Contact Process — H₂SO₄ Manufacturing Mnemonic
Target: mnemonic for contact process sulphuric acid H2SO4
Why is this hard to memorize?
The Contact process manufactures sulphuric acid (H₂SO₄) in three steps: (1) S + O₂ → SO₂ (burning sulphur), (2) 2SO₂ + O₂ ⇌ 2SO₃ (key step — V₂O₅ catalyst, 450°C, 2 atm), (3) SO₃ + H₂SO₄ → H₂S₂O₇ (oleum), then H₂S₂O₇ + H₂O → 2H₂SO₄. Note: SO₃ is NOT dissolved directly in water (violent, forms mist), but in concentrated H₂SO₄ first. NEET and JEE test the three steps, conditions for step 2, and why SO₃ is dissolved in H₂SO₄ rather than water.
Classic mnemonics you should know
"Step 1: S→SO₂ (burn). Step 2: SO₂→SO₃ (V₂O₅, 450°C, 2atm — the "contact" step). Step 3: SO₃→oleum→H₂SO₄"
Step 1 produces SO₂ from sulphur or FeS₂ (pyrite). Step 2 is the equilibrium step — V₂O₅ catalyst speeds up conversion of SO₂ to SO₃. Step 3: SO₃ is absorbed in concentrated H₂SO₄ to form oleum (H₂S₂O₇), then diluted with water.
"SO₃ + H₂O → violent reaction, forms mist of tiny H₂SO₄ droplets (hard to condense, escapes, pollutes). So: dissolve in H₂SO₄ first → oleum → then add water slowly."
The reaction of SO₃ with water is highly exothermic — the heat vaporizes water and creates a fine mist of H₂SO₄ that is nearly impossible to collect. Using concentrated H₂SO₄ as the absorbing medium avoids this problem. The oleum (fuming sulphuric acid) is then safely diluted.
"2SO₂ + O₂ ⇌ 2SO₃, ΔH = −198 kJ. Low T favours product. High P favours product (3→2 moles). But moderate conditions used for rate."
Like Haber, this is a compromise: 450°C (not lower because reaction too slow without reasonable kinetics) + V₂O₅ catalyst + 2 atm (moderate pressure — higher isn't needed because yield is already ~97% with catalyst). Excess O₂ also shifts equilibrium right.
The complete list
- Step 1: S + O₂ → SO₂
- Step 2: 2SO₂ + O₂ ⇌ 2SO₃ (V₂O₅, 450°C)
- Step 3: SO₃ + H₂SO₄ → H₂S₂O₇ (oleum)
- Oleum + H₂O → 2H₂SO₄
- Catalyst: V₂O₅ (vanadium pentoxide)
- SO₃ NOT added to water directly (mist)
- Yield: ~97% conversion in step 2
- ΔH = −198 kJ/mol (exothermic)
Frequently asked questions
Why is V₂O₅ used as the catalyst instead of Pt?
Historically, platinum (Pt) was used as the catalyst but it is (1) extremely expensive, (2) easily poisoned by arsenic impurities in the SO₂ feed gas. Vanadium pentoxide (V₂O₅) is much cheaper and more resistant to poisoning. It gives comparable conversion rates at 450°C. The catalyst works by the Mars-van Krevelen mechanism: V₂O₅ oxidizes SO₂ to SO₃ (becoming V₂O₄), then O₂ re-oxidizes V₂O₄ back to V₂O₅.
What is oleum and why is it formed?
Oleum (fuming sulphuric acid) is H₂S₂O₇ (pyrosulphuric acid), formed by dissolving SO₃ in concentrated H₂SO₄. It is an intermediate step because adding SO₃ directly to water creates an uncontrollable mist. Oleum can be safely diluted with water: H₂S₂O₇ + H₂O → 2H₂SO₄. Oleum itself is also commercially useful — it's a stronger sulfonating agent than H₂SO₄.
Why is excess oxygen used in step 2?
The equilibrium 2SO₂ + O₂ ⇌ 2SO₃ can be shifted right by increasing O₂ concentration (Le Chatelier — adding reactant). Using excess air (oxygen) pushes conversion above 97%. This is cheaper than using higher pressure. The unreacted gases are recycled. In practice, the SO₂:O₂ ratio is about 1:1 (excess O₂ over the stoichiometric 2:1 ratio).
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