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Prentice-Hall © 2002General Chemistry: Chapter 1Slide 1 of 19
Philip DuttonUniversity of Windsor, Canada
Prentice-Hall © 2002
Chapter 1: Matter—Its Properties and Measurement
General ChemistryPrinciples and Modern Applications
Petrucci • Harwood • Herring 8th Edition
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Prentice-Hall © 2002General Chemistry: Chapter 1Slide 2 of 19
Contents
Physical properties and states of matter Système International Units Uncertainty and significant figures Dimensional analysis
http://cwx.prenhall.com/petrucci/chapter1/deluxe.html
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Prentice-Hall © 2002General Chemistry: Chapter 1Slide 3 of 19
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
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States of Matter
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1_15
Matter(materials)
Substances Mixtures
Elements CompoundsHomogeneous
mixtures(solutions)
Heterogeneousmixtures
Physical processes
Chemical
reactions
Classification of Matter
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Separations
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Separating Mixtures
1_17
Substances tobe separateddissolved in liquid
Pureliquid
A B C
mixture
Chromatography
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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
13.4
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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
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Units
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
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Temperature
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Relative Temperatures
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Volume
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Density
= m/V
m=VV=m/
g/mLMass and volume are extensive properties
Density is an intensive property
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ConversionWhat is the mass of a cube of osmium that is 1.25 inches on each side?
Have volume, need density = 22.48g/cm3
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Wrong units
The Gimli Glider, Q86, p30
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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.
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Prentice-Hall © 2002General Chemistry: Chapter 1Slide 19 of 19
End of Chapter Questions
1, 3, 5, 12, 14, 17, 18, 20, 30, 41, 49, 50, 61, 72, 74, 79
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Prentice-Hall © 2002General Chemistry: Chapter 2Slide 20 of 25
Chapter 2: Atoms and the Atomic Theory
Philip DuttonUniversity of Windsor, Canada
Prentice-Hall © 2002
General ChemistryPrinciples and Modern Applications
Petrucci • Harwood • Herring 8th Edition
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Prentice-Hall © 2002General Chemistry: Chapter 2Slide 21 of 25
Contents
• Early chemical discoveries • Electrons and the Nuclear Atom• Chemical Elements• Atomic Masses• The Mole
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Early Discoveries
Lavoisier 1774 Law of conservation of mass
Proust 1799 Law of constant composition
Dalton 1803-1888 Atomic Theory
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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
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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.
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Behavior of charges
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Cathode ray tube
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Properties of cathode rays
Electron m/e = -5.6857 x 10-9 g coulomb-1
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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.
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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.
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The nuclear atom
Geiger and Rutherford1909
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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
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The nuclear atom
Rutherfordprotons 1919
James Chadwickneutrons 1932
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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
1 Å
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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
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Isotopes, atomic numbers and mass numbers
To represent a particular atom we use the symbolism:
A= mass number Z = atomic number
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Measuring atomic masses
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The Periodic tableAlkali Metals
Alkaline Earths
Transition Metals
Halogens
Noble Gases
Lanthanides and Actinides
Main Group
Main Group
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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.
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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
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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
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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)
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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
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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
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Chapter 2 Questions
3, 4, 11, 22, 33, 51, 55, 63, 83.
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Prentice-Hall © 2002General Chemistry: Chapter 3Slide 45 of 37
Philip DuttonUniversity of Windsor, Canada
Prentice-Hall © 2002
Chapter 3: Chemical Compounds
General ChemistryPrinciples and Modern ApplicationsPetrucci • Harwood • Herring 8th Edition
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Prentice-Hall © 2002General Chemistry: Chapter 3Slide 46 of 37
Contents
3-1 Molecular and Ionic Compounds3-2 Molecular Mass3-3 Composition3-4 Oxidation States
3-5 Names and formulas Focus on Mass Spectrometry
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Molecular compounds
1 /inch 0.4 /cm
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Standard color scheme
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Some moleculesH2O2 CH3CH2Cl P4O10
CH3CH(OH)CH3 HCO2H
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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.
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Sodium chloride
Extended array of Na+ and Cl- ions Simplest formula unit is NaCl
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Inorganic molecules
S8P4
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Molecular mass
HOO
H
H
HO
H
OHOHH
H
OH
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
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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
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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(%
g
gC
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Example 3-4
%88.28%10038.197
)00.193(%
%96.17%10038.19745.35%
%48.40%10038.19790.79%
%51.0%10038.197
01.1%
%17.12%10038.197
)01.122(%
ggF
ggCl
ggBr
ggH
ggC
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Prentice-Hall © 2002General Chemistry: Chapter 3Slide 57 of 37
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:
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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
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Step 2: Convert masses to amounts in moles.
OmolOg
OmolOgn
HmolHg
HmolHgn
CmolCg
CmolCgn
O
H
C
737.1999.15
179.27
55.9008.1
163.9
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
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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
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Combustion analysis
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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.
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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
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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.
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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
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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
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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
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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
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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.
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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
2-
perchlorate ClO4- hydrogensulfate HSO4
-
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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.
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Visualizations of some hydrocarbons
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Visualizations of some hydrocarbons
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IsomersIsomers have the same molecular formula but have different arrangements of atoms in space.
H
(c)
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Functional Groups – carboxylic acid
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Functional Groups - alcohol
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Chapter 3 Questions
3, 5, 12, 24, 35, 46, 53, 61, 57, 73, 95, 97
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Prentice-Hall © 2002General Chemistry: Chapter 4Slide 82 of 29
Philip DuttonUniversity of Windsor, Canada
Prentice-Hall © 2002
Chapter 4: Chemical Reactions
General ChemistryPrinciples and Modern Applications
Petrucci • Harwood • Herring 8th Edition
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Prentice-Hall © 2002General Chemistry: Chapter 4Slide 83 of 29
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
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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.
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Chemical Reaction
Nitrogen monoxide + oxygen → nitrogen dioxide
Step 1: Write the reaction using chemical symbols.
NO + O2 → NO2
Step 2: Balance the chemical equation.
2 1 2
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Molecular Representation
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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
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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.
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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.
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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.
6 7
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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.
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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:
2 2
Use the stoichiometric factor or mole ratio in an equation:
nH2O = 2.72 mol H2 × = 2.72 mol H2O2 mol H2O2 mol H2
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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?
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Al + HCl → AlCl3 + H2
Write the Chemical Equation:
Example 4-6
Balance the Chemical Equation:
2 6 2 3
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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:
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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.
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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
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Preparation of a Solution
Weigh the solid sample.Dissolve it in a volumetric flask partially filled with solvent.Carefully fill to the mark.
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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
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Solution Dilution
Mi × Vi = ni
Mi × ViMf × Vf
= nf = Mf × Vf
Mi × ViMf = Vf
= MiVi
Vf
M = nV
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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
Vf
Vi = VfMf
Mi
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4-4 Determining Limiting Reagent
• The reactant that is completely consumed determines the quantities of the products formed.
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Prentice-Hall © 2002General Chemistry: Chapter 4Slide 103 of 29
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.
Page 104
Prentice-Hall © 2002General Chemistry: Chapter 4Slide 104 of 29
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.
nn =
P4
Cl2
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Prentice-Hall © 2002General Chemistry: Chapter 4Slide 105 of 29
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
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Prentice-Hall © 2002General Chemistry: Chapter 4Slide 106 of 29
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.
Page 107
Prentice-Hall © 2002General Chemistry: Chapter 4Slide 107 of 29
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.
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Prentice-Hall © 2002General Chemistry: Chapter 4Slide 108 of 29
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.
Page 109
Prentice-Hall © 2002General Chemistry: Chapter 4Slide 109 of 29
Focus on Industrial Chemistry
Page 110
Prentice-Hall © 2002General Chemistry: Chapter 4Slide 110 of 29
Chapter 4 Questions
1, 6, 12, 25, 39, 45, 53, 65, 69, 75, 84, 94, 83, 112
Page 111
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 111 of 43
Chapter 5: Introduction to Reactions in Aqueous Solutions
Philip DuttonUniversity of Windsor, Canada
Prentice-Hall © 2002
General ChemistryPrinciples and Modern Applications
Petrucci • Harwood • Herring 8th Edition
Page 112
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 112 of 43
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
Page 113
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 113 of 43
5.1 The Nature of Aqueous Solutions
Page 114
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 114 of 43
Electrolytes
• Some solutes can dissociate into ions.
• Electric charge can be carried.
Page 115
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 115 of 43
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.
Page 116
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 116 of 43
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:
Page 117
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 117 of 43
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.
Page 118
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 118 of 43
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:
Page 119
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 119 of 43
[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:
Page 120
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 120 of 43
5-2 Precipitation Reactions
• Soluble ions can combine to form an insoluble compound.
• Precipitation occurs.
Ag+(aq) + Cl-(aq) → AgCl(s)
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Prentice-Hall © 2002General Chemistry: Chapter 5Slide 121 of 43
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
-(aq)
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:
Page 122
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 122 of 43
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
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Prentice-Hall © 2002General Chemistry: Chapter 5Slide 123 of 43
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:
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Prentice-Hall © 2002General Chemistry: Chapter 5Slide 124 of 43
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-,
PO43-
• Except alkali metal and ammonium salts
•Compounds that are insoluble:
Page 125
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 125 of 43
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.
Page 126
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 126 of 43
Acids
• Acids provide H+ in aqueous solution.
• Strong acids:
• Weak acids:
HCl(aq) H+(aq) + Cl-(aq) →
→←CH3CO2H(aq) H+(aq) + CH3CO2-(aq)
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Prentice-Hall © 2002General Chemistry: Chapter 5Slide 127 of 43
Bases
• 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
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Prentice-Hall © 2002General Chemistry: Chapter 5Slide 128 of 43
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)
Page 129
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 129 of 43
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)
Page 130
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 130 of 43
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)
Page 131
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 131 of 43
Limestone and Marble
Page 132
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 132 of 43
Gas Forming Reactions
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Prentice-Hall © 2002General Chemistry: Chapter 5Slide 133 of 43
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.
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Prentice-Hall © 2002General Chemistry: Chapter 5Slide 134 of 43
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.
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Prentice-Hall © 2002General Chemistry: Chapter 5Slide 135 of 43
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.
Page 136
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 136 of 43
Zinc in Copper Sulfate
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Page 137
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 137 of 43
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)
Page 138
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 138 of 43
Balancing Oxidation-Reduction Equations
• Few can be balanced by inspection.• Systematic approach required.
• The Half-Reaction (Ion-Electron) Method
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Prentice-Hall © 2002General Chemistry: Chapter 5Slide 139 of 43
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)
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Prentice-Hall © 2002General Chemistry: Chapter 5Slide 140 of 43
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.
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Prentice-Hall © 2002General Chemistry: Chapter 5Slide 141 of 43
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.
Page 142
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 142 of 43
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!
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Prentice-Hall © 2002General Chemistry: Chapter 5Slide 143 of 43
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.
Page 144
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 144 of 43
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.
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Prentice-Hall © 2002General Chemistry: Chapter 5Slide 145 of 43
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.
Page 146
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 146 of 43
Redox
Page 147
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 147 of 43
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.
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Prentice-Hall © 2002General Chemistry: Chapter 5Slide 148 of 43
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.
Page 149
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 149 of 43
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.
Page 150
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 150 of 43
Indicators
Page 151
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 151 of 43
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)
Page 152
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 152 of 43
Example 5-10
Determine KMnO4 consumed in the reaction:
Determine the concentration:
44
4
42
4
2
10615.511
51
11
847.5511568.0
2
KMnOmolMnOmolKMnOmol
FemolMnOmol
FemolFemol
FegFemolFegn OH
44
4
4 02140.002624.0
10615.5][ KMnOMLKMnOmolKMnO
5 Fe2+(aq) + MnO4-(aq) + 8 H+(aq) → 4 H2O(l) + 5 Fe3+(aq) + Mn2+(aq)
Page 153
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 153 of 43
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.
Page 154
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 154 of 41
Chapter 6: Gases
Philip DuttonUniversity of Windsor, Canada
Prentice-Hall © 2002
General ChemistryPrinciples and Modern Applications
Petrucci • Harwood • Herring 8th Edition
Page 155
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 155 of 41
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
Page 156
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 156 of 41
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
Page 157
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 157 of 41
6-1 Properties of Gases: Gas Pressure
• Gas Pressure
• Liquid Pressure
P (Pa) =
Area (m2)Force (N)
P = g ·h ·d
Page 158
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 158 of 41
Barometric Pressure
Standard Atmospheric Pressure 1.00 atm760 mm Hg, 760 torr101.325 kPa1.01325 bar1013.25 mbar
Page 159
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 159 of 41
Manometers
Page 160
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 160 of 41
6-2 Simple Gas Laws
• Boyle 1662 P 1V PV = constant
Page 161
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 161 of 41
Example 5-6
Relating Gas Volume and Pressure – Boyle’s Law.
P1V1 = P2V2 V2 = P1V1
P2= 694 L Vtank = 644 L
Page 162
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Charles’s Law
Charles 1787Gay-Lussac 1802
V T V = b T
Page 163
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 163 of 41
STP
• Gas properties depend on conditions.
• Define standard conditions of temperature and pressure (STP).
P = 1 atm = 760 mm HgT = 0°C = 273.15 K
Page 164
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 164 of 41
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.
Page 165
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 165 of 41
Formation of Water
Page 166
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 166 of 41
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:
Page 167
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 167 of 41
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
V nTP
Page 168
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 168 of 41
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
Page 169
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 169 of 41
The General Gas Equation
R = = P2V2
n2T2
P1V1
n1T1
= P2
T2
P1
T1
If we hold the amount and volume constant:
Page 170
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 170 of 41
6-4 Applications of the Ideal Gas Equation
Page 171
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 171 of 41
Molar Mass Determination
PV = nRT and n = mM
PV = mM RT
M = mPVRT
Page 172
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 172 of 41
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.
Page 173
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 173 of 41
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
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Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 174 of 41
Example 5-6Example 5-6
Use the Gas Equation:
PV = nRT PV = mM RT M = m
PVRT
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
Page 175
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 175 of 41
Gas Densities
PV = nRT and d = mV
PV = mM RT
MPRTV
m = d =
, n = mM
Page 176
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 176 of 41
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.
Page 177
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 177 of 41
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)
Page 178
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 178 of 41
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
PnRT
V = =(735 mm Hg)
(1.62 mol)(0.08206 L atm mol-1 K-1)(299 K)
760 mm Hg1.00 atm
Page 179
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 179 of 41
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....
Page 180
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 180 of 41
Dalton’s Law of Partial Pressure
Page 181
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 181 of 41
Partial Pressure
Ptot = Pa + Pb +…
Va = naRT/Ptot and Vtot = Va + Vb+…
Va
Vtot
naRT/Ptot
ntotRT/Ptot= =
na
ntot
Pa
Ptot
naRT/Vtot
ntotRT/Vtot= =
na
ntot
na
ntot
= aRecall
Page 182
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 182 of 41
Pneumatic Trough
Ptot = Pbar = Pgas + PH2O
Page 183
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 183 of 41
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.
Page 184
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 184 of 41
Pressure – Assessing Collision Forces
• Translational kinetic energy,
• Frequency of collisions,
• Impulse or momentum transfer,
• Pressure proportional to impulse times frequency
2k mu
21e
VNuv
muI
2muVNP
Page 185
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 185 of 41
Pressure and Molecular Speed
• Three dimensional systems lead to: 2umVN
31P
2u
um is the modal speeduav is the simple averageurms
Page 186
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 186 of 41
Pressure
M3RTu
uM3RT
umRT3
um31PV
rms
2
2A
2A
N
NAssume one mole:
PV=RT so:
NAm = M:
Rearrange:
Page 187
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 187 of 41
Distribution of Molecular Speeds
M3RTu rms
Page 188
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 188 of 41
Determining Molecular Speed
Page 189
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 189 of 41
Temperature
(T)R23e
e32RT
)um21(
32um
31PV
Ak
k
22A
N
N
NN
A
A
Modify:
PV=RT so:
Solve for ek:
Average kinetic energy is directly proportional to temperature!
Page 190
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 190 of 41
6-8 Gas Properties Relating to the Kinetic-Molecular Theory
• Diffusion– Net rate is proportional to
molecular speed.• Effusion
– A related phenomenon.
Page 191
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 191 of 41
Graham’s Law
• Only for gases at low pressure (natural escape, not a jet).• Tiny orifice (no collisions)• Does not apply to diffusion.
A
BA
Brms
Arms
MM
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.
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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.
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Real Gases
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van der Waals Equation
P + n2a V2
V – nb = nRT
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Chapter 6 Questions
9, 13, 18, 31, 45, 49, 61, 63, 71, 82, 85, 97, 104.
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Chapter 7: Thermochemistry
Philip DuttonUniversity of Windsor, Canada
Prentice-Hall © 2002
General ChemistryPrinciples and Modern Applications
Petrucci • Harwood • Herring 8th Edition
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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
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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
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6-1 Getting Started: Some Terminology
• System• Surroundings
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Terminology
• Energy, U– The capacity to do work.
• Work– Force acting through a distance.
• Kinetic Energy– The energy of motion.
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Energy
• Kinetic Energy
ek = 12 mv2 [ek ] = kg m2
s2 = J
w = Fd [w ] = kg ms2 = Jm
• Work
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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.
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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.
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Heat
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).
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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
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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
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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
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Determination of Specific Heat
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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
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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.
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Heats of Reaction
• Exothermic reactions.– Produces heat, qrxn < 0.
• Endothermic reactions.– Consumes heat, qrxn > 0.
• Calorimeter– A device for measuring quantities of heat.
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Bomb Calorimeter
qrxn = -qcal
qcal = qbomb + qwater + qwires +…
Define the heat capacity of the calorimeter: qcal = miciT = CT
all i
heat
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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
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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
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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
qrxn
(a)
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
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Coffee Cup Calorimeter
• A simple calorimeter.– Well insulated and therefore isolated.– Measure temperature change.
qrxn = -qcal
See example 7-4 for a sample calculation.
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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.
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Pressure Volume Work
w = F d = (P A) h = PV
w = -PextV
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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
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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.
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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.
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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
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First Law of Thermodynamics
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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.
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Functions of State
• U is a function of state.– Not easily measured.
U has a unique value between two states.– Is easily measured.
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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
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7-6 Heats of Reaction: U and H
Reactants → Products
Ui Uf
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?
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Heats of Reaction
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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
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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
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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
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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
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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.
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Enthalpy Diagrams
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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
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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
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Hess’s Law Schematically
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• 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.
Hf°
7-8 Standard Enthalpies of Formation
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Standard Enthalpies of Formation
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Standard Enthalpies of Formation
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Standard Enthalpies of Reaction
Hoverall = -2Hf°NaHCO3
+ Hf°Na2CO3
+ Hf
°CO2
+ Hf°H2O
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Enthalpy of Reaction
Hrxn = Hf°products- Hf
°reactants
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Table 7.3 Enthalpies of Formation of Ions in Aqueous Solutions
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7-9 Fuels as Sources of Energy
• Fossil fuels.– Combustion is exothermic.– Non-renewable resource.– Environmental impact.
Page 245
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Chapter 7 Questions
1, 2, 3, 11, 14, 16, 22, 24, 29, 37, 49, 52, 63, 67, 73, 81
Page 246
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Chapter 8: The Atmospheric Gases and Hydrogen
Philip DuttonUniversity of Windsor, Canada
Prentice-Hall © 2002
General ChemistryPrinciples and Modern Applications
Petrucci • Harwood • Herring 8th Edition
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Contents
8-1 The Atmosphere8-3 Nitrogen8-4 Oxygen8-5 The Noble Gases8-6 Hydrogen
Focus on The Carbon Cycle
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8-1 The Atmosophere
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Composition of Dry Air
trace
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Water Vapor
• nH2O PH2O in air.
Relative Humidity =PH2O (actual)
PH2O (max) 100%
Page 251
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Chemicals from the Atmosphere
Page 252
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8-2 Nitrogen
Page 253
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Haber Bosch Process
Page 254
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Anhydrous Ammonia as Fertilizer
Page 255
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Nitrogen Oxides
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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)
Pt
• Oxidizing acid.• Nitration of organic compounds.
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Nitroglycerine
Page 258
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Smog• Sunlight plus products of
combustion – photochemical smog.
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8-3 Oxygen
• Most abundant of elements in Earths crust.
Page 260
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Electrolysis
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Ozone
• 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
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Ozone Depletion
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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
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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.
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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.
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Helium
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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)
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Hemoglobin
Page 269
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Industrial Preparation of CO2
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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.
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Global Warming
• Predict 1.5 to 4.5°C average global temperature increase.
• Computer models.
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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)
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Compounds of Hydrogen
• Covalent hydrides– HCl, NH3
• Ionic Hydrides– CaH2, NaH
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Uses of Hydrogen
• Hydrogenation reactions
Page 275
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Uses of Hydrogen
Page 276
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Focus on The Carbon Cycle
Page 277
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Chapter 8 Questions
1, 2, 5, 9, 10, 23, 29, 35, 41, 45, 53, 60, 63.
Page 278
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Philip DuttonUniversity of Windsor, Canada
Prentice-Hall © 2002
General ChemistryPrinciples and Modern Applications
Petrucci • Harwood • Herring 8th Edition
Chapter 9: Electrons in Atoms
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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
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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
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9-1 Electromagnetic Radiation
• Electric and magnetic fields propagate as waves through empty space or through a medium.
• A wave transmits energy.
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EM Radiation
Low
High
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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
m)
• Velocity (c)—2.997925 108 m s-1.
c = λ λ = c/ = c/λ
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Electromagnetic Spectrum
Page 285
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RedOrange
YellowGreen
BlueIndigo
Violet
Prentice-Hall ©2002 General Chemistry: Chapter 9 Slide 8
ROYGBIV
700 nm 450 nm
Page 286
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Constructive and Destructive Interference
Page 287
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Page 288
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Refraction of Light
Page 289
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9-2 Atomic Spectra
Page 290
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Atomic Spectra
Page 291
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9-3 Quantum Theory
Blackbody Radiation:
Max Planck, 1900: Energy, like matter, is discontinuous.
є = h
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The Photoelectric Effect
• Light striking the surface of certain metals causes ejection of electrons.
> o threshold frequency• e- I• ek
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The Photoelectric Effect
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The Photoelectric Effect
• 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)
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The Photoelectric Effect
Eo = hoEk = eVs o = eVo
h
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
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9-4 The Bohr Atom
E = -RH
n2
RH = 2.179 10-18 J
Page 297
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Energy-Level Diagram
ΔE = Ef – Ei = -RH
nf2
-RH
ni2
–
= RH ( ni2
1nf
2–1
) = h = hc/λ
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Ionization Energy of Hydrogen
ΔE = RH ( ni2
1nf
2–1
) = h
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
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Emission and Absorption Spectroscopy
Page 300
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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.
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deBroglie and Matter Waves
E = mc2
h = mc2
h/c = mc = pp = h/λ
λ = h/p = h/mu
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X-Ray Diffraction
Page 303
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The Uncertainty Principle
Δx Δp ≥ h4π
• Werner Heisenberg
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9-6 Wave Mechanics
2Ln
• Standing waves.– Nodes do not undergo displacement.
λ = , n = 1, 2, 3…
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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
L2ψ
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Probability of Finding an Electron
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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.
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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
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Orbital Energies
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9-8 Interpreting and Representing the Orbitals of the Hydrogen Atom.
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s orbitals
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p Orbitals
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p Orbitals
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d Orbitals
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9-9 Electron Spin: A Fourth Quantum Number
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9-10 Multi-electron Atoms
• Schrödinger equation was for only one e-.• Electron-electron repulsion in multi-
electron atoms.• Hydrogen-like orbitals (by approximation).
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Penetration and Shielding
Zeff is the effective nuclear charge.
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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.
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Orbital Energies
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Orbital Filling
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Aufbau Process and Hunds Rule
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Filling p Orbitals
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Filling the d Orbitals
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Electon Configurations of Some Groups of Elements
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9-12 Electron Configurations and the Periodic Table
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Focus on He-Ne Lasers
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Chapter 9 Questions
1, 2, 3, 4, 12, 15, 17, 19, 22, 25, 34, 35, 41, 67, 69, 71, 83, 85, 93, 98
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Philip DuttonUniversity of Windsor, Canada
Prentice-Hall © 2002
General ChemistryPrinciples and Modern Applications
Petrucci • Harwood • Herring 8th Edition
Chapter 10: The Periodic Table and Some Atomic Properties
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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
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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.
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Periodic Law
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Mendeleev’s Periodic Table1871
— = 44
— = 72— = 68— = 100
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Predicted Elements were Found
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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.
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The Periodic tableAlkali Metals
Alkaline Earths
Transition Metals
Halogens
Noble Gases
Lanthanides and Actinides
Main Group
Main Group
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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.
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Metals Tend to Lose Electrons
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Nonmetals Tend to Gain Electrons
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Electron Configuration of Some Ions
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10-3 The Sizes of Atoms and Ions
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Atomic Radius
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Screening and Penetration
Zeff = Z – S
En = - RH n2
Zeff2
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Cationic Radii
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Anionic Radii
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Atomic and Ionic Radii
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10-4 Ionization Energy
Mg(g) → Mg+(g) + e- I1 = 738 kJ
Mg+(g) → Mg2+(g) + e- I2 = 1451 kJ
I = RH n2Zeff
2
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First Ionization Energy
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Table 10.4 Ionization Energies of the Third-Period Elements (in kJ/mol)
I2 (Mg) vs. I3 (Mg)
7733
1451
I1 (Mg) vs. I1 (Al)
737.7 577.6
I1 (P) vs. I1 (S)
1012 999.6
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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
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First Electron Affinities
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Second Electron Affinities
O(g) + e- → O-(g) EA = -141 kJ
O-(g) + e- → O2-(g) EA = +744 kJ
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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.
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Paramagnetism
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10-7 Periodic Properties of the Elements
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332266
Boiling Point
??
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Melting Points of Elements
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Melting Points of Compounds
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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
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Oxidizing Abilities of the Halogens
2 Na + Cl2 → 2 NaCl
Cl2 + 2 I- → 2 Cl- + I2
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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.
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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.
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Chapter 10 Questions
1, 2, 18, 21, 27, 33, 39, 43, 51, 55