Production of Sulfuric Acid (Oxford AQA IGCSE Chemistry)

The Contact Process

• Sulfuric acid is synthesised by the Contact process

Stage 1

• The first stage is the oxidation of sulfur to form sulfur dioxide:

S (s) + O₂ (g) → SO₂  (g)

sulfur  +  oxygen  →  sulfur dioxide

Stage 2

• The main stage is the oxidation of sulfur dioxide to sulfur trioxide using a vanadium(V) oxide, V₂O₅, catalyst:

2SO₂ (g) + O₂ (g)     2SO₃ (g) 

sulfur dioxide  +  oxygen    sulfur trioxide

• The oxygen used in this stage is obtained from air

• The conditions for this main stage of production are:

  • A temperature of approximately 450 ºC

  • Atmospheric pressure, 2 atm (200 kPa)

Stage 3

• Sulfur trioxide reacts with water to form sulfuric acid

 sulfur trioxide + water → sulfuric acid

2SO₃ (g)  + H₂O (l) → H₂SO₄ (aq)

Why are high temperature and atmospheric pressure used in stage 2?

• The conditions for the second stage of the Contact process have been chosen carefully

2SO₂ + O₂    2SO₃

Temperature: 450ºC

• The forward reaction is exothermic, so increasing the temperature shifts the position of the equilibrium to the left 

  • Therefore, a higher temperature gives a lower yield of sulfur trioxide

• A greater yield of sulfur trioxide would be achieved by using a lower temperature

  • The position of the equilibrium would shift to the right but at low temperatures the rate of reaction is low

• The optimum temperature is a compromise between a higher rate of reaction and a lower yield

Pressure: 2 atm

• An increase in pressure shifts the position of equilibrium to the right because there are fewer molecules of product formed

• However, the position of equilibrium lies far to the right (the equilibrium mixture contains about 96% sulfur trioxide)

• So the reaction is carried out at just above atmospheric pressure because:

  • It is not worth spending the extra energy or money required to produce high pressures

  • A higher pressure also causes the sulfur dioxide to liquefy

Vanadium(V) oxide catalyst

• A catalyst speeds up the rate of reaction without being used up

  • Catalysts do not alter the position of the equilibrium