Question

Write expressions for \(K_{\mathrm{p}}\) for the following reactions. a. \(2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) b. \(\mathrm{CO}_{2}(g)+\mathrm{MgO}(s) \rightleftharpoons \mathrm{MgCO}_{3}(s)\) c. \(\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g)\) d. \(4 \mathrm{KO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{KOH}(s)+3 \mathrm{O}_{2}(g)\)

Short answer

a. Kp = \(\frac{1}{(P_{O_2})^{3/2}}\) b. Kp = \(\frac{1}{P_{CO_2}}\) c. Kp = \(\frac{P_{CO} P_{H_2}}{P_{H_2O}}\) d. Kp = \(\frac{(P_{O_2})^3}{(P_{H_2O})^2}\)

Step-by-step solution

a. 2 Fe(s) + (3/2) O2(g) Fe2O3(s)

For this reaction, the only gas present is O2. Solids are not included in the Kp expressions. Thus, the expression for Kp is: Kp = \(\frac{1}{(P_{O_2})^{3/2}}\)

b. CO2(g) + MgO(s) MgCO3(s)

In this reaction, the only gas present is CO2. Solids are not included in the Kp expressions. Thus, the expression for Kp is: Kp = \(\frac{1}{P_{CO_2}}\)

c. C(s) + H2O(g) CO(g) + H2(g)

In this reaction, there are two gaseous species: H2O and CO. Solids, such as carbon, are not included in the Kp expressions. Thus, the expression for Kp is: Kp = \(\frac{P_{CO} P_{H_2}}{P_{H_2O}}\)

d. 4 KO2(s) + 2 H2O(g) 4 KOH(s) + 3 O2(g)

In this reaction, there are two gaseous species: H2O and O2. Solids are not included in the Kp expressions. Since the reaction coefficients for the gaseous species are non unity, we must raise their pressures to the corresponding stoichiometric coefficients. Thus, the expression for Kp is: Kp = \(\frac{(P_{O_2})^3}{(P_{H_2O})^2}\)

Key concepts

Kp Expression

The equilibrium constant for gas-phase reactions expressed in terms of the partial pressures of the reactants and products is symbolized as \( K_p \). To derive the \( K_p \) expression, only the gaseous species are considered, with each pressure term raised to the power of its stoichiometric coefficient from the balanced equation. For instance, in a simplified reaction \( aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g) \), the \( K_p \) expression takes the form of \( K_p = \frac{(P_{C})^c(P_{D})^d}{(P_{A})^a(P_{B})^b} \), where \( P \) represents the partial pressures of each gas. In cases where solids or liquids are involved, they are omitted from the \( K_p \) expression as their concentrations do not change with pressure and hence, are constant.

Understanding how to form the correct \( K_p \) expression requires identifying the gaseous substances and using their coefficients from the balanced chemical equation. For example, reaction a in the exercise involves one gaseous species, \( O_2 \), and the corresponding \( K_p \) expression is \( K_p = \frac{1}{(P_{O_2})^{3/2}} \). It's important to recognize that solids, like iron and iron oxide, do not appear in the expression.

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction and its reverse reaction proceed at equal rates, resulting in no net change in the concentrations of the reactants and products. It is dynamic because both the forward and reverse reactions are occurring at the same time. At equilibrium, the properties and concentrations of reactants and products remain constant over time, although they need not be equal.

Equilibrium can be quantitatively described by the equilibrium constant \( K \), which is a measure of the tendency of a chemical system to reach this balanced state. For gas-phase reactions, the equilibrium constant is expressed as \( K_p \), using partial pressures. The magnitude of the equilibrium constant gives an indication of the position of equilibrium; a large \( K \) implies a composition heavily favouring products, while a small \( K \) suggests that the reactants are more predominant under equilibrium conditions.

Understanding the concept of equilibrium is critical when determining the \( K_p \) expressions. For example, in reaction c of the exercise, after reaching equilibrium, the system establishes a ratio of the pressures of \( CO \), \( H_2 \), and \( H_2O \) that remains constant, leading to the equilibrium constant expression \( K_p = \frac{P_{CO} P_{H_2}}{P_{H_2O}} \).

Gas Phase Reactions

Gas phase reactions involve reactants and products in the gaseous state and are governed by the kinetic theory of gases. The equilibrium constant expressions for these reactions, such as \( K_p \), are specifically tailored to incorporate the behavior of gases and their partial pressures. One unique characteristic of gases is that they can be compressed or expanded, so their concentrations (or partial pressures) change in response to volume and temperature changes.

When writing equilibrium constant expressions for gas phase reactions, the partial pressures reflect the concentrations of gaseous species. This is unlike liquid and solid phase reactions where concentrations are used as the measure. It's important to note that gases follow the ideal gas law under many conditions, though deviations can occur at high pressures or low temperatures. In exercise d, we consider the gas-phase reactants and products, \( H_2O(g) \) and \( O_2(g) \), when formulating the \( K_p \) expression as \( K_p = \frac{(P_{O_2})^3}{(P_{H_2O})^2} \).

Gaining an understanding of the behavior of gases is crucial for students learning to write accurate \( K_p \) expressions for gas phase reactions, as seen in the provided exercise examples.