Elementary Steps & Intermediates — Identify elementary steps, intermediates, and molecularity - General Chemistry 2

SECTION 1

Historical Context & Motivation

The balanced chemical equation for a reaction tells us the stoichiometric relationship between reactants and products, but it reveals almost nothing about how the transformation actually occurs at the molecular level. By the mid-nineteenth century, chemists recognized that many reactions do not proceed in a single concerted event; instead, they unfold through a sequence of simpler steps. The quest to identify these elementary steps — and the transient species that connect them — became one of the central endeavors of chemical kinetics, forging a bridge between macroscopic rate data and the invisible choreography of molecular collisions.

1850 • Wilhelmy's Rate Measurements

Ludwig Wilhelmy measured the rate of sucrose inversion in acid, providing the first quantitative rate law and hinting that reactions proceed through distinct stages rather than instantaneously.

1884 • van 't Hoff's Études de Dynamique Chimique

Jacobus van 't Hoff systematized reaction orders and proposed that overall reactions could be decomposed into a series of simpler molecular events, laying the conceptual groundwork for reaction mechanisms.

1893 • Bodenstein & the Hydrogen–Iodine System

Max Bodenstein studied H₂ + I₂ → 2 HI and demonstrated that the observed rate law matched a bimolecular elementary step, catalyzing the idea that each elementary step has a rate law dictated directly by its stoichiometry.

1913 • Lindemann–Christiansen Mechanism

Frederick Lindemann proposed that unimolecular decompositions actually begin with a bimolecular activation step, introducing the concept of energized intermediates and multi-step mechanisms for seemingly simple reactions.

1935 • Eyring & Transition-State Theory

Henry Eyring formalized transition-state theory, giving each elementary step a well-defined activation energy and connecting molecular-level events to experimentally measurable rate constants.

The overarching question that this historical arc addresses is deceptively simple: if a balanced equation does not tell us the pathway, what does? The answer lies in dissecting the overall reaction into its elementary steps, identifying the reactive intermediates that shuttle the transformation from one step to the next, and classifying each step by its molecularity — the number of reactant species that collide in a single molecular event.

SECTION 2

Core Principles & Definitions

Understanding reaction mechanisms requires three interlocking concepts: the elementary step as the irreducible unit of chemical change, the intermediate as the ephemeral species generated and consumed within a mechanism, and molecularity as the classification that connects each step to its intrinsic rate law. These ideas are not merely taxonomic; they give chemists predictive power over how reaction rates depend on concentration, temperature, and catalyst identity.

Elementary Step

A reaction that occurs in a single molecular event — one bond-breaking/bond-forming act with no detectable intermediates between reactants and products. Its rate law can be written directly from its stoichiometry.

Reaction Intermediate

A species that is produced in one elementary step and consumed in a subsequent step. It does not appear in the overall balanced equation. Unlike transition states, intermediates occupy energy minima and can sometimes be detected spectroscopically.

Molecularity

The number of reactant molecules (or atoms, ions) that participate in a single elementary step. Classified as unimolecular (1), bimolecular (2), or — rarely — termolecular (3). Molecularity is always a positive integer and applies only to elementary steps, never to overall reactions.

Reaction Mechanism

The complete sequence of elementary steps that converts overall reactants to products. A valid mechanism must satisfy three criteria: the steps must sum to the overall equation, the rate law derived from the mechanism must match experiment, and each step must be physically plausible (molecularity ≤ 3).

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KEY TAKEAWAY

Think of an overall reaction as a cross-country road trip and each elementary step as one leg of the journey between cities. The cities you pass through — places you visit but do not start or end at — are the intermediates. Molecularity is the number of cars that must arrive at the same intersection simultaneously for each leg to proceed. A solo driver is unimolecular; two cars meeting head-on is bimolecular; three cars converging at once is termolecular — possible, but statistically improbable.

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Intermediate vs. Transition State

Students often confuse intermediates with transition states. A transition state (‡) sits at the top of an energy barrier — an energy maximum — and cannot be isolated. An intermediate sits in a local energy minimum between two barriers. It has a finite, though often brief, lifetime and can in principle be detected or even trapped.

SECTION 3

Energy Diagram: Intermediates & Transition States

A reaction energy diagram (potential energy vs. reaction coordinate) is the most powerful tool for distinguishing intermediates from transition states and for visualizing the sequence of elementary steps. In the diagram below, a two-step mechanism is depicted: the overall reaction A → C proceeds through an intermediate B that occupies a local energy minimum between two activation barriers.

The curve traces the potential energy along the reaction coordinate. Transition states (TS₁ ‡ and TS₂ ‡) appear at energy maxima, while the intermediate B occupies the local energy minimum between the two barriers. The taller barrier (E_a,2) identifies the rate-determining step.

Notice that the intermediate B does not appear in the overall equation A → C. It is generated in step 1 and consumed in step 2, existing only transiently during the reaction. The step with the higher activation energy (here, step 2 with E_a,2) is the rate-determining step — the kinetic bottleneck that controls the overall reaction rate. This concept will be pivotal when we derive the predicted rate law from a proposed mechanism.

SECTION 4

Mathematical Framework — Rate Laws from Elementary Steps

The defining mathematical property of an elementary step is that its rate law follows directly from its stoichiometry. This is emphatically not true for overall reactions, whose rate laws must be determined experimentally. Because an elementary step represents a single molecular event, the probability of that event is proportional to the concentrations of the participating species, each raised to the power of its stoichiometric coefficient in that step.

UNIMOLECULAR RATE LAW

A → products Rate = k[A]

One molecule rearranges or decomposes; the rate is first-order in A. Example: isomerization of cyclopropane to propene.

BIMOLECULAR RATE LAW

A + B → products Rate = k[A][B]

Two molecules collide; the rate is first-order in each reactant (second-order overall). If both molecules are the same species: 2A → products, Rate = k[A]².

TERMOLECULAR RATE LAW

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

Three molecules collide simultaneously — a statistically rare event. Termolecular steps are uncommon but not unknown; examples include 2 NO + O₂ → 2 NO₂ (though alternative two-step mechanisms are often debated).

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Critical Distinction

Molecularity is a theoretical property of a single elementary step — it is always a positive integer (1, 2, or 3). Reaction order is an experimentally determined quantity that can be zero, fractional, or even negative. Molecularity and reaction order coincide only for elementary steps. Confusing the two is one of the most common errors in kinetics problems.

To derive the overall rate law from a multi-step mechanism, one typically identifies the rate-determining step and writes its rate law directly. If that rate law contains an intermediate, the steady-state approximation or the pre-equilibrium approximation is applied to express the intermediate's concentration in terms of the starting reactants, yielding a rate law that can be compared to experiment.

SECTION 5

Molecularity — Classification & Examples

Molecularity classifies each elementary step by the number of reactant particles involved in that single collision event. The classification has profound consequences for how we write rate laws and for the physical likelihood of a given step occurring. Below we compare the three categories with representative examples and characteristic rate law forms.

Molecularity	Participants	Rate Law Form	Example
Unimolecular	1 molecule	Rate = k[A]	O₃* → O₂ + O (ozone decomposition)
Bimolecular	2 molecules	Rate = k[A][B] or k[A]²	NO₂ + CO → NO + CO₂
Termolecular	3 molecules	Rate = k[A][B][C] or k[A]²[B]	2 NO + O₂ → 2 NO₂

Classification of elementary steps by molecularity

Schematic depictions of unimolecular, bimolecular, and termolecular elementary steps. Each panel shows the reactant particles, the collision event, and the resulting products, alongside the corresponding rate law form.

Termolecular steps deserve special scrutiny. The probability of three molecules colliding simultaneously with the correct orientation and sufficient energy is exceedingly low. When a proposed termolecular step appears in the literature — such as 2 NO + O₂ → 2 NO₂ — careful analysis often reveals that a two-step bimolecular pathway is kinetically indistinguishable from the termolecular interpretation. For examination purposes, treat termolecular steps as rare but not impossible, and always consider whether an alternative bimolecular mechanism could account for the observed rate law.

SECTION 6

Worked Example — Decomposition of N₂O₅

Consider the overall reaction for the thermal decomposition of dinitrogen pentoxide:

OVERALL REACTION

2 N₂O₅(g) → 4 NO₂(g) + O₂(g)

Experimentally, this reaction is first-order: Rate = k[N₂O₅].

A proposed mechanism consists of three elementary steps:

Step 1 (slow): N₂O₅ → NO₂ + NO₃
Step 2 (fast): NO₂ + NO₃ → NO₂ + NO + O₂
Step 3 (fast): NO + NO₃ → 2 NO₂

Identifying Elementary Steps, Intermediates, and Molecularity

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Step 1 — Verify the Mechanism Sums to the Overall Reaction

Add all three elementary steps and cancel species that appear on both sides. N₂O₅ → NO₂ + NO₃ | NO₂ + NO₃ → NO₂ + NO + O₂ | NO + NO₃ → 2 NO₂. Canceling NO₃ (appears twice as product, twice as reactant), NO (produced in step 2, consumed in step 3), and one NO₂ (produced in step 1, consumed in step 2) gives: 2 N₂O₅ → 4 NO₂ + O₂. This matches the overall equation.

✓ Mechanism sums correctly to 2 N₂O₅ → 4 NO₂ + O₂
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Step 2 — Identify the Intermediates

Intermediates are species produced in one step and consumed in a later step — they do not appear in the overall balanced equation. Scan each species: NO₃ is produced in step 1 and consumed in steps 2 and 3. NO is produced in step 2 and consumed in step 3. Neither NO₃ nor NO appears in the overall equation.

Intermediates: NO₃ and NO
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Step 3 — Classify Each Step by Molecularity

Count the number of reactant particles in each elementary step. Step 1: N₂O₅ → ... (one reactant molecule) → unimolecular. Step 2: NO₂ + NO₃ → ... (two reactant molecules) → bimolecular. Step 3: NO + NO₃ → ... (two reactant molecules) → bimolecular.

Step 1: unimolecular | Step 2: bimolecular | Step 3: bimolecular
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Step 4 — Write the Rate Law for Each Elementary Step

Because each step is elementary, its rate law follows directly from its stoichiometry. Step 1: Rate₁ = k₁[N₂O₅]. Step 2: Rate₂ = k₂[NO₂][NO₃]. Step 3: Rate₃ = k₃[NO][NO₃].

Rate₁ = k₁[N₂O₅] Rate₂ = k₂[NO₂][NO₃] Rate₃ = k₃[NO][NO₃]
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Step 5 — Determine the Predicted Overall Rate Law

Step 1 is the slow (rate-determining) step. The overall rate is governed by this step: Rate = k₁[N₂O₅]. This rate law is first-order in N₂O₅, which matches the experimentally observed rate law. The mechanism is therefore consistent with the kinetic data.

Predicted rate law: Rate = k[N₂O₅] — matches experiment ✓

SECTION 7

Elementary Steps vs. Overall Reactions — Key Differences

A persistent source of confusion in chemical kinetics is the conflation of properties that belong exclusively to elementary steps with those that describe overall reactions. The table below sharpens the distinction across several critical dimensions.

Property	Elementary Step	Overall Reaction
Rate law source	Written directly from stoichiometry	Must be determined experimentally
Molecularity	Defined (1, 2, or 3)	Not applicable — meaningless concept
Reaction order	Equals molecularity (always integer)	Can be zero, fractional, or negative
Intermediates	May appear as products or reactants	Do not appear (cancel when steps are summed)
Transition states	Exactly one per elementary step	One per step in the mechanism
Reversibility	Each step is microscopically reversible	May be treated as irreversible if K >> 1

Elementary step properties vs. overall reaction properties

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KEY TAKEAWAY

The overall balanced equation is like the summary on the back of a novel: it tells you where the story begins and ends, but reveals nothing about the plot twists in between. The mechanism — the sequence of elementary steps — is the full narrative. You can only write a rate law from stoichiometry when you are reading a single chapter (an elementary step), never from the summary (the overall reaction).

SECTION 8

Connections to Advanced Kinetics

The elementary-step framework introduced here is the foundation upon which several more sophisticated treatments are built. As you advance in physical chemistry and biochemistry, you will encounter approximations and theories that refine how intermediates and rate laws are handled in complex mechanisms.

Concept in This Lesson	Advanced Extension	Key Idea
Rate-determining step	Steady-state approximation	When no single step is overwhelmingly slow, set d[intermediate]/dt ≈ 0 to solve for intermediate concentrations algebraically.
Intermediate in equilibrium	Pre-equilibrium approximation	If the first step is fast and reversible, treat reactants and intermediates as being in rapid equilibrium (K = k₁/k₋₁).
Molecularity & collision	Collision theory & TST	Collision theory quantifies bimolecular rate constants from molecular speeds and cross-sections; transition-state theory adds a thermodynamic activation barrier (ΔG‡).
Enzyme–substrate intermediate	Michaelis–Menten kinetics	The ES complex is a biochemical intermediate; applying the steady-state approximation to [ES] yields the Michaelis–Menten equation: v = V_max[S] / ( $K_M$ + [S]).

How elementary-step concepts extend into advanced kinetics

The rigorous treatment of elementary steps also connects to computational chemistry, where quantum mechanical calculations can map out the potential energy surface for each step, locate transition states, and predict activation energies from first principles. The conceptual tools you are developing now — identifying intermediates, assigning molecularity, and writing rate laws from mechanism steps — are precisely the skills required to evaluate whether a computationally proposed mechanism is consistent with experimental kinetic data.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain why molecularity is defined only for elementary steps and not for overall reactions. In your answer, address why the stoichiometric coefficients of an overall reaction cannot be used to write its rate law.

Show answer

Molecularity describes the number of molecules participating in a single collision event — a concept that only has physical meaning for an elementary step, which occurs in one molecular act. An overall reaction is the net result of multiple elementary steps, each with its own molecularity, so the overall stoichiometric coefficients do not correspond to any single collision. The rate law for an overall reaction depends on which step is rate-determining and how intermediates are related to reactant concentrations, not on the overall stoichiometry. For instance, 2 N₂O₅ → 4 NO₂ + O₂ is first-order, not second-order, because the rate-determining step involves only one N₂O₅ molecule.

Question 2: BASIC CALCULATION

Consider the elementary step: 2 NO(g) + Cl₂(g) → 2 NOCl(g). (a) What is the molecularity of this step? (b) Write its rate law. (c) What is the overall reaction order?

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(a) Three reactant molecules participate (two NO and one Cl₂), so this is a termolecular step. (b) Because this is an elementary step, the rate law is written directly from the stoichiometry: Rate = k[NO]²[Cl₂]. Each reactant's concentration is raised to the power of its stoichiometric coefficient in the step. (c) The overall reaction order is the sum of the exponents: 2 + 1 = 3 (third-order overall). Note that termolecular steps are rare because three-body collisions with proper orientation and energy are statistically improbable.

Question 3: INTERMEDIATE

A proposed mechanism for the reaction 2 NO₂(g) + F₂(g) → 2 NO₂F(g) is: Step 1 (slow): NO₂ + F₂ → NO₂F + F | Step 2 (fast): NO₂ + F → NO₂F. (a) Identify all intermediates. (b) Determine the molecularity of each step. (c) Verify that the mechanism sums to the overall reaction. (d) Derive the predicted rate law.

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(a) The fluorine atom F is produced in step 1 and consumed in step 2, so F is the only intermediate. It does not appear in the overall equation. (b) Step 1 has two reactant molecules (NO₂ and F₂), making it bimolecular. Step 2 also has two reactants (NO₂ and F), so it is bimolecular as well. (c) Adding the steps: NO₂ + F₂ + NO₂ + F → NO₂F + F + NO₂F. Canceling F on both sides yields 2 NO₂ + F₂ → 2 NO₂F, which matches. (d) Step 1 is the rate-determining step. Its rate law is Rate = k₁[NO₂][F₂]. This contains no intermediates, so no further substitution is needed. The predicted rate law is first-order in NO₂ and first-order in F₂, second-order overall.

Question 4: APPLIED

The catalytic destruction of ozone in the stratosphere proceeds via the mechanism: Step 1: O₃ + Cl → ClO + O₂ | Step 2: ClO + O → Cl + O₂. (a) What is the overall reaction? (b) Identify the catalyst and the intermediate. (c) Assign molecularity to each step. (d) If step 2 is rate-determining, write the overall rate law. Why might this rate law be difficult to compare directly with experimental data?

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(a) Summing the steps and canceling Cl and ClO: O₃ + O → 2 O₂. (b) Cl is the catalyst — it is consumed in step 1 and regenerated in step 2, appearing unchanged in the net equation. ClO is the intermediate — produced in step 1 and consumed in step 2, absent from the overall equation. (c) Both steps involve two reactant particles, so both are bimolecular. (d) If step 2 is rate-determining: Rate = k₂[ClO][O]. However, ClO is an intermediate. To express the rate law in terms of observable species, one would apply the steady-state or pre-equilibrium approximation to eliminate [ClO]. This illustrates a common complication: rate laws derived from rate-determining steps often contain intermediate concentrations that must be algebraically replaced before comparison with experiment.

Question 5: CRITICAL THINKING

A student proposes the following mechanism for the reaction A + 2B → C + D: Step 1 (fast equilibrium): A + B ⇌ X | Step 2 (slow): X + B → C + D. Another student proposes: Step 1 (slow): A + B → X | Step 2 (fast): X + B → C + D. Both mechanisms sum correctly to the overall equation. (a) Identify the intermediate in both mechanisms. (b) Derive the predicted rate law for each. (c) Explain how an experimentalist could distinguish between the two mechanisms using kinetic data alone.

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(a) In both mechanisms, X is the intermediate — produced in step 1 and consumed in step 2. (b) Mechanism 1: The rate-determining step is step 2, giving Rate = k₂[X][B]. Since step 1 is a fast equilibrium, K₁ = [X]/([A][B]), so [X] = K₁[A][B]. Substituting: Rate = k₂K₁[A][B]² = k_obs[A][B]². The predicted rate law is first-order in A and second-order in B (third-order overall). Mechanism 2: The rate-determining step is step 1, giving Rate = k₁[A][B]. The predicted rate law is first-order in A and first-order in B (second-order overall). (c) An experimentalist can distinguish the mechanisms by measuring the reaction order with respect to B. If doubling [B] while holding [A] constant quadruples the rate, the reaction is second-order in B, supporting Mechanism 1. If the rate only doubles, the reaction is first-order in B, supporting Mechanism 2. This illustrates the power of initial-rate experiments in discriminating among proposed mechanisms.

SUMMARY

Lesson Summary

Summary

A reaction mechanism describes the step-by-step molecular pathway from reactants to products. Each individual step — an elementary step — represents a single molecular event for which the rate law can be written directly from the stoichiometry. Species that are produced in one step and consumed in a subsequent step are called intermediates; they occupy local energy minima on the reaction energy diagram and do not appear in the overall balanced equation. This distinguishes them from transition states, which sit at energy maxima and cannot be isolated.

The molecularity of an elementary step — unimolecular (1 particle), bimolecular (2 particles), or termolecular (3 particles) — directly determines its rate law and is always a positive integer. Critically, molecularity applies only to elementary steps; overall reactions have experimentally determined reaction orders that need not equal any stoichiometric coefficient. A valid mechanism must sum to the overall equation, and the rate law derived from its rate-determining step must match experimental kinetic data.