Problem 30 - Chemistry The Central Science

Question

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.

Step-by-Step Solution

Answer

(a) For the given equilibrium reaction, the equilibrium constant expression in terms of partial pressures is: \(K_p = P_{\mathrm{SO}_2}\). (b) When all compounds are dissolved in water, the equilibrium constant expression in terms of molarities is: \(K_c = \cfrac{[\mathrm{SO}_3^{2-}]}{[\mathrm{SO}_2] \cdot [\mathrm{O}^{2-}]}\).

Step 1: a) Writing the equilibrium constant expression for the reaction in terms of partial pressures

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: \(K_p = P_{\mathrm{SO}_2}\)

Step 2: b) Rewriting the equilibrium constant expression in terms of molarities for the aqueous reaction

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.

Key Concepts

Equilibrium Constant

In the world of chemical reactions, the equilibrium constant, often represented as either \(K_c\) or \(K_p\), plays a vital role in determining the balance point of a reaction. It is an indicator of the extent to which a chemical reaction occurs under specific conditions.
The equilibrium constant expression is derived from the concentrations or partial pressures of the reactants and products involved in the reaction. It is important to note that the equilibrium constant is dependent on the temperature, which is why it is typically stated with a temperature specification.
When dealing with gaseous reactions, the expression \(K_p\) is used, incorporating the partial pressures of the gases. For reactions in solution, \(K_c\) is more applicable as it considers the molar concentrations. Understanding these principles is essential to predict how changes in conditions affect the position of equilibrium.

Partial Pressure

Partial pressure is an important concept when dealing with gases in chemical reactions. It refers to the pressure that each gas in a mixture would exert if it alone occupied the entire volume.
When a reaction at equilibrium involves gases, like the one with \(\mathrm{SO}_2\), the equilibrium constant can be expressed in terms of the partial pressures of the gases involved, known as \(K_p\). This helps chemists understand and predict the behavior of gases in reactions.
It’s crucial to remember that only gases contribute to the \(K_p\) expression, as pure solids and liquids aren't included in the equilibrium constant expression. This is why in the original reaction involving \(\mathrm{Na}_2\mathrm{O}(s)\) and \(\mathrm{Na}_2\mathrm{SO}_3(s)\), these components do not appear in the \(K_p\) expression.

Molarity

Molarity is a measure of concentration used extensively in chemistry, particularly when dealing with reactions happening in solutions. It is defined as the number of moles of a solute present in one liter of solution.
This concept becomes important when the components of an equilibrium reaction are soluble in water, and we seek to express the equilibrium constant as \(K_c\) using molarities.
In the given example, turning the gaseous and solid substances into their aqueous ionic forms allows the use of molarity in the equilibrium constant expression. This conversion is crucial for reactions to be accurately described in aqueous media, ensuring precise calculations in chemistry.

Chemical Reaction

Chemical reactions are transformations that convert one or more substances into new substances. They are the heart of chemical processes and can occur in diverse physical states.
Understanding chemical reactions involves considering the reactants, products, and the conditions under which they occur. The example given shows a system at equilibrium, demonstrated by the double arrow (\(\rightleftharpoons\)).
During equilibrium, although reactions continue to occur, the rates of forward and reverse reactions are equal, leading to constant concentrations of both reactants and products. This is an essential insight when analyzing reactions, as it provides stability to the system at equilibrium and simplifies the calculation of the equilibrium constant.