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 {Kc = [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 Kc = [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)
