Problem 29 - Chemistry The Central Science

Question

Mercury(l) oxide decomposes into elemental mercury and elemental oxygen: \(2 \mathrm{Hg}_{2} \mathrm{O}(s) \rightleftharpoons 4 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\) (a) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) Suppose you run this reaction in a solvent that dissolves elemental mercury and elemental oxygen. Rewrite the equilibrium- constant expression in terms of molarities for the reaction, using (solv) to indicate solvation.

Step-by-Step Solution

Answer

(a) The equilibrium-constant expression in terms of partial pressures for the given reaction is: \[K_p = \frac{P_{Hg}^4 P_{O_2}}{(P_{Hg_2O})^2}\] (b) The equilibrium constant expression in terms of molarities considering solvation for the given reaction is: \[K_c = \frac{[Hg_{(solv)}]^4 [O_{2(solv)}]}{[Hg_2O]^2}\]

Step 1: (a) Writing equilibrium-constant expression in terms of partial pressures

To write the equilibrium constant expression (Kp) for the given reaction, we can use the general formula: \[K_p = \frac{[C]^c[D]^d}{[A]^a[B]^b}\] Where A, B, C, and D are the reaction species in the balanced chemical reaction: aA + bB ⇌ cC + dD, and a, b, c, and d are their respective stoichiometric coefficients. For the given reaction: \[2 Hg_2O(s) \rightleftharpoons 4 Hg(l) + O_2(g)\] The equilibrium constant expression in terms of partial pressures will be: \[K_p = \frac{P_{Hg}^4 P_{O_2}}{(P_{Hg_2O})^2}\]

Step 2: (b) Rewriting the equilibrium-constant expression in terms of molarities considering solvation

To rewrite the equilibrium constant expression in terms of molarities, we'll use the equilibrium constant expression using concentrations (Kc) given by: \[K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\] As the reaction runs in a solvent that dissolves elemental mercury and elemental oxygen, we should represent their solvation with the (solv) index. Thus, for the given reaction: \[2 Hg_2O(s) \rightleftharpoons 4 Hg_{(solv)} + O_{2(solv)}\] The equilibrium constant expression in terms of molarities considering solvation will be: \[K_c = \frac{[Hg_{(solv)}]^4 [O_{2(solv)}]}{[Hg_2O]^2}\]

Key Concepts

Equilibrium Constant

The equilibrium constant is a key concept in chemical equilibrium, representing the ratio of the concentrations of products to reactants at equilibrium. It allows chemists to predict the direction a reaction will shift to reach this state. In the context of our reaction with mercury(I) oxide, the equilibrium constant can be shown in different forms depending on the scenario. One form is based on partial pressures, termed as \( K_p \). The other utilizes concentrations, known as \( K_c \). Both \( K_p \) and \( K_c \) expressions consider the stoichiometric coefficients of the balanced chemical equation.

• \( K_p \) focuses on gaseous reactants and products, hence partial pressures are key.
• \( K_c \) involves solutes in a solution, thus using molar concentrations and indicating solvation if needed.

Understanding which expression to use depends on whether your reaction components are gases or in a solubilized state.

Partial Pressure

Partial pressure is an important aspect when dealing with gases in equilibrium. It refers to the pressure exerted by a single type of gas in a mixture of gases. In the equilibrium expression concerning \( K_p \), the partial pressures of the gaseous products and reactants are used. For our reaction, because mercury(I) oxide decomposes into liquid mercury and gaseous oxygen, we concern ourselves mostly with the partial pressure of oxygen.

• Partial pressure is denoted \( P \), for example, \( P_{O_2} \) for oxygen.
• Incorporates into equilibrium calculations when gases are involved.
• Allows prediction of how changes in conditions affect equilibrium.

Using partial pressures in driving equilibrium constant expressions helps understand how a reaction behaves under various pressures.

Stoichiometry

Stoichiometry involves understanding the relationships between the amounts of reactants and products in a chemical reaction, governed by their balanced equation. Each reaction component has a coefficient expressing how many moles participate or form. For example, in \( 2 Hg_2O(s) \rightarrow 4 Hg(l) + O_2(g) \), two moles of mercury(I) oxide produce four moles of mercury and one mole of oxygen.

• Essential for balancing chemical equations.
• Coefficients guide equilibrium constant expressions by indicating power terms (like \( [Hg]^4 \)).
• Helps in calculating yields and needed reactant quantities.

In equilibrium expressions, stoichiometric coefficients determine the exponents for the concentrations or partial pressures of each substance involved.

Molarity

Molarity is a way of expressing the concentration of a solute in a solution. It's defined as the number of moles of solute per liter of solution, denoted as \( M \). In reactions involving solutions, the equilibrium constant expression may use molarity, represented as \( K_c \), where concentrations of solvated reactants and products are key. For example, in a reaction using the solvent presented, dissolved mercury and oxygen are represented like \([Hg_{(solv)}]\) and \([O_{2(solv)}]\).

• Molarity provides a direct measure of concentration in a solution.
• Allows calculation of equilibria in solution-phase reactions.
• Important in preparing solutions and calculating in chemical analysis.

Thus, when reactions occur in solution, identifying molarity helps in accurately expressing the equilibrium constant, aiding in understanding and predicting reaction behavior.