Textbook Question
Consider the following equilibrium, for which Kp = 0.0752 at 480°C: 2 Cl2(g) + 2 H2O(g) ⇌ 4 HCl(g) + O2(g) (b) What is the value of Kp for the reaction Cl2(g) + H2O(g) ⇌ 2 HCl(g) + 1/2 O2(g)?
Ch.15 - Chemical Equilibrium
Chapter 15, Problem 29
Mercury(I) oxide decomposes into elemental mercury and elemental oxygen: 2 Hg2O(s) ⇌ 4 Hg(l) + O2(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.
Verified step by step guidance1
Step 1: Identify the reaction and the phases of each component. The given reaction is 2 Hg2O(s) ⇌ 4 Hg(l) + O2(g). Note that Hg2O is a solid, Hg is a liquid, and O2 is a gas.
Step 2: Write the equilibrium-constant expression in terms of partial pressures for the gaseous components. Since solids and liquids do not appear in the equilibrium expression, only the gaseous component O2 will be included. The expression is K_p = P_{O2}, where P_{O2} is the partial pressure of O2.
Step 3: Consider the scenario where the reaction occurs in a solvent that dissolves elemental mercury and elemental oxygen. In this case, we need to express the equilibrium constant in terms of molarities.
Step 4: Rewrite the balanced chemical equation considering the solvation: 2 Hg2O(s) ⇌ 4 Hg(solv) + O2(solv). Here, (solv) indicates that the species are dissolved in the solvent.
Step 5: Write the equilibrium-constant expression in terms of molarities for the dissolved species. The expression is K_c = [O2(solv)], where [O2(solv)] is the molarity of dissolved oxygen.
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Equilibrium Constant Expression
The equilibrium constant expression (K) quantifies the ratio of the concentrations or partial pressures of products to reactants at equilibrium. For gaseous reactions, it is expressed in terms of partial pressures, while for reactions in solution, it is expressed in terms of molarities. The general form is K = [products]^[coefficients] / [reactants]^[coefficients], where square brackets denote concentration or pressure.
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Equilibrium Constant Expressions
Le Chatelier's Principle
Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system will adjust to counteract the change and restore a new equilibrium. This principle is crucial for understanding how changes in concentration, pressure, or temperature affect the position of equilibrium in a chemical reaction.
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Solvation and Molarity
Solvation refers to the process of surrounding solute particles with solvent molecules, which can affect the concentration of solutes in a solution. Molarity (M) is a measure of concentration defined as moles of solute per liter of solution. In the context of the given reaction, when elemental mercury and oxygen are dissolved, their concentrations must be expressed in molarity, indicating their effective participation in the equilibrium.
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Related Practice
Textbook Question
The following equilibria were attained at 823 K:
CoO(s) + H2(g) → Co(s) + H2O(g) Kc = 67
CoO(s) + CO(g) → Co(s) + CO2(g) Kc = 490
Based on these equilibria, calculate the equilibrium constant for H2(g) + CO2(g) → CO(g) + H2O(g) at 823 K.
Textbook Question
Consider the equilibrium N2(𝑔) + O2(𝑔) + Br2(𝑔) ⇌ 2 NOBr(𝑔) Calculate the equilibrium constant 𝐾𝑝 for this reaction, given the following information at 298 K:
2 NO(𝑔) + Br2(𝑔) ⇌ 2 NOBr(𝑔) 𝐾𝑐 = 2.02
NO(𝑔) ⇌ N2(𝑔) + O2(𝑔) 𝐾𝑐 = 2.1×1030
Textbook Question
The equilibrium 2 NO(𝑔) + Cl2(𝑔) ⇌ 2 NOCl(𝑔) is established at 500.0 K. An equilibrium mixture of the three gases has partial pressures of 0.095 atm, 0.171 atm, and 0.28 atm for NO, Cl2, and NOCl, respectively. (a) Calculate 𝐾𝑝 for this reaction at 500.0 K.