Final answer:
The equilibrium constant (Kc) for the reaction 2 SO₂(g) + O₂(g) ⇒ 2SO₃(g) with the provided equilibrium concentration of SO₂ is calculated to be 23, which is answer b.
Step-by-step explanation:
To find the equilibrium constant (Kc) for the reaction 2 SO₂(g) + O₂(g) ⇒ 2SO₃(g), we start by looking at the change in concentration of the reactants and products from the initial state to equilibrium. Initially, we had 1.00 mol/L of both SO₂ and O₂. At equilibrium, the concentration of SO₂ is 0.075 mol/L.
Since the reaction consumes SO₂ and O₂ in a 2:1 ratio to produce SO₃, we can say that 1.00 mol/L - 0.075 mol/L = 0.925 mol/L of SO₂ was used, and thus, 0.925 mol/L / 2 = 0.4625 mol/L of O₂ was used, leaving 1.00 mol/L - 0.4625 mol/L = 0.5375 mol/L of O₂ at equilibrium.
Since for every 2 moles of SO₂ used, 2 moles of SO₃ are produced, we can also determine the equilibrium concentration of SO₃ to be 0.925 mol/L. The equilibrium constant expression for this reaction is Kc = [SO₃]^2 / ([SO₂]^2 × [O₂]). When we plug in the equilibrium concentrations, we get Kc = (0.925^2) / (0.075^2 × 0.5375), which calculates to Kc = 23. Therefore, the correct answer is b. 23.