Question

Write expressions for \(K\) and \(K_{\mathrm{p}}\) for the following reactions. a. \(2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{CH}_{4} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) b. \(2 \mathrm{NBr}_{3}(s) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{Br}_{2}(g)\) c. \(2 \mathrm{KClO}_{3}(s) \rightleftharpoons 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) d. \(\mathrm{CuO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Cu}(l)+\mathrm{H}_{2} \mathrm{O}(g)\)

Short answer

The expressions for \(K\) and \(K_{\mathrm{p}}\) for the given reactions are: a. \(K = \frac{[\mathrm{H}_2 \mathrm{O}]}{[\mathrm{NH}_3]^2 [\mathrm{CO}_2]}\) \(K_{\mathrm{p}} = \frac{P_{\mathrm{H}_2\mathrm{O}}}{P_{\mathrm{NH}_3}^2 P_{\mathrm{CO}_2}}\) b. \(K = \frac{[\mathrm{N}_2] [\mathrm{Br}_2]^3}{1}\) \(K_{\mathrm{p}} = \frac{P_{\mathrm{N}_2}P_{\mathrm{Br}_2}^3}{1}\) c. \(K = \frac{[\mathrm{O}_2]^3}{1}\) \(K_{\mathrm{p}} = \frac{P_{\mathrm{O}_2}^3}{1}\) d. \(K = \frac{[\mathrm{H}_2\mathrm{O}]}{[\mathrm{H}_2]}\) \(K_{\mathrm{p}} = \frac{P_{\mathrm{H}_2\mathrm{O}}}{P_{\mathrm{H}_2}}\)

Step-by-step solution

Write the expression for K

The expression for the equilibrium constant for this reaction can be written as: \[K = \frac{[\mathrm{N}_2\mathrm{CH}_4\mathrm{O}] [\mathrm{H}_2\mathrm{O}]^1}{[\mathrm{NH}_3]^2 [\mathrm{CO}_2]^1}\] Since solids and liquids are excluded from the equilibrium constant, the equation becomes: \[K = \frac{1 [\mathrm{H}_2 \mathrm{O}]^1}{[\mathrm{NH}_3]^2 [\mathrm{CO}_2]^1}\]

Write the expression for Kp

The expression for the equilibrium constant in terms of partial pressures for this reaction can be written as: \[K_{\mathrm{p}} = \frac{P_{\mathrm{H}_2\mathrm{O}}}{P_{\mathrm{NH}_3}^2 P_{\mathrm{CO}_2}}\] b. Reaction: \(2 \mathrm{NBr}_{3}(s) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{Br}_{2}(g)\)

Write the expression for K

Given that we don't include pure solids in the equilibrium expressions, we can write the equilibrium constant for this reaction as: \[K = \frac{[\mathrm{N}_2]^1 [\mathrm{Br}_2]^3}{1}\]

Write the expression for Kp

The expression for the equilibrium constant in terms of partial pressures for this reaction can be written as: \[K_{\mathrm{p}} = \frac{P_{\mathrm{N}_2}P_{\mathrm{Br}_2}^3}{1}\] c. Reaction: \(2 \mathrm{KClO}_{3}(s) \rightleftharpoons 2 \mathrm{KCl}(s)+3\mathrm{O}_{2}(g)\)

Write the expression for K

Given that we don't include pure solids in the equilibrium expressions, we can write the equilibrium constant for this reaction as: \[K = \frac{[\mathrm{O}_2]^3}{1}\]

Write the expression for Kp

The expression for the equilibrium constant in terms of partial pressures for this reaction can be written as: \[K_{\mathrm{p}} = \frac{P_{\mathrm{O}_2}^3}{1}\] d. Reaction: \(\mathrm{CuO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons\mathrm{Cu}(l)+\mathrm{H}_{2} \mathrm{O}(g)\)

Write the expression for K

Given that we don't include pure solids and liquids in the equilibrium expressions, we can write the equilibrium constant for this reaction as: \[K = \frac{[\mathrm{H}_2\mathrm{O}]}{[\mathrm{H}_2]}\]

Write the expression for Kp

The expression for the equilibrium constant in terms of partial pressures for this reaction can be written as: \[K_{\mathrm{p}} = \frac{P_{\mathrm{H}_2\mathrm{O}}}{P_{\mathrm{H}_2}}\]

Key concepts

Partial Pressure

Partial pressure is an essential concept in understanding how gases behave in a mixture. Each gas in a mixture exerts its own pressure, as if it were alone in the container. This is the partial pressure. In chemical equilibrium expressions, we often use partial pressures to express the concentration of gaseous reactants and products during a reaction.

The formula to find the partial pressure of a gas can be given as:

• The partial pressure, \(P_i\), of a gas \(i\), is given by \(P_i = X_i \cdot P_{\text{total}}\), where \(X_i\) is the mole fraction of the gas, and \(P_{\text{total}}\) is the total pressure of the gas mixture.

Partial pressures are especially significant when dealing with equilibrium constants, denoted as \(K_p\), for reactions involving gases. For example, in the reaction given above for \(2 \mathrm{NH}_3 + \mathrm{CO}_2 \rightleftharpoons \mathrm{N}_2\mathrm{CH}_4\mathrm{O} + \mathrm{H}_2\mathrm{O}\), we express \(K_p\) using partial pressures as follows:

• The partial pressure of water \(P_{\mathrm{H}_2\mathrm{O}}\) over the partial pressures of ammonia squared, \(P_{\mathrm{NH}_3}^2\), and carbon dioxide, \(P_{\mathrm{CO}_2}\).

Chemical Equilibrium

Chemical equilibrium is a state reached by a reaction where the rates of the forward and reverse reactions are equal. At this point, the concentrations of the reactants and the products remain constant over time, even though both reactions continue to occur.

For example, in the exercise, you see reversible reactions, indicated by the symbol \(\rightleftharpoons\), like

• \(2 \mathrm{NBr}_3(s) \rightleftharpoons \mathrm{N}_2(g) + 3\mathrm{Br}_2(g)\).

This reaction achieves equilibrium when the production rate of \(\mathrm{N}_2\) and \(\mathrm{Br}_2\) matches the reverse reaction rate.

At equilibrium, the equilibrium constant \(K\) is defined as the ratio of the concentrations (or partial pressures) of products to the concentrations (or partial pressures) of reactants, each raised to the power of their stoichiometric coefficients.

• For solids and pure liquids, their concentration expressions do not affect \(K\).

Reaction Expressions

Reaction expressions involve writing equations that describe the equilibrium state of a chemical reaction. These include the equilibrium constant expressions for concentration (\(K\)) and partial pressure (\(K_p\)).

Let's dive into what these expressions represent:

• **Equilibrium Concentration, \(K\):** It is expressed using concentrations \([\text{products}]\) over \([\text{reactants}]\). Each concentration is raised to the power of the coefficient from the balanced equation.


• **Equilibrium Partial Pressure, \(K_p\):** This is analogous to \(K\), but it replaces concentrations with partial pressures, \(P\). Like \(K\), it accounts for the stoichiometric coefficients from the balanced chemical equation, such as \(K_p = \frac{P_{products}}{P_{reactants}}\).

For example, in the given reaction \(\mathrm{CuO}(s) + \mathrm{H}_2(g) \rightleftharpoons \mathrm{Cu}(l) + \mathrm{H}_2 \mathrm{O}(g)\), the equilibrium expression excludes \(\mathrm{CuO}(s)\) and \(\mathrm{Cu}(l)\) because they are a solid and a liquid, respectively. Thus, the expressions are:

• \(K = \frac{\text{[H}_2\text{O]}}{\text{[H}_2\text{]}}\)
• \(K_p = \frac{P_{\mathrm{H}_2\mathrm{O}}}{P_{\mathrm{H}_2}}\)