(missing in your question): 0.66 mol of SO3 is placed in 4.5 L container according to this reaction 2SO3(g) ↔ 2SO2(g) + O2(g) So the answer: for the equilibrium reaction so: Kc = [SO2]^2 * [O2] / [ SO3]^2 when concentration (c) = n / V when n= no.of moles / v (volume per L) ∴ [ SO3]° = 0.66 / 4.5 = 0.147 M by using Ice table we can determine the concentrations: 2SO3(g) ↔ 2SO2(g) + O2(g) 0.147-2X 2x X ∴ [ O2] = 0.17 M / 4.5 L = 0.0378 M So X = 0.0378 ∴ [SO3] = 0.147 - 2 * 0.0378 = 0.0714 M and [ SO2] = 2* 0.0378 = 0.0756 M ∴ by substitution in Kc formula ∴Kc = [0.0756]^2 * [0.0378] / [0.0714] = 0.003
