Question

Write equilibrium constant expressions for the following reactions. For gases use either pressures or concentrations. (a) \(3 \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{O}_{3}(\mathrm{g})\) (b) \(\mathrm{Fe}(\mathrm{s})+5 \mathrm{CO}(\mathrm{g}) \rightleftarrows \mathrm{Fe}(\mathrm{CO})_{5}(\mathrm{g})\) (c) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(\mathrm{s}) \rightleftarrows 2 \mathrm{NH}_{3}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (d) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(\mathrm{s}) \rightleftarrows 2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\)

Short answer

(a) \(K = \frac{[\mathrm{O}_3]^2}{[\mathrm{O}_2]^3}\); (b) \(K = \frac{P_{\mathrm{Fe(CO)_5}}}{P_{\mathrm{CO}}^5}\); (c) \(K = P_{\mathrm{NH}_3}^2 \times P_{\mathrm{CO}_2} \times P_{\mathrm{H}_2\mathrm{O}}\); (d) \(K = [\mathrm{Ag}^+]^2 [\mathrm{SO}_4^{2-}]\).

Step-by-step solution

Understanding the Equilibrium Constant Expression

For a chemical reaction, the equilibrium constant expression (K) is formulated based on the concentrations or partial pressures of the reactants and products. Solids and pure liquids are excluded from this expression. For a general reaction: \(aA + bB \rightleftarrows cC + dD\), the equilibrium constant \(K_c\) for concentrations is \([C]^c[D]^d / [A]^a[B]^b\), and \(K_p\) for partial pressures is \(P_C^cP_D^d / P_A^aP_B^b\).

Writing the Equilibrium Constant for (a)

For the reaction \(3 \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{O}_{3}(\mathrm{g})\), include only gaseous components in the expression. The equilibrium constant expression for this reaction is \(K = \frac{[\mathrm{O}_3]^2}{[\mathrm{O}_2]^3}\).

Writing the Equilibrium Constant for (b)

For the reaction \(\mathrm{Fe}(\mathrm{s}) + 5 \mathrm{CO}(\mathrm{g}) \rightleftarrows \mathrm{Fe}(\mathrm{CO})_{5}(\mathrm{g})\), since Fe is a solid, exclude it from the expression. The equilibrium constant expression is \(K = \frac{P_{\mathrm{Fe(CO)_5}}}{P_{\mathrm{CO}}^5}\). Only pressures of gaseous species are included.

Writing the Equilibrium Constant for (c)

For the reaction \((\mathrm{NH}_4)_{2} \mathrm{CO}_3(\mathrm{s}) \rightleftarrows 2 \mathrm{NH}_3(\mathrm{g}) + \mathrm{CO}_2(\mathrm{g}) + \mathrm{H}_2 \mathrm{O}(\mathrm{g})\), \((\mathrm{NH}_4)_{2} \mathrm{CO}_3\) is solid and should be excluded. The equilibrium constant expression is \(K = P_{\mathrm{NH}_3}^2 \times P_{\mathrm{CO}_2} \times P_{\mathrm{H}_2\mathrm{O}}\).

Writing the Equilibrium Constant for (d)

For the reaction \(\mathrm{Ag}_2 \mathrm{SO}_4(\mathrm{s}) \rightleftarrows 2 \mathrm{Ag}^+(\mathrm{aq}) + \mathrm{SO}_4^{2-}(\mathrm{aq})\), \(\mathrm{Ag}_2 \mathrm{SO}_4\) is solid and excluded. The equilibrium constant expression is \(K = [\mathrm{Ag}^+]^2 [\mathrm{SO}_4^{2-}]\).

Key concepts

Chemical Equilibrium

In chemistry, equilibrium occurs when a reaction reaches a state where the concentrations of reactants and products remain constant over time. This doesn't necessarily mean that the reactants and products are equal in concentration, rather that their rates of formation are balanced. This balance is crucial because it defines how far a reaction will proceed before stopping. To find out how balanced a reaction is, we use the equilibrium constant expressions, denoted as \(K_c\) for concentrations and \(K_p\) for pressures when dealing with gases. These constants provide a snapshot of the reaction's state at equilibrium by comparing the concentration or pressure of products to reactants, using their respective stoichiometric coefficients as exponents in the formula. Understanding chemical equilibrium allows scientists to predict both the extent of a reaction and the choice of conditions that will favor the production of desired products.

Gaseous Reactions

Gaseous reactions involve reactants and products in the gas phase. The equilibrium constant for these reactions can be expressed in terms of partial pressures, \(K_p\). This is particularly useful because the pressure of a gas is directly proportional to its concentration in a closed system. For a gaseous reaction like \(3 \mathrm{O}_{2}( ext{g}) ightleftarrows 2 \mathrm{O}_{3}( ext{g})\), the equilibrium constant \(K_p\) is given by

• \[K_p = \frac{P_{\mathrm{O}_3}^2}{P_{\mathrm{O}_2}^3}\]

This expression emphasizes how changes in the pressure of each component can affect the balance of the reaction. Pressure adjustments can shift the position of equilibrium, a principle known as Le Chatelier's Principle. For example, increasing the pressure typically favors the side of the reaction with fewer gas molecules.

Concentration and Pressure in Equilibrium

Concentration and pressure are key factors that influence equilibrium in chemical reactions, particularly for gases. The state of equilibrium involves both reactants and products achieving a consistent ratio, as dictated by the equilibrium constant. When expressed in terms of concentrations, we use \(K_c\), but when gases are involved, \(K_p\) is more applicable as it reflects partial pressures. For the reaction \(\mathrm{Fe(s)} + 5 \mathrm{CO(g)} \rightleftarrows \mathrm{Fe(CO)}_{5}(\text{g})\), the equilibrium constant in pressures is

• \[K_p = \frac{P_{\mathrm{Fe(CO)_5}}}{P_{\mathrm{CO}}^5}\]

This equation illustrates the dependency of product formation on the initial pressure of \(\mathrm{CO}\). Intensifying the concentration or pressure of reactants typically shifts the equilibrium toward creating more products. Understanding this relationship is critical when manipulating reaction conditions to optimize yields or control reaction rates.

Solids in Equilibrium Expressions

In equilibrium expressions, solids are treated uniquely. They are typically omitted from the equilibrium constant expression because their concentration is constant. Solids have a fixed density and volume, meaning their concentration does not change with the progress of the reaction. This simplifying assumption allows us to focus solely on the components whose concentrations fluctuate, such as gases and aqueous solutions.For example, in the reaction \((\mathrm{NH}_4)_2 \mathrm{CO}_3(\text{s}) \rightleftarrows 2 \mathrm{NH}_3(\text{g}) + \mathrm{CO}_2(\text{g}) + \mathrm{H}_2 \mathrm{O}(\text{g})\), the solid \((\mathrm{NH}_4)_2 \mathrm{CO}_3\) is excluded from the equilibrium constant expression, leaving us with

• \[K_p = P_{\mathrm{NH}_3}^2 \times P_{\mathrm{CO}_2} \times P_{\mathrm{H}_2\mathrm{O}}\]

This principle highlights how solids do not impact the state of equilibrium, allowing for a more manageable calculation of \(K\). Using these concepts, chemists can better predict the equilibrium behavior of reactions and adjust conditions to favor product formation when desired.