Reaction Mechanisms and Elementary Steps

Consider the reaction between nitrogen dioxide and carbon monoxide. The overall equation looks deceptively simple: NO₂ + CO → NO + CO₂. Yet the measured rate law is second order in NO₂ and zero order in CO. No glance at the balanced equation could predict this fact. Something is happening between the reactants that the stoichiometric equation hides completely. This gap between what a chemical equation says and what actually happens, molecule by molecule, is the subject of reaction mechanisms. A mechanism is a proposed sequence of elementary steps. Each step is a real molecular event. Together they add up to the observed overall reaction. Understanding mechanisms lets chemists explain why rate laws take the forms they do. It also lets them predict how a reaction will respond to a catalyst. And it helps identify fleeting intermediates that never appear in the final product mixture. This article builds the vocabulary of mechanistic kinetics from the ground up. You will learn what makes a step "elementary." You will see how molecularity differs from reaction order. You will learn to distinguish a mechanism from overall stoichiometry. Finally, you will meet reaction intermediates. You will see why they matter for everything that follows in kinetics.

Key takeaways

An elementary step is a single molecular event; its rate law follows directly from its molecularity.
Molecularity (unimolecular, bimolecular, termolecular) counts colliding species; reaction order is an empirical, separate quantity measured for the overall reaction.
Overall stoichiometry is a bookkeeping sum of elementary steps; the mechanism is the actual sequence of events.
Intermediates are produced in one step and consumed in another; they never appear in the overall balanced equation.
Termolecular elementary steps are rare because simultaneous three-body collisions are statistically unlikely.
Key idea: A mechanism decomposes an overall reaction into elementary steps whose individual rate laws, written directly from molecularity, must combine to reproduce the experimentally measured rate law.

The core insight

A reaction mechanism is a hypothesis about the sequence of elementary molecular events, each obeying its own rate law fixed by molecularity, whose net sum reproduces both the overall stoichiometric equation and the experimentally observed rate law.

The physical picture

Every macroscopic reaction rate is really a statistical average over enormous numbers of individual molecular events. Most reactions, especially those involving several bonds breaking and forming, cannot happen in one single collision. Instead they proceed through a chain of simpler steps, each one so simple that it corresponds to an actual, physically identifiable event: two molecules colliding and exchanging atoms, one molecule falling apart on its own, or (rarely) three particles meeting at once. An elementary step (or elementary reaction) is exactly this: a single microscopic event that cannot be broken down into simpler chemical steps. Because it represents one real event, its rate law can be written directly from its balanced equation using the number and identity of the colliding species. This is a privilege that the overall reaction almost never enjoys, because the overall equation is merely the net sum of possibly several elementary steps, some fast and some slow, whose rate laws do not simply combine into the overall equation's coefficients. The molecularity of an elementary step is the number of reactant particles that come together in that step. A unimolecular step involves one molecule undergoing a spontaneous change, such as an isomerization or dissociation. A bimolecular step involves a collision between two particles, the most common case at ordinary conditions. A termolecular step requires three particles to meet simultaneously; such steps are rare because the probability of a genuine three-body collision occurring at the same instant is much lower than that of two-body collisions, though a small number of termolecular gas-phase steps (for example, 2 NO + O₂ → 2 NO₂) are well documented. Crucially, the mechanism for NO₂ + CO → NO + CO₂ is now understood to be two elementary steps: Step 1 (slow): NO₂ + NO₂ → NO₃ + NO Step 2 (fast): NO₃ + CO → NO₂ + CO₂ Adding these two elementary equations and cancelling one NO₂ and one NO₃ that appear on both sides reproduces the overall equation NO₂ + CO → NO + CO₂. The species NO₃ is produced in step 1 and completely consumed in step 2: it is a reaction intermediate, present in the reacting mixture but absent from the overall balanced equation and (at appreciable concentration) from the final product mixture.

Figure

Two elementary steps summing to the overall reaction, with NO₃ as the intermediate

The second figure worth visualizing is the energetic profile of a multi-step mechanism, because it makes concrete what "slow step" and "intermediate" mean along a reaction coordinate. An intermediate sits in a local energy well between two transition states; it is a real, if short-lived, chemical species, unlike a transition state, which cannot be isolated even momentarily.

Figure

Energy profile for the two-step NO₂ + CO mechanism

The core relations

For an elementary bimolecular step between A and B:

\text{rate} = k[\mathrm{A}][\mathrm{B}]

where: $k$ is the elementary rate constant and the exponents equal the stoichiometric coefficients of that single step, because molecularity and order coincide only for elementary reactions. For an elementary unimolecular step:

\text{rate} = k[\mathrm{A}]

where: the single-molecule event obeys strict first-order kinetics by definition. For an elementary termolecular step among A, B, C:

\text{rate} = k[\mathrm{A}][\mathrm{B}][\mathrm{C}]

where: all three colliding species appear to the first power; such steps are intrinsically slow because three-body encounters are statistically rare. For the overall reaction rate written for a mechanism of $n$ elementary steps:

\text{overall rate} \neq \sum(\text{stoichiometric coefficients as exponents of overall equation})

where: the overall rate law must instead be derived from the mechanism (via rate-determining step, steady-state, or pre-equilibrium treatments) and matched against experiment.

\sum_{i=1}^{n} (\text{elementary step } i) = \text{overall balanced equation}

where: intermediates appearing on both sides of the summed equation cancel exactly, leaving only the species present in the overall stoichiometry.

Derivation AWhy mechanism, not stoichiometry, fixes the rate law

Goal. Show explicitly why the rate law of an overall reaction cannot generally be read off its balanced equation, using the NO₂ + CO mechanism as the working example, and show how the correct rate law emerges once the mechanism is specified.

Step 1. State the overall equation and the naive guess. The overall reaction is

\mathrm{NO_2 + CO \rightarrow NO + CO_2}

A naive reading of stoichiometry might suggest rate $=k[\mathrm{NO_2}][\mathrm{CO}]$ , first order in each reactant, by analogy with how an elementary bimolecular step would behave. Experimentally, however, the measured rate law is

\text{rate} = k[\mathrm{NO_2}]^2

with no dependence on $[\mathrm{CO}]$ at all. The naive guess is wrong; stoichiometric coefficients of an overall equation carry no kinetic authority.

Step 2. Introduce the proposed mechanism. Chemists propose two elementary steps that sum to the overall equation:

\mathrm{NO_2 + NO_2 \xrightarrow{k_1} NO_3 + NO} \quad (\text{slow})

\mathrm{NO_3 + CO \xrightarrow{k_2} NO_2 + CO_2} \quad (\text{fast})

Because each step is elementary, its rate law follows directly from its molecularity:

\text{rate}_1 = k_1[\mathrm{NO_2}]^2, \qquad \text{rate}_2 = k_2[\mathrm{NO_3}][\mathrm{CO}]

Step 3. Identify the rate-determining step. Step 1 is far slower than step 2 ( $k_1 \ll k_2$ under the conditions studied). Once NO₃ is formed it is consumed almost instantly by the fast step 2. The overall rate of product formation is therefore governed entirely by how fast NO₃ can be produced, since step 2 clears out NO₃ as fast as it appears. This is the rate-determining (or rate-limiting) step, a concept developed fully elsewhere; here it is enough to note that the observed rate equals the rate of the slow step.

Step 4. Match rate law to mechanism. Setting the overall rate equal to the rate of the slow, rate-determining step:

\text{rate} \approx \text{rate}_1 = k_1[\mathrm{NO_2}]^2

This is the experimentally measured rate law: second order in NO₂, zero order in CO. The mechanism, not the overall stoichiometric coefficients, correctly predicts the kinetic behavior.

Step 5. Confirm the intermediate cancels correctly. Summing the two elementary equations:

\mathrm{2NO_2 + NO_3 + CO \rightarrow NO_3 + NO + NO_2 + CO_2}

Canceling one NO₂ from each side and all NO₃ (which appears once on each side) gives back exactly

\boxed{\mathrm{NO_2 + CO \rightarrow NO + CO_2}, \quad \text{rate} = k_1[\mathrm{NO_2}]^2}

The boxed pair, overall stoichiometry plus mechanism-derived rate law, is the complete kinetic description; neither piece alone would have sufficed.

Worked examples

Example 1 — Foundational: classifying molecularity and writing elementary rate laws. Consider three proposed elementary steps drawn from different mechanisms: (a) Cyclopropane → propene (isomerization) (b) O₃ + NO → O₂ + NO₂ (c) 2 NO + O₂ → 2 NO₂ (well-established termolecular gas-phase step) Task: classify the molecularity of each and write its rate law. Step (a) involves one molecule rearranging with no partner. Molecularity = 1 (unimolecular). Since it is elementary, order equals molecularity:

\text{rate}_a = k_a[\text{cyclopropane}]

Step (b) involves two distinct molecules, ozone and nitric oxide, colliding. Molecularity = 2 (bimolecular):

\text{rate}_b = k_b[\mathrm{O_3}][\mathrm{NO}]

Step (c) requires two NO molecules and one O₂ molecule to meet essentially simultaneously. Molecularity = 3 (termolecular):

\text{rate}_c = k_c[\mathrm{NO}]^2[\mathrm{O_2}]

Sanity check: in each case the exponents in the rate law equal the stoichiometric coefficients of that individual elementary equation — this equality is the defining signature of an elementary step, and it is only guaranteed here because none of these equations is claimed to be an overall multi-step reaction. Numerically, if $k_c = 2.0\times10^{9}\ \mathrm{dm^6\,mol^{-2}\,s^{-1}}$ (a typical literature value for this termolecular step) and $[\mathrm{NO}] = 1.0\times10^{-3}\ \mathrm{mol\,dm^{-3}}$ , $[\mathrm{O_2}] = 5.0\times10^{-3}\ \mathrm{mol\,dm^{-3}}$ , then

\text{rate}_c = (2.0\times10^{9})(1.0\times10^{-3})^2(5.0\times10^{-3}) = 1.0\times10^{1}\ \mathrm{mol\,dm^{-3}\,s^{-1}}

The units check: $\mathrm{dm^6\,mol^{-2}\,s^{-1}} \times \mathrm{mol^2\,dm^{-6}} = \mathrm{mol\,dm^{-3}\,s^{-1}}$ , a proper rate unit.

Example 2 — Applied: distinguishing two candidate mechanisms for the same overall reaction using measured kinetics. The gas-phase decomposition of dinitrogen pentoxide follows the overall equation

2\mathrm{N_2O_5} \rightarrow 4\mathrm{NO_2} + \mathrm{O_2}

Experimentally, the rate law is found to be first order overall:

\text{rate} = k_{\text{obs}}[\mathrm{N_2O_5}], \qquad k_{\text{obs}} = 4.8\times10^{-4}\ \mathrm{s^{-1}} \text{ at } 318\ \mathrm{K}

A single-step termolecular guess, "2 N₂O₅ → 4 NO₂ + O₂ happens in one collision of two N₂O₅ molecules," would predict rate $= k[\mathrm{N_2O_5}]^2$ , second order — this contradicts experiment immediately, so a one-step mechanism is ruled out. The accepted multi-step mechanism is:

\mathrm{N_2O_5 \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} NO_2 + NO_3} \quad \text{(fast, reversible)}

\mathrm{NO_2 + NO_3 \xrightarrow{k_2} NO + O_2 + NO_2} \quad \text{(slow)}

\mathrm{NO + NO_3 \xrightarrow{k_3} 2NO_2} \quad \text{(fast)}

Task: verify that summing these three elementary steps reproduces the overall equation, and identify all intermediates. Summing left-hand sides: $\mathrm{N_2O_5 + NO_2 + NO_3 + NO + NO_3}$ . Summing right-hand sides: $\mathrm{NO_2 + NO_3 + NO + O_2 + NO_2 + 2NO_2}$ . Cancel species common to both sides: one NO₂ and one NO₃ cancel from step 1/step 2 boundary; the NO produced in step 2 is consumed in step 3; one NO₃ from step 3 cancels against the NO₃ produced in step 1. Careful bookkeeping (multiplying step 1 and its reverse appropriately, since NO₃ is generated and consumed twice) leaves:

2\mathrm{N_2O_5} \rightarrow 4\mathrm{NO_2} + \mathrm{O_2}

which is exactly the overall equation — confirming the mechanism is stoichiometrically consistent.

Intermediates identified: NO₃ and NO. Both are generated in one step and fully consumed in another; neither appears in the overall equation, and neither would be detected in the bulk product mixture at significant concentration, though NO₃ has in fact been observed spectroscopically as a short-lived transient species in this system, providing direct experimental support for the mechanism. Sanity check on the rate law: applying a steady-state treatment (developed in detail in a later topic) to this mechanism yields

\text{rate} = \frac{k_1 k_2}{k_{-1} + k_2}[\mathrm{N_2O_5}]

which is first order in N₂O₅, matching the experimental $k_{\text{obs}}$ above and confirming that overall reaction order (here, one) can be completely different from the molecularity of the individual elementary steps (here, one and two).

Limiting cases & common traps

Situation	Correct statement
Elementary step	Order = molecularity, always; rate law read directly off stoichiometry
Overall multi-step reaction	Order must be measured experimentally; cannot be inferred from balanced equation
Termolecular elementary steps	Rare but real (e.g., 2NO + O₂ → 2NO₂); most "termolecular-looking" overall reactions are actually multi-step
Reaction intermediate	Produced then fully consumed; concentration typically small and often treated with the steady-state approximation
Catalyst	Consumed in an early step, regenerated in a later one; appears in the mechanism but cancels from the overall equation, like an intermediate — but a catalyst is added externally, not generated internally

The chemist behind it

The modern language of elementary steps and molecularity grew out of collision theory in the 1910s-1920s, but the case that most cleanly demonstrates why mechanism must be distinguished from stoichiometry is the classical work on nitrogen oxide and halogen kinetics carried out by Jacobus van 't Hoff and refined by later kineticists including Max Bodenstein, whose careful gas-phase rate measurements on reactions like 2 HI → H₂ + I₂ and the N₂O₅ decomposition forced chemists to accept that balanced equations are silent about mechanism. Bodenstein's meticulous quantitative studies of intermediates and his introduction of the steady-state concept (further systematized by Irving Langmuir and others) made the intermediate a rigorous, measurable kinetic entity rather than a speculative bystander.

Origins · 1 minMax Bodenstein and how this was discovered

Max Bodenstein

1871-1942 · German physical chemist whose precise gas-phase rate studies on reactions such as H₂ + I₂ and N₂O₅ decomposition established the experimental basis for distinguishing elementary steps from overall stoichiometry and for treating reaction intermediates quantitatively

Ask the tutor: How does the steady-state approximation turn a multi-step mechanism like this one into a single predicted rate law?

What exactly distinguishes an elementary step from an overall reaction?

An elementary step corresponds to one actual molecular event, a single collision or single spontaneous transformation, and it cannot be broken down further into simpler chemical processes. Its rate law is fixed by its molecularity. An overall reaction is simply the net balanced equation obtained by summing all the elementary steps in a mechanism, after any intermediates cancel. The overall equation carries no direct information about the rate law; only the mechanism does. See Rate Laws and Reaction Order for how overall orders are measured experimentally.

Can molecularity ever be a fraction or zero?

No. Molecularity counts actual colliding or transforming particles in one elementary step, so it must be a small positive integer: one, two, or (rarely) three. Reaction order, by contrast, is an empirical exponent extracted from rate data and can be zero, negative, or fractional, because it reflects the net kinetic behavior of a possibly multi-step mechanism, not a single molecular event. Fractional orders are common signatures that a mechanism involves intermediates and pre-equilibria.

How do chemists actually detect a reaction intermediate if it never appears in the products?

Several methods are used together. Spectroscopic techniques such as flash photolysis or fast IR/UV absorption can catch transient species like NO₃ or free radicals in real time. Isotopic labeling can reveal atom connectivity consistent with a proposed intermediate. Kinetic isotope effects and the mathematical form of the rate law, especially fractional or complex orders, offer strong indirect evidence. Trapping experiments, where a scavenger reacts preferentially with the suspected intermediate, provide further confirmation.

Is a catalyst the same thing as a reaction intermediate?

Not quite, though the two are often confused because both appear in the mechanism but cancel from the overall equation. An intermediate is generated internally by the reaction itself, appearing partway through the sequence. A catalyst is added externally at the start, is consumed in an early step, and is regenerated in a later one, ending the reaction unchanged in both amount and chemical form. Catalysis is covered in detail elsewhere in kinetics.

Why can I not just guess the mechanism from the balanced equation?

Because the balanced equation is a statement of atom and mass conservation, nothing more. It has no memory of the path molecules actually took. Two completely different mechanisms, one single-step and one five-step, can produce the exact same overall balanced equation while giving wildly different, experimentally distinguishable rate laws. Only kinetic data, combined with chemical intuition and spectroscopic evidence, can support a proposed mechanism, and even then the mechanism remains a well-supported hypothesis, never a proof.