A mixture of 2.0 moles of SO ₂ and 2.0 moles of O ₂ is placed in a closed vessel of volume 2.0 L and allowed to reach equilibrium according to the reaction: 2 SO ₂( g ) + O ₂( g ) 2 SO ₃( g ) At equilibrium, 1.0 mole of SO ₃ is formed. The value of the equilibrium constant K_ c for this reaction is:

Step-by-step solution

Set up an ICE (Initial, Change, Equilibrium) table for the moles of each species: Initial moles: SO ₂ = 2.0 , O ₂ = 2.0 , SO ₃ = 0 Since 1.0 mole of SO ₃ is formed at equilibrium, the change for SO ₃ is +1.0 mol . From the stoichiometry of the reaction ( 2:1:2 ), the change in SO ₂ is -1.0 mol and the change in O ₂ is -0.5 mol . Equilibrium moles: SO ₂ = 2.0 - 1.0 = 1.0 mol O ₂ = 2.0 - 0.5 = 1.5 mol SO ₃ = 1.0 mol Convert moles to equilibrium concentrations by dividing by the volume of the vessel ( 2.0 L ): [ SO ₂]