Rate-Determining Step and Reaction Mechanisms
A concise revision guide to the rate-determining step: why the slow step fixes the rate equation, deducing a rate equation from a mechanism and a mechanism from a rate equation, intermediates, and how the rate equations for the hydrolysis of halogenoalkanes give evidence for the Sₙ1 and Sₙ2 mechanisms.
- 16.1vii
- 16.9
- 16.10
- 16.11
What these spec points say
- 16.1vii understand the terms: rate-determining step
- 16.9 be able to deduce a rate-determining step from a rate equation and vice versa
- 16.10 be able to deduce a reaction mechanism, using knowledge from a rate equation and the stoichiometric equation for a reaction
- 16.11 understand that knowledge of the rate equations for the hydrolysis of halogenoalkanes can be used to provide evidence for SN1 or SN2 mechanisms for tertiary and primary halogenoalkane hydrolysis
Mechanisms and the Rate-Determining Step
A balanced equation shows what goes in and what comes out; it says nothing about how.
Most reactions happen in a series of simpler steps called the mechanism, each usually involving a collision between just one or two species.
The species formed in one step and used up in a later one are intermediates: they never appear in the overall equation, and they are often too reactive to isolate.
The steps do not all happen at the same speed. One is much slower than the rest, and because everything after it has to wait for it, that step sets the rate of the whole reaction.
It is the rate-determining step. In energy terms the rate-determining step is the one with the largest activation energy, the highest hump on the reaction profile.
Definition: The rate-determining step is the slowest step in a reaction mechanism, and it controls the overall rate of the reaction. An intermediate is a species formed in one step of a mechanism and consumed in a later step.
From Mechanism to Rate Equation
The rate of the overall reaction equals the rate of the rate-determining step, and the rate of any single step depends on the concentrations of the species that collide in it.
So the rate equation is written from the rate-determining step: each species that takes part in the slow step appears in the rate equation, and its order is the number of that species involved.
Species that first appear in a step after the slow step are zero order; they cannot speed up a step they are not part of.
If an intermediate takes part in the slow step, it is replaced by the reactants that form it, because intermediates never appear in a rate equation.
Worked example: nitrogen dioxide and carbon monoxide react at low temperatures, NO₂(g) + CO(g) → NO(g) + CO₂(g), by the mechanism
step 1 (slow): NO₂ + NO₂ → NO + NO₃
step 2 (fast): NO₃ + CO → NO₂ + CO₂
The slow step involves two molecules of NO₂ and nothing else, so rate = k[NO₂]².
Carbon monoxide reacts only in the fast second step, so it is zero order: doubling [CO] has no effect on the rate, even though CO is a reactant.
Adding the two steps cancels the intermediate NO₃ and one of the NO₂ molecules, and gives back the overall equation, which is the check that the mechanism is properly written.
![Reaction profile of a two-step mechanism with the slow rate-determining step, and the worked NO₂ + CO example giving rate = k[NO₂]²](https://www.onlinelearningsystem.net/xyz/wp-content/uploads/2026/09/kinetics-t11-06-rate-determining-step.jpg)
A two-step reaction profile with the intermediate in the dip and the higher barrier of the slow step, beside the worked NO₂ + CO example.
Method: 1 Find the slow step. 2 Write rate = k × the concentration of each species in it, raised to the number of that species. 3 Replace any intermediate by the reactants that form it. 4 Everything that appears only later is zero order.
Check: Rate Equation From a Mechanism
Given a mechanism with its slow step, write the rate equation and pick out the zero-order reactants, for reactions not worked on this page.
From Rate Equation to Mechanism
The reverse problem is more open: given the rate equation and the overall equation, propose a mechanism.
Two rules constrain it. The slow step must contain exactly the species in the rate equation, in the numbers given by the orders. And the steps must add up to the overall equation, with every intermediate cancelling.
Worked example: 2NO(g) + 2H₂(g) → N₂(g) + 2H₂O(g) has the rate equation rate = k[NO]²[H₂]. The slow step therefore involves two NO and one H₂. A three-body collision is unlikely, so a common answer is
step 1 (fast): NO + NO ⇌ N₂O₂
step 2 (slow): N₂O₂ + H₂ → N₂O + H₂O
step 3 (fast): N₂O + H₂ → N₂ + H₂O
The slow step contains the intermediate N₂O₂, which is replaced by the 2NO that form it, giving [NO]²[H₂] as required. The three steps add to the overall equation, with N₂O₂ and N₂O cancelling.
The second molecule of hydrogen reacts after the slow step, which is why the order in H₂ is 1 and not 2.
Other mechanisms can fit the same rate equation, so a question asks for a mechanism consistent with the data, not the mechanism.
An intermediate is formed in one step and used up in a later one; a catalyst is used in one step and regenerated later, so it appears in the mechanism but in neither the reactants nor the products of the overall equation.
Exam focus: Check both things before you write: the species in the slow step (with the earlier fast step feeding it) match the rate equation, and the steps sum to the overall equation.
![Poster building a three-step mechanism for 2NO + 2H₂ from rate = k[NO]²[H₂], with intermediate and catalyst compared.](https://www.onlinelearningsystem.net/xyz/wp-content/uploads/2026/09/t11x-mechbuild.jpg)
A proposed mechanism must put exactly the rate-equation species into the slow step and its steps must add up to the overall equation, as in the three-step route for nitrogen monoxide and hydrogen.
Check: Mechanism From a Rate Equation
Propose or choose steps consistent with a given rate equation and overall equation, and pick out the intermediates, for reactions not used above.
Evidence for Sₙ1 and Sₙ2 in the Hydrolysis of Halogenoalkanes
The hydrolysis of halogenoalkanes by hydroxide ions, R–Br + OH⁻ → R–OH + Br⁻, is the standard case where rate equations distinguish two mechanisms. The overall equation is the same for every halogenoalkanes; the rate equation is not.
Primary: bromoethane. Experiment gives rate = k[CH₃CH₂Br][OH⁻], first order in each and second order overall.
Both species are in the rate-determining step, so the mechanism is a single step. The hydroxide ion attacks the δ+ carbon at the same time as the C–Br bond breaks, passing through a transition state in which the carbon is partly bonded to both.
This is Sₙ2: substitution, nucleophilic, with two species in the slow step (bimolecular).
Tertiary: 2-bromo-2-methylpropane. Experiment gives rate = k[(CH₃)₃CBr], first order overall, with the hydroxide concentration having no effect.
Only the halogenoalkanes is in the rate-determining step, so the mechanism has two steps: a slow breaking of the C–Br bond to give a carbocation, (CH₃)₃C⁺, and Br⁻, then a fast attack by OH⁻ on the carbocation.
This is Sₙ1: one species in the slow step (unimolecular).
The route is available to tertiary compounds because the tertiary carbocation is stabilised by the three alkyl groups, which push electron density towards the positive carbon (a positive inductive effect). A primary carbocation would be far too unstable to form.
| Primary (Sₙ2) | Tertiary (Sₙ1) | |
|---|---|---|
| Rate equation | rate = k[R–Br][OH⁻] | rate = k[R–Br] |
| Overall order | 2 | 1 |
| Number of steps | one, via a transition state | two, via a carbocation intermediate |
| Species in the slow step | halogenoalkane and OH⁻ | halogenoalkane only |
| Effect of doubling [OH⁻] | rate doubles | no change |
Secondary halogenoalkanes can react by either route, depending on the solvent and the nucleophile.
![One-step Sₙ2 mechanism of bromoethane with hydroxide ions through a transition state, with the rate equation rate = k[CH₃CH₂Br][OH⁻], second order overall](https://www.onlinelearningsystem.net/xyz/wp-content/uploads/2026/09/skel-t11-sn2-bromoethane.svg)
![Two-step Sₙ1 mechanism of 2-bromo-2-methylpropane: slow loss of bromide to form a tertiary carbocation, then fast attack by hydroxide ions, with the rate equation rate = k[(CH₃)₃CBr], first order](https://www.onlinelearningsystem.net/xyz/wp-content/uploads/2026/09/skel-t11-sn1-2-bromo-2-methylpropane.svg)
The one-step Sₙ2 hydrolysis of bromoethane and the two-step Sₙ1 hydrolysis of 2-bromo-2-methylpropane, with the rate equation that is evidence for each.
Nucleophilic Substitution of Bromoethane by Hydroxide
Watch a hydroxide ion attack the δ+ carbon of bromoethane from behind, pass through a single transition state and push out the bromide ion, turning the carbon inside out.
© Dr. Mohammed Al-Fatah – onlinelearningsystem.net
Exam wording: “Second order overall, first order in both the halogenoalkanes and OH⁻, so both are in the rate-determining step: Sₙ2.” “First order, independent of [OH⁻], so only the halogenoalkanes is in the rate-determining step, which is the formation of the carbocation: Sₙ1.”
Check: Sₙ1 or Sₙ2 From the Rate Equation
Use rate data for halogenoalkanes not discussed above to decide the mechanism and justify it.
Common Exam Points
Say
“The rate-determining step is the slowest step and controls the overall rate.” “The species in the rate equation are those in the rate-determining step (or in steps before it).” “The steps add up to the overall equation.”
Do not say
“The rate equation is written from the balanced equation.” “A zero-order reactant is not involved in the reaction” (it reacts after the slow step). “First order means Sₙ1” without linking the order to the number of species in the slow step.
Watch for
Mechanisms where the slow step contains an intermediate: substitute the reactants that form it before writing the rate equation. Questions that give a mechanism and ask for the order with respect to a species that appears in the slow step twice: the order is 2.
FAQs
Use these quick answers to check the rate-determining step and mechanisms.
Is the rate-determining step always the first step?
No. It is the slowest step wherever it comes. If a fast equilibrium comes before the slow step, the species in that fast step also affect the rate, because they set the concentration of the intermediate that enters the slow step.
How do I know which species is the intermediate?
It is formed in one step and used up in a later one, so it appears on the product side of an early step and the reactant side of a later step, and it cancels when the steps are added. Intermediates never appear in the overall equation.
Why does a tertiary halogenoalkane react by Sₙ1?
The three alkyl groups block the approach of the nucleophile to the carbon and, more importantly, stabilise the carbocation that forms when the halogen leaves. The molecule can therefore ionise on its own in the slow step, giving a rate equation that depends only on the halogenoalkane.
What does the “2” in Sₙ2 mean?
It means two species are involved in the rate-determining step: the halogenoalkane and the nucleophile collide in a single step, so the rate equation is rate = k[RX][OH⁻]. In Sₙ1 only the halogenoalkane is in the slow step, so the rate equation has one concentration term.
Can a rate equation prove a mechanism?
No. It can rule out any mechanism that does not fit and support one that does, but several mechanisms can give the same rate equation. Say a mechanism is “consistent with” the rate equation, not “proved by” it.
Copyright and author footprint: This OLS revision page was written for Online Learning System by Dr. Mohammed Al-Fatah. It is designed for A Level Chemistry revision and should not be copied or redistributed without permission.
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