ChemistryICSE Class 10 · Chapter 11

Sulphuric Acid ICSE Class 10: Contact Process and Properties

Updated 14 October 2026 · Fact-checked

Sulphuric acid is manufactured by the Contact process, in which sulphur dioxide is oxidised to sulphur trioxide over vanadium pentoxide and absorbed in concentrated acid to form oleum. Dilute sulphuric acid behaves as a dibasic acid; concentrated acid is an oxidising agent, a dehydrating agent and a non-volatile acid.

What Sulphuric Acid is about

Sulphuric acid, H₂SO₄, is called the 'king of chemicals' because so many industries use it. It behaves very differently depending on concentration, and most questions test which behaviour is at work.

The CISCE 2027 syllabus covers: manufacture by the Contact process (equations with conditions); its behaviour as an acid when dilute (with metals, metal oxides, hydroxides, carbonates, hydrogen carbonates, sulphites, hydrogen sulphites and sulphides) and as a dibasic acid; concentrated acid as an oxidising agent (oxidation of carbon and sulphur); as a dehydrating agent (dehydration of sugar and copper(II) sulphate crystals); its non-volatile nature (reaction with sodium or potassium chloride and nitrate); and tests for sulphuric acid.

A useful summary: dilute acid supplies H₃O⁺ ions (acid); concentrated acid supplies oxygen (oxidising), pulls water out of compounds (dehydrating), and stays behind when heated (non-volatile).

Weightage: CISCE publishes no chapter or unit weightage for ICSE Chemistry. Sulphuric Acid is in the 8. Study of Compounds D. Sulphuric Acid part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Chemistry (Science Paper 2) syllabus, examination year 2027

Key concepts

The Contact process

1. Sulphur dioxide is made by burning sulphur (or roasting iron pyrites): $S + O_{2} \to S O_{2}$
2. The gases are purified (dust and arsenic compounds would poison the catalyst) and dried.
3. Catalytic oxidation in the converter: $2 S O_{2} + O_{2} ⇌ 2 S O_{3} + heat$

Catalyst: vanadium pentoxide, V₂O₅
Temperature: about 450-500 °C
Pressure: about 1-2 atmospheres

4. Absorption: SO₃ is dissolved in concentrated sulphuric acid to form oleum (pyrosulphuric acid): $S O_{3} + H_{2} S O_{4} \to H_{2} S_{2} O_{7}$
5. Dilution: oleum is diluted with the right amount of water: $H_{2} S_{2} O_{7} + H_{2} O \to 2 H_{2} S O_{4}$

Why not dissolve SO₃ directly in water? The reaction is so exothermic that the water boils and a dense fog of fine acid droplets forms, which does not condense easily and is lost.

Worked example

Name the catalyst and state the temperature for the conversion of SO₂ to SO₃. Why is SO₃ not absorbed in water?

Try it first, then show the solutionSolution

V₂O₅ at about 450-500 °C.
SO₃ reacts with water with so much heat that a dense acid fog (mist) forms, which is hard to condense.

Answer: Vanadium pentoxide, 450-500 °C; direct absorption gives an acid fog.

Dilute sulphuric acid as an acid; dibasic nature

In water, sulphuric acid ionises in two steps, giving two H₃O⁺ ions per molecule, so it is dibasic:
$H_{2} S O_{4} + H_{2} O \to H_{3} O^{+} + H S O_{4}^{-}$ ; $H S O_{4}^{-} + H_{2} O \to H_{3} O^{+} + S O_{4}^{2 -}$

It therefore forms two series of salts: acid salts (hydrogen sulphates, e.g. NaHSO₄) and normal salts (sulphates, e.g. Na₂SO₄):
$N a O H + H_{2} S O_{4} \to N a H S O_{4} + H_{2} O$ ; $2 N a O H + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H_{2} O$

Typical acid reactions:

Metals: $Z n + H_{2} S O_{4} \to Z n S O_{4} + H_{2}$ ; $F e + H_{2} S O_{4} \to F e S O_{4} + H_{2}$
Oxides and hydroxides: $C u O + H_{2} S O_{4} \to C u S O_{4} + H_{2} O$
Carbonates, hydrogen carbonates: $N a_{2} C O_{3} + H_{2} S O_{4} \to N a_{2} S O_{4} + H_{2} O + C O_{2}$ ; $2 N a H C O_{3} + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H_{2} O + 2 C O_{2}$
Sulphites, hydrogen sulphites: $N a_{2} S O_{3} + H_{2} S O_{4} \to N a_{2} S O_{4} + H_{2} O + S O_{2}$ ; $2 N a H S O_{3} + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H_{2} O + 2 S O_{2}$
Sulphides: $F e S + H_{2} S O_{4} \to F e S O_{4} + H_{2} S$

Worked example

Why is sulphuric acid called a dibasic acid? Give one acid salt and one normal salt it forms with potassium hydroxide.

Try it first, then show the solutionSolution

Each molecule gives two hydronium ions in water (two replaceable H atoms).
Acid salt: KHSO₄; normal salt: K₂SO₄.

Answer: Two replaceable hydrogens; KHSO₄ and K₂SO₄.

Concentrated sulphuric acid as an oxidising agent

Hot concentrated sulphuric acid gives up oxygen and is itself reduced to sulphur dioxide.

Carbon: $C + 2 H_{2} S O_{4} \to C O_{2} + 2 S O_{2} + 2 H_{2} O$

Sulphur: $S + 2 H_{2} S O_{4} \to 3 S O_{2} + 2 H_{2} O$

In each case, sulphuric acid is the oxidising agent (it oxidises C to CO₂ or S to SO₂), and carbon or sulphur is the reducing agent. SO₂ is recognised by its smell of burning sulphur and because it turns acidified potassium dichromate green and decolourises acidified potassium permanganate (purple → colourless).

Worked example

In $S + 2 H_{2} S O_{4} \to 3 S O_{2} + 2 H_{2} O$ , which substance is the oxidising agent? Explain.

Try it first, then show the solutionSolution

Sulphur gains oxygen (S → SO₂): it is oxidised.
Sulphuric acid loses oxygen (H₂SO₄ → SO₂): it is reduced.
The substance that is reduced is the oxidising agent.

Answer: Sulphuric acid.

Concentrated sulphuric acid as a dehydrating agent

A dehydrating agent removes the elements of water (hydrogen and oxygen in the ratio 2 : 1) from a compound, or water of crystallisation from crystals. A drying agent only removes moisture (free water) from a substance without changing it chemically. Concentrated sulphuric acid is both.

Sugar (charring): $C_{12} H_{22} O_{11} \to 12 C + 11 H_{2} O$
Observation: the white sugar turns yellow, then brown, then into a black, spongy mass of carbon that swells up; steam is given off; the reaction is very exothermic.

Copper(II) sulphate crystals: $C u S O_{4} \cdot 5 H_{2} O \to C u S O_{4} + 5 H_{2} O$
Observation: the blue crystals turn white (anhydrous copper(II) sulphate).

Because of its great affinity for water, the concentrated acid is always diluted by adding acid to water slowly, with stirring, never water to acid.

Worked example

Distinguish between a drying agent and a dehydrating agent with an example of each action of concentrated sulphuric acid.

Try it first, then show the solutionSolution

Drying: removes moisture from a gas, e.g. drying HCl or chlorine gas.
Dehydrating: removes the elements of water from a compound, e.g. charring sugar to carbon.

Answer: Drying removes free water; dehydrating removes H and O (as water) from a compound.

Non-volatile nature

Sulphuric acid has a high boiling point (about 338 °C), so it is non-volatile. When heated with salts of volatile acids, it displaces them as vapour:

$N a C l + H_{2} S O_{4} \to N a H S O_{4} + H C l$
$K C l + H_{2} S O_{4} \to K H S O_{4} + H C l$
$N a N O_{3} + H_{2} S O_{4} \to N a H S O_{4} + H N O_{3}$
$K N O_{3} + H_{2} S O_{4} \to K H S O_{4} + H N O_{3}$

This is how HCl and HNO₃ are made in the laboratory.

Worked example

Which property of sulphuric acid is used to prepare hydrogen chloride from common salt? Write the equation.

Try it first, then show the solutionSolution

It is non-volatile; HCl is volatile and escapes.
$N a C l + H_{2} S O_{4} \to N a H S O_{4} + H C l$

Answer: Non-volatile nature; equation above.

Tests for sulphuric acid

Barium chloride solution: a white precipitate of barium sulphate, insoluble in concentrated HCl or HNO₃:

$B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l$

Lead nitrate solution: a white precipitate of lead sulphate:

$P b (N O_{3})_{2} + H_{2} S O_{4} \to P b S O_{4} + 2 H N O_{3}$

The dilute acid with zinc gives hydrogen; the concentrated acid chars sugar.

Hydrochloric acid gives a white precipitate with lead nitrate too (PbCl₂), but PbCl₂ dissolves on heating, while PbSO₄ does not. With barium chloride, only sulphate gives a precipitate.

Worked example

How would you distinguish dilute sulphuric acid from dilute hydrochloric acid?

Try it first, then show the solutionSolution

Add barium chloride solution to each.
Sulphuric acid: white precipitate of BaSO₄, insoluble in acids.
Hydrochloric acid: no precipitate.

Answer: White precipitate with BaCl₂ only for sulphuric acid.

Chemistry in real life: Sulphuric Acid

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.

Dehydration of sugar

Suppose a science teacher pours concentrated sulphuric acid onto a beaker of sugar during a demonstration, and a black column rises out of the beaker with steam.

Explain what happens with an equation and state the property shown.

Try it first, then show the solutionSolution

Concentrated sulphuric acid removes hydrogen and oxygen as water from sugar.
$C_{12} H_{22} O_{11} \to 12 C + 11 H_{2} O$
Carbon is left as a black spongy mass, pushed up by steam; the reaction is very exothermic.

Answer: Dehydration (charring) of sugar; dehydrating property.

Memory hook: Sugar minus water = carbon.

Why SO₃ is not absorbed in water

Suppose a trainee at a sulphuric acid plant suggests saving a step by bubbling sulphur trioxide straight into water.

Explain why this is not done and what is done instead.

Try it first, then show the solutionSolution

SO₃ reacts with water with a great deal of heat.
The water boils and a dense fog of acid droplets forms, which does not condense easily and escapes.
Instead, SO₃ is absorbed in concentrated sulphuric acid to form oleum: $S O_{3} + H_{2} S O_{4} \to H_{2} S_{2} O_{7}$ , which is then diluted: $H_{2} S_{2} O_{7} + H_{2} O \to 2 H_{2} S O_{4}$ .

Answer: Direct absorption forms an acid fog; oleum route is used.

Memory hook: Acid first, water second.

Mass from an industrial equation

Suppose a plant converts 64 kg of sulphur dioxide completely to sulphuric acid by the Contact process.

What mass of sulphuric acid is produced? (S = 32, O = 16, H = 1)

Try it first, then show the solutionSolution

Moles of SO₂ = 64,000 g ÷ 64 = 1000 mol.
Each SO₂ gives one SO₃ and finally one H₂SO₄.
Mass of H₂SO₄ = 1000 × 98 = 98,000 g = 98 kg.

Answer: 98 kg.

Memory hook: One sulphur in, one sulphuric acid out.

Topics in this chapter

Sulphuric Acid formula sheet

Burning sulphur: $S + O_{2} \to S O_{2}$; Contact process step 1.
Catalytic oxidation: $2 S O_{2} + O_{2} ⇌ 2 S O_{3}$; V₂O₅, 450-500 °C, 1-2 atm; exothermic.
Oleum: $S O_{3} + H_{2} S O_{4} \to H_{2} S_{2} O_{7}$; Absorption in conc. acid.
Dilution of oleum: $H_{2} S_{2} O_{7} + H_{2} O \to 2 H_{2} S O_{4}$; Controlled water.
Oxidation of carbon: $C + 2 H_{2} S O_{4} \to C O_{2} + 2 S O_{2} + 2 H_{2} O$; Hot conc. acid.
Dehydration of sugar: $C_{12} H_{22} O_{11} \to 12 C + 11 H_{2} O$; Black spongy carbon.
Test with barium chloride: $B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l$; White precipitate, insoluble in acids.

Sulphuric Acid: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Contact process flow chart

Outline the manufacture of sulphuric acid.

Draw

Box 1: sulphur burner, S + O₂ → SO₂.
Box 2: purification and drying of gases.
Box 3: converter with V₂O₅ at 450-500 °C: 2SO₂ + O₂ ⇌ 2SO₃.
Box 4: absorption tower with conc. H₂SO₄: SO₃ + H₂SO₄ → H₂S₂O₇.
Box 5: dilution: H₂S₂O₇ + H₂O → 2H₂SO₄.

Label

burner
purifier
converter (V₂O₅)
absorption tower
oleum
sulphuric acid

Watch out: Arrows between boxes; conditions written beside the converter.

How the board asks Sulphuric Acid

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Contact process	Equations with conditions; catalyst; why SO₃ is not dissolved in water; table of industrial processes.	Five equations; V₂O₅, 450-500 °C, 1-2 atm; acid fog.
Role of the acid in a reaction	MCQ: in a given equation, does sulphuric acid act as an acid, an oxidising agent, a dehydrating agent or a non-volatile acid?	Products tell you: salt + H₂/CO₂ = acid; SO₂ formed = oxidising; carbon or white CuSO₄ = dehydrating; HCl or HNO₃ released = non-volatile.
Balanced equations	Sulphuric acid with carbon, sulphur, sugar, sodium hydrogen carbonate, sodium hydrogen sulphite.	Balance carefully: hydrogen sulphites and hydrogen carbonates need 2 per acid.
Observations	What is seen when conc. acid is added to sugar or to blue vitriol; which gas decolourises KMnO₄.	Sugar: white → black spongy mass; CuSO₄·5H₂O: blue → white; SO₂ decolourises KMnO₄.
Distinguish and test	Distinguish H₂SO₄ from HCl or HNO₃ with BaCl₂ or Pb(NO₃)₂.	Only sulphate gives an acid-insoluble white precipitate with BaCl₂.

Tricks to remember Sulphuric Acid

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

Dilute: Acid. Concentrated: O-D-N
Concentrated acid is Oxidising, Dehydrating and Non-volatile.
Use it when: Properties at a glance.
Catalysts: Fe-Pt-V
Haber (Fe), Ostwald (Pt), Contact (V₂O₅).
Use it when: Matching processes.
SO₃ never meets water first
SO₃ goes into acid (oleum) first, then oleum meets water.
Use it when: Why oleum is formed.
Spot the role from the product
SO₂ in the products? Oxidising. Carbon or white solid? Dehydrating. HCl or HNO₃ escaping? Non-volatile.
Use it when: Role-of-acid MCQs.

Sulphuric Acid: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Role of the acid · 1 mark · MCQBoard level

In the reaction $C + 2 H_{2} S O_{4} \to C O_{2} + 2 S O_{2} + 2 H_{2} O$ , sulphuric acid acts as:
(a) a dehydrating agent
(b) an oxidising agent
(c) a non-volatile acid
(d) a reducing agent

Try it first, then show the solutionSolution

It oxidises carbon to CO₂ and is itself reduced to SO₂.

Answer: (b)

Example 2: Charring sugar · 1 mark · Assertion-reasonHarder board level

Assertion (A): Concentrated sulphuric acid turns sugar into a black mass of carbon.
Reason (R): Concentrated sulphuric acid is an oxidising agent.
(a) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
(c) (A) is true and (R) is false.
(d) (A) is false and (R) is true.

Try it first, then show the solutionSolution

Both statements are true: sugar is charred, and the hot concentrated acid does oxidise carbon and sulphur.
But charring removes hydrogen and oxygen as water, $C_{12} H_{22} O_{11} \to 12 C + 11 H_{2} O$ : that is the dehydrating property, not the oxidising one.

Answer: (b)

Example 3: Contact process equations · 3 marks · Short answerBoard level

Write the equations, with conditions, for converting sulphur dioxide into sulphuric acid in the Contact process.

Try it first, then show the solutionSolution

$2 S O_{2} + O_{2} ⇌ 2 S O_{3}$ (V₂O₅, 450-500 °C, 1-2 atm)
$S O_{3} + H_{2} S O_{4} \to H_{2} S_{2} O_{7}$
$H_{2} S_{2} O_{7} + H_{2} O \to 2 H_{2} S O_{4}$

Answer: As above.

Example 4: Properties from reactions · 3 marks · Short answerHarder board level

Name the property of concentrated sulphuric acid shown in each: (a) $K N O_{3} + H_{2} S O_{4} \to K H S O_{4} + H N O_{3}$ ; (b) $C u S O_{4} \cdot 5 H_{2} O \to C u S O_{4} + 5 H_{2} O$ ; (c) $S + 2 H_{2} S O_{4} \to 3 S O_{2} + 2 H_{2} O$ .

Try it first, then show the solutionSolution

(a) Volatile HNO₃ is displaced: non-volatile nature.
(b) Water of crystallisation removed: dehydrating.
(c) Sulphur oxidised, acid reduced to SO₂: oxidising.

Answer: (a) Non-volatile (b) Dehydrating (c) Oxidising

Example 5: Dilute acid equations · 3 marks · Short answerHarder board level

Write balanced equations for dilute sulphuric acid with (a) sodium hydrogen carbonate, (b) sodium hydrogen sulphite, (c) iron(II) sulphide.

Try it first, then show the solutionSolution

(a) $2 N a H C O_{3} + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H_{2} O + 2 C O_{2}$
(b) $2 N a H S O_{3} + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H_{2} O + 2 S O_{2}$
(c) $F e S + H_{2} S O_{4} \to F e S O_{4} + H_{2} S$

Answer: As above.

Example 6: Volume of gas from carbon · 3 marks · Short answerHarder board level

1.2 g of carbon is heated with excess concentrated sulphuric acid. Calculate the total volume of gases (CO₂ + SO₂) at S.T.P. and the mass of acid used. (C = 12, H = 1, S = 32, O = 16)

Try it first, then show the solutionSolution

$C + 2 H_{2} S O_{4} \to C O_{2} + 2 S O_{2} + 2 H_{2} O$ ; moles C = 1.2 ÷ 12 = 0.1.
Gases: 0.1 CO₂ + 0.2 SO₂ = 0.3 mol × 22.4 = 6.72 L.
Acid: 0.2 mol × 98 = 19.6 g.

Answer: 6.72 L; 19.6 g.

Example 7: Dehydration of copper sulphate · 2 marks · Short answerHarder board level

25 g of blue copper(II) sulphate crystals are treated with concentrated sulphuric acid. (a) What is seen? (b) What mass of water is removed? (Cu = 64, S = 32, O = 16, H = 1)

Try it first, then show the solutionSolution

(a) The blue crystals turn white.
(b) M of CuSO₄·5H₂O = 250; 25 g = 0.1 mol; water = 0.1 × 90 = 9 g.

Answer: (a) Blue → white (b) 9 g.

Example 8: HOTS: identifying a gas · 4 marks · Long answerHOTS / competency

A gas X, formed when hot concentrated sulphuric acid reacts with sulphur, decolourises purple acidified potassium permanganate. Another gas Y, formed by the action of dilute sulphuric acid on sodium carbonate, does not. (a) Identify X and Y. (b) Both turn lime water milky; how else could you tell them apart? (c) Write the equation forming X.

Try it first, then show the solutionSolution

(a) X = sulphur dioxide; Y = carbon dioxide.
(b) SO₂ turns acidified potassium dichromate green and smells of burning sulphur; CO₂ is odourless and has no effect on dichromate.
(c) $S + 2 H_{2} S O_{4} \to 3 S O_{2} + 2 H_{2} O$
The permanganate test distinguishes them because SO₂ is a reducing agent.

Answer: As above.

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 2 + 1 + 1 + 1 marks.

A Contact process plant

Suppose a flow chart of a Contact plant shows: a sulphur burner → a purifier → a converter at 450 °C → an absorption tower fed with concentrated acid → a dilution tank.

(i) · 2 marks
Write the equation for the reaction in the converter and name the catalyst.
Answer
Answer: $2 S O_{2} + O_{2} ⇌ 2 S O_{3}$ ; vanadium pentoxide (V₂O₅).
(ii) · 1 mark
Why must the gases be purified before the converter?
Answer
Answer: Impurities (dust, arsenic compounds) would poison the catalyst.
(iii) · 1 mark
Name the product of the absorption tower.
Answer
Answer: Oleum (pyrosulphuric acid), H₂S₂O₇.
(iv) · 1 mark
Write the equation in the dilution tank.
Answer
Answer: $H_{2} S_{2} O_{7} + H_{2} O \to 2 H_{2} S O_{4}$

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

With one equation each, show sulphuric acid acting (a) as an acid when dilute and (b) as an oxidising agent when concentrated. (2 marks) (2 marks)

Step	Marks
(a) $Z n + H_{2} S O_{4} \to Z n S O_{4} + H_{2}$ (salt and hydrogen)	1
(b) $C + 2 H_{2} S O_{4} \to C O_{2} + 2 S O_{2} + 2 H_{2} O$ (acid reduced to SO₂)	1

Describe the dehydration of sugar by concentrated sulphuric acid: observation, equation and the difference between dehydrating and drying. (3 marks) (3 marks)

Step	Marks
White sugar turns brown then black, forming a spongy mass; steam evolved	1
$C_{12} H_{22} O_{11} \to 12 C + 11 H_{2} O$	1
Dehydrating removes elements of water from a compound; drying removes moisture only	1

Write the equations of the Contact process from sulphur to sulphuric acid, with the conditions of the catalytic step. (4 marks) (4 marks)

Step	Marks
$S + O_{2} \to S O_{2}$	1
$2 S O_{2} + O_{2} ⇌ 2 S O_{3}$ (V₂O₅, 450-500 °C, 1-2 atm)	1
$S O_{3} + H_{2} S O_{4} \to H_{2} S_{2} O_{7}$	1
$H_{2} S_{2} O_{7} + H_{2} O \to 2 H_{2} S O_{4}$	1

Where students lose marks in Sulphuric Acid

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

Mistake: Choosing SO₂ as the oxidising agent
CISCE examiners' comments, 2024: in $S + 2 H_{2} S O_{4} \to 3 S O_{2} + 2 H_{2} O$ , candidates struggled with the idea of an oxidising agent and chose sulphur dioxide.
Wrong working
Oxidising agent: SO₂
Right working
Oxidising agent: H₂SO₄ (it loses oxygen and is reduced).
Do this: The oxidising agent is a reactant, and it is the one reduced.
Usually costs: 1 mark.
Mistake: Thinking SO₃ can go straight into water
CISCE examiners' comments, 2025: candidates believed SO₃ can be dissolved directly in water and misunderstood the dense fog.
Wrong working
SO₃ + H₂O → H₂SO₄ is the final step.
Right working
SO₃ is absorbed in conc. H₂SO₄ to form oleum, which is then diluted.
Do this: Remember the fog: acid first, water second.
Usually costs: 1 mark.
Mistake: Unbalanced or wrong products
CISCE examiners' comments, 2025: in a 'give balanced equations' question, candidates left equations unbalanced, wrote sulphuric acid as a product of ammonium carbonate with dilute HCl instead of CO₂ and water, and gave wrong products for sulphur with hot concentrated nitric acid.
Wrong working
C + H₂SO₄ → CO₂ + SO₂ + H₂O
Right working
$C + 2 H_{2} S O_{4} \to C O_{2} + 2 S O_{2} + 2 H_{2} O$
Do this: Count each element on both sides before moving on.
Usually costs: 1 mark each.
Mistake: Confusing SO₂ with CO₂
CISCE examiners' comments, 2025: for the gas that decolourises potassium permanganate, candidates chose carbon dioxide.
Wrong working
CO₂ decolourises KMnO₄.
Right working
SO₂ decolourises acidified KMnO₄; CO₂ does not.
Do this: Both turn lime water milky; only SO₂ reacts with KMnO₄ and dichromate.
Usually costs: 1 mark.
Mistake: Oxidising versus reducing role
CISCE examiners' comments, 2025: candidates confused the oxidising and reducing roles of sulphuric acid in a given reaction.
Wrong working
Sulphuric acid acts as a reducing agent when it forms SO₂.
Right working
Forming SO₂ means the acid was reduced: it is the oxidising agent.
Do this: Reduced substance = oxidising agent.
Usually costs: 1 mark.

Practice questions on Sulphuric Acid

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
The catalyst in the Contact process is:
(a) iron
(b) platinum
(c) vanadium pentoxide
(d) nickel
Hint
V for Contact.
Answer
(c)
Q2 · 1 mark · MCQBoard level
Which reaction shows the dehydrating property of concentrated sulphuric acid?
(a) NaCl + H₂SO₄ → NaHSO₄ + HCl
(b) C₁₂H₂₂O₁₁ → 12C + 11H₂O
(c) C + 2H₂SO₄ → CO₂ + 2SO₂ + 2H₂O
(d) Zn + H₂SO₄ → ZnSO₄ + H₂
Hint
Water removed from a compound.
Answer
(b)
Q3 · 1 mark · MCQHarder board level
Statement I: Concentrated sulphuric acid can be used to dry ammonia.
Statement II: Concentrated sulphuric acid is a drying agent.
(a) Both true
(b) Both false
(c) I true, II false
(d) I false, II true
Hint
Acid and base react.
Answer
(d)
Q4 · 3 marks · Short answerBoard level
Write balanced equations for dilute sulphuric acid with (a) zinc, (b) copper(II) oxide, (c) sodium sulphite.
Hint
Salt + gas or water.
Answer
(a) $Z n + H_{2} S O_{4} \to Z n S O_{4} + H_{2}$ (b) $C u O + H_{2} S O_{4} \to C u S O_{4} + H_{2} O$ (c) $N a_{2} S O_{3} + H_{2} S O_{4} \to N a_{2} S O_{4} + H_{2} O + S O_{2}$
Q5 · 2 marks · Short answerHarder board level
Give the conditions (catalyst, temperature, pressure) for the catalytic oxidation of SO₂ and state whether the reaction is exothermic or endothermic.
Hint
Contact converter.
Answer
V₂O₅; about 450-500 °C; about 1-2 atm; exothermic.
Q6 · 3 marks · Short answerHarder board level
Give reasons: (a) Concentrated sulphuric acid is used to prepare nitric acid from potassium nitrate. (b) Water is never added to concentrated sulphuric acid. (c) Sulphuric acid forms two types of salts.
Hint
Non-volatile; exothermic; dibasic.
Answer
(a) It is non-volatile and displaces volatile HNO₃. (b) The mixing is highly exothermic; adding water to acid can make it boil and spurt. (c) It is dibasic: replacing one H gives an acid salt, both gives a normal salt.
Q7 · 2 marks · Short answerHarder board level
Calculate the volume of sulphur dioxide at S.T.P. formed when 3.2 g of sulphur reacts completely with hot concentrated sulphuric acid. (S = 32)
Hint
S : SO₂ = 1 : 3.
Answer
0.1 mol S → 0.3 mol SO₂ = 6.72 L.
Q8 · 2 marks · Short answerHarder board level
How would you distinguish (a) dilute H₂SO₄ from dilute HNO₃ using lead nitrate; (b) dilute H₂SO₄ from dilute HCl using barium chloride?
Hint
Sulphate precipitates.
Answer
(a) H₂SO₄: white precipitate of PbSO₄; HNO₃: no precipitate. (b) H₂SO₄: white precipitate of BaSO₄; HCl: no precipitate.
Q9 · 3 marks · Short answerHOTS / competency
34.2 g of sugar is completely charred by concentrated sulphuric acid. Calculate the mass of carbon formed. (C = 12, H = 1, O = 16)
Hint
M of sucrose = 342.
Answer
Moles = 34.2 ÷ 342 = 0.1; carbon = 0.1 × 12 × 12 = 14.4 g.
Q10 · 2 marks · Short answerHOTS / competency
A student says 'Concentrated sulphuric acid reacts with sodium chloride because it is a strong acid.' Correct the statement and write the equation.
Hint
Volatility, not strength.
Answer
It reacts because it is non-volatile, so it displaces the volatile HCl (which escapes as a gas): $N a C l + H_{2} S O_{4} \to N a H S O_{4} + H C l$ .
Q11 · 2 marks · Very short answerBoard level
Name: (a) the product when SO₃ is absorbed in conc. H₂SO₄; (b) the colour change of blue vitriol with conc. H₂SO₄; (c) a gas that decolourises acidified KMnO₄; (d) the acid salt of sulphuric acid with sodium.
Hint
Half mark each.
Answer
(a) Oleum (H₂S₂O₇) (b) Blue to white (c) Sulphur dioxide (d) Sodium hydrogen sulphate (NaHSO₄)

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.Formula of oleum?Show answer
H₂S₂O₇.
3.Basicity of sulphuric acid?Show answer
2 (dibasic).
4.Observation when conc. H₂SO₄ is added to sugar?Show answer
White sugar turns into a black spongy mass of carbon.
5.Gas formed when conc. H₂SO₄ oxidises carbon?Show answer
CO₂ and SO₂.
6.Test for sulphuric acid?Show answer
White precipitate with BaCl₂, insoluble in acids.
7.Why can H₂SO₄ displace HCl from NaCl?Show answer
It is non-volatile; HCl is volatile.
8.Which gas decolourises acidified KMnO₄?Show answer
Sulphur dioxide.

Best order to study it

Learn the Contact process equations and conditions.
Learn dilute acid reactions, especially hydrogen carbonate and hydrogen sulphite.
Learn the oxidising reactions with carbon and sulphur.
Learn dehydration (sugar, blue vitriol) and drying versus dehydrating.
Learn the non-volatile reactions and the tests.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you write all Contact process equations with V₂O₅, 450-500 °C, 1-2 atm?
Can you explain why oleum is made?
Can you show that sulphuric acid is dibasic with two equations?
Can you write C and S with conc. H₂SO₄, balanced?
Can you describe and write the dehydration of sugar and CuSO₄·5H₂O?
Can you identify the role of the acid from an equation?

Sulphuric Acid: frequently asked questions

Is 'sulphuric acid as a dibasic acid' in the ICSE 2027 syllabus?

Yes. The 2027 syllabus adds it explicitly: show two series of salts (hydrogen sulphates and sulphates).

Do I need the reaction of conc. H₂SO₄ with copper?

The 2027 syllabus names the oxidation of carbon and sulphur. Focus on those two.

What is the difference between a drying and a dehydrating agent?

A drying agent removes moisture (free water) without changing the substance; a dehydrating agent removes the elements of water (H and O in 2 : 1 ratio) from a compound, changing it chemically.

How much detail of the Contact plant is needed?

The syllabus asks for 'equations with conditions where applicable', so learn the equations, catalyst, temperature and pressure.

Based on: CISCE ICSE 2027 Chemistry (Science Paper 2) syllabus; CISCE ICSE 2027 Chemistry (Science Paper 2) specimen question paper and marking scheme; CISCE Analysis of Pupil Performance, ICSE Chemistry (Science Paper 2), 2025; CISCE Analysis of Pupil Performance, ICSE Chemistry (Science Paper 2), 2024