The reaction is depicted as: SO2+½ O2 <-------> SO3. (all species are in gaseous state ) We have initially : SO2= 5 mole O2 = 5 mole SO3= 0 mole. At equilibrium : SO2= 5- 5(0.6) = 2 mole O2 = 5- 5 (0.3) = 3.5 mole SO3 = 5(0.6) = 3 mole. Now , total moles at equilibrium = 8.5 mole. Hence mole fraction of O2 = 3.5/8.5. Total pressure at equilibrium = 1 atm. Hence Partial pressure due to O2 at equilibrium will be = (1atm)×( 3.5/8.5) = 0.41 atm. It was difficult to type it out but I hope you understood