Nitric acid and ammonia synthesis

Why it matters

Ammonia is the gateway to almost all nitrogen fertilisers (urea, ammonium nitrate, DAP) and to nitric acid, which in turn feeds ammonium nitrate, explosives, adipic acid and nitro-aromatics. The two processes together illustrate equilibrium-limited synthesis with a recycle loop and purge (ammonia) and a fast, selectivity-controlled catalytic oxidation (nitric acid).

Key ideas

Ammonia synthesis (Haber–Bosch). N₂ + 3H₂ ⇌ 2NH₃, ΔH ≈ −92 kJ per 2 mol NH₃ at 25 °C.

• Synthesis gas preparation (natural-gas or naphtha based plant): desulphurisation (sulphur poisons the nickel and iron catalysts) → primary steam reforming over Ni, CH₄ + H₂O ⇌ CO + 3H₂, at 750–850 °C → secondary reforming, where air is added: it burns part of the gas to supply heat and brings in exactly the N₂ needed for a 3 : 1 H₂ : N₂ ratio → high- and low-temperature water–gas shift, CO + H₂O ⇌ CO₂ + H₂ → CO₂ removal by absorption (hot potassium carbonate or amines; the CO₂ is a feedstock for urea) → methanation of traces of CO and CO₂ to CH₄, because oxygen compounds poison the iron catalyst.
• Synthesis loop: compressed to about 150–250 bar; reacted at 400–500 °C over a promoted iron catalyst (Fe₃O₄ reduced to α-Fe, promoted with K₂O, Al₂O₃ and CaO). Single-pass conversion is only 15–25 %, so NH₃ is condensed out by refrigeration and unreacted gas is recycled.
• Purge: inerts (Ar from the air, CH₄ from methanation) are not consumed and would accumulate, so a small purge stream is bled from the loop. Purge size is a trade-off between losing H₂ and N₂ and diluting the reactor feed. H₂ is often recovered from the purge.
• Le Chatelier: the reaction is exothermic and decreases moles (4 → 2), so high pressure and low temperature favour equilibrium. Temperature is a compromise with rate; pressure is a compromise with compression cost.

Nitric acid (Ostwald process).

1. Ammonia oxidation: 4NH₃ + 5O₂ → 4NO + 6H₂O, ΔH ≈ −905 kJ, over platinum–rhodium gauze (about 90 % Pt, 10 % Rh) at 850–950 °C with a contact time of about a millisecond. The ammonia–air mixture is kept near 10–11 % NH₃, below the lower explosive limit (about 15 %). High temperature and very short contact time give 93–97 % selectivity to NO; slower or cooler conditions favour N₂ and N₂O. Platinum is slowly lost from the gauze and recovered with palladium catch gauzes.
2. NO oxidation: 2NO + O₂ ⇌ 2NO₂. This is homogeneous, slow and unusual in that its rate increases as temperature falls; high pressure and cooling favour it. Secondary air supplies the extra oxygen.
3. Absorption: 3NO₂ + H₂O → 2HNO₃ + NO in a sieve-tray absorption column with cooling; the NO released is re-oxidised within the column. Product is 55–68 % HNO₃. Water and HNO₃ form a maximum-boiling azeotrope at about 68 % HNO₃, so stronger acid needs extractive distillation with concentrated H₂SO₄ or magnesium nitrate.
4. Pressure options: mono-pressure plants (medium or high pressure throughout) or dual-pressure plants (oxidation at moderate pressure for gauze efficiency, absorption at higher pressure for good absorption and low tail-gas NOx).
5. Tail gas: NOx is removed by selective catalytic reduction with NH₃; N₂O, a strong greenhouse gas formed at the gauze, is decomposed catalytically.

Overall (with NO fully recycled within the absorber): NH₃ + 2O₂ → HNO₃ + H₂O, so 1 mol NH₃ gives at most 1 mol HNO₃.

Formulas

N₂ + 3H₂ ⇌ 2NH₃ Kp = p_NH₃² / (p_N₂ · p_H₂³) (bar⁻²); falls with temperature.

4NH₃ + 5O₂ → 4NO + 6H₂O, 2NO + O₂ → 2NO₂, 3NO₂ + H₂O → 2HNO₃ + NO Overall: NH₃ + 2O₂ → HNO₃ + H₂O

Inert balance on a loop: F · x_I,F = P · x_I,P

• F: fresh feed (kmol/h), P: purge (kmol/h), x_I: inert mole fraction in each. Applies at steady state when inerts leave only in the purge.

Overall conversion = reactant consumed / reactant in fresh feed; single-pass conversion = reactant consumed / reactant entering reactor.

Molar masses: NH₃ 17.03, HNO₃ 63.01 g/mol.

Worked examples

Example 1 (standard): ammonia demand of a nitric acid plant. A plant makes 100 t/day HNO₃ (100 % basis). NH₃ oxidation selectivity to NO is 96 % and absorption recovers 98 % of the NOx as acid. Find NH₃ consumed.

1. Theoretical NH₃ (1 : 1) = 100 × 17.03 / 63.01 = 27.03 t/day.
2. Overall efficiency = 0.96 × 0.98 = 0.9408.
3. Actual NH₃ = 27.03 / 0.9408 = 28.73 t/day.

Answer: about 28.7 t/day of ammonia.

Example 2 (GATE level): ammonia loop with purge. Fresh feed: 100 mol/h containing 24.75 mol N₂, 74.25 mol H₂ and 1 mol Ar. NH₃ is condensed completely; the purge is taken from the recycle gas and may contain 10 mol % Ar. Single-pass conversion of N₂ (and H₂) is 25 %. Find the purge rate, NH₃ produced, overall conversion and recycle rate.

1. Ar balance: 1 = P × 0.10, so P = 10 mol/h.
2. Purge carries 10 × 0.90 = 9 mol/h of N₂ + H₂ (still in 1 : 3 ratio).
3. N₂ + H₂ reacted = 99 − 9 = 90 mol/h (22.5 N₂, 67.5 H₂), giving NH₃ = 2 × 22.5 = 45 mol/h.
4. Overall conversion = 90 / 99 = 0.909.
5. Recycle R has 90 % N₂ + H₂. Reactor feed N₂ + H₂ = 99 + 0.9R. Single pass: 0.25 × (99 + 0.9R) = 90, so 99 + 0.9R = 360 and R = 290 mol/h.
6. Check: Ar at reactor inlet = (1 + 29) / (100 + 290) = 7.7 %.

Answer: P = 10 mol/h, NH₃ = 45 mol/h, overall conversion 90.9 %, R = 290 mol/h.

Common mistakes

• Saying NH₃ is oxidised directly to NO₂ over the catalyst. The gauze gives NO; NO₂ forms afterwards without a catalyst.
• Assuming high temperature helps NO oxidation. That step is faster at lower temperature.
• Confusing the catalysts: iron for NH₃ synthesis, Pt–Rh for NH₃ oxidation, nickel for steam reforming.
• Using 2 mol NO₂ → 2 HNO₃: only 2 of every 3 NO₂ become acid per pass, but with NO recycle the overall yield is 1 HNO₃ per NH₃.
• Mixing single-pass and overall conversion in recycle problems.
• Forgetting that inerts set the purge, not the reactants.

For GATE CH

Expect MCQs on catalysts and promoters, operating conditions and the reasons for them, the role of the secondary reformer, methanation and the purge, the azeotrope limiting nitric acid strength, and why the NH₃–air ratio is fixed. Numericals: recycle–purge balances, overall vs single-pass conversion, stoichiometric NH₃ or air requirement, and Kp-based equilibrium conversion. Practise inert balances and the overall-system approach.

Quick check

1. Why is air added in the secondary reformer?
2. What is the catalyst for NH₃ oxidation, and at what temperature?
3. Why is a purge needed in the ammonia loop?
4. Why can nitric acid not be concentrated beyond about 68 % by simple distillation?

Answers: 1. to supply N₂ for the 3 : 1 ratio and heat by partial combustion; 2. Pt–Rh gauze at about 850–950 °C; 3. to stop inerts (Ar, CH₄) accumulating; 4. HNO₃–water forms a maximum-boiling azeotrope at about 68 %.