Final answer:
The equilibrium constant for the reaction 2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g) is 2.4 x 10⁻³. For the reactions (b) 2 SO₂(g) ⇄ 2 SO₂(g) + O₂(g) and (c) SO₂(g) ⇄ SO₂(g) + 1/2 O₂(g), the equilibrium constants are 1 / 2.4 x 10⁻³ and 2.4 x 10⁻³, respectively.
Step-by-step explanation:
The equilibrium constant, Kc, for the reaction 2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g) is given as 2.4 x 10⁻³ at a given temperature.
To determine the equilibrium constant for the reactions (b) 2 SO₂(g) ⇄ 2 SO₂(g) + O₂(g) and (c) SO₂(g) ⇄ SO₂(g) + 1/2 O₂(g), we can use the stoichiometry of the reactions.
(b) In this reaction, there is only one reactant on the right side, while there are two on the left side. Therefore, the equilibrium constant can be expressed as the reciprocal of the original constant, Kc = 1 / 2.4 x 10⁻³.
(c) In this reaction, there is no change in the number of moles of gas particles. Therefore, the equilibrium constant remains the same as the original constant, Kc = 2.4 x 10⁻³.