Question

Consider the equilibrium \(\mathrm{Na}_{2} \mathrm{O}(s)+\mathrm{SO}_{2}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{SO}_{3}(s)\). (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 equilibriumconstant expression in terms of molarities for the aqueous reaction

Short answer

(a) In terms of partial pressures, the equilibrium constant expression for the reaction \(\mathrm{Na}_{2} \mathrm{O}(s)+\mathrm{SO}_{2}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{SO}_{3}(s)\) is \(K_p = \dfrac{1}{P_{\mathrm{SO}_{2}}}\). (b) For the aqueous reaction, the equilibrium constant expression in terms of molarities is \(K_c = \dfrac{[\mathrm{SO}_3^{2-}]}{[\mathrm{O}^{2-}][\mathrm{SO}_2(g)]}\).

Step-by-step solution

Write the balanced chemical equation for the given reaction.

The balanced chemical equation for the given reaction is already provided: \(\mathrm{Na}_{2} \mathrm{O}(s)+\mathrm{SO}_{2}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{SO}_{3}(s)\)

Write the equilibrium constant expression in terms of partial pressures.

In this step, we need to write the equilibrium constant expression for the reaction in terms of partial pressures (K_p). Only gases are included in the expression, so in this case, the expression will only include the partial pressure of SO2. \(K_p =\dfrac{P_{\mathrm{Na}_{2}\mathrm{SO}_{3}}}{P_{\mathrm{SO}_{2}}}\) However, since Na2O and Na2SO3 compounds are solids, they are not included in the K_p expression. Thus, the final expression becomes: \(K_p = \dfrac{1}{P_{\mathrm{SO}_{2}}}\)

Write the equilibrium constant expression in terms of molarities.

Now, we need to rewrite the equilibrium constant expression in terms of molarities, considering the aqueous reaction (K_c). Solid and liquid reagents are not included in the expression. In an aqueous medium, the reaction can be written as follows: \(\mathrm{2Na^+}(aq)+\mathrm{O^{2-}}(aq)+\mathrm{SO}_2(g) \rightleftharpoons \mathrm{2Na^+}(aq)+\mathrm{SO_3^{2-}}(aq)\) In this case, Na2O and Na2SO3 are soluble in water, transforming into Na+ and O2- ions and Na+ and SO3 2- ions, respectively. The equilibrium constant expression for the aqueous reaction is: \(K_c = \dfrac{[\mathrm{Na}^{+}]^{2} [\mathrm{SO}_3^{2-}]}{[\mathrm{Na}^{+}]^{2} [\mathrm{O}^{2-}][\mathrm{SO}_2(g)]}\) Considering that [Na+] is present in both the numerator and the denominator, we can simplify the equation by canceling the [Na+]^2 term: \(K_c = \dfrac{[\mathrm{SO}_3^{2-}]}{[\mathrm{O}^{2-}][\mathrm{SO}_2(g)]}\)

Key concepts

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, meaning that the concentrations of reactants and products remain constant over time, though not necessarily equal. It is a dynamic state where the reactions are still occurring, but no net change is observed.

When considering chemical equilibrium, it's critical to understand that it doesn't signify that the reactants and products are at equal concentrations. Instead, we are describing a balance of reaction rates. A common misconception is that equilibrium indicates a static or 'frozen' situation, but in reality, the process remains active, continuously oscillating between the reactant and product states.

In the context of the exercise provided, \(\text{Na}_2 \text{O}(s) + \text{SO}_2(g) \rightleftharpoons \text{Na}_2 \text{SO}_3(s)\), we reach equilibrium when the rate at which the sodium oxide (a solid) and sulfur dioxide (a gas) react to form sodium sulfite (another solid) equals the rate at which the reverse reaction occurs, turning sodium sulfite back into sodium oxide and sulfur dioxide. The constant nature of this equilibrium can be represented by the equilibrium constant (\(\text{K}\)).

Partial Pressures

When dealing with gases, the equilibrium constant can also be expressed in terms of partial pressures. The partial pressure of a gas in a mixture is the pressure that gas would exert if it occupied the entire volume of the mixture at the same temperature. This concept is crucial when the reaction involves gases, as in the provided exercise.

For the reaction \(\text{Na}_2 \text{O}(s) + \text{SO}_2(g) \rightleftharpoons \text{Na}_2 \text{SO}_3(s)\), only the sulfur dioxide exists in the gaseous state, so the equilibrium constant in terms of partial pressure, \(\text{K}_p\), considers only \(\text{SO}_2\)'s partial pressure. It's important for students to recognize that solids and liquids do not have partial pressures and are therefore omitted in \(\text{K}_p\) expressions. Hence, the partial pressure equilibrium constant becomes \(\text{K}_p = \frac{1}{P_{\text{SO}_2}}\), maintaining that equilibrium involves understanding how pressure affects the position of equilibrium for gaseous participants.

Molarities

Molarity is a measure of the concentration of a solute in a solution, expressed as the number of moles of solute per liter of solution. In the realm of chemical equilibrium, when reactions occur in an aqueous solution, equilibrium expressions can be written in terms of the molarity of the reactants and products.

The exercise requires rewriting the equilibrium constant expression in terms of molarities, denoted as \(\text{K}_c\). This is particularly relevant when reactants or products are soluble or react in water to form ions. In the given exercise, sodium oxide dissolves in water, releasing sodium ions (\(\text{Na}^+\)) and oxide ions (\(\text{O}^{2-}\)), while sodium sulfite forms sodium ions (\(\text{Na}^+\)) and sulfite ions (\(\text{SO}_3^{2-}\)). This transforms the solid substances into ions within an aqueous solution, incorporating their molar concentrations into the equilibrium expression. Thus, the revised equilibrium constant is represented as \(\text{K}_c = \frac{{[\text{SO}_3^{2-}]}}{{[\text{O}^{2-}][\text{SO}_2(g)]}}\), simplifying the previous formula by canceling the molarity of sodium ions found in both the reactant and product sides of the equation.