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

What does it mean to say that a state of chemical or physical equilibrium is dynamic?

Short Answer

Expert verified
A state of chemical or physical equilibrium is called dynamic because, although there is no apparent change in the system's macroscopic properties, the forward and reverse reactions continue to occur at equal rates on a molecular level. This constant exchange of matter between reactants and products maintains constant concentrations, even though reactant and product molecules are continuously forming and breaking apart.

Step by step solution

01

1. Define dynamic equilibrium

A dynamic equilibrium is a state in which the rates of the forward and reverse reactions in a chemical or physical system are equal, maintaining the concentrations of reactants and products at a constant value over time. In other words, despite the appearance of no change in the macroscopic properties, the system has a continuous exchange of matter at the molecular level.
02

2. Characteristics of dynamic equilibrium

There are a few important features of dynamic equilibrium to consider: - It is only achieved in a closed system. No reactants or products can enter or leave the system. - The concentrations of reactants and products remain constant, even if they are not equal. - The rate of the forward reaction is equal to the rate of the reverse reaction. - It is established when the system reaches its lowest energy state.
03

3. Explain why a state of equilibrium is dynamic

The state of equilibrium is considered dynamic because, at the molecular level, reactions are still occurring. Reactant molecules continue to collide and form product molecules, while product molecules likewise break apart to form reactants. However, because the rates of the forward and reverse reactions are equal, there is no net change in the concentrations of reactants and products. This constant back-and-forth at the molecular level is what makes the equilibrium dynamic. In conclusion, a state of chemical or physical equilibrium is described as dynamic because, although there appears to be no macroscopic change, the forward and reverse reactions continue to occur at equal rates, maintaining a constant exchange of matter at the molecular level.

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

Write the equilibrium expression for each of the following heterogeneous equilibria. a. \(2 \mathrm{H}_{2}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(l)\) b. \(2 \mathrm{K}(s)+\mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{KH}(s)\) c. \(\operatorname{MgO}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{MgCO}_{3}(s)\)

Plants synthesize the sugar dextrose according to the following reaction by absorbing radiant energy from the sun (photosynthesis). $$6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g)$$ Will an increase in temperature tend to favor or discourage the production of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) ?\)

At the point of chemical equilibrium, the rate of the forward reaction ____ the rate of the reverse reaction.

As you learned in Chapter \(7,\) most metal hydroxides are sparingly soluble in water. Write balanced chemical equations describing the dissolving of the following metal hydroxides in water. Write the expression for \(K_{\mathrm{sp}}\) for each process. a. \(\mathrm{Cu}(\mathrm{OH})_{2}(s)\) b. \(\operatorname{Cr}(\mathrm{OH})_{3}(s)\) c. \(\mathrm{Ba}(\mathrm{OH})_{2}(s)\) d. \(\operatorname{Sn}(O H)_{2}(s)\)

For the process $$\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)$$ it is found that the equilibrium concentrations at a particular temperature are \(\left[\mathrm{H}_{2}\right]=1.4 \mathrm{M},\left[\mathrm{CO}_{2}\right]=1.3\) \(M,[\mathrm{CO}]=0.71 \mathrm{M},\) and \(\left[\mathrm{H}_{2} \mathrm{O}\right]=0.66 \mathrm{M} .\) Calculate the equilibrium constant \(K\) for the reaction under these conditions.

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