Elementary Reactions - College Chemistry

SECTION 1

Historical Context & Motivation

The study of how fast chemical reactions proceed — and why — has captivated scientists since the birth of modern chemistry. Early chemists could measure that certain reactions were fast or slow, but they lacked a microscopic picture of what actually happened when molecules transformed into products. The concept of an elementary reaction arose from the desire to decompose complex chemical processes into their simplest, irreducible molecular events. Understanding these fundamental steps became the key to unlocking the entire field of chemical kinetics, connecting macroscopic rate measurements with molecular-level reality.

1864 • Law of Mass Action

Cato Guldberg and Peter Waage proposed that the rate of a chemical reaction is proportional to the product of the concentrations of the reactants, laying the mathematical groundwork for relating stoichiometry to reaction speed.

1889 • Arrhenius Equation

Svante Arrhenius introduced the exponential dependence of rate constants on temperature, providing a quantitative link between molecular energy and the pace of an elementary step.

1935 • Transition State Theory

Henry Eyring, Meredith Gwynne Evans, and Michael Polanyi developed transition state theory, which models elementary reactions as passages over an energy barrier through an activated complex, giving a statistical-mechanical foundation to rate constants.

1986 • Femtochemistry

Ahmed Zewail pioneered the use of ultra-fast laser pulses (on the femtosecond timescale) to observe elementary bond-breaking and bond-forming events in real time, earning the 1999 Nobel Prize in Chemistry.

A central question emerged from this history: when we write a balanced chemical equation, does the reaction actually happen in one concerted step, or does it proceed through a sequence of simpler events? This question distinguishes elementary reactions — those that occur in a single molecular event with no intermediate species — from overall (or net) reactions that may summarize dozens of elementary steps. Grasping this distinction is essential because the rate law for an elementary reaction can be written directly from its stoichiometry, a privilege that does not extend to complex, multi-step mechanisms.

SECTION 2

Core Principles & Definitions

An elementary reaction is defined as a reaction that occurs in a single step at the molecular level: reactant molecules collide, rearrange bonds, and directly form products without passing through any detectable intermediate species. Because no intermediates are involved, the molecularity of an elementary reaction — that is, the number of reactant molecules or atoms that come together in a single reactive encounter — directly determines the form of its rate law. This is the defining advantage of working at the elementary-step level, and it is a principle that cannot be applied to overall stoichiometric equations without mechanistic justification.

Molecularity

The number of reactant species (atoms, molecules, or ions) that participate as reactants in a single elementary step. Molecularity is always a positive integer — 1, 2, or (rarely) 3 — and is a theoretical property of the step, not an experimentally measured quantity.

Rate Law from Stoichiometry

For an elementary reaction, the rate law can be written directly: exponents on concentrations equal the stoichiometric coefficients. This is only valid for elementary steps, never for overall balanced equations unless the mechanism is a single step.

Microscopic Reversibility

Every elementary reaction is, in principle, reversible. At equilibrium, the forward and reverse elementary steps proceed at equal rates. The reverse reaction traverses the same potential energy surface in the opposite direction through the same transition state.

No Detectable Intermediates

An elementary reaction proceeds without forming any species that could, even in principle, be isolated or detected between reactants and products. If intermediates exist, the process is composite and must be broken into multiple elementary steps.

Transition State (Activated Complex)

During an elementary step, reactants pass through a high-energy configuration called the transition state. Unlike an intermediate, the transition state is a saddle point on the potential energy surface and cannot be isolated; it exists only fleetingly.

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KEY TAKEAWAY

Think of a complex reaction mechanism like a cross-country flight with layovers. The overall trip is "New York to Los Angeles," but the actual journey may involve individual legs — New York to Chicago, then Chicago to Denver, then Denver to Los Angeles. Each leg is an elementary reaction: a single, direct event with no hidden stops. The cities where you land between legs are reaction intermediates. You can only understand the total travel time by examining each individual leg, just as you can only derive the overall rate law by analyzing each elementary step.

SECTION 3

Potential Energy Diagram for an Elementary Reaction

A potential energy diagram (also called a reaction coordinate diagram) is the most instructive way to visualize an elementary reaction. On the horizontal axis we plot the reaction coordinate — a generalized measure of the progress from reactants to products — and on the vertical axis we plot potential energy. Because an elementary reaction occurs in a single step, its diagram shows exactly one energy maximum: the transition state. The height of this maximum above the reactant energy is the activation energy (E_a), and the difference between the product and reactant energies is the enthalpy of reaction (ΔH).

A potential energy diagram for an exothermic elementary reaction A + B → C + D. The single maximum corresponds to the transition state (‡). The forward activation energy E_a is the energy gap from reactants to the transition state, and ΔH is the difference between product and reactant energies. Because products are lower in energy, this reaction is exothermic.

Notice that the diagram features exactly one energy maximum. This is the hallmark of an elementary reaction. If a reaction proceeds through two elementary steps, its energy diagram would show two peaks separated by a valley — that valley represents the intermediate, a real (though possibly short-lived) chemical species. The transition state, by contrast, sits at the very top of the barrier and has no finite lifetime; it is a saddle point on the multidimensional potential energy surface, accessible only at the instant of molecular rearrangement.

SECTION 4

Mathematical Framework

The defining mathematical feature of elementary reactions is that their rate laws follow directly from stoichiometry. For an elementary reaction, the exponent on each reactant concentration in the rate expression equals its stoichiometric coefficient. This is a consequence of the molecular-level interpretation: if two molecules of A must collide simultaneously for the reaction to occur, the probability of that event scales with [A]². The overall order of an elementary step equals its molecularity.

Unimolecular Elementary Step

UNIMOLECULAR RATE LAW

A → Products Rate = k[A]

A single molecule undergoes rearrangement or decomposition. The reaction is first order overall. Example: the isomerization of cyclopropane to propene.

Bimolecular Elementary Step

BIMOLECULAR RATE LAW (TWO DIFFERENT SPECIES)

A + B → Products Rate = k[A][B]

Two molecules collide and react. The reaction is second order overall (first order in A, first order in B). This is the most common type of elementary step.

BIMOLECULAR RATE LAW (SAME SPECIES)

2A → Products Rate = k[A]²

Two identical molecules collide. The reaction is second order in A and second order overall. Example: 2 NO₂ → 2 NO + O₂ (if it were elementary, which it is not).

Termolecular Elementary Step

TERMOLECULAR RATE LAW

A + B + C → Products Rate = k[A][B][C]

Three molecules must collide simultaneously — an exceedingly rare event. Termolecular steps are the highest molecularity observed; reactions requiring four-body collisions are considered essentially impossible. Most proposed termolecular steps involve a third body M that carries away excess energy, as in 2 NO + O₂ → 2 NO₂.

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Molecularity vs. Order

Molecularity is a theoretical, integer property of an elementary step and refers to how many reactant particles participate. Reaction order is an experimentally determined quantity that describes how the rate depends on concentrations. For elementary reactions the two coincide, but for overall reactions the order must be determined empirically — it may be fractional, zero, or have no simple relationship to stoichiometric coefficients.

SECTION 5

Classifying Elementary Reactions by Molecularity

Elementary reactions are classified primarily by their molecularity. The following table summarizes the three categories along with their order, probability of occurrence, and representative examples. Note that the probability of a simultaneous collision drops sharply as the number of participants increases, which is why termolecular elementary steps are rare and no verified tetramolecular elementary step has ever been observed.

Molecularity	Participants	Rate Law Form	Overall Order	Example
Unimolecular	1	Rate = k[A]	1st order	Cyclopropane → Propene
Bimolecular	2	Rate = k[A][B] or k[A]²	2nd order	NO₂ + CO → NO + CO₂
Termolecular	3	Rate = k[A][B][C]	3rd order	2 NO + O₂ → 2 NO₂

Classification of elementary reactions by molecularity

Visual comparison of the three molecularity classes. Bimolecular steps dominate because two-body collisions are the most probable reactive events. Termolecular steps are rare because three molecules must collide simultaneously with the correct geometry and sufficient energy. No elementary reaction with molecularity ≥ 4 has ever been confirmed.

An important subtlety concerns unimolecular reactions. Although only one molecule appears in the chemical equation, the molecule must first acquire sufficient energy to cross the activation barrier. In the gas phase, this energy is typically delivered by collisions with other molecules (as described by the Lindemann–Hinshelwood mechanism). At high pressures the rate-limiting step is the unimolecular decomposition itself, so the kinetics appear first order; at very low pressures the rate-limiting step becomes the energization by collisions, and the kinetics can shift toward second order. This fall-off behavior is a rich topic in its own right, but the key point for our purposes is that the elementary step — the actual bond-breaking or rearrangement — remains unimolecular regardless of how energy is delivered.

SECTION 6

Worked Example: Rate Law from a Mechanism

Consider the gas-phase reaction 2 NO₂(g) + F₂(g) → 2 NO₂F(g). The experimentally determined rate law is Rate = k[NO₂][F₂]. A proposed mechanism consists of two elementary steps. Let us verify that the mechanism is consistent with the observed rate law and identify key features.

Deriving the Rate Law from Elementary Steps

1
Step 1 — Write the Proposed Mechanism

The mechanism consists of two elementary steps: Step 1 (slow): NO₂ + F₂ → NO₂F + F Step 2 (fast): NO₂ + F → NO₂F
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Step 2 — Verify the Steps Sum to the Overall Reaction

Adding the two elementary steps: NO₂ + F₂ + NO₂ + F → NO₂F + F + NO₂F. Canceling the intermediate F from both sides gives 2 NO₂ + F₂ → 2 NO₂F, which matches the overall equation.

✓ Steps sum correctly. F is an intermediate (produced in step 1, consumed in step 2).
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Step 3 — Identify the Rate-Determining Step

Step 1 is labeled as the slow step. In a multi-step mechanism where one step is much slower than the others, the rate-determining step (RDS) controls the overall rate. The overall rate law therefore mirrors the rate law of this slowest elementary step.
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Step 4 — Write the Rate Law of the RDS

Because Step 1 is an elementary reaction (bimolecular), we write its rate law directly from its stoichiometry: Rate = k₁[NO₂][F₂]. This expression contains no intermediates — only reactants that appear in the overall equation.

Rate = k[NO₂][F₂] — this matches the experimentally observed rate law (second order overall, first order in each reactant).
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Step 5 — Determine Molecularity and Order

Step 1 is bimolecular (two reactant molecules collide). Step 2 is also bimolecular (NO₂ + F). The overall reaction is second order, consistent with the rate-determining bimolecular elementary step.

Molecularity of RDS = 2 (bimolecular); Overall order = 2.

SECTION 7

Elementary vs. Overall Reactions

One of the most common mistakes in chemical kinetics is to treat an overall balanced equation as though it were an elementary reaction and to write a rate law directly from its coefficients. This shortcut is only valid if the reaction genuinely occurs in a single elementary step. The table below highlights the critical distinctions between elementary and overall reactions, making clear why the mechanistic perspective is essential.

Feature	Elementary Reaction	Overall (Net) Reaction
Definition	Single molecular event — one set of bonds breaks/forms in one step	Sum of all elementary steps; describes net stoichiometric change
Intermediates	None — no detectable species between reactants and products	May involve one or more intermediates produced and consumed in different steps
Rate law	Can be written directly from stoichiometric coefficients	Must be determined experimentally; cannot be deduced from balanced equation alone
Molecularity	Well-defined (1, 2, or 3)	Not applicable — molecularity is undefined for overall reactions
Energy diagram	Single energy maximum (one transition state)	Multiple maxima separated by energy minima (intermediates)
Example	NO₂ + F₂ → NO₂F + F (one step)	2 NO₂ + F₂ → 2 NO₂F (sum of two elementary steps)

Elementary vs. overall reactions — a side-by-side comparison

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KEY TAKEAWAY

Consider the analogy to assembling a product on a factory assembly line. The overall reaction is like the description on the shipping box: 'Raw materials in, finished product out.' It tells you nothing about which station is the bottleneck. Each elementary step corresponds to a single workstation on the line, and the slowest station (the rate-determining step) dictates the throughput of the entire factory. To optimize production, you must understand each individual station — not just the overall input and output.

SECTION 8

Connection to Advanced Kinetic Theory

Elementary reactions serve as the foundation for several advanced theoretical frameworks in chemical kinetics. While this lesson focuses on identifying and characterizing elementary steps, it is instructive to see how these concepts extend into more sophisticated treatments that you will encounter in physical chemistry and beyond.

Concept at This Level	Advanced Extension	Key Idea
Rate law from stoichiometry	Collision Theory	Derives rate constants from molecular speeds, collision cross-sections, and the fraction of collisions with sufficient energy (e⁻ᴱᵃ/ᴿᵀ)
Transition state	Transition State Theory (Eyring)	Models the activated complex as a quasi-equilibrium species; k = (k_BT/h) × e⁻ΔG‡/RT relates the rate constant to Gibbs free energy of activation
Unimolecular steps	RRKM / Lindemann Theory	Accounts for pressure-dependent kinetics of unimolecular reactions using energy-dependent microcanonical rate constants
Mechanism as a series of elementary steps	Steady-State Approximation	Assumes d[intermediate]/dt ≈ 0 to algebraically eliminate intermediate concentrations from rate expressions when no single step is clearly rate-limiting
Microscopic reversibility	Detailed Balance	At equilibrium, each elementary step is individually balanced; the principle constrains the relationship among forward and reverse rate constants (K_eq = $k_f$ / $k_r$ for each step)

How elementary reaction concepts extend into advanced kinetic theory

These connections illustrate that mastery of elementary reactions is not merely an introductory exercise — it is the conceptual bedrock on which quantitative kinetic models are built. When you later derive rate laws using the steady-state approximation or pre-equilibrium assumption, every algebraic step will rely on writing rate laws for individual elementary reactions. Similarly, computational chemistry methods that map out potential energy surfaces are essentially searching for the transition states and intermediates that define the sequence of elementary steps in a mechanism.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain why the rate law for an overall (net) reaction cannot, in general, be written directly from its balanced equation, whereas the rate law for an elementary reaction can. In your answer, clearly distinguish between molecularity and reaction order.

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An elementary reaction describes a single molecular event, so its stoichiometric coefficients literally represent the number of molecules colliding. The probability of that collision is proportional to the concentration of each participant raised to its stoichiometric coefficient, making the rate law derivable by inspection. An overall reaction, by contrast, may consist of multiple elementary steps, and its observed rate law reflects the kinetics of the rate-determining step (or a combination of steps). Molecularity is a property of a single elementary step and must be a positive integer, while reaction order is an experimentally determined exponent that can be fractional, zero, or negative, and applies to any rate law regardless of mechanism.

Question 2: BASIC CALCULATION

The elementary reaction 2 NO(g) + Cl₂(g) → 2 NOCl(g) is proposed as a termolecular step. Write its rate law and determine the overall reaction order.

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Because this is stated to be an elementary reaction, the rate law follows directly from the stoichiometry: Rate = k[NO]²[Cl₂]. The exponent on NO is 2 (its stoichiometric coefficient) and on Cl₂ is 1. The overall order is 2 + 1 = 3, consistent with a termolecular step. Note that termolecular reactions are rare, and many reactions that appear termolecular based on their stoichiometry actually proceed through a sequence of bimolecular steps.

Question 3: INTERMEDIATE

A reaction A₂ + B → C has the following proposed mechanism: Step 1 (fast equilibrium): A₂ ⇌ 2 A (k₁ forward, k₋₁ reverse) Step 2 (slow): A + B → C (k₂) Derive the overall rate law in terms of reactants only.

Show answer

The slow step (Step 2) determines the rate: Rate = k₂[A][B]. However, A is an intermediate and must be eliminated. Because Step 1 is a fast pre-equilibrium, we set the forward and reverse rates equal: k₁[A₂] = k₋₁[A]², giving [A]² = (k₁/k₋₁)[A₂], hence [A] =

(k₁/k₋₁)^{1/2}

× [A₂]^(1/2). Substituting into the rate expression: Rate = k₂

(k₁/k₋₁)^{1/2}

[A₂]^(1/2)[B]. Defining k_obs = k₂

(k₁/k₋₁)^{1/2}

, we get Rate = k_obs[A₂]^(1/2)[B]. The reaction is 1/2 order in A₂ and first order in B, giving an overall order of 3/2 — illustrating that overall reaction orders need not be integers.

Question 4: APPLIED

Ozone decomposition in the stratosphere proceeds by the following mechanism: Step 1: O₃ → O₂ + O (fast equilibrium, k₁/k₋₁) Step 2: O + O₃ → 2 O₂ (slow, k₂) (a) Identify the molecularity of each elementary step. (b) Identify the intermediate. (c) Derive the overall rate law using the pre-equilibrium approach.

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(a) Step 1 forward is unimolecular (O₃ decomposes alone); Step 1 reverse is bimolecular (O₂ + O recombine); Step 2 is bimolecular (O + O₃ collide). (b) The oxygen atom O is the intermediate — it is produced in Step 1 and consumed in Step 2 but does not appear in the overall equation 2 O₃ → 3 O₂. (c) From Step 2: Rate = k₂[O][O₃]. To eliminate [O], use the pre-equilibrium from Step 1: K = k₁/k₋₁ = [O₂][O]/[O₃], giving [O] = K[O₃]/[O₂] = (k₁/k₋₁)[O₃]/[O₂]. Substituting: Rate = k₂(k₁/k₋₁)[O₃]²/[O₂] = k_obs[O₃]²[O₂]⁻¹. The rate is second order in O₃ and inverse first order in O₂, explaining why high O₂ concentrations slow the decomposition.

Question 5: CRITICAL THINKING

A student argues that because the overall reaction H₂(g) + I₂(g) → 2 HI(g) has the experimentally observed rate law Rate = k[H₂][I₂], it must be a single bimolecular elementary step. Critically evaluate this claim. Could a multi-step mechanism also produce the same rate law? Construct such a mechanism and discuss what additional experiments might distinguish between the two possibilities.

Show answer

The student's reasoning is tempting but logically flawed. Agreement between stoichiometric coefficients and rate law exponents is a necessary condition for a single-step mechanism, but it is not sufficient — a multi-step mechanism can yield the same rate law. For example, consider: Step 1 (fast equilibrium): I₂ ⇌ 2 I; Step 2 (slow): H₂ + 2 I → 2 HI. From Step 2: Rate = k₂[H₂][I]². Substituting [I]² = K[I₂]: Rate = k₂K[H₂][I₂] = k_obs[H₂][I₂], identical to the observed law. Historically, this reaction was long considered a textbook example of a single bimolecular step, but work by Sullivan (1967) and subsequent studies provided evidence for the two-step mechanism involving I atoms. To distinguish the mechanisms, one could: (1) search for iodine atoms spectroscopically under reaction conditions; (2) examine the temperature dependence — different mechanisms predict different activation energies; (3) use isotopic labeling or molecular beam experiments to probe the transition state geometry. This example powerfully illustrates that kinetics alone cannot prove a mechanism — it can only rule out inconsistent ones.

SUMMARY

Summary

An elementary reaction is the simplest possible chemical event — a single molecular encounter in which bonds break and/or form without any detectable intermediate species. Elementary steps are classified by their molecularity: unimolecular (one reactant, first-order rate law), bimolecular (two reactants, second-order rate law — the most common type), and termolecular (three reactants, third-order rate law — rare). The defining feature of elementary reactions is that their rate laws can be written directly from their stoichiometric coefficients, a privilege not shared by overall balanced equations.

On a potential energy diagram, an elementary reaction shows exactly one energy maximum — the transition state — unlike multi-step mechanisms that display multiple peaks and valleys. The height of this barrier is the activation energy (E_a). When a mechanism consists of several elementary steps, the rate-determining step (the slowest elementary step) governs the overall rate, and its rate law — written using the elementary-step rule — forms the basis for the experimentally observed rate expression. Mastering elementary reactions is the essential first step toward understanding reaction mechanisms, transition state theory, and the steady-state approximation.