Nitric Acid by the Ostwald Process
Kinetic control, platinum–rhodium gauze and selectivity
Lesson 3578 of 4,500 · Industrial Chemistry: Principles of Major Processes
Learning objectives
- Balance the main ammonia-to-nitric-acid reaction stages
- Explain why ammonia-oxidation selectivity matters
- Distinguish the desired NO route from nitrogen-containing side products
Introduction
Ammonia can be turned into nitric acid, making the Haber and Ostwald processes connected parts of a nitrogen-chemical value chain. The first Ostwald step is a very fast, strongly exothermic oxidation over platinum–rhodium gauze. Its challenge is not merely to react ammonia: the catalyst and conditions should direct nitrogen into nitric oxide rather than unwanted nitrogen or nitrous oxide. Later stages oxidise NO to NO₂ and absorb nitrogen oxides to make acid.
Core explanation
The desired initial reaction is 4NH₃ + 5O₂ → 4NO + 6H₂O. Check nitrogen: four N atoms enter as ammonia and four leave as NO. Hydrogen: 12 atoms make six water molecules. Oxygen: ten atoms from five O₂ molecules become four in NO and six in water. Platinum–rhodium gauze provides an active surface for rapid oxidation. The gauze geometry also exposes gas to a large catalytic area while allowing flow through the converter. U.S. EPA's nitric-acid process description details the platinum–rhodium gauze and main reaction stages.
Nitrogen-containing alternatives compete. Some ammonia nitrogen can become N₂ or N₂O instead of NO, depending on catalyst and conditions. Conversion of ammonia alone is therefore not an adequate performance measure. If all ammonia disappears but much forms N₂, the first-stage NO selectivity is poor and less nitrogen is available for nitric acid. N₂O is also environmentally significant, so selectivity has product and emissions implications. A plant reports desired NO yield or acid yield on a stated feed basis, not just ammonia disappearance.
The gas leaving the first converter is hot. After cooling and heat recovery, NO can be oxidised by oxygen: 2NO + O₂ → 2NO₂. The nitrogen dioxide is then absorbed into water through a coupled set of reactions. A useful balanced representation is 3NO₂ + H₂O → 2HNO₃ + NO. The NO produced by absorption can be reoxidised to NO₂, so absorption is not simply a one-pass consumption of every NO₂ molecule. Gas-liquid contacting, oxygen supply, pressure and temperature affect overall acid recovery.
The overall ideal transformation can be written NH₃ + 2O₂ → HNO₃ + H₂O after intermediates are eliminated and oxygen and water are balanced. It establishes a one-to-one theoretical mole ceiling: one mole of NH₃ nitrogen can make at most one mole of HNO₃ nitrogen. Actual product is lower if nitrogen leaves as by-product, if absorption is incomplete, or if material is lost. The net equation does not show the gauze, NO/NO₂ steps or absorber, so it is unsuitable by itself for explaining the process.
The initial oxidation is exothermic and kinetically sensitive. Gas flow and contact time must be controlled so the mixture reacts rapidly but does not spend unnecessary time at conditions that promote side reactions or catalyst loss. Heat can be recovered from the hot effluent. Subsequent NO oxidation and NO₂ absorption operate under different favourable conditions from the first hot converter, another reason the process is staged.
As a simple nitrogen balance, take 10.0 mol NH₃ entering a first converter. If it all reacts and 90% of its nitrogen follows the NO route, at most 9.0 mol NO forms at that stage on the one-N-to-one-N basis. The remaining 1.0 mol of nitrogen atoms is in other products. If downstream stages were ideal and no additional nitrogen were lost, the maximum HNO₃ from that NO stream would be 9.0 mol. This is not a prediction of actual plant yield; it illustrates that first-stage selectivity sets an upper bound for later acid production.
Step-by-step reasoning
1. Write 4NH₃ + 5O₂ → 4NO + 6H₂O for the desired gauze reaction. 2. Separate ammonia conversion from selectivity of reacted nitrogen toward NO. 3. Write 2NO + O₂ → 2NO₂ and a balanced NO₂ absorption representation. 4. Track the NO regenerated during absorption and its possible reoxidation. 5. Use nitrogen-atom balance to set the theoretical HNO₃ ceiling from ammonia feed. 6. Account for hot-gas heat recovery, by-products and incomplete absorption in a real process.
Visual explanation
Draw three process boxes: platinum–rhodium gauze converter, NO-to-NO₂ oxidation/cooling, and water absorber. Put NH₃ and air at the first inlet. Draw a main N arrow through NO, NO₂ and HNO₃, and smaller branch arrows from the converter to N₂ and N₂O. Add a curved NO arrow from absorber back toward oxidation to show its possible reoxidation.
Real-world analogy
An assembly line may use every raw part but still make many defective objects. Total consumption of parts is like ammonia conversion; the fraction assembled into saleable items is like selectivity toward NO and then acid. Counting only how many parts left the storeroom would overstate useful output.
Real-world example
A fertiliser complex can make ammonia, then use part of it as feedstock for nitric-acid production. The acid may feed further nitrogen-fertiliser manufacture. If first-stage ammonia oxidation sends nitrogen into N₂ or N₂O, that nitrogen no longer becomes acid, and the plant must consider both lost product and emissions treatment.
Why?
Why is platinum–rhodium gauze used rather than simply heating ammonia and oxygen? The catalytic surface accelerates a route that produces useful NO with high selectivity over a short contact time. Heat alone cannot guarantee that the consumed ammonia nitrogen takes the desired pathway.
Common misconception
“The first oxidation makes nitric acid directly.” It makes NO and water in the desired pathway. NO is subsequently oxidised to NO₂, and nitrogen oxides are absorbed into water to make HNO₃. Skipping these stages hides both the nitrogen balance and the need for gas handling.
Worked example
Ten moles of NH₃ enter an idealised first Ostwald converter. Assume all ten react and 90% of ammonia nitrogen becomes NO. Because each NH₃ and NO molecule contains one N atom, the first stage forms 9.0 mol NO; 1.0 mol of N atoms enters side products. Even with perfect downstream oxidation and absorption, no more than 9.0 mol HNO₃ can come from this NO stream. Relative to the 10.0 mol feed-based ceiling, desired nitrogen yield cannot exceed 90% under these assumptions.
Quick check
1. In the desired first Ostwald reaction, how many moles of NO form from four moles of NH₃ at the stoichiometric ceiling? Answer: The balanced reaction makes four moles of NO from four moles of NH₃.
Exam focus
Balance all three teaching equations and keep NO distinct from NO₂. If given ammonia conversion and NO selectivity, multiply them on a compatible nitrogen basis before finding maximum acid. Mention gauze catalysis and competing nitrogen products when explaining process yield.
Advanced insight
The first hot catalytic step and the cooler NO₂-absorption stage have different kinetic and transport demands. Platinum-group metal can be lost or redistributed during service, while N₂O formation may require dedicated abatement. A detailed process model includes catalyst condition, gas composition, heat recovery, absorber efficiency and off-gas treatment. The simple net equation NH₃ + 2O₂ → HNO₃ + H₂O is valuable for atom balance but contains none of those performance variables.
Summary
The Ostwald process converts ammonia to nitric acid through desired NO formation over platinum–rhodium gauze, NO oxidation to NO₂ and absorption of nitrogen oxides in water. Ammonia conversion alone does not measure success because nitrogen can enter N₂ or N₂O. The ideal overall mole ratio is one HNO₃ per NH₃, while selectivity and absorption set the actual result. Staged equipment manages the different reaction and transfer conditions.
Practice questions
1. Balance the desired first reaction of ammonia with oxygen to make NO and water. Answer: 4NH₃ + 5O₂ → 4NO + 6H₂O. 2. Balance the gas-phase oxidation of NO to NO₂. Answer: 2NO + O₂ → 2NO₂. 3. A first converter consumes 20 mol NH₃ and sends 85% of its nitrogen to NO. Find the NO amount on a one-N basis. Answer: It forms 0.85 × 20 = 17 mol NO, with the other 3 mol N atoms in side products. 4. Why is first-stage NO selectivity relevant even if downstream absorption is perfect? Answer: Nitrogen sent to N₂ or N₂O is unavailable for the NO→NO₂→HNO₃ pathway, limiting acid yield.