The rate law for a proposed mechanism is determined by its slowest elementary step, with any intermediate concentrations replaced using earlier steps in the mechanism. You can’t just look at the overall balanced equation to find a rate law. Instead, you need to break the mechanism into its individual steps and apply a few straightforward rules depending on the structure of those steps.
Elementary Steps Dictate the Rate Law
Every proposed mechanism is a sequence of elementary steps, and each elementary step has a rate law you can write directly from its chemical equation. This is the key distinction between an elementary step and an overall reaction: for an elementary step, the exponents in the rate law match the stoichiometric coefficients. If one molecule of A reacts with one molecule of B in a single collision, the rate law for that step is rate = k[A][B]. If two molecules of A collide, it’s rate = k[A]². This works because an elementary step describes a single molecular event.
The common types break down like this:
- Unimolecular (A → products): rate = k[A], first order
- Bimolecular (A + B → products): rate = k[A][B], second order
- Bimolecular (2A → products): rate = k[A]², second order
- Termolecular (2A + B → products): rate = k[A]²[B], third order
Termolecular steps are rare because the odds of three molecules colliding simultaneously with the right orientation and energy are very small. Most mechanisms consist entirely of unimolecular and bimolecular steps.
The Rate-Determining Step Sets the Overall Rate
When one step in a mechanism is significantly slower than all the others, it acts as a bottleneck. The overall reaction can’t proceed any faster than this slowest step, so the rate law for that step becomes the rate law for the entire reaction. Think of it like a highway where traffic funnels into a single lane: no matter how many lanes open up afterward, the flow rate is limited by that one lane.
To write the rate law, you identify the slow step, then write its rate law using the stoichiometric coefficients as exponents. If the slow step is A + B → C with rate constant k₂, the overall rate law is rate = k₂[A][B]. This is the simplest and most common scenario you’ll encounter in general chemistry.
Not every mechanism has a single rate-determining step. When multiple steps proceed at comparable speeds, you need more sophisticated approaches like the steady-state approximation (covered below).
Eliminating Intermediates From the Rate Law
Here’s where it gets tricky. If the slow step involves an intermediate, a species produced in an earlier step that doesn’t appear in the overall reaction, you can’t leave it in the final rate law. Rate laws must be expressed in terms of reactants (and sometimes products) that you can actually measure. So you need to replace the intermediate’s concentration using information from the other steps.
Consider the decomposition of ozone, a classic textbook example:
- Step 1 (fast, reversible): O₃ ⇌ O₂ + O
- Step 2 (slow): O + O₃ → 2 O₂
The slow step gives rate = k₂[O][O₃], but atomic oxygen (O) is an intermediate. To eliminate it, you use the fast equilibrium from Step 1. Because Step 1 reaches equilibrium quickly, the forward and reverse rates are equal: k₁[O₃] = k₋₁[O₂][O]. Solving for [O] gives [O] = k₁[O₃] / (k₋₁[O₂]). Substituting back into the slow step’s rate law produces:
rate = (k₁k₂/k₋₁) × [O₃]² / [O₂]
This final expression contains only observable species: ozone and molecular oxygen. The grouped constants (k₁k₂/k₋₁) act as a single effective rate constant. Notice that [O₂] appears in the denominator, meaning higher oxygen concentrations actually slow the reaction. This kind of product inhibition is a direct consequence of the reversible first step.
The Pre-Equilibrium Approximation
The ozone example above uses the pre-equilibrium approximation, which applies when a fast, reversible step comes before the slow step. The logic is simple: because the first step is fast in both directions, it effectively reaches equilibrium before the slow step consumes much of the intermediate. You can therefore write an equilibrium expression for that first step and solve for the intermediate’s concentration.
The procedure has three parts. First, write the equilibrium expression for the fast step, setting the forward rate equal to the reverse rate. Second, solve that expression for the intermediate concentration. Third, substitute it into the rate law of the slow step. The result is a rate law that contains only measurable species and a composite rate constant built from the individual rate constants of each step.
The Steady-State Approximation
When no single step is clearly the slowest, or when the mechanism has multiple intermediates, the steady-state approximation is more appropriate. The central assumption is that the concentration of a reactive intermediate stays essentially constant over the course of the reaction. It’s being produced and consumed at nearly the same rate, so its concentration never builds up significantly.
Mathematically, you set the rate of change of the intermediate’s concentration equal to zero: d[intermediate]/dt = 0. This converts what would be a difficult differential equation into a simple algebraic one. You write out all the terms that produce the intermediate and all the terms that consume it, set the net rate to zero, and solve for the intermediate’s concentration. Then you substitute that expression into the rate law for the product-forming step.
The steady-state approximation is more general than the pre-equilibrium approach. It doesn’t require any step to be “fast” or “slow” in advance. It only requires that the intermediate is present at much lower concentration than the main reactants and products, which is true for most reactive intermediates like free radicals, carbocations, or atomic fragments.
Checking the Mechanism Against Experiment
A proposed mechanism is only valid if the rate law it predicts matches what’s observed experimentally. This is the fundamental test. You measure how the reaction rate changes when you vary each reactant’s concentration, determine the experimental rate law, and then check whether any proposed mechanism generates that same mathematical form.
It’s worth understanding what this test can and can’t do. A mechanism that predicts the wrong rate law is definitively ruled out. But a mechanism that predicts the correct rate law isn’t proven correct. Multiple different mechanisms can produce the same rate law. Chemists use additional evidence, like detecting intermediates spectroscopically or measuring isotope effects, to distinguish between mechanisms that are kinetically equivalent.
In practice, experimental conditions are often chosen deliberately to simplify the math. If one reactant is present in huge excess, its concentration barely changes during the reaction, and you can treat it as a constant folded into the rate constant. This pseudo-first-order approach lets you isolate the dependence on each reactant one at a time, building up the full rate law piece by piece and comparing it to what the proposed mechanism predicts.
When Catalysts Appear in the Rate Law
A catalyst participates in the mechanism but is regenerated by the end, so it doesn’t appear in the overall balanced equation. It can, however, appear in the rate law. If the catalyst is involved in the rate-determining step, its concentration will show up as a term in the derived rate expression. This is why increasing catalyst concentration speeds up a reaction even though the catalyst isn’t “used up.” Enzymes in biological systems are a common example: the reaction rate depends on enzyme concentration even though the enzyme is recycled with every turnover.