The shape of the lesson
By the end of the lesson, learners can
- Equal and opposite: Start the simulator with only reactant molecules and watch the forward and reverse rates.
- Pushing back: Change one condition at a time in the Le Chatelier simulator, using the Haber process N 2 + 3H 2 ⇌ 2NH 3, ΔH = −92 kJ mol −1.
- Industrial compromises: Theory says low temperature and high pressure give the best yield of ammonia.
- Dynamic: equal rates, constant concentrations, closed system.
- Pressure: count moles of gas on each side.
- Temperature: heating favours the endothermic direction.
- Catalyst: no shift, faster to equilibrium.
One picture every idea returns to
In a closed system a reversible reaction reaches dynamic equilibrium: the forward and reverse reactions continue at equal rates, so the concentrations stay constant. Le Chatelier's principle: if a condition of a system at equilibrium is changed, the position of equilibrium shifts to minimise the change. Adding a reactant shifts it right; raising the pressure shifts it towards fewer gas molecules; raising the temperature shifts it in the endothermic direction. A catalyst speeds up both reactions equally, so it does not move the position; it only gets there faster.
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, Balanced, but never still.. In a sealed flask a reversible reaction settles to a point where nothing seems to change, yet molecules are still reacting in both directions. 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: Equal and opposite | Start the simulator with only reactant molecules and watch the forward and reverse rates. Project tab 1 of the lab (Equilibria lab, dynamic equilibrium) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Pushing back | Change one condition at a time in the Le Chatelier simulator, using the Haber process N 2 + 3H 2 ⇌ 2NH 3, ΔH = −92 kJ mol −1. Project tab 2 of the lab (Equilibria lab, Le Chatelier and Haber) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Industrial compromises | Theory says low temperature and high pressure give the best yield of ammonia. | 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, 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 Industrial compromises, 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 |
|---|---|
| A catalyst does not change the position of equilibrium or the yield; it increases the rate of the forward and reverse reactions equally. | And when explaining a pressure change, always count the moles of gas on each side of the equation; if they are equal, pressure has no effect on the position. |
Dynamic equilibrium and Le Chatelier's principle: worksheet
6 questions, 13 marks. Written in the style of the papers; no past paper question is reproduced.
State two features of a reaction at dynamic equilibrium.
State Le Chatelier's principle.
For N2O4(g) ⇌ 2NO2(g), ΔH = +58 kJ mol−1, N2O4 is pale yellow and NO2 brown. Predict and explain what is seen when the sealed tube is placed in hot water.
Explain why the Haber process uses a pressure of about 200 atm and not a much higher pressure.
Explain why the Contact process operates at only 1 to 2 atm even though higher pressure would shift the equilibrium towards SO3.
An iron catalyst is used in the Haber process. State and explain its effect on the yield of ammonia and on the time taken to reach equilibrium.
Answer key and marking notes
- The rates of the forward and reverse reactions are equal.
- The concentrations of reactants and products remain constant (in a closed system).
If a change is made to a system at equilibrium, the position of equilibrium moves to minimise (oppose) that change.
- The forward reaction is endothermic.
- Raising the temperature shifts the position to the right, absorbing heat.
- More NO2 forms, so the mixture turns darker brown.
- Higher pressure shifts the position right (4 mol of gas to 2 mol), increasing the yield, and increases the rate.
- But very high pressures need thick-walled equipment and a lot of energy for compression, which is expensive.
- They also increase safety risks, so 200 atm is a compromise.
- The position already lies far to the right under these conditions, giving a conversion of about 99%.
- The small extra yield would not justify the cost of high-pressure equipment.
- No effect on the yield: it speeds up the forward and reverse reactions equally, so the position does not change.
- Equilibrium is reached more quickly because the activation energy is lowered.
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.
Which statement describes a reaction that has reached dynamic equilibrium in a closed system?
- Athe forward and reverse reactions continue at equal rates, so the concentrations stay constant
- Bthe forward and reverse reactions have both stopped
- Cthe concentrations of the reactants and products are equal
- Dthe forward reaction is faster than the reverse reaction
In the Haber process, N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ mol−1. What is the effect of increasing the total pressure on the position of equilibrium?
- Ait moves to the right, because there are fewer moles of gas on the right (2) than on the left (4)
- Bit moves to the left, because there are more moles of gas on the left
- Cit does not change, because pressure only affects the rate
- Dit moves to the right, because the forward reaction is exothermic
For the Haber process, N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ mol−1, what happens to the equilibrium yield of ammonia if the temperature is increased?
- Ait decreases, because the equilibrium moves in the endothermic (reverse) direction
- Bit increases, because the rate of reaction increases
- Cit increases, because the forward reaction is exothermic
- Dit stays the same, because temperature only affects the rate
An iron catalyst is used in the Haber process. What effect does the catalyst have on the equilibrium?
- Aequilibrium is reached more quickly, but the position of equilibrium does not change
- Bthe position of equilibrium moves to the right, so the yield of ammonia increases
- Conly the forward reaction is speeded up
- Dthe value of ΔH for the reaction becomes less negative
The graph shows how the concentrations change when N2O4 is sealed in a flask at constant temperature: N2O4(g) ⇌ 2NO2(g). Which statement is true from time te onwards?
- Athe rates of the forward and reverse reactions are equal and not zero
- Bno reaction takes place, because the concentrations do not change
- Cthe concentrations of N2O4 and NO2 are equal
- Dthe rate of the forward reaction is zero
The Haber process is run at about 450 °C, 200 atm (20 MPa) with an iron catalyst. Why is a lower temperature, which would give a higher equilibrium yield, not used?
- Athe rate would be too slow, so 450 °C is a compromise between yield and rate
- Ba lower temperature would move the equilibrium to the left
- Cthe iron catalyst would move the equilibrium further to the right at high temperature
- Dat lower temperatures ammonia would not be formed at all
The Contact process, 2SO2(g) + O2(g) ⇌ 2SO3(g), ΔH = −197 kJ mol−1, is run at about 450 °C with a V2O5 catalyst but only 1 to 2 atm. Why is a high pressure not used, even though it would favour SO3?
- Athe equilibrium already lies well to the right, so the small gain in yield does not justify the cost of high pressure
- Bhigh pressure would move the equilibrium to the left, because there are fewer moles of gas on the right
- Cpressure has no effect on this equilibrium, because all the species are gases
- Dthe V2O5 catalyst only works at low pressure because it changes the position of equilibrium
The equilibrium H2(g) + I2(g) ⇌ 2HI(g) is set up in a sealed syringe. The mixture is compressed to half its volume at constant temperature. Which statement is correct?
- Athe position of equilibrium does not change, although the concentrations of all three gases double
- Bthe position moves to the right, because increasing pressure always favours the products
- Cthe position moves to the left, because the left side has two different gases
- Dthe concentrations do not change, because the position of equilibrium does not change
Quiz key
At dynamic equilibrium both reactions are still going on, but at the same rate, so reactants are made as fast as they are used up and the concentrations no longer change. The reactions have not stopped; that would be a static equilibrium, and particles are still being exchanged. The concentrations are constant, not necessarily equal: at equilibrium there may be far more product than reactant, or the reverse. If the forward reaction were faster, the product concentration would still be rising, so the system would not yet be at equilibrium.
By Le Chatelier's principle the system opposes an increase in pressure by shifting towards the side with fewer gas molecules. The left has 1 + 3 = 4 mol of gas and the right has 2 mol, so the position moves right and the yield of ammonia rises. Moving left would increase the number of gas molecules and raise the pressure further. Pressure does change the position of a gas equilibrium with unequal moles, as well as the rate. The sign of ΔH decides the effect of temperature, not pressure.
Raising the temperature favours the endothermic direction, which absorbs the added heat. Here the forward reaction is exothermic (ΔH negative), so the reverse reaction is favoured and the equilibrium yield of ammonia falls. A higher rate means equilibrium is reached sooner; it does not change where the equilibrium lies. Being exothermic is exactly why higher temperature lowers the yield, not raises it. Temperature does change the position (and the value of Kc) as well as the rate.
A catalyst lowers the activation energy for the forward and reverse reactions by the same amount, so both rates rise by the same factor. Equilibrium is reached faster, but the position (and the yield at equilibrium) is unchanged. Saying it moves the position right is the classic error. It cannot speed up only the forward reaction, because the reverse reaction uses the same pathway backwards. ΔH depends only on the energies of the reactants and products, which the catalyst does not change.
After te both lines are flat, so the system is at dynamic equilibrium: N2O4 molecules still split into NO2 and NO2 molecules still join up, at equal rates. Saying no reaction occurs confuses constant concentrations with stopped reactions. The graph shows different concentrations (0.040 and 0.120 mol dm−3), not equal ones. If the forward rate were zero, the reverse reaction would keep using up NO2 and its concentration would fall.
Because the forward reaction is exothermic, lower temperatures give a higher equilibrium yield, but the rate would be so slow (even with a catalyst) that little ammonia would be made per hour. About 450 °C gives an acceptable yield at an acceptable rate: a compromise. A lower temperature would move the position to the right, not the left. A catalyst never moves the position of equilibrium. Ammonia would still form at lower temperatures; it would simply form too slowly to be economic.
With 3 mol of gas on the left and 2 mol on the right, high pressure would favour SO3, but at 1 to 2 atm the conversion is already very high (about 99% after several passes). The extra yield would not pay for the stronger plant and the energy for compression. High pressure would move the position right, not left, since the right has fewer gas moles. Pressure does affect gas equilibria when the numbers of gas moles differ, as they do here. V2O5 increases the rate; like any catalyst, it does not change the position of equilibrium.
There are 2 mol of gas on each side, so a change in pressure favours neither side and the position is unchanged. However, the same amounts are now in half the volume, so every concentration doubles (the purple iodine colour looks darker). Increasing pressure favours the side with fewer gas moles, not always the products. The number of different gases is irrelevant; only the total moles of gas matter. Saying the concentrations do not change ignores the halved volume: the ratio stays the same, but the concentrations are all higher.
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.