Rate-determining step and reaction mechanismsEdexcel A-Level Chemistry: Revision notes
Section 1
The rate-determining step
Many reactions happen in more than one step. The rate-determining step is the slowest step, and it controls the overall rate. The rate equation shows which species are involved in the rate-determining step, or in earlier steps.
- A species that appears in the rate equation (order 1 or 2) is in the rate-determining step. A second order means two of that species are involved.
- A species that is zero order is not in the rate-determining step. It reacts in a fast step afterwards.
- An intermediate is formed in one step and used up in a later one. It does not appear in the overall equation.
The orders in the rate equation are not the coefficients of the overall equation. They are the number of each species in the rate-determining step.
Section 2
From rate equation to rate-determining step, and back
Rate equation to step: read off the species and their orders. If rate = k[A][B]², the rate-determining step contains one A and two B (or species that give them).
Step to rate equation: the rate-determining step lists the species that appear in the rate equation, each raised to the number of particles in the step. If the slow step is A + B → X + Y, the rate = k[A][B].
For example, if the slow step is NO₂ + F₂ → NO₂F + F, then rate = k[NO₂][F₂].
Write a mechanism with the slow step first, then check it contains exactly the species in the rate equation.
Section 3
Deducing a mechanism
Use the rate equation and the overall equation together. A proposed mechanism is acceptable if:
- the steps add up to the overall equation, once intermediates cancel,
- the rate-determining step contains the species in the rate equation, and
- each step is balanced and involves no more than two particles colliding.
Worked example. For NO₂ + CO → NO + CO₂ with rate = k[NO₂]²: step 1 (slow) 2NO₂ → NO₃ + NO; step 2 (fast) NO₃ + CO → NO₂ + CO₂. Adding the steps gives 2NO₂ + NO₃ + CO → NO₃ + NO + NO₂ + CO₂, which cancels to NO₂ + CO → NO + CO₂. NO₃ is the intermediate, and CO is not in the rate equation because it reacts after the slow step.
Section 4
Acid-catalysed iodination of propanone
The reaction CH₃COCH₃ + I₂ → CH₃COCH₂I + HI is catalysed by acid. Initial-rate experiments show it is first order in propanone, first order in H⁺ and zero order in iodine, so rate = k[CH₃COCH₃][H⁺].
The rate-determining step therefore involves propanone and H⁺ but not iodine. A possible mechanism is:
- slow: propanone is protonated by H⁺ and then loses a proton to form the enol CH₃C(OH)=CH₂
- fast: the enol reacts with I₂ to form CH₃COCH₂I and H⁺
The H⁺ is regenerated, which is why it acts as a catalyst, and it appears in the rate equation even though it is not in the overall equation.
Do not say iodine is in the rate-determining step. Because it is zero order, it reacts after the slow step.
Section 5
Halogenoalkane hydrolysis: SN1 and SN2
Rate equations show how halogenoalkanes react with hydroxide ions.
- Tertiary (for example 2-bromo-2-methylpropane): rate = k[RBr], first order in the halogenoalkane and zero order in OH⁻. This is SN1. Slow step: the C–Br bond breaks heterolytically, forming a carbocation and Br⁻. Fast step: OH⁻ attacks the carbocation. The tertiary carbocation is stabilised by the electron-donating alkyl groups.
- Primary (for example 1-bromobutane): rate = k[RBr][OH⁻], first order in both. This is SN2. In one step, OH⁻ attacks the δ+ carbon from the side opposite the bromine, via a transition state in which the C–OH bond forms as the C–Br bond breaks.
SN1 is unimolecular (one species in the rate-determining step); SN2 is bimolecular (two species).
Link the rate equation, the number of steps and the type of halogenoalkane to name the mechanism: first order in both means SN2 (primary), first order in the halogenoalkane only means SN1 (tertiary).
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Rate-determining step and reaction mechanisms
- Nitrogen dioxide reacts with carbon monoxide: NO₂(g) + CO(g) → NO(g) + CO₂(g). At 500 K the rate equation is rate = k[NO₂]². A proposed mechanism has two steps, and the first step is the slower.Explain why CO does not appear in the rate equation.2 marks
- The hydrolysis of two bromoalkanes by aqueous sodium hydroxide was studied at constant temperature. For 2-bromo-2-methylpropane, doubling the concentration of the bromoalkane doubled the initial rate, but doubling the concentration of hydroxide ions had no effect on the initial rate. For 1-bromobutane, doubling the concentration of either the bromoalkane or the hydroxide ions doubled the initial rate.Explain why the data show that 2-bromo-2-methylpropane is hydrolysed by an SN1 mechanism.2 marks
- The acid-catalysed reaction of propanone with iodine is CH₃COCH₃(aq) + I₂(aq) → CH₃COCH₂I(aq) + HI(aq). Initial rates were measured at constant temperature. Experiment 1: [CH₃COCH₃] = 1.00 mol dm⁻³, [I₂] = 0.0100 mol dm⁻³, [H⁺] = 0.50 mol dm⁻³, initial rate = 2.4 × 10⁻⁶ mol dm⁻³ s⁻¹. Experiment 2: [CH₃COCH₃] = 2.00 mol dm⁻³, [I₂] = 0.0100 mol dm⁻³, [H⁺] = 0.50 mol dm⁻³, initial rate = 4.8 × 10⁻⁶ mol dm⁻³ s⁻¹. Experiment 3: [CH₃COCH₃] = 1.00 mol dm⁻³, [I₂] = 0.0200 mol dm⁻³, [H⁺] = 0.50 mol dm⁻³, initial rate = 2.4 × 10⁻⁶ mol dm⁻³ s⁻¹. Experiment 4: [CH₃COCH₃] = 1.00 mol dm⁻³, [I₂] = 0.0100 mol dm⁻³, [H⁺] = 1.00 mol dm⁻³, initial rate = 4.8 × 10⁻⁶ mol dm⁻³ s⁻¹.Deduce the order of reaction with respect to each of propanone, iodine and hydrogen ions, and write the rate equation.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).