Explain Haber and Contact process conditions
Use equilibrium shifts and reaction rates to explain industrial choices in ammonia and sulfuric-acid manufacture.
AS Level; Cambridge International Chemistry 9701, 2025–2027 syllabus. Equilibrium constants apply at a fixed temperature; include only gaseous or aqueous species in the relevant expressions. For acid–base work, distinguish strength from concentration. Written practice is self-checked
Before you begin
- Balance equations, interpret reversible arrows and use concentration in mol dm⁻³.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does an iron catalyst do in the Haber process?
What you will learn
- Describe conditions used in Haber and Contact processes
- Explain pressure and temperature choices using equilibrium
- Explain the role of catalysts and process trade-offs
Haber conditions balance yield and rate
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is exothermic in the forward direction and reduces the number of gas moles. Higher pressure favours ammonia; lower temperature favours the exothermic direction but slows the reaction.
Industry uses a compromise temperature and high pressure, with an iron catalyst. The catalyst improves the rate without shifting equilibrium; the operating conditions balance yield, speed, energy cost and equipment constraints.
Contact process conditions also use a compromise
In the Contact process, 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is exothermic and goes from three gas moles to two. Lower temperature and higher pressure favour SO₃ at equilibrium.
A vanadium(V) oxide catalyst is used with a moderate temperature and pressure selected for a practical rate and yield. Explain the industrial compromise rather than claiming that the maximum-yield conditions are always the best operating conditions.
Pause and explain
Why does high pressure favour ammonia in the Haber equilibrium?
Worked example
Explain why the Haber process uses a high pressure but not the lowest possible temperature.
Show the worked solution
- Higher pressure favours the side with fewer gas moles, producing more NH₃.
- Lower temperature favours the exothermic forward reaction.
- But very low temperature makes the rate too slow; a practical compromise is used with an iron catalyst.
Answer High pressure improves equilibrium yield because the product side has fewer gas moles. Lower temperature favours NH₃, but the process uses a compromise to maintain an adequate rate; iron catalyst speeds the reaction without changing equilibrium position.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
For N₂ + 3H₂ ⇌ 2NH₃, explain the effects of higher pressure and lower temperature on equilibrium yield.
Give me a hint
Count gas moles and use the forward reaction's enthalpy sign.
Compare my reasoning
- Gas moles fall from four to two, so higher pressure favours products.
- The forward reaction is exothermic, so lower temperature favours products.
- Both changes raise equilibrium ammonia yield.
Higher pressure and lower temperature both favour NH₃ at equilibrium because the product side has fewer gas moles and the forward reaction is exothermic.
Look for these in your work
- Mole count correct
- Temperature reasoning correct
- Yield effect stated
For 2SO₂ + O₂ ⇌ 2SO₃, explain the trade-off when selecting temperature and catalyst.
Give me a hint
Higher yield is favoured by lower temperature; rate requires a practical temperature and catalyst.
Compare my reasoning
- The forward reaction is exothermic, so lower temperature favours SO₃.
- A lower temperature can slow reaction rate.
- V₂O₅ speeds the process, while an operating compromise balances yield and rate.
Lower temperature favours SO₃ but slows the reaction; a V₂O₅ catalyst increases rate, and practical conditions balance yield and throughput.
Look for these in your work
- Equilibrium direction correct
- Rate trade-off recognised
- Catalyst role distinguished
A student claims the catalyst shifts the Haber equilibrium toward products. Correct the statement and identify what the catalyst changes.
Give me a hint
Compare the forward and reverse pathways.
Compare my reasoning
- The catalyst lowers activation barriers for both directions.
- It helps the system reach equilibrium faster.
- Equilibrium composition and K remain governed by conditions, not the catalyst.
The catalyst does not shift equilibrium or change K; it speeds both directions so equilibrium is reached sooner.
Look for these in your work
- Both pathways considered
- Rate effect stated
- Position and K unchanged
Common mistakes
- Saying a catalyst increases equilibrium yield
- Ignoring the exothermic direction when discussing temperature
- Choosing the side with more gas moles at high pressure
- Describing maximum equilibrium yield as the only industrial criterion
What can you explain now?
What does an iron catalyst do in the Haber process?
Compare with the explanation
Speeds the reaction without shifting equilibrium position
It lowers activation energy for both directions.
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Make it stick
How do temperature, pressure and catalysts contribute differently to industrial yield and rate?
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