The Contact process makes sulfuric acid in stages: S(s) + O2(g) → SO2(g), then 2SO2(g) + O2(g) ⇌ 2SO3(g) over vanadium(V) oxide, then SO3 is absorbed into concentrated acid and diluted to sulfuric acid.
The Contact process is the industrial method for manufacturing sulfuric acid, one of the most important chemicals in industry. The syllabus expects you to know its stages, the balanced equations, the catalyst and the operating conditions, and the reason sulfur trioxide is not dissolved directly in water. It is a set-piece question in the Manufactured Substances chapter of 4541.
The balanced equations
The process is carried out in three main stages, each with correct formulae and state symbols.
Stage 1, sulfur (or a sulfide ore) is burned in air to form sulfur dioxide:
S(s) + O2(g) → SO2(g)
Stage 2, sulfur dioxide and more oxygen are passed over a catalyst to form sulfur trioxide. This step is reversible:
2SO2(g) + O2(g) ⇌ 2SO3(g)
Stage 3, sulfur trioxide is absorbed into concentrated sulfuric acid to form oleum, which is then diluted with water to give more sulfuric acid:
SO3(g) + H2SO4(l) → H2S2O7(l)
H2S2O7(l) + H2O(l) → 2H2SO4(l)
Conditions required
The key stage is the reversible conversion of sulfur dioxide to sulfur trioxide. The conditions are a vanadium(V) oxide catalyst (V2O5), a temperature of about 450–500 °C, and a pressure of about 1 atmosphere. A very high pressure is not needed because a high conversion is already achieved, so the extra cost is not worth it. The catalyst speeds up the reaction without being used up.
What you observe and the key reasoning
In the laboratory model, sulfur burns with a blue flame to give the colourless, pungent gas sulfur dioxide. The heart of the exam answer, however, is the reasoning about stage 3: sulfur trioxide is not added straight to water because that reaction is far too vigorous and produces a dense mist of sulfuric acid droplets that will not condense and is wasteful and hazardous. Instead the sulfur trioxide is dissolved in concentrated sulfuric acid to form oleum (H2S2O7), and the oleum is then diluted with a controlled amount of water. The final product is concentrated sulfuric acid, a dense oily liquid.
Where it appears in the SPM exam
In 4541/1 you identify the catalyst or the correct equation for a stage. In 4541/2 the standard question asks you to write all three stages with state symbols, name the catalyst and give the operating conditions, and explain why sulfur trioxide is dissolved in concentrated acid rather than water. You may also be asked for uses of sulfuric acid, making fertilisers, detergents, paints and car batteries, and for the environmental problem of sulfur dioxide escaping and causing acid rain.
How we teach it
Our teachers make sure you memorise the three stages as a chain, get the catalyst name exactly right as vanadium(V) oxide, and give the oleum reasoning in full rather than a one-line answer. Students most often lose marks by writing SO3 + H2O directly, by naming the wrong catalyst, or by forgetting that stage 2 is reversible. Learning the process as a story from sulfur to sulfuric acid keeps every equation and condition in the right place.
Quick summary
The Contact process makes sulfuric acid by burning sulfur to sulfur dioxide, catalytically oxidising that to sulfur trioxide over vanadium(V) oxide at about 450–500 °C and 1 atmosphere, then absorbing the trioxide into concentrated acid to form oleum and diluting it. Learn the four equations, the conditions and the oleum reasoning, and this reliable industrial topic is fully covered.
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