Problem 14 - Chemistry The Molecular Science

Question

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

Step-by-Step Solution

Answer

For (a), \(K_c = \frac{[\mathrm{O}_3]^2}{[\mathrm{O}_2]^3}\); for (b), \(K_c = \frac{[\mathrm{Fe}(\mathrm{CO})_5]}{[\mathrm{CO}]^5}\); for (c), \(K_c = [\mathrm{NH}_3]^2[\mathrm{CO}_2][\mathrm{H}_2\mathrm{O}]\); and for (d), \(K_{sp} = [\mathrm{Ag}^+]^2[\mathrm{SO}_4^{2-}]\).

Step 1: Understanding Equilibrium Constant Expressions

Equilibrium constant expressions represent the ratio of the concentrations (or pressures) of products to reactants at equilibrium, each raised to the power of their coefficients in the balanced equation. Solids and pure liquids are not included in the expression.

Step 2: Writing the Expression for Reaction (a)

For the reaction \(3 \mathrm{O}_{2}( ext{g}) \rightleftharpoons 2 \mathrm{O}_{3}( ext{g})\), the equilibrium constant expression (denoted as \(K_c\) for concentration or \(K_p\) for pressure) is:\[K_c = \frac{[\mathrm{O}_3]^2}{[\mathrm{O}_2]^3}\] (using concentrations) or \[K_p = \frac{P_{\mathrm{O}_3}^2}{P_{\mathrm{O}_2}^3}\] (using pressures).

Step 3: Writing the Expression for Reaction (b)

For \(\mathrm{Fe}(\text{s}) + 5 \mathrm{CO}(\text{g}) \rightleftharpoons \mathrm{Fe}(\mathrm{CO})_{5}(\text{g})\), since \(\mathrm{Fe}(\text{s})\) is a solid, it is not included in the expression:\[ K_c = \frac{[\mathrm{Fe}(\mathrm{CO})_5]}{[\mathrm{CO}]^5} \] or \[ K_p = \frac{P_{\mathrm{Fe}(\mathrm{CO})_5}}{P_{\mathrm{CO}}^5} \].

Step 4: Writing the Expression for Reaction (c)

For \((\mathrm{NH}_{4})_{2}\mathrm{CO}_{3}(\text{s}) \rightleftharpoons 2\mathrm{NH}_3(\text{g}) + \mathrm{CO}_2(\text{g}) + \mathrm{H}_2\mathrm{O}(\text{g})\), since \((\mathrm{NH}_{4})_{2}\mathrm{CO}_{3}(\text{s})\) is a solid, it is not part of the expression:\[ K_c = [\mathrm{NH}_3]^2 [\mathrm{CO}_2] [\mathrm{H}_2\mathrm{O}] \] or \[ K_p = P_{\mathrm{NH}_3}^2 P_{\mathrm{CO}_2} P_{\mathrm{H}_2\mathrm{O}} \].

Step 5: Writing the Expression for Reaction (d)

For \(\mathrm{Ag}_2\mathrm{SO}_4(\text{s}) \rightleftharpoons 2\mathrm{Ag}^+(\text{aq}) + \mathrm{SO}_4^{2-}(\text{aq})\), since \(\mathrm{Ag}_2\mathrm{SO}_4(\text{s})\) is a solid, it is not included:\[ K_{sp} = [\mathrm{Ag}^+]^2 [\mathrm{SO}_4^{2-}] \].

Key Concepts

Gaseous Equilibrium

Gaseous equilibrium refers to a state in chemical reactions involving gases where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of reactants and products over time. This scenario is often represented quantitatively using the equilibrium constant expression, where the concentrations or partial pressures of the gases are used, each raised to the power of their coefficients from the balanced equation of the reaction. At equilibrium, it signifies that the reaction has reached a steady state and concentrations remain constant, although the reaction continues to occur at the molecular level.
In reactions such as the decomposition of ozone, \[3 \text{O}_2(\text{g}) \rightleftharpoons 2 \text{O}_3(\text{g})\], the equilibrium can be expressed in terms of concentrations \[K_c = \frac{[\text{O}_3]^2}{[\text{O}_2]^3}\] or pressures \[K_p = \frac{P_{\text{O}_3}^2}{P_{\text{O}_2}^3}\]. This showcases the balance between the rates of the forward and reverse reactions, noting that solids and liquids are typically excluded from such expressions.

Concentration

Concentration in chemistry refers to the amount of a substance in a given volume. It is a crucial factor in describing equilibria, as the concentration of reactants and products will help determine the position and state of equilibrium in a reversible reaction. In equilibrium constant expressions, only the concentrations of gases and aqueous species are included.The equilibrium constant for reactions using concentrations is denoted as \(K_c\), which provides a snapshot of the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their stoichiometric coefficients. For example, in the following reaction:\[(\text{NH}_4)_2\text{CO}_3(\text{s}) \rightleftharpoons 2\text{NH}_3(\text{g}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{g})\],The equilibrium expression is written as:\[K_c = [\text{NH}_3]^2 [\text{CO}_2] [\text{H}_2\text{O}]\].Products and reactants are raised to the power of their coefficients from the balanced chemical equation. Solids are not included due to their constancy in concentration during the reaction process.

Pressure

In reactions involving gases, pressure is an alternative to concentration for describing equilibria. The equilibrium constant expression can be written in terms of partial pressures, denoted as \(K_p\). Partial pressure indicates the specific pressure each gas contributes in a reaction mixture.For example, in the reaction\[\text{Fe}(\text{s}) + 5 \text{CO}(\text{g}) \rightleftharpoons \text{Fe}(\text{CO})_5(\text{g})\],The equilibrium expression using pressure is:\[K_p = \frac{P_{\text{Fe}(\text{CO})_5}}{P_{\text{CO}}^5}\].The expression involves only the gases, as solids and liquids are omitted. The partial pressures are raised to the power of their coefficients present in the balanced chemical equation.It's essential to note that \(K_p\) and \(K_c\) are related, with their relationship involving the ideal gas constant and temperature: \(K_p = K_c(RT)^{\Delta n}\), where \(\Delta n\) is the change in moles of gas.

Chemical Equation Balancing

Balancing a chemical equation is necessary for correctly expressing equilibrium constant expressions. It ensures the conservation of mass and charge during a chemical reaction, which is crucial when calculating the relationships of concentrations and pressures of the reactants and products at equilibrium.Each species' coefficients in the balanced equation are used as exponents in the equilibrium expression, highlighting their relative amounts involved in reactions. For example, consider:\[3 \text{O}_2(\text{g}) \rightleftharpoons 2 \text{O}_3(\text{g})\].Here, it is important to balance the equation so that when calculating \(K_c\) or \(K_p\), the concentration or pressure of \(\text{O}_2\) is cubed to show its involvement in the reaction, based on its stoichiometry.Balancing accounts for both the conservation of atoms and electrons, giving insight into how changes in one part of the equation might affect the equilibrium position. Thus, without it, a chemistry student would be unable to accurately predict or describe the outcome of a chemical process.