Chemistry: Atoms First V1
Chemistry: Atoms First V1
1st Edition
ISBN: 9781259383120
Author: Burdge
Publisher: McGraw Hill Custom
bartleby

Videos

Textbook Question
Book Icon
Chapter 14, Problem 14.21QP

Calculate ΔSsurr for each of the reactions in Problem 14.15 and determine if each reaction is spontaneous at 25°C.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The entropy change of surroundings (ΔSsurr) in the given set of reactions has to be calculated and the reactions has to be predicted to be spontaneous or non-spontaneous.

Concept introduction:

Entropy is the measure of randomness in the system.   Standard entropy change in system ΔSsys is the difference in entropy of the products and reactants.  ΔSsys can be calculated by the following equation.

 ΔSsys= S°Products- S°reactants

Where

S°reactants is the standard entropy of the reactants

S°Products is the standard entropy of the products

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.

ΔHsystem = ΔH°produdcts- ΔH°reactants

Where,

ΔH°reactants is the standard entropy of the reactants

ΔH°produdcts is the standard entropy of the products

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

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 14.21QP

The given reaction is found to be a spontaneous process with   ΔSsurr=291JK-1mol-1

Explanation of Solution

To record the given data

Given,

S°(Cu)=33.3 JK-1mol-1[H2O(g)]=188.7 JK-1mol-1S°(H2) =131.0J K-1mol-1S°(CuO)=43.5 JK-1mol-1

ΔΗ°f(Cu)=0ΔΗ°f[H2O(g)]=-241.8 kJmol-1ΔΗ°f(H2)=0ΔΗ°f(CuO)=-155.2 kJmol-1T=298K

Explanation:

The given data in the problem statement is recorded as shown above.

To calculate ΔHsystem

Explanation:

The ΔHsystem is the difference in enthalpy of the reactants and products.  Which is calculated by plugging in the values of standard enthalpy of the products and reactants  the given equation.

ΔHsystem = ΔH°products- ΔH°reactants

= ΔΗ°f(Cu)+ΔΗ°f[H2O(g)][ΔΗ°f(H2)+ΔΗ°f(CuO)]= 0+(1)(-241.8KJmol-1)-[0+(1)(-155.2KJmol-1)]= -86.6 kJmol-1

To calculate ΔSsurr

Explanation:

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

ΔSsurr=-ΔΗsysT

= -(-86.6KJmol-1)298K= 0.291 KJK-1mol-1=291JK-1mol-1

To calculate ΔSsys

Explanation:

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

ΔS°reaction = S°products- S°reactants

= S°(Cu)+S°[H2O(g)][S°(H2)+S°(CuO)]=(1)(33.3 JK-1mol-1)+(1)(188.7 JK-1mol-1)-[(1)(131.0J K-1mol-1)+(1)(43.5 JK-1mol-1)]=47.5JK-1mol-1

To calculate ΔSuniv

Explanation:

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

ΔSuniv=ΔSsys+ΔSsurr

= 47.5 JK-1mol-1+291JK-1mol-1=339 JK-1mol-1

Since , ΔSuniv> 0 the given process is spontaneous

Conclusion

The entropy change of surroundings (ΔSsurr) in the given reaction are calculated and each reactions was identified to be spontaneous.

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The entropy change of surroundings (ΔSsurr) in the given set of reactions has to be calculated and the reactions has to be predicted to be spontaneous or non-spontaneous.

Concept introduction:

Entropy is the measure of randomness in the system.   Standard entropy change in system ΔSsys is the difference in entropy of the products and reactants.  ΔSsys can be calculated by the following equation.

 ΔSsys= S°Products- S°reactants

Where

S°reactants is the standard entropy of the reactants

S°Products is the standard entropy of the products

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.

ΔHsystem = ΔH°produdcts- ΔH°reactants

Where,

ΔH°reactants is the standard entropy of the reactants

ΔH°produdcts is the standard entropy of the products

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

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 14.21QP

The given reaction is found to be a spontaneous process with    ΔSsurr=2.10×103JK-1mol-1

Explanation of Solution

Explanation

To record the given data

Given,

S°(Al2O3)=50.99 JK-1mol-1S°(Zn)=41.6JK-1mol-1S°(Al) = 28.3 JK-1mol-1S°(ZnO)=43.9 JK-1mol-1

ΔΗ°f(Al2O3) = -1669.8 KJmol-1ΔΗ°f(Zn) = 0ΔΗ°f(Al)=0ΔΗ°f(ZnO)=-348.0 KJmol-1T=298K

Explanation:

The values are recored as shown above.

To calculate ΔHsystem

Explanation:

The ΔHsystem is the difference in enthalpy of the reactants and products.  Which is calculated by plugging in the values of standard enthalpy of the products and reactants  the given equation.

ΔHsystem = ΔH°products- ΔH°reactants

= ΔΗ°f(Al2O3)+3ΔΗ°f(Zn)- [2ΔΗ°f(Al)+3ΔΗ°f(ZnO)]= (1)(-1669.8 KJmol-1)+0-[0+(3)(-348.0 KJmol-1)]= - 626 KJmol-1

To calculate ΔSsurr

Explanation:

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

ΔSsurr=-ΔΗsysT

= -(- 626 kJmol-1)298K= 2.10 kJK-1mol-1=2.10×103JK1mol-1

To calculate ΔSsys

Explanation:

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

(ΔS°reaction) = S°products- S°reactants

=  [S°(Al2O3)+3S°(Zn)]-[2S°(Al)+3S°(ZnO)]= (1)(50.99 JK-1mol-1)+(3)(41.6JK-1mol-1)-[(2)(28.3 JK-1mol-1)+(3)(43.9 JK-1mol-1)]= -12.5 JK-1mol-1

To calculate ΔSuniv

Explanation:

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

ΔSuniv=ΔSsys+ ΔSsurr

=-12.5 JK-1mol-1+2.10×103JK-1mol-1= 2.09×103JK-1mol-1

Since, ΔSuniv> 0 the given process is spontaneous

Conclusion

The entropy change of surroundings (ΔSsurr) in the given reaction are calculated and each reactions was identified to be spontaneous.

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The entropy change of surroundings (ΔSsurr) in the given set of reactions has to be calculated and the reactions has to be predicted to be spontaneous or non-spontaneous.

Concept introduction:

Entropy is the measure of randomness in the system.   Standard entropy change in system ΔSsys is the difference in entropy of the products and reactants.  ΔSsys can be calculated by the following equation.

 ΔSsys= S°Products- S°reactants

Where

S°reactants is the standard entropy of the reactants

S°Products is the standard entropy of the products

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.

ΔHsystem = ΔH°produdcts- ΔH°reactants

Where,

ΔH°reactants is the standard entropy of the reactants

ΔH°produdcts is the standard entropy of the products

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

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 14.21QP

(c)  The given reaction is found to be a spontaneous process with   ΔSsurr=2.99×103JK-1mol-1

Explanation of Solution

To record the given data

Given,

S°(CO2)=213.6 JK-1mol-1S°(H2O(l)) = 69.9 JK-1mol-1S°(CH4)=186.2 JK-1mol-1S°(O2)=205.0 JK-1mol-1

ΔΗ°f(CO2)=-393.5 KJmol-1ΔΗ°f[H2O(l)]=-285.8 KJmol-1ΔΗ°f(CH4)=-74.85KJmol-1ΔΗ°f(O2)=0T=298K

Explanation:

The given values are recorded as shown above.

To calculate ΔHsystem

Explanation:

The ΔHsystem is the difference in enthalpy of the reactants and products.  Which is calculated by plugging in the values of standard enthalpy of the products and reactants  the given equation.

ΔHsystem = ΔH°produdcts- ΔH°reacttants

= ΔΗ°f(CO2)+2ΔΗ°f[H2O(l)][ΔΗ°f(CH4)+2ΔΗ°f(O2)]= (1)(-393.5 KJmol-1)+(2)(-285.8 KJmol-1)-[(1)(-74.85 KJmol-1)+0]= -890.3 KJmol-1

To calculate ΔSsurr

Explanation:

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

ΔSsurr=-ΔΗsysT

= -(-890.3 KJmol-1)298K= 2.99 KJ K-1mol-1=2.99×103JK-1mol-1

To calculate ΔSsys

Explanation:

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

ΔS°reaction = S°produdcts- S°reactants

= S°(CO2)+2S°(H2O(l)) - [S°(CH4)+2S°(O2)]=(1)(213.6 JK-1mol-1)+(2)(69.9 JK-1mol-1)-[(1)(186.2 JK-1mol-1)+(2)(205.0 JK-1mol-1)]=242.8JK-1mol-1

To calculate ΔSuniv

Explanation:

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

ΔSuniv=ΔSsys+ΔSsurr

Since, ΔSuniv>0 the given process is spontaneous

Conclusion

The entropy change of surroundings (ΔSsurr) in the given reaction are calculated and each reactions was identified to be 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 V1

Ch. 14.3 - Consider the gas-phase reaction of A2 (blue) and...Ch. 14.3 - Prob. 14.3.1SRCh. 14.3 - Prob. 14.3.2SRCh. 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 - Prob. 14.4.1SRCh. 14.4 - Prob. 14.4.2SRCh. 14.4 - Prob. 14.4.3SRCh. 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 - 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 - Prob. 14.12QPCh. 14 - Prob. 14.13QPCh. 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...Ch. 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. 14.4KSP
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 Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
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: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
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
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