Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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
In a closed container with a liquid, the system reaches a state of equilibrium when the rates of evaporation and condensation become equal, leading to a constant vapor pressure above the surface. The opposing processes are evaporation, where liquid particles gain enough energy to change into a vapor, and condensation, where vapor particles lose energy and return to the liquid state. The state of equilibrium is recognized by the constant vapor pressure above the liquid and no net change in the amount of liquid and vapor within the container.

Step by step solution

01

Understanding vapor pressure and equilibrium

To understand why the development of vapor pressure above a liquid in a closed container represents an equilibrium, first, we need to understand the terms vapor pressure and equilibrium. Vapor pressure is the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system. In simpler terms, it means the pressure exerted by the gas particles above a liquid surface. Equilibrium, in the context of chemical reactions, is a state in which the rate of the forward reaction is equal to the rate of the reverse reaction. In such a state, the reactants and products do not undergo any net change with time. Now that we understand the basic concepts, let's discuss the opposing processes and recognize the state of equilibrium.
02

Identifying the opposing processes

In a closed container with a liquid, there are two opposing processes that eventually determine the vapor pressure: 1. Evaporation (or Vaporization): It's the process through which a liquid is transformed into its vapor state. Evaporation increases with an increase in temperature as more energy is provided to the particles to escape the liquid surface. 2. Condensation: It's the process in which the vapor state of a substance is transformed back into its liquid state. In a closed container, some of the vapor particles above the liquid surface lose energy and return to the liquid state. Now let's discuss how these processes contribute to the establishment of equilibrium.
03

Recognizing equilibrium in a closed container

In a closed container containing a liquid, the system will eventually reach a state of equilibrium where the rates of evaporation and condensation become equal. The vapor pressure above the liquid, at this point, remains constant since there is no net change in the number of vapor particles. We can recognize when the system has reached a state of equilibrium by observing these two signs: 1. Constant vapor pressure: When the vapor pressure above the liquid remains constant over time, it indicates that the system has achieved a state of equilibrium. The constant pressure occurs because the rates of evaporation and condensation are balanced, and there is no net change in the number of vapor particles above the liquid surface. 2. No net change in the amount of liquid and vapor: At equilibrium, the amounts of liquid and vapor in the container will remain constant since the rate of vaporization is equal to the rate of condensation. Hence, there will be no net change in the phases of the substance within the closed system. In conclusion, the development of vapor pressure above a liquid in a closed container represents an equilibrium since the rates of evaporation and condensation become equal, and the system achieves a constant vapor pressure and no net change in the phases.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Ammonia, a very important industrial chemical, is produced by the direct combination of the following elements under carefully controlled conditions: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ Suppose, in an experiment, that the reaction mixture is analyzed after equilibrium is reached, and it is found, at a particular temperature, that \(\left[\mathrm{NH}_{3}\right]=0.25 \mathrm{M}\), \(\left[\mathrm{H}_{2}\right]=3.4 \times 10^{-3} M,\) and \(\left[\mathrm{N}_{2}\right]=5.3 \times 10^{-5} M .\) Calculate the value of \(K\) at this temperature.

For the reaction $$ \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ the equilibrium constant, \(K,\) has the value \(5.21 \times 10^{-3}\) at a particular temperature. If the system is analyzed at equilibrium at this temperature, it is found that \([\mathrm{CO}(g)]=4.73 \times 10^{-3} M,\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=5.21 \times 10^{-3} M,\) and \(\left[\mathrm{CO}_{2}(g)\right]=3.99 \times 10^{-2} M .\) What is the equilibrium concentration of \(\mathrm{H}_{2}(g)\) in the system?

Under what circumstances can we compare the solubilities of two salts by directly comparing the values of their solubility products?

Hydrogen gas and chlorine gas in the presence of light react explosively to form hydrogen chloride $$ \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g) $$ The reaction is strongly exothermic. Would an increase in temperature for the system tend to favor or disfavor the production of hydrogen chloride?

Write the equilibrium expression for each of the following reactions. a. \(\mathrm{N}_{2}(g)+3 \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NCl}_{3}(g)\) b. \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\) c. \(\mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{H}_{4}(g)\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free