Smartwork5 Printed Access Card for Use with Chemistry: The Science in Context 5th Edition (SmartWork Access Printed Access Card)
Smartwork5 Printed Access Card for Use with Chemistry: The Science in Context 5th Edition (SmartWork Access Printed Access Card)
5th Edition
ISBN: 9780393615296
Author: Rein V. Kirss (Author), Natalie Foster (Author), Geoffrey Davies (Author) Thomas R. Gilbert (Author)
Publisher: W. W. Norton
bartleby

Videos

Question
Book Icon
Chapter 17, Problem 17.32QP

(a)

Interpretation Introduction

Interpretation: The ΔSο value for each of the given reaction is to be calculated.

Concept introduction: The standard entropy change (ΔSο) is calculated by the formula,

ΔSο=nproducts×Sο(products)mreactants×Sο(reactants)

To determine: The ΔSο value for the given reaction.

(a)

Expert Solution
Check Mark

Answer to Problem 17.32QP

Solution

The ΔSο value for the given reaction is 16.61JK-1_ .

Explanation of Solution

Explanation

Refer Appendix 4 .

The standard molar entropy of N2(g) is 191.5Jmol1K1 .

The standard molar entropy of CH4(g) is 186.2Jmol1K1 .

The standard molar entropy of HCN(g) is 201.81Jmol1K1 .

The standard molar entropy of NH3(g) is 192.5Jmol1K1 .

The given reaction is,

CH4(g)+N2(g)HCN(g)+NH3(g) (1)

The standard entropy change (ΔSο) is calculated by the formula,

ΔSο=nproducts×Sο(products)mreactants×Sο(reactants) (2)

Where,

  • nproducts is the stoichiometry coefficient of products.
  • mreactants is the stoichiometry coefficient of reactants.
  • Sο(products) is the standard molar entropy of products.
  • Sο(reactants) is the standard molar entropy of reactants.

In equation (1),

  • The stoichiometry coefficient of product HCN(g) is 1mol .
  • The stoichiometry coefficient of product NH3(g) is 1mol .
  • The stoichiometry coefficient of reactant N2(g) is 1mol .
  • The stoichiometry coefficient of reactant CH4(g) is 1mol .

The nproducts×Sο(products) is calculated by,

nproducts×Sο(products)=(1mol×SHCN(g)ο)+(1mol×SNH3(g)ο) (3)

Where,

  • SHCN(g)ο is the standard molar entropy of HCN(g) .
  • SNH3(g)ο is the standard molar entropy of NH3(g) .

Substitute the value of SHCN(g)ο and SNH3(g)ο in equation (3).

nproducts×Sο(products)=(1mol×201.81Jmol1K1)+(1mol×192.5Jmol1K1)=201.81JK1+192.5JK1=394.31JK1

The mreactants×Sο(reactants) is calculated by,

mreactants×Sο(reactants)=(1mol×SN2(g)ο)+(1mol×SCH4(g)ο) (4)

Where,

  • SN2(g)ο is the standard molar entropy of N2(g) .
  • SCH4(g)ο is the standard molar entropy of CH4(g) .

Substitute the values of SN2(g)ο and SCH4(g)ο in equation (4).

mreactants×Sο(reactants)=(1mol×191.5Jmol1K1)+(1mol×186.0Jmol1K1)=191.5JK1+186.2JK1=377.7JK1

Substitute the values of mreactants×Sο(reactants) and nproducts×Sο(products) in equation (2).

ΔSο=394.31JK1377.7JK1=16.61JK1

Thus, the ΔSο value for the given reaction is 16.61JK-1_ .

(b)

Interpretation Introduction

To determine: The ΔSο value for the given reaction.

(b)

Expert Solution
Check Mark

Answer to Problem 17.32QP

Solution

The ΔSο value for the given reaction is -11.3JK-1_ .

Explanation of Solution

Explanation

Refer Appendix 4 .

The standard molar entropy of Cu2S(s) is 120.9Jmol1K1 .

The standard molar entropy of O2(g) is 205.0Jmol1K1 .

The standard molar entropy of SO2(g) is 248.2Jmol1K1 .

The standard molar entropy of Cu(s) is 33.2Jmol1K1 .

The given reaction is,

Cu2S(s)+O2(g)2Cu(s)+SO2(g) (5)

In equation (5),

  • The stoichiometry coefficient of reactant Cu2S(s) is 1mol .
  • The stoichiometry coefficient of reactant O2(g) is 1mol .
  • The stoichiometry coefficient of product SO2(g) is 1mol .
  • The stoichiometry coefficient of product Cu(s) is 2mol .

The nproducts×Sο(products) is calculated by,

nproducts×Sο(products)=(1mol×SCu(s)ο)+(1mol×SSO2(g)ο) (6)

Where,

  • SCu(s)ο is the standard molar entropy of Cu(s) .
  • SSO2(g)ο is the standard molar entropy of SO2(g) .

Substitute the values of SCu(s)ο and SSO2(g)ο in equation (6).

nproducts×Sο(products)=(2mol×33.2Jmol1K1)+(1mol×248.2Jmol1K1)=66.4JK1+248.2JK1=314.6JK1

The mreactants×Sο(reactants) is calculated by,

mreactants×Sο(reactants)=(1mol×SCu2S(s)ο)+(1mol×SO2(g)ο) (7)

Where,

  • SCu2S(s)ο is the standard molar entropy of Cu2S(s) .
  • SO2(g)ο is the standard molar entropy of O2(g) .

Substitute the values of SCu2S(s)ο and SO2(g)ο in equation (7).

mreactants×Sο(reactants)=(1mol×120.9Jmol1K1)+(1mol×205.0Jmol1K1)=120.9JK1+205.0JK1=325.9JK1

Substitute the values of mreactants×Sο(reactants) and nproducts×Sο(products) in equation (2).

ΔSο=314.6JK1325.9JK1=11.3JK1

Thus, the ΔSο value for the given atmospheric reaction is -11.3JK-1_ .

(c)

Interpretation Introduction

To determine: The ΔSο value for the given reaction.

(c)

Expert Solution
Check Mark

Answer to Problem 17.32QP

Solution

The ΔSο value for the given reaction is -306.6JK-1_ .

Explanation of Solution

Explanation

Refer Appendix 4 .

The standard molar entropy of SO3(g) is 256.8Jmol1K1 .

The standard molar entropy of H2O(l) is 69.9Jmol1K1 .

The standard molar entropy of H2SO4(aq) is 20.1Jmol1K1 .

The given reaction is,

SO3(g)+H2O(l)H2SO4(aq) (8)

In equation (8),

  • The stoichiometry coefficient of reactant SO3(g) is 1mol .
  • The stoichiometry coefficient of reactant H2O(l) is 1mol .
  • The stoichiometry coefficient of product H2SO4(aq) is 1mol .

The nproducts×Sο(products) is calculated by,

nproducts×Sο(products)=1mol×SH2SO4(aq)ο (9)

Where,

  • SH2SO4(aq)ο is the standard molar entropy of H2SO4(aq) .

Substitute the value of SH2SO4(aq)ο in equation (9).

nproducts×Sο(products)=1mol×20.1Jmol1K1=20.1JK1

The mreactants×Sο(reactants) is calculated by,

mreactants×Sο(reactants)=(1mol×SSO3(g)ο)+(1mol×SH2O(l)ο) (10)

Where,

  • SSO3(g)ο is the standard molar entropy of SO3(g) .
  • SH2O(l)ο is the standard molar entropy of H2O(l)

Substitute the values of SSO3(g)ο and SH2O(l)ο in equation (10). mreactants×Sο(reactants)=(1mol×256.8Jmol1K1)+(1mol×69.9Jmol1K1)=256.8JK1+69.9JK1=326.7JK1

Substitute the values of mreactants×Sο(reactants) and nproducts×Sο(products) in equation (2).

ΔSο=20.1JK1326.7JK1=306.6JK1

Thus, the ΔSο value for the given atmospheric reaction is -306.6JK-1_ .

(d)

Interpretation Introduction

To determine: The ΔSο value for the given reaction.

(d)

Expert Solution
Check Mark

Answer to Problem 17.32QP

Solution

The ΔSο value for the given reaction is -124.6JK-1_ .

Explanation of Solution

Explanation

Refer Appendix 4 .

The standard molar entropy of S(g) is 167.8Jmol1K1 .

The standard molar entropy of O2(g) is 205.0Jmol1K1 .

The standard molar entropy of SO2(g) is 248.2Jmol1K1 .

The given reaction is,

S(g)+O2(g)SO2(g) (11)

In equation (11),

  • The stoichiometry coefficient of product SO2(g) is 1mol .
  • The stoichiometry coefficient of reactant S(g) is 1mol .
  • The stoichiometry coefficient of reactant O2(g) is 1mol .

The nproducts×Sο(products) is calculated by,

nproducts×Sο(products)=1mol×SSO2(g)ο (12)

Where,

  • SSO2(g)ο is the standard molar entropy of SO2(g) .

Substitute the value of SSO2(g)ο in equation (12).

nproducts×Sο(products)=1mol×248.2Jmol1K1=248.2JK1

The mreactants×Sο(reactants) is calculated by,

mreactants×Sο(reactants)=(1mol×SS(g)ο)+(1mol×SO2(g)ο) (13)

Where,

  • SS(g)ο is the standard molar entropy of S(g) .
  • SO2(g)ο is the standard molar entropy of O2(g) .

Substitute the values of SS(g)ο and SO2(g)ο in equation (13).

mreactants×Sο(reactants)=(1mol×167.8Jmol1K1)+(1mol×205.0Jmol1K1)=167.8JK1+205.0JK1=372.8JK1

Substitute the values of mreactants×Sο(reactants) and nproducts×Sο(products) in equation (2).

ΔSο=248.2JK1372.8JK1=124.6JK1

Thus, the ΔSο value for the given atmospheric reaction is -124.6JK-1_ .

Conclusion

  1. a. The ΔSο value for the given reaction is 16.61JK-1_ .
  2. b. The ΔSο value for the given reaction is -11.3JK-1_ .
  3. c. The ΔSο value for the given reaction is -306.6JK-1_ .
  4. d. The ΔSο value for the given reaction is -124.6JK-1_

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 17 Solutions

Smartwork5 Printed Access Card for Use with Chemistry: The Science in Context 5th Edition (SmartWork Access Printed Access Card)

Ch. 17 - Prob. 17.3VPCh. 17 - Prob. 17.4VPCh. 17 - Prob. 17.5VPCh. 17 - Prob. 17.6VPCh. 17 - Prob. 17.7VPCh. 17 - Prob. 17.8VPCh. 17 - Prob. 17.9QPCh. 17 - Prob. 17.10QPCh. 17 - Prob. 17.11QPCh. 17 - Prob. 17.12QPCh. 17 - Prob. 17.13QPCh. 17 - Prob. 17.14QPCh. 17 - Prob. 17.15QPCh. 17 - Prob. 17.16QPCh. 17 - Prob. 17.17QPCh. 17 - Prob. 17.18QPCh. 17 - Prob. 17.19QPCh. 17 - Prob. 17.20QPCh. 17 - Prob. 17.21QPCh. 17 - Prob. 17.22QPCh. 17 - Prob. 17.23QPCh. 17 - Prob. 17.24QPCh. 17 - Prob. 17.25QPCh. 17 - Prob. 17.26QPCh. 17 - Prob. 17.27QPCh. 17 - Prob. 17.28QPCh. 17 - Prob. 17.29QPCh. 17 - Prob. 17.30QPCh. 17 - Prob. 17.31QPCh. 17 - Prob. 17.32QPCh. 17 - Prob. 17.33QPCh. 17 - Prob. 17.34QPCh. 17 - Prob. 17.35QPCh. 17 - Prob. 17.36QPCh. 17 - Prob. 17.37QPCh. 17 - Prob. 17.38QPCh. 17 - Prob. 17.39QPCh. 17 - Prob. 17.40QPCh. 17 - Prob. 17.41QPCh. 17 - Prob. 17.42QPCh. 17 - Prob. 17.43QPCh. 17 - Prob. 17.44QPCh. 17 - Prob. 17.45QPCh. 17 - Prob. 17.46QPCh. 17 - Prob. 17.47QPCh. 17 - Prob. 17.48QPCh. 17 - Prob. 17.49QPCh. 17 - Prob. 17.50QPCh. 17 - Prob. 17.51QPCh. 17 - Prob. 17.52QPCh. 17 - Prob. 17.53QPCh. 17 - Prob. 17.54QPCh. 17 - Prob. 17.55QPCh. 17 - Prob. 17.56QPCh. 17 - Prob. 17.57QPCh. 17 - Prob. 17.58QPCh. 17 - Prob. 17.59QPCh. 17 - Prob. 17.60QPCh. 17 - Prob. 17.61QPCh. 17 - Prob. 17.62QPCh. 17 - Prob. 17.63QPCh. 17 - Prob. 17.64QPCh. 17 - Prob. 17.65QPCh. 17 - Prob. 17.66QPCh. 17 - Prob. 17.67QPCh. 17 - Prob. 17.68QPCh. 17 - Prob. 17.69QPCh. 17 - Prob. 17.70QPCh. 17 - Prob. 17.71QPCh. 17 - Prob. 17.72QPCh. 17 - Prob. 17.73QPCh. 17 - Prob. 17.74QPCh. 17 - Prob. 17.75QPCh. 17 - Prob. 17.76QPCh. 17 - Prob. 17.77QPCh. 17 - Prob. 17.78QPCh. 17 - Prob. 17.79QPCh. 17 - Prob. 17.80QPCh. 17 - Prob. 17.81QPCh. 17 - Prob. 17.82QPCh. 17 - Prob. 17.83QPCh. 17 - Prob. 17.84QPCh. 17 - Prob. 17.85QPCh. 17 - Prob. 17.86QPCh. 17 - Prob. 17.87QPCh. 17 - Prob. 17.88APCh. 17 - Prob. 17.89APCh. 17 - Prob. 17.90APCh. 17 - Prob. 17.91APCh. 17 - Prob. 17.92APCh. 17 - Prob. 17.93APCh. 17 - Prob. 17.94APCh. 17 - Prob. 17.95APCh. 17 - Prob. 17.96APCh. 17 - Prob. 17.97APCh. 17 - Prob. 17.98APCh. 17 - Prob. 17.99APCh. 17 - Prob. 17.100APCh. 17 - Prob. 17.101APCh. 17 - Prob. 17.102APCh. 17 - Prob. 17.103APCh. 17 - Prob. 17.104APCh. 17 - Prob. 17.105AP
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
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
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