Lesson 3 of 3
Equilibrium in industry
Explain the choices in an ammonia plant.
A-Level 8873, revised syllabus (2026-2027)
I can apply Le Chatelier's Principle qualitatively.Syllabus
Syllabus 8873, 8(b). Apply Le Chatelier's Principle qualitatively.
I can identify changes that affect the equilibrium constant.Syllabus
Syllabus 8873, 8(c). Identify changes that affect the equilibrium constant.
I can explain the Haber process conditions.Syllabus
Syllabus 8873, 8(g). Explain the Haber process conditions.
Choose a useful production rate and yield
Industrial conditions combine equilibrium, kinetics and cost.
Make a guess. You are not marked.
Ammonia forms exothermically: N2 + 3H2 <=> 2NH3. What does a lower temperature do to the equilibrium yield and the rate?
Show the answer
Yield rises, but the rate falls
The exothermic forward direction is favoured at lower temperature, but fewer collisions have energy above Ea, so equilibrium is reached more slowly.
The Haber process makes ammonia: N2(g) + 3H2(g) ⇌ 2NH3(g), with an exothermic forward reaction. A representative school description uses an iron catalyst, about 450 °C and a pressure of about 200 atm. Actual plants vary; the reasoning behind the conditions matters more than treating one pair of numbers as universal.
| Choice | Chemical reason | Practical trade-off |
|---|---|---|
| High pressure | Compression favours 2 mol of product gas over 4 mol of reactant gas; it also increases gas concentrations. | Compression consumes energy and stronger pressure-resistant equipment costs more. Indefinitely high pressure is not economical. |
| Moderately high temperature | A higher temperature increases reaction rate, but decreases Kc for the exothermic forward reaction. | A compromise gives an acceptable rate while retaining a useful equilibrium yield. Very low temperature gives a favourable equilibrium but a slow approach to it. |
| Iron catalyst | Lowers activation barriers and accelerates approach to equilibrium. | Allows an economically useful rate; it does not increase the equilibrium yield at fixed temperature. |
| Cool the outlet and separate ammonia | Ammonia is condensed and removed from the gas stream. | Separation allows product collection and leaves unreacted nitrogen and hydrogen for recycling. |
| Recycle unreacted gases | Gases not converted on one pass return to the reactor. | Raises the overall conversion of fresh feed, although conversion in one pass remains limited by the operating equilibrium. |
- Feed and compress
Supply purified nitrogen and hydrogen in an approximately 1:3 mole ratio.
- React over iron
Use high pressure and a moderately high temperature; the outlet still contains reactants as well as ammonia.
- Cool and separate
Condense ammonia for collection.
- Recycle
Return unreacted nitrogen and hydrogen to the reaction loop.
Worked exampleRead an industrial data comparison
Two runs use the same catalyst and pressure. At 400 °C the equilibrium NH3 proportion is higher than at 500 °C, but the initial production rate is lower. Explain both observations.
Try it first. Then show the worked stepsHide the worked steps
- Ammonia formation is exothermic. The lower temperature favours the forward equilibrium and gives a larger Kc.
- At the lower temperature, a smaller fraction of collisions has enough energy to react, so the rate constant is smaller.
- The manufacturer must compare the amount made per unit time, equipment and energy costs, not maximise the equilibrium percentage in isolation.
Higher equilibrium yield and faster reaction are separate advantages that can favour different temperatures.
Check your understanding
A proposal says, "More iron catalyst increases the equilibrium yield of ammonia." Which correction is right?
Show the answer
More catalyst helps equilibrium be reached sooner, but the yield at equilibrium is unchanged.
A catalyst speeds up the forward and reverse reactions equally. It does not change Kc or the equilibrium yield at the same temperature, pressure and feed.
Exam-style question
Paper 2 Section B styleThis question draws on Chemical Equilibria and Data and Experimental Skills.
Ethanoic acid and ethanol react reversibly to form ethyl ethanoate and water.
CH3COOH(l) + C2H5OH(l) ⇌ CH3COOC2H5(l) + H2O(l)
1.00 mol of ethanoic acid and 1.00 mol of ethanol were sealed in a tube with a few drops of acid catalyst and kept at 60 °C until equilibrium was reached. The mixture then contained 0.333 mol of ethanoic acid. Ignore any water in the catalyst.
Ammonia is made by the Haber process. N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = -92 kJ mol-1. Table 1 shows the percentage of ammonia in the equilibrium mixture under different conditions.
Table 1. Percentage of ammonia at equilibrium pressure / atm 300 °C 400 °C 500 °C 100 52 25 10 200 63 36 18 400 74 48 30
- (a) State what is meant by dynamic equilibrium. [2]
- (b) Write the expression for Kc for the esterification. [1]
- (c) Calculate Kc at 60 °C. [3]
- (d) Explain why Kc for this reaction has no units, and why the volume of the mixture is not needed. [1]
- (e) A student repeats the experiment with 1.00 mol of ethanoic acid and 3.00 mol of ethanol. She claims that 0.90 mol of ethyl ethanoate forms. Use your value of Kc to test her claim, and explain the effect of the extra ethanol. [3]
- (f) Use Table 1 to describe and explain the effect of temperature on the percentage of ammonia. [4]
- (g) Use Table 1 to explain the effect of pressure on the percentage of ammonia. [2]
- (h) A student says: "An ammonia plant should run at 300 °C and 400 atm, because these conditions give the highest yield." Evaluate this claim. [3]
- (i) State the effect of the iron catalyst on the percentage of ammonia at equilibrium. Explain your answer. [1]
[20]
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