Manufacture of ammonia

Fertiliser factories, not chemistry labs, are where almost all the world's ammonia is made. A farmer's field needs nitrogen in tonnes, and while the air above that field is nearly 80% nitrogen gas, plants cannot use it directly — the triple bond holding N2 together is far too strong for roots to break. Haber's Process solves this by forcing nitrogen and hydrogen gases to combine over a hot iron catalyst under enormous pressure, locking atmospheric nitrogen into a usable compound, ammonia, on an industrial scale. This article extends the descriptive picture of raw materials and operating conditions already covered, adding the mole-ratio arithmetic that turns the balanced equation into real masses of ammonia a plant can weigh and sell, plus the exam traps and quick-fire practice to make the topic exam-ready.

The core insight

Haber's Process combines nitrogen and hydrogen gases (1:3 by volume) over a promoted iron catalyst at about 450-500 degrees C and 200-350 atmospheres to form ammonia reversibly, and the balanced equation's mole ratio lets us calculate the mass of ammonia obtainable from a given mass of nitrogen.

The core idea

Haber's Process is the industrial route to ammonia, standing apart from the small-scale laboratory preparation (heating an ammonium salt with a base) in almost every respect — scale, apparatus, and the chemistry itself.

The equation and why it is reversible. Nitrogen and hydrogen combine directly:

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

where: N2 and H2 are the reactant gases, NH3 is ammonia, and the double arrow shows the reaction is reversible — ammonia can decompose back into nitrogen and hydrogen, so only part of the feed gas converts on any single pass through the reactor.

Raw materials and conditions. Nitrogen is drawn from the fractional distillation of liquid air; hydrogen comes from natural gas or water gas. Both gases are purified of sulphur compounds before use, since these poison the iron catalyst (the same poisoning risk the Contact Process for sulphuric acid faces). The gases are compressed and passed, in a 1:3 volume ratio matching the equation's coefficients, over finely divided iron with molybdenum as a promoter, at roughly 450-500 degrees C and 200-350 atmospheres.

Recycling makes it economical. Because conversion per pass is only partial, the ammonia formed is liquefied by cooling (it condenses far more readily than N2 or H2) and drawn off, while the unreacted nitrogen and hydrogen are compressed again and sent back into the reaction chamber with fresh feed gas.

Figure

Flow diagram of the Haber process showing compressed nitrogen and hydrogen entering a heated catalyst chamber then a condenser separating liquid ammonia from recycled unreacted gas

Derivation AMass of ammonia obtainable from a given mass of nitrogen

The balanced equation for Haber's Process fixes not just a volume ratio for gases but a mole ratio that holds for any measured mass of reactant, and the goal here is to turn a given mass of nitrogen gas into the mass of ammonia it can theoretically produce by combining the mole-ratio method with molar masses, the same reasoning used across every stoichiometry problem in the syllabus.

Step 1. Write the balanced equation and read off the mole ratio directly from its coefficients: $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ , so 1 mole of N2 reacts with 3 moles of H2 to give 2 moles of NH3.

Step 2. Convert the given mass of nitrogen into moles using its molar mass, $M(\text{N}_2) = 28\ \text{g/mol}$ (2 x 14, from atomic mass of nitrogen 14):

n(\text{N}_2) = \frac{\text{mass of N}_2}{28}

where: mass of N2 is in grams and $n(\text{N}_2)$ is the number of moles of nitrogen supplied.

Step 3. Apply the mole ratio from Step 1: since 1 mole of N2 gives 2 moles of NH3, the moles of ammonia formed are $n(\text{NH}_3) = 2 \times n(\text{N}_2)$ .

Step 4. Convert moles of ammonia back to mass using $M(\text{NH}_3) = 17\ \text{g/mol}$ (14 + 3, nitrogen plus three hydrogens):

\text{mass of NH}_3 = n(\text{NH}_3) \times 17 = \left(\frac{\text{mass of N}_2}{28}\right) \times 2 \times 17 = \text{mass of N}_2 \times \frac{34}{28}

where: the factor $34/28 \approx 1.214$ is the theoretical mass of ammonia obtainable per unit mass of nitrogen fed in, assuming complete conversion. The same method converts the required mass of hydrogen: since 1 mole N2 needs 3 moles H2, $\text{mass of H}_2 = \text{mass of N}_2 \times \dfrac{3 \times 2}{28} = \text{mass of N}_2 \times \dfrac{6}{28}$ .

Worked example

Problem: 140 kg of nitrogen gas is fed into a Haber plant along with the correct mass of hydrogen. Calculate (a) the mass of hydrogen required, and (b) the theoretical mass of ammonia produced, assuming complete conversion.

Step 1. Moles of nitrogen supplied: $n(\text{N}_2) = 140\ \text{kg} / 28\ \text{kg/kmol} = 5\ \text{kmol}$ (working in kilomoles keeps the kilogram units consistent).

Step 2. Hydrogen required, from the 1:3 mole ratio: $n(\text{H}_2) = 3 \times 5 = 15\ \text{kmol}$ , giving a mass of $15 \times 2 = 30\ \text{kg}$ of hydrogen.

Step 3. Ammonia formed, from the 1:2 mole ratio between N2 and NH3: $n(\text{NH}_3) = 2 \times 5 = 10\ \text{kmol}$ , giving a mass of $10 \times 17 = 170\ \text{kg}$ .

Step 4. Check using the shortcut formula from the derivation: $\text{mass of NH}_3 = 140 \times 34/28 = 170\ \text{kg}$ , which matches. Answer: 30 kg of hydrogen is required, and the theoretical mass of ammonia produced is 170 kg.

Competitive stretch

A real reactor rarely converts completely in one pass. If only 15% of the feed converts to ammonia per pass, the actual mass produced per pass is $0.15 \times 170 = 25.5\ \text{kg}$ , and the remaining $170 - 25.5 = 144.5\ \text{kg}$ worth of unreacted N2 and H2 (by the same mole accounting) is compressed and recycled back into the reactor rather than wasted, which is exactly why recycling — not a higher single-pass conversion — is what makes the process economical.

Common traps

Using the molar mass of atomic nitrogen (14) instead of molecular nitrogen N2 (28) when converting mass to moles — always check whether the substance is monoatomic or diatomic before dividing.
Applying the mole ratio (1:3:2) to masses directly instead of converting to moles first — mole ratios only work with moles, never with raw grams or kilograms.
Confusing this industrial mass-yield calculation with the descriptive conditions question (catalyst, temperature, pressure) — both are examinable, but they test different skills.
Forgetting that the 170 kg answer is a theoretical maximum assuming 100% conversion; real plants get much less per pass and rely on recycling to approach it over time.

Ask the tutor: Want to practice converting a given mass of hydrogen (instead of nitrogen) into the mass of ammonia produced, or see how the recycling percentage changes the number of passes needed to use up all the feed gas?

Why do we need to convert mass to moles before using the mole ratio from the equation

The coefficients in a balanced equation — 1, 3, 2 here — count particles (or moles of particles), not grams. Nitrogen and ammonia have different molar masses, so equal masses of each do not contain equal numbers of moles. Converting mass to moles first puts both substances on the same particle-counting footing, so the ratio from the equation can be applied correctly; skipping this step and dividing masses directly gives a wrong answer almost every time.

Does the 170 kg of ammonia from 140 kg of nitrogen actually come out of a real plant

No, not in a single pass. That figure is the theoretical maximum assuming every molecule of nitrogen reacts, which does not happen because the reaction is reversible and only partially converts each time gas passes over the catalyst. A real plant recovers a much smaller mass of ammonia per pass, but recycles the unreacted nitrogen and hydrogen repeatedly, so that over many passes the plant's overall yield approaches the theoretical value calculated from the mole ratio.

Is the equilibrium reasoning behind choosing high pressure and moderate temperature examinable

No. The syllabus asks you to state the conditions — catalyst (iron with a molybdenum promoter), temperature around 450-500 degrees C, and pressure around 200-350 atmospheres — and to know that unreacted gases are recycled because conversion is incomplete. Explaining these choices using Le Chatelier's principle or equilibrium shift belongs to a later stage of chemistry and is explicitly outside this syllabus; stick to stating and using the conditions, not justifying them through equilibrium theory.

Manufacture of ammonia

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads