Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Problem 99

(a) For each of the following reactions, predict the sign of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) and discuss briefly how these factors determine the magnitude of \(K\). (b) Based on your general chemical knowledge, predict which of these reactions will have \(K>0 .\) (c) In each case indicate whether \(K\) should increase or decrease with increasing temperature. (i) \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{MgO}(s)\) (ii) \(2 \mathrm{KI}(s) \rightleftharpoons 2 \mathrm{~K}(g)+\mathrm{I}_{2}(g)\) (iii) \(\mathrm{Na}_{2}(g) \rightleftharpoons 2 \mathrm{Na}(g)\) (iv) \(2 \mathrm{~V}_{2} \mathrm{O}_{5}(s) \rightleftharpoons 4 \mathrm{~V}(s)+5 \mathrm{O}_{2}(g)\)

Short Answer

Expert verified
(i) \(\Delta H^{\circ} < 0\), \(\Delta S^{\circ} < 0\), \(K>0\) at low temperatures, and \(K\) decreases with increasing temperature. (ii) \(\Delta H^{\circ} > 0\), \(\Delta S^{\circ} > 0\), \(K>0\) at high temperatures, and \(K\) increases with increasing temperature. (iii) \(\Delta H^{\circ} > 0\), \(\Delta S^{\circ} > 0\), \(K>0\) at high temperatures, and \(K\) increases with increasing temperature. (iv) \(\Delta H^{\circ} > 0\), \(\Delta S^{\circ} > 0\), \(K>0\) at high temperatures, and \(K\) increases with increasing temperature.

Step by step solution

01

Determine \(\Delta H^{\circ}\) sign

For this reaction, the formation of a solid product (MgO) occurs from the solid reactants (Mg) and gaseous reactant (O2). Since energy is released when compounds are formed from their elements, the enthalpy change, \(\Delta H^{\circ}\), should be negative.
02

Determine \(\Delta S^{\circ}\) sign

The reaction is a combination of solid and gaseous reactants forming a solid product. There will be a decrease in the number of gas molecules in the process, resulting in a decrease in entropy. Therefore, \(\Delta S^{\circ}\) is negative.
03

Determine \(K\) range and temperature dependence

Since both \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are negative, this reaction would be favorable at lower temperatures, where the exothermic aspect dominates. Hence, \(K > 0\) at lower temperatures. However, for higher temperatures, the positive contribution from \(\Delta S^{\circ}T\) would make \(K\) decrease with increasing temperature. (ii) \(2 \mathrm{KI}(s) \rightleftharpoons 2 \mathrm{~K}(g)+\mathrm{I}_{2}(g)\)
04

Determine \(\Delta H^{\circ}\) sign

When solid KI decomposes into its gaseous elements, energy is needed to break the ionic bonds. Therefore, \(\Delta H^{\circ}\) is positive.
05

Determine \(\Delta S^{\circ}\) sign

In this reaction, solid KI is converted into gaseous K and I2, resulting in an increase in entropy since the number of gas molecules increases. Therefore, \(\Delta S^{\circ}\) is positive.
06

Determine \(K\) range and temperature dependence

Since both \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are positive, this reaction is favored at higher temperatures where the contribution from \(\Delta S^{\circ}T\) is significant. Therefore, \(K>0\) at high temperatures, and \(K\) will increase with increasing temperature. (iii) \(\mathrm{Na}_{2}(g) \rightleftharpoons 2 \mathrm{Na}(g)\)
07

Determine \(\Delta H^{\circ}\) sign

In this reaction, energy is needed to break the bond between Na atoms. Therefore, \(\Delta H^{\circ}\) is positive.
08

Determine \(\Delta S^{\circ}\) sign

Since the number of gas molecules increases from one to two, the entropy increases. Therefore, \(\Delta S^{\circ}\) is positive.
09

Determine \(K\) range and temperature dependence

With both \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) being positive, this reaction is favored at higher temperatures. Therefore, \(K>0\) at high temperatures, and \(K\) will increase with increasing temperature. (iv) \(2 \mathrm{~V}_{2} \mathrm{O}_{5}(s) \rightleftharpoons 4 \mathrm{~V}(s)+5 \mathrm{O}_{2}(g)\)
10

Determine \(\Delta H^{\circ}\) sign

The decomposition of V2O5 into its elements requires the input of energy to break the chemical bonds. Therefore, \(\Delta H^{\circ}\) is positive.
11

Determine \(\Delta S^{\circ}\) sign

The reaction involves the formation of gaseous O2 from solid V2O5, leading to an increase in the number of gas molecules and an increase in entropy. Therefore, \(\Delta S^{\circ}\) is positive.
12

Determine \(K\) range and temperature dependence

Since both \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are positive, this reaction is favored at higher temperatures. Therefore, \(K>0\) at high temperatures and \(K\) will increase with increasing temperature.

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) Give two examples of endothermic processes that are spontaneous. (b) Give an example of a process that is spontaneous at one temperature but nonspontaneous at a different temperature.

When most elastomeric polymers (e.g., a rubber band) are stretched, the molecules become more ordered, as illustrated here: Suppose you stretch a rubber band. (a) Do you expect the entropy of the system to increase or decrease? (b) If the rubber band were stretched isothermally, would heat need to be absorbed or emitted to maintain constant temperature? (c) Try this experiment: Stretch a rubber band and wait a moment. Then place the stretched rubber band on your upper lip, and let it return suddenly to its unstretched state (remember to keep holding on). What do you observe? Are your observations consistent with your answer to part (b)?

The \(K_{b}\) for methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is given in Appendix D. (a) Write the chemical equation for the equilibrium that corresponds to \(K_{b} .\) (b) By using the value of \(K_{b},\) calculate \(\Delta G^{\circ}\) for the equilibrium in part (a). (c) What is the value of \(\Delta G\) at equilibrium? (d) What is the value of \(\Delta G\) when \(\left[\mathrm{H}^{+}\right]=6.7 \times 10^{-9} \mathrm{M},\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]=2.4 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=0.098 \mathrm{M} ?\)

For the isothermal expansion of a gas into a vacuum, \(\Delta E=0, q=0\), and \(w=0\). (a) Is this a spontaneous process? (b) Explain why no work is done by the system during this process. (c) In thermodynamics, what is the "driving force" for the expansion of the gas?

A standard air conditioner involves a refrigerant that is typically now a fluorinated hydrocarbon, such as \(\mathrm{CH}_{2} \mathrm{~F}_{2}\). An air- conditioner refrigerant has the property that it readily vaporizes at atmospheric pressure and is easily compressed to its liquid phase under increased pressure. The operation of an air conditioner can be thought of as a closed system made up of the refrigerant going through the two stages shown here (the air circulation is not shown in this diagram). During expansion, the liquid refrigerant is released into an expansion chamber at low pressure, where it vaporizes. The vapor then undergoes compression at high pressure back to its liquid phase in a compression chamber. (a) What is the sign of \(q\) for the expansion? (b) What is the sign of \(q\) for the compression? (c) In a central air-conditioning system, one chamber is inside the home and the other is outside. Which chamber is where, and why? (d) Imagine that a sample of liquid refrigerant undergoes expansion followed by compression, so that it is back to its original state. Would you expect that to be a reversible process? (e) Suppose that a house and its exterior are both initially at \(31^{\circ} \mathrm{C}\). Some time after the air conditioner is turned on, the house is cooled to \(24^{\circ} \mathrm{C}\). Is this process spontaneous or nonspontaneous?
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

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