Problem 2 - Chemistry and Chemical Reactivity

Question

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

Step-by-Step Solution

Answer

K expressions: (a) \(\frac{P_{\mathrm{O}_3}^2}{P_{\mathrm{O}_2}^3}\), (b) \(\frac{P_{\mathrm{Fe(CO)_5}}}{P_{\mathrm{CO}}^5}\), (c) \(P_{\mathrm{NH}_3}^2 \cdot P_{\mathrm{CO}_2} \cdot P_{\mathrm{H}_2O}\), (d) \([\mathrm{Ag}^+]^2 \cdot [\mathrm{SO}_4^{2-}]\).

Step 1: Understanding Equilibrium Constant Expressions

The equilibrium constant expression, either in terms of concentration \( K_c \) or pressure \( K_p \), is written based on the reaction types in question. Solids and pure liquids are omitted because their concentrations do not change. For gases, \( K_p \) can be used if pressure is considered.

Step 2: Write the Equilibrium Expression for Reaction (a)

For the reaction \( 3 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{O}_{3}(\mathrm{g}) \), the equilibrium constant expression \( K_p \) using partial pressures is given by:\[ K_p = \frac{P_{\mathrm{O}_3}^2}{P_{\mathrm{O}_2}^3} \] where \( P_{\mathrm{O}_3} \) and \( P_{\mathrm{O}_2} \) are the partial pressures of \( \mathrm{O}_3 \) and \( \mathrm{O}_2 \) respectively.

Step 3: Write the Equilibrium Expression for Reaction (b)

For \( \mathrm{Fe}(\mathrm{s}) + 5 \mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{Fe}(\mathrm{CO})_5(\mathrm{g}) \), write the equilibrium expression ignoring the solid \( \mathrm{Fe} \):\[ K_p = \frac{P_{\mathrm{Fe(CO)_5}}}{P_{\mathrm{CO}}^5} \] where \( P_{\mathrm{Fe(CO)_5}} \) and \( P_{\mathrm{CO}} \) are the partial pressures of \( \mathrm{Fe(CO)_5} \) and \( \mathrm{CO} \).

Step 4: Write the Equilibrium Expression for Reaction (c)

For the reaction \( \left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3(\mathrm{s}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{g}) + \mathrm{CO}_2(\mathrm{g}) + \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \), write the equilibrium expression, omitting the solid:\[ K_p = P_{\mathrm{NH}_3}^2 \cdot P_{\mathrm{CO}_2} \cdot P_{\mathrm{H}_2O} \] where \( P_{\mathrm{NH}_3} \), \( P_{\mathrm{CO}_2} \), and \( P_{\mathrm{H}_2O} \) are the partial pressures of \( \mathrm{NH}_3 \), \( \mathrm{CO}_2 \), and \( \mathrm{H}_2O \) respectively.

Step 5: Write the Equilibrium Expression for Reaction (d)

For \( \mathrm{Ag}_2 \mathrm{SO}_4(\mathrm{s}) \rightleftharpoons 2 \mathrm{Ag}^+(\mathrm{aq}) + \mathrm{SO}_4^{2-}(\mathrm{aq}) \), write the equilibrium expression excluding the solid:\[ K_{sp} = [\mathrm{Ag}^+]^2 \cdot [\mathrm{SO}_4^{2-}] \] where \([\mathrm{Ag}^+]\) and \([\mathrm{SO}_4^{2-}]\) are the molar concentrations of the ions in the solution.

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a fascinating and essential concept in chemistry, describing a state where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products over time. This balance means that the reaction is not static but dynamic, with molecules continuously being converted back and forth. In a chemical equation showing equilibrium, the double-arrow symbol (\( \rightleftharpoons\)) indicates this balance in progress.

At equilibrium, the relationship between the concentrations of reactants and products is expressed mathematically through the equilibrium constant, denoted as \( K \). This constant gives us insight into the position of equilibrium and helps predict how the system will respond to changes in conditions according to Le Chatelier's principle.

It's important to note that not all the components of a reaction affect the equilibrium constant. Solids and pure liquids, for instance, are not included in the expression for \( K \), because their concentrations are constant and do not affect the position of equilibrium.

Partial Pressures

Partial pressures are crucial when dealing with gases in a chemical equilibrium context. Each gas in a mixture contributes to the total pressure, and its contribution is known as its partial pressure. For reactions involving gases, we use these partial pressures to write the equilibrium constant expression \( K_p \), where \( P \) stands for the partial pressure.

For instance, in the reaction \( 3 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{O}_{3}(\mathrm{g}) \), the equilibrium expression using partial pressures is:

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

Here, \( P_{\mathrm{O}_3} \) and \( P_{\mathrm{O}_2} \) represent the partial pressures of \( \mathrm{O}_3 \) and \( \mathrm{O}_2 \), demonstrating how changes in pressure impact the position of equilibrium.

In practice, measuring these partial pressures helps chemists predict how a gaseous equilibrium system will respond to changes in conditions like volume or temperature.

Molar Concentrations

Molar concentrations describe the amount of solute per volume of solution, often expressed as moles per liter (M). In equilibrium expressions for reactions occurring in solution, these concentrations help show the distribution of substances between reactants and products.

For equilibrium systems involving aqueous solutions, we use the equilibrium constant expression \( K_c \). For example, in the dissolution of silver sulfate:

• \( \mathrm{Ag}_2 \mathrm{SO}_4(\mathrm{s}) \rightleftharpoons 2 \mathrm{Ag}^+(\mathrm{aq}) + \mathrm{SO}_4^{2-}(\mathrm{aq}) \)

The equilibrium expression in terms of molar concentrations is:

• \[ K_{sp} = [\mathrm{Ag}^+]^2 \cdot [\mathrm{SO}_4^{2-}] \]

where \([\mathrm{Ag}^+]\) and \([\mathrm{SO}_4^{2-}]\) are concentrations of ions in solution. These concentrations, paired with \( K_{sp} \), the solubility product constant, can determine the extent of dissolution in a solution.

In equilibrium reactions involving solutions, adjusting concentrations allows chemists to manipulate the system's position and yield.

Solubility Product

The solubility product, denoted \( K_{sp} \), is a special type of equilibrium constant applicable to the solubility of ionic compounds. It provides insight into how much of a solid can dissolve in a solution to form a saturated solution, beyond which any additional solid remains undissolved.

For example, in the reaction:

• \( \mathrm{Ag}_2 \mathrm{SO}_4(\mathrm{s}) \rightleftharpoons 2 \mathrm{Ag}^+(\mathrm{aq}) + \mathrm{SO}_4^{2-}(\mathrm{aq}) \)

the solubility product expression is:

• \[ K_{sp} = [\mathrm{Ag}^+]^2 \cdot [\mathrm{SO}_4^{2-}] \]

In this case, \( K_{sp} \) represents the maximum product of the ion concentrations in a saturated solution. If the product of the ion concentrations exceeds this value, precipitation occurs.

This concept is particularly important in predicting whether a precipitate will form under particular conditions and in designing systems where desirable precipitates are needed, such as in chemical separation processes or materials synthesis.