Chemistry: Atoms First 2e · Kinetics

Reaction Mechanisms

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In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

A balanced chemical equation is a summary: it shows what starts a reaction and what ends it, but not how the atoms get from one side to the other. A is the step-by-step sequence of molecular events that actually occurs — which bonds break, which bonds form, and in what order. Each step is an elementary reaction (): a single molecular event such as a molecule splitting or two molecules colliding.

The mechanism matters because it determines the rate law. When a rate law like rate = k[NO]²[Br₂] does not match the coefficients of the overall equation, the mismatch is evidence that the reaction proceeds through several steps, one slower than the others. That slow step — the — controls the overall speed, and its dictates the rate law's concentration dependence.

Why this matters

Mechanisms are where kinetics becomes chemistry: knowing the mechanism tells a chemist which bonds to weaken, what to stabilize, or what catalyst to design to make a reaction faster or more selective. The Haber synthesis of ammonia was refined for decades by improving the picture of its mechanism, and enzyme mechanisms explain drug action, metabolism, and disease. For exams, mechanism problems test the central idea: the rate law is set by the slow elementary step, and intermediates never appear in the overall rate law.

The college version

Core Concepts

Elementary steps and molecularity

An elementary reaction describes what happens in a single molecular event; it cannot be broken into simpler steps. Each step is classified by its molecularity — the number of reactant particles that come together:

Unimolecular (one molecule): a single molecule rearranges or splits, e.g., cyclobutane → 2 ethylene; rate law = k[A].
Bimolecular (two particles): two molecules or atoms collide, e.g., NO₂ + CO → NO + CO₂; rate law = k[A][B].
Termolecular (three particles): three particles collide at once, e.g., 2NO + O₂ → 2NO₂; rate law = k[A]²[B]. Termolecular steps are rare because three particles must meet simultaneously.

For an elementary step, the rate law is written directly from its stoichiometry: exponents equal molecularities. This rule applies only to elementary steps — never to the overall balanced equation.

Intermediates and transition states

An intermediate is a species produced in one elementary step and consumed in a later one, so it does not appear in the overall equation — for example, NO₃ produced in step 1 and consumed in step 2 of a classic mechanism. Intermediates are real molecular species with finite lifetimes. Do not confuse them with transition states — fleeting, maximum-energy configurations at the top of the activation barrier that exist for only one bond vibration and can never be isolated.

The rate-determining step

When one elementary step is much slower than the others, the overall reaction cannot proceed faster than that step allows — it is the rate-determining step (rate-limiting step). The overall rate law is the slow step's rate law, after substituting any concentrations set by fast steps that precede it. The slow step is often, but not always, the first step.

Fast equilibrium before the slow step

If a fast, reversible step precedes the slow step, it establishes a . Solving its equilibrium constant expression for the intermediate's concentration and substituting into the slow step's rate law eliminates the intermediate — the standard technique for deriving rate laws for multi-step mechanisms.

Testing a proposed mechanism

A mechanism is accepted only if (1) its steps sum to the overall balanced equation, (2) its derived rate law matches the experimentally measured one, and (3) its steps are chemically reasonable. If the derived rate law disagrees with experiment, the mechanism is wrong no matter how elegant it looks.

How It Works / Step-by-Step Process

Write the proposed elementary steps and check that they sum to the overall balanced equation.
Identify the rate-determining (slowest) step.
Write its rate law from its molecularity: rate = k times each reactant concentration raised to its coefficient in that step.
If the slow step contains an intermediate, replace its concentration using the fast preceding step's equilibrium expression.
Compare the derived rate law with the experimental one; if they disagree, revise the mechanism.

Common Confusions

Do Not Confuse	With	Difference
Coefficients in the overall equation	Exponents in the rate law	Exponents equal molecularities only for elementary steps; overall rate laws must be measured.
Intermediate	Transition state	An intermediate is a real species with a finite lifetime, formed in one step and used in the next; a transition state is the fleeting energy maximum of a single step and can never be isolated.
Intermediate	Catalyst	An intermediate is produced and then consumed within the reaction; a catalyst is added, participates, and is regenerated. Both vanish from the overall equation, but the catalyst is never consumed net.
The slow step being the first step	The slow step being rate-determining	The rate-determining step can sit anywhere in the sequence; if it is not first, fast preceding steps set up a pre-equilibrium that feeds it.
A mechanism that sums correctly	A mechanism that is correct	It must also reproduce the experimental rate law; summing correctly is necessary but not sufficient.

Eli explains

The same idea, in plain words

Explain it like I’m 10

A reaction mechanism is like a cake recipe: the overall equation only tells you the ingredients you start with and the cake you end with, but the recipe gives each step — mix, add eggs, bake. The slowest step, like waiting for the oven, controls how long everything takes.

Worked example

Worked example 1: the classic two-step mechanism

The reaction 2 NO₂ + F₂ → 2 NO₂F has the experimental rate law rate = k[NO₂][F₂]. Proposed mechanism:

Step 1 (slow): NO₂ + F₂ → NO₂F + F Step 2 (fast): NO₂ + F → NO₂F

First, check that the steps sum to the overall equation: adding step 1 and step 2 gives 2NO₂ + F₂ → 2NO₂F, with the F atom canceling as an intermediate. Next, write the rate law from the slow step's molecularity:

$rate = k 1 [NO 2][F 2]$

The slow step is bimolecular in NO₂ and F₂ and contains no intermediate, so the derived rate law matches the experimental one exactly. Note that the coefficient 2 on NO₂ in the overall equation is irrelevant — the slow step is first order in NO₂.

Worked example 2: a fast equilibrium before the slow step

The reaction 2 NO + Br₂ → 2 NOBr has the experimental rate law rate = k[NO]²[Br₂]. Proposed mechanism:

Step 1 (fast, reversible): NO + Br₂ ⇌ NOBr₂ Step 2 (slow): NOBr₂ + NO → 2 NOBr

The steps sum to the overall equation (NOBr₂ is an intermediate). The slow step is bimolecular, so its rate law is:

$rate = k 2 [NOBr 2][NO]$

This contains the intermediate NOBr₂, which must be eliminated. The fast step reaches equilibrium, so solve its equilibrium expression for the intermediate:

$K 1 = [NOBr 2] [NO][Br 2] \Rightarrow [NOBr 2] = K 1 [NO][Br 2]$

Substitute back into the slow-step rate law:

$rate = k 2 K 1 [NO] 2 [Br 2] = k[NO] 2 [Br 2]$

where the composite constant k = k₂ K₁ collects the elementary rate constant and the equilibrium constant. The derived rate law matches the experimental one, so the mechanism is consistent with the data — and it shows why intermediates vanish from rate laws: the fast step controls their concentrations.

Key takeaways

A mechanism is a sequence of elementary steps; each step's rate law comes from its molecularity, not from the overall equation.
The rate-determining step controls the rate; its rate law, after substituting pre-equilibrium concentrations, is the overall rate law.
Intermediates cancel from the overall equation and never appear in the rate law.
A valid mechanism must sum to the overall equation, give the correct experimental rate law, and be chemically plausible.
Unimolecular and bimolecular steps are common, termolecular steps rare; when a fast step precedes the slow step, solve its equilibrium expression for the intermediate and substitute.

Check yourself

5 review questions from the chapter. Try each one, then open the answer.

What is the rate law for the elementary step 2A + B → products?
Show answer
rate = k[A]²[B], because the step is termolecular and exponents come from the molecularities of the elementary step.
Why do intermediates never appear in an overall rate law?
Show answer
An intermediate is produced in one step and consumed in another, so its concentration is determined by the steps that create and destroy it; substituting those relationships eliminates it from the final rate law.
In a mechanism where a fast equilibrium precedes the slow step, how is the intermediate removed from the rate law?
Show answer
Use the fast step's equilibrium constant expression to solve for the intermediate's concentration, then substitute it into the slow step's rate law.
A proposed mechanism sums correctly but predicts a rate law different from the experiment. What can you conclude?
Show answer
The mechanism is inconsistent with the data and must be revised — summing correctly is necessary but not sufficient.
What is the difference between a unimolecular and a bimolecular elementary step?
Show answer
A unimolecular step involves one molecule rearranging or splitting; a bimolecular step involves two particles colliding. Molecularity sets the rate law exponents (1 and 2).

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Reaction mechanism: The sequence of elementary steps describing how a reaction occurs.
Elementary step: A single molecular event that cannot be subdivided.
Molecularity: The number of particles that collide in an elementary step (1, 2, or 3).
Intermediate: A species made in one step and used up in a later step.
Rate-determining step: The slowest elementary step, which limits the overall rate.
Pre-equilibrium: A fast, reversible step that establishes equilibrium before the slow step.

Sources & references

openstax.org — Chemistry Atoms First 2e

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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