Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Mercury(I) 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.

Short Answer

Expert verified
(a) The equilibrium constant expression in terms of partial pressures for the given reaction is: \[K_P = \frac{P_{\mathrm{Hg}}^4 \times P_{\mathrm{O}_2}}{P_{\mathrm{Hg}_2\mathrm{O}}^2}\] (b) The equilibrium constant expression in terms of molarities for the given reaction in a solvent is: \[K_C = \frac{[\mathrm{Hg(solv)}]^4 \times [\mathrm{O_2(solv)}]}{[\mathrm{Hg_2O(solv)}]^2}\]

Step by step solution

01

(a) Equilibrium constant expression in terms of partial pressures

To write the equilibrium constant expression for the decomposition reaction of Mercury(I) oxide in terms of partial pressures, we can use the formula for K_P, the equilibrium constant for gases in terms of partial pressures: \[K_P = \frac{(\text{Partial pressure of products})^{\text{stoichiometric coefficients}}}{(\text{Partial pressure of reactants})^{\text{stoichiometric coefficients}}}\] For the given reaction: \[2 \mathrm{Hg}_2\mathrm{O}(s) \rightleftharpoons 4\mathrm{Hg}(l) + \mathrm{O}_2(g)\] The equilibrium constant expression in terms of partial pressures can be written as: \[K_P = \frac{P_{\mathrm{Hg}}^4 \times P_{\mathrm{O}_2}}{P_{\mathrm{Hg}_2\mathrm{O}}^2}\] Keep in mind that since Mercury(I) oxide is a solid and elemental mercury is a liquid, their partial pressures are not included in the equilibrium constant expression. The only relevant partial pressure is for elemental oxygen gas.
02

(b) Equilibrium constant expression in terms of molarities

To rewrite the equilibrium constant expression in terms of molarities, we can use the formula for K_C, the equilibrium constant for the reaction in terms of concentrations: \[K_C = \frac{[\text{Products}]^{\text{stoichiometric coefficients}}}{[\text{Reactants}]^{\text{stoichiometric coefficients}}}\] Let's use the (solv) notation for solute concentrations. For the given reaction in a solvent: \[2 \mathrm{Hg}_2\mathrm{O}(s) \rightleftharpoons 4\mathrm{Hg(solv)} + \mathrm{O}_2(solv)\] The equilibrium constant expression in terms of molarities can be written as: \[K_C = \frac{[\mathrm{Hg(solv)}]^4 \times [\mathrm{O_2(solv)}]}{[\mathrm{Hg_2O(solv)}]^2}\] Since Mercury(I) oxide is still a solid in the solvent, its concentration remains constant and does not appear in the equilibrium constant expression. The only relevant concentrations are for solvated elemental mercury and elemental oxygen.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant
In chemical reactions, not all reactants are completely converted to products. Instead, a dynamic equilibrium is established where the forward and backward reactions occur at the same rate. The equilibrium constant, denoted as either \(K_P\) for gases or \(K_C\) for solutions, quantifies this balance.
The expression for the equilibrium constant is derived based on the reaction's stoichiometry. For reactions involving gases, partial pressures are used in \(K_P\), while concentrations expressed as molarity are used in \(K_C\). A larger equilibrium constant means the reaction favors the products at equilibrium, while a smaller one indicates that reactants are favored.
In the decomposition reaction of mercury(I) oxide, the focus is on products that are either gases or dissolved in a solvent, as these contribute to the equilibrium expression.
Partial Pressures
Gases exert pressure when they occupy a container. This pressure is known as partial pressure in the context of a mixture of gases, and each gas contributes to the total pressure proportionally to its amount present.
In the equilibrium constant expression for gases, such as \(K_P\), partial pressures replace concentrations. Only gaseous components appear in the expression. For example, in the mercury(I) oxide decomposition, the solid and liquid forms do not factor into the equilibrium expression since their partial pressures, by definition, are not applicable.
Partial pressures are crucial for reactions involving gases because they directly affect the position of equilibrium and the value of \(K_P\). Adjusting the volume or temperature of the system can change these pressures, thereby shifting the equilibrium position.
Molarity
Molarity is a measure of the concentration of a solute in a solution and is defined as moles of solute per liter of solution. It's a fundamental concept in chemistry for describing how "crowded" a solute is within a solution.
In reactions carried out in a solvent, equilibrium constants are expressed in terms of molarity, denoted \(K_C\).
This involves solute concentrations—such as our dissolved elemental mercury and oxygen. Molarity helps us to predict how changes in concentration of the solutes will affect the equilibrium of the system.
  • Higher molarity implies more solute per unit volume.
  • Lower molarity may mean either less solute or a greater volume.
Understanding molarity is essential for interpreting solvation effects on chemical equilibrium.
Mercury Compound Reactions
Mercury compounds exhibit unique behaviors because mercury can exist in multiple oxidation states, influencing their reactivity and equilibrium states.
The decomposition of mercury(I) oxide is a classic reaction where heating causes the solid compound to yield liquid mercury and gaseous oxygen.
This particular reaction, when placed in a solvent system, focuses on the dissolved species, allowing mercury and oxygen to participate in further chemical processes or analysis.
  • Mercury(I) oxide is stable as a solid, making it energetically favorable to decompose only upon energy input.
  • The resulting mercury and oxygen, when solvated, engage differently depending on the solvent utilized.
Understanding the behavior of mercury in such reactions is vital for applications in chemical synthesis and industrial processes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Both the forward reaction and the reverse reaction in the following equilibrium are believed to be elementary steps: $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}(g)+\mathrm{Cl}(g) $$ At \(25^{\circ} \mathrm{C}\), the rate constants for the forward and reverse reactions are \(1.4 \times 10^{-28} M^{-1} \mathrm{~s}^{-1}\) and \(9.3 \times 10^{10} \mathrm{M}^{-1} \mathrm{~s}^{-1}\), respectively. (a) What is the value for the equilibrium constant at \(25^{\circ} \mathrm{C}\) ? (b) Are reactants or products more plentiful at equilibrium?

Two different proteins \(\mathrm{X}\) and \(\mathrm{Y}\) are dissolved in aqueous solution at \(37^{\circ} \mathrm{C}\). The proteins bind in a \(1: 1\) ratio to form XY. A solution that is initially \(1.00 \mathrm{mM}\) in each protein is allowed to reach equilibrium. At equilibrium, \(0.20 \mathrm{~m} M\) of free \(\mathrm{X}\) and \(0.20 \mathrm{~m} M\) of free \(\mathrm{Y}\) remain. What is \(K_{c}\) for the reaction?

If \(K_{c}=0.042\) for \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)\) at \(500 \mathrm{~K}\), what is the value of \(K_{p}\) for this reaction at this temperature?

Consider the reaction \(\mathrm{IO}_{4}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{4} \mathrm{IO}_{6}{ }^{-}(a q)\); \(K_{c}=3.5 \times 10^{-2}\). If you start with \(25.0 \mathrm{~mL}\) of a \(0.905 \mathrm{M}\) solution of \(\mathrm{NaIO}_{4}\), and then dilute it with water to \(500.0 \mathrm{~mL}\), what is the concentration of \(\mathrm{H}_{4} \mathrm{IO}_{6}{ }^{-}\)at equilibrium?

The reaction \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)\) has \(K_{p}=\) \(0.0870\) at \(300^{\circ} \mathrm{C}\). A flask is charged with \(0.50\) atm \(\mathrm{PCl}_{3}, 0.50\) atm \(\mathrm{Cl}_{2}\), and \(0.20 \mathrm{~atm} \mathrm{PCl}_{5}\) at this temperature. (a) Use the reaction quotient to determine the direction the reaction must proceed to reach equilibrium. (b) Calculate the equilibrium partial pressures of the gases. (c) What effect will increasing the volume of the system have on the mole fraction of \(\mathrm{Cl}_{2}\) in the equilibrium mixture? (d) The reaction is exothermic. What effect will increasing the temperature of the system have on the mole fraction of \(\mathrm{Cl}_{2}\) in the equilibrium mixture?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free