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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
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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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
Prentice-Hall © 2002General Chemistry: Chapter 1Slide 7 of 19
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
Prentice-Hall © 2002General Chemistry: Chapter 1Slide 9 of 19
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
Prentice-Hall © 2002General Chemistry: Chapter 1Slide 10 of 19
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.
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
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
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.
Prentice-Hall © 2002General Chemistry: Chapter 2Slide 29 of 25
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)
Prentice-Hall © 2002General Chemistry: Chapter 2Slide 42 of 25
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
Prentice-Hall © 2002General Chemistry: Chapter 2Slide 43 of 25
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
Prentice-Hall © 2002General Chemistry: Chapter 2Slide 44 of 25
Chapter 2 Questions
3, 4, 11, 22, 33, 51, 55, 63, 83.
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
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
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.
Prentice-Hall © 2002General Chemistry: Chapter 3Slide 65 of 37
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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Prentice-Hall © 2002General Chemistry: Chapter 3Slide 69 of 37
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
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
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.
Prentice-Hall © 2002General Chemistry: Chapter 4Slide 89 of 29
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.
Prentice-Hall © 2002General Chemistry: Chapter 4Slide 90 of 29
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.
Prentice-Hall © 2002General Chemistry: Chapter 4Slide 97 of 29
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
Prentice-Hall © 2002General Chemistry: Chapter 4Slide 101 of 29
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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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.
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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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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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.
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.
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.
Prentice-Hall © 2002General Chemistry: Chapter 4Slide 109 of 29
Focus on Industrial Chemistry
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
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
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
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 113 of 43
5.1 The Nature of Aqueous Solutions
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 114 of 43
Electrolytes
• Some solutes can dissociate into ions.
• Electric charge can be carried.
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.
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:
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.
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:
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:
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)
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:
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
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:
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:
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.
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)
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
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)
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)
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)
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 131 of 43
Limestone and Marble
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 132 of 43
Gas Forming Reactions
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.
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.
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.
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 136 of 43
Zinc in Copper Sulfate
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
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)
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
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)
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.
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.
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!
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.
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.
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.
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 146 of 43
Redox
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.
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.
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.
Prentice-Hall © 2002General Chemistry: Chapter 5Slide 150 of 43
Indicators
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)
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)
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.
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
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
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
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
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
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 159 of 41
Manometers
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 160 of 41
6-2 Simple Gas Laws
• Boyle 1662 P 1V PV = constant
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
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 162 of 41
Charles’s Law
Charles 1787Gay-Lussac 1802
V T V = b T
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
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.
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 165 of 41
Formation of Water
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:
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
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
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:
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 170 of 41
6-4 Applications of the Ideal Gas Equation
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 171 of 41
Molar Mass Determination
PV = nRT and n = mM
PV = mM RT
M = mPVRT
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.
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
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
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
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.
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)
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
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....
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 180 of 41
Dalton’s Law of Partial Pressure
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
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 182 of 41
Pneumatic Trough
Ptot = Pbar = Pgas + PH2O
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.
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
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
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:
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 187 of 41
Distribution of Molecular Speeds
M3RTu rms
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 188 of 41
Determining Molecular Speed
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!
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.
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.
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 192 of 41
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.
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 193 of 41
Real Gases
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 194 of 41
van der Waals Equation
P + n2a V2
V – nb = nRT
Prentice-Hall © 2002 General Chemistry: Chapter 6 Slide 195 of 41
Chapter 6 Questions
9, 13, 18, 31, 45, 49, 61, 63, 71, 82, 85, 97, 104.
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 196 of 50
Chapter 7: Thermochemistry
Philip DuttonUniversity of Windsor, Canada
Prentice-Hall © 2002
General ChemistryPrinciples and Modern Applications
Petrucci • Harwood • Herring 8th Edition
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 197 of 50
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
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 198 of 50
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
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 199 of 50
6-1 Getting Started: Some Terminology
• System• Surroundings
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 200 of 50
Terminology
• Energy, U– The capacity to do work.
• Work– Force acting through a distance.
• Kinetic Energy– The energy of motion.
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 201 of 50
Energy
• Kinetic Energy
ek = 12 mv2 [ek ] = kg m2
s2 = J
w = Fd [w ] = kg ms2 = Jm
• Work
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 202 of 50
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.
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 203 of 50
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).
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 205 of 50
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
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 209 of 50
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
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 210 of 50
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.
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 211 of 50
Heats of Reaction
• Exothermic reactions.– Produces heat, qrxn < 0.
• Endothermic reactions.– Consumes heat, qrxn > 0.
• Calorimeter– A device for measuring quantities of heat.
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 212 of 50
Bomb Calorimeter
qrxn = -qcal
qcal = qbomb + qwater + qwires +…
Define the heat capacity of the calorimeter: qcal = miciT = CT
all i
heat
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 213 of 50
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
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 214 of 50
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
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 215 of 50
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
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 216 of 50
Coffee Cup Calorimeter
• A simple calorimeter.– Well insulated and therefore isolated.– Measure temperature change.
qrxn = -qcal
See example 7-4 for a sample calculation.
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 217 of 50
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.
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 221 of 50
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.
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 222 of 50
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
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 223 of 50
First Law of Thermodynamics
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 224 of 50
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.
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 225 of 50
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
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 227 of 50
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?
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 228 of 50
Heats of Reaction
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 229 of 50
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
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 230 of 50
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
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 231 of 50
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
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 232 of 50
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
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 233 of 50
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
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 235 of 50
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
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 238 of 50
• 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
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 240 of 50
Standard Enthalpies of Formation
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 241 of 50
Standard Enthalpies of Reaction
Hoverall = -2Hf°NaHCO3
+ Hf°Na2CO3
+ Hf
°CO2
+ Hf°H2O
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 242 of 50
Enthalpy of Reaction
Hrxn = Hf°products- Hf
°reactants
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 243 of 50
Table 7.3 Enthalpies of Formation of Ions in Aqueous Solutions
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 244 of 50
7-9 Fuels as Sources of Energy
• Fossil fuels.– Combustion is exothermic.– Non-renewable resource.– Environmental impact.
Prentice-Hall © 2002 General Chemistry: Chapter 7 Slide 245 of 50
Chapter 7 Questions
1, 2, 3, 11, 14, 16, 22, 24, 29, 37, 49, 52, 63, 67, 73, 81
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 246 of 32
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
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 247 of 32
Contents
8-1 The Atmosphere8-3 Nitrogen8-4 Oxygen8-5 The Noble Gases8-6 Hydrogen
Focus on The Carbon Cycle
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 248 of 32
8-1 The Atmosophere
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 249 of 32
Composition of Dry Air
trace
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 250 of 32
Water Vapor
• nH2O PH2O in air.
Relative Humidity =PH2O (actual)
PH2O (max) 100%
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 251 of 32
Chemicals from the Atmosphere
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 252 of 32
8-2 Nitrogen
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 253 of 32
Haber Bosch Process
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 254 of 32
Anhydrous Ammonia as Fertilizer
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 255 of 32
Nitrogen Oxides
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 256 of 32
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.
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 257 of 32
Nitroglycerine
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 258 of 32
Smog• Sunlight plus products of
combustion – photochemical smog.
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 259 of 32
8-3 Oxygen
• Most abundant of elements in Earths crust.
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 260 of 32
Electrolysis
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 261 of 32
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
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 262 of 32
Ozone Depletion
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 263 of 32
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
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 264 of 32
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.
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 265 of 32
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.
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 266 of 32
Helium
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 267 of 32
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)
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 268 of 32
Hemoglobin
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 269 of 32
Industrial Preparation of CO2
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 270 of 32
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.
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 271 of 32
Global Warming
• Predict 1.5 to 4.5°C average global temperature increase.
• Computer models.
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 272 of 32
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)
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 273 of 32
Compounds of Hydrogen
• Covalent hydrides– HCl, NH3
• Ionic Hydrides– CaH2, NaH
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 274 of 32
Uses of Hydrogen
• Hydrogenation reactions
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 275 of 32
Uses of Hydrogen
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 276 of 32
Focus on The Carbon Cycle
Prentice-Hall © 2002 General Chemistry: Chapter 8 Slide 277 of 32
Chapter 8 Questions
1, 2, 5, 9, 10, 23, 29, 35, 41, 45, 53, 60, 63.
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 278 of 50
Philip DuttonUniversity of Windsor, Canada
Prentice-Hall © 2002
General ChemistryPrinciples and Modern Applications
Petrucci • Harwood • Herring 8th Edition
Chapter 9: Electrons in Atoms
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 279 of 50
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
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 280 of 50
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
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 281 of 50
9-1 Electromagnetic Radiation
• Electric and magnetic fields propagate as waves through empty space or through a medium.
• A wave transmits energy.
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 282 of 50
EM Radiation
Low
High
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 283 of 50
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
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 284 of 50
Electromagnetic Spectrum
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 285 of 50
RedOrange
YellowGreen
BlueIndigo
Violet
Prentice-Hall ©2002 General Chemistry: Chapter 9 Slide 8
ROYGBIV
700 nm 450 nm
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 286 of 50
Constructive and Destructive Interference
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 287 of 50
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 288 of 50
Refraction of Light
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 289 of 50
9-2 Atomic Spectra
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 290 of 50
Atomic Spectra
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 291 of 50
9-3 Quantum Theory
Blackbody Radiation:
Max Planck, 1900: Energy, like matter, is discontinuous.
є = h
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 292 of 50
The Photoelectric Effect
• Light striking the surface of certain metals causes ejection of electrons.
> o threshold frequency• e- I• ek
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 293 of 50
The Photoelectric Effect
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 294 of 50
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)
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 295 of 50
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
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 296 of 50
9-4 The Bohr Atom
E = -RH
n2
RH = 2.179 10-18 J
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 297 of 50
Energy-Level Diagram
ΔE = Ef – Ei = -RH
nf2
-RH
ni2
–
= RH ( ni2
1nf
2–1
) = h = hc/λ
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 298 of 50
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
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 299 of 50
Emission and Absorption Spectroscopy
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 300 of 50
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.
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 301 of 50
deBroglie and Matter Waves
E = mc2
h = mc2
h/c = mc = pp = h/λ
λ = h/p = h/mu
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 302 of 50
X-Ray Diffraction
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 303 of 50
The Uncertainty Principle
Δx Δp ≥ h4π
• Werner Heisenberg
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 304 of 50
9-6 Wave Mechanics
2Ln
• Standing waves.– Nodes do not undergo displacement.
λ = , n = 1, 2, 3…
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 305 of 50
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
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 306 of 50
Probability of Finding an Electron
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 307 of 50
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.
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 308 of 50
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
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 309 of 50
Orbital Energies
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 310 of 50
9-8 Interpreting and Representing the Orbitals of the Hydrogen Atom.
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 311 of 50
s orbitals
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 312 of 50
p Orbitals
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 313 of 50
p Orbitals
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 314 of 50
d Orbitals
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 315 of 50
9-9 Electron Spin: A Fourth Quantum Number
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 316 of 50
9-10 Multi-electron Atoms
• Schrödinger equation was for only one e-.• Electron-electron repulsion in multi-
electron atoms.• Hydrogen-like orbitals (by approximation).
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 317 of 50
Penetration and Shielding
Zeff is the effective nuclear charge.
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 318 of 50
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.
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 319 of 50
Orbital Energies
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 320 of 50
Orbital Filling
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 321 of 50
Aufbau Process and Hunds Rule
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 322 of 50
Filling p Orbitals
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 323 of 50
Filling the d Orbitals
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 324 of 50
Electon Configurations of Some Groups of Elements
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 325 of 50
9-12 Electron Configurations and the Periodic Table
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 326 of 50
Focus on He-Ne Lasers
Prentice-Hall © 2002 General Chemistry: Chapter 9 Slide 327 of 50
Chapter 9 Questions
1, 2, 3, 4, 12, 15, 17, 19, 22, 25, 34, 35, 41, 67, 69, 71, 83, 85, 93, 98
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
Prentice-Hall © 2002 General Chemistry: Chapter 10 Slide 329 of 35
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