/*! 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 91 Ammonium nitrate dissolves spont... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 19: Problem 91

Ammonium nitrate dissolves spontaneously and endothermally in water at room temperature. What can you deduce about the sign of \(\Delta S\) for this solution process?

Short Answer

Expert verified
The dissolution of ammonium nitrate in water is spontaneous and endothermic, so \(\Delta G < 0\) and \(\Delta H > 0\). Using the Gibbs free energy equation (\(\Delta G = \Delta H - T \Delta S\)), we deduce that the change in entropy (\(\Delta S\)) must be positive for the inequality to hold true. Therefore, \(\Delta S > 0\) for this solution process.

Step by step solution

01

Recall the Gibbs free energy equation

The Gibbs free energy equation relates the enthalpy, entropy, and temperature of a process to its spontaneity. The equation is given by: \[ \Delta G = \Delta H - T \Delta S\] where \(\Delta G\) is the change in Gibbs free energy, \(\Delta H\) is the change in enthalpy, \(T\) is the temperature in Kelvin, and \(\Delta S\) is the change in entropy. A spontaneous process is characterized by a negative value of \(\Delta G\), i.e., \(\Delta G < 0\).
02

Analyze the given information

We are given that the dissolution of ammonium nitrate is spontaneous and endothermic. This means that \(\Delta G < 0\) and \(\Delta H > 0\) (endothermic implies an increase in enthalpy or a positive value of \(\Delta H\)). From the Gibbs free energy equation, we have: \[\Delta G = \Delta H - T \Delta S < 0\]
03

Deduce the sign of \(\Delta S\)

From the inequality \(\Delta G = \Delta H - T \Delta S < 0\), we have: \[T \Delta S > \Delta H\] Since the temperature \(T\) is positive (room temperature) and \(\Delta H > 0\), we can deduce the sign of \(\Delta S\). For the inequality to hold true, the change in entropy (\(\Delta S\)) must be positive. Thus, the dissolution of ammonium nitrate in water has a positive change in entropy (\(\Delta S > 0\)).

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

Consider the reaction 2 \(\mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g) .(\mathbf{a})\) Using data from Appendix \(\mathrm{C},\) calculate \(\Delta G^{\circ}\) at 298 \(\mathrm{K}\) . (b) Calculate \(\Delta G\) at 298 \(\mathrm{K}\) if the partial pressures of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) are 0.40 atm and 1.60 atm, respectively.

(a) Can endothermic chemical reactions be spontaneous? (b) Can a process be spontaneous at one temperature and nonspontaneous at a different temperature? (c) Water can be decomposed to form hydrogen and oxygen, and the hydrogen and oxygen can be recombined to form water. Does this mean that the processes are thermodynamically reversible? (d) Does the amount of work that a system can do on its surroundings depend on the path of the process?

Using data from Appendix \(\mathrm{C}\) , calculate \(\Delta G^{\circ}\) for the following reactions. Indicate whether each reaction is spontaneous at 298 \(\mathrm{K}\) under standard conditions. (a) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (b) \(\mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 3 \mathrm{NO}(g)\) (c) \(6 \mathrm{Cl}_{2}(g)+2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \rightarrow 4 \mathrm{FeCl}_{3}(s)+3 \mathrm{O}_{2}(g)\) (d) \(\mathrm{SO}_{2}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)\)

A system goes from state 1 to state 2 and back to state \(1 .\) (a) Is \(\Delta E\) the same in magnitude for both the forward and reverse processes? (b) Without further information, can you conclude that the amount of heat transferred to the system as it goes from state 1 to state 2 is the same or different as compared to that upon going from state 2 back to state 1\(?(\mathbf{c})\) Suppose the changes in state are reversible processes. Is the work done by the system upon going from state 1 to state 2 the same or different as compared to that upon going from state 2 back to state 1\(?\)

A certain constant-pressure reaction is barely nonspontaneous at \(45^{\circ} \mathrm{C}\) . The entropy change for the reaction is 72 \(\mathrm{J} / \mathrm{K} .\) Estimate \(\Delta H .\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Chemistry Branches

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.