At 1000 K, 𝐾_𝑝 = 1.85 for the reaction SO₂(𝑔) + 1/2 O₂(𝑔) ⇌ SO₃(𝑔) (a) What is the value of K_p for the reaction SO₃(𝑔) ⇌ SO₂(𝑔) + 1/2 O₂(𝑔)? (b) What is the value of K_p for the reaction 2 SO₂(𝑔) + O₂(𝑔) ⇌ 2 SO₃(𝑔)?

Verified step by step guidance

1. To find the value of Kp for the reverse reaction SO3(g) ⇌ SO2(g) + 1/2 O2(g), we need to take the reciprocal of the given Kp for the forward reaction. This is because the equilibrium constant for a reverse reaction is the inverse of the equilibrium constant for the forward reaction.
2. For part (b), we need to determine the Kp for the reaction 2 SO2(g) + O2(g) ⇌ 2 SO3(g). This reaction is a multiple of the original reaction, specifically, it is doubled. When a reaction is multiplied by a factor, the equilibrium constant is raised to the power of that factor.
3. Since the original reaction is SO2(g) + 1/2 O2(g) ⇌ SO3(g) with Kp = 1.85, and the new reaction is doubled, we raise the original Kp to the power of 2.
4. Thus, the Kp for the reaction 2 SO2(g) + O2(g) ⇌ 2 SO3(g) is (Kp_original)^2.
5. Remember, when dealing with equilibrium constants, the direction and stoichiometry of the reaction significantly affect the value of Kp. Always consider these factors when manipulating reactions.