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Chapter 15: Problem 29

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

Expert verified
(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: \[K_c = [\mathrm{Hg(solv)}]^{4} [\mathrm{O}_{2}(solv)]\]

Step by step solution

01

Identify the Balanced Reaction

First, we identify the balanced reaction given in the exercise: \[2 \mathrm{Hg}_{2} \mathrm{O}(s) \rightleftharpoons 4 \mathrm{Hg}(l) + \mathrm{O}_{2}(g)\]
02

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:
03

Write the expression for the equilibrium constant in terms of molarities

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

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Most popular questions from this chapter

For thereaction, at $\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g), K_{c}=55.3\( at \)700 \mathrm{~K}$. In a 10.0-L flask containing an equilibrium mixture of the three gases, there are $1.30 \mathrm{~g} \mathrm{H}_{2}\( and \)21.0 \mathrm{~g} \mathrm{I}_{2}$. What is the mass of HI in the flask?

If \(K_{c}=0.013 \mathrm{~L} / \mathrm{mol}\) for $2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$ at \(1000 \mathrm{~K}\), what is the value of \(K_{p}\) for this reaction at this temperature?

As shown in Table \(15.2, K_{p}\) for the equilibrium $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ is \(4.39 \times 10^{-9}\) at \(450^{\circ} \mathrm{C}\). For each of the mixtures listed here, indicate whether the mixture is at equilibrium at $450^{\circ} \mathrm{C}$. If it is not at equilibrium, indicate the direction (toward product or toward reactants) in which the mixture must shift to achieve equilibrium. (a) $9.93 \mathrm{MPa} \mathrm{NH}_{3}, 4.56 \mathrm{MPa} \mathrm{N}_{2}, 5.57 \mathrm{MPa} \mathrm{H}_{2}$ (b) \(5.78 \mathrm{MPa} \mathrm{NH}_{3}, 14.49 \mathrm{MPa} \mathrm{N}_{2},\) no \(\mathrm{H}_{2}\) (c) $1.32 \mathrm{MPa} \mathrm{NH}_{3}, 2.74 \mathrm{MPa} \mathrm{N}_{2}, 8.31 \mathrm{Mpa} \mathrm{H}_{2}$

Which of the following reactions lies to the right, favoring the formation of products, and which lies to the left, favoring formation of reactants? (a) $\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\( at \)1300 \mathrm{~K} K_{c}=0.57$ (b) \(2 \mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{C}(s)\) at \(900 \mathrm{~K} K_{p}=0.0572\)

For the equilibrium $$2 \operatorname{IBr}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\operatorname{Br}_{2}(g)$$ \(K_{p}=8.5 \times 10^{-3}\) at \(150^{\circ} \mathrm{C}\). If \(5.07 \mathrm{kPa}\) of IBr is placed in a 10.0 -L container, what is the partial pressure of all substances after equilibrium is reached?
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