Consider the equilibrium \(\mathrm{Na}_{2} \mathrm{O}(s)+\mathrm{SO}_{2}(g) \rightleftharpoons\) \(\mathrm{Na}_{2} \mathrm{SO}_{3}(s) .(\mathbf{a})\) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) All the compounds in this reaction are soluble in water. Rewrite the equilibrium-constant expression in terms of molarities for the aqueous reaction.

We are given the following equilibrium reaction: \(\mathrm{Na}_2\mathrm{O}(s) + \mathrm{SO}_2(g) \rightleftharpoons \mathrm{Na}_2\mathrm{SO}_3(s)\) To write the equilibrium constant expression in terms of partial pressures, we use the following formula: \(K_p = \cfrac{{P_{\mathrm{Na}_2\mathrm{SO}_3}}^{\mathrm{c}}}{{P_{\mathrm{Na}_2\mathrm{O}}}^{\mathrm{a}} \cdot {P_{\mathrm{SO}_2}}^{\mathrm{b}}}\) where a, b, and c are the stoichiometric coefficients of Na2O, SO2, and Na2SO3 in the balanced reaction. Since Na2O and Na2SO3 are solids, their partial pressures do not affect the equilibrium constant. Thus, the expression simplifies to: \(K_p = P_{\mathrm{SO}_2}\)

As all compounds in the reaction are soluble in water, we write the reaction in its ionic form with aqueous ions: \(\mathrm{2Na}^{+}(aq) + \mathrm{SO}_2(g) + \mathrm{O}^{2-}(aq) \rightleftharpoons \mathrm{2Na}^{+}(aq) + \mathrm{SO}_3^{2-}(aq)\) Now, we write the equilibrium constant expression in terms of molarities using the formula: \(K_c = \cfrac{\{[M^{c}][M^{d}]...\}}{\{[M^{a}][M^{b}]...\}}\) Here, M represents the molar concentration of the species, and a, b, c, d, ... are stoichiometric coefficients. In this reaction, the stoichiometric coefficients are all equal to 1, so the expression simplifies to: \(K_c = \cfrac{[\mathrm{SO}_3^{2-}]}{[\mathrm{SO}_2] \cdot [\mathrm{O}^{2-}]}\) This is the equilibrium constant expression for the given reaction in terms of molarities when all compounds are dissolved in water.