ChemistryICSE Class 10 · Sulphuric Acid

Contact Process: Manufacture of Sulphuric Acid (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

In the Contact process, sulphur is burnt to sulphur dioxide, which is purified and oxidised to sulphur trioxide over vanadium pentoxide at about 450–500 °C and 1–2 atm. The sulphur trioxide is absorbed in concentrated sulphuric acid to form oleum, which is diluted with water to give sulphuric acid.

Follow the arrows: SO₂ is made and purified, turned to SO₃ over V₂O₅ at 450–500 °C, absorbed as oleum, then diluted to H₂SO₄.

Illustration drawn for QuestionBuddy.

Understand Contact Process

The syllabus asks for the Contact process as equations with conditions.

Step 1: Making sulphur dioxide. Sulphur (or iron pyrites) is burnt in air:

$S + O_{2} \to S O_{2}$

Step 2: Purification. The gases are freed from dust, arsenic compounds and moisture, because these would poison the catalyst (reduce its activity).

Step 3: Catalytic oxidation (the 'contact' step). Sulphur dioxide and air are passed over vanadium pentoxide (V₂O₅) in a converter:

$2 S O_{2} + O_{2} ⇌ 2 S O_{3}$ + heat

Catalyst: V₂O₅ (platinum also works but is costly and easily poisoned).
Temperature: about 450–500 °C. The reaction is exothermic and reversible, so a lower temperature would favour more SO₃ but would be too slow; this is a compromise.
Pressure: about 1–2 atmospheres. Although 3 volumes become 2, the conversion is already high at this pressure, so very high pressures are not needed.

The name 'Contact' comes from this step: the gases react on contact with the catalyst surface.

Step 4: Absorption. SO₃ is not dissolved directly in water. That reaction is so exothermic that the water boils and a dense fog (mist) of acid droplets forms, which is hard to condense and is lost. Instead, SO₃ is absorbed in concentrated sulphuric acid to form oleum (fuming sulphuric acid, pyrosulphuric acid):

$S O_{3} + H_{2} S O_{4} \to H_{2} S_{2} O_{7}$

Step 5: Dilution. Oleum is diluted with the correct amount of water to give sulphuric acid of the required strength:

$H_{2} S_{2} O_{7} + H_{2} O \to 2 H_{2} S O_{4}$

Bookkeeping. Each S atom ends up in one H₂SO₄: S → SO₂ → SO₃ → H₂SO₄. So 32 g of sulphur can give at most 98 g of sulphuric acid, and 64 g of SO₂ gives 98 g of acid.

Contact Process: formulas and key facts

Burning sulphur: $S + O_{2} \to S O_{2}$
Catalytic oxidation: $2 S O_{2} + O_{2} 450 - 500^{\circ} C, 1 - 2 atm ⇌ V_{2} O_{5} 2 S O_{3}$
Absorption to oleum: $S O_{3} + H_{2} S O_{4} \to H_{2} S_{2} O_{7}$
Dilution of oleum: $H_{2} S_{2} O_{7} + H_{2} O \to 2 H_{2} S O_{4}$
Conditions: Catalyst vanadium pentoxide, V₂O₅
Temperature about 450 to 500 °C
Pressure about 1 to 2 atm
Absorbent concentrated sulphuric acid, giving oleum
Reason for purifying impurities poison the catalyst
Mole link: $1 S \to 1 S O_{2} \to 1 S O_{3} \to 1 H_{2} S O_{4}$

Catalyst	vanadium pentoxide, V₂O₅
Temperature	about 450 to 500 °C
Pressure	about 1 to 2 atm
Absorbent	concentrated sulphuric acid, giving oleum
Reason for purifying	impurities poison the catalyst

How to solve Contact Process questions

Write the five equations in order and attach each condition to the right step. For numericals, use the 1 : 1 : 1 : 1 link from sulphur to acid.

1Step 1: burn sulphur to SO₂.
2Step 2: purify (say why: catalyst poisoning).
3Step 3: $2 S O_{2} + O_{2} ⇌ 2 S O_{3}$ over V₂O₅ at about 450–500 °C, 1–2 atm.
4Step 4: absorb SO₃ in concentrated H₂SO₄ to give oleum (explain why not in water).
5Step 5: dilute oleum with water.
6For masses: moles of S = moles of SO₂ = moles of SO₃ = moles of H₂SO₄.

Contact Process in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Acid from a sulphur stockpile

Suppose a fertiliser company has 1.6 tonnes of sulphur and plans to turn all of it into sulphuric acid by the Contact process. (S = 32, H = 1, O = 16)

What is the maximum mass of sulphuric acid it can make?

Try it first, then show the solutionSolution

Moles of S $= 1600 000/32 = 50000$ mol.
Each S gives one H₂SO₄, so 50 000 mol of acid.
Mass $= 50000 \times 98 = 4900000$ g $= 4.9$ tonnes.

Answer: 4.9 tonnes of sulphuric acid at most.

Memory hook: 32 g of sulphur becomes 98 g of acid: roughly three times the mass.

Tricks to remember Contact Process

Short memory aids. Say each one aloud once, then use it on a question the same day.

V for Vanadium, V for Victory over SO₂
V₂O₅ is the Contact catalyst; Pt is for Ostwald; Fe for Haber.
Use it when: Matching catalysts.
Burn, Clean, Contact, Absorb, Dilute
'Big Cats Can Attack Deer': the five steps in order.
Use it when: Writing the process in sequence.
Acid absorbs, water fogs
SO₃ into water makes fog; SO₃ into acid makes oleum.
Use it when: Reason for absorbing in concentrated acid.

Contact Process: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Catalyst · 1 mark · MCQBoard level

The catalyst used in the conversion of sulphur dioxide to sulphur trioxide in the Contact process is:
(a) iron
(b) nickel
(c) vanadium pentoxide
(d) molybdenum

Try it first, then show the solutionSolution

Iron (with molybdenum) is for Haber's process; nickel for hydrogenation.
V₂O₅ catalyses the oxidation of SO₂.

Answer: (c) vanadium pentoxide.

Example 2: The contact step · 2 marks · Short answerBoard level

Write the balanced equation for the catalytic oxidation of sulphur dioxide and state the temperature and pressure used.

Try it first, then show the solutionSolution

$2 S O_{2} + O_{2} ⇌ 2 S O_{3}$ over V₂O₅.
About 450–500 °C and 1–2 atm.

Answer: 2SO₂ + O₂ ⇌ 2SO₃; about 450–500 °C, 1–2 atm.

Example 3: From SO₃ to acid · 3 marks · Short answerHarder board level

(a) Why is sulphur trioxide not dissolved directly in water? (b) Write the equations for the two steps actually used to convert SO₃ into sulphuric acid. (c) Name the intermediate product.

Try it first, then show the solutionSolution

(a) The reaction is highly exothermic; a dense fog of acid droplets forms, which is difficult to condense.
(b) $S O_{3} + H_{2} S O_{4} \to H_{2} S_{2} O_{7}$ ; $H_{2} S_{2} O_{7} + H_{2} O \to 2 H_{2} S O_{4}$ .
(c) Oleum (pyrosulphuric acid), H₂S₂O₇.

Answer: (a) Dense acid fog. (b) As above. (c) Oleum.

Example 4: Gas volumes in the converter · 3 marks · Short answerHarder board level

40 L of sulphur dioxide and 30 L of oxygen enter a converter, and the SO₂ is completely oxidised: $2 S O_{2} + O_{2} \to 2 S O_{3}$ . Find the volume and composition of the gas leaving (all volumes at the same temperature and pressure, SO₃ still a gas).

Try it first, then show the solutionSolution

O₂ used $= 40/2 = 20$ L; O₂ left $= 30 - 20 = 10$ L.
SO₃ formed $= 40$ L.
Total $= 10 + 40 = 50$ L.

Answer: 50 L: 40 L SO₃ and 10 L O₂.

Example 5: Masses through the process · 4 marks · Long answerHOTS / competency

8 kg of sulphur trioxide is absorbed in concentrated sulphuric acid and the oleum is then diluted. Calculate (a) the mass of oleum formed, (b) the mass of sulphuric acid finally obtained, (c) the mass of water needed for dilution. (S = 32, O = 16, H = 1)

Try it first, then show the solutionSolution

Moles of SO₃ $= 8000/80 = 100$ mol.
(a) H₂S₂O₇ (M = 178): 100 mol $= 17800$ g $= 17.8$ kg.
(b) Each H₂S₂O₇ gives 2H₂SO₄: 200 mol $\times 98 = 19600$ g $= 19.6$ kg.
(c) Water $= 100$ mol $\times 18 = 1800$ g $= 1.8$ kg (check: 17.8 + 1.8 = 19.6).

Answer: (a) 17.8 kg (b) 19.6 kg (c) 1.8 kg.

Example 6: Explaining the conditions · 4 marks · Long answerHOTS / competency

In the Contact process: (a) why is the SO₂ purified before entering the converter? (b) why is a temperature of 450–500 °C used rather than a lower one? (c) 12.8 g of SO₂ is converted fully to sulphuric acid: what mass of acid forms and what volume of oxygen at S.T.P. is needed in the contact step? (S = 32, O = 16, H = 1)

Try it first, then show the solutionSolution

(a) Dust and arsenic impurities poison the V₂O₅ catalyst.
(b) The reaction is exothermic, so lower temperatures favour SO₃ but react too slowly; 450–500 °C is a compromise.
(c) Moles of SO₂ $= 12.8/64 = 0.2$ ; H₂SO₄ $= 0.2 \times 98 = 19.6$ g.
O₂ $= 0.2/2 = 0.1$ mol $= 2.24$ L at S.T.P.

Answer: (a) Prevent catalyst poisoning. (b) Rate–yield compromise. (c) 19.6 g; 2.24 L.

Where students lose marks in Contact Process

The slips examiners see most often, and what to write instead.

Mistake: Mixing up industrial processes
In 2024 (Q3(iv)) examiners noted the Contact process confused with Ostwald's or Baeyer's process.
Do this: Tabulate: Haber (NH₃, Fe), Ostwald (HNO₃, Pt), Contact (H₂SO₄, V₂O₅), Baeyer (pure alumina, NaOH).
Usually costs: 1 mark per item.
Mistake: Thinking SO₃ is dissolved in water
In 2025 (Q1(x)) many believed SO₃ can be dissolved directly in water and misunderstood the dense fog.
Do this: SO₃ + water gives an acid mist that is hard to condense; absorb in concentrated H₂SO₄ instead.
Usually costs: 1 mark.
Mistake: One-way arrow and no conditions
Equations without the catalyst, temperature or reversibility lose the condition mark.
Wrong working
2SO₂ + O₂ → 2SO₃
Right working
2SO₂ + O₂ ⇌ 2SO₃ (V₂O₅, about 450–500 °C, 1–2 atm)
Do this: Show ⇌ and write the conditions.
Usually costs: ½ to 1 mark.
Mistake: Wrong oleum equation
Students write SO₃ + H₂O → H₂S₂O₇.
Do this: Oleum forms from SO₃ and H₂SO₄: SO₃ + H₂SO₄ → H₂S₂O₇.
Usually costs: 1 mark.

Exam tips

Learn the Contact process as five short equations with conditions; no plant diagram is needed.
Remember the pressure is low (1–2 atm), unlike Haber's process.
In reason questions, write 'dense fog of acid droplets, difficult to condense'.
For masses, the S → H₂SO₄ ratio is 1 : 1.

Practice questions on Contact Process

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
SO₃ is absorbed in concentrated sulphuric acid to form:
(a) dilute sulphuric acid
(b) oleum
(c) sulphurous acid
(d) sulphur dioxide
Hint
H₂S₂O₇.
Answer
(b) oleum.
Q2 · 1 mark · Very short answerBoard level
Write the formula of oleum.
Hint
Pyrosulphuric acid.
Answer
H₂S₂O₇.
Q3 · 2 marks · Short answerBoard level
In an industrial plant, gases leaving a converter packed with vanadium pentoxide are passed into concentrated sulphuric acid. Name the process, the gas formed in the converter and the product formed in the absorber.
Hint
V₂O₅ is the clue to the process.
Answer
Contact process; sulphur trioxide (SO₃) forms in the converter; oleum (H₂S₂O₇) forms in the absorber.
Q4 · 2 marks · Short answerHarder board level
Is the catalytic oxidation of SO₂ exothermic or endothermic? How does this affect the choice of temperature?
Hint
Yield versus rate.
Answer
Exothermic. A lower temperature would increase the yield of SO₃ but slow the reaction; about 450–500 °C gives a good yield at a useful rate.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The contact step of the Contact process is run at only about 1–2 atm.
Reason (R): The reaction $2 S O_{2} + O_{2} ⇌ 2 S O_{3}$ involves no change in the number of gas molecules.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Count the gas molecules on each side.
Answer
(c). 3 molecules of gas give 2, so R is false; a low pressure is used because the conversion is already high at 1–2 atm.
Q6 · 2 marks · Short answerHarder board level
What mass of sulphuric acid can be obtained from 8 kg of sulphur by the Contact process? (S = 32, H = 1, O = 16)
Hint
1 S → 1 H₂SO₄.
Answer
Moles of S $= 8000/32 = 250$ ; H₂SO₄ $= 250 \times 98 = 24500$ g $= 24.5$ kg.
Q7 · 3 marks · Short answerHarder board level
Give the conditions and catalyst for (a) Haber's process and (b) the contact step of the Contact process, and name one feature the two reactions share.
Hint
Both are reversible.
Answer
(a) Iron (Mo promoter), about 450–500 °C, about 200–900 atm. (b) V₂O₅, about 450–500 °C, 1–2 atm. Shared: both are reversible, exothermic reactions in which fewer gas molecules form, run at a compromise temperature with a catalyst.
Q8 · 2 marks · Short answerHOTS / competency
Suppose a plant's SO₂ feed is not purified and the conversion to SO₃ steadily falls over a few days. Explain the likely cause and why replacing V₂O₅ with platinum would not solve the problem.
Hint
Catalyst poisoning.
Answer
Impurities such as arsenic compounds and dust poison the catalyst, lowering its activity. Platinum is also poisoned by these impurities (and is more expensive), so the gas must be purified instead.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Catalyst in the Contact process?Show answer
Vanadium pentoxide (V₂O₅).
2.Pressure used in the converter?Show answer
About 1–2 atm.
3.Product of SO₃ with concentrated H₂SO₄?Show answer
Oleum, H₂S₂O₇.
4.Why purify SO₂?Show answer
Impurities poison the catalyst.
5.Mass of H₂SO₄ from 32 g S?Show answer
98 g.

Contact Process: frequently asked questions

What are the steps of the Contact process?

Burn sulphur to SO₂; purify the gas; oxidise SO₂ to SO₃ over V₂O₅ at about 450–500 °C; absorb SO₃ in concentrated H₂SO₄ to form oleum; dilute oleum with water.

Why is it called the Contact process?

Because SO₂ and O₂ react on contact with the surface of the solid catalyst.

Why is SO₃ absorbed in concentrated sulphuric acid?

Dissolving it in water is so exothermic that a fog of acid droplets forms which cannot be condensed easily; absorbing in acid avoids this.

Why is a low pressure used, unlike Haber's process?

The conversion of SO₂ to SO₃ is already high at 1–2 atm, so high pressure is not worth its cost.