Sulfuric acid and the Contact process

This page covers one Form 4 content standard from the Manufactured Substances in Industry chapter: sulfuric acid and the Contact process. Sulfuric acid is one of the most important industrial chemicals in the world, and the exam asks you both for its uses and for the three stages of the Contact process, including the chemical equations, the catalyst and the reason oleum is made instead of adding water directly. Learn the equations and conditions precisely, because this is a favourite structured-question topic.

Why sulfuric acid matters

Sulfuric acid, H₂SO₄, is used to manufacture fertilisers such as ammonium sulfate and superphosphate, to make detergents, paints and pigments, synthetic fibres and plastics, and as the electrolyte in lead-acid accumulators (car batteries). Concentrated sulfuric acid is also a powerful drying agent and dehydrating agent. Because so many industries depend on it, the amount of sulfuric acid a country produces is often taken as a rough measure of its industrial activity.

The Contact process: three stages

The Contact process manufactures sulfuric acid in three linked stages.

Stage 1, Production of sulfur dioxide. Molten sulfur is burned in dry air, or sulfide ores are roasted, to give sulfur dioxide:

S + O₂ → SO₂

Stage 2, Conversion of sulfur dioxide to sulfur trioxide. Sulfur dioxide and excess oxygen (from air) are passed over a vanadium(V) oxide, V₂O₅, catalyst at about 450–500 °C and 1 atmosphere of pressure. This step is reversible:

2SO₂ + O₂ ⇌ 2SO₃

The vanadium(V) oxide is a catalyst, it speeds the reaction up and is not used up. The moderate temperature and normal pressure are a compromise that gives a good yield of sulfur trioxide at a reasonable rate and cost.

Stage 3, Formation of sulfuric acid. Sulfur trioxide is not added directly to water, because that reaction is so violent that it forms a fine mist of sulfuric acid that is hard to condense. Instead, sulfur trioxide is dissolved in concentrated sulfuric acid to form oleum, H₂S₂O₇:

SO₃ + H₂SO₄ → H₂S₂O₇

The oleum is then diluted with the correct amount of water to give sulfuric acid:

H₂S₂O₇ + H₂O → 2H₂SO₄

The pollution problem

The sulfur dioxide made in Stage 1, and any that escapes, is a major air pollutant. It dissolves in rainwater to form acids and is a chief cause of acid rain, which damages buildings, corrodes metals, lowers the pH of rivers and lakes and harms plants and aquatic life. Factories therefore control and scrub their sulfur dioxide emissions.

Worked example

Question. In the Contact process, sulfur is the raw material for sulfuric acid. Calculate the maximum mass of sulfuric acid, H₂SO₄, that can be produced from 64 g of sulfur, assuming every stage goes to completion. [Relative atomic mass: H = 1, O = 16, S = 32]

Step 1, Trace the sulfur through the process. Each sulfur atom passes through the chain S → SO₂ → SO₃ → H₂SO₄, so 1 mol of S gives 1 mol of H₂SO₄.

Step 2, Find the moles of sulfur. Molar mass of S = 32 g mol⁻¹. Moles of S = mass ÷ molar mass = 64 g ÷ 32 g mol⁻¹ = 2 mol.

Step 3, Use the 1 : 1 ratio. Moles of H₂SO₄ = 2 mol.

Step 4, Convert moles of acid to mass. Molar mass of H₂SO₄ = (2 × 1) + 32 + (4 × 16) = 2 + 32 + 64 = 98 g mol⁻¹. Mass of H₂SO₄ = moles × molar mass = 2 mol × 98 g mol⁻¹ = 196 g.

Answer. The maximum mass of sulfuric acid that can be produced is 196 g.

Practice question

Question. (a) Name the catalyst used in Stage 2 of the Contact process and state the temperature commonly used. (b) Write the balanced equation for Stage 2. (c) Explain why sulfur trioxide is dissolved in concentrated sulfuric acid rather than in water.

Answer. (a) The catalyst is vanadium(V) oxide, V₂O₅, and the temperature is about 450–500 °C. (b) 2SO₂ + O₂ ⇌ 2SO₃. (c) Adding sulfur trioxide directly to water is extremely exothermic and violent and produces a large cloud of sulfuric acid mist that is difficult to condense and collect; dissolving it in concentrated sulfuric acid to form oleum, which is then diluted, controls the reaction and avoids the mist.

Exam tip

Three things score marks almost every time this topic appears: the three balanced equations, the catalyst (vanadium(V) oxide) with the reversible arrow in Stage 2, and the reason for making oleum. Write the Stage 2 equation with the ⇌ sign and remember the catalyst is not used up. When you are asked why water is not used directly, always give the full reason, the reaction is too vigorous and forms a mist of sulfuric acid, not just “it is dangerous.” A one-word safety answer loses the explanation mark.

Reasoning about industrial conditions

Examiners often ask why particular conditions are chosen. In Stage 2, a very high temperature would speed the reaction up but, because the forward reaction is exothermic, it would lower the yield of sulfur trioxide; a very low temperature would raise the yield but make the reaction too slow. The chosen temperature of about 450–500 °C is therefore a compromise between rate and yield, and the vanadium(V) oxide catalyst gives an acceptable rate without needing an even higher temperature. Normal pressure is used because the yield is already high, so the extra cost of high-pressure equipment is not justified. Being able to give this “rate versus yield versus cost” reasoning is exactly what higher-mark questions reward, and the same logic reappears in the Haber process for ammonia.

Where this fits

This is content standard 8.1 of the Manufactured Substances in Industry chapter. It builds on the catalyst and rate ideas from Rate of Reaction and pairs naturally with the Haber process in the next standard. Reinforce it with the chapter revision notes and practise the equation and calculation questions in the worked examples. In our online 1-to-1 SPM Chemistry lessons, taught in English, from RM50/hr, with a paid one-hour trial, our teachers drill the three equations and the oleum reasoning until they are automatic, because those are the parts of 4541 that students most often muddle.

Quick recap

• Sulfuric acid is used in fertilisers, detergents, paints, batteries and as a drying/dehydrating agent.
• Stage 1: S + O₂ → SO₂.
• Stage 2: 2SO₂ + O₂ ⇌ 2SO₃, catalyst V₂O₅, about 450–500 °C, 1 atm.
• Stage 3: SO₃ + H₂SO₄ → H₂S₂O₇ (oleum), then H₂S₂O₇ + H₂O → 2H₂SO₄.
• Sulfur dioxide causes acid rain, so emissions are controlled.