The shape of the lesson
By the end of the lesson, learners can
- Rate equations and the units of k: Measure the rate of a reaction in the simulator at several starting concentrations and see how the initial gradient changes.
- Reading orders from graphs: Use the simulator to plot concentration against time, then compare the shapes with Fig. 1.
- Initial rates: ×2 conc. gives rate ×1, ×2, ×4 for order 0, 1, 2.
- Units of k: rearrange and cancel; order 2 gives dm3 mol−1 s−1.
- Constant half-life means first order; k = 0.693 / t½.
- Rate-[A] graphs: flat, straight through origin, upward curve.
One picture every idea returns to
A rate equation has the form rate = k[A]m[B]n. The powers m and n are the orders with respect to each reactant (0, 1 or 2 at this level) and their sum is the overall order. Orders are found by experiment, never from the stoichiometric equation. In an initial-rates table, doubling a concentration while the others stay fixed multiplies the rate by 1 (zero order), 2 (first order) or 4 (second order). A first-order reaction has a constant half-life, and k = 0.693 / t½. The rate constant k changes only with temperature (or a catalyst), and its units depend on the overall order.
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, How rate depends on concentration.. A rate equation says exactly how the rate responds when you change each concentration. 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: Rate equations and the units of k | Measure the rate of a reaction in the simulator at several starting concentrations and see how the initial gradient changes. Project tab 1 of the lab (Rate equations and reaction kinetics lab, initial rates) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Reading orders from graphs | Use the simulator to plot concentration against time, then compare the shapes with Fig. 1. Project tab 2 of the lab (Rate equations and reaction kinetics lab, graph shapes and half-life) and let a student drive it while the class predicts. | Numbered Heads Together |
| 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, 16 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 Reading orders from graphs, 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.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
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 |
|---|---|
| Orders come from experimental data, not from the coefficients in the balanced equation. | Doubling a concentration and seeing the rate quadruple means second order, not fourth. A constant half-life proves first order only; do not use t ½ = 0.693/ k for other orders. Always give units for k, worked out from the overall order, and remember k does not change with concentration. |
Orders of reaction and rate equations: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
For the reaction 2NO + 2H2 → N2 + 2H2O at constant temperature:
| Expt | [NO] / mol dm−3 | [H2] / mol dm−3 | initial rate / mol dm−3 s−1 |
|---|---|---|---|
| 1 | 0.0100 | 0.0100 | 1.20 × 10−5 |
| 2 | 0.0200 | 0.0100 | 4.80 × 10−5 |
| 3 | 0.0100 | 0.0300 | 3.60 × 10−5 |
Deduce the rate equation and calculate k with units.
Using your answer to question 1, calculate the initial rate when [NO] = 0.0300 mol dm−3 and [H2] = 0.0200 mol dm−3.
Explain why the order of a reaction cannot be deduced from the balanced equation.
The concentration of a reactant falls from 0.800 to 0.400 mol dm−3 in 35.0 s and from 0.400 to 0.200 mol dm−3 in a further 35.0 s. State the order and calculate k.
Sketch and describe the rate against concentration graphs for a reactant that is (a) zero order, (b) first order, (c) second order.
A reaction is second order overall with rate = k[A]2. When [A] = 0.150 mol dm−3 the rate is 4.50 × 10−3 mol dm−3 s−1. Calculate k with units.
Answer key and marking notes
- Expts 1 and 2: [NO] doubles, rate ×4, so second order in NO.
- Expts 1 and 3: [H2] ×3, rate ×3, so first order in H2.
- Rate = k[NO]2[H2].
- k = 1.20 × 10−5 / (0.01002 × 0.0100) = 12.0 dm6 mol−2 s−1.
- Rate = 12.0 × 0.03002 × 0.0200.
- = 2.16 × 10−4 mol dm−3 s−1.
- Most reactions happen in several steps; the rate depends only on the species in or before the rate-determining step.
- So orders must be found by experiment, and they need not match the stoichiometric coefficients (2 for H2 in the equation above, but first order).
- Successive half-lives are equal, so the reaction is first order.
- k = 0.693 / t½ = 0.693 / 35.0.
- k = 0.0198 s−1.
- (a) horizontal line: rate is independent of [A].
- (b) straight line through the origin: rate ∝ [A].
- (c) curve upwards through the origin, getting steeper: rate ∝ [A]2.
- k = 4.50 × 10−3 / 0.1502.
- k = 0.200 dm3 mol−1 s−1.
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.
The rate equation for a reaction is rate = k[P][Q]2. What are the order with respect to Q and the overall order?
- Asecond order in Q; third order overall
- Bsecond order in Q; second order overall
- Cfirst order in Q; third order overall
- Dthird order in Q; third order overall
A reaction is second order overall. What are the units of the rate constant k?
- Adm3 mol−1 s−1
- Bs−1
- Cmol dm−3 s−1
- Ddm6 mol−2 s−1
In two experiments at the same temperature, [B] is kept constant and [A] is doubled. The initial rate also doubles. What is the order with respect to A?
- Afirst order
- Bzero order
- Csecond order
- Dit cannot be found without knowing [B]
The graph shows how the concentration of reactant A changes with time. What is the order with respect to A, and what is the rate constant?
- Afirst order; k = 0.0347 s−1
- Bfirst order; k = 0.0500 s−1
- Czero order; k = 0.0200 mol dm−3 s−1
- Dsecond order; k = 0.0347 dm3 mol−1 s−1
A graph of initial rate against [X] is a straight line through the origin. What is the order with respect to X?
- Afirst order
- Bzero order
- Csecond order
- Dit depends on the temperature
For rate = k[A][B]2, an experiment with [A] = 0.0500 mol dm−3 and [B] = 0.100 mol dm−3 has an initial rate of 3.00 × 10−4 mol dm−3 s−1. What is k?
- A0.600 dm6 mol−2 s−1
- B0.0600 dm3 mol−1 s−1
- C6.00 dm6 mol−2 s−1
- D0.0600 dm6 mol−2 s−1
Using rate = k[A][B]2 with k = 0.600 dm6 mol−2 s−1, what is the initial rate when [A] = 0.100 mol dm−3 and [B] = 0.300 mol dm−3?
- A5.40 × 10−3 mol dm−3 s−1
- B1.80 × 10−2 mol dm−3 s−1
- C1.80 × 10−3 mol dm−3 s−1
- D5.40 × 10−2 mol dm−3 s−1
A first-order reaction has k = 0.0231 s−1. How long does it take for the concentration of the reactant to fall to 12.5% of its starting value?
- A90.0 s
- B30.0 s
- C45.0 s
- D60.0 s
Quiz key
The power on [Q] is 2, so the reaction is second order in Q; P is first order, so the overall order is 1 + 2 = 3. Second order overall ignores the first-order P term. First order in Q confuses Q with P. Third order in Q mistakes the overall order for the order in Q.
k = rate / [A]2 = (mol dm−3 s−1) / (mol dm−3)2 = dm3 mol−1 s−1. s−1 is for first order, mol dm−3 s−1 for zero order (the same as the rate) and dm6 mol−2 s−1 for third order overall.
Rate ∝ [A]n, and 2n = 2 gives n = 1. Zero order would leave the rate unchanged; second order would make it four times larger. [B] does not matter because it was kept constant, which is exactly why the experiments were designed that way.
[A] halves every 20 s (0.80 to 0.40 to 0.20 to 0.10), so the half-life is constant and the reaction is first order. k = 0.693 / t½ = 0.693 / 20 = 0.0347 s−1. 0.0500 s−1 is 1/20, which forgets the ln 2 factor. A zero-order plot would be a straight line, not a curve. For a second-order reaction each half-life would be twice as long as the one before.
Rate = k[X] gives a straight line through the origin with gradient k. Zero order gives a horizontal line, since the rate does not change with [X]. Second order gives a curve that gets steeper. Temperature changes k, which changes the gradient, but not the shape of the graph or the order.
k = 3.00 × 10−4 / (0.0500 × 0.1002) = 3.00 × 10−4 / 5.00 × 10−4 = 0.600, and the reaction is third order overall, so the units are dm6 mol−2 s−1. 0.0600 comes from not squaring [B], whichever units are attached. 6.00 is a power-of-ten slip in the division.
Rate = 0.600 × 0.100 × 0.0900 = 5.40 × 10−3 mol dm−3 s−1. 1.80 × 10−2 forgets to square [B]. 1.80 × 10−3 squares [A] instead of [B]. 5.40 × 10−2 is a power-of-ten slip.
t½ = 0.693 / 0.0231 = 30.0 s, constant for a first-order reaction. 100% to 50% to 25% to 12.5% is three half-lives, so the time is 3 × 30.0 = 90.0 s. 30.0 s is only one half-life. 45.0 s treats the fall as linear. 60.0 s is two half-lives, which only reaches 25%.
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.