At a particular temperature, 12.0 moles of \(\mathrm{SO}_{3}\) is placed into a 3.0-L rigid container, and the \(\mathrm{SO}_{3}\) dissociates by the reaction $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$. At equilibrium, 3.0 moles of \(\mathrm{SO}_{2}\) is present. Calculate \(K\) for this reaction.

At the beginning, we have 12.0 moles of \(\mathrm{SO}_3\) and 0 moles of \(\mathrm{SO}_2\) and \(\mathrm{O}_2\). When the reaction reaches equilibrium, we know there are 3.0 moles of \(\mathrm{SO}_2\). Since the stoichiometry of the reaction is \(2\mathrm{SO}_3 \rightleftharpoons 2\mathrm{SO}_2 + 1\mathrm{O}_2\), let's find out how many moles of \(\mathrm{SO}_3\) have reacted and also the moles of \(\mathrm{O}_2\) at equilibrium: 1. Moles of \(\mathrm{SO}_{2}\) formed = 3.0 moles 2. Moles of \(\mathrm{SO}_{3}\) reacted = Moles of \(\mathrm{SO}_{2}\) formed = 3.0 moles 3. Moles of \(\mathrm{SO}_{3}\) remaining = 12.0 (initial) - 3.0 (reacted) = 9.0 moles 4. Moles of \(\mathrm{O}_{2}\) formed = 0.5 × Moles of \(\mathrm{SO}_{2}\) formed = 0.5 × 3.0 = 1.5 moles

Now, we can convert moles to concentrations by dividing with the volume of the container (3.0 liters). Initial concentrations: 1. [\(\mathrm{SO}_3\)] = 12.0 moles / 3.0 L = 4.0 M 2. [\(\mathrm{SO}_2\)] = 0 moles / 3.0 L = 0 M 3. [\(\mathrm{O}_2\)] = 0 moles / 3.0 L = 0 M Equilibrium concentrations: 1. [\(\mathrm{SO}_3\)] = 9.0 moles / 3.0 L = 3.0 M 2. [\(\mathrm{SO}_2\)] = 3.0 moles / 3.0 L = 1.0 M 3. [\(\mathrm{O}_2\)] = 1.5 moles / 3.0 L = 0.5 M

Knowing the equilibrium concentrations, we can use the reaction quotient formula: $$K = \frac{[\mathrm{SO}_{2}]^2[\mathrm{O}_{2}]}{[\mathrm{SO}_{3}]^2}$$ Substitute the equilibrium concentrations: $$K = \frac{(1.0 \,\mathrm{M})^2 (0.5\, \mathrm{M})}{(3.0\, \mathrm{M})^2}$$ Simplify the expression: $$K = \frac{0.5}{9}$$ Calculate the value: $$K = 0.056$$ The equilibrium constant for the given reaction is 0.056.