Ch 11 20 general chemistry 8e petrucci

Philip DuttonUniversity of Windsor, Canada

N9B 3P4

General ChemistryPrinciples and Modern Applications

Petrucci • Harwood • Herring 8th Edition

Chapter 11: Chemical Bonding I:Basic Concepts

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

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.

Lewis Symbols

• A chemical symbol represents the nucleus and the core e-.

• Dots around the symbol represent valence e-.

••

N••

••

• P••

••

• As••

••

• Sb••

••

• Bi••

••

••Al••

• Se•• •••

Ar••

••

••I •••

••

Lewis Structures for Ionic Compounds

• O•••

•••

••O••

••

••Ba

Cl•••

••

••Cl•••

••••

••Cl••

••

••Mg

11-2 Covalent Bonding

Coordinate Covalent Bonds

HN ••H

Cl ••Cl••

••

Multiple Covalent Bonds

C••

••

O••

• •

• •O••

• •

• •CO O

••

•••

••

•CO O ••

•••

••

••CO O

••

Multiple Covalent Bonds

N••

••

•N N

••

•N••

••

N N•

•• ••

•N N •• ••

11-3 Polar Covalent Bonds

H Clδ+ δ-

Analogy to Population

Electronegativity

Percent Ionic Character

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.

Skeletal Structure

• Identify central and terminal atoms.

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.

Strategy for Writing Lewis

Structures

Formal Charge

FC = #valence e- - #lone pair e- - #bond pair e-21

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

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

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 ••

••

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 ••

••

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+ - +-

11-5 Resonance

O O O O O O••••

••

••••

••

••+ +- -

O O O••••••••

••+ -½ -½

11-6 Exceptions to the Octet Rule

• Odd e- species.

N=O ••

••

•••

H—C—H

•O—H ••

••

Exceptions to the Octet Rule

• Incomplete octets.

••••

••

•••• ••••

••

•••• ••

••

•• ••

••••

••

•••• ••

•••• -

•• ••

••

Exceptions to the Octet Rule

• Expanded octets.

••••

••

••••

•• P

••••Cl

••••

•••• •• ••

••

••••

••Cl•• S

••

••F

••••

••

•••• •• ••

••F

••••••F••

F••

••

Expanded Valence Shell

11-7 The Shapes of Molecules

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.

Balloon Analogy

Methane, Ammonia and Water

Table 11.1 Molecular Geometry as a Function of Electron Group Geometry

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.

Dipole Moments

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

Bond Length

Bond Energies

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

Focus on Polymers – Macromolecular Substances

Chapter 11 Questions

1, 4, 6, 8, 10, 11, 15, 27, 33, 37, 53, 57, 65 (also calculate formal charges), 71, 86, 94

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Chapter 12: Chemical Bonding II:Additional Aspects

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

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.

Potential Energy Diagram

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.

Bonding in H2S

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:

Example 12-1Sketch the orbitals:

Overlap the orbitals:

Describe the shape: Trigonal pyramidal

12-3 Hybridization of Atomic Orbitals

sp3 Hybridization

Bonding in Methane

sp3 Hybridization in Nitrogen

Bonding in Nitrogen

sp2 Hybridization

Orbitals in Boron

sp Hybridization

Orbitals in Beryllium

sp3d and sp3d2 Hybridization

Hybrid Orbitals and VSEPR

• Write a plausible Lewis structure.• Use VSEPR to predict electron geometry.• Select the appropriate hybridization.

12-4 Multiple Covalent Bonds

• Ethylene has a double bond in its Lewis structure.• VSEPR says trigonal planar at carbon.

Ethylene

Acetylene

• Acetylene, C2H2, has a triple bond.• VSEPR says linear at carbon.

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

Combining Atomic Orbitals

Molecular Orbitals of Hydrogen

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

Bond Order

• Stable species have more electrons in bonding orbitals than antibonding.

Bond Order = No. e- in bonding MOs - No. e- in antibonding MOs

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

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).

Molecular Orbitals of the Second Period

Combining p orbitals

Expected MO Diagram of C2

Modified MO Diagram of C2

MO Diagrams of 2nd Period Diatomics

MO Diagrams of Heteronuclear Diatomics

12-6 Delocalized Electrons

Benzene

12-7 Bonding in Metals

• Electron sea model– Nuclei in a sea of e-.– Metallic lustre.– Malleability.

Force applied

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.

Band Theory

Semiconductors

Photovoltaic Cells

Focus on Photoelectron Spectroscopy

1, 3, 8, 10, 16, 29, 33, 39, 45, 59, 68, 72, 76

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Chapter 13: Liquids, Solids and Intermolecular Forces

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

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

Intermolecular Forces

13-2 Vaporization of Liquids:Vapor Pressure

Enthalpy of Vaporization

ΔHvap = Hvapor – Hliquid = - ΔHcondensation

Boiling Point

Mercury manometer

Vapor pressure of liquid

Pvap independent

of Vliq

Pvap independent

of Vgas

Pvap dependent on

Vapor Pressure and Boiling Point

(e) (d) (c) (b) (a)

Ln P = -A ( ) + B1T

A = ΔHvap

Clausius-Clapeyron Equation

Ln P = -A ( ) + B1T

Ln = - ( - ) P2P1

ΔHvap

13-3 Some Properties of Solids

Freezing Point Melting Point

ΔHfus(H2O) = +6.01 kJ/mol

Sublimation

ΔHsub = ΔHfus + ΔHvap

= -ΔHdeposition

13-4 Phase Diagrams

Iodine

Phase Diagrams

Carbon dioxide

Supercritical Fluids

The Critical Point

Critical Temperatures and Pressures

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.

Phenomenon of Induction

Instantaneous and Induced Dipoles

Dipole Dipole Interactions

13-6 Hydrogen Bonding

Hydrogen Bonding in HF(g)

Hydrogen Bonding in Water

around a molecule in the solid in the liquid

Other examples of H-Bonds

13-7 Chemical Bonds as Intermolecular Forces

Other Carbon Allotropes

Interionic Forces

13-8 Crystal Structures

Unit Cells in the Cubic Crystal System

Holes in Crystals

Hexagonal Close Packed (hcp)

Coordination Number

Counting Cell Occupancy

X-Ray Diffraction

Cesium Chloride

Atomic Radii from Crystal Structures

Sodium Chloride

Holes in Crystals

13-9 Energy Changes in the Formation of Ionic Crystals

1, 3, 4, 13, 24, 26, 31, 45, 52, 61, 94, 107

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Chapter 14: Solutions and Their Physical Properties

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

Contents

14-8 Freezing-Point Depression and Boiling-Point Elevation of Nonelectrolyte solutions.

14-9 Solutions of Electrolytes14-10 Colloidal Mixtures

Focus on Chromatography

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.

Table 14.1 Some Common Solutions

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)

10% Ethanol Solution (v/v)

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.

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%

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)

14-3 Intermolecular Forces and the Solution Process

ΔHb ΔHc

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

Ideal Solution

Non-ideal Solutions

• Adhesive forces greater than cohesive forces.

ΔHsoln < 0

Non-ideal Solutions

• Adhesive forces are less than cohesive forces.

• At the limit these solutions are heterogeneous.

ΔHsoln > 0

Ionic Solutions

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

14-4 Solution Formation and Equilibrium

saturated

Solubility Curves

Supersaturated Unsaturated

14-5 Solubility of Gases

• Most gases are less soluble in water as temperature increases.

• In organic solvents the reverse is often true.

Henry’s Law

• Solubility of a gas increases with increasing pressure. C = k Pgas

=23.54 mL1.00 atm

= 23.54 ml N2/atm

CPgas = =

100 mL = 4.25 atm23.54 ml N2/atm

Henry’s Law

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

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

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

Liquid-Vapor Equilibrium

Fractional Distillation

Non-ideal behavior

14-7 Osmotic Pressure

Osmotic Pressure

πV = nRT

π = RTnV = M RT

For dilute solutions of electrolytes:

Osmotic Pressure

Isotonic Saline 0.92% m/V

Hypertonic > 0.92% m/Vcrenation

Hypotonic > 0.92% m/V

rupture

Reverse Osmosis - Desalination

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.

Vapor Pressure Lowering

ΔTf = -Kf m

ΔTb = -Kb m

Practical Applications

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.

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

Interionic Interactions

• Arrhenius theory does not correctly predict the conductivity of concentrated electrolytes.

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.

14-10 Colloidal Mixtures

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.

Dialysis

Focus on Chromatography

Stationary Phasesilicon gumaluminasilica

Mobile Phasesolventgas

Chromatography

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.

N9B 3P4

General ChemistryPrinciples and Modern ApplicationsPetrucci • Harwood • Herring 8th Edition

Chapter 15: Chemical Kinetics

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

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

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+]

0.0010 M

38.5 s

Rates of Chemical Reaction

Δ[Sn4+]Δt

2 Fe3+(aq) + Sn2+ → 2 Fe2+(aq) + Sn4+(aq)

Δ[Fe2+] Δt

Δ[Fe3+] Δt

= - 12

General Rate of Reaction

a A + b B → c C + d D

Rate of reaction = rate of disappearance of reactants

=Δ[C]

=Δ[D]

Δ[A] Δt

= -Δ[B]

Δt1b= -

= rate of appearance of products

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)

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]

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

- 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

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 + ….

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.

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.

Example 15-3

R2 = k[HgCl2]2m[C2O4

R3 = k[HgCl2]3m[C2O4

k(2[HgCl2]3)m[C2O42-]3

n k[HgCl2]3

m[C2O42-]3

2m = 2.0 therefore m = 1.0

k2m[HgCl2]3m[C2O4

k[HgCl2]3m[C2O4

= = 2.0=2mR3

= k(2[HgCl2]3)m[C2O42-]3

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

k(0.105)(0.30)n k(0.105)(0.15)n

7.110-5

1.810-5= 3.94

(0.30)n (0.15)n

= = 2n =

2n = 3.98 therefore n = 2.0

+ = Third Order

R2 = k[HgCl2]2 [C2O4

First order

Example 15-3

Second order

15-4 Zero-Order Reactions

A → products

Rrxn = k [A]0

Rrxn = k

[k] = mol L-1 s-1

Integrated Rate Law

- dt= kd[A] [A]0

-[A]t + [A]0 = kt

[A]t = [A]0 - kt

-Δ[A]dt

= k-d[A]Move to the

infinitesimal= k

And integrate from 0 to time t

15-5 First-Order Reactions

H2O2(aq) → H2O(l) + ½ O2(g)

= -k [H2O2] d[H2O2 ]

= - k dt[H2O2] d[H2O2 ]

= -ktln [A]t

ln[A]t = -kt + ln[A]0

[k] = s-1

First-Order Reactions

Half-Life

• t½ is the time taken for one-half of a reactant to be consumed.

= -ktln [A]t

= -kt½ ln ½[A]0

- ln 2 = -kt½

t½ = ln 2k

0.693k

Half-LifeButOOBut(g) → 2 CH3CO(g) + C2H4(g)

Some Typical First-Order Processes

15-6 Second-Order Reactions

• Rate law where sum of exponents m + n +… = 2.

A → products

dt= - kd[A][A]2

= kt +1[A]0[A]t

dt= -k[A]2

d[A][k] = M-1 s-1 = L mol-1 s-1

Second-Order Reaction

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.

Testing for a Rate Law

Plot [A] vs t.

Plot ln[A] vs t.

Plot 1/[A] vs t.

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.

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.

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.

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

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.

Activation Energy

Kinetic Energy

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.

Collision Theory

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.

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

Arrhenius Plot

N2O5(CCl4) → N2O4(CCl4) + ½ O2(g)

= -1.2104 KR

-Ea = 1.0102 kJ mol-1

Arrhenius Equation

k = Ae-Ea/RT ln k = + ln AR

ln k2– ln k1 = + ln A - - ln AR

ln = - R

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.

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.

A Rate Determining Step

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]

dt= k[HI][ICl]

dt= k[H2][ICl]

Slow Step Followed by a Fast Step

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

slow N2O2(g) + O2(g) 2NO2(g)k2

dt= k2 [NO]2[O2]

k12NO(g) + O2(g) → 2 NO2(g)

K =k-1

k1 =[NO]

[N2O2]

= K [NO]2

k1= [NO]2[N2O2]

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]

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])

Kinetic Consequences of Assumptions

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)

d[NO2] k1k3[NO]2[O2] =( k3[O2])

k1[NO]2 =

d[NO2] k1k3[NO]2[O2] =( k2)

[NO]2[O2] =k1k3

Let k2 << k3

Let k2 >> k3

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.

11-5 Catalysis

Catalysis on a Surface

Enzyme Catalysis

E + S ESk1

ES → E + Pk2

Saturation Kinetics

E + S ESk1

→ 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] =

Michaelis-Menten

(k-1+k2 ) + k1[S] k1k2[E]0 [S]

(k-1+k2 ) + [S]k2[E]0 [S]

KM + [S]k2[E]0 [S]

d[P]k2[E]0

k2 [E]0 [S]

N9B 3P4

General ChemistryPrinciples and Modern ApplicationsPetrucci • Harwood • Herring 8th Edition

Chapter 16: Principles of Chemical Equilibrium

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

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

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)

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=

CO(g) + 2 H2(g) CH3OH(g)k1

Three Approaches to Equilibrium

Three Approaches to the Equilibrium

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

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

0 is a standard reference state

= 1 mol L-1 (ideal conditions)

= B[B]

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.

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

1Kc(3)= = 1.710-13

[N2][O2] [NO]2

[N2][O2]½ [N2O]

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

[O2] = V

nO2 = RT

The Equilibrium Constant, KP

2 SO2(g) + O2(g) 2 SO3(g)

Kc = [SO2]2[O2]

[SO3] RT

= RTPSO3

2PSO2PO2

Kc = KP(RT) KP = Kc(RT)-1

KP = Kc(RT)Δn

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

PCOPH2 (RT)1

Burnt Lime

CaCO3(s) CaO(s) + CO2(g)

Kc = [CO2] KP = PCO2(RT)

16-4 The Significance of the Magnitude of the Equilibrium Constant.

16-5 The Reaction Quotient, Q: Predicting the Direction of Net Change.

• 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

Reaction Quotient

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.

Le Châtellier’s Principle

Q = = Kc [SO2]2[O2][SO3]2

Q > Kc

2 SO2(g) + O2(g) 2 SO3(g)k1

Kc = 2.8102 at 1000K

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

= VnSO3

2nSO2nO2

Effect of change in volume

Kc = [C]c[D]d [G]g[H]h

= V(a+b)-(g+h)nG

anA nB

V-ΔnnG

anA nB

• 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.

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.

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.

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.

Focus on the Nitrogen Cycle and the Synthesis of Nitrogen Compounds.

N2(g) + O2(g) NO(g)k1

KP = 4.7 10-31 at 298K and 1.3 x 10-4 at 1800K

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.

N9B 3P4

Chapter 17: Acids and Bases

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.

17-1 The Arrhenius Theory: A Brief Review

HCl(g) → H+(aq) + Cl-(aq)

NaOH(s) → Na+(aq) + OH-(aq)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.

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

Base Ionization Constant

NH3 + H2O NH4+ + OH-

Kc= [NH3][H2O][NH4

+][OH-]

Kb= Kc[H2O] = [NH3]

[NH4+][OH-]

= 1.810-5

base acidconjugate

conjugate

Acid Ionization Constant

CH3CO2H + H2O CH3CO2- + H3O+

Kc= [CH3CO2H][H2O][CH3CO2

-][H3O+]

Ka= Kc[H2O] = = 1.810-5

[CH3CO2H][CH3CO2

-][H3O+]

baseacidconjugate

conjugate

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

17-3 The Self-Ionization of Water and the pH Scale

Ion Product of Water

Kc= [H2O][H2O][H3O+][OH-]

H2O + H2O H3O+ + OH-

base acidconjugate

conjugate

KW= Kc[H2O][H2O] = = 1.010-14[H3O+][OH-]

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

pH and pOH Scales

17-4 Strong Acids and Bases

HCl CH3CO2H

Thymol Blue Indicator

pH < 1.2 < pH < 2.8 < pH

17-5 Weak Acids and Bases

Acetic Acid HC2H3O2 or CH3CO2H

Weak Acids

Ka= = 1.810-5

[CH3CO2H][CH3CO2

-][H3O+]

pKa= -log(1.810-5) = 4.74

glycine H2NCH2CO2H

lactic acid CH3CH(OH) CO2H

Table 17.3 Ionization Constants of Weak Acids and Bases

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.

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

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.

Percent Ionization

HA + H2O H3O+ + A-

Degree of ionization =[H3O+] from HA

[HA] originally

Percent ionization =[H3O+] from HA

[HA] originally 100%

Percent Ionization

Ka =[H3O+][A-]

Ka =n H3O+ A-

17-6 Polyprotic Acids

H3PO4 + H2O H3O+ + H2PO4-

H2PO4- + H2O H3O+ + HPO4

HPO42- + H2O H3O+ + PO4

Phosphoric acid:

A triprotic acid.

Ka = 7.110-3

Ka = 6.310-8

Ka = 4.210-13

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.

Table 17.4 Ionization Constants of Some Polyprotic Acids

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

Eqlbrm conc. (3.0-x) M x M x M

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

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

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

Sulfuric Acid

Sulfuric acid:

A diprotic acid.

H2SO4 + H2O H3O+ + HSO4-

HSO4- + H2O H3O+ + SO4

Ka = very large

Ka = 1.96

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.

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

+] = ?

= 1.010-14

1.810-5= 5.610-10

Ka Kb = Kw

Hydrolysis

• Water (hydro) causing cleavage (lysis) of a bond.

Na+ + H2O → Na+ + H2O

NH4+ + H2O → NH3 + H3O+

Cl- + H2O → Cl- + H2O

No reaction

Hydrolysis

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.

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

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

H H····

····

···· ···· ··

H H····

····

···· ··

H H····

····

·· ···· ··

H H····

···· ··

·· ··

Ka 103 Ka =1.310-2

Strengths of Organic Acids

H H····

·· ··H

OCH H····

Ka = 1.810-5 Ka =1.310-16

acetic acid ethanol

Focus on the Anions Formed

OCH····

·· ··

····

-···· ··

····

Structural Effects

···· ··

····

Ka = 1.810-5

Ka = 1.310-5

Structural Effects

···· ··

····

Ka = 1.810-5

Ka = 1.410-3

·· ··

····

Strengths of Amines as Bases

·· NBr

pKb = 4.74 pKa = 7.61

ammonia bromamine

Strengths of Amines as Bases

pKb = 4.74 pKa = 3.38 pKb = 3.37

methylamine ethylamine propylamine

NH2 NH2 NH2

Resonance Effects

Inductive Effects

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

Showing Electron Movement

Focus On Acid Rain

CO2 + H2O H2CO3

H2CO3 + H2O HCO3- + H3O+

3 NO2 + H2O 2 HNO3 + NO

N9B 3P4

Chapter 18: Additional Aspects of Acid-Base Equilibria

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

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.

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

Acetic Acid and Hydrochloric Acid

0.1 M HCl 0.1 M CH3CO2H 0.1 M HCl +0.1 M CH3CO2H

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

Example 18-1

Initial concs.

weak acid 0.100 M 0 M 0 M

strong acid 0 M 0.100 M 0 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

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

[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

Suppression of Ionization of a Weak Acid

Suppression of Ionization of a Weak Base

Solutions of Weak Acids and Their Salts

Solutions of Weak Bases and Their Salts

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

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

How A Buffer Works

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-]

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]

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]

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

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

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

Six Methods of Preparing Buffer Solutions

Calculating Changes in Buffer Solutions

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

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.

Indicator Colors and Ranges

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.

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

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.

Titration of a Strong Base with a Strong Acid

Titration of a Weak Acid with a Strong Base

Titration of a Weak Polyprotic Acid

H3PO4 H2PO4- HPO4

2- PO43-

NaOHNaOH

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

Kb = Kw/Ka = 2.410-2

Example 18-9

Kb = 2.410-2 PO43- + H2O → OH- + HPO4

Initial concs. 1.0 M 0 M 0 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?

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.

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

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.

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-]

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.

N9B 3P4

Chapter 19: Solubility and Complex-Ion Equilibria

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

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.

Table 19-1 Several Solubility Product Constants at 25°C

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

19-3 The Common-Ion Effect in Solubility Equilibria

The Common-Ion Effect and Le Chatelliers Principle

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.

Effects on the Solubility of Ag2CrO4

Ion Pairs

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.

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.

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

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.

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-

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

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.

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

19-8 Equilibria Involving Complex Ions

AgCl(s) + 2 NH3(aq) → [Ag(NH3)2]+(aq) + Cl-(aq)

Complex Ions

• Coordination compounds.– Substances which contain complex ions.

• Complex ions.– A polyatomic cation or anion

composed of:• A central metal ion.• Ligands

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

Table 19.2 Formation Constants for Some Complex Ions

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

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 =

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.

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.

Qualitative Cation Analysis

Chloride Group Precipitates

(a) Group precipitate

Wash ppt with hot water PbCl2 is slightly soluble. Test aqueous solution with CrO4

(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)

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.

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 =

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)

A Sensitive Test for Copper(II)

[Cu(H2O)4]2+(aq) + 4 NH3(aq) → [Cu(NH3)4]2+(aq) + 4 H2O(l)

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)

N9B 3P4

Chapter 20: Spontaneous Change: Entropy and Free Energy

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

20-1 Spontaneity: The Meaning of Spontaneous Change

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)

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.

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

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

Boltzmann Distribution

Entropy Change

ΔS = qrev

20-3 Evaluating Entropy and Entropy Changes

• Phase transitions.– Exchange of heat can be carried out reversibly.

ΔS = ΔH

H2O(s, 1 atm) H2O(l, 1 atm) ΔHfus° = 6.02 kJ at 273.15 K

ΔSfus = ΔHfus

6.02 kJ mol-1

273.15 K = 2.2010-2 kJ mol-1 K-1

Trouton’s Rule

ΔS = ΔHvap

87 kJ mol-1 K-1

Raoult’s Law

°PA = APA

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)]

Entropy as a Function of Temperature

Vibrational Energy and Entropy

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.

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.

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

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.

Table 20.1 Criteria for Spontaneous Change

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)]

20-6 Free Energy Change and Equilibrium

Free Energy Change and Equilibrium

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

Relationship Between S and S°

qrev = -w = RT lnVf

ΔS = qrev

T= R ln

ΔS = Sf – Si = R lnVf

= R lnPi

= -R lnPf

S = S° - R ln PP°

= S° - R ln P1

= S° - R ln P

2 N2(g) + 3 H2(g) 2 NH3(g)SNH3

= SNH3

– Rln PNH3

SN2 – Rln PN2

SH2 – Rln PH2

ΔSrxn = 2(SNH3 – Rln PNH3

) – 2(SN2 – Rln PN2

) –3(SH2 – Rln PH2

ΔSrxn = 2 SNH3 – 2SN2

–3SH2+ Rln

PN2PH2

ΔSrxn = ΔS°rxn + RlnPN2

ΔG Under Non-standard Conditions

ΔG = ΔH° - TΔS ΔSrxn = ΔS°rxn + RlnPN2

ΔG = ΔH° - TΔS°rxn – TR lnPN2

ΔG = ΔG° + RT lnPN2

ΔG = ΔG° + RT ln Q

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

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.

Significance of the Magnitude of ΔG

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.

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)

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…

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°

Temperature Dependence of Keq

ln Keq = -ΔH°RT

ΔS°R

slope = -ΔH°

-ΔH° = R slope

= -8.3145 J mol-1 K-1 2.2104 K

= -1.8102 kJ mol-1

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.

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

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:

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

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