✓.