For the exothermic formation of sulphur trioxide from sulphur dioxide and oxygen in gas phase
          

(i) write the expression for the equilibrium constant for the reaction;
(ii) at room temperature (300 K) will K_p be greater than, less than or equal to K_p at 900 K;
(iii) how will the equilibrium be affected if the volume of the vessel containing the three gases is reduced, keeping the temperature constant; what happens;
(iv) what is the effect of adding 1 mole He(g) to a flask containing SO₂,O₂ and SO₃ at equilibrium at constant temperature?

(i) The chemical equation for the reaction is
             
The expression for the equilibrium constant is

(ii) The forward reaction is exothermic in nature i.e. accompanied by the evolution of heat. By decreasing the temperature (from 900K to 300K), then according to Le-Chatelier’s principle, the equilibrium shifts towards the forward direction where the evolution of heat takes place in order to nullify the effect of a decrease in temperature. This means that partial pressure of SO₃(g) will increase while those of SO₂(g) and O₂(g) decreases. Hence K_p will increase by decreasing the temperature.
(iii) By decreasing the volume, the number of molecules per unit volume will increase. According to Le-Chatelier’s principle, the equilibrium shifts towards that side where the number of moles per unit volume decreases. Thus, the speed of forward reaction will increase. Therefore, the molar concentration of SO₃(g) will increase.
(iv) Helium is an inert gas and does not react with either the reactants or products. Therefore, the partial pressures of the gases will remain same on adding one mole of helium as the volume of the flask is fixed. Hence the equilibrium will remain unaffected.