Contact process for H₂SO₄ — steps, conditions, and catalyst role

medium CBSE JEE-MAIN NEET NCERT Class 12 3 min read

Question

Describe the Contact process for the manufacture of sulphuric acid. Give the reaction at each step, the conditions used, and explain the role of V₂O₅ catalyst. Why is concentrated H₂SO₄ used in the absorption step instead of water?

(NCERT Class 12, p-Block Elements)


Solution — Step by Step

Sulphur or sulphide ores are burned in air:

S+O2SO2\text{S} + \text{O}_2 \rightarrow \text{SO}_2

or from pyrite: 4FeS2+11O22Fe2O3+8SO24\text{FeS}_2 + 11\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 + 8\text{SO}_2

The SO₂ gas is purified (dust removed, dried) before the next step.

2SO2+O2V2O5450°C,2 atm2SO3ΔH=196 kJ/mol2\text{SO}_2 + \text{O}_2 \xrightleftharpoons[\text{V}_2\text{O}_5]{450°\text{C}, 2 \text{ atm}} 2\text{SO}_3 \quad \Delta H = -196 \text{ kJ/mol}

Conditions: Temperature 400-450°C, pressure 1-2 atm, catalyst V₂O₅ (vanadium pentoxide).

This is the rate-determining step and the reason the process is called the “Contact” process — SO₂ and O₂ come into contact with the solid catalyst surface.

SO₃ is not dissolved directly in water. Instead, it is absorbed in concentrated H₂SO₄ to form oleum (fuming sulphuric acid):

SO3+H2SO4H2S2O7 (oleum)\text{SO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{S}_2\text{O}_7 \text{ (oleum)}

Oleum is then carefully diluted with water:

H2S2O7+H2O2H2SO4\text{H}_2\text{S}_2\text{O}_7 + \text{H}_2\text{O} \rightarrow 2\text{H}_2\text{SO}_4

If SO₃ is added directly to water, the reaction is so exothermic that it produces a fine mist of H₂SO₄ droplets. This acid mist is very difficult to condense and causes severe pollution. Using concentrated H₂SO₄ as the absorbing medium gives a controlled, liquid-phase reaction without mist formation.


Why This Works

The key step (SO₂ → SO₃) is an equilibrium reaction. Le Chatelier’s principle guides the choice of conditions:

  • Low temperature favours product (exothermic reaction), but too low = too slow. 450°C is the compromise — fast enough with V₂O₅ catalyst while giving ~97% conversion.
  • High pressure favours product (3 moles gas → 2 moles gas), but the yield at 1-2 atm is already good enough. Higher pressures would increase equipment costs without significant benefit.
  • Excess O₂ shifts equilibrium right — typically a 1:1 ratio of SO₂:air is used (giving excess O₂).

V₂O₅ works by temporarily oxidizing and reducing: SO₂ reduces V₂O₅ to V₂O₄, then O₂ re-oxidizes V₂O₄ back to V₂O₅. This redox cycle lowers the activation energy without being consumed.


Alternative Method

For a quick-recall framework, remember the three C’s of the Contact process: Combustion (make SO₂) → Conversion (SO₂ to SO₃ with catalyst) → Capture (absorb SO₃ in H₂SO₄).

CBSE and NEET frequently ask: “Why is SO₃ not dissolved directly in water?” This is a 2-mark question. The answer centres on acid mist formation. They also ask for the catalyst name (V₂O₅) and the temperature (450°C). These are direct-recall marks — don’t lose them.


Common Mistake

Students write that “high pressure is used in the Contact process” — many textbooks mention 2 atm, which is barely above atmospheric. This is NOT a high-pressure process (unlike the Haber process which uses 200 atm). The equilibrium is already favourable at low pressures because the conversion at 450°C with V₂O₅ is ~97%. Writing “high pressure” in a board exam can cost you marks.

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