Solutions Manual
Chemistry & Chemical Reactivity
10th EDITION
John C. Kotz, Paul M. Treichel, John R. Townsend
David A. Treichel & Alton J. Banks
Part 1: Chapter 18-25
Part 2: Chapter 9-17
Part 3: Chapter 1-8
All Chapters Arranged Reverse.
, Table of Contents
Chapter 1: Basic Concepts of Chemistry ..........................................................................1
Let’s Review- The Tools of Quantitative Chemistry....................................13
Chapter 2: Atoms, Molecules and Ions ...........................................................................32
Chapter 3: Chemical Reactions.......................................................................................77
Chapter 4: Stoichiometry: Quantitative Information about Chemical Reactions .........104
Chapter 5: Principles of Chemical Reactivity: Energy and Chemical Reactions .........158
Chapter 6: The Structure of Atoms ...............................................................................204
Chapter 7: The Structure of Atoms and Periodic Trends ..............................................227
Chapter 8: Bonding and Molecular Structure ...............................................................254
Chapter 9: Orbital Hybridization and Molecular Orbitals ............................................292
Chapter 10: Gases & Their Properties ..........................................................................317
Chapter 11: Intermolecular Forces and Liquids............................................................359
Chapter 12: The Solid State..........................................................................................379
Chapter 13: Solutions and Their Behavior ...................................................................407
Chapter 14: Chemical Kinetics: The Rates of Chemical Reactions .............................447
Chapter 15: Principles of Chemical Reactivity: Equilibria...........................................488
Chapter 16: Principles of Chemical Reactivity:The Chemistry of Acids and Bases ....524
Chapter 17: Principles of Chemical Reactivity:Other Aspects of Aqueous Equilibria 566
Chapter 18: Thermodynamics-Entropy and Free Energy .............................................627
Chapter 19: Principles of Chemical Reactivity: Electron Transfer Reactions ..............670
Chapter 20: Environmental Chemistry: Environment, Energy, & Sustainability.........719
Chapter 21: The Chemistry of the Main Group Elements ............................................742
Chapter 22: The Chemistry of the Transition Elements ...............................................781
Chapter 23: Carbon: Not Just Another Element ...........................................................808
Chapter 24: Biochemistry .............................................................................................842
Chapter 25: Nuclear Chemistry ....................................................................................857
, Chapter 18
Thermodynamics-Entropy and Free Energy
Applying Chemical Principles
Thermodynamics and Living Things
18.1.1. Which is the favored process?
Consider the reactions:
(1) Creatine phosphate + H2O→ Creatine + HPi ∆G = -43.3 kJ/mol
(2) Adenosine + HPi → Adenosine-5-monophosphate + H2O ∆G = +9.2 kJ/mol
(3) Creatine phosphate + Adenosine → Creatine + Adenosine-5-monophosphate -34.1 kJ/mol
The negative net ∆G (reaction 3) indicates that the transfer of phosphate from creatine to
adenosine is product-favored.
18.1.2. Relationship between ∆rG˚′ and ∆rG˚:
∆rG˚′ = ∆rG˚ + RT ln K and substituting the K expression: K =
[C] ⎡⎣ H 3O+ ⎤⎦
[ A ][ B]
[C] ⎡⎣ H O ⎤⎦
+
Δ r G 0' = Δ r G 0 + 8.31 × 10 −3 ( 298 K ) ln
3
[ A ][ B]
[1] ⎡1 × 10 ⎤ −7
Δ r G 0' = Δ r G 0 + 8.31 × 10 −3 ( 298 K ) ln ⎣ 1 1 ⎦
[ ][ ]
Δ r G 0' = Δ r G 0 + 8.31 × 10 −3 ( 298 K )( −16.12 ) = - 39.9 kJ/mol
Δ r G 0' = Δ r G 0 - 39.9 kJ/mol
Are Diamonds Forever?
18.2.1. (a) Use ∆fG° values from Appendix L to calculate ∆rG° and Keq for the reaction under
standard conditions and 298.15 K.
∆rG° = ∆fG° (graphite) - ∆fG° (diamond)
∆rG° = 0.0 kJ/mol(1mol) – 2.900 kJ/mol(1mol) = -2.9 kJ (2 sf)
∆rG˚ = - RTln Keq and -2.9 × 103 J = - (8.3145 J/K·mol)(298.15 K)ln Keq
- 2.9 × 10 3J
Solving for ln Keq: ln K eq = = 1.1698 and K = 3.22
− ( 8.3145 J/K ⋅ mol) ( 298.15 K )
, Chapter 18 Thermodynamics - Entropy and Free Energy
(b) Use ∆fH° and S° values from Appendix L to calculate ∆rG° and Keq for the reaction at
1000K. Assume that enthalpy and entropy values are valid at these temperatures. Does
heating shift the equilibrium toward the formation of diamond or graphite?
∆rH° = ∆fH°(graphite) - ∆fH°(diamond)
∆rH° = 0.0 kJ/mol(1mol) – 1.8 kJ/mol(1mol) = -1.8 kJ
∆rS° = S° (graphite) - S° (diamond)
∆rS° = 5.6 J/K·mol (1mol) - 2.377 J/K·mol (1mol) = 3.2 J/K
∆rG˚= ∆fH˚ - T∆rS˚ = -1.8 kJ(1000 J/1 kJ) – (1000 K)(3.2 J/K) = - 5000 J or -5.0 kJ
∆rG˚ = - RTln Keq
-5000 J = - (8.3145 J/K·mol)(1000 K)ln Keq and
- 5000 J
ln K eq = = 0.6013 and K eq = 1.8
− ( 8.3145 J/K ⋅ mol) (1000 K )
So heating shifts the equilibrium toward the formation of diamond.
(c) Why is the formation of diamond favored at high pressures?
Examine the phase diagram and you’ll note that as P increases, diamond is favored for
any given T.
(d) Why is the conversion done at much higher T and P?
Higher temperatures will increase the RATE (kinetics, right?) for the process!
18.2.2. Regarding the conversion of buckminsterfullerene into diamond:
(a) Calculate ∆rH˚ for conversion of C60(s) to C(diamond) at standard state at 298.2 K
∆rH˚ = 60 mol • ∆Hf˚ C(diamond)(s) - 1 mol • ∆Hf˚ C60(s)
∆rH˚ = (60 mol • 1.8 kJ/mol) – (1 mol • 2320 kJ/mol) = -2212 kJ
(b) Is conversion of buckminsterfullerene to diamond product-favored at room T?
∆rG˚ = ∆rH˚ - T∆rS
We are told to assume that ∆rS = 0
∆rG˚ = -2210 kJ – (298.2 K • 0) = -2210 kJ, so conversion is product-favored
Units for thermodynamic processes are typically expressed for the balanced equation given.
Hence the equation for the formation of HCl: H2 + Cl2 → 2 HCl has a ∆rG°, ∆rH°, and ∆rS° that
628
Chemistry & Chemical Reactivity
10th EDITION
John C. Kotz, Paul M. Treichel, John R. Townsend
David A. Treichel & Alton J. Banks
Part 1: Chapter 18-25
Part 2: Chapter 9-17
Part 3: Chapter 1-8
All Chapters Arranged Reverse.
, Table of Contents
Chapter 1: Basic Concepts of Chemistry ..........................................................................1
Let’s Review- The Tools of Quantitative Chemistry....................................13
Chapter 2: Atoms, Molecules and Ions ...........................................................................32
Chapter 3: Chemical Reactions.......................................................................................77
Chapter 4: Stoichiometry: Quantitative Information about Chemical Reactions .........104
Chapter 5: Principles of Chemical Reactivity: Energy and Chemical Reactions .........158
Chapter 6: The Structure of Atoms ...............................................................................204
Chapter 7: The Structure of Atoms and Periodic Trends ..............................................227
Chapter 8: Bonding and Molecular Structure ...............................................................254
Chapter 9: Orbital Hybridization and Molecular Orbitals ............................................292
Chapter 10: Gases & Their Properties ..........................................................................317
Chapter 11: Intermolecular Forces and Liquids............................................................359
Chapter 12: The Solid State..........................................................................................379
Chapter 13: Solutions and Their Behavior ...................................................................407
Chapter 14: Chemical Kinetics: The Rates of Chemical Reactions .............................447
Chapter 15: Principles of Chemical Reactivity: Equilibria...........................................488
Chapter 16: Principles of Chemical Reactivity:The Chemistry of Acids and Bases ....524
Chapter 17: Principles of Chemical Reactivity:Other Aspects of Aqueous Equilibria 566
Chapter 18: Thermodynamics-Entropy and Free Energy .............................................627
Chapter 19: Principles of Chemical Reactivity: Electron Transfer Reactions ..............670
Chapter 20: Environmental Chemistry: Environment, Energy, & Sustainability.........719
Chapter 21: The Chemistry of the Main Group Elements ............................................742
Chapter 22: The Chemistry of the Transition Elements ...............................................781
Chapter 23: Carbon: Not Just Another Element ...........................................................808
Chapter 24: Biochemistry .............................................................................................842
Chapter 25: Nuclear Chemistry ....................................................................................857
, Chapter 18
Thermodynamics-Entropy and Free Energy
Applying Chemical Principles
Thermodynamics and Living Things
18.1.1. Which is the favored process?
Consider the reactions:
(1) Creatine phosphate + H2O→ Creatine + HPi ∆G = -43.3 kJ/mol
(2) Adenosine + HPi → Adenosine-5-monophosphate + H2O ∆G = +9.2 kJ/mol
(3) Creatine phosphate + Adenosine → Creatine + Adenosine-5-monophosphate -34.1 kJ/mol
The negative net ∆G (reaction 3) indicates that the transfer of phosphate from creatine to
adenosine is product-favored.
18.1.2. Relationship between ∆rG˚′ and ∆rG˚:
∆rG˚′ = ∆rG˚ + RT ln K and substituting the K expression: K =
[C] ⎡⎣ H 3O+ ⎤⎦
[ A ][ B]
[C] ⎡⎣ H O ⎤⎦
+
Δ r G 0' = Δ r G 0 + 8.31 × 10 −3 ( 298 K ) ln
3
[ A ][ B]
[1] ⎡1 × 10 ⎤ −7
Δ r G 0' = Δ r G 0 + 8.31 × 10 −3 ( 298 K ) ln ⎣ 1 1 ⎦
[ ][ ]
Δ r G 0' = Δ r G 0 + 8.31 × 10 −3 ( 298 K )( −16.12 ) = - 39.9 kJ/mol
Δ r G 0' = Δ r G 0 - 39.9 kJ/mol
Are Diamonds Forever?
18.2.1. (a) Use ∆fG° values from Appendix L to calculate ∆rG° and Keq for the reaction under
standard conditions and 298.15 K.
∆rG° = ∆fG° (graphite) - ∆fG° (diamond)
∆rG° = 0.0 kJ/mol(1mol) – 2.900 kJ/mol(1mol) = -2.9 kJ (2 sf)
∆rG˚ = - RTln Keq and -2.9 × 103 J = - (8.3145 J/K·mol)(298.15 K)ln Keq
- 2.9 × 10 3J
Solving for ln Keq: ln K eq = = 1.1698 and K = 3.22
− ( 8.3145 J/K ⋅ mol) ( 298.15 K )
, Chapter 18 Thermodynamics - Entropy and Free Energy
(b) Use ∆fH° and S° values from Appendix L to calculate ∆rG° and Keq for the reaction at
1000K. Assume that enthalpy and entropy values are valid at these temperatures. Does
heating shift the equilibrium toward the formation of diamond or graphite?
∆rH° = ∆fH°(graphite) - ∆fH°(diamond)
∆rH° = 0.0 kJ/mol(1mol) – 1.8 kJ/mol(1mol) = -1.8 kJ
∆rS° = S° (graphite) - S° (diamond)
∆rS° = 5.6 J/K·mol (1mol) - 2.377 J/K·mol (1mol) = 3.2 J/K
∆rG˚= ∆fH˚ - T∆rS˚ = -1.8 kJ(1000 J/1 kJ) – (1000 K)(3.2 J/K) = - 5000 J or -5.0 kJ
∆rG˚ = - RTln Keq
-5000 J = - (8.3145 J/K·mol)(1000 K)ln Keq and
- 5000 J
ln K eq = = 0.6013 and K eq = 1.8
− ( 8.3145 J/K ⋅ mol) (1000 K )
So heating shifts the equilibrium toward the formation of diamond.
(c) Why is the formation of diamond favored at high pressures?
Examine the phase diagram and you’ll note that as P increases, diamond is favored for
any given T.
(d) Why is the conversion done at much higher T and P?
Higher temperatures will increase the RATE (kinetics, right?) for the process!
18.2.2. Regarding the conversion of buckminsterfullerene into diamond:
(a) Calculate ∆rH˚ for conversion of C60(s) to C(diamond) at standard state at 298.2 K
∆rH˚ = 60 mol • ∆Hf˚ C(diamond)(s) - 1 mol • ∆Hf˚ C60(s)
∆rH˚ = (60 mol • 1.8 kJ/mol) – (1 mol • 2320 kJ/mol) = -2212 kJ
(b) Is conversion of buckminsterfullerene to diamond product-favored at room T?
∆rG˚ = ∆rH˚ - T∆rS
We are told to assume that ∆rS = 0
∆rG˚ = -2210 kJ – (298.2 K • 0) = -2210 kJ, so conversion is product-favored
Units for thermodynamic processes are typically expressed for the balanced equation given.
Hence the equation for the formation of HCl: H2 + Cl2 → 2 HCl has a ∆rG°, ∆rH°, and ∆rS° that
628
