# Industrial Equilibria: Haber and Contact Processes

> Industrial equilibria for JEE Chemistry: Le Chatelier applied to ammonia synthesis and sulphuric acid manufacture, optimum conditions and catalyst choices.

- Canonical URL: https://prepelephant.com/topics/jee/chemistry/industrial-equilibria-haber-contact
- Exam / course: JEE · Subject: Chemistry
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Industrial Equilibria: Haber and Contact Processes", PrepElephant, https://prepelephant.com/topics/jee/chemistry/industrial-equilibria-haber-contact

## Direct answer

Ammonia synthesis runs near 700 K and about 200 atm because equilibrium yield and reaction rate pull in opposite directions: N2 + 3H2 giving 2NH3 releases 92 kJ per mole of ammonia, so low temperature favours yield while forbidding speed, and the four moles to two contraction makes high pressure a yield booster. The Contact process mirrors the logic on 2SO2 + O2 giving 2SO3 (about -196 kJ), choosing 720 K, a modest 2 atm and a V2O5 catalyst, with excess air supplying the shift. Le Chatelier's principle and the reaction quotient Q against K rule every decision: products leave the equilibrium sign when Q is pushed below K, catalysts never move K, and continuous removal of product does the work that temperature cannot.

## What you must remember

- **Haber conditions:** 700 K, about 200 atm, iron catalyst promoted by Mo and K2O-Al2O3; per pass conversion is modest, so unreacted N2-H2 is recycled — the economically honest answer.
- **Pressure logic:** delta n(g) = -2 for ammonia, so pressure raises yield; delta n(g) = -1 for SO3, and since K is already huge, only about 2 atm is spent.
- **Temperature compromise:** exothermic both — yield wants low T, rate wants high T; 700 K (Haber) and 720 K (Contact) are the negotiated settlements.
- **Product removal:** liquefying NH3 out shifts equilibrium forward; absorbing SO3 in concentrated H2SO4 as oleum (never water — acid mist) does the same for Contact.
- **Catalyst's honest role:** Fe and V2O5 speed both directions and lower the activation barrier; K, delta G and equilibrium composition stay untouched.
- **Q versus K:** Q < K drives the forward reaction, Q > K the reverse, Q = K nothing — the quantitative replacement for hand-waving Le Chatelier statements.
- **Catalyst poison vigilance:** sulphur and arsenic impurities poison iron and V2O5 respectively; feed gas purification precedes both processes.
- **Excess air rationale:** extra O2 raises SO2 conversion both by raising a reactant's partial pressure and by keeping Q below K for longer.

## Pushing Le Chatelier on the Haber bed

Walk an engineer's reasoning. Compress to 200 atm: the system counters by making fewer gas moles, so ammonia fraction rises. Cool toward 700 K: yield climbs further but the nitrogen triple bond gets sluggish, so a catalyst earns its place. Now the masterstroke — chill the exit gases so ammonia liquefies off and recycle the unreacted N2 + H2 back over the catalyst. Each pass converts perhaps a tenth of the nitrogen, yet the loop achieves overall conversion above 95 per cent, which is why "recycling of reactants" is the answer to every "how is the yield improved industrially" question. Check the same logic on Contact: excess air (reactant pressure up), 720 K (compromise), oleum absorption (product withdrawal), V2O5 (rate, not K). The two processes teach one idea from opposite ends — manipulate Q, respect K.

## How the exam frames it

Chemical equilibrium with its industrial applications remains a JEE Main staple, usually one assertion-reason or condition-identification question; JEE Advanced prefers the Q-versus-K arithmetic and the subtle inert-gas cases. Two traps dominate. First, "catalyst increases the yield of ammonia" is false — it shortens the time to reach the same equilibrium. Second, adding an inert gas at constant volume changes nothing (partial pressures unchanged), while at constant pressure it dilutes the mixture and shifts equilibrium toward more gas moles — exactly backwards for the Haber reaction. One more: raising temperature in an exothermic synthesis lowers K, so the same Q suddenly exceeds K and the reaction runs backward.

## Frequently asked questions

### Why does the Haber process not use a lower temperature for better yield?

Thermodynamically it should, but the rate becomes commercially useless; 700 K with promoted iron is the compromise between yield per pass and throughput.

### What does liquefying ammonia achieve in the Haber loop?

Removing the product drops its partial pressure, pushing Q below K so the equilibrium shifts forward, while recycled reactants conserve feedstock.

### What role does V2O5 play in the Contact process?

It provides an alternate low-activation-energy path for SO2 oxidation, accelerating approach to equilibrium without changing K or the yield ceiling.

### What happens when an inert gas is added at constant volume?

Nothing — partial pressures of the reacting gases are unchanged, so Q stays equal to K and the composition does not shift.

### Why is SO3 absorbed in concentrated H2SO4 rather than water?

Direct hydration forms a troublesome acid mist, while oleum absorbs SO3 smoothly; product withdrawal also pulls the oxidation equilibrium forward.