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Chapter 13: Problem 11

Explain why the development of a vapor pressure above a liquid in a closed container represents an equilibrium. What are the opposing processes? How do we recognize when the system has reached a state of equilibrium?

Short Answer

Expert verified
The development of vapor pressure above a liquid in a closed container represents an equilibrium because the opposing processes of evaporation and condensation occur at equal rates. Evaporation is when liquid molecules gain enough energy to overcome intermolecular forces and enter the gas phase, while condensation is when gas molecules lose energy and reattach to the liquid's surface. We can recognize a system has reached equilibrium when the vapor pressure remains constant over time, which can be measured using a pressure gauge or manometer, or by observing no changes in the amounts of liquid or gas within the system.

Step by step solution

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1. Understanding Vapor Pressure Equilibrium

Vapor pressure develops when molecules in a liquid escape its surface and become gas particles in the space above the liquid. In a closed container, there is limited space for these gas particles to disperse, and the pressure created by these gas particles is referred to as the vapor pressure. Over time, the vapor pressure will increase as more liquid molecules escape into the gas phase. However, eventually, an equilibrium will be reached in which the rate of liquid-to-gas conversion equals the rate of gas-to-liquid conversion. This equilibrium represents vapor pressure equilibrium.
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2. The Opposing Processes

The opposing processes in vapor pressure equilibrium involve two major participants: evaporation and condensation. In the case of evaporation, liquid molecules gain sufficient energy to overcome the intermolecular forces between the molecules and escape the liquid's surface, entering the gas phase. On the other hand, condensation involves gas molecules losing energy, resulting in a decrease in their kinetic energy, which allows them to re-attach to the liquid's surface. When the rate of evaporation equals the rate of condensation, these processes are considered to be in equilibrium.
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3. Recognizing Equilibrium

We can recognize when the system has reached a state of equilibrium by monitoring the vapor pressure. Once the vapor pressure of a system remains constant over time, it is an indication that the rate of evaporation has become equal to the rate of condensation, and the system is in equilibrium. To measure the vapor pressure, we can use a pressure gauge or a manometer. Another way to recognize equilibrium is by observing that there is no observable change in the amount of liquid or gas in the system, indicating that the processes are occurring at equal rates and equilibrium has been achieved.

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Most popular questions from this chapter

Ammonia is produced by the Haber process, in which nitrogen and hydrogen are reacted directly using an iron mesh impregnated with oxides as a catalyst. For the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ equilibrium constants $\left(K_{\mathrm{p}} \text { values ) as a function of temperature }\right.\( are\)\begin{array}{ll}{300^{\circ} \mathrm{C},} & {4.34 \times 10^{-3}} \\ {500^{\circ} \mathrm{C},} & {1.45 \times 10^{-5}} \\\ {600^{\circ} \mathrm{C},} & {2.25 \times 10^{-6}}\end{array}$ Is the reaction exothermic or endothermic?

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. $2 \mathrm{PBr}_{3}(g)+3 \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{PCl}_{3}(g)+3 \mathrm{Br}_{2}(g)$

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.)

Solid \(\mathrm{NH}_{4} \mathrm{HS}\) decomposes by the following endothermic process: $$\mathrm{NH}_{4} \mathrm{HS}(s) \leftrightharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g)$$ a. What effect will adding more \(\mathrm{NH}_{3}(g)\) have on the equilibrium? b. What effect will adding more \(\mathrm{NH}_{4} \mathrm{HS}(s)\) have on the equilibrium? c. What effect will increasing the volume of the container have on the equilibrium? d. What effect will decreasing the temperature have on the equilibrium?

The value of the equilibrium constant \(K\) depends on which of the following (more than one answer may be correct)? a. the initial concentrations of the reactants b. the initial concentrations of the products c. the temperature of the system d. the nature of the reactants and products Explain.
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