Sulfur trioxide is produced in the Contact process according to the following reversible reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) At a certain temperature, the equilibrium constant expression for this reaction is: Kc = [SO₃]² / ([SO₂]² × [O₂]) Explain what the value of Kc tells us about the position of equilibrium, and describe how the value of Kc changes when (i) the temperature is increased and (ii) a catalyst is added.

A large value of Kc indicates that, at equilibrium, the concentration of SO₃ is high compared with that of SO₂ and O₂, so the equilibrium lies to the right and products are favoured.
A small value of Kc indicates the opposite – the equilibrium lies to the left and reactants are favoured.
(i) Because the forward reaction is exothermic, increasing the temperature supplies heat to the system. The equilibrium shifts to the endothermic side (the reverse reaction) to absorb the added heat, so the concentration of SO₃ falls relative to SO₂ and O₂. Consequently Kc decreases.
(ii) A catalyst provides an alternative pathway with a lower activation energy, increasing the rate of both the forward and reverse reactions equally. It does not affect the relative concentrations of products and reactants at equilibrium, so Kc remains unchanged.