Final answer:
The equilibrium constant (Ke) for the reaction 2 SO2(g) + O2(g) ⇒ 2 SO3(g), given the concentrations of reactants and products at equilibrium, is calculated to be 280.
Step-by-step explanation:
The given equilibrium reaction is 2 SO2(g) + O2(g) ⇒ 2 SO3(g).
To find the equilibrium constant (Ke), we use the equilibrium concentrations. Initially, 1.00 mol/L of SO2 and O2 were present, and at equilibrium, the concentration of SO2 was 0.075 mol/L.
Since 2 moles of SO2 produce 2 moles of SO3, and the reaction goes to completion by 0.925 mol/L (1.00 - 0.075), the amount of SO3 formed at equilibrium will be 0.925 mol/L.
The reduction in O2 will be half of that in SO2 since only 1 mole of O2 is needed for 2 moles of SO2, resulting in an O2 equilibrium concentration of 0.5375 mol/L (1.00 - 0.925/2).
The equilibrium constant expression is
Ke = \frac{[SO3]^2}{[SO2]^2[O2]},
which gives us Ke = \frac{(0.925)^2}{(0.075)^2 × 0.5375}.
After calculating, we find that Ke = 280.