Final answer:
The value of the equilibrium constant, K, at 1100K is 0.5098. The equilibrium constant, denoted as K, is related to Kp by the equation Kp = K(RT)^(Δn).
Step-by-step explanation:
At 1100 K, the equilibrium constant, Kp, is given as 0.25 for the reaction 2SO₂(g) + O₂(g) ⇔ 2SO₃(g). The equilibrium constant, denoted as K, is related to Kp by the equation Kp = K(RT)^(Δn), where R is the gas constant, T is the temperature in kelvins, and Δn is the change in the number of moles of gas between reactants and products.
In this reaction, there are three moles of gas on the left side (2SO₂ + O₂) and two moles on the right side (2SO₃), resulting in Δn = 2 - 3 = -1. Since the reaction proceeds in the reverse direction (from right to left) to reach equilibrium, Δn is negative. As a result, K at 1100 K is lower than Kp, and its value is 0.25 divided by the term (RT)^(Δn), which is less than 1. Therefore, K at 1100 K is less than 0.25, reflecting the shift towards the reactants at this temperature.
The equilibrium constant, K, can be calculated using the formula:
K = (PSO₃)²/ (PSO₂)²(PO₂)
Given that Kp = 0.25 at 1100K, we can plug the values into the formula to find the value of K:
K = (0.25)^(1/2) = 0.5098