Sulfuric acid is manufactured by the Contact process, outlined in Fig. 14.1. Fig. 14.1 (a) Write balanced equations for the three main reaction stages: burn...

This question tests the Contact process for making sulfuric acid: the three reaction stages, the conditions and their reasons, and a mass calculation.

(a) The three main stages, each balanced [1] each:

Burning sulfur: \( \text{S} + \text{O}_2 \rightarrow \text{SO}_2 \)

Oxidation of sulfur dioxide: \( 2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3 \)

Formation of oleum: \( \text{SO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{S}_2\text{O}_7 \)

(b) For the SO₂ oxidation stage the temperature is about 450°C [1]: the forward reaction is exothermic, so a lower temperature would give a higher yield, but 450°C keeps the rate fast enough, making it a compromise [1]. The pressure is only about 2 atm [1]: the yield is already about 98% at low pressure, so a higher, more expensive pressure is not worthwhile [1].

(c) The catalyst is vanadium(V) oxide, V₂O₅ [1].

(d) Adding SO₃ straight into water produces a large amount of heat and a dangerous, choking mist (fog) of sulfuric acid that is hard to condense [1]. Dissolving SO₃ in concentrated sulfuric acid to form oleum, then carefully adding water, is safe and controlled [1].

(e) Work in tonne-moles. The relative formula masses are SO₂ = 32 + (2 × 16) = 64 and H₂SO₄ = (2 × 1) + 32 + (4 × 16) = 98.

The stages carry through one mole of sulfur atoms, so 1 mol SO₂ gives 1 mol H₂SO₄ [1]. Using M(H₂SO₄) = 98 [1]:

(f) Any two uses of sulfuric acid: making fertilisers, detergents, paints, use as car battery acid, or making other chemicals [1][1].

Exam reminder: in tonne-mole calculations you can keep working in tonnes as long as you divide by the relative formula mass; the mole ratio from the balanced equations is what links reactant to product.