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Chapter 13: Problem 81

Calculate a value for the equilibrium constant for the reaction $$\mathrm{O}_{2}(g)+\mathrm{O}(g) \rightleftharpoons \mathrm{O}_{3}(g)$$ given $$\mathrm{NO}_{2}(g) \stackrel{h \nu}{\rightleftharpoons} \mathrm{NO}(g)+\mathrm{O}(g) \quad K=6.8 \times 10^{-49}$$ $$\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \quad K=5.8 \times 10^{-34}$$ (Hint: When reactions are added together, the equilibrium expressions are multiplied.)

Short Answer

Expert verified
The equilibrium constant for the reaction \(\mathrm{O}_{2}(g)+\mathrm{O}(g) \rightleftharpoons \mathrm{O}_{3}(g)\) is approximately \(3.944 \times 10^{-82}\).

Step by step solution

01

Adding the first two reactions together

Add reaction 2 and reaction 3: $$\mathrm{NO}_{2}(g) \stackrel{h \nu}{\rightleftharpoons} \mathrm{NO}(g)+\mathrm{O}(g)$$ $$\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$$ To add the reactions, we need to sum them: $$\mathrm{NO}_{2}(g) - \mathrm{NO}_{2}(g) + \mathrm{O}_{3}(g) \rightleftharpoons \mathrm{O}(g) + \mathrm{NO}(g) - \mathrm{NO}(g) + \mathrm{O}_{2}(g)$$
02

Simplify the reaction

Simplify the reaction by canceling out the common terms on both sides: $$\mathrm{O}_{2}(g) + \mathrm{O}(g) \rightleftharpoons \mathrm{O}_{3}(g)$$ This is our desired reaction.
03

Calculate the equilibrium constant for the combined reaction

To find the equilibrium constant for the combined reaction, multiply the equilibrium constants of reactions 2 and 3: $$K = K_1 \times K_2 = (6.8 \times 10^{-49}) \times (5.8 \times 10^{-34})$$
04

Calculate the final value for the equilibrium constant

Calculate the equilibrium constant for the combined reaction: $$K = (6.8 \times 5.8) \times 10^{-49-34} = 39.44 \times 10^{-83}$$ The equilibrium constant for the reaction $$\mathrm{O}_{2}(g)+\mathrm{O}(g) \rightleftharpoons \mathrm{O}_{3}(g)$$ is approximately $$3.944 \times 10^{-82}$$.

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Most popular questions from this chapter

Write expressions for \(K\) and \(K_{\mathrm{p}}\) for the following reactions. a. $2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{CH}_{4} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)$ b. $2 \mathrm{NBr}_{3}(s) \Longrightarrow \mathrm{N}_{2}(g)+3 \mathrm{Br}_{2}(g)$ c. $2 \mathrm{KClO}_{3}(s) \Longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)$ d. $\mathrm{CuO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Cu}(l)+\mathrm{H}_{2} \mathrm{O}(g)$

The hydrocarbon naphthalene was frequently used in mothballs until recently, when it was discovered that human inhalation of naphthalene vapors can lead to hemolytic anemia. Naphthalene is 93.71% carbon by mass, and a 0.256-mole sample of naphthalene has a mass of 32.8 g. What is the molecular formula of naphthalene? This compound works as a pesticide in mothballs by sublimation of the solid so that it fumigates enclosed spaces with its vapors according to the equation Naphthalene \((s) \rightleftharpoons\) naphthalene \((g)\) $K=4.29 \times 10^{-6}(\( at \)298 \mathrm{~K})\( If \)3.00 \mathrm{~g}$ solid naphthalene is placed into an en with a volume of \(5.00 \mathrm{~L}\) at $25^{\circ} \mathrm{C},$ what percentage thalene will have sublimed once equilibriur estahlished?

At \(900^{\circ} \mathrm{C}, K_{\mathrm{p}}=1.04\) for the reaction $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ At a low temperature, dry ice (solid \(\mathrm{CO}_{2} ),\) calcium oxide, and calcium carbonate are introduced into a \(50.0-\mathrm{L}\) reaction chamber. The temperature is raised to \(900^{\circ} \mathrm{C},\) resulting in the dry ice converting to gaseous \(\mathrm{CO}_{2} .\) For the following mixtures, will the initial amount of calcium oxide increase, decrease, or remain the same as the system moves toward equilibrium at \(900^{\circ} \mathrm{C} ?\) a. \(655 \mathrm{g} \mathrm{CaCO}_{3}, 95.0 \mathrm{g}\) CaO, \(P_{\mathrm{CO}_{2}}=2.55 \mathrm{atm}\) b. $780 \mathrm{g} \mathrm{CaCO}_{3}, 1.00 \mathrm{g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=1.04 \mathrm{atm}$ c. $0.14 \mathrm{g} \mathrm{CaCO}_{3}, 5000 \mathrm{g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=1.04 \mathrm{atm}$ d. $715 \mathrm{g} \mathrm{CaCO}_{3}, 813 \mathrm{g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=0.211 \mathrm{atm}$

Write the equilibrium expression (K) for each of the following gas-phase reactions. a. \(N_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)\) b. \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) c. $\operatorname{SiH}_{4}(g)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \operatorname{SiCl}_{4}(g)+2 \mathrm{H}_{2}(g)$ d. $2 \mathrm{PBr}_{3}(g)+3 \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{PCl}_{3}(g)+3 \mathrm{Br}_{2}(g)$

Consider the following reaction: $$\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)$$ Amounts of \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CO}, \mathrm{H}_{2},\) and \(\mathrm{CO}_{2}\) are put into a flask so that the composition corresponds to an equilibrium position. If the CO placed in the flask is labeled with radioactive \(^{14} \mathrm{C}\) will \(^{14} \mathrm{C}\) be found only in \(\mathrm{CO}\) molecules for an indefinite period of time? Explain.
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