The shape of the lesson
By the end of the lesson, learners can
- Matching a mechanism to the data: Work through each example: write the steps, check they add up to the overall equation, then read the rate equation off the slow step.
- SN1 or SN2? Let the rate equation decide: Compare the one-step and two-step hydrolysis mechanisms in the simulator and note which species is present in the slow step.
- Rate-determining step: the slowest step.
- Orders = number of each particle in (or before) the slow step.
- Zero order: joins after the slow step.
- Intermediate made then used; catalyst used then regenerated.
One picture every idea returns to
The rate-determining step is the slowest step in a mechanism. Only species that take part in the rate-determining step, or in fast steps before it, appear in the rate equation, and the order of each matches the number of its particles involved. A species with zero order joins the reaction after the slow step. An intermediate is made in one step and used up in a later one, so it never appears in the overall equation; a catalyst is used in one step and regenerated later, so it can appear in the rate equation but not in the overall equation. A mechanism is consistent with the data if its steps add up to the overall equation and its slow step matches the rate equation.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, The slowest step sets the pace.. Most reactions happen in a series of steps, and the overall rate can be no faster than the slowest one. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Matching a mechanism to the data | Work through each example: write the steps, check they add up to the overall equation, then read the rate equation off the slow step. Project tab 3 of the lab (Rate equations and reaction kinetics lab, mechanism and rate-determining step) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: SN1 or SN2? Let the rate equation decide | Compare the one-step and two-step hydrolysis mechanisms in the simulator and note which species is present in the slow step. Project tab 3 of the lab (Rate equations and reaction kinetics lab, mechanism and rate-determining step) and let a student drive it while the class predicts. | Corners |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 15 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in SN1 or SN2? Let the rate equation decide, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Corners
Put one answer or viewpoint in each corner of the room. Students move to the corner they agree with, justify the choice to a partner there, then hear one voice from each corner before the reveal.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| The rate equation tells you about the slow step, not the overall equation, so do not copy coefficients from the overall equation. | A reactant that is zero order is still needed: it reacts in a fast step after the rate-determining step. An intermediate is formed then used up; a catalyst is used then regenerated. Do not call the transition state an intermediate: a transition state is a peak on the energy profile, an intermediate sits in a dip. |
Rate-determining step and mechanisms: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
For NO2 + CO → NO + CO2, rate = k[NO2]2. Propose a two-step mechanism and identify the intermediate.
Bromoethane is hydrolysed by aqueous OH− with rate = k[CH3CH2Br][OH−]. Explain how this supports an SN2 mechanism.
2-bromo-2-methylpropane reacts with OH−(aq). Predict the rate equation and give the two steps of the mechanism.
Distinguish between an intermediate and a catalyst in a reaction mechanism.
For X + 2Y → Z, rate = k[X][Y]. A student suggests a one-step mechanism. Explain why it is wrong.
On an energy profile for a two-step reaction, how can you identify (a) the rate-determining step, (b) the intermediate?
Answer key and marking notes
- Step 1 (slow): NO2 + NO2 → NO3 + NO (two NO2, so second order).
- Step 2 (fast): NO3 + CO → NO2 + CO2 (CO after the slow step, so zero order).
- NO3 is the intermediate; the steps add up to the overall equation.
- Both species are first order, so one of each is in the rate-determining step.
- In SN2 the OH− attacks the δ+ carbon as the C–Br bond breaks, in a single step.
- So the only step contains both species, matching the rate equation.
- Rate = k[(CH3)3CBr].
- Step 1 (slow): (CH3)3CBr → (CH3)3C+ + Br−.
- Step 2 (fast): (CH3)3C+ + OH− → (CH3)3COH.
- An intermediate is formed in one step and used up in a later step.
- A catalyst is used up in one step and regenerated in a later step, so it is present at the start and the end.
- A one-step mechanism would need one X and two Y colliding, giving rate = k[X][Y]2.
- The data show first order in Y, so the slow step has only one Y and the second Y reacts in a later fast step.
- (a) The step with the higher activation energy (the higher hump measured from its starting level).
- (b) The intermediate sits in the dip (local minimum) between the two transition states.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
What is meant by the rate-determining step of a reaction?
- Athe slowest step in the mechanism, which limits the overall rate
- Bthe first step in the mechanism
- Cthe step with the most exothermic enthalpy change
- Dthe step that produces the final product
Bromoethane, a primary halogenoalkane, is hydrolysed by aqueous hydroxide ions by the SN2 mechanism. What is the rate equation?
- Arate = k[CH3CH2Br][OH−]
- Brate = k[CH3CH2Br]
- Crate = k[OH−]
- Drate = k[CH3CH2Br]2
A mechanism has two steps: H2O2 + I− → IO− + H2O, then IO− + 2H+ + I− → I2 + H2O. What is IO−?
- Aan intermediate
- Ba catalyst
- Ca reactant in the overall equation
- Da transition state
The acid-catalysed iodination of propanone is zero order with respect to iodine. What does this tell you?
- Aiodine is involved only in a step after the rate-determining step
- Biodine does not take part in the reaction
- Ciodine is the catalyst
- Diodine is present in the rate-determining step
The energy profile is for a two-step reaction. Which statement is correct?
- Astep 2 is rate-determining because it has the larger activation energy, measured from the intermediate
- Bstep 1 is rate-determining because it always happens first
- Cstep 1 is rate-determining because it starts from the reactants
- Dneither step is rate-determining because the overall reaction is exothermic
For 2NO + O2 → 2NO2, rate = k[NO]2[O2]. Which mechanism is consistent with this?
- ANO + NO ⇌ N2O2 (fast); N2O2 + O2 → 2NO2 (slow)
- BNO + O2 → NO3 (slow); NO3 + NO → 2NO2 (fast)
- CNO + NO → N2O2 (slow); N2O2 + O2 → 2NO2 (fast)
- DO2 → 2O (slow); 2O + 2NO → 2NO2 (fast)
For the hydrolysis of 2-bromo-2-methylpropane, [OH−] is doubled at constant [(CH3)3CBr] and temperature. What happens, and why?
- Athe rate is unchanged, because OH− reacts only with the carbocation in a fast step after the slow ionisation
- Bthe rate doubles, because OH− attacks the carbon as Br− leaves
- Cthe rate quadruples, because two OH− ions are needed
- Dthe rate halves, because OH− stabilises the carbocation
For CH3COCH3 + I2 → CH3COCH2I + HI, doubling [H+] doubles the rate, yet H+ does not appear in the overall equation. What is the best explanation?
- AH+ is a catalyst: it takes part in the rate-determining step and is regenerated later
- BH+ is an intermediate formed in the slow step
- CH+ is a spectator ion and the observation must be an experimental error
- DH+ is a reactant that is used up, so it should be in the overall equation
Quiz key
The rate-determining step is the slowest step; the species in it (and in any fast steps before it) appear in the rate equation. It is often, but not always, the first step. Its enthalpy change does not decide its speed: the activation energy does. The last step forms the product but may be very fast.
In SN2 the OH− attacks the δ+ carbon while the C-Br bond breaks, in one step, so both species are first order. Rate = k[CH3CH2Br] is the SN1 rate equation for tertiary halogenoalkanes. Hydroxide alone ignores the halogenoalkane, which is certainly in the only step. Second order in bromoethane would need two bromoethane molecules to collide.
IO− is formed in step 1 and used up in step 2, so it is an intermediate; it cancels when the steps are added. A catalyst is the reverse: used first, then regenerated. IO− does not appear in the overall equation H2O2 + 2I− + 2H+ → I2 + 2H2O. A transition state is a peak on the energy profile, which cannot be isolated; an intermediate sits in a dip.
Rate = k[CH3COCH3][H+]. Iodine is used up (it is in the overall equation), but only in a fast step after the slow step, so its concentration does not affect the rate. It is a reactant, not a catalyst; H+ is the catalyst. If iodine were in the rate-determining step, it would appear in the rate equation.
Step 1 climbs only a little from the reactants, but step 2 climbs much further from the intermediate, so step 2 has the larger activation energy and is the slow, rate-determining step. The first step is not automatically the slowest. Whether the overall reaction is exothermic does not decide which step is slow. The species in step 2 (and those in step 1, which form the intermediate) appear in the rate equation.
In the correct mechanism, N2O2 is made from two NO in a fast step before the slow step, and the slow step uses it with one O2, so two NO and one O2 are involved up to the slow step: rate = k[NO]2[O2]. A slow NO + O2 step gives rate = k[NO][O2]. A slow 2NO step gives rate = k[NO]2, zero order in O2. A slow O2 splitting step gives rate = k[O2].
SN1: the slow step is (CH3)3CBr → (CH3)3C+ + Br−; OH− reacts with the carbocation afterwards, so rate = k[(CH3)3CBr] and the rate does not change. Doubling describes the SN2 behaviour of primary halogenoalkanes. Only one OH− is used per molecule and it is zero order. OH− is not involved in stabilising the carbocation; the three alkyl groups do that.
Rate = k[CH3COCH3][H+]: H+ is first order, so it is in the slow step, but it is regenerated, so it is a homogeneous catalyst and cancels from the overall equation (HI is formed, but that comes from the iodine). An intermediate is made and then used up, so it would not be added at the start. Spectator ions do not affect the rate. H+ is not used up overall.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.