Skip to main content
ExamExplained
SA · Chemistry
Chemistry study scene
§-Syllabus dot point
SAChemistrySyllabus dot point

How do industrial conditions for the Haber process balance reaction rate against equilibrium yield?

Explain how temperature, pressure and catalyst conditions in the Haber process are chosen to compromise between reaction rate and equilibrium yield.

How the Haber process balances rate against equilibrium yield: the effect of pressure, temperature and the iron catalyst, the compromise temperature, recycling, and worked SACE-style yield and economic-trade-off analysis.

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. Lead worked calculation
  3. The equilibrium and the dilemma
  4. Pressure
  5. Temperature: the compromise
  6. The iron catalyst
  7. Removal and recycling
  8. Why it matters for managing processes

What this dot point is asking

SACE expects you to apply Le Chatelier and rate reasoning together, explain each chosen condition as a trade-off, and recognise the role of removal and recycling.

Lead worked calculation

The equilibrium and the dilemma

The forward reaction is exothermic and reduces the number of gas moles (424 \rightarrow 2). Le Chatelier therefore says the highest equilibrium yield of ammonia comes from high pressure and low temperature. But low temperature also means a very slow rate, so the plant would reach that high yield far too slowly to be economic. This tension between yield (thermodynamics) and rate (kinetics) is the heart of the process.

Pressure

Temperature: the compromise

Because the forward reaction is exothermic, lowering the temperature raises the equilibrium yield and KcK_c, but slows the rate. Raising it speeds the rate but lowers the yield. The chosen temperature, around 400400 to 450 C450\ ^\circ\text{C}, is a deliberate compromise: high enough for a commercially fast rate (helped by the catalyst), low enough that the yield is still worthwhile. Neither the maximum-yield nor the maximum-rate condition is used alone.

The iron catalyst

A finely divided iron catalyst (with promoters) lowers the activation energy, letting equilibrium be reached quickly at the compromise temperature. Critically, the catalyst speeds the forward and reverse reactions equally, so it does not change the equilibrium position, the value of KcK_c, or the maximum yield; it only reduces the time taken to get there. Without it, even 450 C450\ ^\circ\text{C} would be too slow.

Removal and recycling

The single-pass conversion is low (around 15%15\%), but the process is made efficient by removing ammonia and recycling unreacted gases. The reaction mixture is cooled so ammonia (higher boiling point) liquefies and is separated, while unreacted N2\text{N}_2 and H2\text{H}_2 are pumped back into the reactor. Removing the product also shifts the equilibrium forward (Le Chatelier), and recycling means almost all the feedstock is eventually converted, giving a high overall yield.

Why it matters for managing processes

The Haber process is the textbook case of managing a chemical process: every condition is chosen to balance competing demands of equilibrium yield, reaction rate, energy cost and safety. The same compromise thinking, supported by Le Chatelier and kinetics, applies across industrial chemistry, from the Contact process to methanol synthesis.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20215 marksThe Haber process is N2(g)+3H2(g)2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g), ΔH=92 kJ mol1\Delta H = -92\ \text{kJ mol}^{-1}. Industrial plants operate at about 450 C450\ ^\circ\text{C} and 200 atm200\ \text{atm} with an iron catalyst. Explain, using Le Chatelier's principle and rate considerations, why these conditions are a compromise rather than the conditions that give the highest equilibrium yield.
Show worked answer →

Pressure: there are 44 gas moles on the left and 22 on the right, so high pressure shifts the equilibrium to the right, increasing yield, and also increases rate by raising concentration. Very high pressure is used but limited by the cost and safety of building plant to withstand it. (2 marks)

Temperature: the forward reaction is exothermic, so a low temperature would give the highest equilibrium yield, but it would make the rate impractically slow. A high temperature gives a fast rate but a low yield. About 450 C450\ ^\circ\text{C} is a compromise: fast enough rate with an acceptable yield. (2 marks)

Catalyst: the iron catalyst increases the rate (lowers EaE_a) so equilibrium is reached quickly, but does not change the position or the yield. (1 mark)

SACE 20194 marksIn a Haber reactor, 50.0 mol50.0\ \text{mol} of N2\text{N}_2 and excess H2\text{H}_2 pass through, and the reaction reaches 15%15\% conversion of nitrogen per pass before the gases are cooled to remove ammonia and the unreacted gases are recycled. (a) Calculate the moles of NH3\text{NH}_3 formed per pass. (b) Explain why recycling raises the overall yield despite the low conversion per pass.
Show worked answer →

(a) N2\text{N}_2 converted =0.15×50.0=7.5 mol= 0.15 \times 50.0 = 7.5\ \text{mol}. From the 1:21:2 ratio, n(NH3)=2×7.5=15 moln(\text{NH}_3) = 2 \times 7.5 = 15\ \text{mol} per pass. (2 marks)

(b) Although only 15%15\% converts each pass, the ammonia is removed by cooling and the unreacted N2\text{N}_2 and H2\text{H}_2 are fed back into the reactor. Over many passes almost all the nitrogen and hydrogen is eventually converted, so the overall yield is high even though the single-pass yield is low. (2 marks)

ExamExplained