Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN: 9781305079243
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
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Chapter 19, Problem 48E
Interpretation Introduction

Interpretation: Reaction of commercial production of NO is given. The value of ΔH°,ΔG° and ΔS° is to be calculated for the given reaction. The explanation of the fact that NO formed in an automobile engine but does not readily decompose back to N2 and O2 in the atmosphere is to be stated.

Concept introduction: The expression to calculate ΔH° is,

ΔH°=npΔH°(product)nfΔH°(reactant)

The expression to calculate ΔS° is,

ΔS°=npΔS°(product)nfΔS°(reactant)

A reaction is said to be favored if the value of ΔG° is negative.

To determine: The value of ΔH°,ΔG° and ΔS° for the given reaction; the explanation of the fact that NO is formed in an automobile engine but does not readily decompose back to N2 and O2 in the atmosphere.

Expert Solution & Answer
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Answer to Problem 48E

Answer

The value of ΔH° for the given reaction is 180kJ_ .

The value of ΔS° for the given reaction is 25J/mol_ .

The value of ΔG° for the given reaction is 87kJ_ .

The molecule NO does not readily decompose back to N2 and O2 in the atmosphere because rate of reaction becomes very slow at reduced temperature.

Explanation of Solution

Explanation

The value of ΔH° for the given reaction is 180kJ_ .

The stated reaction is,

N2(g)+O2(g)2NO(g)

Refer to Appendix 4 .

The value of ΔH°(kJ/mol) for the given reactant and product is,

Molecules ΔH°(kJ/mol)
NO(g) 90
N2(g) 0
O2(g) 0

The formula of ΔH° is,

ΔH°=npΔH°(product)nfΔH°(reactant)

Where,

  • ΔH° the standard enthalpy of reaction.
  • np is the number of moles of each product.
  • nr is the number of moles each reactant.
  • ΔH°(product) is the standard enthalpy of product at a pressure of 1atm .
  • ΔH°(reactant) is the standard enthalpy of reactant at a pressure of 1atm .

Substitute all values from the table in the above equation.

ΔH°=npΔH°(product)nfΔH°(reactant)=[2(90){(0)+(0)}]kJ=180kJ_

The value of ΔS° for the given reaction is 25J/mol_ .

Refer to Appendix 4

The stated reaction is,

N2(g)+O2(g)2NO(g)

The value of ΔS°(J/Kmol) for the given reactant and product is,

Molecules ΔS°(J/Kmol)
NO(g) 211
N2(g) 192
O2(g) 205

The formula of ΔS° is,

ΔS°=npΔS°(product)nfΔS°(reactant)

Where,

  • ΔS° is the standard enthalpy of reaction.
  • np is the number of moles of each product.
  • nr is the number of moles each reactant.
  • ΔS°(product) is the standard enthalpy of product at a pressure of 1atm .
  • ΔS°(reactant) is the standard enthalpy of reactant at a pressure of 1atm .

Substitute all values from the table in the above equation.

ΔS°=npΔS°(product)nfΔS°(reactant)=[2(211){(192)+(205)}]J/mol=25J/mol_

The value of ΔG° for the given reaction is 87kJ_ .

The stated reaction is,

N2(g)+O2(g)2NO(g)

Refer to Appendix 4 .

The value of ΔG°(kJ/mol) for the given reactant and product is,

Molecules ΔG°(kJ/mol)
NO(g) 87
N2(g) 0
O2(g) 0

The formula of ΔG° is,

ΔG°=npΔG°(product)nfΔG°(reactant)

Where,

  • np is the number of moles of each product.
  • nr is the number of moles each reactant.
  • ΔG°(product) is the free energy of product at a pressure of 1atm .
  • ΔG°(reactant) is the free energy of reactant at a pressure of 1atm .

Substitute all values from the table in the above equation.

ΔG°=npΔG°(product)nfΔG°(reactant)=[(87){12(0)+12(0)}]kJ=87kJ_

The molecule NO does not readily decompose back to N2 and O2 in the atmosphere because rate of reaction becomes very slow at reduced temperature.

The reaction between N2 and O2 to form NO at room temperature is unfavorable. It is formed in the combustion chamber of an automobile engine at very high temperature. When it escape into the atmosphere, the temperature decreases so rapidly that NO does not readily decompose back to N2 and O2 in the atmosphere because rate of reaction becomes very slow at reduced temperature.

Conclusion

Conclusion

The required value of ΔS° for the given reaction is 25J/mol_ , ΔG° is 87kJ_ and ΔH° is 180kJ_ . The molecule NO does not readily decompose back to N2 and O2 in the atmosphere because rate of reaction becomes very slow at reduced temperature

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Chapter 19 Solutions

Chemistry: An Atoms First Approach

Ch. 19 - Prob. 1QCh. 19 - Prob. 2QCh. 19 - Prob. 3QCh. 19 - Diagonal relationships in the periodic table exist...Ch. 19 - Prob. 5QCh. 19 - Prob. 6QCh. 19 - Prob. 7QCh. 19 - Prob. 8QCh. 19 - Prob. 9QCh. 19 - Prob. 10QCh. 19 - Prob. 11ECh. 19 - Prob. 12ECh. 19 - Prob. 13ECh. 19 - Prob. 14ECh. 19 - Prob. 15ECh. 19 - Prob. 16ECh. 19 - Prob. 17ECh. 19 - Prob. 18ECh. 19 - Prob. 19ECh. 19 - Prob. 20ECh. 19 - Prob. 21ECh. 19 - Prob. 22ECh. 19 - Prob. 23ECh. 19 - Prob. 24ECh. 19 - Consider element 113. What is the expected...Ch. 19 - Prob. 26ECh. 19 - Prob. 27ECh. 19 - Prob. 28ECh. 19 - Prob. 29ECh. 19 - Prob. 30ECh. 19 - Prob. 31ECh. 19 - Prob. 32ECh. 19 - Prob. 33ECh. 19 - Prob. 34ECh. 19 - The following illustration shows the orbitals used...Ch. 19 - Prob. 36ECh. 19 - Silicon is produced for the chemical and...Ch. 19 - Prob. 38ECh. 19 - Prob. 39ECh. 19 - Prob. 40ECh. 19 - Prob. 41ECh. 19 - Prob. 42ECh. 19 - Prob. 43ECh. 19 - Prob. 44ECh. 19 - Prob. 45ECh. 19 - Prob. 46ECh. 19 - Prob. 47ECh. 19 - Prob. 48ECh. 19 - Prob. 49ECh. 19 - Prob. 50ECh. 19 - Prob. 51ECh. 19 - Prob. 52ECh. 19 - Use bond energies to estimate the maximum...Ch. 19 - Prob. 54ECh. 19 - Prob. 55ECh. 19 - Prob. 56ECh. 19 - Prob. 57ECh. 19 - Prob. 58ECh. 19 - Prob. 59ECh. 19 - Describe the bonding in SO2 and SO3 using the...Ch. 19 - Prob. 61ECh. 19 - Prob. 62ECh. 19 - Prob. 63ECh. 19 - Prob. 64ECh. 19 - Prob. 65ECh. 19 - Prob. 66ECh. 19 - Prob. 67ECh. 19 - Prob. 68ECh. 19 - Prob. 69ECh. 19 - Prob. 70ECh. 19 - Prob. 71ECh. 19 - Prob. 72ECh. 19 - Prob. 73AECh. 19 - The inert-pair effect is sometimes used to explain...Ch. 19 - Prob. 75AECh. 19 - Prob. 76AECh. 19 - Prob. 77AECh. 19 - Prob. 78AECh. 19 - Prob. 79AECh. 19 - Draw Lewis structures for the AsCl4+ and AsCl6...Ch. 19 - Prob. 81AECh. 19 - Prob. 82AECh. 19 - Prob. 83AECh. 19 - Prob. 84AECh. 19 - Prob. 85AECh. 19 - Prob. 86AECh. 19 - Prob. 87CWPCh. 19 - Prob. 88CWPCh. 19 - Prob. 89CWPCh. 19 - Prob. 90CWPCh. 19 - What is the hybridization of the underlined...Ch. 19 - Prob. 92CWPCh. 19 - What is the hybridization of the central atom in...Ch. 19 - Prob. 94CWPCh. 19 - Prob. 95CWPCh. 19 - Prob. 96CWPCh. 19 - Prob. 97CPCh. 19 - Prob. 98CPCh. 19 - Prob. 99CPCh. 19 - Prob. 100CPCh. 19 - Prob. 101CPCh. 19 - Prob. 102CPCh. 19 - Prob. 103CPCh. 19 - Prob. 104CPCh. 19 - Prob. 105CPCh. 19 - Prob. 106IPCh. 19 - Prob. 107IPCh. 19 - Prob. 108IPCh. 19 - Prob. 109IPCh. 19 - Prob. 110MPCh. 19 - Prob. 111MP
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