Contact process for H₂SO₄ manufacture — flowchart with conditions at each step

Question

Describe the Contact process for manufacturing sulphuric acid ( $H_2SO_4$ ). What are the conditions at each step and why?

Solution — Step by Step

Sulphur (or iron pyrites) is burned in excess air:

\text{S} + \text{O}_2 \xrightarrow{} \text{SO}_2

Or from pyrites: $4\text{FeS}_2 + 11\text{O}_2 \to 2\text{Fe}_2\text{O}_3 + 8\text{SO}_2$

The $\text{SO}_2$ gas is purified (dust, arsenic compounds removed) because impurities poison the catalyst in the next step.

2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3 \quad \Delta H = -196 \text{ kJ/mol}

Conditions: $V_2O_5$ catalyst (vanadium pentoxide), 450-degree C, 1-2 atm pressure.

Why these specific conditions? The reaction is exothermic and reversible. By Le Chatelier’s principle:

Lower temperature favours forward reaction (exothermic), but too low makes it impractically slow. 450-degree C is the optimum compromise.
Higher pressure favours forward reaction (3 mol gas $\to$ 2 mol gas), but 1-2 atm is enough with the catalyst.
$V_2O_5$ replaced the older platinum catalyst because it is cheaper and less susceptible to poisoning.

$\text{SO}_3$ is absorbed in concentrated $H_2SO_4$ (98%) to form oleum (fuming sulphuric acid):

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

Then oleum is diluted with water to get the desired concentration:

\text{H}_2\text{S}_2\text{O}_7 + \text{H}_2\text{O} \to 2\text{H}_2\text{SO}_4

Why not dissolve $\text{SO}_3$ directly in water? Because $\text{SO}_3 + \text{H}_2\text{O}$ is highly exothermic and produces a dense, dangerous acid mist that is difficult to condense. Absorbing in $\text{H}_2\text{SO}_4$ avoids this.

graph TD
    A[Sulphur or FeS2] -->|Burn in air| B[SO2 gas]
    B -->|Purification| C[Clean SO2]
    C -->|V2O5, 450C, 1-2 atm| D[SO3]
    D -->|Absorb in conc. H2SO4| E[Oleum H2S2O7]
    E -->|Dilute with water| F[H2SO4 product]

Why This Works

The entire process is an application of Le Chatelier’s principle to an industrial equilibrium. The exothermic, gas-volume-reducing reaction ( $2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3$ ) is pushed forward by moderate temperature and a catalyst that speeds up equilibrium attainment without changing its position.

The absorption step is the cleverest part — by avoiding direct water contact with $\text{SO}_3$ , we get a controlled, safe process with near-100% absorption efficiency.

Alternative Method

For exam questions asking “why not dissolve $\text{SO}_3$ in water directly,” the answer has two parts:

The reaction $\text{SO}_3 + \text{H}_2\text{O} \to \text{H}_2\text{SO}_4$ is so exothermic that it produces acid mist
Acid mist is difficult to condense and causes pollution

This is a 1-mark CBSE favourite and appears in NEET as an assertion-reason question.

Common Mistake

Students often write that high pressure is used in the Contact process. The actual pressure is only 1-2 atm (near atmospheric). While high pressure would thermodynamically favour SO3 formation, it is not economically justified because the $V_2O_5$ catalyst already gives 97-98% conversion at 1-2 atm. Contrast this with the Haber process where 200 atm is needed — there the equilibrium conversion is much lower without pressure.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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