/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 The enthalpy of solution of \(\m... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 22

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 high solubility of \(\mathrm{KBr}\) in water, despite its endothermic enthalpy of solution (\(螖H = +198\, \mathrm{kJ/mol}\)), can be explained by the significant increase in entropy (disorder) that occurs as the \(\mathrm{KBr}\) dissociates into its component ions. This increase in entropy, along with the strong ion-dipole attractions between the \(\mathrm{KBr}\) ions and water molecules, offset the positive enthalpy change and make the overall Gibbs Free Energy (螖G) favorable for \(\mathrm{KBr}\) to dissolve spontaneously in water, resulting in high solubility.

Step by step solution

01

Understanding Enthalpy and Entropy

The solubility of a substance depends not only on the enthalpy change but also the entropy change. Enthalpy change, denoted by 螖H, is a measure of energy absorbed or released during a process. A positive 螖H indicates an endothermic process (energy is absorbed), while a negative 螖H indicates an exothermic process (energy is released). Entropy, denoted by 螖S, is a measure of the degree of disorder in a system. An increase in entropy means an increase in disorder, while a decrease in entropy means a decrease in disorder.
02

Recall Gibbs Free Energy

The Gibbs Free Energy (螖G) is the main factor that determines whether a process is spontaneous or not. It is calculated using the formula: \[螖G = 螖H - T螖S\] Where 螖G is Gibbs Free Energy, 螖H is the Enthalpy change, T is the temperature in Kelvin, and 螖S is the Entropy change. A process is spontaneous if 螖G is negative, non-spontaneous if 螖G is positive.
03

Enthalpy Change of KBr in Water

We are given that the enthalpy change (螖H) of \(\mathrm{KBr}\) in water is about \(+198\, \mathrm{kJ/mol}\), which means the process is endothermic, and energy is absorbed when KBr dissolves in water. At first glance, this may lead us to believe that the solubility should be low because the process requires energy input.
04

Entropy Change of KBr in Water

When KBr dissolves in water, it breaks into its component ions - \(K^+\) and \(Br^-\). This increases the number of particles in the solution, leading to an increase in disorder or entropy (螖S). Entropy is an important factor in determining solubility, and the increase in entropy can help offset the positive enthalpy change, making the overall process more favorable.
05

Explaining High Solubility

The high solubility of KBr in water can be explained by the fact that the increase in entropy outweighs the endothermic enthalpy change. This means that although energy is absorbed when KBr dissolves in water, the resulting increase in disorder (entropy) is significant enough to make the overall process (Gibbs Free Energy) favorable, resulting in a high solubility. Additionally, the strong ion-dipole attractions between the KBr ions and the water molecules further contribute to the solubility.

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 91影视!

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

Indicate whether each statement is true or false: (a) \(\mathrm{NaCl}\) dissolves in water but not in benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) because benzene is denser than water. (b) \(\mathrm{NaCl}\) dissolves in water but not in benzene because water has a large dipole moment and benzene has zero dipole moment. (c) \(\mathrm{NaCl}\) dissolves in water but not in benzene because the water-ion interactions are stronger than benzene-ion interactions.

Most fish need at least \(4 \mathrm{ppm}\) dissolved \(\mathrm{O}_{2}\) in water for survival. (a) What is this concentration in \(\mathrm{mol} / \mathrm{L}\) ? (b) What partial pressure of \(\mathrm{O}_{2}\) above water is needed to obtain \(4 \mathrm{ppm} \mathrm{O}_{2}\) in water at \(10^{\circ} \mathrm{C}\) ? (The Henry's law constant for \(\mathrm{O}_{2}\) at this temperature is \(1.71 \times 10^{-3} \mathrm{~mol} / \mathrm{L}\)-atm.)

Compounds like sodium stearate, called "surfactants" in general, can form structures known as micelles in water, once the solution concentration reaches the value known as the critical micelle concentration (cmc). Micelles contain dozens to hundreds of molecules. The cme depends on the substance, the solvent, and the temperature. At and above the cmc, the properties of the solution vary drastically. (a) The turbidity (the amount of light scattering) of solutions increases dramatically at the cmc. Suggest an explanation. (b) The ionic conductivity of the solution dramatically changes at the cme. Suggest an explanation. (c) Chemists have developed fluorescent dyes that glow brightly only when the dye molecules are in a hydrophobic environment. Predict how the intensity of such fluorescence would relate to the concentration of sodium stearate as the sodium stearate concentration approaches and then increases past the cme.

Commercial aqueous nitric acid has a density of \(1.42 \mathrm{~g} / \mathrm{mL}\) and is \(16 \mathrm{M}\). Calculate the percent \(\mathrm{HNO}_{3}\) by mass in the solution.

A solution is made containing \(20.8 \mathrm{~g}\) of phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\) in \(425 \mathrm{~g}\) of ethanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right)\). Calculate (a) the mole fraction of phenol, (b) the mass percent of phenol, (c) the molality of phenol.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

Inorganic Chemistry

Read Explanation
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.