Reaction mechanism

A balanced chemical equation is a receipt, not a story. It tells you the ingredients that went in and the products that came out, but says nothing about how the reaction actually got there. Real reactions rarely happen in one clean collision — they usually creep forward through a hidden sequence of simpler steps, called the reaction mechanism. Chemists care about mechanisms because the mechanism, not the tidy overall equation, is what fixes the experimentally observed rate law: the powers of concentration in the rate expression often do not match the stoichiometric coefficients at all, and the reason is always buried in the mechanism's slowest step. This article builds the tools to read that story — elementary steps, intermediates, molecularity, and the rate-determining step — and to predict a rate law directly from a proposed mechanism, exactly as ISC Class 12 Unit 3 (Chemical Kinetics) expects, at the conceptual level, without steady-state calculus.

The core insight

The observed rate law of a multi-step reaction is fixed by its slowest elementary step (the rate-determining step), applying the Law of Mass Action to that step alone, which is why the rate law's powers can differ from the overall equation's stoichiometric coefficients.

The core idea

Elementary reaction versus overall reaction. An elementary reaction is a single molecular event — one collision (or one decomposition) that occurs exactly as written, in a single step, with no hidden intermediate stages. Its molecularity is simply the number of reacting species colliding in that one step: unimolecular for one species, bimolecular for two, and so on. An overall reaction is the net, balanced equation a chemist writes down after summing everything that happens — but that equation may hide a mechanism, a sequence of two or more elementary steps that together add up to the observed products and reactants.

Intermediates. A reaction intermediate is a species that is produced in one elementary step and fully consumed in a later step. Because it cancels out of the addition, it never appears in the overall balanced equation — yet it is a genuine, if often short-lived, chemical entity that can, in favourable cases, be detected or even isolated. This makes it fundamentally different from the activated complex (or transition state), which is not a stable species at all but a fleeting, high-energy arrangement of atoms sitting at the very top of an energy barrier, existing for roughly one bond-vibration and never isolable.

The rate-determining step (RDS). In almost every real multi-step mechanism, the elementary steps do not proceed at equal speeds — one step is dramatically slower than the rest. That slow step acts as a bottleneck: however fast the other steps are individually, the overall reaction can never outrun its slowest link. This slowest elementary step is called the rate-determining step, and because nothing downstream of it can happen faster than it supplies material, its own rate law — obtained by applying the Law of Mass Action to that one step, using the step's own molecularity as the powers — becomes the observed rate law of the entire reaction.

Why order can differ from stoichiometry. This single idea — "look only at the RDS" — explains one of the most common surprises in kinetics: the powers in an experimentally measured rate law frequently do not match the coefficients of the overall balanced equation. A species that appears with a large coefficient in the overall equation may still appear with a smaller power (or not at all) in the rate law, if it is only consumed in a step that comes after the RDS. Once the RDS has already happened, the rate is already "locked in" — later fast steps just mop up the remaining reactants without slowing anything further.

Figure

A two step reaction mechanism drawn as a bottleneck diagram with a wide slow first step and a narrow fast second step

Molecularity, order, and the RDS

Because the RDS is a genuine elementary step, the Law of Mass Action applies to it directly: the rate of that one step equals a rate constant times the concentration of each reacting species in the step, raised to the power of its coefficient in that step alone.

\text{Rate}_{\text{overall}} = \text{Rate}_{\text{RDS}} = k\,[\text{species}_1]^{a}[\text{species}_2]^{b}\cdots

where: $k$ is the rate constant of the rate-determining step, and $a, b, \ldots$ are the molecularity-based powers of each species exactly as it appears in the RDS (not necessarily its coefficient in the overall equation). Any reactant that appears only in a step after the RDS, or only as part of an intermediate that itself does not survive to the final rate expression, contributes nothing extra to this rate law.

Derivation APredicting a rate law from a proposed two-step mechanism

The goal is to take a proposed multi-step mechanism for the general reaction $A + 2B \rightarrow C$ , identify which elementary step controls the overall speed, and derive the exact rate law a chemist would report from experiment — showing concretely why the coefficient of $B$ in the balanced equation (2) need not equal its power in the rate law, which is the single most tested idea in this topic.

Step 1. Write down the proposed mechanism as two elementary steps that add up to the overall equation. Step 1 (slow): $A + B \rightarrow X$ . Step 2 (fast): $X + B \rightarrow C$ . Adding these two steps and cancelling the intermediate $X$ , which appears once as a product and once as a reactant, recovers the overall equation $A + 2B \rightarrow C$ exactly, confirming the mechanism is stoichiometrically consistent.

Step 2. Identify the rate-determining step. Step 1 is labelled "slow," so it is the bottleneck: the overall reaction cannot proceed faster than $A$ and $B$ can come together to form $X$ in Step 1, no matter how quickly Step 2 then consumes $X$ .

Step 3. Apply the Law of Mass Action to the RDS alone. Step 1, $A + B \rightarrow X$ , is elementary and bimolecular, with one $A$ and one $B$ colliding. Its rate law is therefore $\text{Rate} = k[A][B]$ , using the coefficients of $A$ and $B$ exactly as they appear in this one step.

Step 4. Check that Step 2 adds nothing further. Step 2 consumes a second molecule of $B$ , but it does so only after the rate has already been set by Step 1; being fast, Step 2 never becomes the bottleneck, so this second $B$ does not contribute an extra power to the observed rate law.

Step 5. State the predicted rate law and compare it with stoichiometry. The overall predicted rate law is $\text{Rate} = k[A][B]$ — first order in $A$ , first order in $B$ , second order overall — even though the balanced equation shows $B$ with coefficient $2$ . This is the general result: a species' power in the rate law equals its coefficient in the RDS, which can be smaller than (or absent from) its coefficient in the overall balanced equation.

Worked example

Question. For the mechanism above, $A + B \xrightarrow{\text{slow}} X$ followed by $X + B \xrightarrow{\text{fast}} C$ , the rate constant of the rate-determining step is found experimentally to be $k = 0.050\ \text{dm}^3\text{mol}^{-1}\text{s}^{-1}$ . Calculate the initial rate of the reaction when $[A] = 0.20\ \text{mol dm}^{-3}$ and $[B] = 0.40\ \text{mol dm}^{-3}$ .

Step 1. Identify the correct rate law to use. From the derivation above, the RDS is Step 1, so the observed rate law is $\text{Rate} = k[A][B]$ — not any expression built from the overall stoichiometric coefficients of $A + 2B \rightarrow C$ .

Step 2. Substitute the given rate constant and concentrations directly into this rate law: $\text{Rate} = (0.050)(0.20)(0.40)$ .

Step 3. Carry out the arithmetic: $0.050 \times 0.20 = 0.010$ , and $0.010 \times 0.40 = 0.0040$ .

Step 4. Attach units. Since $k$ carries units of $\text{dm}^3\text{mol}^{-1}\text{s}^{-1}$ and each concentration is in $\text{mol dm}^{-3}$ , the product has units of $\text{mol dm}^{-3}\text{s}^{-1}$ , correct for a rate. The initial rate is $4.0\times10^{-3}\ \text{mol dm}^{-3}\text{s}^{-1}$ .

Competitive stretch

If a student mistakenly used the overall stoichiometry and wrote $\text{Rate} = k[A][B]^2$ , the same numbers would give $0.050 \times 0.20 \times (0.40)^2 = 0.0016\ \text{mol dm}^{-3}\text{s}^{-1}$ — a different, wrong answer, and a good self-check: whenever a rate-law question gives a mechanism, the coefficients of the overall equation are a trap, not an input.

Common traps

Writing the rate law directly from the overall balanced equation's coefficients — it must come from the rate-determining step of the actual (or proposed) mechanism, or from experiment, never from stoichiometry alone.
Confusing a reaction intermediate with the activated complex — an intermediate is a real, if short-lived, chemical species that can in principle be detected; the activated complex is a fleeting transition-state arrangement that is never isolable.
Assuming every reactant in the overall equation must appear in the rate law — species consumed only in fast steps after the RDS often drop out of the observed rate expression entirely.
Believing a mechanism is "proved" simply because it adds up to the correct overall equation — a mechanism is only validated when the rate law it predicts matches the rate law measured experimentally.

Ask the tutor: if predicting a rate law from a mechanism with a fast pre-equilibrium step (like the NO plus O2 example) still feels shaky, ask your AI tutor to walk through one more mechanism with you step by step.

Why doesn't the rate law just come from the balanced equation the way it does for gas laws or stoichiometry problems?

Stoichiometric coefficients tell you the ratio of moles consumed and produced overall, once the reaction is complete — they are bookkeeping, not a record of how the molecules actually collided. The rate law, by contrast, describes the mechanism of the slowest step, which is a real physical collision (or set of collisions) with its own molecularity. These two things happen to coincide only for a genuinely single-step (elementary) overall reaction; for any multi-step mechanism they can, and usually do, differ, which is exactly why chemists cannot guess a rate law from a balanced equation and must determine it experimentally or from a proposed mechanism.

How do you actually tell which step is the rate-determining step if the question does not label one as slow?

At this syllabus level, ISC questions will always tell you explicitly which step is slow (the RDS) either by labelling it or by giving you the experimentally observed rate law and asking you to check whether a mechanism is consistent with it, as in the NO plus O2 example. You are not expected to derive which step is rate-determining from first principles using steady-state mathematics — that lies beyond this syllabus. Your job is either to apply the Law of Mass Action to the given slow step, or to verify that a proposed slow step's predicted rate law matches a given experimental one.

If an intermediate appears in the slow step, does the rate law from that step just get reported as the final answer?

Only if the intermediate does not appear in the rate-determining step itself. If the RDS's own reactants happen to be original reactants (as in the A + 2B example above), you can report that rate law directly. But if the slow step is itself written in terms of an intermediate formed in an earlier fast, reversible step (as in the NO plus O2 example, where the slow step consumes the intermediate N2O2), the intermediate's concentration must be re-expressed in terms of the original reactants before the rate law can be compared with experiment, since a valid rate law is written only in species present in the overall balanced equation.