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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

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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Evaporation
Evaporation is a process where molecules from a liquid gain enough energy to escape into the gas phase. This happens because some molecules on the surface acquire sufficient kinetic energy to overcome the intermolecular forces holding them in the liquid.
  • In every liquid, molecules are constantly moving and exchanging energy through collisions.
  • Only those molecules with enough energy can break free from the surface, causing evaporation.
As evaporation continues, it leads to the formation of vapor above the liquid.
This process is crucial for understanding vapor pressure equilibrium in a closed system because it is one of the forces driving the system towards equilibrium. Outward evaporation from the liquid into the vapor phase competes with condensation, a key opposing process.
Condensation
Condensation is the opposite of evaporation. It occurs when gas phase molecules lose their energy and transition back into the liquid phase by attaching themselves to the liquid's surface.
  • In the gas phase, molecules constantly move around, colliding and losing or exchanging energy.
  • When they lose enough kinetic energy, they become slower and are able to attach to the surface of the liquid, undergoing condensation.
As a closed system approaches equilibrium, the rate of condensation matches the rate of evaporation.
This is a defining characteristic of reaching vapor pressure equilibrium. The unchanging level of liquid indicates that dynamic balance is achieved between these two processes.
Closed System
A closed system is essential for achieving equilibrium in terms of vapor pressure. In such a system, no matter enters or escapes, allowing vapor pressure to stabilize without being influenced by external factors.
  • A closed container prevents the exchange of gases with the external environment.
  • This helps maintain a constant number of gas molecules in the system as evaporation and condensation occur.
Over time, these stable conditions allow the system to reach equilibrium where the rate of evaporation equals the rate of condensation.
Constant vapor pressure over time indicates the achievement of equilibrium within this enclosed space.
Intermolecular Forces
Intermolecular forces are the attractions between molecules that hold them together in the liquid phase. These forces play a significant role in the processes of evaporation and condensation.
  • In evaporation, molecules need enough kinetic energy to overcome these forces and escape into the gas phase.
  • During condensation, intermolecular forces help gas molecules attach back to the liquid surface by overcoming their kinetic energy.
The balance of intermolecular forces and kinetic energy determines whether molecules remain in the liquid or escape into the vapor phase.
Understanding these forces is vital because they dictate how easily a liquid can transition to vapor and vice versa, affecting the equilibrium state in a closed system.

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Most popular questions from this chapter

At \(125^{\circ} \mathrm{C}, K_{\mathrm{p}}=0.25\) for the reaction $$2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ A \(1.00-\mathrm{L}\) flask containing 10.0 \(\mathrm{g} \mathrm{NaHCO}_{3}\) is evacuated and heated to \(125^{\circ} \mathrm{C} .\) a. Calculate the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) after equilibrium is established. b. Calculate the masses of \(\mathrm{NaHCO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) present at equilibrium. c. Calculate the minimum container volume necessary for all of the \(\mathrm{NaHCO}_{3}\) to decompose.

Suppose a reaction has the equilibrium constant \(K=1.7 \times 10^{-8}\) at a particular temperature. Will there be a large or small amount of unreacted starting material present when this reaction reaches equilibrium? Is this reaction likely to be a good source of products at this temperature?

Consider the following reactions: \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) \quad\) and \(\quad \mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \rightleftharpoons 2 \mathrm{HI}(g)\) List two property differences between these two reactions that relate to equilibrium.

Suppose a reaction has the equilibrium constant \(K=1.3 \times 10^{8} .\) What does the magnitude of this constant tell you about the relative concentrations of products and reactants that will be present once equilibrium is reached? Is this reaction likely to be a good source of the products?

The gas arsine, \(\mathrm{AsH}_{3},\) decomposes as follows: $$2 \mathrm{AsH}_{3}(g) \rightleftharpoons 2 \mathrm{As}(s)+3 \mathrm{H}_{2}(g)$$ In an experiment at a certain temperature, pure \(\mathrm{AsH}_{3}(g)\) was placed in an empty, rigid, sealed flask at a pressure of 392.0 torr. After 48 hours the pressure in the flask was observed to be constant at 488.0 torr. a. Calculate the equilibrium pressure of \(\mathrm{H}_{2}(g)\) b. Calculate \(K_{\mathrm{p}}\) for this reaction.

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