0.04 moles of sulfur trioxide is placed in a flask (1.50dm^3) and allowed to reach equilibrium at 600 degrees. If 30% of the sulfur trioxide decomposes to sulfur dioxide and oxygen - what is the equilibrium constant?

The easiest way to start your answer is by writing out the equation for the reaction {2SO₃ (g) <-> 2SO₂ (g) + O₂ (g) } and considering what the equilibrium constant is {K_c = [products] / [reactants] } - where the concentrations shown are the equilibrium constants and stoichiometries of the reactants and products need to be taken into account. Then construct a table with row titles; "initial moles", "change in moles", "moles at equilibrium" and "concentration at equilibrium" and column headings of the different reactant and product species. SO₃: Initial moles = 0.04, Change in moles = -0.3 x 0.04 = -0.012 (because only 30% of the SO₃ decomposes), Moles at eqm. = 0.04 - 0.012 = 0.028, Conc. at eqm. (mol dm^-3) = 0.028 mol / 1.5 dm³ = 0.0187 mol dm^-3 SO₂: Initial moles = 0, Change in moles = + 0.012, Moles at eqm. = 0.012, Conc. at eqm. (mol dm^-3) = 0.012 / 1.5 = 0.008 mol dm^-3 O₂: Initial moles = 0, Change in moles = 1/2 x + 0.012 = +0.006 (remember stoichiometries in the equation), Moles at eqm. = 0.006, Conc. at eqm. (mol dm^-3) = 0.006 / 1.5 = 0.004 mol dm^-3 Therefore K_c = [SO₂]² [O₂] / [SO₃]² = (0.008 mol dm^-3)²(0.004 mol dm^-3) / (0.0187 mol dm^-3)² = 7.32 x 10^-4 mol dm^-3 (always check your answer and check that the units are right as well)