The shape of the lesson
By the end of the lesson, learners can
- Only some collisions count: In the collision simulator, count the collisions and the successful ones.
- Measuring a rate: Choose a method in the simulator that suits the reaction: something you can measure changing with time.
- Rate from a graph: The rate at any moment is the gradient of the curve at that moment.
- Rate = change in concentration ÷ time.
- Initial rate: tangent at t = 0.
- Collision theory: energy ≥ Ea and correct orientation.
- Concentration, pressure, surface area: more frequent collisions.
One picture every idea returns to
The rate of reaction is the change in amount or concentration of a reactant or product per unit time, often in mol dm−3 s−1. It is found from the gradient of a concentration-time (or volume-time) graph; the gradient of the tangent at t = 0 is the initial rate. By collision theory, particles react only if they collide with at least the activation energy, Ea, and in the right orientation. Raising the concentration, the pressure of a gas or the surface area of a solid makes collisions more frequent, so successful collisions happen more often.
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 fast is fast?. Some reactions are over in a flash; others take years. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Only some collisions count | In the collision simulator, count the collisions and the successful ones. Project tab 1 of the lab (Reaction kinetics lab, collision theory) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Measuring a rate | Choose a method in the simulator that suits the reaction: something you can measure changing with time. Project tab 3 of the lab (Reaction kinetics lab, measuring a rate) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Rate from a graph | The rate at any moment is the gradient of the curve at that moment. Project tab 5 of the lab (Reaction kinetics lab, rate factors) and let a student drive it while the class predicts. | Think-Pair-Share |
| 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, 13 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 Rate from a graph, 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.
Think-Pair-Share
Pose the question, give thirty seconds of silent thinking, then a minute to compare with a partner before a few pairs share with the class. The silent minute is the part that cannot be cut.
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 |
|---|---|
| Increasing the concentration does not give the particles more energy; it only increases the frequency of collisions, so the number of successful collisions per second rises. | Say “more frequent successful collisions”, not just “more collisions”. Temperature is the factor that changes the energy of the particles, covered in the next lesson. |
Rate of reaction: worksheet
6 questions, 13 marks. Written in the style of the papers; no past paper question is reproduced.
State the two conditions needed for a collision between particles to result in a reaction.
Explain, using collision theory, why increasing the pressure increases the rate of a reaction between gases.
Suggest a method of following the rate of each reaction: (a) Zn + H2SO4 → ZnSO4 + H2; (b) Na2S2O3 + 2HCl → 2NaCl + S + SO2 + H2O.
In an experiment 0.150 g of magnesium is lost from a 1.00 g ribbon in 30.0 s. Calculate the mean rate of reaction of magnesium in mol s−1. (Ar: Mg 24.3)
A tangent drawn to a volume-time curve at t = 0 passes through (0 s, 0 cm3) and (25 s, 60 cm3). Calculate the initial rate.
Explain the shape of the volume-time graph for the reaction of excess marble chips with hydrochloric acid.
Answer key and marking notes
- The particles must collide with energy equal to or greater than the activation energy.
- They must collide with the correct orientation.
- The same number of molecules occupy a smaller volume, so the concentration is higher.
- Collisions are more frequent, so there are more successful collisions per unit time.
- (a) Collect the hydrogen in a gas syringe and record the volume at regular times.
- (b) Time how long it takes for the precipitate of sulfur to hide a cross under the flask.
- n(Mg) = 0.150 ÷ 24.3 = 6.17 × 10−3 mol.
- Rate = 6.17 × 10−3 ÷ 30.0 = 2.06 × 10−4 mol s−1.
- Gradient = 60 ÷ 25.
- Initial rate = 2.4 cm3 s−1.
- It is steepest at the start, where the acid concentration is highest and collisions most frequent.
- The gradient falls as the acid is used up, because collisions become less frequent.
- It becomes horizontal when all the acid (the limiting reagent) has reacted.
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.
According to collision theory, which condition must be met for a collision between two reactant particles to lead to a reaction?
- Athe particles must collide with energy greater than or equal to the activation energy and with the correct orientation
- Bthe particles only need to collide; every collision between reactant particles leads to reaction
- Cthe particles must collide with energy equal to the enthalpy change of the reaction, ΔH
- Dthe particles must collide with energy greater than or equal to the activation energy; orientation does not matter
Rate of reaction is the change in amount or concentration of a reactant or product per unit time. When magnesium reacts with dilute hydrochloric acid, 48.0 cm3 of hydrogen is collected in the first 2.00 minutes. What is the mean rate of hydrogen production in cm3 s−1?
- A0.400
- B24.0
- C2.50
- D0.800
The rate of hydrolysis of an ester by aqueous acid is followed by removing samples at intervals, adding each to ice-cold water and titrating quickly with standard alkali. Why is each sample added to ice-cold water?
- Ato quench the sample: cooling and diluting slow the reaction almost to a stop, so the titration measures the composition at the moment of sampling
- Bto make the reaction in the sample go to completion before titrating
- Cto neutralise the acid catalyst so it does not affect the titre
- Dto raise the activation energy so that no further collisions occur
The pressure of a mixture of two reacting gases is doubled at constant temperature. Why does the rate increase?
- Athe molecules are closer together, so there are more collisions per unit time, and so more successful collisions per unit time
- Bthe molecules have more kinetic energy, so a greater proportion have energy at least equal to Ea
- Cthe activation energy is lowered because the molecules are squeezed together
- Dthe reaction becomes more exothermic, releasing heat that speeds it up
The graph shows the volume of gas collected during a reaction. A tangent has been drawn to the curve at t = 0. What is the initial rate of reaction in cm3 s−1?
- A1.50
- B0.300
- C0.667
- D0.948
Marble chips react with excess hydrochloric acid in an open flask on a balance:
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
The mass falls by 0.880 g in the first 50.0 s. What is the mean rate of production of CO2 in mol s−1? (Mr CO2 = 44.0)
- A4.00 × 10−4
- B8.00 × 10−4
- C1.76 × 10−2
- D2.00 × 10−2
In the reaction Na2S2O3(aq) + 2HCl(aq) → 2NaCl(aq) + S(s) + SO2(g) + H2O(l), the time t for a cross under the flask to disappear is measured, and the rate is taken as proportional to 1/t. With 0.150 mol dm−3 sodium thiosulfate, t = 40.0 s. Assuming the rate is proportional to the thiosulfate concentration, what is t when 0.0600 mol dm−3 sodium thiosulfate is used under the same conditions?
- A100 s
- B16.0 s
- C40.0 s
- D250 s
Two experiments use the same mass of calcium carbonate with the same volume and concentration of hydrochloric acid; the acid is the limiting reagent. Experiment 1 uses large lumps and experiment 2 uses powder. How does the volume-time curve for experiment 2 compare with experiment 1?
- Ait is steeper at the start and levels off sooner, at the same final volume
- Bit is steeper at the start and levels off at a larger final volume
- Cit has the same initial gradient but levels off sooner
- Dit is less steep at the start because the powder clumps together, but reaches the same final volume
Quiz key
A successful collision needs enough energy to start breaking bonds (at least the activation energy, Ea) and the right orientation so the reacting parts of the particles meet. Most collisions do not lead to reaction, so 'every collision reacts' is wrong; if it were true, reactions in gases would be almost instantaneous. ΔH is the overall energy difference between reactants and products, not the energy barrier that must be overcome. Orientation does matter: a collision with enough energy but the wrong alignment does not react, which is why only a fraction of energetic collisions succeed.
Mean rate = volume / time = 48.0 cm3 / (2.00 × 60 s) = 48.0 / 120 = 0.400 cm3 s−1. 24.0 comes from dividing by 2.00 without converting to seconds, so it is the rate in cm3 min−1, not per second. 2.50 is time divided by volume (120 / 48.0), which is upside down. 0.800 divides by 60 s, treating 2.00 minutes as 1 minute.
Titration takes time, so the reaction in the sample must be 'frozen' (quenched). Cooling reduces the proportion of molecules with energy at least Ea and dilution lowers the collision frequency, so the composition stays close to that at the sampling time. Letting the reaction go to completion would give the same titre for every sample, making the method useless. Water does not neutralise the acid; the acid is in fact part of what is titrated. Cooling does not change Ea; it changes how many molecules can overcome it, and collisions still occur.
At higher pressure there are more molecules per unit volume, so collisions happen more often and the number of successful collisions per second rises; the fraction of collisions that succeed is unchanged. The temperature is constant, so the energy distribution of the molecules does not change and the proportion with energy at least Ea is the same. Pressure does not change Ea; only a catalyst provides a route with a different activation energy. ΔH is fixed by the reactants and products and is not affected by pressure in this way.
The tangent passes through (0 s, 0 cm3) and (40 s, 60 cm3), so gradient = 60 / 40 = 1.50 cm3 s−1. 0.300 is the final volume divided by the total time (60 / 200), a mean rate over the whole run, which is much smaller than the initial rate. 0.667 is time divided by volume (40 / 60), upside down. 0.948 is the mean rate over the first 40 s read from the curve (37.9 / 40); a chord always gives a lower value than the tangent at the start because the rate is falling.
Mass loss = mass of CO2 escaping. Amount = 0.880 / 44.0 = 0.0200 mol, so rate = 0.0200 / 50.0 = 4.00 × 10−4 mol s−1. 8.00 × 10−4 is the rate at which HCl is used up (2 mol HCl per mol CO2), not the rate of CO2 production. 1.76 × 10−2 is the rate in g s−1 (0.880 / 50.0), not converted to moles. 2.00 × 10−2 is the amount of CO2 formed, not divided by the time.
The cross disappears when the same amount of sulfur has formed, so rate is proportional to 1/t. The concentration falls by a factor of 0.150 / 0.0600 = 2.5, so the rate falls by 2.5 and the time rises by 2.5: 40.0 × 2.5 = 100 s. 16.0 s (40.0 / 2.5) treats time as proportional to concentration, which would mean a weaker solution reacts faster. 40.0 s assumes concentration has no effect. 250 s applies the factor of 2.5 twice (40.0 × 2.52), which would only be right if the rate were proportional to the square of the concentration.
Powder has a much larger surface area, so more acid particles can collide with the solid per second and the initial rate (initial gradient) is greater; the reaction also finishes sooner. The acid is limiting and its amount is the same, so the final volume of CO2 is the same. A larger final volume would need more limiting reagent, not a larger surface area. The initial gradient cannot be the same, because the gradient is the rate and the rate depends on the surface area. Powder reacts faster, not slower; any clumping is minor compared with the huge increase in exposed surface.
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.