Sulphuric Acid (H2SO4)
Preparation, Properties, Uses, Structure and PYQs
1. Preparation
Sulphuric acid is primarily prepared by the Contact Process on an industrial scale. The process involves the following steps:
- Step 1: Burning of Sulphur or Pyrites
S + O2 → SO2 (Sulphur dioxide)
Or, 4FeS2 + 11O2 → 2Fe2O3 + 8SO2 - Step 2: Oxidation of SO2 to SO3
2SO2 + O2 ⇌ 2SO3 (Catalyst: V2O5 at 450-500°C, exothermic reaction)
Conditions: Low temperature (450°C) and high pressure favor forward reaction (Le Chatelier's principle). - Step 3: Absorption of SO3
SO3 is absorbed in 98% H2SO4 to form oleum (H2S2O7).
SO3 + H2SO4 → H2S2O7 (Oleum or Pyrosulphuric acid) - Step 4: Dilution of Oleum
H2S2O7 + H2O → 2H2SO4
Because the reaction is highly exothermic, leading to the formation of a corrosive sulfuric acid mist that is difficult to condense. Hence absorbed in 98% H2SO4 first.
2. Properties
Physical Properties
- Colorless, odorless, viscous (oily) liquid.
- Molecular weight: 98 g/mol.
- Density: 1.84 g/cm³ (concentrated).
- Boiling point: 337°C; Melting point: 10°C.
- Highly hygroscopic (absorbs moisture) and miscible with water in all proportions (exothermic dissolution).
Chemical Properties
- Acidic Nature: Strong dibasic acid.
H2SO4 ⇌ H⁺ + HSO4⁻ (First ionization strong)
HSO4⁻ ⇌ H⁺ + SO4²⁻ (Second ionization weak)
Reacts with bases: 2NaOH + H2SO4 → Na2SO4 + 2H2O
With metals: Zn + H2SO4 → ZnSO4 + H2 (dilute) - Dehydrating Agent: Removes water from compounds.
C12H22O11 (sucrose) → 12C + 11H2O (charred sugar)
Used to dry gases like HCl, SO2. - Oxidizing Agent: Hot concentrated H2SO4 acts as oxidant.
Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O
C + 2H2SO4 → CO2 + 2SO2 + 2H2O - Reaction with Salts: Displaces volatile acids.
NaCl + H2SO4 → NaHSO4 + HCl
3. Uses
- Fertilizers: Manufacture of superphosphate (Ca(H2PO4)2) and ammonium sulphate ((NH4)2SO4).
- Petroleum Refining: Removal of impurities.
- Detergents and Soaps: Sulphonation reactions.
- Pigments and Dyes: Production of TiO₂, paints.
- Explosives: Nitration for TNT, etc.
- Batteries: Electrolyte in lead-acid batteries.
- Metallurgy: Pickling of metals (removing oxides).
- Laboratory Reagent: Dehydrating agent, acid in titrations.
4. Structure of Sulphuric Acid
- Molecular formula: H2SO4
- Central atom: Sulphur (S) in +6 oxidation state
- Hybridization of S: sp³
- Geometry around S: Tetrahedral
- Sulphur is bonded to:
- Two oxygen atoms by double bonds (S=O)
- Two hydroxyl groups (–OH) by single bonds
- Bond lengths:
Bond Length (pm) S=O 142 pm S–OH 157 pm - Actual structure shows resonance: All four S–O bonds have double bond character due to delocalization (pπ–dπ bonding).
- Two O–H bonds are responsible for acidic nature (easily donate H⁺).
• Structure → Tetrahedral, sp³, +6 oxidation state of S
• Catalyst in Contact process → V2O5 (not Pt now)
• Temperature compromise in Contact process (not too low to kill rate)
• Difference in reactions of dilute vs conc. H2SO4
• Dehydrating action on sugar, formic acid, oxalic acid
• Oxidising action products with Cu, C, S, P, etc.
Sulphuric Acid Previous Year Questions
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In the Contact process, the catalyst used for the conversion of SO2 to SO3 is:
(A) V2O5 (B) Fe2O3 (C) CuCl2 (D) Ni
View Detailed Answer
Answer: (A) V2O5Vanadium pentoxide (V2O5) is the commercially used catalyst (1–2% on silica support) at 450–500 °C. It works via redox cycle (V⁵⁺ ⇌ V⁴⁺). Earlier platinum was used but it is costly and easily poisoned by arsenic impurities.
Sulphuric acid is manufactured by Contact process. Which of the following statements is incorrect?
(A) SO3 is absorbed in conc. H2SO4 to form oleum
(B) Oleum is diluted with water to get H2SO4
(C) SO3 is directly dissolved in water
(D) Catalyst used is V2O5
View Detailed Answer
Answer: (C)SO3 is never directly absorbed in water because it forms a dense fog/mist of tiny H2SO4 droplets which does not condense easily. Hence SO3 + H2SO4 → H2S2O7 (oleum), then oleum is diluted carefully with water to get 98% H2SO4.
The oxidation state of sulphur in H2SO4 and SO3 respectively are:
(A) +6, +6 (B) +4, +6 (C) +6, +4 (D) +4, +4
View Detailed Answer
Answer: (A)Calculation:
H2SO4 → 2(+1) + x + 4(−2) = 0 ⇒ x = +6
SO3 → x + 3(−2) = 0 ⇒ x = +6
Sulphur exhibits its highest oxidation state (+6) in both.
The geometry and hybridization of sulphur in H2SO4 molecule are:
(A) Tetrahedral, sp³ (B) Trigonal planar, sp² (C) Square planar, dsp² (D) Octahedral, d²sp³
View Detailed Answer
Answer: (A) Tetrahedral, sp³Sulphur is surrounded by four oxygen atoms (two S=O and two S–OH). Electron pairs = 4 → sp³ hybridization → tetrahedral geometry. Bond angle ≈ 109.5° (slightly distorted due to double bonds).
Hot concentrated H2SO4 reacts with carbon according to the reaction:
C + 2H2SO4 → CO2 + 2SO2 + 2H2O
The change in oxidation number of sulphur in this reaction is:
View Detailed Answer
Answer: From +6 to +4 (decrease by 2)In H2SO4, S = +6
In SO2, S = +4
Each H2SO4 molecule gains 2 electrons (reduction). Carbon is oxidized from 0 to +4 (oxidation). Hence hot conc. H2SO4 acts as an oxidising agent.
Which of the following is formed when conc. H2SO4 is treated with P4O10?
(A) H4P2O7 (B) HPO3 (C) H3PO4 (D) SO3
View Detailed Answer
Answer: (D) SO3Conc. H2SO4 acts as a strong dehydrating agent:
P4O10 + 6H2SO4(conc, hot) → 4H3PO4 + 6SO3↑
Commonly asked dehydration reaction.
Oil of vitriol is:
(A) HNO3 (B) H2SO4 (C) HCl (D) H3PO4
View Detailed Answer
Answer: (B) H2SO4Concentrated sulphuric acid is highly viscous and oily in appearance, hence historically called “oil of vitriol”.
The correct order of acid strength of H2SO4, H2SeO4 and H2TeO4 is:
View Detailed Answer
Answer: H2SO4 > H2SeO4 > H2TeO4Acid strength of oxyacids increases with electronegativity of central atom. Electronegativity order: S > Se > Te ⇒ H₂SO₄ is the strongest.
In the Contact process, the temperature is maintained around 450 °C because:
(A) Reaction is endothermic
(B) Higher temperature deactivates catalyst
(C) It is a compromise between favourable equilibrium and reasonable rate
(D) SO3 decomposes above 500 °C
View Detailed Answer
Answer: (C)2SO2 + OV ⇌ 2SO3 ΔH = −196 kJ/mol (exothermic)
Low temperature favours equilibrium but rate becomes too slow. Hence 450–500 °C is the optimum compromise for good yield and sufficient rate.
The product obtained when copper reacts with cold dilute H2SO4 is:
(A) CuSO4 (B) CuS (C) No reaction (D) SO2
View Detailed Answer
Answer: (C) No reactionCold dilute H2SO4 does not oxidise copper (Cu lies below hydrogen in electrochemical series). Copper reacts only with hot concentrated H2SO4 to give SO2.
Concentrated H2SO4 cannot be used for drying H2 because:
(A) It oxidises H₂ (B) It decomposes H₂ (C) It reacts with H₂ (D) It absorbs H2
Answer & Explanation
Answer: (C)H2 + H2SO4(conc, hot) → SO2 + 2H2O (H2 is oxidised to H2O)
Hence conc. H2SO4 cannot be used for drying hydrogen (it reacts instead of just absorbing moisture).
Hot concentrated sulphuric acid is a moderately strong oxidising agent. Which of the following reactions does not show oxidising behaviour?
(A) Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O
(B) C + 2H2SO4 → CO2 + 2SO2 + 2H2O
(C) BaCl2 + H2SO4 → BaSO4 + 2HCl
(D) None of the above
Answer & Explanation
Answer: (C)Reaction (C) is simple double displacement (no change in oxidation state). Reactions (A) & (B) show reduction of S from +6 → +4 (oxidising nature).
The correct statement(s) about H2SO4 is/are:
(I) It acts as a dehydrating agent
(II) It acts as an oxidising agent
(III) It behaves as a reducing agent
(IV) It acts as a Lewis acid
(A) I, II (B) I, II, IV (C) II, III, IV (D) All
Answer & Explanation
Answer: (B) I, II, IVH2SO4 never acts as reducing agent (S is already in +6 state). It is an excellent dehydrating, oxidising (hot conc.) and Lewis acid (SO3 part accepts electrons).
The product obtained when formic acid is treated with conc. H2SO4 is:
(A) CO (B) CO2 (C) HCHO (D) CH3COOH
Answer & Explanation
Answer: (A) COHCOOH → CO + H2O (conc. H2SO4 dehydrates formic acid to carbon monoxide)
Oxalic acid gives CO + CO2 + H2O.
Among the following, the species in which sulphur has the highest oxidation state is:
(A) H2SO5 (B) SO3 (C) H2SO4 (D) SO2Cl2
Answer & Explanation
Answer: (A) H2SO5 (Peroxomonosulphuric acid/Carot’s acid)Oxidation state: H2SO5 → +8 (peroxide linkage), H2SO5/SO3/SO2Cl2 → +6.
The reaction of conc. H2SO4 with NaBr produces:
(A) Br2 (B) HBr (C) SO2 (D) Na2SO4 only
Answer & Explanation
Answer: (A) Br22NaBr + 3H2SO4 → 2NaHSO4 + Br2 + SO2 + 2H2O
Conc. H2SO4 oxidises bromide to bromine (unlike chloride, which gives only HCl).
Sulphur dioxide is obtained when conc. H2SO4 reacts with:
(A) Cu (B) Hg (C) Both (A) and (B) (D) None
Answer & Explanation
Answer: (C) BothCu + 2H2SO4 → CuSO4 + SO2 + 2H2O
Hg + 2H2SO4 → HgSO4 + SO2 + 2H2O
Classic oxidising action of hot conc. H2SO4.
Which of the following is known as oil of vitriol?
(A) H3PO4 (B) H2SO3 (C) H2SO4 (D) H2S2O7
Answer & Explanation
Answer: (C) H2SO4Historical name due to its oily viscous nature.
In the structure of H2SO4, the total number of σ and π bonds are respectively:
Answer & Explanation
Answer: 6σ and 4πStructure: S is bonded to 4 oxygen atoms → 4 S–O σ bonds + 2 O–H σ bonds = 6 σ bonds
Two S=O contributes 2 π bonds each → 2 × 2 = 4 π bonds (due to resonance).
The catalyst used in the Contact process is poisoned by:
(A) As2O3 (B) P2O5 (C) CO2 (D) N2
Answer & Explanation
Answer: (A) As2O3Arsenic compounds irreversibly poison V2O5 catalyst; that's why sulphur or pyrite must be purified before burning.
Which of the following is incorrect statement for Sulphuric acid?
(A) Higher viscosity than water. (B) Solutes dissolve slowly in it. (C) High vapour pressure. (D) Difficult to remove from the crystallised materials.
Answer & Explanation
Answer: (C) High vapour pressure.Sulphuric acid exhibits strong intermolecular hydrogen bonding. These strong forces mean that a lot of energy is required for its molecules to escape into the gas phase, resulting in a low vapour pressure and high boiling point.