Chemistry: Atoms First
Chemistry: Atoms First
3rd Edition
ISBN: 9781259638138
Author: Julia Burdge, Jason Overby Professor
Publisher: McGraw-Hill Education
bartleby

Videos

Textbook Question
Book Icon
Chapter 14, Problem 14.23QP

Using data from Appendix 2, calculate ΔS°rxn and ΔSsurr for each of the reactions in Problem 14.11 and determine if each reaction is spontaneous at 25°C.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The standard entropy change of the reaction (ΔS°rxn) and entropy change in the surroundings (ΔS°surr) has to be calculated for the given set of reactions and the reactions has to be predicted to be spontaneous on non-spontaneous.

Concept introduction:

 Entropy is the measure of randomness in the system.   Standard entropy change in a reaction is the difference in entropy of the products and reactants.  (ΔS°rxn)   can be calculated by the following equation.

ΔS°rxn = S°Products- S°reactants

Where,

S°reactants is the standard entropy of the reactants

S°Products is the standard entropy of the products

Standard entropy change in a reaction and entropy change in the system are same. Enthalpy is the amount energy absorbed or released in a process.

 The enthalpy change in a system Ηsys) can be calculated by the following equation.

ΔHrxn = ΔH°produdcts- ΔH°reactants

Where,

ΔH°reactants is the standard entropy of the reactants

ΔH°produdcts is the standard entropy of the products

Standard enthalpy change in a reaction and entropy change in the system are same.

Using the value for the change in enthalpy in a system and the temperature, we can calculate ΔSsurr for an isothermal reaction.

ΔSsurr=-ΔΗsysT

The summation of the change in entropy of the system and surroundings will give the value for the change in enthalpy in the universe( ΔSuniv ). If the ΔSuniv is positive, then the reaction will be spontaneous.  If the ΔSuniv is negative then the reaction will be non-spontaneous.

Answer to Problem 14.23QP

The standard entropy of formation, ΔS°rxn=-75.9 JK-1mol-1

The entropy of surroundings, ΔSsurr=185JK-1mol-1

The given reaction is spontaneous

Explanation of Solution

Given,

S°(PCl5)=364.5 JK-1mol-1S°(PCl3(g))=217.1 JK-1mol-1S°(Cl2)=223.0JK-1mol-1

ΔΗf°(PCl5)=-374.9 kJmol-1ΔΗf°(PCl3(g))=-319.7kJmol-1ΔΗf°(Cl2)=0T=298K

To calculate ΔS°reaction

The ΔS°reaction is the difference in entropy of the reactants and products.  This is calculated by plugging in the values of standard entropy of the products and reactants in the given equation.

ΔS°rxn = S°Products- S°reactants

= S°(PCl5) -[S°(PCl3(g))+S°(Cl2)]= 364.5 JK-1mol-1-[217.1 JK-1mol-1+223.0JK-1mol-1]= -75.6JK-1mol-1

To calculate ΔH°rxn

The ΔH°rxn is the difference in enthalpy of the reactants and products.  This is calculated by plugging in the values of standard enthalpy of the products and reactants in the given equation.

ΔHrxn = ΔH°produdcts- ΔH°reacttants

= ΔΗf°(PCl5) -[ΔΗf°(PCl3(g)) +ΔΗf°(Cl2)] = -374.9 kJmol-1 -[-319.7kJmol-1+0]= -55.2 KJmol-1

To calculate ΔSsurr

ΔSsurr is calculated by plugging in the values of standard enthalpy change of the system and the temperature in the given equation.

ΔSsurr=-ΔΗsysT

= -(-55.2 kJmol-1)298K= 0.185 kJK-1mol-1=185 JK-1mol

To calculate ΔSsurr

ΔSuniv is calculated by plugging in the values of ΔSsys and ΔSsurr in the given equation.

ΔSuniv=ΔSsys+ΔSsurr

= -75.6 JK-1mol-1+185 JK-1mol-1= 109JK-1mol-1

Since , ΔSuniv> 0 the given process is spontaneous

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The standard entropy change of the reaction (ΔS°rxn) and entropy change in the surroundings (ΔS°surr) has to be calculated for the given set of reactions and the reactions has to be predicted to be spontaneous on non-spontaneous.

Concept introduction:

 Entropy is the measure of randomness in the system.   Standard entropy change in a reaction is the difference in entropy of the products and reactants.  (ΔS°rxn)   can be calculated by the following equation.

ΔS°rxn = S°Products- S°reactants

Where,

S°reactants is the standard entropy of the reactants

S°Products is the standard entropy of the products

Standard entropy change in a reaction and entropy change in the system are same. Enthalpy is the amount energy absorbed or released in a process.

 The enthalpy change in a system Ηsys) can be calculated by the following equation.

ΔHrxn = ΔH°produdcts- ΔH°reactants

Where,

ΔH°reactants is the standard entropy of the reactants

ΔH°produdcts is the standard entropy of the products

Standard enthalpy change in a reaction and entropy change in the system are same.

Using the value for the change in enthalpy in a system and the temperature, we can calculate ΔSsurr for an isothermal reaction.

ΔSsurr=-ΔΗsysT

The summation of the change in entropy of the system and surroundings will give the value for the change in enthalpy in the universe( ΔSuniv ). If the ΔSuniv is positive, then the reaction will be spontaneous.  If the ΔSuniv is negative then the reaction will be non-spontaneous.

Answer to Problem 14.23QP

The standard entropy of formation, ΔS°rxn= 215.8 JK-1mol-1

The entropy of surroundings, ΔSsurr=-609 JK-1mol-1

The given reaction is non-spontaneous

Explanation of Solution

Given,

S°[Hg]=77.4JK-1mol-1[O2]=205.0 JK-1mol1S°(HgO)=72.0JK-1mol-1

ΔΗf°(Hg)=0ΔΗf°[O2]=0ΔΗf°(HgO)=-90.7kJmol-1T=298K

To calculate ΔS°rxn

The ΔS°rxn is the difference in entropy of the reactants and products.  This is calculated by plugging in the values of standard entropy of the products and reactants in the given equation.

ΔS°rxn = S°products- S°reactants

= 2S°[Hg]+ S°[O2]-2S°(HgO)= (2)(77.4JK-1mol-1)+205.0 JK-1mol-[(2)(72.0JK-1mol-1] = 215.8JK-1mol-1

To calculate ΔH°rxn

The ΔH°rxn is the difference in enthalpy of the reactants and products.  This is calculated by plugging in the values of standard enthalpy of the products and reactants in the given equation.

ΔHrxn = ΔH°products- ΔH°reactants

= 2ΔΗf°(Hg)+ ΔΗf°[O2]-2ΔΗf°(HgO)= 0+0 - [2(-90.7kJmol-1)]=181.4kJmol-1

To calculate ΔSsurr

ΔSsurr is calculated by plugging in the values of standard enthalpy change of the system and the temperature in the given equation.

ΔSsurr=-ΔΗsysT

= -(181.4 kJmol-1)298K= -0.609 kJK-1mol-1=-609 JK-1mol-1

To calculate ΔSuniv

ΔSuniv is calculated by plugging in the values of ΔSsys and ΔSsurr in the given equation.

ΔSuniv=ΔSsys+ΔSsurr

= 215.8 JK-1mol-1+(-609JK-1mol-1)= -393.2 JK-1mol-1

Since, ΔSuniv< 0 the given process is spontaneous.

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The standard entropy change of the reaction (ΔS°rxn) and entropy change in the surroundings (ΔS°surr) has to be calculated for the given set of reactions and the reactions has to be predicted to be spontaneous on non-spontaneous.

Concept introduction:

 Entropy is the measure of randomness in the system.   Standard entropy change in a reaction is the difference in entropy of the products and reactants.  (ΔS°rxn)   can be calculated by the following equation.

ΔS°rxn = S°Products- S°reactants

Where,

S°reactants is the standard entropy of the reactants

S°Products is the standard entropy of the products

Standard entropy change in a reaction and entropy change in the system are same. Enthalpy is the amount energy absorbed or released in a process.

 The enthalpy change in a system Ηsys) can be calculated by the following equation.

ΔHrxn = ΔH°produdcts- ΔH°reactants

Where,

ΔH°reactants is the standard entropy of the reactants

ΔH°produdcts is the standard entropy of the products

Standard enthalpy change in a reaction and entropy change in the system are same.

Using the value for the change in enthalpy in a system and the temperature, we can calculate ΔSsurr for an isothermal reaction.

ΔSsurr=-ΔΗsysT

The summation of the change in entropy of the system and surroundings will give the value for the change in enthalpy in the universe( ΔSuniv ). If the ΔSuniv is positive, then the reaction will be spontaneous.  If the ΔSuniv is negative then the reaction will be non-spontaneous.

Answer to Problem 14.23QP

The standard entropy of formation, ΔS°rxn=98.2JK-1mol-1

The entropy of surroundings, ΔSsurr=-1.46×1023JK-1mol-1

The given reaction is nonspontaneous

Explanation of Solution

To record the given data

ΔΗ°f[H2(g)]=436.4kJmol-1T=298K

To calculate ΔS°rxn

Explanation:

The ΔS°rxn is the difference in entropy of the reactants and products.  This is calculated by plugging in the values of standard entropy of the products and reactants in the given equation.

ΔS°rxn = S°products- S°reactants

= 2S°[H(g)]-S°[H2(g)]=(2)(114.6JK-1mol-1) - (1)(131.0JK-1mol-1)=98.2JK-1mol-1

To calculate ΔH°rxn

The ΔH°rxn is the difference in enthalpy of the reactants and products.  This is calculated by plugging in the values of standard enthalpy of the products and reactants in the given equation.

ΔHrxn = ΔH°products- ΔH°reactants

= 2ΔΗ°f[H(g)]-ΔΗ°f[H2(g)]=(2)(436.4kJmol-1) - [(1)(436.4kJmol-1)]=436.6kJmol-1

To calculate ΔSsurr

ΔSsurr is calculated by plugging in the values of standard enthalpy change of the system and the temperature in the given equation.

ΔSsurr=-ΔΗsysT

= -(436.4 kJmol-1)298K= -1.46 kJK-1mol-1=-1.46×103 JK-1mol-1

To calculate ΔSuniv

ΔSuniv is calculated by plugging in the values of ΔSsys and ΔSsurr in the given equation.

ΔSuniv=ΔSsys+ΔSsurr

= 98.2 JK-1mol-1+(-1.46×103 JK-1mol-1)= -1.36×103×103 JK-1mol-1

Since, ΔSuniv<0 the given process is spontaneous.

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The standard entropy change of the reaction (ΔS°rxn) and entropy change in the surroundings (ΔS°surr) has to be calculated for the given set of reactions and the reactions has to be predicted to be spontaneous on non-spontaneous.

Concept introduction:

 Entropy is the measure of randomness in the system.   Standard entropy change in a reaction is the difference in entropy of the products and reactants.  (ΔS°rxn)   can be calculated by the following equation.

ΔS°rxn = S°Products- S°reactants

Where,

S°reactants is the standard entropy of the reactants

S°Products is the standard entropy of the products

Standard entropy change in a reaction and entropy change in the system are same. Enthalpy is the amount energy absorbed or released in a process.

 The enthalpy change in a system Ηsys) can be calculated by the following equation.

ΔHrxn = ΔH°produdcts- ΔH°reactants

Where,

ΔH°reactants is the standard entropy of the reactants

ΔH°produdcts is the standard entropy of the products

Standard enthalpy change in a reaction and entropy change in the system are same.

Using the value for the change in enthalpy in a system and the temperature, we can calculate ΔSsurr for an isothermal reaction.

ΔSsurr=-ΔΗsysT

The summation of the change in entropy of the system and surroundings will give the value for the change in enthalpy in the universe( ΔSuniv ). If the ΔSuniv is positive, then the reaction will be spontaneous.  If the ΔSuniv is negative then the reaction will be non-spontaneous.

Answer to Problem 14.23QP

The standard entropy of formation, ΔS°rxn=-282 JK-1mol-1

The entropy of surroundings, ΔSsurr=7.20×1023JK-1mol-1

The given reaction is spontaneous

Explanation of Solution

To record the given data

S°(UF6)=378JK-1mol-1S°(U)=50.21JK-1mol-1S°(F2)=203.34JK-1mol-1

ΔΗf°(UF6)=-2147 kJmol-1ΔΗf°(U)=0ΔΗf°(F2)=0T=298K

To calculate ΔS°rxn

The ΔS°rxn is the difference in entropy of the reactants and products.  This is calculated by plugging in the values of standard entropy of the products and reactants in the given equation.

ΔS°rxn = S°produdcts- S°reactants

= 2S°(UF6)-[S°(U)+3S°(F2)]= 378JK-1mol-1-[50.21JK-1mol-1+(3)(203.34JK-1mol-1)] =-282JK-1mol-1

To calculate ΔH°rxn

The ΔH°rxn is the difference in enthalpy of the reactants and products.  This is calculated by plugging in the values of standard enthalpy of the products and reactants in the given equation.

ΔHrxn = ΔH°products- ΔH°reactants

= ΔΗf°(UF6)-[ΔΗf°(U)+3ΔΗf°(F2)= -2147 kJmol-1 - (0+0)= -2147 kJmol-1

To calculate ΔSsurr

ΔSsurr is calculated by plugging in the values of standard enthalpy change of the system and the temperature in the given equation.

ΔSsurr=-ΔΗsysT

= -(-2147 kJmol-1)298K= 7.20 kJK-1mol-1=7.20×103 JK-1mol-1

To calculate ΔSuniv

ΔSuniv is calculated by plugging in the values of ΔSsys and ΔSsurr in the given equation.

ΔSuniv= ΔSsys+ΔSsurr

= -282JK-1mol-1+7.20×103 JK-1mol-16.92×103 JK-1mol-1

Since, ΔSuniv> 0 the given process is spontaneous.

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!

Chapter 14 Solutions

Chemistry: Atoms First

Ch. 14.3 - Consider the gas-phase reaction of A2 (blue) and...Ch. 14.3 - Prob. 14.3.1SRCh. 14.3 - For which of the following chemical reactions is S...Ch. 14.3 - Prob. 14.3.3SRCh. 14.4 - Determine if each of the following is a...Ch. 14.4 - For each of the following, calculate Suniv and...Ch. 14.4 - (a) Calculate Suniv and determine if the reaction...Ch. 14.4 - The following table shows the signs of Ssys,...Ch. 14.4 - Using data from Appendix calculate S (in J/K mol)...Ch. 14.4 - Prob. 14.4.2SRCh. 14.4 - Prob. 14.4.3SRCh. 14.4 - Prob. 14.4.4SRCh. 14.4 - Prob. 14.4.5SRCh. 14.5 - According to Table 14 4, a reaction will be...Ch. 14.5 - A reaction will be spontaneous only at low...Ch. 14.5 - Given that the reaction 4Fe(s) + 3O2(g) + 6H2O(l) ...Ch. 14.5 - Prob. 5PPCCh. 14.5 - Prob. 14.6WECh. 14.5 - Prob. 6PPACh. 14.5 - For each reaction, determine the value of Gf that...Ch. 14.5 - Prob. 6PPCCh. 14.5 - Prob. 14.7WECh. 14.5 - Prob. 7PPACh. 14.5 - Prob. 7PPBCh. 14.5 - Prob. 7PPCCh. 14.5 - Prob. 14.5.1SRCh. 14.5 - Prob. 14.5.2SRCh. 14.5 - Prob. 14.5.3SRCh. 14 - Using Gf values from Appendix 2, calculate the...Ch. 14 - Prob. 14.2KSPCh. 14 - Using Grxnvalues from Appendix 2, calculate the...Ch. 14 - Prob. 14.4KSPCh. 14 - Explain what is meant by a spontaneous process....Ch. 14 - Which of the following processes are spontaneous...Ch. 14 - Prob. 14.3QPCh. 14 - Prob. 14.4QPCh. 14 - Prob. 14.5QPCh. 14 - Prob. 14.6QPCh. 14 - Prob. 14.7QPCh. 14 - Consider two gas samples at STP: one consisting of...Ch. 14 - Now consider the reaction F2(g)2F(g)at constant...Ch. 14 - Which of the following best describes why entropy...Ch. 14 - Which of the following best explains why entropy...Ch. 14 - How does the entropy of a system change for each...Ch. 14 - How does the entropy of a system change for each...Ch. 14 - Predict whether the entropy change is positive or...Ch. 14 - Prob. 14.11QPCh. 14 - Calculate Ssys for (a) the isothermal expansion of...Ch. 14 - Calculate Ssys for (a) the isothermal compression...Ch. 14 - Using the data in Appendix 2, calculate the...Ch. 14 - Using the data in Appendix 2, calculate the...Ch. 14 - For each pair of substances listed here, choose...Ch. 14 - Arrange the following substances (1 mole each) in...Ch. 14 - State the second law of thermodynamics in words,...Ch. 14 - State the third law of thermodynamics in words,...Ch. 14 - Calculate Ssurr for each of the reactions in...Ch. 14 - Calculate Ssurr for each of the reactions in...Ch. 14 - Using data from Appendix 2, calculate Srxn and...Ch. 14 - Using data from Appendix 2, calculate Srxn and...Ch. 14 - When a folded protein in solution is heated to a...Ch. 14 - Define free energy. What are its units?Ch. 14 - Why is it more convenient to predict the direction...Ch. 14 - What is the significance of the sign of Gsys?Ch. 14 - From the following combinations of H and S,...Ch. 14 - Prob. 14.29QPCh. 14 - Calculate G for the following reactions at 25C....Ch. 14 - Calculate G for the following reactions at 25C....Ch. 14 - From the values of H and S, predict which of the...Ch. 14 - Find the temperatures at which reactions with the...Ch. 14 - The molar heats of fusion and vaporization of...Ch. 14 - The molar heats of fusion and vaporization of...Ch. 14 - Use the values listed in Appendix 2 to calculate G...Ch. 14 - Certain bacteria in the soil obtain the necessary...Ch. 14 - What is a coupled reaction? What is its importance...Ch. 14 - What is the role of ATP in biological reactions?Ch. 14 - Prob. 14.40QPCh. 14 - Predict the signs of H, S, and G of the system for...Ch. 14 - A student placed 1 g of each of three compounds A,...Ch. 14 - The enthalpy change in the denaturation of a...Ch. 14 - Consider the following facts: Water freezes...Ch. 14 - Ammonium nitrate (NH4NO3) dissolves spontaneously...Ch. 14 - The standard enthalpy of formation and the...Ch. 14 - (a) Troutons rule states that the ratio of the...Ch. 14 - Referring to Problem 14.47, explain why the ratio...Ch. 14 - Prob. 14.49QPCh. 14 - Prob. 14.50QPCh. 14 - Prob. 14.51QPCh. 14 - Prob. 14.52QPCh. 14 - Prob. 14.53QPCh. 14 - The molar heat of vaporization of ethanol is 39 3...Ch. 14 - As an approximation, we can assume that proteins...Ch. 14 - When a native protein in solution is heated to a...Ch. 14 - A 74.6-g ice cube floats in the Arctic Sea. The...Ch. 14 - A reaction for which H and S are both negative is...Ch. 14 - The sublimation of carbon dioxide at 78C is given...Ch. 14 - Many hydrocarbons exist as structural isomers,...Ch. 14 - Consider the following reaction at 298 K. 2H2(s) +...Ch. 14 - Which of the following is not accompanied by an...Ch. 14 - Which of the following are not state functions: S,...Ch. 14 - Give a detailed example of each of the following,...Ch. 14 - Hydrogenation reactions (e.g., the process of...Ch. 14 - At 0 K. the entropy of carbon monoxide crystal is...Ch. 14 - Which of the following thermodynamic functions are...
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.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
Text book image
Principles of Modern Chemistry
Chemistry
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:Cengage Learning
Text book image
Chemistry for Engineering Students
Chemistry
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
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
General Chemistry - Standalone book (MindTap Cour...
Chemistry
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Cengage Learning
Text book image
Chemistry by OpenStax (2015-05-04)
Chemistry
ISBN:9781938168390
Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Publisher:OpenStax
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