The Haber process: rate vs yield (SACE Stage 2 Chemistry)

Syllabus dot point

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

9 min readUpdated 2 Jun 2026

Quick answer

The Haber process, $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ ( $\Delta H = -92\ \text{kJ mol}^{-1}$ ), is run under conditions that balance rate against yield. High pressure (about $200\ \text{atm}$ ) raises both yield (fewer gas moles on the right) and rate. A low temperature would maximise yield (exothermic forward reaction) but be too slow, so a compromise temperature near $450\ ^\circ\text{C}$ is used. An iron catalyst speeds the approach to equilibrium without changing yield. Ammonia is removed and unreacted gas recycled.

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What this dot point is asking
Lead worked calculation
The equilibrium and the dilemma
Pressure
Temperature: the compromise
The iron catalyst
Removal and recycling
Why it matters for managing processes
Exam-style questions

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

Yield trade-off across temperatures

Two reactor settings give the following single-pass results for the same feed: at $350\ ^\circ\text{C}$ , equilibrium yield $= 40\%$ but time to reach equilibrium $= 60\ \text{min}$ ; at $450\ ^\circ\text{C}$ , yield $= 25\%$ but time $= 5\ \text{min}$ . Compare the ammonia produced per hour to show why the hotter setting is preferred.

Production at 350 C

In one hour the reactor completes $1$ cycle at $40\%$ , giving a relative output of $1 \times 0.40 = 0.40$ units of yield per hour.

Production at 450 C

In one hour the reactor completes $\dfrac{60}{5} = 12$ cycles at $25\%$ , giving $12 \times 0.25 = 3.0$ units per hour.

Interpret

Even though each batch at $450\ ^\circ\text{C}$ has a lower yield, the much faster rate means roughly $7.5$ times more ammonia is made per hour. This is exactly why the industrial compromise favours a higher temperature: throughput, not single-pass yield, drives the economics.

The equilibrium and the dilemma

The forward reaction is exothermic and reduces the number of gas moles ( $4 \rightarrow 2$ ). 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

Why high pressure helps both yield and rate

High pressure shifts the equilibrium toward the side with fewer gas moles (the product side, $2$ moles versus $4$ ), increasing yield, and it also increases the concentration of the gases, increasing the rate. Both effects favour high pressure. The practical ceiling (about $200\ \text{atm}$ ) is set by the cost and safety of building reactors and compressors strong enough, not by the chemistry.

Temperature: the compromise

Because the forward reaction is exothermic, lowering the temperature raises the equilibrium yield and $K_c$ , but slows the rate. Raising it speeds the rate but lowers the yield. The chosen temperature, around $400$ to $450\ ^\circ\text{C}$ , 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 $K_c$ , or the maximum yield; it only reduces the time taken to get there. Without it, even $450\ ^\circ\text{C}$ would be too slow.

Removal and recycling

The single-pass conversion is low (around $15\%$ ), 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 $\text{N}_2$ and $\text{H}_2$ 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.

Recommending a very low temperature for maximum yield

The classic SACE trap is to argue, correctly from Le Chatelier, that low temperature maximises yield, and then recommend it, ignoring rate. A very low temperature makes the rate impractically slow, so equilibrium is never reached in useful time. The answer must present the temperature as a compromise between yield and rate. A second error is claiming the catalyst increases yield; it only speeds the approach to the same equilibrium.

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 questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2021-style5 marks

The Haber process is

\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)

\Delta H = -92\ \text{kJ mol}^{-1}

. Industrial plants operate at about

450\ ^\circ\text{C}

and

200\ \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 $4$ gas moles on the left and $2$ 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 $450\ ^\circ\text{C}$ is a compromise: fast enough rate with an acceptable yield. (2 marks)

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

SACE 2019-style4 marks

In a Haber reactor,

50.0\ \text{mol}

\text{N}_2

and excess

\text{H}_2

pass through, and the reaction reaches

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

\text{NH}_3

formed per pass. (b) Explain why recycling raises the overall yield despite the low conversion per pass.

Show worked answer →

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

(b) Although only $15\%$ converts each pass, the ammonia is removed by cooling and the unreacted $\text{N}_2$ and $\text{H}_2$ 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)

Practise this

Sources & how we know this

Stage 2 Chemistry subject outline (2025 onwards) — SACE Board
Stage 2 Chemistry: external assessment — SACE Board