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
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Chapter 17, Problem 17.31QP

(a)

Interpretation Introduction

Interpretation: The ΔSο value for each of the given atmospheric reaction that contributes to the formation of photochemical smog 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 atmospheric reaction that contributes to the formation of photochemical smog.

(a)

Expert Solution
Check Mark

Answer to Problem 17.31QP

Solution

The ΔSο value for the given atmospheric reaction is 24.9JK-1_ .

Explanation of Solution

Explanation

Refer Appendix 4 .

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

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

The standard molar entropy of NO(g) is 210.7Jmol1K1 .

The given reaction is,

N2(g)+O2(g)2NO(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 NO(g) is 2mol .
  • The stoichiometry coefficient of reactant N2(g) is 1mol .
  • The stoichiometry coefficient of reactant O2(g) is 1mol .

The nproducts×Sο(products) is calculated by,

nproducts×Sο(products)=2mol×SNO(g)ο (3)

Where,

  • SNO(g)ο is the standard molar entropy of N2(g) .

Substitute the value of SNO(g)ο in equation (3).

nproducts×Sο(products)=2mol×210.7Jmol1K1=421.4JK1

The mreactants×Sο(reactants) is calculated by,

mreactants×Sο(reactants)=(1mol×SN2(g)ο)+(1mol×SO2(g)ο) (4)

Where,

  • SN2(g)ο is the standard molar entropy of N2(g) .
  • SO2(g)ο is the standard molar entropy of O2(g) .

Substitute the values of SN2(g)ο and SO2(g)ο in equation (4).

mreactants×Sο(reactants)=(1mol×191.5Jmol1K1)+(1mol×205.0Jmol1K1)=191.5JK1+205.0JK1=396.5JK1

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

ΔSο=421.4JK1396.5JK1=24.9JK1

Thus, the ΔSο value for the given atmospheric reaction is 24.9JK-1_ .

(b)

Interpretation Introduction

To determine: The ΔSο value for the given atmospheric reaction that contributes to the formation of photochemical smog.

(b)

Expert Solution
Check Mark

Answer to Problem 17.31QP

Solution

The ΔSο value for the given atmospheric reaction is -146.4JK-1_ .

Explanation of Solution

Explanation

Refer Appendix 4 .

The standard molar entropy of NO2(g) is 240.0Jmol1K1 .

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

The standard molar entropy of NO(g) is 210.7Jmol1K1 .

The given reaction is,

2NO(g)+O2(g)2NO2(g) (5)

In equation (5),

  • The stoichiometry coefficient of reactant NO(g) is 2mol .
  • The stoichiometry coefficient of reactant O2(g) is 1mol .
  • The stoichiometry coefficient of product NO2(g) is 2mol .

The nproducts×Sο(products) is calculated by,

nproducts×Sο(products)=2mol×SNO2(g)ο (6)

Where,

  • SNO2(g)ο is the standard molar entropy of NO2(g) .

Substitute the value of SNO2(g)ο in equation (6).

nproducts×Sο(products)=2mol×240.0Jmol1K1=480.0JK1

The mreactants×Sο(reactants) is calculated by,

mreactants×Sο(reactants)=(2mol×SNO(g)ο)+(1mol×SO2(g)ο) (7)

Where,

  • SNO(g)ο is the standard molar entropy of NO(g) .
  • SO2(g)ο  is the standard molar entropy of O2(g) .

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

mreactants×Sο(reactants)=(2mol×210.7Jmol1K1)+(1mol×205.0Jmol1K1)=421.4JK1+205.0JK1=626.4JK1

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

ΔSο=480.0JK1626.4JK1=146.4JK1

Thus, the ΔSο value for the given atmospheric reaction is -146.4JK-1_ .

(c)

Interpretation Introduction

To determine: The ΔSο value for the given atmospheric reaction that contributes to the formation of photochemical smog.

(c)

Expert Solution
Check Mark

Answer to Problem 17.31QP

Solution

The ΔSο value for the given atmospheric reaction is -73.2JK-1_ .

Explanation of Solution

Explanation

Refer Appendix 4 .

The standard molar entropy of NO2(g) is 240.0Jmol1K1 .

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

The standard molar entropy of NO(g) is 210.7Jmol1K1 .

The given reaction is,

NO(g)+12O2(g)NO2(g) (8)

In equation (8),

  • The stoichiometry coefficient of reactant NO(g) is 1mol .
  • The stoichiometry coefficient of reactant O2(g) is 12mol .
  • The stoichiometry coefficient of product NO2(g) is 1mol .

The nproducts×Sο(products) is calculated by,

nproducts×Sο(products)=1mol×SNO2(g)ο (9)

Where,

  • SNO2(g)ο is the standard molar entropy of NO2(g) .

Substitute the value of SNO2(g)ο in equation (9).

nproducts×Sο(products)=1mol×240.0Jmol1K1=240.0JK1

The mreactants×Sο(reactants) is calculated by,

mreactants×Sο(reactants)=(1mol×SNO(g)ο)+(12mol×SO2(g)ο) (10)

Where,

  • SNO(g)ο is the standard molar entropy of NO(g) .
  • SO2(g)ο is the standard molar entropy of O2(g)

Substitute the values of SNO(g)ο and SO2(g)ο in equation (10).

mreactants×Sο(reactants)=(1mol×210.7Jmol1K1)+(12mol×205.0Jmol1K1)=210.7JK1+102.5JK1=313.2JK1

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

ΔSο=240.0JK1313.2JK1=73.2JK1

Thus, the ΔSο value for the given atmospheric reaction is -73.2JK-1_ .

(d)

Interpretation Introduction

To determine: The ΔSο value for the given atmospheric reaction that contributes to the formation of photochemical smog.

(d)

Expert Solution
Check Mark

Answer to Problem 17.31QP

Solution

The ΔSο value for the given atmospheric reaction is -175.8JK-1_ .

Explanation of Solution

Explanation

Refer Appendix 4 .

The standard molar entropy of NO2(g) is 240.0Jmol1K1 .

The standard molar entropy of N2O4(g) is 304.2Jmol1K1 .

The given reaction is,

2NO2(g)N2O4(g) (11)

In equation (11),

  • The stoichiometry coefficient of reactant NO2(g) is 2mol .
  • The stoichiometry coefficient of product N2O4(g) is 1mol .

The nproducts×Sο(products) is calculated by,

nproducts×Sο(products)=1mol×SN2O4(g)ο (12)

Where,

  • SN2O4(g)ο is the standard molar entropy of N2O4(g) .

Substitute the value of SN2O4(g)ο in equation (12).

nproducts×Sο(products)=1mol×304.2Jmol1K1=304.2JK1

The mreactants×Sο(reactants) is calculated by,

mreactants×Sο(reactants)=(2mol×SNO2(g)ο) (13)

Where,

  • SNO2(g)ο is the standard molar entropy of NO2(g) .

Substitute the value of SNO2(g)ο in equation (13).

mreactants×Sο(reactants)=2mol×240.0Jmol1K1=480.0JK1

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

ΔSο=304.2JK1480.0JK1=175.8JK1

Thus, the ΔSο value for the given atmospheric reaction is -175.8JK-1_ .

Conclusion

  1. a. The ΔSο value for the given atmospheric reaction is 24.9JK-1_ .
  2. b. The ΔSο value for the given atmospheric reaction is -146.4JK-1_ .
  3. c. The ΔSο value for the given atmospheric reaction is -73.2JK-1_ .
  4. d. The ΔSο value for the given atmospheric reaction is -175.8JK-1_

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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
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