Logout Open In App
Open In App
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

Chapter 18: Problem 5

How does the entropy of a system change for each of the following processes? (a) A solid melts. (b) A liquid freezes. (c) A liquid boils. (d) A vapor is converted to a solid. (e) A vapor condenses to a liquid. (f) A solid sublimes. (g) Urea dissolves in water.

Short Answer

Expert verified
Entropy increases when a solid melts, a liquid boils, a solid sublimes, and when urea dissolves in water. Entropy decreases when a liquid freezes, a vapor condenses to a liquid, and a vapor is converted to a solid.

Step by step solution

01

Understanding Entropy Change During Melting

When a solid melts to a liquid, the entropy increases. This is because the particles in a liquid are more disordered and have more freedom of movement than in a solid.
02

Understanding Entropy Change During Freezing

The reverse process, when a liquid freezes into a solid, the entropy decreases. The particles in a solid are more ordered and have less freedom of movement than in a liquid.
03

Understanding Entropy Change During Boiling

When a liquid boils to form a gas, the entropy increases. The particles in a gas are much more disordered and have a lot more freedom of movement than in a liquid.
04

Understanding Entropy Change During Vapor to Solid Conversion

When a vapor is converted to a solid, the entropy decreases significantly. The particles in a vapor are very unordered and have high freedom of movement. In a solid, the opposite is true.
05

Understanding Entropy Change During Condensation

When a vapor condenses to a liquid, the entropy decreases. The particles in a liquid are more ordered and have less freedom of movement compared to a gas.
06

Understanding Entropy Change During Sublimation

When a solid sublimes to a gas, the entropy increases greatly. The path from solid to gas bypasses the liquid state, resulting in a large increase in disorder.
07

Understanding Entropy Change During Dissolution of Urea in Water

When urea dissolves in water, the entropy increases. The dissolved particles of urea increase the disorder of the solution.

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.

Start your free trial

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

A student looked up the \(\Delta G_{\mathrm{f}}^{\circ}, \Delta H_{\mathrm{f}}^{\circ}\), and \(S^{\circ}\) values for \(\mathrm{CO}_{2}\) in Appendix 2. Plugging these values into Equation \((18.10),\) he found that \(\Delta G_{\mathrm{f}}^{\circ} \neq \Delta H_{\mathrm{f}}^{\circ}-T S^{\circ}\) at \(298 \mathrm{~K}\). What is wrong with his approach?

Comment on the correctness of the analogy sometimes used to relate a student's dormitory room becoming untidy to an increase in entropy.

Derive the following equation $$ \Delta G=R T \ln (Q / K) $$ where \(Q\) is the reaction quotient and describe how you would use it to predict the spontaneity of a reaction.

In the metabolism of glucose, the first step is the conversion of glucose to glucose 6 -phosphate: glucose \(+\mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow\) glucose 6 -phosphate \(+\mathrm{H}_{2} \mathrm{O}\) $$ \Delta G^{\circ}=13.4 \mathrm{~kJ} / \mathrm{mol} $$ Because \(\Delta G^{\circ}\) is positive, this reaction does not favor the formation of products. Show how this reaction can be made to proceed by coupling it with the hydrolysis of ATP. Write an equation for the coupled reaction and estimate the equilibrium constant for the coupled process.

Consider two carboxylic acids (acids that contain the \(-\mathrm{COOH}\) group \(): \mathrm{CH}_{3} \mathrm{COOH}\) (acetic acid, \(K_{\mathrm{a}}=1.8 \times 10^{-5}\) ) and \(\mathrm{CH}_{2} \mathrm{ClCOOH}\) (chloroacetic acid, \(K_{\mathrm{a}}=1.4 \times 10^{-3}\) ). (a) Calculate \(\Delta G^{\circ}\) for the ionization of these acids at \(25^{\circ} \mathrm{C}\) (b) From the equation \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ},\) we see that the contributions to the \(\Delta G^{\circ}\) term are an enthalpy term \(\left(\Delta H^{\circ}\right)\) and a temperature times entropy term \(\left(T \Delta S^{\circ}\right)\). These contributions are listed below for the two acids: Which is the dominant term in determining the value of \(\Delta G^{\circ}\) (and hence \(K_{\mathrm{a}}\) of the acid)? (c) What processes contribute to \(\Delta H^{\circ} ?\) (Consider the ionization of the acids as a Bronsted acid-base reaction.) (d) Explain why the \(T \Delta S^{\circ}\) term is more negative for \(\mathrm{CH}_{3} \mathrm{COOH}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Organic 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.

Sign-up for free