/*! 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 11 How is the entropy change that a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 11

How is the entropy change that accompanies a reaction related to the entropy change that happens when the reaction runs in reverse?

Short Answer

Expert verified
Answer: The entropy change of a reaction running in reverse is the negative of the entropy change that accompanies the forward reaction. The magnitudes of their entropy changes are the same, but with reversed signs.

Step by step solution

01

Understanding entropy as a state function

Entropy (S) is a state function, which implies that its value depends only on the initial and final states of a system, regardless of the path taken to go from the initial state to the final state. That means that if we reverse the direction of a reaction, the difference between final and initial states remains the same, just the signs will flip.
02

Entropy change accompanying a forward reaction

In a forward reaction, we have an initial state and a final state. The entropy change for the forward reaction, 螖S_forward, can be calculated using the following formula: 螖S_forward = S_final - S_initial
03

Entropy change accompanying a reverse reaction

If we reverse the direction of the reaction, then the initial and final states will also be reversed. In other words, the initial state of the reverse reaction is the final state of the forward reaction and the final state of the reverse reaction is the initial state of the forward reaction. Hence, the entropy change for the reverse reaction, 螖S_reverse, can be calculated using: 螖S_reverse = S_initial - S_final
04

Relationship between forward and reverse entropy changes

Now that we have the expressions for the entropy change in both forward and reverse reactions, we can relate these two values: 螖S_forward = -(螖S_reverse) This relationship means that the change in entropy of a reaction running in reverse is the negative of the entropy change that accompanies the forward reaction. It also highlights that the magnitudes of their entropy changes are the same but with reversed signs.

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

Use the free energies of formation from Appendix 4 to calculate the standard free-energy change for the decomposition of ammonia in the following reaction: $$2 \mathrm{NH}_{3}(g) \rightarrow \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)$$ Is the reaction spontaneous under standard conditions?

Above what temperature does nitrogen monoxide form from nitrogen and oxygen? $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{NO}(g)$$ Assume that the values of \(\Delta H_{\mathrm{rxn}}^{\circ}\) and \(\Delta S_{\mathrm{rxn}}^{\circ}\) do not change appreciably with temperature.

The values of \(\Delta H_{\mathrm{rxn}}^{\circ}\) and \(\Delta S_{\mathrm{rxn}}^{\circ}\) for the reaction $$2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{NO}_{2}(g)$$ are \(-12 \mathrm{kJ}\) and \(-146 \mathrm{J} / \mathrm{K}\) a. Use these values to calculate \(\Delta G_{\text {rxn }}^{\circ}\) at \(298 \mathrm{K}\) b. Explain why the value of \(\Delta S_{\text {rxn }}^{\circ}\) is negative.

Two allotropes ( \(A\) and \(B\) ) of sulfur interconvert at \(369 \mathrm{K}\) and 1 atm pressure:$$\mathrm{S}_{\mathrm{s}}(s, \mathrm{A}) \rightarrow \mathrm{S}_{\mathrm{g}}(s, \mathrm{B})$$.The enthalpy change in this transition is \(297 \mathrm{J} / \mathrm{mol}\). What is the entropy change?

Which of the following combinations of entropy changes for a process are mathematically possible? a. \(\Delta S_{\text {sys }}>0, \Delta S_{\text {surr }}>0, \Delta S_{\text {univ }}>0\) b. \(\Delta S_{\text {sys }}>0, \Delta S_{\text {surr }}<0, \Delta S_{\text {univ }}>0\) c. \(\Delta S_{\text {sys }}>0, \Delta S_{\text {surr }}>0, \Delta S_{\text {univ }}<0\)
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

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.