Haber process for ammonia — conditions, equilibrium, Le Chatelier optimization

Question

Explain how the Haber process conditions (temperature, pressure, catalyst) are optimised using Le Chatelier’s principle. Why is 450°C used when a lower temperature would give better yield?

(JEE Main 2024 tested Le Chatelier application; CBSE 11 boards ask the full process)

Solution — Step by Step

\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \quad \Delta H = -92 \text{ kJ/mol}

Key facts: the reaction is exothermic (negative $\Delta H$ ), involves a decrease in moles of gas (4 moles → 2 moles), and is reversible.

Le Chatelier says: for an exothermic reaction, low temperature favours the forward direction. But the reaction is painfully slow at low temperatures — even with a catalyst.

The compromise: 450°C. High enough for the iron catalyst to work efficiently, low enough to give a reasonable equilibrium yield (~15-20%).

4 moles of gas → 2 moles of gas. Le Chatelier says high pressure shifts equilibrium towards fewer moles = more NH₃.

Industrially, 200 atm is used. Higher pressures would give better yield, but the cost of building equipment that withstands extreme pressures is not worth the marginal gain.

Finely divided iron with molybdenum promoter (Fe + Mo) is used. The catalyst does NOT change the equilibrium — it only makes the system reach equilibrium faster.

NH₃ is continuously removed by cooling and condensation. Removing the product shifts equilibrium right (Le Chatelier), driving more conversion. Unreacted N₂ and H₂ are recycled back.

flowchart TD
    A["N₂ + 3H₂ feed<br/>(1:3 ratio)"] --> B["Compressor<br/>200 atm"]
    B --> C["Reactor<br/>Fe catalyst + Mo promoter<br/>450°C"]
    C --> D["Cooler/Condenser"]
    D --> E["Liquid NH₃<br/>separated"]
    D --> F["Unreacted N₂ + H₂"]
    F -->|"Recycled"| B
    style C fill:#ffeb3b,stroke:#333

Why This Works

The Haber process is the textbook case study for applying Le Chatelier’s principle to industrial chemistry. Every condition is a compromise between thermodynamic yield and kinetic feasibility:

Temperature: low is better thermodynamically, but kinetically useless. 450°C is the sweet spot.
Pressure: high is better both thermodynamically and kinetically, but equipment cost limits it.
Catalyst: does not change yield but makes the compromise temperature actually work.
Product removal: the cleverest trick — by continuously removing NH₃, the system never reaches equilibrium, so the forward reaction keeps running.

Alternative Method

For numerical problems, use the expression for $K_p$ :

K_p = \frac{P_{NH_3}^2}{P_{N_2} \cdot P_{H_2}^3}

At a given temperature, $K_p$ is fixed. Increasing total pressure increases all partial pressures, but the denominator (with $P^4$ overall) increases more than the numerator ( $P^2$ ), so to maintain $K_p$ , more NH₃ must form. This is Le Chatelier in mathematical form.

Common Mistake

Students write that the catalyst “shifts the equilibrium towards products.” This is fundamentally wrong. A catalyst speeds up both forward and reverse reactions equally — it helps the system reach equilibrium faster but does not change the equilibrium position or $K_p$ . The yield at equilibrium with or without a catalyst is identical. The catalyst only ensures we get that yield in a reasonable time at 450°C instead of waiting years.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next