The shape of the lesson
By the end of the lesson, learners can
- The Haber process: Change the temperature and pressure in the simulator and read the equilibrium yield of ammonia.
- The Contact process: Follow sulfur from the furnace to sulfuric acid in the simulator.
- Why these conditions?: Extended.
- Haber: 450 °C, 20 000 kPa, iron; N2 from air, H2 from methane.
- Contact: 450 °C, 200 kPa, V2O5; SO2 from burning sulfur.
- Temperature: compromise between yield and rate.
- Catalyst: faster, not more.
One picture every idea returns to
Haber process: N2(g) + 3H2(g) ⇌ 2NH3(g). Nitrogen comes from the air and hydrogen from methane. Conditions: 450 °C, 20 000 kPa (200 atm), iron catalyst. Contact process: 2SO2(g) + O2(g) ⇌ 2SO3(g). Sulfur dioxide comes from burning sulfur or roasting sulfide ores; oxygen from the air. Conditions: 450 °C, 200 kPa (2 atm), vanadium(V) oxide catalyst. Extended: both forward reactions are exothermic and reduce the number of gas molecules, so the conditions are a compromise.
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, Equilibrium on an industrial scale.. Ammonia for fertilisers and sulfuric acid for almost everything are made by reversible reactions. 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: The Haber process | Change the temperature and pressure in the simulator and read the equilibrium yield of ammonia. Project tab 2 of the lab (Equilibria lab, Le Chatelier and Haber) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: The Contact process | Follow sulfur from the furnace to sulfuric acid in the simulator. Project tab 5 of the lab (Nitrogen and sulfur lab, Contact process) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Why these conditions? | Extended. | Jigsaw |
| 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, 18 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 Why these conditions?, 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.
Jigsaw
Split the content into expert pieces. Expert groups master one piece each, then regroup so every home team has one expert per piece, and each expert teaches their piece to the team.
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 catalyst does not increase the yield. | And 450 °C is not chosen for the best yield: the best yield would be at a low temperature. It is a compromise between yield and rate. Name both in your answer. |
The Haber and Contact processes: worksheet
6 questions, 18 marks. Written in the style of the papers; no past paper question is reproduced.
For the Haber process, write the equation with state symbols and the reversible sign, and state the sources of the two raw materials and the three conditions used.
For the Contact process, state the equation for the reversible step, the catalyst, and two sources of sulfur dioxide.
(Extended) The forward reaction in the Contact process is exothermic. Explain why a temperature of 450 °C is used.
(Extended) Explain why a pressure of 20 000 kPa is used in the Haber process, rather than a lower or a much higher pressure.
Describe how ammonia is separated from the unreacted gases in the Haber process, and what happens to those gases.
(Extended) Explain why a catalyst is used in the Haber process even though it does not increase the yield.
Answer key and marking notes
- N2(g) + 3H2(g) ⇌ 2NH3(g).
- Nitrogen from the air; hydrogen from methane.
- 450 °C and 20 000 kPa (200 atm).
- Iron catalyst.
- 2SO2 + O2 ⇌ 2SO3.
- Vanadium(V) oxide.
- Burning sulfur; roasting sulfide ores.
- A lower temperature would move the equilibrium to the right and give a higher yield of sulfur trioxide.
- But at a lower temperature the rate would be too slow.
- 450 °C is a compromise giving a reasonable yield at an acceptable rate.
- There are 4 gas molecules on the left and 2 on the right, so higher pressure moves the equilibrium to the right and increases the yield.
- Higher pressure also increases the rate, because the gas particles collide more frequently.
- But very high pressures are expensive to produce and need stronger, costlier equipment,
- and they are more dangerous; 20 000 kPa is a compromise.
- The mixture is cooled; ammonia condenses to a liquid and is removed.
- The unreacted nitrogen and hydrogen are recycled back to the reactor.
- It increases the rate, so equilibrium is reached faster and more ammonia is made per hour.
- It allows a reasonable rate at a lower temperature, which gives a better yield and saves energy costs.
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 set of conditions is used in the Contact process?
- A450 °C, 200 kPa, vanadium(V) oxide catalyst
- B450 °C, 20 000 kPa, vanadium(V) oxide catalyst
- C450 °C, 200 kPa, iron catalyst
- D200 °C, 450 kPa, vanadium(V) oxide catalyst
Which raw materials provide the sulfur dioxide and oxygen used in the Contact process?
- Asulfur and air
- Bsulfur and water
- Cmethane and air
- Dsulfuric acid and air
What is most of the ammonia made by the Haber process used for?
- Amaking fertilisers
- Bmaking sulfuric acid
- Cas a fuel for power stations
- Dpurifying drinking water
In the Haber process, N2(g) + 3H2(g) ⇌ 2NH3(g). What volume of hydrogen reacts completely with 50 dm3 of nitrogen, both measured at the same temperature and pressure?
- A150 dm3
- B50 dm3
- C100 dm3
- D16.7 dm3
An engineer suggests running the Haber process at 250 °C instead of 450 °C, with the same pressure and catalyst. What would happen?
- AThe equilibrium yield of ammonia would rise, but the rate would be much slower
- BThe equilibrium yield of ammonia would fall, and the rate would be slower
- CThe equilibrium yield of ammonia would rise, and the rate would be faster
- DThe equilibrium yield would be unchanged, but the rate would be faster
Which statement is true for the reversible reaction in BOTH the Haber process and the Contact process?
- AThe forward reaction is exothermic and produces fewer gas molecules
- BThe forward reaction is endothermic and produces fewer gas molecules
- CThe reaction uses an iron catalyst
- DThe reaction is carried out at 20 000 kPa
In the Contact process, what would be the effect on the equilibrium yield of sulfur trioxide of raising the temperature from 450 °C to 600 °C?
- AThe yield would decrease, because the equilibrium moves in the endothermic direction, to the left
- BThe yield would increase, because the rate increases
- CThe yield would increase, because the equilibrium moves in the endothermic direction, to the right
- DThe yield would not change, because the catalyst controls the yield
The high pressure in the Haber process increases the rate of reaction as well as the yield. Why does it increase the rate?
- AThe gas particles are closer together, so they collide more frequently
- BThe particles have more kinetic energy, so more collisions are successful
- CThe activation energy is lowered
- DThe equilibrium moves to the right, so the reaction goes faster
Quiz key
The Contact process runs at 450 °C and 200 kPa (2 atm) with a vanadium(V) oxide catalyst. 20 000 kPa is the Haber process pressure, and iron is the Haber process catalyst. The last option swaps the numbers for temperature and pressure.
Burning sulfur (or roasting sulfide ores) in air gives sulfur dioxide, and air supplies the oxygen. Water provides no oxygen gas for this step, methane is a raw material for the hydrogen in the Haber process, and sulfuric acid is the final product of the whole process, not a raw material.
Most ammonia is turned into nitrogen-containing fertilisers such as ammonium nitrate and ammonium sulfate. Sulfuric acid comes from the Contact process, ammonia is not used as a power-station fuel, and water is purified with chlorine, not ammonia.
The equation shows 1 molecule of nitrogen reacting with 3 molecules of hydrogen, so for gases the volume ratio is 1 : 3 and 50 × 3 = 150 dm3 of hydrogen is needed. 50 dm3 assumes a 1 : 1 ratio, 100 dm3 uses the 2 from the ammonia, and 16.7 dm3 divides by 3 instead of multiplying.
Lowering the temperature moves the equilibrium in the exothermic direction, which is towards ammonia, so the yield rises. But the particles have less energy, so fewer collisions reach the activation energy and the rate is much slower, which is why 450 °C is kept as a compromise. The yield cannot fall with cooling for an exothermic forward reaction, and cooling never makes a reaction faster.
Both forward reactions are exothermic, and both reduce the number of gas molecules: 4 to 2 in the Haber process and 3 to 2 in the Contact process. That is why a lower temperature and a higher pressure would favour the products in each. Neither forward reaction is endothermic. Iron and 20 000 kPa belong to the Haber process only; the Contact process uses vanadium(V) oxide and 200 kPa.
The forward reaction is exothermic, so the reverse reaction is endothermic. Raising the temperature moves the equilibrium in the endothermic direction, to the left, so less sulfur trioxide is present at equilibrium. A faster rate gets to equilibrium sooner but does not raise the yield. The endothermic direction is to the left here, not the right, and a catalyst never affects the equilibrium yield.
At high pressure the same gas is squeezed into a smaller volume, so there are more particles per unit volume and collisions are more frequent, which increases the rate. The temperature, and so the energy of the particles, is set separately. Pressure does not change the activation energy, and the shift in equilibrium position explains the higher yield, not the higher rate.
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.