Lesson 5 · 25 min

The Haber process: an industrial compromise

At room temperature K is enormous but the reaction is hopelessly slow; heating speeds it up but pushes the equilibrium back. Compute the yield from thermodynamic data and see where 450 °C and 200 bar come from.

In this lesson

Compute ΔG° and K for ammonia synthesis from ΔH° and ΔS°, and explain why K falls steeply with temperature
Use K to find the equilibrium mole fraction of NH₃ at a given temperature and pressure
Separate equilibrium yield from rate, and explain why industrial conditions are a compromise between them
Say where the ideal-gas model and the measurements part company, and why

On 13 October 1908 Fritz Haber filed his patent on the “synthesis of ammonia from its elements”; a few years later Carl Bosch scaled it up into an industrial plant. Today roughly half of the world’s population is fed thanks to nitrogen fertiliser made from synthetic ammonia (Erisman and co-workers estimated 48 % in 2008). The reaction looks disarmingly simple:

\ce{N2(g) + 3H2(g) <=> 2NH3(g)}\qquad \Delta_r H^\circ = -91.8\ \mathrm{kJ\,mol^{-1}}

The difficulty: at room temperature it lies almost entirely on the side of ammonia, yet it is so slow that nothing happens — and whatever speeds it up pushes the equilibrium back.

Predict first

Ammonia synthesis is exothermic and K is about $5\times10^5$ at 25 °C. Yet plants run at 400–500 °C, where K is smaller. Why?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

Heating shifts the equilibrium right and raises the yield At room temperature the reaction is far too slow; it must be heated to go at a useful rate To boil the ammonia off Because K increases with temperature

Thermodynamics: where K comes from

\Delta_r G^\circ = \Delta_r H^\circ - T\Delta_r S^\circ = -RT\ln K

At 298 K, ΔH° = −91.8 kJ/mol and ΔS° = −198.1 J/(mol K): four gas molecules become two, so entropy falls.

$\Delta_r H^\circ$: 2 × ΔfH°(NH₃) = 2 × (−45.9) kJ/mol
$\Delta_r S^\circ$: 2(192.77) − 191.61 − 3(130.68) = −198.1 J/(mol K)
$K$: equilibrium constant with 1 bar standard states

At 25 °C, $\Delta_r G^\circ = -91.8 - 298.15\times(-0.1981) = -32.7$ kJ/mol and $K = e^{32700/(8.314\times298.15)} = 5.4\times10^5$ . Both ΔH° and ΔS° are negative, so the $-T\Delta_r S^\circ$ term grows with temperature until it overwhelms ΔH°: treating both as constant, $K = 1$ at $T = 91\,800/198.1 = 463$ K (190 °C). In reality ΔH° becomes more negative at high temperature (the products have a smaller heat capacity than the reactants); with NIST heat-capacity data, at 450 °C $\Delta_r H^\circ = -105.7$ kJ/mol and $K = 4.9\times10^{-5}\ \mathrm{bar^{-2}}$ — about four times smaller than the constant-value estimate.

With $\Delta n = -2$ , the pressure appears in the equilibrium condition:

K = \frac{y_{\ce{NH3}}^2}{y_{\ce{N2}}\,y_{\ce{H2}}^3}\left(\frac{p^\circ}{P}\right)^2

At fixed K, doubling the total pressure P forces the mole-fraction ratio $y_{\ce{NH3}}^2/(y_{\ce{N2}}y_{\ce{H2}}^3)$ up fourfold — the quantitative reason high pressure raises the yield.

Worked example

Equilibrium yield at 450 °C and 200 bar

Feed $\ce{N2}$ and $\ce{H2}$ in a 1 : 3 ratio; at 450 °C, $K = 4.90\times10^{-5}\ \mathrm{bar^{-2}}$ . Find the equilibrium mole fraction of $\ce{NH3}$ at 200 bar (ideal gases).

Equilibrium NH₃ mole percent for a 3 : 1 feed: measurements versus this site’s ideal-gas calculation
Temperature	Pressure	Measured / %	Computed / %
450 °C	10 atm	2.04	2.20
450 °C	100 atm	16.35	16.19
450 °C	300 atm	35.5	32.0
500 °C	600 atm	42.1	34.0

Source: Measured: A. T. Larson, R. L. Dodge, J. Am. Chem. Soc. 45, 2918 (1923); A. T. Larson, J. Am. Chem. Soc. 46, 367 (1924). Computed: NIST Shomate data, ideal gases.

Up to 100 atm the calculation matches the measurements to within one percentage point; at higher pressures the measured yields are clearly larger. At high pressure ammonia is far from an ideal gas: intermolecular attraction makes its fugacity coefficient less than 1, as if its “effective pressure” were lower than its partial pressure, so the equilibrium leans further towards ammonia.

The rate–yield compromise: lower the temperature and the equilibrium yield rises (about 36 % at 400 °C and 200 bar) but the reaction slows dramatically; raise it and the reaction speeds up while the equilibrium yield falls to 17 % (500 °C). In a finite contact time the ammonia actually obtained peaks at some intermediate temperature — the teaching model in the lower half of the bench shows exactly that. Real plants add two more tricks: condense out the ammonia and recycle the unreacted $\ce{N2}$ and $\ce{H2}$ to the converter, and use a promoted iron catalyst to raise the rate.

Equilibrium yield from thermodynamic data · checked against 1923–1924 measurements

Haber process: yield and conditions

$\ce{N2(g)} + \ce{3H2(g)} \mathrel{\ce{<=>}} \ce{2NH3(g)},\quad\allowbreak \Delta_r H^\circ_{298} = -91.8\ \mathrm{kJ\,mol^{-1}}$

Temperature · 400 °C Total pressure · 100 bar Feed H₂ : N₂ · 3.0 : 1 Inert gas (per mol N₂) · 0.00 mol

Show measurements (Larson & Dodge)

Equilibrium NH₃24.7 %

N₂ converted at equilibrium39.6 %

K / bar⁻²1.80 × 10⁻⁴

Thermodynamics at this temperature: ΔrH° = −104.6 kJ/mol · ΔrS° = −227.1 J/(mol·K) · ΔrG° = 48.3 kJ/mol

Equilibrium NH₃ mole fraction vs total pressure

350 °C
400 °C
450 °C
500 °C
550 °C
Measured, 1923–1924 (hollow points)

One pass through the converter: thermodynamic limit vs kinetics

Apparent activation energy Ea · 100 kJ/mol kτ at 400 °C (contact time) · 1.00

Equilibrium limit
Reached in a finite contact time

Temperature of best single-pass yield424 °C

NH₃ after one pass15.6 % (max 17.0 %)

k(T) / k(400 °C)1

Look at the lower half: teal is the equilibrium limit, orange the ammonia actually reached in a finite contact time. Raise the contact time kτ (a bigger converter, slower gas) and the best temperature moves lower; raise the apparent activation energy and the peak sharpens. Then change the feed from 3 : 1 to 2 : 1, or add inert gas, and watch the equilibrium yield fall.

Concept check

What does the iron catalyst in an ammonia converter change?

The reaction’s K How fast equilibrium is reached The reaction’s ΔH° The equilibrium mole fraction of NH₃

Try the calculation

Treating ΔH° = −91.8 kJ/mol and ΔS° = −198.1 J/(mol K) as constant, estimate the temperature at which K = 1 for ammonia synthesis.

Try the calculation

Use the bench (or the method of the worked example) to find the equilibrium mole percent of NH₃ at 400 °C and 200 bar with a 3 : 1 feed.

Key takeaways

Ammonia synthesis is exothermic with falling entropy: K ≈ 5 × 10⁵ at room temperature but drops steeply, reaching 1 near 190 °C.
With Δn = −2, high pressure raises the yield: at 450 °C, 0.2 % at 1 bar but about 25 % at 200 bar.
The industrial temperature is a compromise between rate and yield; the catalyst only speeds things up, while condensing and recycling make up for the modest single-pass conversion.
Ideal-gas calculations match the measurements up to about 100 atm and underestimate the yield at higher pressures.

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Data and references

J. W. Erisman et al., “How a century of ammonia synthesis changed the world”, Nature Geoscience 1, 636–639 (2008).
NIST Chemistry WebBook, SRD 69: Shomate parameters for N₂, H₂ and NH₃ (from M. W. Chase, NIST-JANAF Thermochemical Tables, 1998).
A. T. Larson, R. L. Dodge, “The ammonia equilibrium”, J. Am. Chem. Soc. 45, 2918–2930 (1923); A. T. Larson, J. Am. Chem. Soc. 46, 367–372 (1924).
P. Atkins, J. de Paula, J. Keeler, Atkins’ Physical Chemistry, 11th ed. (2018), Topic 6B (response of equilibria to conditions).