ch 11 20 general chemistry 8e petrucci
TRANSCRIPT
Philip DuttonUniversity of Windsor, Canada
N9B 3P4
Prentice-Hall © 2002
General ChemistryPrinciples and Modern Applications
Petrucci • Harwood • Herring 8th Edition
Chapter 11: Chemical Bonding I:Basic Concepts
Prentice-Hall © 2002 General Chemistry: Chapter 11 Slide 2 of 43
Contents11-1 Lewis Theory: An Overview11-2 Covalent Bonding: An Introduction11-3 Polar Covalent Bonds11-4 Writing Lewis Structures11-5 Resonance11-6 Exceptions to the Octet Rule11-7 The Shapes of Molecules11-8 Bond Order and Bond Lengths11-9 Bond Energies Focus on Polymers—
Macromolecular Substances
Prentice-Hall © 2002 General Chemistry: Chapter 11 Slide 3 of 43
11-1 Lewis Theory: An Overview
• Valence e- play a fundamental role in chemical bonding.
• e- transfer leads to ionic bonds.
• Sharing of e- leads to covalent bonds.
• e- are transferred of shared to give each atom a noble gas configuration – the octet.
Prentice-Hall © 2002 General Chemistry: Chapter 11 Slide 4 of 43
Lewis Symbols
• A chemical symbol represents the nucleus and the core e-.
• Dots around the symbol represent valence e-.
Si•
••
•
N••
••
• P••
••
• As••
••
• Sb••
••
• Bi••
••
•
••Al••
• Se•• •••
Ar••
••
••I •••
••
••
Prentice-Hall © 2002 General Chemistry: Chapter 11 Slide 5 of 43
Lewis Structures for Ionic Compounds
Ba•
• O•••
•••
••O••
••
••Ba
2+ 2-
Mg•
•
Cl•••
••
••Cl•••
••••
••Cl••
••
••Mg
2+ -2
BaO
MgCl2
Prentice-Hall © 2002 General Chemistry: Chapter 11 Slide 6 of 43
11-2 Covalent Bonding
Prentice-Hall © 2002 General Chemistry: Chapter 11 Slide 7 of 43
Coordinate Covalent Bonds
HN ••H
HH
NH
HH
H
+
Cl ••Cl••
••
••
-
Prentice-Hall © 2002 General Chemistry: Chapter 11 Slide 8 of 43
Multiple Covalent Bonds
C••
••
O••
• •
• •O••
• •
• •CO O
•
••
•••
••
••
••
•CO O ••
•••
••
••
••CO O
••
••
••
••
Prentice-Hall © 2002 General Chemistry: Chapter 11 Slide 9 of 43
Multiple Covalent Bonds
N••
••
•N N
•
••
•
•
••
•N••
••
•
N N•
•• ••
•N N •• ••
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11-3 Polar Covalent Bonds
H Clδ+ δ-
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Analogy to Population
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Electronegativity
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Percent Ionic Character
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Writing Lewis Structures
• All the valence e- of atoms must appear.• Usually, the e- are paired.• Usually, each atom requires an octet.
– H only requires 2 e-.• Multiple bonds may be needed.
– Readily formed by C, N, O, S, and P.
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Skeletal Structure
• Identify central and terminal atoms.
CH
HH
HCH
H
O
Prentice-Hall © 2002 General Chemistry: Chapter 11 Slide 16 of 43
Skeletal Structure
• Hydrogen atoms are always terminal atoms.• Central atoms are generally those with the lowest
electronegativity.• Carbon atoms are always central atoms.• Generally structures are compact and symmetrical.
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Strategy for Writing Lewis
Structures
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Formal Charge
FC = #valence e- - #lone pair e- - #bond pair e-21
Prentice-Hall © 2002 General Chemistry: Chapter 11 Slide 19 of 43
Example 11-6
Writing a lewis Structure for a Polyatomic Ion.
Write the Lewis structure for the nitronium ion, NO2+.
Step 1: Total valence e- = 5 + 6 + 6 – 1 = 16 e-
Step 2: Plausible structure: O—N—O
Step 3: Add e- to terminal atoms: O—N—O ••
•• ••
•• ••
••
Step 4: Determine e- left over: 16 – 4 – 12 = 0
Prentice-Hall © 2002 General Chemistry: Chapter 11 Slide 20 of 43
Example 11-6
Step 5: Use multiple bonds to satisfy octets.
••
•• •••• ••
••O—N—O •• •••• ••O=N=O
Step 6: Determine formal charges:
FC(O) = 6 - 4 – (4) = 021
FC(N) = 5 - 0 – (8) = +121
+
Prentice-Hall © 2002 General Chemistry: Chapter 11 Slide 21 of 43
Alternative Lewis Structure
••
•• •••• ••
••O—N—O + -+
FC(O≡) = 6 - 2 – (6) = +121
FC(N) = 5 - 0 – (8) = +121
FC(O—) = 6 - 6 – (2) = -121
••O N O ••
••
••
Prentice-Hall © 2002 General Chemistry: Chapter 11 Slide 22 of 43
Alternative Lewis Structures
• Sum of FC is the overall charge.• FC should be as small as possible.• Negative FC usually on most electronegative elements.• FC of same sign on adjacent atoms is unlikely.
+
••O≡N—O ••
••
••
-+
Prentice-Hall © 2002 General Chemistry: Chapter 11 Slide 23 of 43
Example 11-7Using the Formal Charge Concept in Writing Lewis Structures.
Write the most plausible Lewis structure of nitrosyl chloride, NOCl, one of the oxidizing agents present in aqua regia.
2+ 2- - 2+ - +-
Prentice-Hall © 2002 General Chemistry: Chapter 11 Slide 24 of 43
11-5 Resonance
O O O O O O••••
••
••
••••
••••
••••
••
••+ +- -
O O O••••••••
••+ -½ -½
Prentice-Hall © 2002 General Chemistry: Chapter 11 Slide 25 of 43
11-6 Exceptions to the Octet Rule
• Odd e- species.
N=O ••
••
•••
H—C—H
H
•
•O—H ••
••
Prentice-Hall © 2002 General Chemistry: Chapter 11 Slide 26 of 43
Exceptions to the Octet Rule
• Incomplete octets.
B
F
FF
••••
••
••
•••• ••••
••
B
F
FF
-
•••• ••
••
••
••
+
••
•• ••
••••
B
F
FF
••
••
•••• ••
•••• -
+
•• ••
Prentice-Hall © 2002 General Chemistry: Chapter 11 Slide 27 of 43•• ••
••
Exceptions to the Octet Rule
• Expanded octets.
P
Cl
ClCl
••••
••
••
••••
•• P
Cl
Cl
••••Cl
••••
••••
•••• •• ••
••
Cl
••••
••Cl•• S
F
F
••
••F
••••
••
••
•••• •• ••
••F
••••••F••
F••
••
••
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Expanded Valence Shell
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11-7 The Shapes of Molecules
H O H
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Terminology
• Bond length – distance between nuclei.• Bond angle – angle between adjacent bonds.• VSEPR Theory
– Electron pairs repel each other whether they are in chemical bonds (bond pairs) or unshared (lone pairs). Electron pairs assume orientations about an atom to minimize repulsions.
• Electron group geometry – distribution of e- pairs.• Molecular geometry – distribution of nuclei.
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Balloon Analogy
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Methane, Ammonia and Water
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Table 11.1 Molecular Geometry as a Function of Electron Group Geometry
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Applying VSEPR Theory
• Draw a plausible Lewis structure.• Determine the number of e- groups and identify
them as bond or lone pairs.• Establish the e- group geometry.• Determine the molecular geometry.
• Multiple bonds count as one group of electrons.• More than one central atom can be handled
individually.
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Dipole Moments
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Dipole Moments
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Bond Order and Bond Length
• Bond Order– Single bond, order = 1– Double bond, order = 2
• Bond Length – Distance between two nuclei
• Higher bond order– Shorter bond– Stronger bond
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Bond Length
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Bond Energies
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Bond Energies
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Bond Energies and Enthalpy of Reaction
ΔHrxn = ΔH(product bonds) - ΔH(reactant bonds)
= ΔH bonds formed - ΔH bonds broken
= -770 kJ/mol – (657 kJ/mol) = -114 kJ/mol
Prentice-Hall © 2002 General Chemistry: Chapter 11 Slide 42 of 43
Focus on Polymers – Macromolecular Substances
Prentice-Hall © 2002 General Chemistry: Chapter 11 Slide 43 of 43
Chapter 11 Questions
1, 4, 6, 8, 10, 11, 15, 27, 33, 37, 53, 57, 65 (also calculate formal charges), 71, 86, 94
Philip DuttonUniversity of Windsor, Canada
N9B 3P4
Prentice-Hall © 2002
General ChemistryPrinciples and Modern Applications
Petrucci • Harwood • Herring 8th Edition
Chapter 12: Chemical Bonding II:Additional Aspects
Prentice-Hall © 2002 General Chemistry: Chapter 12 Slide 45 of 47
Contents12-1 What a Bonding Theory Should Do12-2 Introduction to the Valence-Bond Method12-3 Hybridization of Atomic Orbitals12-4 Multiple Covalent Bonds12-5 Molecular Orbital Theory12-6 Delocalized Electrons: Bonding in the
Benzene Molecule12-7 Bonding in Metals Focus on Photoelectron Spectroscopy
Prentice-Hall © 2002 General Chemistry: Chapter 12 Slide 46 of 47
12-1 What a Bonding Theory Should Do
• Bring atoms together from a distance.– e- are attracted to both nuclei.– e- are repelled by each other.– Nuclei are repelled by each other.
• Plot the total potential energy verses distance.– -ve energies correspond to net attractive forces.– +ve energies correspond to net repulsive forces.
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Potential Energy Diagram
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12-2 Introduction to the Valence-Bond Method
• Atomic orbital overlap describes covalent bonding.
• Area of overlap of orbitals is in phase. • A localized model of bonding.
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Bonding in H2S
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Example 12-1Using the Valence-Bond Method to Describe a Molecular Structure.
Describe the phosphine molecule, PH3, by the valence-bond method..
Identify valence electrons:
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Example 12-1Sketch the orbitals:
Overlap the orbitals:
Describe the shape: Trigonal pyramidal
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12-3 Hybridization of Atomic Orbitals
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sp3 Hybridization
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sp3 Hybridization
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Bonding in Methane
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sp3 Hybridization in Nitrogen
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Bonding in Nitrogen
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sp2 Hybridization
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Orbitals in Boron
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sp Hybridization
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Orbitals in Beryllium
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sp3d and sp3d2 Hybridization
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Hybrid Orbitals and VSEPR
• Write a plausible Lewis structure.• Use VSEPR to predict electron geometry.• Select the appropriate hybridization.
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12-4 Multiple Covalent Bonds
• Ethylene has a double bond in its Lewis structure.• VSEPR says trigonal planar at carbon.
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Ethylene
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Acetylene
• Acetylene, C2H2, has a triple bond.• VSEPR says linear at carbon.
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12-5 Molecular Orbital Theory
• Atomic orbitals are isolated on atoms.• Molecular orbitals span two or more atoms.• LCAO
– Linear combination of atomic orbitals.
Ψ1 = φ1 + φ2 Ψ2 = φ1 - φ2
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Combining Atomic Orbitals
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Molecular Orbitals of Hydrogen
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Basic Ideas Concerning MOs
• Number of MOs = Number of AOs.• Bonding and antibonding MOs formed from AOs.• e- fill the lowest energy MO first.• Pauli exclusion principle is followed.• Hund’s rule is followed
Prentice-Hall © 2002 General Chemistry: Chapter 12 Slide 71 of 47
Bond Order
• Stable species have more electrons in bonding orbitals than antibonding.
Bond Order = No. e- in bonding MOs - No. e- in antibonding MOs
2
Prentice-Hall © 2002 General Chemistry: Chapter 12 Slide 72 of 47
Diatomic Molecules of the First-Period
BO = (1-0)/2 = ½ H2+
BO = (2-0)/2 = 1 H2+
BO = (2-1)/2 = ½ He2+
BO = (2-2)/2 = 0 He2+
BO = (e-bond - e-
antibond )/2
Prentice-Hall © 2002 General Chemistry: Chapter 12 Slide 73 of 47
Molecular Orbitals of the Second Period
• First period use only 1s orbitals.• Second period have 2s and 2p orbitals available.
• p orbital overlap:– End-on overlap is best – sigma bond (σ).– Side-on overlap is good – pi bond (π).
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Molecular Orbitals of the Second Period
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Combining p orbitals
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Expected MO Diagram of C2
Prentice-Hall © 2002 General Chemistry: Chapter 12 Slide
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Modified MO Diagram of C2
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MO Diagrams of 2nd Period Diatomics
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MO Diagrams of Heteronuclear Diatomics
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12-6 Delocalized Electrons
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Benzene
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Benzene
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Ozone
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12-7 Bonding in Metals
• Electron sea model– Nuclei in a sea of e-.– Metallic lustre.– Malleability.
Force applied
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Bonding in Metals
Band theory.• Extension of MO theory.
N atoms give N orbitals thatare closely spaced in energy.
• N/2 are filled.The valence band.
• N/2 are empty.The conduction band.
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Band Theory
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Semiconductors
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Photovoltaic Cells
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Focus on Photoelectron Spectroscopy
Prentice-Hall © 2002 General Chemistry: Chapter 12 Slide 90 of 47
Chapter 12 Questions
1, 3, 8, 10, 16, 29, 33, 39, 45, 59, 68, 72, 76
Philip DuttonUniversity of Windsor, Canada
N9B 3P4
Prentice-Hall © 2002
General ChemistryPrinciples and Modern Applications
Petrucci • Harwood • Herring 8th Edition
Chapter 13: Liquids, Solids and Intermolecular Forces
Prentice-Hall © 2002 General Chemistry: Chapter 13 Slide 92 of 35
Contents13-1 Intermolecular Forces and some Properties of Liquids13-2 Vaporization of Liquids: Vapor Pressure13-3 Some Properties of Solids13-4 Phase Diagrams13-5 Van der Waals Forces13-6 Hydrogen Bonding13-7 Chemical Bonds as Intermolecular Forces13-8 Crystal structures13-8 Energy Changes in the Formation of Ionic Crystals
Focus on Liquid Crystals
Prentice-Hall © 2002 General Chemistry: Chapter 13 Slide 93 of 35
13-1 Intermolecular Forces and Some Properties of Liquids
• Cohesive Forces– Intermolecular forces between like molecules.
• Adhesive Forces– Intermolecular forces between unlike molecules.
• Surface Tension– Energy or work required to increase the surface area
of a liquid.• Viscosity
– A liquids resistance to flow
Prentice-Hall © 2002 General Chemistry: Chapter 13 Slide 94 of 35
Intermolecular Forces
Prentice-Hall © 2002 General Chemistry: Chapter 13 Slide 95 of 35
Intermolecular Forces
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13-2 Vaporization of Liquids:Vapor Pressure
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Enthalpy of Vaporization
ΔHvap = Hvapor – Hliquid = - ΔHcondensation
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Boiling Point
Mercury manometer
Vapor pressure of liquid
Pvap independent
of Vliq
Pvap independent
of Vgas
Pvap dependent on
T
Prentice-Hall © 2002 General Chemistry: Chapter 13 Slide 99 of 35
Vapor Pressure and Boiling Point
(e) (d) (c) (b) (a)
Ln P = -A ( ) + B1T
A = ΔHvap
R
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Clausius-Clapeyron Equation
Ln P = -A ( ) + B1T
Ln = - ( - ) P2P1
1T2
1T1
ΔHvap
R
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13-3 Some Properties of Solids
Freezing Point Melting Point
ΔHfus(H2O) = +6.01 kJ/mol
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Sublimation
ΔHsub = ΔHfus + ΔHvap
= -ΔHdeposition
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13-4 Phase Diagrams
Iodine
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Phase Diagrams
Carbon dioxide
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Supercritical Fluids
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The Critical Point
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Critical Temperatures and Pressures
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Water
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13-5 Van der Waals Forces
• Instantaneous dipoles.– Electrons move in an orbital to cause a polarization.
• Induced dipoles.– Electrons move in response to an outside force.
• Dispersion or London forces.– Instaneous dipole – induced dipole attraction.– Related to polarizability.
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Phenomenon of Induction
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Instantaneous and Induced Dipoles
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Dipole Dipole Interactions
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13-6 Hydrogen Bonding
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Hydrogen Bonding in HF(g)
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Hydrogen Bonding in Water
around a molecule in the solid in the liquid
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Other examples of H-Bonds
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13-7 Chemical Bonds as Intermolecular Forces
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Other Carbon Allotropes
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Interionic Forces
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13-8 Crystal Structures
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Unit Cells in the Cubic Crystal System
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Holes in Crystals
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Hexagonal Close Packed (hcp)
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Coordination Number
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Counting Cell Occupancy
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X-Ray Diffraction
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X-Ray Diffraction
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Cesium Chloride
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Atomic Radii from Crystal Structures
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Sodium Chloride
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Holes in Crystals
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13-9 Energy Changes in the Formation of Ionic Crystals
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Chapter 13 Questions
1, 3, 4, 13, 24, 26, 31, 45, 52, 61, 94, 107
Philip DuttonUniversity of Windsor, Canada
N9B 3P4
Prentice-Hall © 2002
General ChemistryPrinciples and Modern Applications
Petrucci • Harwood • Herring 8th Edition
Chapter 14: Solutions and Their Physical Properties
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 136 of 46
Contents
14-1 Types of Solutions: Some Terminology14-2 Solution Concentration14-3 Intermolecular Forces and the Solution Process14-4 Solution Formation and Equilibrium14-5 Solubilities of Gases14-6 Vapor Pressure of Solutions14-7 Osmotic Pressure
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 137 of 46
Contents
14-8 Freezing-Point Depression and Boiling-Point Elevation of Nonelectrolyte solutions.
14-9 Solutions of Electrolytes14-10 Colloidal Mixtures
Focus on Chromatography
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13-1 Types of Solution: Some Terminology
• Solutions are homogeneous mixtures.– Uniform throughout.
• Solvent.– Determines the state of matter in which the solution
exists.– Is the largest component.
• Solute.– Other solution components said to be dissolved in the
solution.
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Table 14.1 Some Common Solutions
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14-2 Solution Concentration.
• Mass percent. (m/m)• Volume percent. (v/v)• Mass/volume percent. (m/v)
• Isotonic saline is prepared by dissolving 0.9 g of NaCl in 100 mL of water and is said to be:
0.9% NaCl (mass/volume)
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10% Ethanol Solution (v/v)
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ppm, ppb and ppt
• Very low solute concentrations are expressed as:
ppm: parts per million (g/g, mg/L)ppb: parts per billion (ng/g, g/L)ppt: parts per trillion (pg/g, ng/L)
note that 1.0 L 1.0 g/mL = 1000 gppm, ppb, and ppt are properly m/m or v/v.
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Mole Fraction and Mole Percent
= Amount of component i (in moles)
Total amount of all components (in moles)
1 + 2 + 3 + …n = 1
Mole % i = i 100%
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Molarity and Molality
Molarity (M) = Amount of solute (in moles)
Volume of solution (in liters)
Molality (m) = Amount of solute (in moles)
Mass of solvent (in kilograms)
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14-3 Intermolecular Forces and the Solution Process
ΔHa
ΔHb ΔHc
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Intermolecular Forces in Mixtures
• Magnitude of ΔHa, ΔHb, and ΔHc depend on intermolecular forces.
• Ideal solution– Forces are similar between all
combinations of components. ΔHsoln = 0
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Ideal Solution
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Non-ideal Solutions
• Adhesive forces greater than cohesive forces.
ΔHsoln < 0
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Non-ideal Solutions
• Adhesive forces are less than cohesive forces.
• At the limit these solutions are heterogeneous.
ΔHsoln > 0
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Ionic Solutions
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Hydration Energy
NaCl(s) → Na+(g) + Cl-(g) ΔHlattice > 0
Na+(g) + xs H2O(l) → Na+(aq) ΔHhydration < 0
Cl-(g) + xs H2O(l) → Cl-(aq) ΔHhydration < 0
ΔHsoln > 0 but ΔGsolution < 0
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14-4 Solution Formation and Equilibrium
saturated
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Solubility Curves
Supersaturated Unsaturated
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14-5 Solubility of Gases
• Most gases are less soluble in water as temperature increases.
• In organic solvents the reverse is often true.
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Henry’s Law
• Solubility of a gas increases with increasing pressure. C = k Pgas
k = C
Pgas
=23.54 mL1.00 atm
= 23.54 ml N2/atm
k
CPgas = =
100 mL = 4.25 atm23.54 ml N2/atm
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 156 of 46
Henry’s Law
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 157 of 46
14-6 Vapor Pressures of Solutions
• Roault, 1880s.– Dissolved solute lowers vapor pressure of solvent.– The partial pressure exerted by solvent vapor above an
ideal solution is the product of the mole fraction of solvent in the solution and the vapor pressure of the pure solvent at a given temperature.
PA = A P°A
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 158 of 46
Example 14-6
Predicting vapor pressure of ideal solutions.
The vapor pressures of pure benzene and toluene at 25°C are 95.1 and 28.4 mm Hg, respectively. A solution is prepared in which the mole fractions of benzene and toluene are both 0.500. What are the partial pressures of the benzene and toluene above this solution? What is the total vapor pressure?
Balanced Chemical Equation:
Pbenzene = benzene P°benzene = (0.500)(96.1 mm Hg) = 47.6 mm Hg
Ptoluene = toluene P°toluene = (0.500)(28.4 mm Hg) = 14.2 mm Hg
Ptotal = Pbenzene + Ptoluene = 61.8 mm Hg
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 159 of 46
Example 14-7
Calculating the Composition of Vapor in Equilibrium with a Liquid Solution.
What is the composition of the vapor in equilibrium with the benzene-toluene solution?
Partial pressure and mole fraction:
benzene = Pbenzene/Ptotal = 47.6 mm Hg/61.89 mm Hg = 0.770
toluene = Ptoluene/Ptotal = 14.2 mm Hg/61.89 mm Hg = 0.230
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 160 of 46
Liquid-Vapor Equilibrium
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 161 of 46
Fractional Distillation
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 162 of 46
Fractional Distillation
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 163 of 46
Non-ideal behavior
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 164 of 46
14-7 Osmotic Pressure
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 165 of 46
Osmotic Pressure
πV = nRT
π = RTnV = M RT
For dilute solutions of electrolytes:
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 166 of 46
Osmotic Pressure
Isotonic Saline 0.92% m/V
Hypertonic > 0.92% m/Vcrenation
Hypotonic > 0.92% m/V
rupture
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 167 of 46
Reverse Osmosis - Desalination
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 168 of 46
14-8 Freezing-Point Depression and Boiling Point Elevation of Nonelectrolyte
Solutions
• Vapor pressure is lowered when a solute is present.– This results in boiling point elevation.– Freezing point is also effected and is lowered.
• Colligative properties.– Depends on the number of particles present.
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 169 of 46
Vapor Pressure Lowering
ΔTf = -Kf m
ΔTb = -Kb m
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 170 of 46
Practical Applications
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 171 of 46
14-9 Solutions of Electrolytes
• Svante Arrhenius– Nobel Prize 1903.– Ions form when electrolytes dissolve in solution.– Explained anomalous colligative properties
ΔTf = -Kf m = -1.86°C m-1 0.0100 m = -0.0186°C
Compare 0.0100 m aqueous urea to 0.0100 m NaCl (aq)
Freezing point depression for NaCl is -0.0361°C.
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 172 of 46
van’t Hoff
ΔTf = -i Kf m
i = = = 1.98measured ΔTf
ΔTb = -i Kb m
expected ΔTf
0.0361°C
0.0186°C
π = -i M RT
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 173 of 46
Interionic Interactions
• Arrhenius theory does not correctly predict the conductivity of concentrated electrolytes.
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 174 of 46
Debye and Hückel
• 1923– Ions in solution do not
behave independently.– Each ion is surrounded
by others of opposite charge.
– Ion mobility is reduced by the drag of the ionic atmosphere.
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 175 of 46
14-10 Colloidal Mixtures
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 176 of 46
Colloids
• Particles of 1-1000 nm size.– Nanoparticles of various
shapes: rods, discs, spheres.– Particles can remain
suspended indefinitly.• Milk is colloidal.
• Increasing ionic strength can cause precipitation.
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 177 of 46
Dialysis
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 178 of 46
Focus on Chromatography
Stationary Phasesilicon gumaluminasilica
Mobile Phasesolventgas
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 179 of 46
Chromatography
Prentice-Hall © 2002 General Chemistry: Chapter 14 Slide 180 of 46
Chapter 14 Questions
Develop problem solving skills and base your strategy not on solutions to specific problems but on understanding.
Choose a variety of problems from the text as examples.
Practice good techniques and get coaching from people who have been here before.
Philip DuttonUniversity of Windsor, Canada
N9B 3P4
Prentice-Hall © 2002
General ChemistryPrinciples and Modern ApplicationsPetrucci • Harwood • Herring 8th Edition
Chapter 15: Chemical Kinetics
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 182 of 55
Contents
15-1 The Rate of a Chemical Reaction15-2 Measuring Reaction Rates15-3 Effect of Concentration on Reaction Rates:
The Rate Law15-4 Zero-Order Reactions15-5 First-Order Reactions15-6 Second-Order Reactions15-7 Reaction Kinetics: A Summary
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 183 of 55
Contents
15-8 Theoretical Models for Chemical Kinetics15-9 The Effect of Temperature on Reaction Rates15-10 Reaction Mechanisms15-11 Catalysis
Focus On Combustion and Explosions
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 184 of 55
15-1 The Rate of a Chemical Reaction
• Rate of change of concentration with time.
2 Fe3+(aq) + Sn2+ → 2 Fe2+(aq) + Sn4+(aq)
t = 38.5 s [Fe2+] = 0.0010 M
Δt = 38.5 s Δ[Fe2+] = (0.0010 – 0) M
Rate of formation of Fe2+= = = 2.610-5 M s-1Δ[Fe2+]
Δt
0.0010 M
38.5 s
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 185 of 55
Rates of Chemical Reaction
Δ[Sn4+]Δt
2 Fe3+(aq) + Sn2+ → 2 Fe2+(aq) + Sn4+(aq)
Δ[Fe2+] Δt
=12
Δ[Fe3+] Δt
= - 12
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 186 of 55
General Rate of Reaction
a A + b B → c C + d D
Rate of reaction = rate of disappearance of reactants
=Δ[C]
Δt1c
=Δ[D]
Δt1d
Δ[A] Δt
1a
= -Δ[B]
Δt1b= -
= rate of appearance of products
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 187 of 55
15-2 Measuring Reaction Rates
H2O2(aq) → H2O(l) + ½ O2(g)
2 MnO4-(aq) + 5 H2O2(aq) + 6 H+ →
2 Mn2+ + 8 H2O(l) + 5 O2(g)
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 188 of 55
H2O2(aq) → H2O(l) + ½ O2(g)
Example 15-2
-(-1.7 M / 2600 s) =
6 10-4 M s-1
-(-2.32 M / 1360 s) = 1.7 10-3 M s-1
Determining and Using an Initial Rate of Reaction.
Rate = -Δ[H2O2]
Δt
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 189 of 55
Example 15-2
-Δ[H2O2] = -([H2O2]f - [H2O2]i) = 1.7 10-3 M s-1 Δt
Rate = 1.7 10-3 M s-1
Δt=
- Δ[H2O2]
[H2O2]100 s – 2.32 M = -1.7 10-3 M s-1 100 s
= 2.17 M
= 2.32 M - 0.17 M [H2O2]100 s
What is the concentration at 100s?
[H2O2]i = 2.32 M
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 190 of 55
15-3 Effect of Concentration on Reaction Rates: The Rate Law
a A + b B …. → g G + h H ….
Rate of reaction = k [A]m[B]n ….
Rate constant = k
Overall order of reaction = m + n + ….
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 191 of 55
Example 15-3 Method of Initial RatesEstablishing the Order of a reaction by the Method of Initial Rates.
Use the data provided establish the order of the reaction with respect to HgCl2 and C2O2
2- and also the overall order of the reaction.
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 192 of 55
Example 15-3
Notice that concentration changes between reactions are by a factor of 2.
Write and take ratios of rate laws taking this into account.
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 193 of 55
Example 15-3
R2 = k[HgCl2]2m[C2O4
2-]2n
R3 = k[HgCl2]3m[C2O4
2-]3n
R2 R3
k(2[HgCl2]3)m[C2O42-]3
n k[HgCl2]3
m[C2O42-]3
n =
2m = 2.0 therefore m = 1.0
R2 R3
k2m[HgCl2]3m[C2O4
2-]3n
k[HgCl2]3m[C2O4
2-]3n
= = 2.0=2mR3
R3
= k(2[HgCl2]3)m[C2O42-]3
n
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 194 of 55
Example 15-3
R2 = k[HgCl2]21[C2O4
2-]2n = k(0.105)(0.30)n
R1 = k[HgCl2]11[C2O4
2-]1n = k(0.105)(0.15)n
R2 R1
k(0.105)(0.30)n k(0.105)(0.15)n
=
7.110-5
1.810-5= 3.94
R2 R1
(0.30)n (0.15)n
= = 2n =
2n = 3.98 therefore n = 2.0
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 195 of 55
+ = Third Order
R2 = k[HgCl2]2 [C2O4
2-]2
First order
Example 15-3
1
Second order
2
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 196 of 55
15-4 Zero-Order Reactions
A → products
Rrxn = k [A]0
Rrxn = k
[k] = mol L-1 s-1
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 197 of 55
Integrated Rate Law
- dt= kd[A] [A]0
[A]t
0
t
-[A]t + [A]0 = kt
[A]t = [A]0 - kt
Δt
-Δ[A]dt
= k-d[A]Move to the
infinitesimal= k
And integrate from 0 to time t
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 198 of 55
15-5 First-Order Reactions
H2O2(aq) → H2O(l) + ½ O2(g)
= -k [H2O2] d[H2O2 ]
dt
= - k dt[H2O2] d[H2O2 ]
[A]0
[A]t
0
t
= -ktln [A]t
[A]0
ln[A]t = -kt + ln[A]0
[k] = s-1
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 199 of 55
First-Order Reactions
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 200 of 55
Half-Life
• t½ is the time taken for one-half of a reactant to be consumed.
= -ktln [A]t
[A]0
= -kt½ ln ½[A]0
[A]0
- ln 2 = -kt½
t½ = ln 2k
0.693k
=
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 201 of 55
Half-LifeButOOBut(g) → 2 CH3CO(g) + C2H4(g)
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 202 of 55
Some Typical First-Order Processes
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 203 of 55
15-6 Second-Order Reactions
• Rate law where sum of exponents m + n +… = 2.
A → products
dt= - kd[A][A]2
[A]0
[A]t
0
t
= kt +1[A]0[A]t
1
dt= -k[A]2
d[A][k] = M-1 s-1 = L mol-1 s-1
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 204 of 55
Second-Order Reaction
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 205 of 55
Pseudo First-Order Reactions
• Simplify the kinetics of complex reactions• Rate laws become easier to work with.
CH3CO2C2H5 + H2O → CH3CO2H + C2H5OH
• If the concentration of water does not change appreciably during the reaction.– Rate law appears to be first order.
• Typically hold one or more reactants constant by using high concentrations and low concentrations of the reactants under study.
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 206 of 55
Testing for a Rate Law
Plot [A] vs t.
Plot ln[A] vs t.
Plot 1/[A] vs t.
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 207 of 55
15-7 Reaction Kinetics: A Summary
• Calculate the rate of a reaction from a known rate law using:
• Determine the instantaneous rate of the reaction by:
Rate of reaction = k [A]m[B]n ….
Finding the slope of the tangent line of [A] vs t or,
Evaluate –Δ[A]/Δt, with a short Δt interval.
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 208 of 55
Summary of Kinetics
• Determine the order of reaction by:
Using the method of initial rates.
Find the graph that yields a straight line.
Test for the half-life to find first order reactions.
Substitute data into integrated rate laws to find the rate law that gives a consistent value of k.
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 209 of 55
Summary of Kinetics
• Find the rate constant k by:
• Find reactant concentrations or times for certain conditions using the integrated rate law after determining k.
Determining the slope of a straight line graph.
Evaluating k with the integrated rate law.
Measuring the half life of first-order reactions.
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 210 of 55
15-8 Theoretical Models for Chemical Kinetics
• Kinetic-Molecular theory can be used to calculate the collision frequency.– In gases 1030 collisions per second.– If each collision produced a reaction, the rate would be
about 106 M s-1.– Actual rates are on the order of 104 M s-1.
• Still a very rapid rate.– Only a fraction of collisions yield a reaction.
Collision Theory
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 211 of 55
Activation Energy
• For a reaction to occur there must be a redistribution of energy sufficient to break certain bonds in the reacting molecule(s).
• Activation Energy is:– The minimum energy above the average kinetic energy
that molecules must bring to their collisions for a chemical reaction to occur.
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 212 of 55
Activation Energy
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 213 of 55
Kinetic Energy
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 214 of 55
Collision Theory
• If activation barrier is high, only a few molecules have sufficient kinetic energy and the reaction is slower.
• As temperature increases, reaction rate increases.
• Orientation of molecules may be important.
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 215 of 55
Collision Theory
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 216 of 55
Transition State Theory
• The activated complex is a hypothetical species lying between reactants and products at a point on the reaction profile called the transition state.
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 217 of 55
15-9 Effect of Temperature on Reaction Rates
• Svante Arrhenius demonstrated that many rate constants vary with temperature according to the equation:
k = Ae-Ea/RT
ln k = + ln AR
-Ea
T
1
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 218 of 55
Arrhenius Plot
N2O5(CCl4) → N2O4(CCl4) + ½ O2(g)
= -1.2104 KR
-Ea
-Ea = 1.0102 kJ mol-1
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 219 of 55
Arrhenius Equation
k = Ae-Ea/RT ln k = + ln AR
-Ea
T
1
ln k2– ln k1 = + ln A - - ln AR
-Ea
T2
1
R
-Ea
T1
1
ln = - R
-Ea
T2
1k2
k1
T1
1
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 220 of 55
15-10 Reaction Mechanisms
• A step-by-step description of a chemical reaction.• Each step is called an elementary process.
– Any molecular event that significantly alters a molecules energy of geometry or produces a new molecule.
• Reaction mechanism must be consistent with:– Stoichiometry for the overall reaction.– The experimentally determined rate law.
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 221 of 55
Elementary Processes
• Unimolecular or bimolecular.• Exponents for concentration terms are the same as
the stoichiometric factors for the elementary process.
• Elementary processes are reversible.• Intermediates are produced in one elementary
process and consumed in another. • One elementary step is usually slower than all the
others and is known as the rate determining step.
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 222 of 55
A Rate Determining Step
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 223 of 55
Slow Step Followed by a Fast Step
H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g)dt
= k[H2][ICl]d[P]
Postulate a mechanism:
H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g)
slowH2(g) + ICl(g) HI(g) + HCl(g)
fastHI(g) + ICl(g) I2(g) + HCl(g)
dt= k[H2][ICl]
d[HI]
dt= k[HI][ICl]
d[I2]
dt= k[H2][ICl]
d[P]
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 224 of 55
Slow Step Followed by a Fast Step
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 225 of 55
Fast Reversible Step Followed by a Slow Step
2NO(g) + O2(g) → 2 NO2(g)dt
= -kobs[NO2]2[O2]d[P]
Postulate a mechanism:
dt = k2[N2O2][O2]
d[NO2]
fast 2NO(g) N2O2(g)k1
k-1
slow N2O2(g) + O2(g) 2NO2(g)k2
dt= k2 [NO]2[O2]
d[I2]
k-1
k12NO(g) + O2(g) → 2 NO2(g)
K =k-1
k1 =[NO]
[N2O2]
= K [NO]2
k-1
k1= [NO]2[N2O2]
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 226 of 55
The Steady State Approximation
dt= k1[NO]2 – k2[N2O2] – k3[N2O2][O2] = 0
d[N2O2]
N2O2(g) + O2(g) 2NO2(g)k3
2NO(g) N2O2(g)k-1
k12NO(g) N2O2(g)
N2O2(g) + O2(g) 2NO2(g)k3
N2O2(g) 2NO(g) k2
k12NO(g) N2O2(g)
dt = k3[N2O2][O2]
d[NO2]
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 227 of 55
The Steady State Approximation
dt= k1[NO]2 – k2[N2O2] – k3[N2O2][O2] = 0
d[N2O2]
k1[NO]2 = [N2O2](k2 + k3[O2])
k1[NO]2
[N2O2] =(k2 + k3[O2])
dt = k3[N2O2][O2]
d[NO2] k1k3[NO]2[O2] =(k2 + k3[O2])
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 228 of 55
Kinetic Consequences of Assumptions
dt
d[NO2] k1k3[NO]2[O2] =(k2 + k3[O2])
N2O2(g) + O2(g) 2NO2(g)k3
N2O2(g) 2NO(g) k2
k12NO(g) N2O2(g)
dt
d[NO2] k1k3[NO]2[O2] =( k3[O2])
k1[NO]2 =
dt
d[NO2] k1k3[NO]2[O2] =( k2)
[NO]2[O2] =k1k3
k2
Let k2 << k3
Let k2 >> k3
Or
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 229 of 55
11-5 Catalysis
• Alternative reaction pathway of lower energy.• Homogeneous catalysis.
– All species in the reaction are in solution. • Heterogeneous catalysis.
– The catalyst is in the solid state.– Reactants from gas or solution phase are adsorbed.– Active sites on the catalytic surface are important.
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 230 of 55
11-5 Catalysis
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 231 of 55
Catalysis on a Surface
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 232 of 55
Enzyme Catalysis
E + S ESk1
k-1
ES → E + Pk2
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 233 of 55
Saturation Kinetics
E + S ESk1
k-1
→ E + Pk2
dt= k1[E][S] – k-1[ES] – k2[ES]= 0
d[P]dt
= k2[ES]d[P]
k1[E][S] = (k-1+k2 )[ES]
[E] = [E]0 – [ES]
k1[S]([E]0 –[ES]) = (k-1+k2 )[ES]
(k-1+k2 ) + k1[S] k1[E]0 [S]
[ES] =
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 234 of 55
Michaelis-Menten
dt=
d[P]
(k-1+k2 ) + k1[S] k1k2[E]0 [S]
dt=
d[P]
(k-1+k2 ) + [S]k2[E]0 [S]
k1
dt=
d[P]
KM + [S]k2[E]0 [S]
dt=
d[P]k2[E]0
dt=
d[P]
KM
k2 [E]0 [S]
Prentice-Hall © 2002 General Chemistry: Chapter 15 Slide 235 of 55
Chapter 15 Questions
Develop problem solving skills and base your strategy not on solutions to specific problems but on understanding.
Choose a variety of problems from the text as examples.
Practice good techniques and get coaching from people who have been here before.
Philip DuttonUniversity of Windsor, Canada
N9B 3P4
Prentice-Hall © 2002
General ChemistryPrinciples and Modern ApplicationsPetrucci • Harwood • Herring 8th Edition
Chapter 16: Principles of Chemical Equilibrium
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 237 of 27
Contents
16-1 Dynamic Equilibrium16-2 The Equilibrium Constant Expression16-3 Relationships Involving Equilibrium Constants16-4 The Significance of the Magnitude of an
Equilibrium Constant16-5 The Reaction Quotient, Q: Predicting the
Direction of a Net Change
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 238 of 27
Contents
16-6 Altering Equilibrium Conditions: Le Châtelliers Principle
16-7 Equilibrium Calculations: Some Illustrative ExamplesFocus On The Nitrogen Cycle and the Synthesis of Nitrogen Compounds
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 239 of 27
16-1 Dynamic Equilibrium
• Equilibrium – two opposing processes taking place at equal rates.
H2O(l) H2O(g)
NaCl(s) NaCl(aq)H2O I2(H2O) I2(CCl4)
CO(g) + 2 H2(g) CH3OH(g)
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 240 of 27
16-2 The Equilibrium Constant Expression
Forward: CO(g) + 2 H2(g) → CH3OH(g)
Reverse: CH3OH(g) → CO(g) + 2 H2(g)
At Equilibrium:
Rfwrd = k1[CO][H2]2
Rrvrs = k-1[CH3OH]
Rfwrd = Rrvrs
k1[CO][H2]2 = k-1[CH3OH]
[CH3OH]
[CO][H2]2=
k1
k-1
= Kc
CO(g) + 2 H2(g) CH3OH(g)k1
k-1
k1
k-1
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 241 of 27
Three Approaches to Equilibrium
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 242 of 27
Three Approaches to the Equilibrium
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 243 of 27
Three Approaches to Equilibrium
[CH3OH]
[CO][H2]2
Kc(1) = 14.2 M-2
Kc(2) = 14.2 M-2
Kc(3) = 14.2 M-2
[CH3OH]
[CO][H2]Kc =
[CH3OH]
[CO](2[H2])
0.596 M-1
1.09 M-1
1.28 M-1
1.19 M-1
2.17 M-1
2.55 M-1
CO(g) + 2 H2(g) CH3OH(g)k1
k-1
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 244 of 27
General Expressions
a A + b B …. → g G + h H ….
Equilibrium constant = Kc= [A]m[B]n ….
[G]g[H]h ….
Thermodynamic
Equilibrium constant = Keq= (aG)g(aH)h ….
(aA)a(aB)b ….
aB = [B]cB
0cB
0 is a standard reference state
= 1 mol L-1 (ideal conditions)
= B[B]
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 245 of 27
16-3 Relationships Involving the Equilibrium Constant
• Reversing an equation causes inversion of K.• Multiplying by coefficients by a common factor
raises the equilibrium constant to the corresponding power.
• Dividing the coefficients by a common factor causes the equilibrium constant to be taken to that root.
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 246 of 27
Combining Equilibrium Constant Expressions
N2O(g) + ½O2 2 NO(g) Kc= ?
N2(g) + ½O2 N2O(g) Kc(2)= 2.710+18
N2(g) + O2 2 NO(g) Kc(3)= 4.710-31
Kc= [N2O][O2]½ [NO]2
=[N2][O2]½
[N2O][N2][O2] [NO]2
Kc(2)
1Kc(3)= = 1.710-13
[N2][O2] [NO]2
=
[N2][O2]½ [N2O]
=
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 247 of 27
Gases: The Equilibrium Constant, KP
• Mixtures of gases are solutions just as liquids are.• Use KP, based upon partial pressures of gases.
2 SO2(g) + O2(g) 2 SO3(g) Kc = [SO2]2[O2]
[SO3]2
[SO3]=V
nSO3 = RT
PSO3 [SO2]=V
nSO2 = RT
PSO2
[O2] = V
nO2 = RT
PO2
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 248 of 27
The Equilibrium Constant, KP
2 SO2(g) + O2(g) 2 SO3(g)
Kc = [SO2]2[O2]
[SO3] RT
PSO3
2
RT
PSO2
RT
PO2
=
2
= RTPSO3
2PSO2PO2
2
Kc = KP(RT) KP = Kc(RT)-1
KP = Kc(RT)Δn
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 249 of 27
Pure Liquids and Solids
• Equilibrium constant expressions do not contain concentration terms for solid or liquid phases of a single component (that is, pure solids or liquids).
C(s) + H2O(g) CO(g) + H2(g)
Kc = [H2O]2
[CO][H2] =PH2O
2
PCOPH2 (RT)1
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 250 of 27
Burnt Lime
CaCO3(s) CaO(s) + CO2(g)
Kc = [CO2] KP = PCO2(RT)
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 251 of 27
16-4 The Significance of the Magnitude of the Equilibrium Constant.
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 252 of 27
16-5 The Reaction Quotient, Q: Predicting the Direction of Net Change.
CO(g) + 2 H2(g) CH3OH(g)k1
k-1
• Equilibrium can be approached various ways.• Qualitative determination of change of initial
conditions as equilibrium is approached is needed.
At equilibrium Qc = Kc Qc = [A]t
m[B]tn
[G]tg[H]t
h
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 253 of 27
Reaction Quotient
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 254 of 27
16-6 Altering Equilibrium Conditions: Le Châtellier’s Principle
• When an equilibrium system is subjected to a change in temperature, pressure, or concentration of a reacting species, the system responds by attaining a new equilibrium that partially offsets the impact of the change.
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 255 of 27
Le Châtellier’s Principle
Q = = Kc [SO2]2[O2][SO3]2
Q > Kc
2 SO2(g) + O2(g) 2 SO3(g)k1
k-1
Kc = 2.8102 at 1000K
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 256 of 27
Effect of Condition Changes
• Adding a gaseous reactant or product changes Pgas.• Adding an inert gas changes the total pressure.
– Relative partial pressures are unchanged.• Changing the volume of the system causes a change
in the equilibrium position.
Kc = [SO2]2[O2]
[SO3] V
nSO3
2
V
nSO2
V
nO2
=
2
= VnSO3
2nSO2nO2
2
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 257 of 27
Effect of change in volume
Kc = [C]c[D]d [G]g[H]h
= V(a+b)-(g+h)nG
anA nB
g nHh
a
V-ΔnnG
anA nB
g nHh
a=
• When the volume of an equilibrium mixture of gases is reduced, a net change occurs in the direction that produces fewer moles of gas. When the volume is increased, a net change occurs in the direction that produces more moles of gas.
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 258 of 27
Effect of Temperature on Equilibrium
• Raising the temperature of an equilibrium mixture shifts the equilibrium condition in the direction of the endothermic reaction.
• Lowering the temperature causes a shift in the direction of the exothermic reaction.
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 259 of 27
Effect of a Catalyst on Equilibrium
• A catalyist changes the mechanism of a reaction to one with a lower activation energy.
• A catalyst has no effect on the condition of equilibrium.– But does affect the rate at which equilibrium is
attained.
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 260 of 27
16-7 Equilibrium Calculations: Some Illustrative Examples.
• Five numerical examples are given in the text that illustrate ideas that have been presented in this chapter.
• Refer to the “comments” which describe the methodology. These will help in subsequent chapters.
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 261 of 27
Focus on the Nitrogen Cycle and the Synthesis of Nitrogen Compounds.
N2(g) + O2(g) NO(g)k1
k-1
KP = 4.7 10-31 at 298K and 1.3 x 10-4 at 1800K
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 262 of 27
Synthesis of Ammonia
The optimum conditions are only for the equilibrium position and do not take into account the rate at which equilibrium is attained.
Prentice-Hall © 2002 General Chemistry: Chapter 16 Slide 263 of 27
Chapter 16 Questions
Develop problem solving skills and base your strategy not on solutions to specific problems but on understanding.
Choose a variety of problems from the text as examples.
Practice good techniques and get coaching from people who have been here before.
Philip DuttonUniversity of Windsor, Canada
N9B 3P4
Prentice-Hall © 2002
General ChemistryPrinciples and Modern Applications
Petrucci • Harwood • Herring 8th Edition
Chapter 17: Acids and Bases
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 265 of 47
Contents
17-1 The Arrhenius Theory: A Brief Review17-2 Brønsted-Lowry Theory of Acids and Bases17-3 The Self-Ionization of Water and the pH Scale17-4 Strong Acids and Strong Bases17-5 Weak Acids and Weak Bases17-6 Polyprotic Acids17-7 Ions as Acids and Bases17-8 Molecular Structure and Acid-Base Behavior17-8 Lewis Acids and Bases
Focus On Acid Rain.
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 266 of 47
17-1 The Arrhenius Theory: A Brief Review
HCl(g) → H+(aq) + Cl-(aq)
NaOH(s) → Na+(aq) + OH-(aq)H2O
H2O
Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) → H2O(l) + Na+(aq) + Cl-(aq)
H+(aq) + OH-(aq) → H2O(l)
Arrhenius theory did not handle non OH- bases such as ammonia very well.
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 267 of 47
17-2 Brønsted-Lowry Theory of Acids and Bases
• An acid is a proton donor.• A base is a proton acceptor.
NH3 + H2O NH4+ + OH-
NH4+ + OH- NH3 + H2O
base acid
baseacid
conjugate acid
conjugate base
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 268 of 47
Base Ionization Constant
NH3 + H2O NH4+ + OH-
Kc= [NH3][H2O][NH4
+][OH-]
Kb= Kc[H2O] = [NH3]
[NH4+][OH-]
= 1.810-5
base acidconjugate
acid
conjugate
base
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 269 of 47
Acid Ionization Constant
CH3CO2H + H2O CH3CO2- + H3O+
Kc= [CH3CO2H][H2O][CH3CO2
-][H3O+]
Ka= Kc[H2O] = = 1.810-5
[CH3CO2H][CH3CO2
-][H3O+]
baseacidconjugate
acid
conjugate
base
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 270 of 47
Table 17.1 Relative Strengths of Some Brønsted-Lowry Acids and Bases
HClO4 + H2O ClO4- + H3O+
NH4+ + CO3
2- NH3 + HCO3-HCl + OH- Cl- + H2O
H2O + I- OH- + HIH2O + I- OH- + HI
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 271 of 47
17-3 The Self-Ionization of Water and the pH Scale
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 272 of 47
Ion Product of Water
Kc= [H2O][H2O][H3O+][OH-]
H2O + H2O H3O+ + OH-
base acidconjugate
acid
conjugate
base
KW= Kc[H2O][H2O] = = 1.010-14[H3O+][OH-]
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 273 of 47
pH and pOH
• The potential of the hydrogen ion was defined in 1909 as the negative of the logarithm of [H+].
pH = -log[H3O+] pOH = -log[OH-]
-logKW = -log[H3O+]-log[OH-]= -log(1.010-14)
KW = [H3O+][OH-]= 1.010-14
pKW = pH + pOH= -(-14)
pKW = pH + pOH = 14
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 274 of 47
pH and pOH Scales
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 275 of 47
17-4 Strong Acids and Bases
HCl CH3CO2H
Thymol Blue Indicator
pH < 1.2 < pH < 2.8 < pH
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 276 of 47
17-5 Weak Acids and Bases
Acetic Acid HC2H3O2 or CH3CO2H
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 277 of 47
Weak Acids
Ka= = 1.810-5
[CH3CO2H][CH3CO2
-][H3O+]
pKa= -log(1.810-5) = 4.74
glycine H2NCH2CO2H
lactic acid CH3CH(OH) CO2H
C
OH
O
R
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 278 of 47
Table 17.3 Ionization Constants of Weak Acids and Bases
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 279 of 47
Example 17-5Determining a Value of KA from the pH of a Solution of a Weak Acid.
Butyric acid, HC4H7O2 (or CH3CH2CH2CO2H) is used to make compounds employed in artificial flavorings and syrups. A 0.250 M aqueous solution of HC4H7O2 is found to have a pH of 2.72. Determine KA for butyric acid.
HC4H7O2 + H2O C4H77O2 + H3O+ Ka = ?
Solution:
For HC4H7O2 KA is likely to be much larger than KW. Therefore assume self-ionization of water is unimportant.
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 280 of 47
Example 17-5
HC4H7O2 + H2O C4H7O2 + H3O+
Initial conc. 0.250 M 0 0
Changes -x M +x M +x M
Eqlbrm conc. (0.250-x) M x M x M
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 281 of 47
Example 17-5
Log[H3O+] = -pH = -2.72
HC4H7O2 + H2O C4H77O2 + H3O+
[H3O+] = 10-2.72 = 1.910-3 = x
[H3O+] [C4H7O2-]
[HC4H7O2] Ka=
1.910-3 · 1.910-3
(0.250 – 1.910-3)=
Ka= 1.510-5 Check assumption: Ka >> KW.
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 282 of 47
Percent Ionization
HA + H2O H3O+ + A-
Degree of ionization =[H3O+] from HA
[HA] originally
Percent ionization =[H3O+] from HA
[HA] originally 100%
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 283 of 47
Percent Ionization
Ka =[H3O+][A-]
[HA]
Ka =n H3O+ A-
n
HAn
1 V
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 284 of 47
17-6 Polyprotic Acids
H3PO4 + H2O H3O+ + H2PO4-
H2PO4- + H2O H3O+ + HPO4
2-
HPO42- + H2O H3O+ + PO4
3-
Phosphoric acid:
A triprotic acid.
Ka = 7.110-3
Ka = 6.310-8
Ka = 4.210-13
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 285 of 47
Phosphoric Acid
• Ka1 >> Ka2
• All H3O+ is formed in the first ionization step.
• H2PO4- essentially does not ionize further.
• Assume [H2PO4-] = [H3O+].
• [HPO42-] Ka2
regardless of solution molarity.
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 286 of 47
Table 17.4 Ionization Constants of Some Polyprotic Acids
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 287 of 47
Example 17-9Calculating Ion Concentrations in a Polyprotic Acid Solution.
For a 3.0 M H3PO4 solution, calculate:
(a) [H3O+]; (b) [H2PO4-]; (c) [HPO4
2-] (d) [PO43-]
H3PO4 + H2O H2PO4- + H3O+
Initial conc. 3.0 M 0 0
Changes -x M +x M +x M
Eqlbrm conc. (3.0-x) M x M x M
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 288 of 47
Example 17-9H3PO4 + H2O H2PO4
- + H3O+
[H3O+] [H2PO4-]
[H3PO4] Ka=
x · x
(3.0 – x)=
Assume that x << 3.0
= 7.110-3
x2 = (3.0)(7.110-3) x = 0.14 M
[H2PO4-] = [H3O+] = 0.14 M
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 289 of 47
Example 17-9
H2PO4- + H2O HPO4
2- + H3O+
[H3O+] [HPO42-]
[H2PO4-]
Ka=y · (0.14 + y) (0.14 - y)
= = 6.310-8
Initial conc. 0.14 M 0 0.14 M
Changes -y M +y M +y M
Eqlbrm conc. (0.14 - y) M y M (0.14 +y) M
y << 0.14 M y = [HPO42-] = 6.310-8
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 290 of 47
Example 17-9
HPO4- + H2O PO4
3- + H3O+
[H3O+] [HPO42-]
[H2PO4-]
Ka=(0.14)[PO4
3-]6.310-8
= = 4.210-13 M
[PO43-] = 1.910-19 M
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 291 of 47
Sulfuric Acid
Sulfuric acid:
A diprotic acid.
H2SO4 + H2O H3O+ + HSO4-
HSO4- + H2O H3O+ + SO4
2-
Ka = very large
Ka = 1.96
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 292 of 47
General Approach to Solution Equilibrium Calculations
• Identify species present in any significant amounts in solution (excluding H2O).
• Write equations that include these species.– Number of equations = number of unknowns.
• Equilibrium constant expressions.• Material balance equations.• Electroneutrality condition.
• Solve the system of equations for the unknowns.
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 293 of 47
17-7 Ions as Acids and Bases
NH4+ + H2O NH3 + H3O+
baseacid
CH3CO2- + H2O CH3CO2H + OH-
base acid
[NH3] [H3O+] [OH-] Ka= [NH4
+] [OH-]
[NH3] [H3O+] Ka= [NH4
+] = ?
=KW
Kb
= 1.010-14
1.810-5= 5.610-10
Ka Kb = Kw
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 294 of 47
Hydrolysis
• Water (hydro) causing cleavage (lysis) of a bond.
Na+ + H2O → Na+ + H2O
NH4+ + H2O → NH3 + H3O+
Cl- + H2O → Cl- + H2O
No reaction
No reaction
Hydrolysis
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 295 of 47
17-8 Molecular Structure and Acid-Base Behavior
• Why is HCl a strong acid, but HF is a weak one?• Why is CH3CO2H a stronger acid than CH3CH2OH?
• There is a relationship between molecular structure and acid strength.
• Bond dissociation energies are measured in the gas phase and not in solution.
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 296 of 47
Strengths of Binary Acids
HI HBr HCl HF
160.9 > 141.4 > 127.4 > 91.7 pm
297 < 368 < 431 < 569 kJ/mol
Bond length
Bond energy
109 > 108 > 1.3106 >> 6.610-4 Acid strength
HF + H2O → [F-·····H3O+] F- + H3O+
ion pairH-bonding
free ions
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 297 of 47
Strengths of Oxoacids
• Factors promoting electron withdrawal from the OH bond to the oxygen atom:– High electronegativity (EN) of the central atom.– A large number of terminal O atoms in the molecule.
H-O-Cl H-O-Br
ENCl = 3.0 ENBr= 2.8
Ka = 2.910-8 Ka = 2.110-9
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 298 of 47
S OO
O
O
H H····
····
-
2+
··
···· ···· ··
-
S OO
O
H H····
····
-
+··
···· ··
S OO
O
O
H H····
····
·· ···· ··
S OO
O
H H····
···· ··
·· ··
Ka 103 Ka =1.310-2
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 299 of 47
Strengths of Organic Acids
C OC
O
H H····
·· ··H
H
OCH H····
H
H
C
H
H
Ka = 1.810-5 Ka =1.310-16
acetic acid ethanol
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 300 of 47
Focus on the Anions Formed
OCH····
H
H
C
H
H
C
O
C
O
H-
··
·· ··
H
H
····
C
O
C
O
H
-···· ··
H
H
····
··
-
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 301 of 47
Structural Effects
C
H
H
C
O
C
O
H-
···· ··
H
H
····
CH
H
H
C
O
O
-··
····
····
C
H
H
C
H
H
C
H
H
C
H
H
C
H
H
Ka = 1.810-5
Ka = 1.310-5
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 302 of 47
Structural Effects
C
O
C
O
H-
···· ··
H
H
····
Ka = 1.810-5
Ka = 1.410-3
C
O
C
O
H-
··
·· ··
H
Cl
····
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 303 of 47
Strengths of Amines as Bases
NH
H
H
·· NBr
H
H
··
pKb = 4.74 pKa = 7.61
ammonia bromamine
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 304 of 47
Strengths of Amines as Bases
CH
H
H
C
H
H
CH
H
H
C
H
H
CH
H
H
C
H
H
pKb = 4.74 pKa = 3.38 pKb = 3.37
methylamine ethylamine propylamine
NH2 NH2 NH2
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 305 of 47
Resonance Effects
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 306 of 47
Inductive Effects
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 307 of 47
17-9 Lewis Acids and Bases
• Lewis Acid– A species (atom, ion or molecule) that is an electron
pair acceptor.• Lewis Base
– A species that is an electron pair donor.
base acid adduct
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 308 of 47
Showing Electron Movement
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 309 of 47
Focus On Acid Rain
CO2 + H2O H2CO3
H2CO3 + H2O HCO3- + H3O+
3 NO2 + H2O 2 HNO3 + NO
Prentice-Hall © 2002 General Chemistry: Chapter 17 Slide 310 of 47
Chapter 17 Questions
Develop problem solving skills and base your strategy not on solutions to specific problems but on understanding.
Choose a variety of problems from the text as examples.
Practice good techniques and get coaching from people who have been here before.
Philip DuttonUniversity of Windsor, Canada
N9B 3P4
Prentice-Hall © 2002
General ChemistryPrinciples and Modern Applications
Petrucci • Harwood • Herring 8th Edition
Chapter 18: Additional Aspects of Acid-Base Equilibria
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 312 of 42
Contents
18-1 The Common-Ion Effect in Acid-Base Equilibria18-2 Buffer Solutions18-3 Acid-Base Indicators18-4 Neutralization Reactions and Titration Curves18-5 Solutions of Salts of Polyprotic Acids18-6 Acid-Base Equilibrium Calculations: A Summary
Focus On Buffers in Blood
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 313 of 42
18-1 The Common-Ion Effect in Acid-Base Equilibria
• The Common-Ion Effect describes the effect on an equilibrium by a second substance that furnishes ions that can participate in that equilibrium.
• The added ions are said to be common to the equilibrium.
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 314 of 42
Solutions of Weak Acids and Strong Acids
• Consider a solution that contains both 0.100 M CH3CO2H and 0.100 M HCl.
CH3CO2H + H2O CH3CO2- + H3O+
HCl + H2O Cl- + H3O+
(0.100-x) M x M x M
0.100 M 0.100 M
[H3O+] = (0.100 + x) M essentially all due to HCl
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 315 of 42
Acetic Acid and Hydrochloric Acid
0.1 M HCl 0.1 M CH3CO2H 0.1 M HCl +0.1 M CH3CO2H
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 316 of 42
Example 18-1Demonstrating the Common-Ion Effect:
A Solution of a weak Acid and a Strong Acid.
(a) Determine [H3O+] and [CH3CO2-] in 0.100 M CH3CO2H.
(b) Then determine these same quantities in a solution that is 0.100 M in both CH3CO2H and HCl.
CH3CO2H + H2O → H3O+ + CH3CO2-
Recall Example 17-6 (p 680):
[H3O+] = [CH3CO2-] = 1.310-3 M
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 317 of 42
Example 18-1
CH3CO2H + H2O → H3O+ + CH3CO2-
Initial concs.
weak acid 0.100 M 0 M 0 M
strong acid 0 M 0.100 M 0 M
Changes -x M +x M +x M
Eqlbrm conc. (0.100 - x) M (0.100 + x) M x M
Assume x << 0.100 M, 0.100 – x 0.100 + x 0.100 M
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 318 of 42
Example 18-1
Eqlbrm conc. (0.100 - x) M (0.100 + x) M x M
Assume x << 0.100 M, 0.100 – x 0.100 + x 0.100 M
CH3CO2H + H2O → H3O+ + CH3CO2-
[H3O+] [CH3CO2-]
[C3CO2H]Ka=
x · (0.100 + x) (0.100 - x)
=
x · (0.100) (0.100)
= = 1.810-5
[CH3CO2-] = 1.810-5 M compared to 1.310-3 M.
Le Chatellier’s Principle
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 319 of 42
Suppression of Ionization of a Weak Acid
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 320 of 42
Suppression of Ionization of a Weak Base
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 321 of 42
Solutions of Weak Acids and Their Salts
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 322 of 42
Solutions of Weak Bases and Their Salts
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 323 of 42
18-2 Buffer Solutions
• Two component systems that change pH only slightly on addition of acid or base.– The two components must not neutralize each other but
must neutralize strong acids and bases.
• A weak acid and it’s conjugate base.• A weak base and it’s conjugate acid
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 324 of 42
Buffer Solutions
• Consider [CH3CO2H] = [CH3CO2-] in a solution.
[H3O+] [CH3CO2-]
[C3CO2H]Ka= = 1.810-5
= 1.810-5 [CH3CO2
-]
[C3CO2H]Ka[H3O+] =
pH = -log[H3O+] = -logKa = -log(1.810-5) = 4.74
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 325 of 42
How A Buffer Works
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 326 of 42
The Henderson-Hasselbalch Equation
• A variation of the ionization constant expression.• Consider a hypothetical weak acid, HA, and its
salt NaA:
HA + H2O A- + H3O+[H3O+] [A-]
[HA]Ka=
[H3O+] [HA]
Ka=[A-]
-log[H3O+]-log [HA]
-logKa=[A-]
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 327 of 42
Henderson-Hasselbalch Equation
-log[H3O+] - log [HA]
-logKa=[A-]
pH - log [HA]
pKa =[A-]
pKa + log [HA]
pH =[A-]
pKa + log [acid]
pH =[conjugate base]
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 328 of 42
Henderson-Hasselbalch Equation
• Only useful when you can use initial concentrations of acid and salt.– This limits the validity of the equation.
• Limits can be met by:
0.1 < [HA]
< 10[A-]
[A-] > 10Ka and [HA] > 10Ka
pKa + log [acid]
pH=[conjugate base]
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 329 of 42
Example 18-5Preparing a Buffer Solution of a Desired pH.
What mass of NaC2H3O2 must be dissolved in 0.300 L of 0.25 M HC2H3O2 to produce a solution with pH = 5.09? (Assume that the solution volume is constant at 0.300 L)
HC2H3O2 + H2O C2H3O2- + H3O+
Equilibrium expression:
[H3O+] [HC2H3O2]
Ka=[C2H3O2
-]= 1.810-5
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 330 of 42
Example 18-5
[H3O+] [HC2H3O2]
Ka=[C2H3O2
-]= 1.810-5
[H3O+] = 10-5.09 = 8.110-6
[HC2H3O2] = 0.25 M
Solve for [C2H3O2-]
[H3O+] [HC2H3O2]= Ka
[C2H3O2-] = 0.56 M
8.110-6
0.25= 1.810-5
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 331 of 42
Example 18-5
1 mol NaC2H3O2
82.0 g NaC2H3O2
mass C2H3O2- = 0.300 L
[C2H3O2-] = 0.56 M
1 L 0.56 mol
1 mol C2H3O2-
1 mol NaC2H3O2
= 14 g NaC2H3O2
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 332 of 42
Six Methods of Preparing Buffer Solutions
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 333 of 42
Calculating Changes in Buffer Solutions
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 334 of 42
Buffer Capacity and Range
• Buffer capacity is the amount of acid or base that a buffer can neutralize before its pH changes appreciably.– Maximum buffer capacity exists when [HA] and [A-]
are large and approximately equal to each other.• Buffer range is the pH range over which a buffer
effectively neutralizes added acids and bases.– Practically, range is 2 pH units around pKa
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 335 of 42
18-3 Acid-Base Indicators
• Color of some substances depends on the pH.
HIn + H2O In- + H3O+
>90% acid form the color appears to be the acid color
>90% base form the color appears to be the base color
Intermediate color is seen in between these two states.
Complete color change occurs over 2 pH units.
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 336 of 42
Indicator Colors and Ranges
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 337 of 42
18-4 Neutralization Reactions and Titration Curves
• Equivalence point:– The point in the reaction at which both acid and base have been
consumed.– Neither acid nor base is present in excess.
• End point:– The point at which the indicator changes color.
• Titrant:– The known solution added to the solution of unknown concentration.
• Titration Curve:– The plot of pH vs. volume.
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 338 of 42
The millimole
• Typically:– Volume of titrant added is less than 50 mL.– Concentration of titrant is less than 1 mol/L.– Titration uses less than 1/1000 mole of acid and base.
L/1000 mol/1000
= M = L
molmL
mmol=
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 339 of 42
Titration of a Strong Acid with a Strong Base
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 340 of 42
Titration of a Strong Acid with a Strong Base
• The pH has a low value at the beginning.• The pH changes slowly
– until just before the equivalence point.• The pH rises sharply
– perhaps 6 units per 0.1 mL addition of titrant.• The pH rises slowly again.• Any Acid-Base Indicator will do.
– As long as color change occurs between pH 4 and 10.
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 341 of 42
Titration of a Strong Base with a Strong Acid
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 342 of 42
Titration of a Weak Acid with a Strong Base
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 343 of 42
Titration of a Weak Acid with a Strong Base
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 344 of 42
Titration of a Weak Polyprotic Acid
H3PO4 H2PO4- HPO4
2- PO43-
NaOHNaOH
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 345 of 42
18-5 Solutions of Salts of Polyprotic Acids
• The third equivalence point of phosphoric acid can only be reached in a strongly basic solution.
• The pH of this third equivalence point is not difficult to caluclate.– It corresponds to that of Na3PO4 (aq) and PO4
3- can ionize only as a base.
PO43- + H2O → OH- + HPO4
2-
Kb = Kw/Ka = 2.410-2
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 346 of 42
Example 18-9
Kb = 2.410-2 PO43- + H2O → OH- + HPO4
2-
Initial concs. 1.0 M 0 M 0 M
Changes -x M +x M +x M
Eqlbrm conc. (1.00 - x) M x M x M
Determining the pH of a Solution Containing the Anion (An-) of a Polyprotic Acid.
Sodium phosphate, Na3PO4, is an ingredient of some preparations used to clean painted walls before they are repainted. What is the pH of 1.0 M Na3PO4?
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 347 of 42
Example 18-9
x2 + 0.024x – 0.024 = 0 x = 0.14 M
pOH = +0.85 pH = 13.15
[OH-] [HPO42-]
[PO43-]
Kb=x · x
(1.00 - x)= = 2.410-2
It is more difficult to calculate the pH values of NaH2PO4 and Na2HPO4 because two equilibria must be considered simultaneously.
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 348 of 42
Concentrated Solutions of Polyprotic Acids
• For solutions that are reasonably concentrated (> 0.1 M) the pH values prove to be independent of solution concentrations.
for H2PO4-
for HPO42-
pH = 0.5 (pKa1 + pKa2) = 0.5 (2.15 + 7.20) = 4.68
pH = 0.5 (pKa1 + pKa2) = 0.5 (7.20 + 12.38) = 9.79
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 349 of 42
18-6 Acid-Base Equilibrium Calculations:A Summary
• Determine which species are potentially present in solution, and how large their concentrations are likely to be.
• Identify possible reactions between components and determine their stoichiometry.
• Identify which equilibrium equations apply to the particular situation and which are most significant.
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 350 of 42
Focus On Buffers in Blood
CO2(g) + H2O H2CO3(aq)
H2CO3(aq) + H2O(l) HCO3-(aq)
Ka1 = 4.410-7 pKa1 = 6.4
pH = 7.4 = 6.4 +1.0
pH = pKa1 + log [H2CO3]
[HCO3-]
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 351 of 42
Buffers in Blood
• 10/1 buffer ratio is somewhat outside maximum buffer capacity range but…
• The need to neutralize excess acid (lactic) is generally greater than the need to neutralize excess base.
• If additional H2CO3 is needed CO2 from the lungs can be utilized.
• Other components of the blood (proteins and phosphates) contribute to maintaining blood pH.
Prentice-Hall © 2002 General Chemistry: Chapter 18 Slide 352 of 42
Chapter 18 Questions
Develop problem solving skills and base your strategy not on solutions to specific problems but on understanding.
Choose a variety of problems from the text as examples.
Practice good techniques and get coaching from people who have been here before.
Philip DuttonUniversity of Windsor, Canada
N9B 3P4
Prentice-Hall © 2002
General ChemistryPrinciples and Modern Applications
Petrucci • Harwood • Herring 8th Edition
Chapter 19: Solubility and Complex-Ion Equilibria
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 354 of 34
Contents
19-1 The Solubility Product Constant, Ksp
19-2 The Relationship Between Solubility and Ksp
19-3 The Common-Ion Effect in Solubility Equilibria19-4 Limitations of the Ksp Concept
19-5 Criteria for Precipitation and Its Completeness19-6 Fractional Precipitation19-7 Solubility and pH19-8 Equilibria Involving Complex Ions19-9 Qualitative Cation Analysis
Focus On Shells, Teeth, and Fossils
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 355 of 34
19-1 The Solubility Product Constant, Ksp
CaSO4(s) Ca2+(aq) + SO42-(aq)
Ksp = [Ca2+][SO42-] = 9.110-6 at 25°C
• The equilibrium constant for the equilibrium established between a solid solute and its ions in a saturated solution.
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 356 of 34
Table 19-1 Several Solubility Product Constants at 25°C
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 357 of 34
The Relationship Between Solubility and Ksp
• Molar solubility.– The molarity in a saturated
aqueous solution.– Related to Ksp
g BaSO4/100 mL → mol BaSO4/L
→ [Ba2+] and [SO42-]
→ Ksp = 1.110-10
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 358 of 34
19-3 The Common-Ion Effect in Solubility Equilibria
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 359 of 34
The Common-Ion Effect and Le Chatelliers Principle
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 360 of 34
19-4 Limitations of the Ksp Concept
• Ksp is usually limited to slightly soluble solutes.– For more soluble solutes we must use ion activities
• Activities (effective concentrations) become smaller than the measured concentrations.
• The Salt Effect (or diverse ion effect).– Ionic interactions are important even when an ion is
not apparently participating in the equilibrium.• Uncommon ions tend to increase solublity.
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 361 of 34
Effects on the Solubility of Ag2CrO4
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 362 of 34
Ion Pairs
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 363 of 34
Incomplete Dissociation
• Assumption that all ions in solution are completely dissociated is not valid.
• Ion Pair formation occurs.– Some solute “molecules” are present in solution.– Increasingly likely as charges on ions increase.
Ksp (CaSO4) = 2.310-4 by considering solubility in g/100 mL
Table 19: Ksp = 9.110-6
Activities take into account ion pair formation and must be used.
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 364 of 34
Simultaneous Equilibria
• Other equilibria are usually present in a solution.– Kw
for example.– These must be taken into account if they affect the
equilibrium in question.
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 365 of 34
19-5 Criteria for Precipitation and Its Completeness
AgI(s) Ag+(aq) + I-(aq)
Mix AgNO3(aq) and KI(aq) to obtain a solution that is 0.010 M in Ag+ and 0.015 M in I-.
Saturated, supersaturated or unsaturated?
Q = [Ag+][Cl-] = (0.010)(0.015) = 1.510-4 > Ksp
Ksp = [Ag+][Cl-] = 8.510-17
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 366 of 34
The Ion Product
Q is generally called the ion product.
Q > Ksp Precipitation should occur.
Q = Ksp The solution is just saturated.
Q < Ksp Precipitation cannot occur.
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 367 of 34
Example 19-5Applying the Criteria for Precipitation of a Slightly Soluble Solute.
Three drops of 0.20 M KI are added to 100.0 mL of 0.010 M Pb(NO3)2. Will a precipitate of lead iodide form? (1 drop = 0.05 mL)
PbI2(s) → Pb2+(aq) + 2 I-(aq) Ksp= 7.110-9
Determine the amount of I- in the solution:
= 310-5 mol I-
nI- = 3 drops 1 drop
0.05 mL
1000 mL1 L
1 L0.20 mol KI
1 mol KI1 mol I-
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 368 of 34
Example 19-5
[I-] = 0.1000 L
310-5 mol I-
= 310-4 mol I-
Determine the concentration of I- in the solution:
Apply the Precipitation Criteria:
Q = [Pb2+][I-]2 = (0.010)(310-4)2
= 910-10 < Ksp = 7.110-9
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 369 of 34
19-6 Fractional Precipitation
• A technique in which two or more ions in solution are separated by the proper use of one reagent that causes precipitation of both ions.
• Significant differences in solubilities are necessary.
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 370 of 34
19-7 Solubility and pH
Mg(OH)2 (s) Mg2+(aq) + 2 OH-(aq) Ksp = 1.810-11
OH-(aq) + H3O+(aq) H2O(aq) K = 1/Kw = 1.01014
2 OH-(aq) + 2 H3O+(aq) 2 H2O(aq) K' = (1/Kw)2 = 1.01028
Mg(OH)2 (s) + H3O+(aq) Mg2+(aq) + 2 OH-(aq)
K = Ksp(1/Kw)2 = (1.810-11)(1.010-14) = 1.81017
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 371 of 34
19-8 Equilibria Involving Complex Ions
AgCl(s) + 2 NH3(aq) → [Ag(NH3)2]+(aq) + Cl-(aq)
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 372 of 34
Complex Ions
• Coordination compounds.– Substances which contain complex ions.
• Complex ions.– A polyatomic cation or anion
composed of:• A central metal ion.• Ligands
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 373 of 34
Formation Constant of Complex Ions
AgCl(s) + 2 NH3(aq) → [Ag(NH3)2]+(aq) + Cl-(aq)
AgCl(s) → Ag+(aq) + Cl-(aq)
Ag+(aq) + 2 NH3(aq) → [Ag(NH3)2]+(aq)
Ksp = 1.810-11
Kf = = 1.6107[Ag(NH3)2]+
[Ag+][NH3]2
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 374 of 34
Table 19.2 Formation Constants for Some Complex Ions
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 375 of 34
Example 19-11Determining Whether a Precipitate will Form in a Solution Containing Complex Ions.
A 0.10 mol sample of AgNO3 is dissolved in 1.00 L of 1.00 M NH3. If 0.010 mol NaCl is added to this solution, will AgCl(s) precipitate?
Ag+(aq) + 2 NH3(aq) → [Ag(NH3)2]+(aq)Assume Kf is large:
Initial conc. 0.10 M 1.00 M 0 M
Change -0.10 M -0.20 M +0.10 M
Eqlbrm conc. (0) M 0.80 M 0.10 M
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 376 of 34
Example 19-11[Ag+] is small but not 0, use Kf to calculate [Ag+]:
Ag+(aq) + 2 NH3(aq) → [Ag(NH3)2]+(aq)
Initial concs. 0 M 0.80 M 0.10 M
Changes +x M +2x M -x M
Eqlbrm conc. x M 0.80 + 2x M 0.10 - x M
0.10(1.6 107)(0.80)2
x = [Ag+] = = 9.810-9 M
= 1.6107[Ag(NH3)2]+
[Ag+][NH3]2
0.10-xx(0.80 + 2x)2
0.10x(0.80)2
= Kf =
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 377 of 34
Example 19-11Compare Qsp to Ksp and determine if precipitation will occur:
= (9.810-9)(1.010-2) = 9.810-11[Ag+][Cl-]Qsp =
Ksp = 1.810-10
Qsp < Ksp
AgCl does not precipitate.
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 378 of 34
19-9 Qualitative Cation Analysis
• An analysis that aims at identifying the cations present in a mixture but not their quantities.
• Think of cations in solubility groups according to the conditions that causes precipitation
chloride group hydrogen sulfide group ammonium sulfide group carbonate group.
–Selectively precipitate the first group of cations then move on to the next.
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 379 of 34
Qualitative Cation Analysis
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 380 of 34
Chloride Group Precipitates
(a) Group precipitate
Wash ppt with hot water PbCl2 is slightly soluble. Test aqueous solution with CrO4
2-.
(c) Pb2+(aq) + CrO42- → PbCrO4(s)
Test remaining precipitate with ammonia.(b) AgCl(s) + 2 NH3(aq) →
Ag(NH3)2 (aq) + Cl-(aq)
(b) Hg2Cl2(a) + 2 NH3 → Hg(l) + HgNH2Cl(s) + NH4
+(aq) + Cl-(aq)
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 381 of 34
Hydrogen Sulfide Equilibria
H2S(aq) + H2O(l) HS-(aq) + H3O+(aq) Ka1 = 1.010-7
HS-(aq) + H2O(l) S2-(aq) + H3O+(aq) Ka2 = 1.010-19
S2- is an extremely strong base and is unlikely to be theprecipitating agent for the sulfide groups.
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 382 of 34
Lead Sulfide Equilibria
PbS(s) + H2O(l) Pb2+(aq) + HS-(aq) + OH-(aq)
Ksp = 310-28
H3O+(aq) + HS-(aq) H2S(aq) + H2O(aq) 1/Ka1 = 1.0/1.010-7
H3O+(aq) + OH-(aq) H2O(l) + H2O(l) 1/Kw = 1.0/1.010-14
PbS(s) + 2 H3O(l) Pb2+(aq) + H2S(aq) + 2 H2O(l)
Kspa = = 310-7Ksp
Ka1 Kw
310-28 1.010-7
1.010-14 =
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 383 of 34
Dissolving Metal Sulfides
• Several methods exist to re-dissolve precipitated metal sulfides.– React with an acid.
• FeS readily soluble in strong acid but PbS and HgS are not because their Ksp values are too low.
– React with an oxidizing acid.3 CuS(aq) + 8 H+(aq) + 2 NO3
-(aq) →
3 Cu2+(aq) + 3 S(s) + 2 NO(g) + 4 H2O(l)
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 384 of 34
A Sensitive Test for Copper(II)
[Cu(H2O)4]2+(aq) + 4 NH3(aq) → [Cu(NH3)4]2+(aq) + 4 H2O(l)
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 385 of 34
Focus On Shells, Teeth and Fossils
Ca2+(aq) + 2 HCO3-(aq) →
CaCO3(s) + H2O(l) + CO2(g)
Calcite
Ca5(PO4)3OH(s) + 4 H3O+(aq) → 5 Ca2+(s) + 5 H2O(l) + 3 HPO42-(aq)
Fluoroapatite
Ca5(PO4)3F(s)
Hydroxyapatite
Ca5(PO4)3OH(s)
Prentice-Hall © 2002 General Chemistry: Chapter 19 Slide 386 of 34
Chapter 19 Questions
Develop problem solving skills and base your strategy not on solutions to specific problems but on understanding.
Choose a variety of problems from the text as examples.
Practice good techniques and get coaching from people who have been here before.
Philip DuttonUniversity of Windsor, Canada
N9B 3P4
Prentice-Hall © 2002
General ChemistryPrinciples and Modern Applications
Petrucci • Harwood • Herring 8th Edition
Chapter 20: Spontaneous Change: Entropy and Free Energy
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 388 of 37
Contents
20-1 Spontaneity: The Meaning of Spontaneous Change20-2 The Concept of Entropy20-3 Evaluating Entropy and Entropy Changes20-4 Criteria for Spontaneous Change:
The Second Law of Thermodynamics
20-5 Standard Fee Energy Change, ΔG°20-6 Free Energy Change and Equilibrium20-7 ΔG° and Keq
as Functions of Temperature
20-8 Coupled ReactionsFocus On Coupled Reactions in Biological Systems
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 389 of 37
20-1 Spontaneity: The Meaning of Spontaneous Change
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 390 of 37
Spontaneous Process
• A process that occurs in a system left to itself.– Once started, no external actions is necessary to make
the process continue.• A non-spontaneous process will not occur without
external action continuously applied.
4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)
H2O(s) H2O(l)
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 391 of 37
Spontaneous Process
• Potential energy decreases.• For chemical systems the internal energy U is
equivalent to potential energy.
• Berthelot and Thomsen 1870’s– Spontaneous change occurs in the direction in which
the enthalpy of a system decreases.– Mainly true but there are exceptions.
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 392 of 37
20-2 The Concept of Entropy
• Entropy, S.– The greater the number
of configurations of the microscopic particles among the energy levels in a particular system, the greater the entropy of the system.
ΔS > 0 spontaneous
ΔU = ΔH = 0
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 393 of 37
The Boltzmann Equation for Entropy
• States, S.– The microscopic energy levels
available in a system.• Microstates, W.
– The particular way in which particles are distributed amongst the states. Number of microstates = W.
• The Boltzmann constant, k.– Effectively the gas constant per molecule = R/NA.
S = k lnW
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 394 of 37
Boltzmann Distribution
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 395 of 37
Entropy Change
ΔS = qrev
T
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 396 of 37
20-3 Evaluating Entropy and Entropy Changes
• Phase transitions.– Exchange of heat can be carried out reversibly.
ΔS = ΔH
Ttr
H2O(s, 1 atm) H2O(l, 1 atm) ΔHfus° = 6.02 kJ at 273.15 K
ΔSfus = ΔHfus
Ttr
°=
6.02 kJ mol-1
273.15 K = 2.2010-2 kJ mol-1 K-1
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 397 of 37
Trouton’s Rule
ΔS = ΔHvap
Tbp
87 kJ mol-1 K-1
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 398 of 37
Raoult’s Law
°PA = APA
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 399 of 37
Absolute Entropies
• Third law of thermodynamics.– The entropy of a pure perfect crystal at 0 K is zero.
• Standard molar entropy.– Tabulated in Appendix D.
ΔS = [ pS°(products) - rS°(reactants)]
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 400 of 37
Entropy as a Function of Temperature
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 401 of 37
Vibrational Energy and Entropy
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 402 of 37
20-4 Criteria for Spontaneous Change:The Second Law of Thermodynamics.
ΔStotal = ΔSuniverse = ΔSsystem + ΔSsurroundings
The Second Law of Thermodynamics:
ΔSuniverse = ΔSsystem + ΔSsurroundings > 0
All spontaneous processes produce an increase in the entropy of the universe.
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 403 of 37
Free Energy and Free Energy Change
• Hypothetical process:– only pressure-volume work, at constant T and P.
qsurroundings = -qp = -ΔHsys
• Make the enthalpy change reversible.– large surroundings, infinitesimal change in temperature.
• Under these conditions we can calculate entropy.
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 404 of 37
Free Energy and Free Energy Change
TΔSuniv. = TΔSsys – ΔHsys = -(ΔHsys – TΔSsys)
-TΔSuniv. = ΔHsys – TΔSsys
G = H - TS
ΔG = ΔH - TΔS
For the universe:
For the system:
ΔGsys = - TΔSuniverse
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 405 of 37
Criteria for Spontaneous Change
ΔGsys < 0 (negative), the process is spontaneous.
ΔGsys = 0 (zero), the process is at equilibrium.
ΔGsys > 0 (positive), the process is non-spontaneous.
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 406 of 37
Table 20.1 Criteria for Spontaneous Change
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 407 of 37
20-5 Standard Free Energy Change, ΔG°
• The standard free energy of formation, ΔGf°.– The free energy change for a reaction in which a
substance in its standard state is formed from its elements in reference forms in their standard states.
• The standard free energy of reaction, ΔG°.
ΔG° = [ p ΔGf°(products) - r ΔGf°(reactants)]
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 408 of 37
20-6 Free Energy Change and Equilibrium
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 409 of 37
Free Energy Change and Equilibrium
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 410 of 37
Relationship of ΔG° to ΔG for Nonstandard Conditions
2 N2(g) + 3 H2(g) 2 NH3(g)
ΔG = ΔH - TΔS ΔG° = ΔH° - TΔS°
For ideal gases ΔH = ΔH°
ΔG = ΔH° - TΔS
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 411 of 37
Relationship Between S and S°
qrev = -w = RT lnVf
Vi
ΔS = qrev
T= R ln
Vf
Vi
ΔS = Sf – Si = R lnVf
Vi
= R lnPi
Pf
= -R lnPf
Pi
S = S° - R ln PP°
= S° - R ln P1
= S° - R ln P
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 412 of 37
2 N2(g) + 3 H2(g) 2 NH3(g)SNH3
= SNH3
– Rln PNH3
SN2 =
SN2 – Rln PN2
SH2 =
SH2 – Rln PH2
ΔSrxn = 2(SNH3 – Rln PNH3
) – 2(SN2 – Rln PN2
) –3(SH2 – Rln PH2
)
ΔSrxn = 2 SNH3 – 2SN2
–3SH2+ Rln
PN2PH2
PNH3
2
2
3
ΔSrxn = ΔS°rxn + RlnPN2
PH2
PNH3
2
2
3
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 413 of 37
ΔG Under Non-standard Conditions
ΔG = ΔH° - TΔS ΔSrxn = ΔS°rxn + RlnPN2
PH2
PNH3
2
2
3
ΔG = ΔH° - TΔS°rxn – TR lnPN2
PH2
PNH3
2
2
3
ΔG = ΔG° + RT lnPN2
PH22
PNH3
2
3
ΔG = ΔG° + RT ln Q
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 414 of 37
ΔG and the Equilibrium Constant Keq
ΔG = ΔG° + RT ln Q
ΔG = ΔG° + RT ln Keq= 0
If the reaction is at equilibrium then:
ΔG° = -RT ln Keq
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 415 of 37
Criteria for Spontaneous Change
Every chemical reaction consists of both a forward and a reverse reaction.
The direction of spontaneous change is the direction in which the free energy decreases.
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 416 of 37
Significance of the Magnitude of ΔG
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 417 of 37
The Thermodynamic Equilibrium Constant: Activities
S = S° - R ln PP°
= S° - R ln P1
For ideal gases at 1.0 bar:
S = S° - R ln cc°
= S° - R ln aTherefore, in solution:
PV=nRT or P=(n/V)RT, pressure is an effective concentration
The effective concentration in the standard state for an ideal solution is c° = 1 M.
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 418 of 37
Activities
• For pure solids and liquids:» a = 1
• For ideal gases: » a = P (in bars, 1 bar = 0.987
atm)• For ideal solutes in aqueous solution:
» a = c (in mol L-1)
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The Thermodynamic Equilibrium Constant, Keq
• A dimensionless equilibrium constant expressed in terms of activities.
• Often Keq = Kc
• Must be used to determine ΔG.
ΔG = ΔG° + RT ln agah…aaab…
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20-7 ΔG° and Keq as Functions of Temperature
ΔG° = ΔH° -TΔS° ΔG° = -RT ln Keq
ln Keq = -ΔG°RT
=-ΔH°RT
TΔS°RT
+
ln Keq = -ΔH°RT
ΔS°R
+
ln = -ΔH°RT2
ΔS°R
+-ΔH°RT1
ΔS°R
+- =-ΔH°
R1T2
1T1
-Keq1
Keq2
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 421 of 37
Temperature Dependence of Keq
ln Keq = -ΔH°RT
ΔS°R
+
slope = -ΔH°
R
-ΔH° = R slope
= -8.3145 J mol-1 K-1 2.2104 K
= -1.8102 kJ mol-1
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 422 of 37
20-8 Coupled Reactions
• In order to drive a non-spontaneous reactions we changed the conditions (i.e. temperature or electrolysis)
• Another method is to couple two reactions.– One with a positive ΔG and one with a negative ΔG.– Overall spontaneous process.
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 423 of 37
Smelting Copper Ore
Cu2O(s) → 2 Cu(s) + ½ O2(g)ΔG°673K = +125 kJ Δ
Cu2O(s) + C(s) → 2 Cu(s) + CO(g)
Spontaneous reaction!
Cu2O(s) → 2 Cu(s) + ½ O2(g)Non-spontaneous reaction: +125 kJ
C(s) + ½ O2(g) → CO(g) Spontaneous reaction: -175 kJ
-50 kJ
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Focus On Coupled Reactions in Biological Systems
ADP3- + HPO42- + H+ → ATP4- + H2O
ΔG° = -9.2 kJ mol-1
ΔG = ΔG° + RT ln aATPaH2O
aADPaPiaH3O+
But [H3O+] = 10-7 M not 1.0 M.
ΔG = -9.2 kJ mol-1 + 41.6 kJ mol-1
= +32.4 kJ mol-1 = ΔG°'The biological standard state:
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Focus On Coupled Reactions in Biological Systems
Glucose → 2 lactate + 2 H+ -218 kJ2 ADP3- + 2 HPO4
2- + 2 H+ → 2 ATP4- + 2 H2O +64 kJ
2 ADP3- + 2 HPO42- + glucose → 2 ATP4- + 2 H2O + 2 lactate -153 kJ
Prentice-Hall © 2002 General Chemistry: Chapter 20 Slide 426 of 37
Chapter 20 Questions
Develop problem solving skills and base your strategy not on solutions to specific problems but on understanding.
Choose a variety of problems from the text as examples.
Practice good techniques and get coaching from people who have been here before.