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Chapter 14: Problem 8

Write equilibrium constant expressions for \(K_{\mathrm{c}},\) and for \(K_{P}\), if applicable, for the following processes: (a) \(2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)\) (b) \(3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g)\) (c) \(\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g)\) (d) \(\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{C}(s) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g)\) (e) \(\mathrm{HCOOH}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{HCOO}^{-}(a q)\) (f) \(2 \mathrm{HgO}(s) \rightleftharpoons 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\)

Short Answer

Expert verified
The equilibrium constant expressions for the listed reactions are - (a) Kc = \([CO]^2 [O2]\) / \([CO2]^2\), Kp = \(P_{CO}^2 P_{O2}\) / \(P_{CO2}^2\),(b) Kc = \([O3]^2\) / \([O2]^3\), Kp = \(P_{O3}^2\) / \(P_{O2}^3\),(c) Kc = \([COCl2]\) / \([CO][Cl2]\), Kp = \(P_{COCl2}\) / \(P_{CO}P_{Cl2}\),(d) Kc = \([CO][H2]\) / \([H2O]\), Kp = \(P_{CO} P_{H2}\) / \(P_{H2O}\),(e) Kc = \([H+][HCOO^-]\) / \([HCOOH]\),(f) Kc = \([O2]\), Kp = \(P_{O2}\).

Step by step solution

01

Express Equilibrium Constant for 2 CO2(g) ⇌ 2 CO(g) + O2(g)

The equilibrium constant expression for this reaction in terms of concentrations (Kc) is: Kc = \([CO]^2 [O2]\) / \([CO2]^2\). The equilibrium constant expression in terms of partial pressure (Kp) is identical, just with partial pressures in place of concentrations: Kp = \(P_{CO}^2 P_{O2}\) / \(P_{CO2}^2\).
02

Express Equilibrium Constant for 3 O2(g) ⇌ 2 O3(g)

The equilibrium constant expression for this reaction in terms of concentrations (Kc) is: Kc = \([O3]^2\) / \([O2]^3\). Similarly, the equilibrium constant expression in terms of partial pressure (Kp) is: Kp = \(P_{O3}^2\) / \(P_{O2}^3\).
03

Express Equilibrium Constant for CO(g) + Cl2(g) ⇌ COCl2(g)

The equilibrium constant expression for this reaction in terms of concentrations (Kc) is: Kc = \([COCl2]\) / \([CO][Cl2]\). And in terms of partial pressure (Kp): Kp = \(P_{COCl2}\) / \(P_{CO}P_{Cl2}\).
04

Express Equilibrium Constant for H2O(g) + C(s) ⇌ CO(g) + H2(g)

Only gases are included in the equilibrium expression, so the reaction in terms of concentrations (Kc) is: Kc = \([CO][H2]\) / \([H2O]\). And in terms of partial pressure (Kp): Kp = \(P_{CO} P_{H2}\) / \(P_{H2O}\).
05

Express Equilibrium Constant for HCOOH(aq) ⇌ H+(aq) + HCOO-(aq)

This equilibrium expression includes the aqueous solutions. In terms of concentration (Kc): Kc = \([H+][HCOO^-]\) / \([HCOOH]\). There will be no Kp for this as there is no contribution of gases.
06

Express Equilibrium Constant for 2 HgO(s) ⇌ 2 Hg(l) + O2(g)

Only gases are included in the equilibrium expression, hence in this reaction the liquids and solids are not considered. The equilibrium expression in terms of concentration (Kc) is Kc = \([O2]\). The equilibrium constant expression in terms of partial pressure (Kp) is Kp = \(P_{O2}\).

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Most popular questions from this chapter

Define equilibrium. Give two examples of a dynamic equilibrium.

The equilibrium constant \(\left(K_{P}\right)\) for the formation of the air pollutant nitric oxide (NO) in an automobile engine at \(530^{\circ} \mathrm{C}\) is \(2.9 \times 10^{-11}\) : $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$ (a) Calculate the partial pressure of NO under these conditions if the partial pressures of nitrogen and oxygen are 3.0 atm and 0.012 atm, respectively. (b) Repeat the calculation for atmospheric conditions where the partial pressures of nitrogen and oxygen are 0.78 atm and 0.21 atm and the temperature is \(25^{\circ} \mathrm{C}\). (The \(K_{P}\) for the reaction is \(4.0 \times 10^{-31}\) at this temperature.) (c) Is the formation of NO endothermic or exothermic? (d) What natural phenomenon promotes the formation of NO? Why?

Baking soda (sodium bicarbonate) undergoes thermal decomposition as follows: $$2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ Would we obtain more \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) by adding extra baking soda to the reaction mixture in (a) a closed vessel or (b) an open vessel?

Photosynthesis can be represented by $$\begin{array}{r}6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \\\\\Delta H^{\circ}=2801 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ Explain how the equilibrium would be affected by the following changes: (a) Partial pressure of \(\mathrm{CO}_{2}\) is increased. (b) \(\mathrm{O}_{2}\) is removed from the mixture. (c) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (glucose) is removed from the mixture. (d) More water is added. (e) A catalyst is added. (f) Temperature is decreased.

Consider the reaction $$2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ At \(430^{\circ} \mathrm{C},\) an equilibrium mixture consists of 0.020 mole of \(\mathrm{O}_{2}, 0.040\) mole of \(\mathrm{NO},\) and 0.96 mole of \(\mathrm{NO}_{2}\). Calculate \(K_{P}\) for the reaction, given that the total pressure is 0.20 atm.
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