Ch 01 10 general chemistry 8e petrucci

Philip DuttonUniversity of Windsor, Canada

Chapter 1: Matter—Its Properties and Measurement

General ChemistryPrinciples and Modern Applications

Petrucci • Harwood • Herring 8th Edition

Contents

Physical properties and states of matter Système International Units Uncertainty and significant figures Dimensional analysis

http://cwx.prenhall.com/petrucci/chapter1/deluxe.html

Properties of Matter

Matter: Occupies space, has mass and inertia

Composition: Parts or componentsex. H2O, 11.9% H and 88.81% O

Properties: Distinguishing features physical and chemical properties

States of Matter

Matter(materials)

Substances Mixtures

Elements CompoundsHomogeneous

mixtures(solutions)

Heterogeneousmixtures

Physical processes

Chemical

reactions

Classification of Matter

Separations

Separating Mixtures

Substances tobe separateddissolved in liquid

Pureliquid

mixture

Chromatography

Significant Figures

Number

6.29 g0.00348 g9.0 1.0 10-8

100 eggs100 g = 3.14159

Count from left from first non-zero digit. Adding and subtracting.

Use the number of decimal places in the number with thefewest decimal places.

1.14 0.611.67613.416

SignificantFigures

3322infinitebad notationvarious

Significant figures

Multiplying and dividing.

Use the fewest significant figures.

0.01208 0.236

Rounding Off

3rd digit is increased if4th digit 5

Report to 3 significant figures.

10.235 12.4590 19.75 15.651

10.212.519.815.7

= 0.512

= 5.12 10-3

S.I. UnitsLength metre, mMass Kilogram, kgTime second, sTemperature Kelvin, KQuantity Mole, 6.022×1023 mol-1

Derived QuantitiesForce Newton, kg m s-2

Pressure Pascal, kg m-1 s-2

Eenergy Joule, kg m2 s-2

Other Common UnitsLength Angstrom, Å, 10-8 cmVolume Litre, L, 10-3 m3

Energy Calorie, cal, 4.184 JPressure 1 Atm = 1.064 x 102 kPa 1 Atm = 760 mm Hg

Temperature

Relative Temperatures

Volume

Density

m=VV=m/

g/mLMass and volume are extensive properties

Density is an intensive property

ConversionWhat is the mass of a cube of osmium that is 1.25 inches on each side?

Have volume, need density = 22.48g/cm3

Wrong units

The Gimli Glider, Q86, p30

Uncertainties

• Systematic errors.– Thermometer constantly 2°C too low.

• Random errors– Limitation in reading a scale.

• Precision– Reproducibility of a measurement.

• Accuracy– How close to the real value.

End of Chapter Questions

1, 3, 5, 12, 14, 17, 18, 20, 30, 41, 49, 50, 61, 72, 74, 79

Chapter 2: Atoms and the Atomic Theory

Contents

• Early chemical discoveries • Electrons and the Nuclear Atom• Chemical Elements• Atomic Masses• The Mole

Early Discoveries

Lavoisier 1774 Law of conservation of mass

Proust 1799 Law of constant composition

Dalton 1803-1888 Atomic Theory

Dalton’s Atomic Theory

Each element is composed of small particles called atoms.

Atoms are neither created nor destroyed in chemical reactions.

All atoms of a given element are identical

Compounds are formed when atoms of more than one element

combine

Consequences of Dalton’s theory

In forming carbon monoxide, 1.33 g of oxygen combines with 1.0 g of carbon.

In the formation of hydrogen peroxide 2.66 g of oxygen combines with 1.0 g of hydrogen.

Law of Definite Proportions: combinations of elements are in ratios of small whole numbers.

Behavior of charges

Cathode ray tube

Properties of cathode rays

Electron m/e = -5.6857 x 10-9 g coulomb-1

Charge on the electron

From 1906-1914 Robert Millikan showed ionized oil drops can be balanced against the pull of gravity by an electric field.

The charge is an integral multiple of the electronic charge, e.

Radioactivity

Radioactivity is the spontaneous emission of radiation from a substance.

X-rays and -rays are high-energy light.

-particles are a stream of helium nuclei, He2+.

-particles are a stream of high speed electrons

that originate in the nucleus.

The nuclear atom

Geiger and Rutherford1909

The -particle experiment Most of the mass and all of the

positive charge is concentrated in a small region called the nucleus .

There are as many electrons outside

the nucleus as there are units of positive charge on the nucleus

The nuclear atom

Rutherfordprotons 1919

James Chadwickneutrons 1932

Atomic Diameter 10-8 cm Nuclear diameter 10-13 cm

Nuclear Structure

Particle Mass Chargekg amu Coulombs (e)

Electron 9.109 x 10-31 0.000548 –1.602 x 10-19 –1Proton 1.673 x 10-27 1.00073 +1.602 x 10-19 +1Neutron 1.675 x 10-27 1.00087 0 0

Scale of Atoms

Useful units:

1 amu (atomic mass unit) = 1.66054 x 10-24 kg 1 pm (picometer) = 1 x 10-12 m 1 Å (Angstrom) = 1 x 10-10 m = 100 pm = 1 x 10-8 cm

The heaviest atom has a mass of only 4.8 x 10-22 g

and a diameter of only 5 x 10-10 m.

Biggest atom is 240 amu and is 50 Å across.Typical C-C bond length 154 pm (1.54 Å)

Molecular models are 1 Å /inch or about 0.4 Å /cm

Isotopes, atomic numbers and mass numbers

To represent a particular atom we use the symbolism:

A= mass number Z = atomic number

Measuring atomic masses

The Periodic tableAlkali Metals

Alkaline Earths

Transition Metals

Halogens

Noble Gases

Lanthanides and Actinides

Main Group

The Periodic Table

• Read atomic masses.• Read the ions formed by main group elements.• Read the electron configuration.• Learn trends in physical and chemical properties.

We will discuss these in detail in Chapter 10.

The Mole

• Physically counting atoms is impossible.• We must be able to relate measured mass to

numbers of atoms.– buying nails by the pound.– using atoms by the gram

Avogadro’s number

The mole is an amount of substance that contains the same number of elementary entities as there are carbon-12 atoms in exactly 12 g of carbon-12.

NA = 6.02214199 x 1023 mol-1

Molar Mass

• The molar mass, M, is the mass of one mole of a substance.

M (g/mol 12C) = A (g/atom 12C) x NA (atoms 12C /mol 12C)

Combining Several Factors in a Calculation—Molar Mass, the Avogadro Constant, Percent Abundance.Potassium-40 is one of the few naturally occurring radioactive isotopes of elements of low atomic number. Its percent natural abundance among K isotopes is 0.012%. How many 40K atoms do you ingest by drinking one cup of whole milk containing 371 mg of K?

Want atoms of 40K, need atoms of K,Want atoms of K, need moles of K,Want moles of K, need mass and M(K).

Example 2-9

Convert strategy to plan

mK(mg) x (1g/1000mg) mK (g) x 1/MK (mol/g) nK(mol)

Convert mass of K(mg K) into moles of K (mol K)

Convert moles of K into atoms of 40K

nK(mol) x NA atoms K x 0.012% atoms 40K

nK = (371 mg K) x (10-3 g/mg) x (1 mol K) / (39.10 g K)

= 9.49 x 10-3 mol K

and plan into action

atoms 40K = (9.49 x 10-3 mol K) x (6.022 x 1023 atoms K/mol K) x (1.2 x 10-4 40K/K)

= 6.9 x 1017 40K atoms

Chapter 2 Questions

3, 4, 11, 22, 33, 51, 55, 63, 83.

Chapter 3: Chemical Compounds

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

Contents

3-1 Molecular and Ionic Compounds3-2 Molecular Mass3-3 Composition3-4 Oxidation States

3-5 Names and formulas Focus on Mass Spectrometry

Molecular compounds

1 /inch 0.4 /cm

Standard color scheme

Some moleculesH2O2 CH3CH2Cl P4O10

CH3CH(OH)CH3 HCO2H

Ionic compounds

Atoms of almost all elements can gain or lose electrons to form charged species called ions.

Compounds composed of ions are known as ionic compounds.

¾ Metals tend to lose electrons to form positively charged ions called cations.

Ö Non-metals tend to gain electrons to form negatively charged ions called anions.

Sodium chloride

Extended array of Na+ and Cl- ions Simplest formula unit is NaCl

Inorganic molecules

Molecular mass

Molecular formula C6H12O6

Empirical formula CH2O

Glucose

6 x 12.01 + 12 x 1.01 + 6 x 16.00

Molecular Mass:Use the naturally occurring mixture of isotopes,

= 180.18

Exact Mass:Use the most abundant isotopes,

6 x 12.000000 + 12 x 1.007825 + 6 x 15.994915= 180.06339

Halothane C2HBrClF3

M(C2HBrClF3) = 2MC + MH + MBr + MCl + 3MF

= (2 12.01) + 1.01 + 79.90 + 35.45 + (3 19.00)

= 197.38 g/mol

Chemical Composition

Mole ratio nC/nhalothane

Mass ratio mC/mhalothane

Example 3.4Calculating the Mass Percent Composition of a Compound

Calculate the molecular massM(C2HBrClF3

) = 197.38 g/mol

For one mole of compound, formulate the mass ratio and convert to percent:

%17.12%10038.197

)01.122(%

Example 3-4

%88.28%10038.197

)00.193(%

%96.17%10038.19745.35%

%48.40%10038.19790.79%

%51.0%10038.197

%17.12%10038.197

)01.122(%

Empirical formula

1. Choose an arbitrary sample size (100g).2. Convert masses to amounts in moles.3. Write a formula.4. Convert formula to small whole numbers.5. Multiply all subscripts by a small whole number

to make the subscripts integral.

5 Step approach:

Determining the Empirical and Molecular Formulas of a Compound from Its Mass Percent Composition.

Dibutyl succinate is an insect repellent used against household ants and roaches. Its composition is 62.58% C, 9.63% H and 27.79% O. Its experimentally determined molecular mass is 230 u. What are the empirical and molecular formulas of dibutyl succinate?

Step 1: Determine the mass of each element in a 100g sample.

C 62.58 g H 9.63 g O 27.79 g

Dibutyl succinate is an insect repellent used against household ants and roaches. Its composition is 62.58% C, 9.63% H and 27.79% O. Its experimentally determined molecular mass is 230 u. What are the empirical and molecular formulas of dibutyl succinate?

Example 3-5

Step 2: Convert masses to amounts in moles.

OmolOg

OmolOgn

HmolHg

HmolHgn

CmolCg

CmolCgn

737.1999.15

179.27

55.9008.1

210.5011.12

158.62

Step 3: Write a tentative formula.

Step 4: Convert to small whole numbers.

C5.21H9.55O1.74

C2.99H5.49O

Example 3-5

Step 5: Convert to a small whole number ratio.

Multiply 2 to get C5.98H10.98O2

The empirical formula is C6H11O2

Step 6: Determine the molecular formula.

Empirical formula mass is 115 u.Molecular formula mass is 230 u.

The molecular formula is C12H22O4

Example 3-5

Combustion analysis

Oxidation States

Metals tend to lose electrons.

Na Na+ + e-

Non-metals tend to gain electrons.

Cl + e- Cl-

Reducing agents

Oxidizing agents

We use the Oxidation State to keep track of the number of electrons that have been gained or lost by an element.

Rules for Oxidation States

1. The oxidation state (OS) of an individual atom in a free element is 0.

2. The total of the OS in all atoms in: i. Neutral species is 0.ii. Ionic species is equal to the charge on the ion.

3. In their compounds, the alkali metals and the alkaline earths have OS of +1 and +2 respectively.

4. In compounds the OS of fluorine is always –1

Rules for Oxidation States

6. In compounds, the OS of hydrogen is usually +1

7. In compounds, the OS of oxygen is usually –2.

8. In binary (two-element) compounds with metals:i. Halogens have OS of –1,ii. Group 16 have OS of –2 andiii. Group 15 have OS of –3.

Assigning Oxidation States.

What is the oxidation state of the underlined element in each of the following? a) P4; b) Al2O3; c) MnO4

-; d) NaH

a) P4 is an element. P OS = 0

b) Al2O3: O is –2. O3 is –6. Since (+6)/2=(+3), Al OS = +3.

c) MnO4-: net OS = -1, O4 is –8. Mn OS = +7.

d) NaH: net OS = 0, rule 3 beats rule 5, Na OS = +1 and H OS = -1.

Example 3-7

Naming Compounds

Trivial names are used for common compounds.

A systematic method of naming compounds is known as a system of nomenclature.

Organic compoundsInorganic compounds

Inorganic Nomenclature

Binary Compounds of Metals and Nonmetals

NaCl = sodium chloride

name is unchanged

“ide” endingelectrically neutral

MgI2 = magnesium iodide

Al2O3 = aluminum oxide

Na2S = sodium sulfide

Binary Compounds of Two Non-metals

Molecular compoundsusually write the positive OS element first.HCl hydrogen chloride

mono 1 penta 5

di 2 hexa 6

tri 3 hepta 7

tetra 4 octa 8

Some pairs form more than one compound

Binary Acids

Emphasize the fact that a molecule is an acid by altering the name.

HCl hydrogen chloride hydrochloric acid

HF hydrogen fluoride hydrofluoric acid

Acids produce H+ when dissolved in water.

They are compounds that ionize in water.

Polyatomic Ions

Polyatomic ions are very common.

Table 3.3 gives a list of some of them. Here are a few:

ammonium ion NH4+ acetate ion C2H3O2

carbonate ion CO32- hydrogen carbonate HCO3

hypochlorite ClO- phosphate PO43-

chlorite ClO2- hydrogen phosphate

HPO42-

chlorate ClO3- sulfate SO4

perchlorate ClO4- hydrogensulfate HSO4

Naming Organic Compounds

Organic compounds abound in natureFats, carbohydrates and proteins are foods.

Propane, gasoline, kerosene, oil.

Drugs and plastics

Carbon atoms form chains and rings and act as the framework of molecules.

Visualizations of some hydrocarbons

IsomersIsomers have the same molecular formula but have different arrangements of atoms in space.

Functional Groups – carboxylic acid

Functional Groups - alcohol

Chapter 3 Questions

3, 5, 12, 24, 35, 46, 53, 61, 57, 73, 95, 97

Chapter 4: Chemical Reactions

Contents

4-1 Chemical Reactions and Chemical Equations4-2 Chemical Equations and Stoichiometry4-3 Chemical Reactions in Solution4-4 Determining the Limiting reagent4-5 Other Practical Matters in Reaction

Stoichiometry Focus on Industrial Chemistry

4-1 Chemical Reactions and Chemical Equations

As reactants are converted to products we observe:– Color change– Precipitate formation– Gas evolution– Heat absorption or evolution

Chemical evidence may be necessary.

Chemical Reaction

Nitrogen monoxide + oxygen → nitrogen dioxide

Step 1: Write the reaction using chemical symbols.

NO + O2 → NO2

Step 2: Balance the chemical equation.

Molecular Representation

Balancing Equations

• Never introduce extraneous atoms to balance.

NO + O2 → NO2 + O

• Never change a formula for the purpose of balancing an equation.

NO + O2 → NO3

Balancing Equation Strategy

• Balance elements that occur in only one compound on each side first.

• Balance free elements last.

• Balance unchanged polyatomics as groups.

• Fractional coefficients are acceptable and can be cleared at the end by multiplication.

Example 4-2Writing and Balancing an Equation: The Combustion of a Carbon-Hydrogen-Oxygen Compound.Liquid triethylene glycol, C6H14O4, is used a a solvent and plasticizer for vinyl and polyurethane plastics. Write a balanced chemical equation for its complete combustion.

152 6 7C6H14O4 + O2 → CO2 + H2O 6

2. Balance H.

2 C6H14O4 + 15 O2 → 12 CO2 + 14 H2O

4. Multiply by two

Example 4-2

3. Balance O.

and check all elements.

Chemical Equation:

1. Balance C.

4-2 Chemical Equations and Stoichiometry

• Stoichiometry includes all the quantitative relationships involving:– atomic and formula masses– chemical formulas.

• Mole ratio is a central conversion factor.

Example 4-3Relating the Numbers of Moles of Reactant and Product.How many moles of H2O are produced by burning 2.72 mol H2 in an excess of O2?

H2 + O2 → H2O

Write the Chemical Equation:

Balance the Chemical Equation:

Use the stoichiometric factor or mole ratio in an equation:

nH2O = 2.72 mol H2 × = 2.72 mol H2O2 mol H2O2 mol H2

Example 4-6Additional Conversion Factors ina Stoichiometric Calculation: Volume, Density, and Percent Composition.An alloy used in aircraft structures consists of 93.7% Al and 6.3% Cu by mass. The alloy has a density of 2.85 g/cm3. A 0.691 cm3 piece of the alloy reacts with an excess of HCl(aq). If we assume that all the Al but none of the Cu reacts with HCl(aq), what is the mass of H2 obtained?

Al + HCl → AlCl3 + H2

Write the Chemical Equation:

Example 4-6

Balance the Chemical Equation:

2 6 2 3

2 Al + 6 HCl → 2 AlCl3 + 3 H2

Example 4-6

Plan the strategy:

cm3 alloy → g alloy → g Al → mole Al → mol H2 → g H2

We need 5 conversion factors!

Write the Equation

mH2 = 0.691 cm3 alloy × × ×2.85 g alloy

1 cm397.3 g Al

100 g alloy

1 mol Al26.98 g Al

3 mol H22 mol Al

2.016 g H21 mol H2

= 0.207 g H2

and Calculate:

4-3 Chemical Reactions in Solution

• Close contact between atoms, ions and molecules necessary for a reaction to occur.

• Solvent– We will usually use aqueous (aq) solution.

• Solute– A material dissolved by the solvent.

Molarity

Molarity (M) = Volume of solution (L) Amount of solute (mol solute)

If 0.444 mol of urea is dissolved in enough water to make 1.000 L of solution the concentration is:

curea = 1.000 L0.444 mol urea = 0.444 M CO(NH2)2

Preparation of a Solution

Weigh the solid sample.Dissolve it in a volumetric flask partially filled with solvent.Carefully fill to the mark.

Calculating the mass of Solute in a solution of Known Molarity.We want to prepare exactly 0.2500 L (250 mL) of an 0.250 M K2CrO4 solution in water. What mass of K2CrO4 should we use?

Plan strategy:

Example 4-6

Volume → moles → mass

We need 2 conversion factors!Write equation and calculate:

mK2CrO4 = 0.2500 L × × = 12.1 g

0.250 mol 1.00 L

194.02 g1.00 mol

Solution Dilution

Mi × Vi = ni

Mi × ViMf × Vf

= nf = Mf × Vf

Mi × ViMf = Vf

= MiVi

M = nV

Preparing a solution by dilution.A particular analytical chemistry procedure requires 0.0100 M K2CrO4. What volume of 0.250 M K2CrO4 should we use to prepare 0.250 L of 0.0100 M K2CrO4?

Calculate:

VK2CrO4 = 0.2500 L × × = 0.0100 L

0.0100 mol 1.00 L

1.000 L0.250 mol

Example 4-10

Plan strategy: Mf = MiVi

Vi = VfMf

4-4 Determining Limiting Reagent

• The reactant that is completely consumed determines the quantities of the products formed.

Determining the Limiting Reactant in a Reaction.Phosphorus trichloride , PCl3, is a commercially important compound used in the manufacture of pesticides, gasoline additives, and a number of other products. It is made by the direct combination of phosphorus and chlorine

P4 (s) + 6 Cl2 (g) → 4 PCl3 (l)What mass of PCl3 forms in the reaction of 125 g P4 with 323 g Cl2?

Example 4-12

Strategy: Compare the actual mole ratio to the required mole ratio.

Example 4-12

nCl2 = 323 g Cl2 × = 4.56 mol Cl2

1 mol Cl2

70.91 g Cl2

nP4 = 125 g P4 × = 1.01 mol P4

1 mol P4

123.9 g P4

actual = 4.55 mol Cl2/mol P4

theoretical = 6.00 mol Cl2/mol P4

Chlorine gas is the limiting reagent.

4-5 Other Practical Matters in Reaction Stoichiometry

Theoretical yield is the expected yield from a reactant.Actual yield is the amount of product actually produced.

Percent yield = × 100%Actual yieldTheoretical Yield

Theoretical, Actual and Percent Yield

• When actual yield = % yield the reaction is said to be quantitative.

• Side reactions reduce the percent yield.• By-products are formed by side reactions.

Consecutive Reactions, Simultaneous Reactions and

Overall Reactions

• Multistep synthesis is often unavoidable.• Reactions carried out in sequence are called

consecutive reactions.• When substances react independently and at

the same time the reaction is a simultaneous reaction.

Overall Reactions and Intermediates

• The Overall Reaction is a chemical equation that expresses all the reactions occurring in a single overall equation.

• An intermediate is a substance produced in one step and consumed in another during a multistep synthesis.

Focus on Industrial Chemistry

Chapter 4 Questions

1, 6, 12, 25, 39, 45, 53, 65, 69, 75, 84, 94, 83, 112

Chapter 5: Introduction to Reactions in Aqueous Solutions

Contents

5-1 The Nature of Aqueous Solutions5-2 Precipitation Reactions5-3 Acid-Base Reactions5-4 Oxidation-Reduction: Some General Principles 5-5 Balancing Oxidation-Reduction Equations5-6 Oxidizing and Reducing Agents5-7 Stoichiometry of Reactions in Aqueous

Solutions: Titrations Focus on Water Treatment

5.1 The Nature of Aqueous Solutions

Electrolytes

• Some solutes can dissociate into ions.

• Electric charge can be carried.

Types of Electrolytes

• Weak electrolyte partially dissociates.– Fair conductor of electricity.

• Non-electrolyte does not dissociate. – Poor conductor of electricity.

• Strong electrolyte dissociates completely.– Good electrical conduction.

Representation of Electrolytes using Chemical Equations

MgCl2(s) → Mg2+(aq) + 2 Cl-(aq)

A strong electrolyte:

A weak electrolyte:

CH3CO2H(aq) ← CH3CO2-(aq) + H+(aq)→

CH3OH(aq)

A non-electrolyte:

MgCl2(s) → Mg2+(aq) + 2 Cl-(aq)MgCl2(s) → Mg2+(aq) + 2 Cl-(aq)

[Mg2+] = 0.0050 M [Cl-] = 0.0100 M [MgCl2] = 0 M [Mg2+] = 0.0050 M [Cl-] = 0.0100 M [MgCl2] = 0 M

Notation for Concentration

In 0.0050 M MgCl2:

Stoichiometry is important.

Example 5-1Calculating Ion concentrations in a Solution of a Strong Electolyte.What are the aluminum and sulfate ion concentrations in 0.0165 M Al2(SO4)3?.

Al2(SO4)3 (s) → 2 Al3+(aq) + 3 SO42-(aq)

Balanced Chemical Equation:

[Al] = × =1 L

2 mol Al3+

1 mol Al2(SO4)3

0.0165 mol Al2(SO4)3 0.0330 M Al3+

Example 5-1

0.0495 M SO42-[SO4

2-] = × =1 mol Al2(SO4)3

Sulfate Concentration:

1 L3 mol SO4

2-0.0165 mol Al2(SO4)3

Aluminum Concentration:

5-2 Precipitation Reactions

• Soluble ions can combine to form an insoluble compound.

• Precipitation occurs.

Ag+(aq) + Cl-(aq) → AgCl(s)

Ag+(aq) + NO3-(aq) + Na+(aq) + I-(aq) →

AgI(s) + Na+(aq) + NO3-(aq)

Spectator ionsAg+(aq) + NO3

-(aq) + Na+(aq) + I-(aq) → AgI(s) + Na+(aq) + NO3

Net Ionic Equation

AgNO3(aq) +NaI (aq) → AgI(s) + NaNO3(aq)Overall Precipitation Reaction:

Complete ionic equation:

Ag+(aq) + I-(aq) → AgI(s)

Net ionic equation:

Solubility Rules

• Compounds that are soluble:

Li+, Na+, K+, Rb+, Cs+ NH4+

NO3- ClO4

- CH3CO2-

– Alkali metal ion and ammonium ion salts

– Nitrates, perchlorates and acetates

Solubility Rules

– Chlorides, bromides and iodides Cl-, Br-, I-

• Except those of Pb2+, Ag+, and Hg22+.

– Sulfates SO42-

• Except those of Sr2+, Ba2+, Pb2+ and Hg22+.

• Ca(SO4) is slightly soluble.

•Compounds that are mostly soluble:

Solubility Rules

– Hydroxides and sulfides HO-, S2-

• Except alkali metal and ammonium salts• Sulfides of alkaline earths are soluble• Hydroxides of Sr2+ and Ca2+ are slightly soluble.

– Carbonates and phosphates CO32-,

• Except alkali metal and ammonium salts

•Compounds that are insoluble:

5-3 Acid-Base Reactions

• Latin acidus (sour)– Sour taste

• Arabic al-qali (ashes of certain plants)– Bitter taste

• Svante Arrhenius 1884 Acid-Base theory.

• Acids provide H+ in aqueous solution.

• Strong acids:

• Weak acids:

HCl(aq) H+(aq) + Cl-(aq) →

→←CH3CO2H(aq) H+(aq) + CH3CO2-(aq)

• Bases provide OH- in aqueous solution.

• Strong bases:

• Weak bases:

→←NH3(aq) + H2O(l) OH-(aq) + NH4+(aq)

NaOH(aq) Na+(aq) + OH-(aq) →H2O

Recognizing Acids and Bases.

• Acids have ionizable hydrogen ions.– CH3CO2H or HC2H3O2

• Bases have OH- combined with a metal ion. KOH

or are identified by chemical equations Na2CO3(s) + H2O(l)→ HCO3

-(aq) + 2 Na+(aq) + OH-(aq)

More Acid-Base Reactions

• Milk of magnesia Mg(OH)2

Mg(OH)2(s) + 2 H+(aq) → Mg2+(aq) + 2 H2O(l)

Mg(OH)2(s) + 2 CH3CO2H(aq) → Mg2+(aq) + 2 CH3CO2

-(aq) + 2 H2O(l)

More Acid-Base Reactions

• Limestone and marble.

CaCO3(s) + 2 H+(aq) → Ca2+(aq) + H2CO3(aq)

But: H2CO3(aq) → H2O(l) + CO2(g)

CaCO3(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) + CO2(g)

CaCO3(s) + 2 H+(aq) → Ca2+(aq) + H2CO3(aq)

CaCO3(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) + CO2(g)

Limestone and Marble

Gas Forming Reactions

Fe2O3(s) + 3 CO(g) → 2 Fe(l) + 3 CO2(g)

• Hematite is converted to iron in a blast furnace.

Fe2O3(s) + 3 CO(g) → 2 Fe(l) + 3 CO2(g)Fe2O3(s) + 3 CO(g) → 2 Fe(l) + 3 CO2(g)

CO(g) is oxidized to carbon dioxide.

Fe3+ is reduced to metallic iron.

5-4 Oxidation-Reduction: SomeGeneral Principles

• Oxidation and reduction always occur together.

Oxidation State Changes

Fe2O3(s) + 3 CO(g) → 2 Fe(l) + 3 CO2(g)3+ 2- 2+ 2- 4+ 2-0

• Assign oxidation states:

CO(g) is oxidized to carbon dioxide.

Fe3+ is reduced to metallic iron.

Oxidation and Reduction

• Oxidation– O.S. of some element increases in the reaction.– Electrons are on the right of the equation

• Reduction – O.S. of some element decreases in the reaction.– Electrons are on the left of the equation.

Zinc in Copper Sulfate

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

Half-Reactions

• Represent a reaction by two half-reactions.

Oxidation:

Reduction:

Overall:

Zn(s) → Zn2+(aq) + 2 e-

Cu2+(aq) + 2 e- → Cu(s)

Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)

Balancing Oxidation-Reduction Equations

• Few can be balanced by inspection.• Systematic approach required.

• The Half-Reaction (Ion-Electron) Method

Example 5-6

Balancing the Equation for a Redox Reaction in Acidic Solution. The reaction described below is used to determine the sulfite ion concentration present in wastewater from a papermaking plant. Write the balanced equation for this reaction in acidic solution..

SO32-(aq) + MnO4

-(aq) → SO42-(aq) + Mn2+(aq)

Example 5-6

SO32-(aq) + MnO4

-(aq) → SO42-(aq) + Mn2+(aq)

Determine the oxidation states:

4+ 6+7+ 2+

SO32-(aq) → SO4

2-(aq) + 2 e-(aq)

Write the half-reactions:

5 e-(aq) +MnO4-(aq) → Mn2+(aq)

Balance atoms other than H and O:Already balanced for elements.

Example 5-6Balance O by adding H2O:

H2O(l) + SO32-(aq) → SO4

2-(aq) + 2 e-(aq)

5 e-(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l)

Balance hydrogen by adding H+:

H2O(l) + SO32-(aq) → SO4

2-(aq) + 2 e-(aq) + 2 H+(aq)

8 H+(aq) + 5 e-(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l)

Check that the charges are balanced: Add e- if necessary.

Example 5-6Multiply the half-reactions to balance all e-:

5 H2O(l) + 5 SO32-(aq) → 5 SO4

2-(aq) + 10 e-(aq) + 10 H+(aq)

16 H+(aq) + 10 e-(aq) + 2 MnO4-(aq) → 2 Mn2+(aq) + 8 H2O(l)

Add both equations and simplify:

5 SO32-(aq) + 2 MnO4

-(aq) + 6H+(aq) → 5 SO4

2-(aq) + 2 Mn2+(aq) + 3 H2O(l)

Check the balance!

Balancing in Acid

• Write the equations for the half-reactions.– Balance all atoms except H and O.– Balance oxygen using H2O.– Balance hydrogen using H+.– Balance charge using e-.

• Equalize the number of electrons.• Add the half reactions.• Check the balance.

Balancing in Basic Solution

• OH- appears instead of H+.

• Treat the equation as if it were in acid.– Then add OH- to each side to neutralize H+.– Remove H2O appearing on both sides of

equation.• Check the balance.

5-6 Oxidizing and Reducing Agents.

• An oxidizing agent (oxidant ):– Contains an element whose oxidation state

decreases in a redox reaction

• A reducing agent (reductant):– Contains an element whose oxidation state

increases in a redox reaction.

Example 5-8

Identifying Oxidizing and Reducing Agents. Hydrogen peroxide, H2O2, is a versatile chemical. Its uses include bleaching wood pulp and fabrics and substituting for chlorine in water purification. One reason for its versatility is that it can be either an oxidizing or a reducing agent. For the following reactions, identify whether hydrogen peroxide is an oxidizing or reducing agent.

5 H2O2(aq) + 2 MnO4-(aq) + 6 H+ → 8 H2O(l) + 2 Mn2+(aq) + 5 O2(g)

Example 5-8

H2O2(aq) + 2 Fe2+(aq) + 2 H+ → 2 H2O(l) + 2 Fe3+(aq)

Iron is oxidized and peroxide is reduced.

Manganese is reduced and peroxide is oxidized.

5-7 Stoichiometry of Reactions in Aqueous Solutions: Titrations.

• Titration– Carefully controlled addition of one solution to

another.• Equivalence Point

– Both reactants have reacted completely.• Indicators

– Substances which change colour near an equivalence point.

Indicators

Example 5-10

Standardizing a Solution for Use in Redox Titrations. A piece of iron wire weighing 0.1568 g is converted to Fe2+(aq) and requires 26.42 mL of a KMnO4(aq) solution for its titration. What is the molarity of the KMnO4(aq)?

5 Fe2+(aq) + MnO4-(aq) + 8 H+(aq) →

4 H2O(l) + 5 Fe3+(aq) + Mn2+(aq)

Example 5-10

Determine KMnO4 consumed in the reaction:

Determine the concentration:

10615.511

847.5511568.0

KMnOmolMnOmolKMnOmol

FemolMnOmol

FemolFemol

FegFemolFegn OH

4 02140.002624.0

10615.5][ KMnOMLKMnOmolKMnO

5 Fe2+(aq) + MnO4-(aq) + 8 H+(aq) → 4 H2O(l) + 5 Fe3+(aq) + Mn2+(aq)

Chapter 5 Questions

1, 2, 3, 5, 6, 8, 14, 17, 19, 24, 27, 33, 37, 41, 43, 51, 53, 59, 68, 71, 82, 96.

Chapter 6: Gases

Contents

6-1 Properties of Gases: Gas Pressure6-2 The Simple Gas Laws6-3 Combining the Gas Laws:

The Ideal Gas Equation and The General Gas Equation

6-4 Applications of the Ideal Gas Equation 6-5 Gases in Chemical Reactions6-6 Mixtures of Gases

Contents

6-6 Mixtures of Gases6-7 Kinetic—Molecular Theory of Gases6-8 Gas Properties Relating to the

Kinetic—Molecular Theory6-9 Non-ideal (real) Gases Focus on The Chemistry of Air-Bag

Systems

6-1 Properties of Gases: Gas Pressure

• Gas Pressure

• Liquid Pressure

P (Pa) =

Area (m2)Force (N)

P = g ·h ·d

Barometric Pressure

Standard Atmospheric Pressure 1.00 atm760 mm Hg, 760 torr101.325 kPa1.01325 bar1013.25 mbar

Manometers

6-2 Simple Gas Laws

• Boyle 1662 P 1V PV = constant

Example 5-6

Relating Gas Volume and Pressure – Boyle’s Law.

P1V1 = P2V2 V2 = P1V1

P2= 694 L Vtank = 644 L

Charles’s Law

Charles 1787Gay-Lussac 1802

V T V = b T

• Gas properties depend on conditions.

• Define standard conditions of temperature and pressure (STP).

P = 1 atm = 760 mm HgT = 0°C = 273.15 K

Avogadro’s Law

• Gay-Lussac 1808– Small volumes of gases react in the ratio of

small whole numbers.

• Avogadro 1811– Equal volumes of gases have equal numbers of

molecules and– Gas molecules may break up when they react.

Formation of Water

Avogadro’s Law

V n or V = c n

At STP

1 mol gas = 22.4 L gas

At an a fixed temperature and pressure:

6-3 Combining the Gas Laws: The Ideal Gas Equation and the General Gas

Equation

• Boyle’s law V 1/P• Charles’s law V T• Avogadro’s law V n

PV = nRT

The Gas Constant

R = PVnT

= 0.082057 L atm mol-1 K-1

= 8.3145 m3 Pa mol-1 K-1

PV = nRT

= 8.3145 J mol-1 K-1

= 8.3145 m3 Pa mol-1 K-1

The General Gas Equation

R = = P2V2

If we hold the amount and volume constant:

6-4 Applications of the Ideal Gas Equation

Molar Mass Determination

PV = nRT and n = mM

PV = mM RT

M = mPVRT

Example 6-10

Determining a Molar Mass with the Ideal Gas Equation.Polypropylene is an important commercial chemical. It is used in the synthesis of other organic chemicals and in plastics production. A glass vessel weighs 40.1305 g when clean, dry and evacuated; it weighs 138.2410 when filled with water at 25°C (δ=0.9970 g cm-3) and 40.2959 g when filled with propylene gas at 740.3 mm Hg and 24.0°C. What is the molar mass of polypropylene?

Strategy:

Determine Vflask. Determine mgas. Use the Gas Equation.

Example 5-6

Determine Vflask:

Vflask = mH2O dH2O = (138.2410 g – 40.1305 g) (0.9970 g cm-3)

Determine mgas:

= 0.1654 g

mgas = mfilled - mempty = (40.2959 g – 40.1305 g)

= 98.41 cm3 = 0.09841 L

Example 5-6Example 5-6

Use the Gas Equation:

PV = nRT PV = mM RT M = m

M = (0.9741 atm)(0.09841 L)

(0.6145 g)(0.08206 L atm mol-1 K-1)(297.2 K)

M = 42.08 g/mol

Gas Densities

PV = nRT and d = mV

PV = mM RT

m = d =

, n = mM

6-5 Gases in Chemical Reactions

• Stoichiometric factors relate gas quantities to quantities of other reactants or products.

• Ideal gas equation used to relate the amount of a gas to volume, temperature and pressure.

• Law of combining volumes can be developed using the gas law.

Example 6-10

Using the Ideal gas Equation in Reaction Stoichiometry Calculations.The decomposition of sodium azide, NaN3, at high temperatures produces N2(g). Together with the necessary devices to initiate the reaction and trap the sodium metal formed, this reaction is used in air-bag safety systems. What volume of N2(g), measured at 735 mm Hg and 26°C, is produced when 70.0 g NaN3 is decomposed.

2 NaN3(s) → 2 Na(l) + 3 N2(g)

Example 6-10

Determine moles of N2:

Determine volume of N2:

nN2 = 70 g N3 1 mol NaN3

65.01 g N3/mol N3

3 mol N2

2 mol NaN3

= 1.62 mol N2

= 41.1 L

V = =(735 mm Hg)

(1.62 mol)(0.08206 L atm mol-1 K-1)(299 K)

760 mm Hg1.00 atm

6-6 Mixtures of Gases

• Partial pressure– Each component of a gas mixture exerts a

pressure that it would exert if it were in the container alone.

• Gas laws apply to mixtures of gases.• Simplest approach is to use ntotal, but....

Dalton’s Law of Partial Pressure

Partial Pressure

Ptot = Pa + Pb +…

Va = naRT/Ptot and Vtot = Va + Vb+…

naRT/Ptot

ntotRT/Ptot= =

naRT/Vtot

ntotRT/Vtot= =

= aRecall

Pneumatic Trough

Ptot = Pbar = Pgas + PH2O

6-7 Kinetic Molecular Theory

• Particles are point masses in constant, random, straight line motion.

• Particles are separated by great

distances.

• Collisions are rapid and elastic.

• No force between particles.

• Total energy remains constant.

Pressure – Assessing Collision Forces

• Translational kinetic energy,

• Frequency of collisions,

• Impulse or momentum transfer,

• Pressure proportional to impulse times frequency

2muVNP

Pressure and Molecular Speed

• Three dimensional systems lead to: 2umVN

um is the modal speeduav is the simple averageurms

Pressure

um31PV

NAssume one mole:

PV=RT so:

NAm = M:

Rearrange:

Distribution of Molecular Speeds

M3RTu rms

Determining Molecular Speed

Temperature

(T)R23e

)um21(

Modify:

PV=RT so:

Solve for ek:

Average kinetic energy is directly proportional to temperature!

6-8 Gas Properties Relating to the Kinetic-Molecular Theory

• Diffusion– Net rate is proportional to

molecular speed.• Effusion

– A related phenomenon.

Graham’s Law

• Only for gases at low pressure (natural escape, not a jet).• Tiny orifice (no collisions)• Does not apply to diffusion.

3RT/MB3RT/M

)(u)(u

BofeffusionofrateAofeffusionofrate

• Ratio used can be:– Rate of effusion (as above)– Molecular speeds– Effusion times

– Distances traveled by molecules– Amounts of gas effused.

6-9 Real Gases

• Compressibility factor PV/nRT = 1• Deviations occur for real gases.

– PV/nRT > 1 - molecular volume is significant.– PV/nRT < 1 – intermolecular forces of attraction.

Real Gases

van der Waals Equation

P + n2a V2

V – nb = nRT

Chapter 6 Questions

9, 13, 18, 31, 45, 49, 61, 63, 71, 82, 85, 97, 104.

Chapter 7: Thermochemistry

Contents

7-1 Getting Started: Some Terminology7-2 Heat7-3 Heats of Reaction and Calorimetry7-4 Work7-5 The First Law of Thermodynamics7-6 Heats of Reaction: U and H7-7 The Indirect Determination of H: Hess’s Law

Contents

7-7 The Indirect Determination of H, Hess’s Law7-8 Standard Enthalpies of Formation7-9 Fuels as Sources of Energy Focus on Fats, Carbohydrates, and

Energy Storage

6-1 Getting Started: Some Terminology

• System• Surroundings

Terminology

• Energy, U– The capacity to do work.

• Work– Force acting through a distance.

• Kinetic Energy– The energy of motion.

Energy

• Kinetic Energy

ek = 12 mv2 [ek ] = kg m2

s2 = J

w = Fd [w ] = kg ms2 = Jm

• Work

Energy

• Potential Energy– Energy due to condition, position, or

composition.– Associated with forces of attraction or

repulsion between objects.• Energy can change from potential to

kinetic.

Energy and Temperature

• Thermal Energy– Kinetic energy associated with random

molecular motion.– In general proportional to temperature.– An intensive property.

• Heat and Work– q and w.– Energy changes.

Energy transferred between a system and its surroundings as a result of a temperature difference.

• Heat flows from hotter to colder.– Temperature may change.– Phase may change (an isothermal process).

Units of Heat

• Calorie (cal)– The quantity of heat required to change the

temperature of one gram of water by one degree Celsius.

• Joule (J)– SI unit for heat

1 cal = 4.184 J

Heat Capacity

• The quantity of heat required to change the temperature of a system by one degree.

– Molar heat capacity.• System is one mole of substance.

– Specific heat capacity, c.• System is one gram of substance

– Heat capacity• Mass specific heat.

q = mcT

q = CT

Conservation of Energy

• In interactions between a system and its surroundings the total energy remains constant— energy is neither created nor destroyed.

qsystem + qsurroundings = 0

qsystem = -qsurroundings

Determination of Specific Heat

Example 7-2

Determining Specific Heat from Experimental Data.Use the data presented on the last slide to calculate the specific heat of lead.

qlead = -qwater

qwater = mcT = (50.0 g)(4.184 J/g °C)(28.8 - 22.0)°C

qwater = 1.4x103 J

qlead = -1.4x103 J = mcT = (150.0 g)(c)(28.8 - 100.0)°C

clead = 0.13 Jg-1°C-1

7-3 Heats of Reaction and Calorimetry

• Chemical energy. – Contributes to the internal energy of a system.

• Heat of reaction, qrxn.

– The quantity of heat exchanged between a system and its surroundings when a chemical reaction occurs within the system, at constant temperature.

Heats of Reaction

• Exothermic reactions.– Produces heat, qrxn < 0.

• Endothermic reactions.– Consumes heat, qrxn > 0.

• Calorimeter– A device for measuring quantities of heat.

Bomb Calorimeter

qrxn = -qcal

qcal = qbomb + qwater + qwires +…

Define the heat capacity of the calorimeter: qcal = miciT = CT

Using Bomb Calorimetry Data to Determine a Heat of Reaction.The combustion of 1.010 g sucrose, in a bomb calorimeter,

causes the temperature to rise from 24.92 to 28.33°C. The heat capacity of the calorimeter assembly is 4.90 kJ/°C.

(a) What is the heat of combustion of sucrose, expressed in kJ/mol C12H22O11

(b) Verify the claim of sugar producers that one teaspoon of sugar (about 4.8 g) contains only 19 calories.

Example 7-3

Calculate qcalorimeter:

qcal = CT = (4.90 kJ/°C)(28.33-24.92)°C = (4.90)(3.41) kJ= 16.7 kJ

Calculate qrxn:

qrxn = -qcal = -16.7 kJ

per 1.010 g

Example 7-3

Calculate qrxn in the required units:

qrxn = -qcal = -16.7 kJ1.010 g

= -16.5 kJ/g

343.3 g1.00 mol

= -16.5 kJ/g

= -5.65 103 kJ/mol

Calculate qrxn for one teaspoon:

4.8 g1 tsp

= (-16.5 kJ/g)(qrxn (b))( )= -19 cal/tsp1.00 cal4.184 J

Example 7-3

Coffee Cup Calorimeter

• A simple calorimeter.– Well insulated and therefore isolated.– Measure temperature change.

qrxn = -qcal

See example 7-4 for a sample calculation.

7-4 Work

• In addition to heat effects chemical reactions may also do work.

• Gas formed pushes against the atmosphere.

• Volume changes.

• Pressure-volume work.

Pressure Volume Work

w = F d = (P A) h = PV

w = -PextV

Example 7-3

Assume an ideal gas and calculate the volume change:

Vi = nRT/P = (0.100 mol)(0.08201 L atm mol-1 K-1)(298K)/(2.40 atm)= 1.02 L

Vf = 1.88 L

Example 7-5

Calculating Pressure-Volume Work.Suppose the gas in the previous figure is 0.100 mol He at 298 K. How much work, in Joules, is associated with its expansion at constant pressure.

V = 1.88-1.02 L = 0.86 L

Example 7-3

Calculate the work done by the system:

w = -PV = -(1.30 atm)(0.86 L)(= -1.1 102 J

Example 7-5

) 101 J1 L atm

Where did the conversion factor come from?

Compare two versions of the gas constant and calculate.

8.3145 J/mol K ≡ 0.082057 L atm/mol K1 ≡ 101.33 J/L atm

Hint: If you use pressure in kPa you get Joules directly.

7-5 The First Law of Thermodynamics

• Internal Energy, U.– Total energy (potential and kinetic) in a system.

•Translational kinetic energy.•Molecular rotation.•Bond vibration.•Intermolecular attractions.•Chemical bonds.•Electrons.

First Law of Thermodynamics

• A system contains only internal energy.– A system does not contain heat or work.– These only occur during a change in the system.

• Law of Conservation of Energy– The energy of an isolated system is constant

U = q + w

First Law of Thermodynamics

State Functions

• Any property that has a unique value for a specified state of a system is said to be a State Function.

• Water at 293.15 K and 1.00 atm is in a specified state. • d = 0.99820 g/mL• This density is a unique function of the state.• It does not matter how the state was established.

Functions of State

• U is a function of state.– Not easily measured.

U has a unique value between two states.– Is easily measured.

Path Dependent Functions

• Changes in heat and work are not functions of state.– Remember example 7-5, w = -1.1 102 J in a one step

expansion of gas:– Consider 2.40 atm to 1.80 atm and finally to 1.30 atm.

w = (-1.80 atm)(1.30-1.02)L – (1.30 atm)(1.88-1.36)L= -0.61 L atm – 0.68 L atm = -1.3 L atm= 1.3 102 J

7-6 Heats of Reaction: U and H

Reactants → Products

U = Uf - Ui

U = qrxn + w

In a system at constant volume:

U = qrxn + 0 = qrxn = qv

But we live in a constant pressure world! How does qp relate to qv?

Heats of Reaction

Heats of ReactionqV = qP + w

We know that w = - PV and U = qP, therefore:U = qP - PVqP = U + PV

These are all state functions, so define a new function.Let H = U + PVThen H = Hf – Hi = U + PV If we work at constant pressure and temperature:

H = U + PV = qP

Comparing Heats of Reaction

qP = -566 kJ/mol = H

PV = P(Vf – Vi)= RT(nf – ni)= -2.5 kJ

U = H - PV= -563.5 kJ/mol= qV

Changes of State of Matter

H2O (l) → H2O(g) H = 44.0 kJ at 298 K

Molar enthalpy of vaporization:

Molar enthalpy of fusion:

H2O (s) → H2O(l) H = 6.01 kJ at 273.15 K

Example 7-3Example 7-8

Break the problem into two steps: Raise the temperature of the liquid first then completely vaporize it. The total enthalpy change is the sum of the changes in each step.

Enthalpy Changes Accompanying Changes in States of Matter.Calculate H for the process in which 50.0 g of water is converted from liquid at 10.0°C to vapor at 25.0°C.

= (50.0 g)(4.184 J/g °C)(25.0-10.0)°C + 50.0 g18.0 g/mol 44.0 kJ/mol

Set up the equation and calculate:qP = mcH2OT + nHvap

= 3.14 kJ + 122 kJ = 125 kJ

Standard States and Standard Enthalpy Changes

• Define a particular state as a standard state.• Standard enthalpy of reaction, H°

– The enthalpy change of a reaction in which all reactants and products are in their standard states.

• Standard State– The pure element or compound at a pressure of 1

bar and at the temperature of interest.

Enthalpy Diagrams

7-7 Indirect Determination of H:Hess’s Law

H is an extensive property.– Enthalpy change is directly proportional to the amount of

substance in a system.

N2(g) + O2(g) → 2 NO(g) H = +180.50 kJ

½N2(g) + ½O2(g) → NO(g) H = +90.25 kJ

H changes sign when a process is reversed

NO(g) → ½N2(g) + ½O2(g) H = -90.25 kJ

Hess’s Law

• Hess’s law of constant heat summation– If a process occurs in stages or steps (even

hypothetically), the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps.

½N2(g) + O2(g) → NO2(g) H = +33.18 kJ

½N2(g) + ½O2(g) → NO(g) H = +90.25 kJ

NO(g) + ½O2(g) → NO2(g) H = -57.07 kJ

Hess’s Law Schematically

• The enthalpy change that occurs in the formation of one mole of a substance in the standard state from the reference forms of the elements in their standard states.

• The standard enthalpy of formation of a pure element in its reference state is 0.

7-8 Standard Enthalpies of Formation

Standard Enthalpies of Formation

Standard Enthalpies of Reaction

Hoverall = -2Hf°NaHCO3

+ Hf°Na2CO3

+ Hf°H2O

Enthalpy of Reaction

Hrxn = Hf°products- Hf

°reactants

Table 7.3 Enthalpies of Formation of Ions in Aqueous Solutions

7-9 Fuels as Sources of Energy

• Fossil fuels.– Combustion is exothermic.– Non-renewable resource.– Environmental impact.

Chapter 7 Questions

1, 2, 3, 11, 14, 16, 22, 24, 29, 37, 49, 52, 63, 67, 73, 81

Chapter 8: The Atmospheric Gases and Hydrogen

Contents

8-1 The Atmosphere8-3 Nitrogen8-4 Oxygen8-5 The Noble Gases8-6 Hydrogen

Focus on The Carbon Cycle

8-1 The Atmosophere

Composition of Dry Air

Water Vapor

• nH2O PH2O in air.

Relative Humidity =PH2O (actual)

PH2O (max) 100%

Chemicals from the Atmosphere

8-2 Nitrogen

Haber Bosch Process

Anhydrous Ammonia as Fertilizer

Nitrogen Oxides

Nitric Acid Production

4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l)2 NO(g) + O2(g) → 2 NO2(g)

3NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)

• Oxidizing acid.• Nitration of organic compounds.

Nitroglycerine

Smog• Sunlight plus products of

combustion – photochemical smog.

8-3 Oxygen

• Most abundant of elements in Earths crust.

Electrolysis

• O3 is an allotrope of oxygen.• An excellent oxidizing agent.

3 O2(g) → 2 O3(g) H° = +285 kJ

O2 + UV radiation → 2 O

M + O2 + O → O3 + M*

O3 + UV radiation → O2 + O O3 + O → 2 O2 H° = -389.8 kJ

Ozone Depletion

O3 + NO → NO2 + O2

NO2 + O → NO + O2

O3 + O → 2 O2

Natural:

O3 + Cl → ClO + O2

ClO + O → Cl + O2

O3 + O → 2 O2

Human activity:

CCl2F2 + UV radiation → CClF2 + Cl

8-4 The Noble Gases

• In 1785 Cavendish could not get all the material in air to react in an electric discharge.

• 100 years later Rayleigh and Ramsay isolated argon.– Greek argos—the lazy one.

Noble Gases

• Used in light bulbs, lasers and flash bulbs.• He and Ar are used as “blanket” materials to

keep air out of certain systems.• He is used as a breathing mixture for deep

diving applications.• Superconducting magnets use He(l) as coolant.

Helium

8-5 Oxides of Carbon

• 370 ppm CO2 in air. CO only minor.

• Rich combustion:

• Lean combustion:

C8H18(l) + 12.5 O2 → 8CO2(g) + 9 H2O(l)

C8H18(l) + 12 O2 → 7CO2(g) + CO(g) + 9 H2O(l)

Hemoglobin

Industrial Preparation of CO2

Greenhouse Effect

a) Incoming sunlight hits the earths surface.

b) Earths surface emits infrared light.

c) IR absorbed in atmosphere by CO2 and other greenhouse gases. Atmosphere warms up.

Global Warming

• Predict 1.5 to 4.5°C average global temperature increase.

• Computer models.

8-6 Hydrogen

• Minor component of atmosphere.• 90% of atoms and 75% of universe mass.• Produced using the water—gas reactions:

C(s) + H2O(g) → CO(g) + H2(g)CO(g) + H2O(g) → CO2(g) + H2(g)

Or by the reforming of methane:

CH4(g) + H2O(g) → CO(g) + 3 H2(g)

Compounds of Hydrogen

• Covalent hydrides– HCl, NH3

• Ionic Hydrides– CaH2, NaH

Uses of Hydrogen

• Hydrogenation reactions

Uses of Hydrogen

Focus on The Carbon Cycle

Chapter 8 Questions

1, 2, 5, 9, 10, 23, 29, 35, 41, 45, 53, 60, 63.

Chapter 9: Electrons in Atoms

Contents

9-1 Electromagnetic Radiation9-2 Atomic Spectra9-3 Quantum Theory9-4 The Bohr Atom9-5 Two Ideas Leading to a New Quantum Mechanics9-6 Wave Mechanics9-7 Quantum Numbers and Electron Orbitals

Contents

9-8 Quantum Numbers9-9 Interpreting and Representing Orbitals of the

Hydrogen Atom9-9 Electron Spin9-10 Multi-electron Atoms9-11 Electron Configurations9-12 Electron Configurations and the Periodic Table Focus on Helium-Neon Lasers

9-1 Electromagnetic Radiation

• Electric and magnetic fields propagate as waves through empty space or through a medium.

• A wave transmits energy.

EM Radiation

Frequency, Wavelength and Velocity

• Frequency () in Hertz—Hz or s-1.• Wavelength (λ) in meters—m.

• cm m nm pm (10-2 m) (10-6 m) (10-9 m) (10-10 m) (10-12

• Velocity (c)—2.997925 108 m s-1.

c = λ λ = c/ = c/λ

Electromagnetic Spectrum

RedOrange

YellowGreen

BlueIndigo

Violet

ROYGBIV

700 nm 450 nm

Constructive and Destructive Interference

Refraction of Light

9-2 Atomic Spectra

Atomic Spectra

9-3 Quantum Theory

Blackbody Radiation:

Max Planck, 1900: Energy, like matter, is discontinuous.

є = h

The Photoelectric Effect

• Light striking the surface of certain metals causes ejection of electrons.

> o threshold frequency• e- I• ek

• At the stopping voltage the kinetic energy of the ejected electron has been converted to potential.

mu2 = eVs12

• At frequencies greater than o:

Vs = k ( - o)

Eo = hoEk = eVs o = eVo

eVo, and therefore o, are characteristic of the metal.Conservation of energy requires that:

h = mu2 + eVo21

mu2 = h - eVo eVs = 21

Ephoton = Ek + Ebinding

Ek = Ephoton - Ebinding

9-4 The Bohr Atom

E = -RH

RH = 2.179 10-18 J

Energy-Level Diagram

ΔE = Ef – Ei = -RH

= RH ( ni2

) = h = hc/λ

Ionization Energy of Hydrogen

ΔE = RH ( ni2

As nf goes to infinity for hydrogen starting in the ground state:

h = RH ( ni2

1 ) = RH

This also works for hydrogen-like species such as He+ and Li2+.

h = -Z2 RH

Emission and Absorption Spectroscopy

9-5 Two Ideas Leading to a New Quantum Mechanics

• Wave-Particle Duality.– Einstein suggested particle-like properties of

light could explain the photoelectric effect.– But diffraction patterns suggest photons are

wave-like.• deBroglie, 1924

– Small particles of matter may at times display wavelike properties.

deBroglie and Matter Waves

E = mc2

h = mc2

h/c = mc = pp = h/λ

λ = h/p = h/mu

X-Ray Diffraction

The Uncertainty Principle

Δx Δp ≥ h4π

• Werner Heisenberg

9-6 Wave Mechanics

• Standing waves.– Nodes do not undergo displacement.

λ = , n = 1, 2, 3…

Wave Functions

• ψ, psi, the wave function.– Should correspond to a

standing wave within the boundary of the system being described.

• Particle in a box.

Lxnsin

Probability of Finding an Electron

Wave Functions for Hydrogen

• Schrödinger, 1927 Eψ = H ψ

– H (x,y,z) or H (r,θ,φ)

ψ(r,θ,φ) = R(r) Y(θ,φ)

R(r) is the radial wave function.Y(θ,φ) is the angular wave function.

Principle Shells and Subshells

• Principle electronic shell, n = 1, 2, 3…• Angular momentum quantum number,

l = 0, 1, 2…(n-1)

l = 0, sl = 1, pl = 2, dl = 3, f

• Magnetic quantum number, ml= - l …-2, -1, 0, 1, 2…+l

Orbital Energies

9-8 Interpreting and Representing the Orbitals of the Hydrogen Atom.

s orbitals

p Orbitals

d Orbitals

9-9 Electron Spin: A Fourth Quantum Number

9-10 Multi-electron Atoms

• Schrödinger equation was for only one e-.• Electron-electron repulsion in multi-

electron atoms.• Hydrogen-like orbitals (by approximation).

Penetration and Shielding

Zeff is the effective nuclear charge.

9-11 Electron Configurations

• Aufbau process.– Build up and minimize energy.

• Pauli exclusion principle.– No two electrons can have all four quantum

numbers alike.• Hund’s rule.

– Degenerate orbitals are occupied singly first.

Orbital Energies

Orbital Filling

Aufbau Process and Hunds Rule

Filling p Orbitals

Filling the d Orbitals

Electon Configurations of Some Groups of Elements

9-12 Electron Configurations and the Periodic Table

Focus on He-Ne Lasers

Chapter 9 Questions

1, 2, 3, 4, 12, 15, 17, 19, 22, 25, 34, 35, 41, 67, 69, 71, 83, 85, 93, 98

Chapter 10: The Periodic Table and Some Atomic Properties

Contents

10-1 Classifying the Elements: The Periodic Law and the Periodic Table

10-2 Metals and Nonmetals and Their Ions10-3 The Sizes of Atoms and Ions10-4 Ionization Energy10-5 Electron Affinity10-6 Magnetic Properties10-7 Periodic Properties of the Elements

Focus on The Periodic Law and Mercury

10-1 Classifying the Elements: The Periodic Law and the Periodic Table

• 1869, Dimitri Mendeleev Lother Meyer

When the elements are arranged in order of increasing atomic mass, certain sets of properties recur periodically.

Periodic Law

Mendeleev’s Periodic Table1871

— = 44

— = 72— = 68— = 100

Predicted Elements were Found

X-Ray Spectra

• Moseley 1913–X-ray emission is

explained in terms of transitions in which e- drop into orbits close to the atomic nucleus.

–Correlated frequencies to nuclear charges.

= A (Z – b)2

–Used to predict new elements (43, 61, 75) later discovered.

The Periodic tableAlkali Metals

Alkaline Earths

Transition Metals

Halogens

Noble Gases

Lanthanides and Actinides

Main Group

10-2 Metals and Nonmetals and Their Ions

• Metals– Good conductors of heat and electricity.– Malleable and ductile.– Moderate to high melting points.

• Nonmetals– Nonconductors of heat and electricity.– Brittle solids.– Some are gases at room temperature.

Metals Tend to Lose Electrons

Nonmetals Tend to Gain Electrons

Electron Configuration of Some Ions

10-3 The Sizes of Atoms and Ions

Atomic Radius

Screening and Penetration

Zeff = Z – S

En = - RH n2

Cationic Radii

Anionic Radii

Atomic and Ionic Radii

10-4 Ionization Energy

Mg(g) → Mg+(g) + e- I1 = 738 kJ

Mg+(g) → Mg2+(g) + e- I2 = 1451 kJ

I = RH n2Zeff

First Ionization Energy

Table 10.4 Ionization Energies of the Third-Period Elements (in kJ/mol)

I2 (Mg) vs. I3 (Mg)

I1 (Mg) vs. I1 (Al)

737.7 577.6

I1 (P) vs. I1 (S)

1012 999.6

10-5 Electron Affinity

F(g) + e- → F-(g) EA = -328 kJ

F(1s22s22p5) + e- → F-(1s22s22p5)

Li(g) + e- → Li-(g) EA = -59.6 kJ

First Electron Affinities

Second Electron Affinities

O(g) + e- → O-(g) EA = -141 kJ

O-(g) + e- → O2-(g) EA = +744 kJ

10-6 Magnetic Properties

• Diamagnetic atoms or ions:– All e- are paired.– Weakly repelled by a magnetic field.

• Paramagnetic atoms or ions:– Unpaired e-.– Attracted to an external magnetic field.

Paramagnetism

10-7 Periodic Properties of the Elements

332266

Boiling Point

Melting Points of Elements

Melting Points of Compounds

Reducing Ability of Group 1 and 2 Metals

2 K(s) + 2 H2O(l) → 2 K+ + 2 OH- + H2(g)

Ca(s) + 2 H2O(l) → Ca2+ + 2 OH- + H2(g)

I1 = 419 kJ

I1 = 590 kJI2 = 1145 kJ

Oxidizing Abilities of the Halogens

2 Na + Cl2 → 2 NaCl

Cl2 + 2 I- → 2 Cl- + I2

Acid Base Nature of Element Oxides

• Basic oxides or base anhydrides:Li2O(s) + H2O(l) → 2 Li+(aq) + 2 OH-(aq)

• Acidic oxides or acid anhyhydrides:SO2 (g) + H2O(l) → H2SO3(aq)

• Na2O and MgO yield basic solutions

• Cl2O, SO2 and P4O10 yield acidic solutions

• SiO2 dissolves in strong base, acidic oxide.

Focus on The Periodic Law and Mercury

• Should be a solid.

• Relativistic shrinking of s-orbitals affects all heavy metals but is maximum with Hg.

Chapter 10 Questions

1, 2, 18, 21, 27, 33, 39, 43, 51, 55

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