Chemistry · General chemistry II · Concept
Testing a reaction mechanism with its rate law
Decide whether a proposed mechanism fits an observed rate law with a short if–then–else test: add the steps, write the slow step’s rate law, replace any intermediate, then compare.
What a mechanism claims
An overall equation tells you what reacts and what forms. A mechanism is a proposal for how it happens: a sequence of elementary steps, each a single collision or rearrangement that occurs exactly as written. That difference matters for kinetics. You cannot read a rate law from an overall equation, but you can read one from an elementary step, because its rate depends on how often those particular particles meet.
The test, one decision at a time
Every mechanism question is the same four checks. Work through them in order and stop as soon as one fails.
| Check | If | Then | Else |
|---|---|---|---|
| 1. Add the steps | They add up to the overall reaction (intermediates cancel) | Go to check 2 | Ruled out: it describes a different reaction |
| 2. Slow step | Exactly one step is slow | Its rate law is k × [each reactant]^coefficient; go to check 3 | Mark the rate-determining step before predicting |
| 3. Intermediates | The slow-step rate law has no intermediate | Go to check 4 | Replace the intermediate using a fast equilibrium before the slow step, then look again |
| 4. Compare | Predicted orders match the observed rate law | Consistent: possible, not proved | Ruled out as written |
Why the slow step sets the rate
Steps after the slow step can only pass along what the slow step supplies, so the overall rate cannot exceed it. Steps before it are fast enough to keep up; if a fast first step is reversible, its forward and reverse reactions keep pace with each other and stay near equilibrium while the slow step drains a little product. That is why the slow step’s rate law, not the overall equation, predicts the observed orders.
Why can’t the overall coefficients be orders?
Coefficients count how many particles are consumed in total. Orders describe which particles are present in the slowest collision. The two agree only when the overall reaction happens to be one elementary step, which is the claim you are testing, not something you may assume.
Handling a fast first step
When the slow step uses an intermediate, its rate law contains a concentration you cannot measure directly. If a fast, reversible step forms that intermediate, set its forward rate equal to its reverse rate and solve for the intermediate. Substituting leaves only species from the overall reaction, and the constants combine into one observed k.
Reading clues in an observed rate law
Orders are fingerprints of the rate-determining step. A reactant with a higher order than its coefficient suggests it enters more than once before or during the slow step. A negative order in a product suggests a fast equilibrium that releases that product before the slow step. A half order suggests a molecule splitting into two fragments in a fast equilibrium, such as Cl₂ ⇌ 2Cl.
What “consistent” does and doesn’t mean
Passing all four checks makes a mechanism possible, not proven: a different sequence can predict the same rate law. Failing either the sum check or the rate-law check rules the mechanism out as written. Extra evidence, such as detecting a proposed intermediate, is what strengthens one consistent mechanism over another.
Try it on your own work
In a Chemistry box, write the overall reaction, each step with the word slow or fast after it, and the observed rate law, such as rate = k[NO]^2[Br2]. With Check on, the rate-law row reports whether the steps add up, which intermediates were replaced and whether the predicted orders match.
Common mistakes
- Using the overall equation’s coefficients as the reaction orders.
- Leaving an intermediate in the predicted rate law.
- Writing the rate law from a fast step when a slow step exists.
- Calling a mechanism proven because its rate law matches.
- Accepting steps that don’t add up exactly to the overall reaction.
Work through an example
Nitric oxide and bromine form nitrosyl bromide, 2NO(g) + Br₂(g) → 2NOBr(g). Experiments give rate = k[NO]²[Br₂]. Which of these two-step mechanisms is consistent with that rate law? Mechanism A: NO + Br₂ → NOBr₂ (slow), then NOBr₂ + NO → 2NOBr (fast). Mechanism B: NO + Br₂ ⇌ NOBr₂ (fast), then NOBr₂ + NO → 2NOBr (slow).
Rule a mechanism in or out for 2NO + Br₂ →Sources and scope
Authored study material. Tool results depend on the stated inputs and model assumptions.
- Tro, Chemistry: A Molecular Approach, 4th ed., §14.6 Reaction Mechanisms, pp. 648–653 (a valid mechanism: steps sum to the overall reaction and predict the observed rate law, p. 651)
- OpenStax Chemistry 2e — Reaction mechanisms