Rate-determining step and reaction mechanismsAQA A-Level Chemistry: Revision notes
Section 1
Mechanisms and the rate-determining step
Many reactions happen in more than one step. A mechanism is the series of steps by which reactants become products. The steps must add up to the overall equation, with any species that appear on both sides cancelling.
A species that is formed in one step and used up in a later step is an intermediate. It does not appear in the overall equation.
In a multi-step reaction, one step is much slower than the others. This is the rate-determining step (the slowest step). The overall rate cannot be faster than the slowest step, just as a queue moves at the speed of its slowest stage. Changing the rate of a fast step has little effect on the overall rate.
The rate-determining step is not always the first step. It is whichever step is slowest.
Section 2
What the orders tell you about the rate-determining step
The rate equation contains the reactants (and any catalyst) whose concentrations affect the rate. These are the species involved in the rate-determining step or in steps before it.
- First order in a species: one molecule or ion of it is in the rate-determining step (or before it)
- Second order: two molecules are involved
- Zero order: the species is not in the rate-determining step or any step before it; it reacts in a fast step after the slow step
The rate equation is therefore built from the species in the slow step. If the slow step is A + B → products, the rate equation is rate = k[A][B].
Count the species in the slow step. The number of each species equals its order, so the rate equation can be written straight from the slow step.
Section 3
Worked example: testing a mechanism
The reaction NO₂ + CO → NO + CO₂ has rate = k[NO₂]². A proposed mechanism is:
- Step 1 (slow): 2NO₂ → NO₃ + NO
- Step 2 (fast): NO₃ + CO → NO₂ + CO₂
Check 1: adding the steps gives 2NO₂ + NO₃ + CO → NO₃ + NO + NO₂ + CO₂, which simplifies to NO₂ + CO → NO + CO₂, so the mechanism matches the overall equation. NO₃ is an intermediate.
Check 2: the slow step contains two NO₂ and no CO, so it predicts rate = k[NO₂]². This matches the experiment, and CO is zero order because it reacts after the slow step.
The mechanism is therefore consistent with the data. It is not proved, because a different mechanism could also fit the rate equation.
Do not write the rate equation from the overall equation. For this reaction that would predict rate = k[NO₂][CO], which does not match the data.
Section 4
Catalysts and intermediates
A catalyst is used up in one step and regenerated in a later step, so it does not appear in the overall equation. If the catalyst takes part in or before the slow step, it appears in the rate equation. For example, in the iodination of propanone, H⁺ is a catalyst and the rate equation is rate = k[CH₃COCH₃][H⁺]. Iodine is zero order, so it is not involved until after the slow step.
An intermediate (such as NO₃ above) does not appear in the overall equation. At this level, the rate equation is written in terms of the reactants and catalysts.
Increasing the rate of a fast step has almost no effect on the overall rate. To speed up the reaction, you must speed up the rate-determining step.
If a species appears in the rate equation but not in the overall equation, it is a catalyst.
Section 5
Halogenoalkane hydrolysis: orders and mechanisms
Orders can tell different mechanisms apart.
2-bromo-2-methylpropane, (CH₃)₃CBr: rate = k[(CH₃)₃CBr]. Only the halogenoalkane is in the slow step: the C–Br bond breaks to form a carbocation and Br⁻. Hydroxide ions react with the carbocation in a fast step afterwards, so they are zero order.
Bromoethane, CH₃CH₂Br: rate = k[CH₃CH₂Br][OH⁻]. One halogenoalkane molecule and one hydroxide ion are in the rate-determining step: the OH⁻ attacks the δ+ carbon as the C–Br bond breaks, in a single step.
Must know
- The slowest step is the rate-determining step; the other steps do not affect the overall rate
- The steps of a mechanism must add up to the overall equation
- An intermediate is formed in one step and used in a later one; it is not in the overall equation
- Order shows how many molecules of that species are in or before the slow step
- Zero order: the species is not in the slow step or any step before it
- The rate equation can be written from the species in the slow step
- A mechanism that fits the rate equation is consistent with it, but not proved
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Rate-determining step and reaction mechanisms
- The reaction NO₂(g) + CO(g) → NO(g) + CO₂(g) has the experimentally determined rate equation rate = k[NO₂]². A two-step mechanism is proposed. Step 1 (slow): 2NO₂ → NO₃ + NO. Step 2 (fast): NO₃ + CO → NO₂ + CO₂.Explain why CO does not appear in the rate equation even though it is a reactant.2 marks
- In acidic solution, propanone reacts with iodine: CH₃COCH₃ + I₂ → CH₃COCH₂I + HI. The experimentally determined rate equation is rate = k[CH₃COCH₃][H⁺]. The reaction is zero order with respect to iodine. Hydrogen ions are not used up.Explain why H⁺ appears in the rate equation although it does not appear in the overall equation.2 marks
- Two halogenoalkanes are hydrolysed by aqueous sodium hydroxide at constant temperature. For 2-bromo-2-methylpropane, (CH₃)₃CBr, the reaction is first order with respect to the halogenoalkane and zero order with respect to hydroxide ions. For bromoethane, CH₃CH₂Br, the reaction is first order with respect to the halogenoalkane and first order with respect to hydroxide ions.Use the orders of reaction for 2-bromo-2-methylpropane to explain what they show about its mechanism.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).