Organic Chemistry
Organic Chemistry
4th Edition
ISBN: 9780073402772
Author: Janice G. Smith
Publisher: MCG
bartleby

Videos

Question
Book Icon
Chapter 6, Problem 6.35P
Interpretation Introduction

(a)

Interpretation: The starting material or product which is favored at equilibrium is to be identified.

Concept introduction: The change in Gibbs free energy is represented by ΔG°. It is a state function. The value of ΔG° depends on Keq. The free energy change is calculated as,

ΔG°=2.303RTlogKeq

If the ΔG° is greater than zero and Keq is smaller than one, then the formation of starting material is favored at equilibrium. However, if the ΔG° is smaller than zero and Keq is greater than one, the formation of product is favored at equilibrium.

Expert Solution
Check Mark

Answer to Problem 6.35P

The formation of starting material is favored at the given value of Keq.

Explanation of Solution

Given

The value of Keq is 0.5.

The given value of Keq is smaller than 1. It implies that the concentration of the starting material is comparatively higher than that of the product at equilibrium. Therefore, the formation of starting material is favored at equilibrium.

Conclusion

(a) The formation of starting material is favored at the given value of Keq.

Interpretation Introduction

(b)

Interpretation: The starting material or product which is favored at equilibrium is to be identified.

Concept introduction: The change in Gibbs free energy, enthalpy and entropy is represented by ΔG°, ΔH° and ΔS° respectively. They are state functions. The relation of ΔG°, ΔH° and ΔS° is shown as,

ΔG°=ΔH°TΔS°

The change in Gibbs free energy describes the spontaneity of the reaction. The change in enthalpy describes the relative bond strength in the substance, whereas the change in entropy describes the randomness in the system.

Expert Solution
Check Mark

Answer to Problem 6.35P

The formation of the productis favored at the given value of ΔG°.

Explanation of Solution

Given:

The value of ΔG° is 100kJ/mol.

The given value of ΔG° is negative. The reaction becomes spontaneous if ΔG° has negative value. Therefore, for the given value of ΔG°, the formation of the product is favored at equilibrium.

Conclusion

The formation of the product is favored at the given value of ΔG°.

Interpretation Introduction

(c)

Interpretation: The starting material or product which is favored at equilibrium is to be identified.

Concept introduction: The change in Gibbs free energy, enthalpy and entropy is represented by ΔG°, ΔH° and ΔS° respectively. They are state functions. The relation of ΔG°, ΔH° and ΔS° is shown as,

ΔG°=ΔH°TΔS°

The change in Gibbs free energy describes the spontaneity of the reaction. The change in enthalpy describes the relative bond strength in the substance, whereas the change in entropy describes the randomness in the system.

Expert Solution
Check Mark

Answer to Problem 6.35P

The formation of the starting material is favoredat the given values of ΔH°.

Explanation of Solution

Given

The values of ΔH° are 8.0kJ/mol and 200kJ/mol.

The given values of ΔH° are positive. It indicates that bond strength in starting material is stronger than bond strength in the product. Therefore, the breaking of bonds in the starting material becomes difficult. Therefore, the formation of the starting material is favored at equilibrium.

Conclusion

The formation of the starting material is favouredat the given values of ΔH°.

Interpretation Introduction

(d)

Interpretation: The starting material or product which is favored at equilibrium is to be identified.

Concept introduction: The change in Gibbs free energy is represented by ΔG°. It is a state function. The value of ΔG° depends on Keq. The free energy change is calculated as,

ΔG°=2.303RTlogKeq

If the ΔG° is greater than zero and Keq is smaller than one, then the formation of starting material is favored at equilibrium. However, if the ΔG° is smaller than zero and Keq is greater than one, the formation of product is favored at equilibrium.

Expert Solution
Check Mark

Answer to Problem 6.35P

The formation of the productis favored at the given value of Keq.

Explanation of Solution

Given

The value of Keq is 16.

The given value of Keq is greater than 1. It implies that the concentration of the starting material is comparatively lower than that of the product. Therefore, the formation of product is favored at equilibrium.

Conclusion

The formation of the product is favored at the given value of Keq.

Interpretation Introduction

(e)

Interpretation: The starting material or product which is favored at equilibrium is to be identified.

Concept introduction: The change in Gibbs free energy, enthalpy and entropy is represented by ΔG°, ΔH° and ΔS° respectively. They are state functions. The relation of ΔG°, ΔH° and ΔS° is shown as,

ΔG°=ΔH°TΔS°

The change in Gibbs free energy describes the spontaneity of the reaction. The change in enthalpy describes the relative bond strength in the substance, whereas the change in entropy describes the randomness in the system.

Expert Solution
Check Mark

Answer to Problem 6.35P

The formation of the starting materialis favored at the given value of ΔG°.

Explanation of Solution

Given

The value of ΔG° is 2.0kJ/mol.

The given value of ΔG° is positive. The reaction becomes spontaneous if ΔG° has negative Value. Therefore, at the given value of ΔG° the formation of the starting material is favored at equilibrium.

Conclusion

The formation of the starting material is favored at the given value of ΔG°.

Interpretation Introduction

(f)

Interpretation: The starting material or product which is favored at equilibrium is to be identified.

Concept introduction: The change in Gibbs free energy, enthalpy and entropy is represented by ΔG°, ΔH° and ΔS° respectively. They are state functions. The relation of ΔG°, ΔH° and ΔS° is shown as,

ΔG°=ΔH°TΔS°

The change in Gibbs free energy describes the spontaneity of the reaction. The change in enthalpy describes the relative bond strength in the substance, whereas the change in entropy describes the randomness in the system.

Expert Solution
Check Mark

Answer to Problem 6.35P

The formation of the productis favored at the given value of ΔS°.

Explanation of Solution

Given

The value of ΔS° is 8J/(Kmol).

For the spontaneous reaction, the value of ΔG° and ΔS° must be negative and positive respectively. The given value of ΔS° is positive. Therefore, the formation of product is favored at equilibrium.

Conclusion

The formation of the product is favored at the given value of ΔS°.

Interpretation Introduction

(g)

Interpretation: The starting material or product which isfavored at equilibrium is to be identified.

Concept introduction: The change in Gibbs free energy, enthalpy and entropy is represented by ΔG°, ΔH° and ΔS° respectively. They are state functions. The relation of ΔG°, ΔH° and ΔS° is shown as,

ΔG°=ΔH°TΔS°

The change in Gibbs free energy describes the spontaneity of the reaction. The change in enthalpy describes the relative bond strength in the substance, whereas the change in entropy describes the randomness in the system.

Expert Solution
Check Mark

Answer to Problem 6.35P

The formation of the starting material is favored at the given value of ΔS°.

Explanation of Solution

Given

The value of ΔS° is 8J/(Kmol).

The value of ΔG° and ΔS° must be negative and positive respectively for the spontaneous reaction. The given value of ΔS° is negative. Hence, the reaction is non spontaneous and therefore, the formation of starting material is favored at equilibrium

Conclusion

The formation of the starting material is favored at the given value of ΔS°.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Differentiate between single links and multicenter links.
I need help on my practice final, if you could explain how to solve this that would be extremely helpful for my final thursday. Please dumb it down chemistry is not my strong suit. If you could offer strategies as well to make my life easier that would be beneficial
None

Chapter 6 Solutions

Organic Chemistry

Ch. 6 - Prob. 6.11PCh. 6 - For a reaction with H=40kJ/mol, decide which of...Ch. 6 - For a reaction with H=20kJ/mol, decide which of...Ch. 6 - Draw an energy diagram for a reaction in which the...Ch. 6 - Prob. 6.15PCh. 6 - Prob. 6.16PCh. 6 - Problem 6.19 Consider the following energy...Ch. 6 - Draw an energy diagram for a two-step reaction,...Ch. 6 - Which value if any corresponds to a faster...Ch. 6 - Prob. 6.20PCh. 6 - Problem 6.23 For each rate equation, what effect...Ch. 6 - Prob. 6.22PCh. 6 - Identify the catalyst in each equation. a....Ch. 6 - Draw the products of homolysis or heterolysis of...Ch. 6 - Explain why the bond dissociation energy for bond...Ch. 6 - Classify each transformation as substitution,...Ch. 6 - Prob. 6.27PCh. 6 - Draw the products of each reaction by following...Ch. 6 - Prob. 6.29PCh. 6 - Prob. 6.30PCh. 6 - Prob. 6.31PCh. 6 - Prob. 6.32PCh. 6 - Prob. 6.33PCh. 6 - Prob. 6.34PCh. 6 - Prob. 6.35PCh. 6 - 6.39. a. Which value corresponds to a negative...Ch. 6 - Prob. 6.37PCh. 6 - At 25 C, the energy difference Go for the...Ch. 6 - For which of the following reaction is S a...Ch. 6 - Prob. 6.40PCh. 6 - Prob. 6.41PCh. 6 - 6.44 Consider the following reaction: . Use curved...Ch. 6 - Prob. 6.43PCh. 6 - Draw an energy diagram for the Bronsted-Lowry...Ch. 6 - Prob. 6.45PCh. 6 - Prob. 6.46PCh. 6 - Prob. 6.47PCh. 6 - Prob. 6.48PCh. 6 - The conversion of acetyl chloride to methyl...Ch. 6 - Prob. 6.50PCh. 6 - Prob. 6.51PCh. 6 - 6.54 Explain why is more acidic than , even...Ch. 6 - Prob. 6.53PCh. 6 - Prob. 6.54PCh. 6 - Prob. 6.55PCh. 6 - Although Keq of equation 1 in problem 6.57 does...Ch. 6 - Prob. 6.57P
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
  • Text book image
    Chemistry
    Chemistry
    ISBN:9781305957404
    Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
    Publisher:Cengage Learning
    Text book image
    Chemistry: An Atoms First Approach
    Chemistry
    ISBN:9781305079243
    Author:Steven S. Zumdahl, Susan A. Zumdahl
    Publisher:Cengage Learning
    Text book image
    Chemistry
    Chemistry
    ISBN:9781133611097
    Author:Steven S. Zumdahl
    Publisher:Cengage Learning
  • Text book image
    Chemistry: The Molecular Science
    Chemistry
    ISBN:9781285199047
    Author:John W. Moore, Conrad L. Stanitski
    Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry: An Atoms First Approach
Chemistry
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781133611097
Author:Steven S. Zumdahl
Publisher:Cengage Learning
Text book image
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
The Laws of Thermodynamics, Entropy, and Gibbs Free Energy; Author: Professor Dave Explains;https://www.youtube.com/watch?v=8N1BxHgsoOw;License: Standard YouTube License, CC-BY