Chemistry: The Science in Context (Fifth Edition)
Chemistry: The Science in Context (Fifth Edition)
5th Edition
ISBN: 9780393614046
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Stacey Lowery Bretz, Geoffrey Davies
Publisher: W. W. Norton & Company
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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_

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Chapter 17 Solutions

Chemistry: The Science in Context (Fifth Edition)

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
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