Question

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 equilibriumconstant expression in terms of molarities for the reaction, using (solv) to indicate solvation.

Short answer

(a) The equilibrium constant (K_p) in terms of partial pressures for the given reaction is: $$K_p=P_{{\mathrm{O}}_2}$$ (b) The equilibrium constant (K_c) in terms of molarities with solvation for the given reaction is: $$K_c=[\mathrm{Hg}(\mathrm{solv}){]}^4[{\mathrm{O}}_2(solv)]$$

Step-by-step solution

Identify the Balanced Reaction

First, we identify the balanced reaction given in the exercise: $$2{\mathrm{Hg}}_2\mathrm{O}(s)\rightleftharpoons 4\mathrm{Hg}(l)+{\mathrm{O}}_2(g)$$

Write the expression for the equilibrium constant in terms of partial pressures

The equilibrium constant (K_p) in terms of partial pressures is given by: $$K_p=\frac{P_{\mathrm{Hg}}^4{P}_{{\mathrm{O}}_2}}{(P_{{\mathrm{Hg}}_2\mathrm{O}}{)}^2}$$ Since the mercury(I) oxide is in solid state and elemental mercury is in liquid state, their partial pressures do not affect the equilibrium constant, so the expression simplifies to: $$K_p=P_{{\mathrm{O}}_2}$$ (b) Rewriting the equilibrium constant expression in terms of molarities for the reaction, using (solv) to indicate solvation:

Write the expression for the equilibrium constant in terms of molarities

The equilibrium constant (K_c) in terms of molarities is given by: $$K_c=\frac{[\mathrm{Hg}(\mathrm{solv}){]}^4[{\mathrm{O}}_2(solv)]}{[{\mathrm{Hg}}_2\mathrm{O}(s){]}^2}$$ Since the mercury(I) oxide is in solid state, its concentration does not affect the equilibrium constant, so the expression simplifies to: $$K_c=[\mathrm{Hg}(\mathrm{solv}){]}^4[{\mathrm{O}}_2(solv)]$$

Key concepts

Equilibrium Constant

The equilibrium constant is a vital concept in chemical equilibrium, which helps us understand the balance achieved between forward and reverse reactions at a certain point. For any chemical reaction, the equilibrium constant quantifies the ratio of the concentrations or partial pressures of products to reactants at equilibrium. It is denoted by either $K_c$, when concentrations are used, or $K_p$ when partial pressures are involved.
For reversible reactions, writing the equilibrium expression correctly is essential:

• Include only gaseous or aqueous species since solids and liquids don't affect the equilibrium constant.
• The concentrations or pressures of the products are raised to the power of their coefficients in the balanced equation.
• The expression reflects the dynamism of the reaction, as it indicates the "position" of equilibrium.

Understanding $K$ aids in predicting the extent of reactions and connector conditions, making it a fundamental tool in chemical analysis.

Partial Pressure

Partial pressure refers to the pressure exerted by a single gas in a mixture of gases. In a closed system like a reaction vessel, each gas contributes to the total pressure proportional to its quantity. When reactions involve gases, partial pressures become crucial to determining equilibrium conditions.
In the context of equilibrium reactions, partial pressure is used when gaseous components are involved:

• The expression $K_p$ relates to these pressures for reactions in the gas phase.
• Partial pressure is calculated using the ideal gas law $PV=nRT$.
• Substances in different physical states (solids/liquids) don't alter $K_p$ since they lack measurable pressures.

This makes partial pressure a handy measure to express equilibrium in gases, as seen in the mercury oxide decomposition reaction.

Molarity

Molarity is a common way to express solute concentration in a solution, calculated by dividing the moles of solute by the volume of the solvent in liters. It is represented as $mol/L$. When considering reactions happening in solutions, molarity becomes significant for expressing concentration terms for equilibrium calculations.
Key points to note about molarity in equilibrium contexts:

• Equilibrium constants expressed in molarity are denoted by $K_c$.
• Changes in concentration can shift equilibrium via Le Chatelier's principle.
• It captures how much solute effectively participates in the reaction within the solution.

Recognizing this measure's physical representation ensures accurate equilibrium constant expressions such as those involving solvated mercury and oxygen in reactions.

Solvation

Solvation refers to the process by which solvent molecules surround and interact with solute ions or molecules. In chemical equilibria involving solutions, solvation influences how reactants and products stabilize energetically and structurally.
Why solvation matters in equilibrium:

• Solvation can affect solute reactivity and solubility, impacting equilibrium positions.
• In aqueous solutions, water molecules frequently act as the solvent shell.
• Solvated species appear differently in equilibrium expressions, e.g., $[\text{Hg(solv)}]$.

Understanding solvation provides context for why specific forms of reactants/products predominate at equilibrium, enhancing the chemist's ability to manipulate and predict outcomes in dissolved systems.