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

The enthalpy of solution of \(\mathrm{KBr}\) in water is about \(+198 \mathrm{~kJ} / \mathrm{mol}\). Nevertheless, the solubility of \(\mathrm{KBr}\) in water is relatively high. Why does the solution process occur even though it is endothermic?

Short Answer

Expert verified
The solubility of KBr in water is high even though its enthalpy of solution is endothermic because the positive entropy change associated with the dissolution process offsets the positive enthalpy of solution. At normal environmental temperatures, the T∆S term in the Gibbs Free Energy equation (∆G = ∆H - T∆S) becomes large enough to counteract the endothermic enthalpy, resulting in a negative ∆G. This makes the dissolution of KBr in water a spontaneous and favorable process.

Step by step solution

01

Define the enthalpy of solution

Enthalpy of solution is the amount of heat absorbed or released when a solute dissolves in a solvent. A positive enthalpy of solution value means an endothermic process—the dissolution requires energy input. In this case, KBr's enthalpy of solution in water is +198 kJ/mol, which means the dissolving process is endothermic.
02

Discuss entropy and its role in the dissolution process

Entropy is a measure of the disorder or randomness of a system. The dissolution of a solute in a solvent generally increases the entropy because the solute particles become more dispersed. This results in a positive value for the entropy change of solution. In this case, the dissolution of KBr in water leads to an increase in entropy due to the dispersion of KBr particles in the water.
03

Introduce Gibbs Free Energy and its relation with enthalpy and entropy

Gibbs Free Energy (G) is the overall driving force of a process, which combines both the enthalpy (H) and the entropy (S) of the reaction. The change in Gibbs Free Energy of a process (∆G) is given by the following equation: ∆G = ∆H - T∆S where ∆H is the change in enthalpy, T is the temperature in Kelvin, and ∆S is the change in entropy. For a process to be spontaneous and favorable, the Gibbs Free Energy change needs to be negative.
04

Apply the Gibbs Free Energy equation to KBr dissolution

In the case of KBr dissolution, we know that the enthalpy change (∆H) is positive (+198 kJ/mol), indicating an endothermic process. However, the entropy change (∆S) is also positive, as the dissolution increases the disorder in the system. When evaluating the Gibbs Free Energy equation, the temperature term (T) can have a significant impact. At higher temperatures, the T∆S term may outweigh the positive ∆H, resulting in a negative ∆G. In such cases, the dissolution process becomes spontaneous and favorable.
05

Explain the high solubility of KBr in water

Despite the endothermic nature of KBr dissolution, its solubility in water remains high. This is because, at normal environmental temperatures, the positive entropy change from the dissolution process offsets the endothermic enthalpy of solution. As a result, the overall Gibbs Free Energy change becomes negative, making the dissolution of KBr in water a spontaneous and favorable process.

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

Describe how you would prepare each of the following aqueous solutions: (a) \(1.50 \mathrm{~L}\) of \(0.110 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) solution, starting with solid \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} ;\) (b) \(120 \mathrm{~g}\) of a solution that is \(0.65 \mathrm{~m}\) in \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), starting with the solid solute; (c) \(1.20 \mathrm{~L}\) of a solution that is \(15.0 \% \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}\) by mass (the density of the solution is \(1.16 \mathrm{~g} / \mathrm{mL}\) ), starting with solid solute; (d) a \(0.50 \mathrm{M}\) solution of \(\mathrm{HCl}\) that would just neutralize \(5.5 \mathrm{~g}\) of \(\mathrm{Ba}(\mathrm{OH})_{2}\) starting with \(6.0\) M HCl.

Two beakers are placed in a sealed box at \(25^{\circ} \mathrm{C}\). One beaker contains \(30.0 \mathrm{~mL}\) of a \(0.050 \mathrm{M}\) aqueous solution of a nonvolatile nonelectrolyte. The other beaker contains \(30.0 \mathrm{~mL}\) of a \(0.035 \mathrm{M}\) aqueous solution of \(\mathrm{NaCl}\). The water vapor from the two solutions reaches equilibrium. (a) In which beaker does the solution level rise, and in which one does it fall? (b) What are the volumes in the two beakers when equilibrium is attained, assuming ideal behavior?

Commercial concentrated aqueous ammonia is \(28 \% \mathrm{NH}_{3}\) by mass and has a density of \(0.90 \mathrm{~g} / \mathrm{mL}\). What is the molarity of this solution?

Acetonitrile (CH \(_{3} \mathrm{CN}\) ) is a polar organic solvent that dissolves a wide range of solutes, including many salts. The density of a \(1.80 \mathrm{M}\) LiBr solution in acetonitrile is \(0.826 \mathrm{~g} / \mathrm{cm}^{3}\). Calculate the concentration of the solution in (a) molality, (b) mole fraction of LiBr, (c) mass percentage of \(\mathrm{CH}_{3} \mathrm{CN}\).

(a) Explain why carbonated beverages must be stored in sealed containers. (b) Once the beverage has been opened, why does it maintain more carbonation when refrigerated than at room temperature?

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