AP Chapter 3_Combine

1

Atomic Mass

� Atoms are so small, it is difficult to discuss how much they weigh in grams

� Use atomic mass units.

� an atomic mass unit (amu) is one twelfth the mass of a carbon-12 atom

� This gives us a basis for comparison

� The decimal numbers on the table are atomic masses in amu

They are not whole numbers

� Because they are based on averages of atoms and of isotopes.

� can figure out the average atomic mass from the mass of the isotopes and their relative abundance.

� add up the percent as decimals times the masses of the isotopes.

Examples

� There are two isotopes of carbon 12C with a mass of 12.00000 amu(98.892%), and 13C with a mass of 13.00335 amu (1.108%)

� There are two isotopes of nitrogen , one with an atomic mass of 14.0031 amu and one with a mass of 15.0001 amu. What is the percent abundance of each?

The Mole

� The mole is a number

� a very large number, but still, just a number

� 6.022 x 1023 of anything is a mole

� a large dozen

� The number of atoms in exactly 12 grams of carbon-12

The Mole

� Makes the numbers on the table the mass of the average atom

� Average atomic mass

� Just atomic mass

2

Molar mass

� mass of 1 mole of a substance

� often called molecular weight.

� To determine the molar mass of an element, look on the table.

� To determine the molar mass of a compound, add up the molar masses of the elements that make it up.

Find the molar mass of

� CH4

� Mg3P2

� Ca(NO3)2

� Al2(Cr2O7)3

� CaSO4 � 2H2O

Percent Composition� Percent of each element a compound is

composed of.

� Find the mass of each element, divide by the total mass, multiply by a 100.

� Easiest if you use a mole of the compound.

� find the percent composition of CH4� Al2(Cr2O7)3

� CaSO4 � 2H2O

Working backwards

� From percent composition, you can determine the empirical formula.

� Empirical Formula the lowest ratio of atoms in a molecule

� Based on mole ratios

� A sample is 59.53% C, 5.38%H, 10.68%N, and 24.40%O what is its empirical formula.

Pure O2 in CO2 is absorbed

H2O is absorbed

Sample is burned completely to form CO2 and H2O

� A 0.2000 gram sample of a compound (vitamin C) composed of only C, H, and O is burned completely with excess O2 . 0.2998 g of CO2 and 0.0819 g of H2O are produced. What is the empirical formula?

3

Empirical To Molecular Formulas

� Empirical is lowest ratio

� Molecular is actual molecule

� Need Molar mass

� Ratio of empirical to molar mass will tell you the molecular formula

� Must be a whole number because...

Example

� A compound is made of only sulfur and nitrogen. It is 69.6% S by mass. Its molar mass is 184 g/mol. What is its formula?

Chemical Equations

� Are sentences.

� Describe what happens in a chemical reaction.

� Reactants → Products

� Equations should be balanced

� Have the same number of each kind of atoms on both sides because ...

Balancing equations

CH4 + O2 → CO2 + H2OReactantsReactants ProductsProducts

C1 1

O2 3

H4 2

Balancing equationsBalancing equations

CH4 + O2 → CO2 + 2 H2OReactantsReactants ProductsProducts

C1 1

O2 3

H4 2 4


CH4 + O2 → CO2 + 2 H2OReactantsReactants ProductsProducts

C1 1

O2 3

H4 2 4

4

4


CH4 + 2O2 → CO2 + 2 H2OReactantsReactants ProductsProducts

C1 1

O2 3

H4 2 4

44

Abbreviations

� (s) , ↓ (for product)

� (g) , ↑ (for product)

� (aq)

� heat

� ∆

� catalyst

Practice� Ca(OH)2 + H3PO4 → H2O + Ca3(PO4)2

� KClO3(s) → Cl2(g) + O2(g)

� Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron(III) chloride and dihydrogen monosulfide gas.

� Fe2O3(s) + Al(s) → Fe(s) + Al2O3(s)

Meaning� A balanced equation can be used to

describe a reaction in molecules and atoms.

� Not grams.

� Chemical reactions happen molecules at a time

� or dozens of molecules at a time

� or moles of molecules.

Stoichiometry

� Given an amount of either starting material or product, determining the other quantities.

� use conversion factors from

– molar mass (g - mole)

– balanced equation (mole - mole)

� keep track

Examples

� One way of producing O2(g) involves the decomposition of potassium chlorate into potassium chloride and oxygen gas. A 25.5 g sample of Potassium chlorate is decomposed. How many moles of O2(g) are produced?

� How many grams of potassium chloride?

� How many grams of oxygen?

5

Examples

� A piece of aluminum foil 5.11 in x 3.23 in x 0.0381 in is dissolved in excess HCl(aq). How many grams of H2(g) are produced?

� How many grams of each reactant are needed to produce 15 grams of iron from the following reaction?

Fe2O3(s) + Al(s) → Fe(s) + Al2O3(s)

Examples

� K2PtCl4(aq) + NH3(aq) →Pt(NH3)2Cl2 (s)+ KCl(aq)

� what mass of Pt(NH3)2Cl2 can be produced from 65 g of K2PtCl4 ?

� How much KCl will be produced?

� How much from 65 grams of NH3?

Yield

How much you get from an chemical reaction

Limiting Reagent

� Reactant that determines the amount of product formed.

� The one you run out of first.

� Makes the least product.

� Book shows you a ratio method.

� It works.

� So does mine

Limiting reagent

� To determine the limiting reagent requires that you do two stoichiometry problems.

� Figure out how much product each reactant makes.

� The one that makes the least is the limiting reagent.

Example

� Ammonia is produced by the following reaction

N2 + H2 → NH3 What mass of ammonia can be produced from a mixture of 500. g N2and 100. g H2 ?

� How much unreacted material remains?

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Example

� A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. Which is the limiting reactant and how much excess reactant remains after the reaction has stopped?

� 4 NH3(g) + 5 O2(g)→4 NO(g) + 6 H2O(g)

Excess Reagent

� The reactant you don’t run out of.

� The amount of stuff you make is the yield.

� The theoretical yield is the amount you would make if everything went perfect.

� The actual yield is what you make in the lab.

Percent Yield

� % yield = Actual x 100%

Theoretical

� % yield = what you got x 100%

what you could have got

Examples

� Aluminum burns in bromine producing aluminum bromide. In a laboratory 6.0 g of aluminum reacts with excess bromine. 50.3 g of aluminum bromide are produced. What are the three types of yield.

Examples� Years of experience have proven that the

percent yield for the following reaction is 74.3%

Hg + Br2 → HgBr2If 10.0 g of Hg and 9.00 g of Br2 are reacted, how much HgBr2 will be produced?

� If the reaction did go to completion, how much excess reagent would be left?

Examples

� Commercial brass is an alloy of Cu and Zn. It reacts with HCl by the following reaction Zn(s) + 2HCl(aq) → ZnCl2 (aq) + H2(g) Cu does not react. When 0.5065 g of brass is reacted with excess HCl, 0.0985 g of ZnCl2 are eventually isolated. What is the composition of the brass?

1

More Preliminaries

Scientific Method

Metric SystemUncertainty

Complex sig figs

● What if it uses both addition and multiplication rules?

● Round when you change rules.

9.23 8.44.53

1.882

− × =

( )2.5 1.8 (3.3 2.7)× + × =

Scientific method.

● A way of solving problems● Observation- what is seen or measured● Hypothesis- educated guess of why

things behave the way they do. (possible explanation)

● Experiment- designed to test hypothesis● leads to new observations,● and the cycle goes on

Scientific method.● After many cycles, a broad, generalizable

explanation is developed for why things behave the way they do

● Theory● Also regular patterns of how things

behave the same in different systems emerges

● Law● Laws are summaries of observations● Often mathematical relationship

Scientific method.

● Theories have predictive value.● The true test of a theory is if it can

predict new behaviors.● If the prediction is wrong, the theory

must be changed.● Theory- why● Law – how● Law – equation of how things change

Observations

Hypothesis

Experiment

Law

Theory(Model)

Prediction

Experiment

Modify

2

Metric System

● Every measurement has two parts● Number● Scale (unit)● SI system (le Systeme International)

based on the metric system● Prefix + base unit● Prefix tells you the power of 10 to

multiply by - decimal system -easy conversions

Metric System

● Base Units● Mass - kilogram (kg)

● Length- meter (m)● Time - second (s)● Temperature- Kelvin (K)● Electric current- ampere (amp, A)● Amount of substance- mole (mol)

Prefixes● giga- G 1,000,000,000 109

● mega - M 1,000,000 106

● kilo - k 1,000 103

● deci- d 0.1 10-1

● centi- c 0.01 10-2

● milli- m 0.001 10-3

● micro- µ 0.000001 10-6

● nano- n 0.000000001 10-9

Deriving the Liter

● Liter is defined as the volume of 1 dm3

● gram is the mass of 1 cm3

Mass and Weight

● Mass is measure of resistance to change in motion

● Weight is force of gravity.● Sometimes used interchangeably● Mass can’t change, weight can

Uncertainty● Basis for significant figures ● All measurements are uncertain to

some degree● Precision- how repeatable ● Accuracy- how correct - closeness to

true value.● Random error - equal chance of being

high or low- addressed by averaging measurements - expected

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Uncertainty● Systematic error- same direction each

time● Want to avoid this● Bad equipment or bad technique.● Better precision implies better accuracy● You can have precision without

accuracy● You can’t have accuracy without

precision (unless you’re really lucky).

Dimensional Analysis

Using the units to solve problems

Dimensional Analysis● Use conversion factors to change the units● Conversion factors = 1

● 1 foot = 12 inches (equivalence statement)● 12 in = 1 = 1 ft.

1 ft. 12 in● 2 conversion factors● multiply by the one that will give you the

correct units in your answer.

Examples

● 11 yards = 2 rod● 40 rods = 1 furlong● 8 furlongs = 1 mile● The Kentucky Derby race is 1.25 miles.

How long is the race in rods, furlongs, meters, and kilometers?

● A marathon race is 26 miles, 385 yards. What is this distance in rods and kilometers?

Examples

● Science fiction often uses nautical analogies to describe space travel. If the starship U.S.S. Enterprise is traveling at warp factor 1.71, what is its speed in knots?

● Warp 1.71 = 5.00 times the speed of light● speed of light = 3.00 x 108 m/s● 1 knot = 2000 yd/h exactly

● Because you never learned dimensional analysis, you have been working at a fast food restaurant for the past 35 years wrapping hamburgers. Each hour you wrap 184 hamburgers. You work 8 hours per day. You work 5 days a week. you get paid every 2 weeks with a salary of $840.34. How many hamburgers will you have to wrap to make your first one million dollars?

Examples

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● A senior was applying to college and wondered how many applications she needed to send. Her counselor explained that with the excellent grade she received in chemistry she would probably be accepted to one school out of every three to which she applied. She immediately realized that for each application she would have to write 3 essays, and each essay would require 2 hours work. Of course writing essays is no simple matter. For each hour of serious essay writing, she would need to expend 500 calories which she could derive from her mother's apple pies. Every three times she cleaned her bedroom, her mother would made her an apple pie. How many times would she have to clean her room in order to gain acceptance to 10 colleges?

Units to a Power

● How many m3 is 1500 cm3?

1500 cm3 1 m100 cm

1 m100 cm

1 m100 cm

1500 cm3 1 m100 cm

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Units to a Power

● How many cm2 is 15 m2?● 36 cm3 is how many mm3?

Multiple units

● The speed limit is 65 mi/hr. What is this in m/s?– 1 mile = 1760 yds– 1 meter = 1.094 yds

65 mihr

1760 yd1 mi 1.094 yd

1 m 1 hr60 min

1 min60 s

Multiple units

● Lead has a density of 11.4 g/cm3. What is this in pounds per quart?

– 454 g = 1 lb– 1 L = 1.094 qt Temperature and Density

5

Temperature

● A measure of the average kinetic energy

● Different temperature scales, all are talking about the same height of mercury.

● Derive a equation for converting ºF toºC

0ºC 32ºF

0ºC = 32ºF

100ºC 212ºF

100ºC = 212ºF

0ºC = 32ºF

0ºC 32ºF 100ºC 212ºF0ºC 32ºF

100ºC = 212ºF

0ºC = 32ºF

100ºC = 180ºF

How much it changes

100ºC 212ºF0ºC 32ºF

100ºC = 212ºF

0ºC = 32ºF

100ºC = 180ºF

1ºC = (180/100)ºF

1ºC = 9/5ºF

How much it changes

ºC

ºF 9

5

0ºC is not 0ºF

6

ºC

ºF

(0,32)= (C1,F1)

ºC

ºF

(0,32) = (C1,F1)

(100,212) = (C2,F2)

Density

● Ratio of mass to volume● D = m/V● Useful for identifying a compound● Useful for predicting weight● An intrinsic property- does depend on

what the material is

Density Problem

● An empty container weighs 121.3 g. Filled with carbon tetrachloride (density 1.53 g/cm3 ) the container weighs 283.2 g. What is the volume of the container?

Density Problem

● A 55.0 gal drum weighs 75.0 lbs. when empty. What will the total mass be when filled with ethanol?

density 0.789 g/cm3

1 gal = 3.78 L1 lb = 454 g

Physical Changes

● A change that changes appearances, without changing the composition.

● Chemical changes - a change where a new form of matter is formed.

● Also called chemical reaction.● Not phase changes

– Ice is still water.

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Mixtures

● Made up of two substances.● Variable composition.● Heterogeneous- mixture is not the same

from place to place.● Chocolate chip cookie, gravel, soil.● Homogeneous- same composition

throughout.● Kool-aid, air.● Every part keeps its properties.

Separating mixtures

● Only a physical change- no new matter● Filtration- separate solids from liquids

with a barrier● Distillation- separate because of

different boiling points– Heat mixture– Catch vapor in cooled area

● Chromatography- different substances are attracted to paper or gel, so move at different speeds

Chromatography Phases

● A part of a sample with uniform composition, therefore uniform properties

● Homogeneous- 1 phase

● Heterogeneous – more than 1

Solutions

● Homogeneous mixture● Mixed molecule by molecule● Can occur between any state of matter.● Solid in liquid- Kool-aid● Liquid in liquid- antifreeze● Gas in gas- air● Solid in solid - brass● Liquid in gas- water vapor

Solutions

● Like all mixtures, they keep the properties of the components.

● Can be separated by physical means● Not easily separated- can be separated

without creating anything new.

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Substances

● Elements- simplest kind of matter● Cannot be broken down into simpler● All one kind of atom.● Compounds are substances that can be

broken down by chemical methods● When they are broken down, the pieces

have completely different properties than the compound. Salt

● Made of molecules- two or more atoms stuck together

Compound or Mixture

Compound Mixture

One kind of piece-Molecules

More than one kind -Molecule or atoms

Making is a chemical change

Making is a physical change

Only one kind Variable composition

Which is it?

ElementCompoundMixture

1

Liquids and solids

They are similar to each other�Different than gases.�They are incompressible.�Their density doesn’t change much

with temperature.�These similarities are due

• to the molecules staying close together in solids and liquids

• and far apart in gases�What holds them close together?

Intermolecular forces� Inside molecules (intramolecular) the

atoms are bonded to each other.� Intermolecular refers to the forces

between the molecules.�Holds the molecules together in the

condensed states.

Intermolecular forces�Strong

• covalent bonding• ionic bonding

�Weak• Dipole dipole• London dispersion forces

�During phase changes the molecules stay intact.

�Energy used to overcome forces.

Dipole - Dipole�Remember where the polar definition

came from?�Molecules line up in the presence of a

electric field. The opposite ends of the dipole can attract each other so the molecules stay close together.

�1% as strong as covalent bonds�Weaker with greater distance.�Small role in gases.

+- +-

+-

+

-

+

-

+-

+-+-

+-

+-

2

Hydrogen Bonding�Especially strong dipole-dipole forces

when H is attached to F, O, or N�These three because-

• They have high electronegativity.• They are small enough to get close.

�Effects boiling point.

CH4

SiH4

GeH4SnH4

PH3

NH3 SbH3

AsH3

H2O

H2SH2Se

H2TeHF

HI

HBrHCl

Boiling Points

0ºC

100

-100

200

Water

δ+δ-

δ+

Each water molecule can make up to four H-bonds

London Dispersion Forces�Non - polar molecules also exert

forces on each other.�Otherwise, no solids or liquids.�Electrons are not evenly distributed at

every instant in time.�Have an instantaneous dipole.� Induces a dipole in the atom next to it.� Induced dipole- induced dipole

interaction.

Example

H H H HH H H H

δ+ δ-

H H H H

δ+ δ- δ+ δ−

London Dispersion Forces�Weak, short lived.�Lasts longer at low temperature.�Eventually long enough to make liquids.�More electrons, more polarizable.�Bigger molecules, higher melting and

boiling points.�Weaker than other forces.

3

Van der Waal’s forces�London dispersion forces and Dipole

interactions�Order of increasing strength

• LDF• Dipole• H-bond• Real bonds

Liquids�Many of the properties due to

internal attraction of atoms.• Beading• Surface tension • Capillary action• Viscosity

�Stronger intermolecular forces cause each of these to increase.

Surface tension

�Molecules in the middle are attracted in all directions.

�Molecules at the the top are only pulled inside.

�Minimizes surface area.

Capillary Action�Liquids spontaneously rise in a

narrow tube.� Intermolecular forces are cohesive,

connecting like things.�Adhesive forces connect to

something else.�Glass is polar.

• It attracts water molecules.

Beading� If a polar substance

is placed on a non-polar surface. • There are cohesive,• But no adhesive

forces.

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Viscosity�How much a liquid resists flowing.�Large forces, more viscous.�Large molecules can get tangled up.�Cyclohexane has a lower viscosity

than hexane.�Because it is a circle- more compact.

How much of these?�Stronger forces, bigger effect.

• Hydrogen bonding• Dipole-dipole• LDF

� In that order

•H next to O,N, or F•Polar molecules•All molecules

Model of a Liquid�Can’t see molecules so picture them

as-� In motion but attracted to each other�With regions arranged like solids but

• with higher disorder.• with fewer holes than a gas.• Highly dynamic, regions changing

between types.

Phases�The phase of a substance is

determined by three things.• The temperature.• The pressure.• The strength of intermolecular forces.

Solids�Two major types.�Amorphous- those with much

disorder in their structure.�Crystalline- have a regular

arrangement of components in their structure.

Crystals�Lattice- a three dimensional grid that

describes the locations of the pieces in a crystalline solid.

�Unit Cell-The smallest repeating unit in of the lattice.

�Three common types.

5

Cubic Body-Centered Cubic

Face-Centered Cubic The book drones on about�Using diffraction patterns to identify

crystal structures.�Talks about metals and the closest

packing model.� It is interesting, but trivial.�We need to focus on metallic bonding.�Why do metal atoms stay together?�How their bonding affects their

properties.

Solids�There are many amorphous solids.�Like glass.�We tend to focus on crystalline solids.� two types.

• Ionic solids have ions at the lattice points.

• Molecular solids have molecules.�Sugar vs. Salt.

Metallic Bonds�How atoms are held together in the

solid.�Metals hold onto their valence

electrons very weakly.�Think of them as positive ions

floating in a sea of electrons.

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Sea of Electrons

+ + + ++ + + +

+ + + +

�Electrons are free to move through the solid.

�Metals conduct electricity.

Metals are Malleable�Hammered into shape (bend).�Ductile - drawn into wires.�Because of mobile valence electrons

Malleable

+ + + ++ + + +

+ + + +

Malleable

+ + + +

+ + + ++ + + +

�Electrons allow atoms to slide by but still be attracted.

Metallic bonding

1s

2s

2p

3s

3p

Filled Molecular Orbitals

Empty Molecular Orbitals

Magnesium Atoms



The 1s, 2s, and 2p electrons are close to

nucleus, so they are not able to move

around.

1s

2s

2p

3s

3p

Magnesium Atoms

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1s

2s

2p

3s

3p

Magnesium Atoms

The 3s and 3p orbitals overlap and form

molecular orbitals.



1s

2s

2p

3s

3p

Magnesium Atoms

Electrons in these energy levels can

travel freely throughout the crystal.



1s

2s

2p

3s

3p

Magnesium Atoms

This makes metals conductors

Malleable because the bonds are flexible.

Carbon- A Special Atomic Solid�There are three types of solid carbon.�Amorphous- soot - uninteresting.�Diamond- hardest natural substance

on earth, insulates both heat and electricity.

�Graphite- slippery, conducts electricity.

�How the atoms in these network solids are connected explains why.

Diamond- each Carbon is sp3

hybridized, connected to four other carbons.

�Carbon atoms are locked into tetrahedral shape.

�Strong σσσσ bonds give the huge molecule its hardness.

Why is it an insulator?

All the electrons need to be shared in the covalent bonds

Can’t move around

8

�Each carbon is connected to threeother carbons and sp2 hybridized.

�The molecule is flat with 120º angles in fused 6 member rings.

�The ππππ bonds extend above and below the plane.

Graphite is different. This π bond overlap forms a huge π bonding network.

�Electrons are free to move throughout these delocalized orbitals.

�Conductselectricity

�The layers slideby each other.

�Lubricant

Molecular solids.�Molecules occupy the corners of the

lattices.�Different molecules have different

forces between them.�These forces depend on the size of

the molecule.�They also depend on the strength

and nature of dipole moments.

Those without dipoles.

� Most are gases at 25ºC.�The only forces are London Dispersion

Forces.�These depend on number of electrons.�Large molecules (such as I 2 ) can be

solids even without dipoles. (LDF)

Those with dipoles.�Dipole-dipole forces are generally

stronger than L.D.F.�Hydrogen bonding is stronger than

Dipole-dipole forces.�No matter how strong the

intermolecular force, it is always much, much weaker than the forces in bonds.

�Stronger forces lead to higher melting and freezing points.

Water is special

HO

H

δδδδ++++

δδδδ++++

δδδδ-

�Each molecule has two polar O-H bonds.

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Water is special

HO

H

δδδδ++++

δδδδ++++

�Each molecule has two polar O-H bonds.

�Each molecule has two lone pair on its oxygen.

Water is special�Each molecule has two polar

O-H bonds.�Each molecule has two lone

pair on its oxygen.�Each oxygen can interact with

2 hydrogen atoms.

HO

H

δδδδ++++

δδδδ++++

Water is special

H

O

H

δδδδ++++

δδδδ++++

H

O

H

δδδδ++++

δδδδ++++

H

O

H

δδδδ++++

δδδδ++++

�This gives water an especially high melting and boiling point.

Ionic Solids�The extremes in dipole-dipole forces-

atoms are actually held together by opposite charges.

�Huge melting and boiling points.�Atoms are locked in lattice so hard and

brittle.�Every electron is accounted for so they

are poor conductors-good insulators.�Until melted or dissolved.

Phase Changes

Vapor Pressure� Vaporization - change from

liquid to gas at boiling point.� Evaporation - change from

liquid to gas below boiling point

� Heat (or Enthalpy) of

Vaporization ( ∆∆∆∆Hvap )- the

energy required to vaporize

1 mol at 1 atm.

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�Vaporization is an endothermic process - it requires heat.

�Energy is required to overcome intermolecular forces.

�Responsible for cool beaches.�Why we sweat.

Condensation�Change from gas to liquid.�Achieves a dynamic equilibrium with

vaporization in a closed system.�What is a closed system?�A closed system means matter

can’t go in or out. �Put a cork in it.�What the heck is a “dynamic

equilibrium?”

�When first sealed the molecules gradually escape the surface of the liquid

Dynamic equilibrium�When first sealed the

molecules gradually escape the surface of the liquid

�As the molecules build up above the liquid some condense back to a liquid.

Dynamic equilibrium

�As time goes by the rate of vaporization remains constant

� but the rate of condensation increases because thereare more molecules to

condense.�Equilibrium is reached

when

Dynamic equilibriumRate of Vaporization =

Rate of Condensation�Molecules are constantly changing

phase “Dynamic”�The total amount of liquid and vapor

remains constant “Equilibrium”

Dynamic equilibrium

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Vapor pressure�The pressure above the liquid at

equilibrium.�Liquids with high vapor pressures

evaporate easily. �They are called volatile.�Decreases with increasing

intermolecular forces. • Bigger molecules (bigger LDF)• More polar molecules (dipole-dipole)

Vapor pressure� Increases with increasing

temperature.�Easily measured in a barometer.

Dish of Hg

Vacuum

Patm=

760 torr

A barometer will hold a column of mercury 760 mm high at one atm

Dish of Hg

Vacuum

Patm=

760 torr

A barometer will hold a column of mercury 760 mm high at one atm.

If we inject a volatile liquid in the barometer it will rise to the top of the mercury.

Dish of Hg

Patm=

760 torr

A barometer will hold a column of mercury 760 mm high at one atm.

If we inject a volatile liquid in the barometer it will rise to the top of the mercury.

There it will vaporize and push the column of mercury down.

Water

Dish of Hg

736 mm Hg

Water Vapor

�The mercury is pushed down by the vapor pressure.

�Patm = PHg + Pvap

�Patm - PHg = Pvap

� 760 - 736 = 24 torr

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Temperature Effect

Kinetic energy

# of

mol

ecul

es

T1

Energy needed to overcome intermolecular forces

Kinetic energy

# of

mol

ecul

es

T1

Energy needed to overcome intermolecular forcesT1

T2

� At higher temperature more molecules have enough energy - higher vapor pressure.

Energy needed to overcome intermolecular forces

Mathematical relationship

� ln is the natural logarithm• ln = Log base e• e = Euler’s number an irrational number

like π�∆∆∆∆Hvap is the heat of vaporization in J/mol

1

2

vapT vap

vapT 2 1

P H 1 1ln = -

P R T T

∆

�R = 8.3145 J/K mol.�Surprisingly this is the graph of a

straight line. � If you graph ln P vs 1/T

Mathematical relationship1

2

vapT vap

vapT 2 1

P H 1 1ln = -

P R T T

∆

�The vapor pressure of water is 23.8 torr at 25°C. The heat of vaporization of water is 43.9 kJ/mol. Calculate the vapor pressure at 50°C

�At what temperature would it have a vapor pressure of 760 torr?

Mathematical relationship1

2

vapT vap

vapT 2 1

P H 1 1ln = -

P R T T

∆

Changes of state�The graph of temperature versus

heat applied is called a heating curve.

�The temperature a solid turns to a liquid is the melting point.

�The energy required to accomplish this change is called the Heat (or Enthalpy) of Fusion ∆∆∆∆Hfus

13

-40

-20

0

20

40

60

80

100

120

140

0 10 90 190 730 740

Heating Curve for Water

IceWater and Ice

Water

Water and Steam

Steam

-40

-20

0

20

40

60

80

100

120

140

0 10 90 190 730 740

Heating Curve for Water

Heat of Fusion

Heat ofVaporization

1/ Slope is Heat Capacity

Melting Point�Melting point is determined by the

vapor pressure of the solid and the liquid.

�At the melting point the vapor pressure of the solid =

vapor pressure of the liquidSolid Water

Liquid Water

Water Vapor Vapor

Solid Water

Liquid Water

Water Vapor Vapor

� If the vapor pressure of the solid is higher than that of the liquid the solid will release molecules to achieve equilibrium.

Solid Water

Liquid Water

Water Vapor Vapor

�While the molecules of condense to a liquid.

14

�This can only happen if the temperature is above the freezing point since solid is turning to liquid.

Solid Water

Liquid Water

Water Vapor Vapor

� If the vapor pressure of the liquid is higher than that of the solid, the liquid will release molecules to achieve equilibrium.

Solid Water

Liquid Water

Water Vapor Vapor

Solid Water

Liquid Water

Water Vapor Vapor

�While the molecules condense to a solid.

�The temperature must be below the freezing point since the liquid is turning to a solid.

Solid Water

Liquid Water

Water Vapor Vapor

� If the vapor pressure of the solid and liquid are equal, the solid and liquid are vaporizing and condensing at the same rate. The Melting point .

Solid Water

Liquid Water

Water Vapor Vapor

Boiling Point�Reached when the vapor pressure

equals the external pressure.�Normal boiling point is the boiling

point at 1 atm pressure.�Superheating - Heating above the

boiling point.�Supercooling - Cooling below the

freezing point.

15

Phase Diagrams.�A plot of temperature versus

pressure for a closed system, with lines to indicate where there is a phase change.

Temperature

SolidLiquid

Gas

1 Atm

AA

BB

CCD

D D

Pre

ssu

re

D

SolidLiquid

Gas

Triple Point

Critical Point

Temperature

Pre

ssu

re SolidLiquid

Gas

�This is the phase diagram for water.�The density of liquid water is higher

than solid water.

Temperature

Pre

ssu

re

Solid Liquid

Gas

1 Atm

�This is the phase diagram for CO 2

�The solid is more dense than the liquid�The solid sublimes at 1 atm.

Temperature

Pre

ssu

re

16

1

Page 1

Solutions Occur in all phases�The solvent does the dissolving.�The solute is dissolved.�There are examples of all types of

solvents dissolving all types of solvent.

�We will focus on aqueous solutions.

Ways of Measuring

�Molarity = moles of soluteLiters of solute

�% mass = Mass of solute x 100Mass of solution

�Mole fraction of component AχχχχA = NA

NA + NB

�Molality = moles of solute Kilograms of solvent

�Molality is abbreviated m�Normality - read but don’t focus on it.� It is molarity x number of active

pieces

Ways of Measuring

Energy of Making Solutions�Heat of solution ( ∆∆∆∆Hsoln ) is the energy

change for making a solution.�Most easily understood if broken into

steps.�1.Break apart solvent �2.Break apart solute�3. Mixing solvent and solute

1. Break apart Solvent�Have to overcome attractive forces.

∆∆∆∆H1 >0

2. Break apart Solute.�Have to overcome attractive forces.

∆∆∆∆H2 >0

2

Page 2

3. Mixing solvent and solute�∆∆∆∆H3 depends on what you are mixing.� If molecules can attract each other

∆∆∆∆H3 is large and negative.�Molecules can’t attract- ∆∆∆∆H3 is small

and negative.�This explains the rule “Like dissolves

Like”

Energy

Reactants

Solution

∆∆∆∆H1

∆∆∆∆H2

∆∆∆∆H3

Solvent

Solute and Solvent

�Size of ∆∆∆∆H3 helps determine whether a solution will form

∆∆∆∆H3

Solution

Types of Solvent and solutes

� If ∆∆∆∆Hsoln is small and positive, a solution will still form because of entropy.

�There are many more ways for them to become mixed than there is for them to stay separate.

Structure and Solubility�Water soluble molecules must have

dipole moments -polar bonds.�To be soluble in nonpolar solvents

the molecules must be non polar.�Read Vitamin A - Vitamin C

discussion pg. 509

Soap

P O-

CH3

CH2CH2

CH2CH2

CH2

CH2

CH2

O-

O-

Soap

�Hydrophobic non-polar end

P O-

CH3

CH2CH2

CH2CH2

CH2

CH2

CH2

O-

O-

3

Page 3

Soap

�Hydrophilic polar end

P O-

CH3

CH2CH2

CH2CH2

CH2

CH2

CH2

O-

O-

P O-

CH3

CH2CH2

CH2CH2

CH2

CH2

CH2

O-

O-

_

�A drop of grease in water�Grease is non-polar�Water is polar�Soap lets you dissolve the non-polar

in the polar.

Hydrophobic ends dissolve in

grease

Hydrophilic ends dissolve in water

�Water molecules can surround and dissolve grease.

�Helps get grease out of your way.

4

Page 4

Pressure effects�Changing the pressure doesn’t affect

the amount of solid or liquid that dissolves

�They are incompressible.� It does affect gases.

Dissolving Gases�Pressure affects the

amount of gas that can dissolve in a liquid.

�The dissolved gas is at equilibrium with the gas above the liquid.

�The gas is at equilibrium with the dissolved gas in this solution.

�The equilibrium is dynamic.

� If you increase the pressure the gas molecules dissolve faster.

�The equilibrium is disturbed.

�The system reaches a new equilibrium with more gas dissolved.

�Henry’s Law.P= kC

Pressure = constant x Concentration

of gasThe stronger the attraction

of the two, the higher the constant.

Temperature Effects

� Increased temperature increases the rate at which a solid dissolves.

�We can’t predict whether it will increase the amount of solid that dissolves.

�We must read it from a graph of experimental data.

5

Page 5

20 40 60 80 100

Gases are predictable

�As temperature increases, solubility decreases.

�Gas molecules can move fast enough to escape.

�Thermal pollution.

Vapor Pressure of Solutions�A nonvolatile solvent lowers the

vapor pressure of the solution.�The molecules of the solvent

must overcome the force of both the other solvent molecules and thesolute molecules.

Raoult’s Law:�Psoln = χχχχsolvent x Psolvent�Vapor pressure of the solution =

mole fraction of solvent xvapor pressure of the pure solvent

�Applies only to an ideal solution where the solute doesn’t contribute to the vapor pressure.

Aqueous Solution

Pure water

�Water has a higher vapor pressure than a solution

Aqueous Solution

Pure water

�Water evaporates faster from for water than solution

6

Page 6

�The water condenses faster in the solution so it should all end up there.

Aqueous Solution

Pure water

Practice Problem�A solution of cyclopentane with a

nonvolatile compound has vapor pressure of 211 torr. If vapor pressure of the pure liquid is 313 torr, what is the mole fraction of the cyclopentane?

Please enter your answer�Determine the vapor pressure of a

solution at 25 °°°° C that has 45 grams of C6H12O6, glucose, dissolved in 72 grams of H 2O. The vapor pressure of pure water at 25 °°°° C is 23.8 torr.

�What is the composition of a pentane-hexane solution that has a vapor pressure of 350 torr at 25ºC ?

�The vapor pressures at 25ºC are• pentane 511 torr• hexane 150 torr.

�What is the composition of the vapor?

Practice Question

�Liquid-liquid solutions where both are volatile.

�Modify Raoult’s Law to

�Ptotal = PA + PB = χχχχAPA0 + χχχχBPB

0

�Ptotal = vapor pressure of mixture� PA

0= vapor pressure of pure A� If this equation works then the solution

is ideal.

Ideal solutions

χb

χA

Va

por

Pre

ssur

e

P of pure A

P of pure B

Vapor Pressure of solution

7

Page 7

Deviations� If solvent has a strong affinity for

solute (H bonding).�Lowers solvent’s ability to escape.�Lower vapor pressure than expected.�Negative deviation from Raoult’s law.�∆∆∆∆Hsoln is large and negative

exothermic.�Endothermic ∆∆∆∆Hsoln indicates positive

deviation.χb

χA

Va

por

Pre

ssur

e

Positive deviations-

Weak attraction between solute and solvent

Positive ∆Hsoln

χb

χA

Va

por

Pre

ssur

e

Negative deviations-

Strong attraction between solute and solvent

Negative ∆Hsoln

Colligative Properties�Because dissolved particles affect

vapor pressure - they affect phase changes.

�Colligative properties depend only on the number - not the kind of solute particles present

�Useful for determining molar mass

Boiling point Elevation

�Because a non-volatile solute lowers the vapor pressure it raises the boiling point.

� The equation is: ∆∆∆∆T = Kbmmmmsolute�∆∆∆∆T is the change in the boiling point�Kb is a constant determined by the

solvent.�mmmmsolute is the molality of the solute

Freezing point Depression

�Because a non-volatile solute lowers the vapor pressure of the solution it lowers the freezing point.

� The equation is: ∆∆∆∆T = -Kfmmmmsolute�∆∆∆∆T is the change in the freezing point�Kf is a constant determined by the

solvent�mmmmsolute is the molality of the solute

8

Page 8

1 atm

Vapor Pressure of solution

Vapor Pressure of pure water

1 atm

Freezing and boiling points of solvent

1 atm

Freezing and boiling points of solvent

1 atm

∆Tf∆Tb

Electrolytes in solution�Since colligative properties only

depend on the number of molecules.� Ionic compounds should have a

bigger effect.�When they dissolve they dissociate.� Individual Na and Cl ions fall apart.�1 mole of NaCl makes 2 moles of ions.�1mole Al(NO 3)3 makes 4 moles ions.

�Electrolytes have a bigger impact on on melting and freezing points per mole because they make more pieces.

�Relationship is expressed using the van’t Hoff factor i

i = Moles of particles in solutionMoles of solute dissolved

�The expected value can be determined from the formula of the compound.

9

Page 9

�The actual value is usually less because

�At any given instant some of the ions in solution will be paired up.

� Ion pairing increases with concentration.

� i decreases with increasing concentration.

�We can change our formulas to∆Τ∆Τ∆Τ∆Τ = iKm

LAB�Purpose: to experimentally

determine the van’t Hoff factor for sodium chloride

�Materials and equipment• Sodium chloride Water• Food coloring• Beakers Thermometer• Graduated cylinder Ice cube tray• Foam cup

Lab�1. Make approximately 0.50 m , 1.0 m,

and 1.5 m NaCl solutions�2. Add a different color of food

coloring for each�3. Put in labeled ice tray�4. Freeze overnight�5. Melt the ice cubes in their own

solutions and determine the freezing point depression

Lab�Calculations�1. Determine the van’t Hoff factor for

sodium chloride in each solution.�Error analysis and conclusion

1

Page 1

Kinetics

� The study of reaction rates.� Spontaneous reactions are reactions

that will happen - but we can’t tell how fast.

� Diamond will spontaneously turn to graphite – eventually.

� Reaction mechanism- the steps by which a reaction takes place.

Review- Collision Theory� Particles have to collide to react.� Have to hit hard enough� Things that increase this increase rate� High temp – faster reaction� High concentration – faster reaction� Small particles = greater surface area

means faster reaction

Reaction Rate

� Rate = Conc. of A at t2 -Conc. of A at t1t2- t1

� Rate = ∆∆∆∆[A]∆t

� Change in concentration per unit time� For this reaction� N2 + 3H2 2NH3

� As the reaction progresses the concentration H2 goes down

Concentration

Time

[H2]

N2 + 3H2 → 2NH3

� As the reaction progresses the concentration N2 goes down 1/3 as fast

Concentration

Time

[H2]

[N2]

N2 + 3H2 → 2NH3

� As the reaction progresses the concentration NH3 goes up 2/3 times

Concentration

Time

[H2]

[N2]

[NH3]

N2 + 3H2 → 2NH3

2

Page 2

Calculating Rates

� Average rates are taken over long intervals

� Instantaneous rates are determined by finding the slope of a line tangent to the curve at any given point because the rate can change over time

� Derivative.

� Average slope method

Concentration

Time

∆[H2]

∆t

� Instantaneous slope method.

Concentration

Time

∆[H2]∆ t

d[H2]dt

Defining Rate�We can define rate in terms of the

disappearance of the reactant or in terms of the rate of appearance of the product.

� In our exampleN2 + 3H2 2NH3

�∆[H2] = 3∆[N2] ∆t ∆t

�∆[NH3] = -2∆[N2] ∆t ∆t

Rate Laws

� Reactions are reversible.� As products accumulate they can begin

to turn back into reactants.� Early on the rate will depend on only the

amount of reactants present.� We want to measure the reactants as

soon as they are mixed.� This is called the Initial rate method.

� Two key points� The concentration of the products do

not appear in the rate law because this is an initial rate.

� The order (exponent) must be determined experimentally,

� can’t be obtained from the equation

Rate Laws

3

Page 3

� You will find that the rate will only depend on the concentration of the reactants. (Initially)

� Rate = k[NO2]n

� This is called a rate law expression.� k is called the rate constant.� n is the order of the reactant -usually a

positive integer.

2 NO2 2 NO + O2

� The rate of appearance of O2 can be said to be.

� Rate' = ∆[O2] = k'[NO2]∆t

� Because there are 2 NO2 for each O2

� Rate = 2 x Rate'� So k[NO2]

n = 2 x k'[NO2]n

� So k = 2 x k'

2 NO2 2 NO + O2

Types of Rate Laws� Differential Rate law - describes how rate

depends on concentration.� Integrated Rate Law - Describes how

concentration depends on time.� For each type of differential rate law

there is an integrated rate law and vice versa.

� Rate laws can help us better understand reaction mechanisms.

Determining Rate Laws� The first step is to determine the form of

the rate law (especially its order).� Must be determined from experimental

data.� For this reaction

2 N2O5 (aq) 4NO2 (aq) + O2(g)� The reverse reaction won’t play a role

because the gas leaves

[N2O5] (mol/L) Time (s)

1.00 0

0.88 200

0.78 400

0.69 600

0.61 800

0.54 1000

0.48 1200

0.43 1400

0.38 1600

0.34 1800

0.30 2000

Now graph the data

00.10.20.30.40.50.60.70.80.9

1

0 200

400

600

800

1000

1200

1400

1600

1800

2000

� To find rate we have to find the slope at two points

� We will use the tangent method.

4

Page 4

00.10.20.30.40.50.60.70.80.9

1

0 200

400

600

800

1000

1200

1400

1600

1800

2000

At .80 M the rate is (.88 - .68) = 0.20 =- 5.0x 10 -4

(200-600) -400

00.10.20.30.40.50.60.70.80.9

1

0 200

400

600

800

1000

1200

1400

1600

1800

2000

At .40 M the rate is (.52 - .32) = 0.20 =- 2.5 x 10 -4

(1000-1800) -800

� Since the rate at twice as fast when the concentration is twice as big the rate law must be..

� First power

� Rate = -∆[N2O5] = k[N2O5]1 = k[N2O5]∆t

� We say this reaction is first order in N2O5� The only way to determine order is to run

the experiment.

The method of Initial Rates

� This method requires that a reaction be run several times.

� The initial concentrations of the reactants are varied.

� The reaction rate is measured just after the reactants are mixed.

� Eliminates the effect of the reverse reaction.

An example

� For the reactionBrO3

- + 5 Br- + 6H+ 3Br2 + 3 H2O

� The general form of the Rate Law is

Rate = k[BrO3-]n[Br-]m[H+]p

� We use experimental data to determine the values of n,m,and p

Initial concentrations (M)

Rate (M/s)

BrO3- Br- H+

0.10 0.10 0.10 8.0 x 10-4

0.20 0.10 0.10 1.6 x 10-3

0.20 0.20 0.10 3.2 x 10-3

0.10 0.10 0.20 3.2 x 10-3

Now we have to see how the rate changes with concentration

5

Page 5

Integrated Rate Law

� Expresses the reaction concentration as a function of time.

� Form of the equation depends on the order of the rate law (differential).

� Changes Rate = ∆[A]n

∆t� We will only work with n=0, 1, and 2

First Order� For the reaction 2N2O5 4NO2 + O2

� We found the Rate = k[N2O5]1

� If concentration doubles rate doubles.� If we integrate this equation with respect

to time we get the Integrated Rate Law� ln[N2O5] = - kt + ln[N2O5]0� ln is the natural ln� [N2O5]0 is the initial concentration.

� General form Rate = ∆[A] / ∆t = k[A]� ln[A] = - kt + ln[A]0� In the form y = mx + b� y = ln[A] m = -k� x = t b = ln[A]0� A graph of ln[A] vs time is a straight

line.

First Order

� By getting the straight line you can prove it is first order

� Often expressed in a ratio

First Order

� By getting the straight line you can prove it is first order

� Often expressed in a ratio

First Order

[ ][ ]lnA

A = kt0

Half Life� The time required to reach half the

original concentration.� If the reaction is first order� [A] = [A]0/2 when t = t1/2

6

Page 6

Half Life• The time required to reach half the

original concentration.• If the reaction is first order• [A] = [A]0/2 when t = t1/2

[ ]

[ ]ln

A

A = kt0

01 2

2

�ln(2) = kt1/2

Half Life� t1/2 = 0.693 / k� The time to reach half the original

concentration does not depend on the starting concentration.

� An easy way to find k

� The highly radioactive plutonium in nuclear waste undergoes first-order decay with a half-life of approximately 24,000 years. How many years must pass before the level of radioactivity due to the plutonium falls to 1/128th (about 1%) of its original potency?

Second Order

� Rate = -∆[A]/∆t = k[A]2

� integrated rate law � 1/[A] = kt + 1/[A]0� y= 1/[A] m = k� x= t b = 1/[A]0� A straight line if 1/[A] vs t is graphed� Knowing k and [A]0 you can calculate [A]

at any time t

Second Order Half Life

� [A] = [A]0 /2 at t = t1/2

1

20

2[ ]A = kt + 1

[A]10

22[ [A]

- 1A]

= kt0 0

1

tk[A]1 =

1

02

1

[A] = kt

01 2

Zero Order Rate Law� Rate = k[A]0 = k� Rate does not change with concentration.� Integrated [A] = -kt + [A]0� When [A] = [A]0 /2 t = t1/2

� t1/2 = [A]0 /2k

7

Page 7

� Most often when reaction happens on a surface because the surface area stays constant.

� Also applies to enzyme chemistry.

Zero Order Rate Law

Time

Concentration

Time

Concentration

∆[∆[∆[∆[A]/∆∆∆∆t

∆∆∆∆t

k =

∆[∆[∆[∆[A]

Summary of Rate Laws

More Complicated Reactions� BrO3

- + 5 Br- + 6H+ 3Br2 + 3 H2O

� For this reaction we found the rate law to be

� Rate = k[BrO3-][Br-][H+]2

� To investigate this reaction rate we need to control the conditions

Rate = k[BrO3-][Br-][H+]2

� We set up the experiment so that two of the reactants are in large excess.

� [BrO3-]0= 1.0 x 10-3 M

� [Br-]0 = 1.0 M� [H+]0 = 1.0 M� As the reaction proceeds [BrO3

-] changes noticeably

� [Br-] and [H+] don’t

8

Page 8

� This rate law can be rewritten� Rate = k[BrO3

-][Br-]0[H+]0

2

� Rate = k[Br-]0[H+]0

2[BrO3-]

� Rate = k’[BrO3-]

� This is called a pseudo first order rate law.

�k = k’[Br-]0[H

+]02

Rate = k[BrO3-][Br-][H+]2

Reaction Mechanisms� The series of steps that actually occur in

a chemical reaction.� Kinetics can tell us something about the

mechanism� A balanced equation does not tell us

how the reactants become products.

� 2NO2 + F2 2NO2F � Rate = k[NO2][F2]� The proposed mechanism is� NO2 + F2 NO2F + F (slow)� F + NO2 NO2F (fast) � F is called an intermediate It is formed

then consumed in the reaction

Reaction Mechanisms� Each of the two reactions is called an

elementary step .� The rate for a reaction can be written

from its molecularity .� Molecularity is the number of pieces

that must come together.� Elementary steps add up to the

balanced equation

Reaction Mechanisms

� Unimolecular step involves one molecule - Rate is first order.

� Bimolecular step - requires two molecules - Rate is second order

� Termolecular step- requires three molecules - Rate is third order

� Termolecular steps are almost never heard of because the chances of three molecules coming into contact at the same time are miniscule.

� A products Rate = k[A]� A+A products Rate= k[A]2

� 2A products Rate= k[A]2

� A+B products Rate= k[A][B]� A+A+B Products Rate= k[A]2[B]� 2A+B Products Rate= k[A]2[B]� A+B+C Products Rate= k[A][B][C]

Molecularity and Rate Laws

9

Page 9

How to get rid of intermediates� They can’t appear in the rate law.� Slow step determines the rate and the rate

law� Use the reactions that form them� If the reactions are fast and irreversible

the concentration of the intermediate is based on stoichiometry.

� If it is formed by a reversible reaction set the rates equal to each other.

Formed in reversible reactions� 2 NO + O2 2 NO2

� Mechanism� 2 NO N2O2 (fast)� N2O2 + O2 2 NO2 (slow)� rate = k2[N2O2][O2]� k1[NO]2 = k-1[N2O2] (equilibrium)� Rate = k2 (k1/ k-1)[NO]2[O2]� Rate =k [NO]2[O2]

Formed in fast reactions� 2 IBr I2+ Br2

� Mechanism� IBr I + Br (fast)� IBr + Br I + Br2 (slow)� I + I I2 (fast)� Rate = k[IBr][Br] but [Br]= [IBr] because

the first step is fast� Rate = k[IBr][IBr] = k[IBr]2

� 2 NO2Cl → 2 NO2 + Cl2(balanced equation)

NO2Cl → NO2 + Cl (slow)NO2Cl + Cl → NO2 + Cl2 (fast)

Collision theory� Molecules must collide to react.� Concentration affects rates because

collisions are more likely.� Must collide hard enough.� Temperature and rate are related.� Only a small number of collisions

produce reactions.

Potential

Energy

Reaction Coordinate

Reactants

Products

10

Page 10

Potential

Energy

Reaction Coordinate

Reactants

Products

Activation Energy Ea

Potential

Energy

Reaction Coordinate

Reactants

Products

Activated complex

Potential

Energy

Reaction Coordinate

Reactants

Products∆∆∆∆E}

Potential

Energy

Reaction Coordinate

2BrNO

2NO + Br

Br---NO

Br---NO

2

Transition State

Terms� Activation energy - the minimum energy

needed to make a reaction happen.� Activated Complex or Transition State -

The arrangement of atoms at the top of the energy barrier.

Arrhenius� Said the at reaction rate should

increase with temperature.� At high temperature more molecules

have the energy required to get over the barrier.

� The number of collisions with the necessary energy increases exponentially.

11

Page 11

Arrhenius� Number of collisions with the required

energy = ze-Ea/RT

� z = total collisions

� e is Euler’s number (inverse of ln)� Ea = activation energy� R = ideal gas constant (in J/K mol)

� T is temperature in Kelvin

Problem with this� Observed rate is too small � Due to molecular orientation- the have

to be facing the right way� written into equation as p the steric

factor.

ON

Br

ON

Br

O N Br ONBr ONBr

O NBr

O N BrONBr No Reaction

O NBr

O NBr

Arrhenius Equation�k = zpe-Ea/RT = Ae-Ea/RT

� A is called the frequency factor = zp� k is the rate constant� ln k = -(Ea/R)(1/T) + ln A� Another line !!!!� ln k vs 1/T is a straight line� With slope Ea/R so we can find Ea

� And intercept ln A

� A reaction is found to have a rate constant of 8.60x10-1sec-1 at 523 K and an activation energy of 120.8 kJ/mol. What is the value of the rate constant at 270 K?

Which statement is true concerning the plot of rate constants at various temperatures for a particular reaction?

A) A steep slope of the ln k versus 1/T plot is indicative of small changes in the rate constant for a given increase in temperature.

B) Different sections of the ln k versus 1/T plot show different Ea values.

C) The plot of k versus T shows a linear increase in k as the temperature increases.

D) A steep slope of the ln k versus 1/T plot is indicative of a large Ea.

E) The y-intercept of the ln k versus 1/T plot is the Ea value for that reaction

12

Page 12

Activation Energy and Rates

The final saga

Mechanisms and rates � There is an activation energy for each

elementary step.� Activation energy determines k.

� k = Ae- (Ea/RT)

� k determines rate� Slowest step (rate determining) must

have the highest activation energy.

• This reaction takes place in three steps

☺

Ea

First step is fastLow activation energy

Second step is slow

High activation energy

Ea

☺

Ea

Third step is fast

Low activation energy

☺

13

Page 13

Second step is rate determining

Intermediates are present

Activated Complexes or Transition States

Catalysts� Speed up a reaction without being used

up in the reaction.� Enzymes are biological catalysts.� Homogenous Catalysts are in the same

phase as the reactants.� Heterogeneous Catalysts are in a

different phase as the reactants.

How Catalysts Work� Catalysts allow reactions to proceed by a

different mechanism - a new pathway.� New pathway has a lower activation

energy.� More molecules will have this activation

energy.� Does not change ∆E� Show up as a reactant in one step and a

product in a later stepPt surface

HH

HH

HH

HH

� Hydrogen bonds to surface of metal.

� Break H-H bonds

Heterogenous Catalysts

14

Page 14

Pt surface

HH

HH


C HH C

HH

Pt surface

HH

HH


C HH C

HH

� The double bond breaks and bonds to the catalyst.

Pt surface

HH

HH


C HH C

HH

� The hydrogen atoms bond with the carbon

Pt surfaceH


C HH C

HH

H HH

Homogenous Catalysts� Chlorofluorocarbons (CFCs) catalyze

the decomposition of ozone.� Enzymes regulating the body

processes. (Protein catalysts)

Catalysts and rate� Catalysts will speed up a reaction but

only to a certain point.� Past a certain point adding more

reactants won’t change the rate.� Zero Order

15

Page 15

Catalysts and rate.

Concentration of reactants

Rate

� Rate increases until the active sites of catalyst are filled.

� Then rate is independent of concentration

1

Equilibrium

Reactions are reversible�A + B C + D ( forward)�C + D A + B (reverse)�Initially there is only A and B so only

the forward reaction is possible�As C and D build up, the reverse

reaction speeds up while the forward reaction slows down.

�Eventually the rates are equal

Rea

ctio

n R

ate

Time

Forward Reaction

Reverse reaction

Equilibrium

What is equal at Equilibrium?�Rates are equal�Concentrations are not.�Rates are determined by

concentrations and activation energy.�The concentrations do not change at

equilibrium.�or if the reaction is verrrry slooooow.

Law of Mass Action�For any reaction � jA + kB lC + mD

� K = [C]l[D]m PRODUCTSpower

[A]j[B]k REACTANTSpower

�K is called the equilibrium constant.

� is how we indicate a reversible reaction

Playing with K�If we write the reaction in reverse.�lC + mD jA + kB�Then the new equilibrium constant is

�K’ = [A]j[B]k = 1/K[C]l[D]m

2

Playing with K�If we multiply the equation by a

constant�njA + nkB nlC + nmD�Then the equilibrium constant is

�K’ =[C]nl[D]nm ([C]l[D]m)n= Kn

[A]nj[B]nk= ([A] j[B]k)n

The units for K�Are determined by the various

powers and units of concentrations.�They depend on the reaction.

K is CONSTANT�At any temperature.�Temperature affects rate.�The equilibrium concentrations don’t

have to be the same, only K.�Equilibrium position is a set of

concentrations at equilibrium.�There are an unlimited number.

Equilibrium Constant

One for each Temperature

Calculate K�N2 + 3H2 2NH3�Initial At Equilibrium�[N2]0 =1.000 M [N2] = 0.921M�[H2]0 =1.000 M [H2] = 0.763M�[NH3]0 =0 M [NH3] = 0.157M

Calculate K�N2 + 3H2 2NH3�Initial At Equilibrium�[N2]0 = 0 M [N2] = 0.399 M�[H2]0 = 0 M [H2] = 1.197 M�[NH3]0 = 1.000 M [NH3] = 0.203M�K is the same no matter what the

amount of starting materials

3

Equilibrium and Pressure�Some reactions are gaseous�PV = nRT�P = (n/V)RT�P = CRT�C is a concentration in moles/Liter�C = P/RT

Equilibrium and Pressure�2SO2(g) + O2(g) 2SO3(g)

� Kp = (PSO3)2

(PSO2)2 (PO2)

� K = [SO3]2

[SO2]2 [O2]

Equilibrium and Pressure

� K = (PSO3/RT)2

(PSO2/RT)2(PO2/RT)

� K = (PSO3)2 (1/RT)2

(PSO2)2(PO2) (1/RT)3

� K = Kp (1/RT)2= Kp RT

(1/RT)3

General Equation�jA + kB lC + mD

�Kp= (PC)l (PD)m= (CCxRT)l (CDxRT)m

(PA)j (PB)k (CAxRT)j(CBxRT)k

�Kp= (CC)l (CD)mx(RT)l+m

(CA)j(CB)kx(RT)j+k

�Kp = K (RT)(l+m)-(j+k) = K (RT)∆n

�∆n=(l+m)-(j+k)=Change in moles of gas

Homogeneous Equilibria�So far every example dealt with

reactants and products where all were in the same phase.

�We can use K in terms of either concentration or pressure.

�Units depend on reaction.

Heterogeneous Equilibria�If the reaction involves pure solids or

pure liquids the concentration of the solid or the liquid doesn’t change.

�As long as they are not used up we can leave them out of the equilibrium expression.

�For example

4

For Example�H2(g) + I2(s) 2HI(g)

�K = [HI]2

[H2][I2]

�But the concentration of I2 does not change.

�K[I2]= [HI]2 = K’[H2]

Write the equilibrium constant for the heterogeneous reaction

2 2CO H OC. P P

[ ][ ]2 2

1A.

CO H O

[ ] [ ]22 3 CO 2B. Na CO P H O

3 2 3 2 22NaHCO (s) Na CO (s) + CO (g) + H O(g). ⇌

[ ][ ][ ][ ]

2 3 2 2

3

Na CO CO H O D.

NaHCO

[ ][ ][ ][ ]

2 3 2 22

3

Na CO CO H O E.

NaHCO

The Reaction Quotient�Tells you the direction the reaction

will go to reach equilibrium�Calculated the same as the

equilibrium constant, but for a system not at equilibrium

�Q = [Products]coefficient

[Reactants] coefficient

�Compare value to equilibrium constant

What Q tells us�If Q<K

�Not enough products�Shift to right

�If Q>K �Too many products�Shift to left

�If Q=K system is at equilibrium

Example�for the reaction�2NOCl(g) 2NO(g) + Cl2(g) �K = 1.55 x 10-5 M at 35ºC�In an experiment 0.10 mol NOCl,

0.0010 mol NO(g) and 0.00010 mol Cl2 are mixed in 2.0 L flask.

�Which direction will the reaction proceed to reach equilibrium?

Solving Equilibrium Problems�Given the starting concentrations and

one equilibrium concentration.�Use stoichiometry to figure out other

concentrations and K.�Learn to create a table of initial and

final conditions.

5

�Consider the following reaction at 600ºC

�2SO2(g) + O2(g) 2SO3(g)�In a certain experiment 2.00 mol of

SO2, 1.50 mol of O2 and 3.00 mol of SO3 were placed in a 1.00 L flask. At equilibrium 3.50 mol of SO3 were found to be present. Calculate

�The equilibrium concentrations of O2and SO2, K and KP

�Consider the same reaction at 600ºC2SO2(g) + O2(g) 2SO3(g)

�In a different experiment .500 mol SO2and .350 mol SO3 were placed in a 1.000 L container. When the system reaches equilibrium 0.045 mol of O2are present.

�Calculate the final concentrations of SO2 and SO3 and K

Solving Equilibrium Problems

Type 1

What if you’re not given equilibrium concentration?

�The size of K will determine what approach to take.

�First let’s look at the case of a LARGE value of K ( >100).

�Allows us to make simplifying assumptions.

Example�H2(g) + I2(g) 2HI(g)�K = 7.1 x 102 at 25ºC�Calculate the equilibrium

concentrations if a 5.00 L container initially contains 15.8 g of H2 294 g I2 .

� [H2]0 = (15.8g/2.02)/5.00 L = 1.56 M � [I2]0 = (294g/253.8)/5.00L = 0.232 M� [HI]0 = 0

6

�Q= 0<K so more product will be formed.

�Set up table of initial, final and change in concentrations.

�Assumption since K is large- reaction will almost go to completion.

�Stoichiometry tells us I2 is LR, it will be smallest at equilibrium let it be x

�Choose X so it is small.�For I2 the change in X must be

X-.232 M�Final must = initial + change

H2(g) I2(g) 2 HI(g) initial 1.56 M 0.232 M 0 Mchange final X

�Using to stoichiometry we can find�Change in H2 = X-0.232 M�Change in HI = -twice change in H2 �Change in HI = 0.464-2X

H2(g) I2(g) 2 HI(g) initial 1.56 M 0.232 M 0 Mchange X-0.232 M final X

�Now we can determine the final concentrations by adding.

H2(g) I2(g) 2 HI(g) initial 1.56 M 0.232 M 0 Mchange X-0.232 M X-0.232 M 0.464-2X final X

�Now plug these values into the equilibrium expression

�K = (0.464-2X)2 = 7.1 x 102

(1.328+X)(X)

H2(g) I2(g) 2 HI(g) initial 1.56 M 0.232 M 0 Mchange X-0.232 M X-0.232 M 0.464-2X final 1.328+X X 0.464-2X

Why we chose X�K = (0.464-2X)2 = 7.1 x 102

(1.328+X)(X)�Since X is going to be small, we can

ignore it in relation to 0.464 and 1.328�So we can rewrite the equation�7.1 x 102 = (0.464)2

(1.328)(X)�Makes the algebra easy

7

�When we solve for X we get 2.3 x 10-4

�So we can find the other concentrations

� I2 = 2.3 x 10-4 M

�H2 = 1.328 M �HI = 0.464 M

H2(g) I2(g) 2 HI(g) initial 1.56 M 0.232 M 0 Mchange X-0.232 M X-0.232 M 0.464-2X final 1.328+X X 0.464-2X

Checking the assumption�The rule of thumb is that if the value

of X is less than 5% of all the smallest concentrations, our assumption was valid.

�If not we would have had to use the quadratic equation

�More on this later.�Our assumption was valid.

Practice�For the reaction Cl2 + O2 2ClO(g)

K = 156� In an experiment 0.100 mol ClO, 1.00

mol O2 and 0.0100 mol Cl2 are mixed in a 4.00 L flask.

� If the reaction is not at equilibrium, which way will it shift?

�Calculate the equilibrium concentrations.

At an elevated temperature, the reaction:

has a value of Keq = 944. If 0.234 mol IBris placed in a 1.00 L. flask and allowed to reach equilibrium, what is the equilibrium concentration in M. of I2?

2 2I (g) + Br (g) 2IBr(g)⇌

Type 2Problems with small K

K< .01

Process is the same�Set up table of initial, change, and

final concentrations.�Choose X to be small.�For this case it will be a product.�For a small K the product

concentration is small.

8

For example�For the reaction

2NOCl 2NO +Cl2�K= 1.6 x 10-5

� If 1.20 mol NOCl, 0.45 mol of NO and 0.87 mol Cl2 are mixed in a 1 L container

�What are the equilibrium concentrations

�Q = [NO]2[Cl2] = (0.45)2(0.87) = 0.15 M[NOCl]2 (1.20)2

�Choose X to be small�NO will be LR�Choose NO to be X

2NOCl 2NO + Cl2Initial 1.20 0.45 0.87

Change

Final

�Figure out change in NO

�Change = final - initial�change = X-0.45


Change

Final X

�Now figure out the other changes�Use stoichiometry�Change in Cl2 is 1/2 change in NO�Change in NOCl is - change in NO


Change X-.45

Final X

� Now we can determine final concentrations

� Add


Change 0.45-X X-.45 0.5X -.225

Final X

� Now we can write equilibrium constant� K = (X)2(0.5X+0.645)

(1.65-X)2

� Now we can test our assumption X is small ignore it in + and -


Change 0.45-X X-.45 0.5X -.225

Final 1.65-X X 0.5 X +0.645

9

� K = (X)2(0.645) = 1.6 x 10-5

(1.65)2

� X= 8.2 x 10-3

� Figure out final concentrations


Change 0.45-X X-.45 0.5X -.225

Final 1.65-X X 0.5 X +0.645

� [NOCl] = 1.64� [Cl2] = 0.649� Check assumptions� .0082/.649 = 1.2 % OKAY!!!


Change 0.45-X X-.45 0.5X -.225

Final 1.65-X X 0.5 X +0.645

Practice Problem�For the reaction

2ClO(g) Cl2 (g) + O2 (g)�K = 6.4 x 10-3

�In an experiment 0.100 mol ClO(g), 1.00 mol O2 and 1.00 x 10-2 mol Cl2are mixed in a 4.00 L container.

�What are the equilibrium concentrations?

Type 3Mid-range K’s

.01<K<10

No Simplification�Choose X to be small.�Can’t simplify so we will have to solve

the quadratic (we hope)�H2(g) + I2 (g) 2HI(g) K=38.6�What is the equilibrium

concentrations if 1.800 mol H2, 1.600 mol I2 and 2.600 mol HI are mixed in a 2.000 L container?

Problems Involving Pressure�Solved exactly the same, with same

rules for choosing X depending on KP�For the reaction N2O4(g) 2NO2(g)

KP = .131 atm. What are the equilibrium pressures if a flask initially contains 1.000 atm N2O4?

10

Le Châtelier’s Principle�If a stress is applied to a system at

equilibrium, the position of the equilibrium will shift to reduce the stress.

�3 Types of stress�Concentration�Pressure�Temperature

Change amounts of reactants and/or products

�Adding product makes Q>K�Removing reactant makes Q>K�Adding reactant makes Q<K�Removing product makes Q<K �Determine the effect on Q, will tell

you the direction of shift

Change Pressure�By changing volume�System will move in the direction that

has the least moles of gas.�Because partial pressures (and

concentrations) change, a new equilibrium must be reached.

�System tries to minimize the moles of gas if volume is reduced

�And visa versa

Change in Pressure�By adding an inert gas�Partial pressures of reactants and

product are not changed�No effect on equilibrium position

Change in Temperature�Affects the rates of both the forward

and reverse reactions.�Doesn’t just change the equilibrium

position, changes the equilibrium constant.

�The direction of the shift depends on whether it is exo- or endothermic

Exothermic�∆H<0�Releases heat�Think of heat as a product�Raising temperature push toward

reactants.�Shifts to left.

11

Endothermic�∆H>0�Produces heat�Think of heat as a reactant�Raising temperature push toward

products.�Shifts to right.

1

Arrhenius Definition� Acids produce hydrogen ions in

aqueous solution.� Bases produce hydroxide ions when

dissolved in water.� Limits to aqueous solutions.� Only one kind of base.� NH3 ammonia could not be an

Arrhenius base.

Bronsted-Lowry Definitions� And acid is an proton (H+) donor and a

base is a proton acceptor.� Acids and bases always come in pairs.� HCl is an acid.� When it dissolves in water it gives its

proton to water.

� HCl(g) + H2O(l) H3O+ + Cl-

� Water is a base -makes hydronium ion

Pairs� General equation � HA(aq) + H2O(l) H3O+(aq) + A-(aq)� Acid + Base Conjugate acid +

Conjugate base� This is an equilibrium.� Competition for H+ between H2O and A-

� The stronger base controls direction.� If H2O is a stronger base it takes the H+

� Equilibrium moves to right.

Acid dissociation constant Ka� The equilibrium constant for the general

equation.� HA(aq) + H2O(l) H3O+(aq) + A-(aq)

� Ka = [H3O+][A-][HA]

� H3O+ is often written H+ ignoring the water in equation (it is implied).

Acid dissociation constant Ka� HA(aq) H+(aq) + A-(aq)

� Ka = [H+][A-][HA]

� We can write the expression for any acid.

� Strong acids dissociate completely.� Equilibrium far to right.� Conjugate base must be weak.

2

Back to Pairs� Strong acids

� Ka is large

� [H+] is equal to [HA]

� A- is a weaker base than water

� Weak acids

� Ka is small

� [H+] <<< [HA]� A- is a stronger

base than water

Types of Acids� Polyprotic Acids- more than 1 acidic

hydrogen (diprotic, triprotic).� Oxyacids - Proton is attached to the

oxygen of an ion.� Organic acids contain the Carboxyl

group -COOH with the H attached to O� Generally very weak.

Amphoteric� Behave as both an acid and a base.� Water autoionizes� 2H2O(l) H3O+(aq) + OH-(aq)� KW= [H3O+][OH-]=[H+][OH-]� At 25ºC KW = 1.0 x10-14

� In EVERY aqueous solution.� Neutral solution [H+] = [OH-]= 1.0 x10-7

� Acidic solution [H+] > [OH-]� Basic solution [H+] < [OH-]

pH� pH= -log[H+]� Used because [H+] is usually very small� As pH decreases, [H+] increases

exponentially� Sig figs only the digits after the decimal

place of a pH are significant� [H+] = 1.0 x 10-8 pH= 8.00 2 sig figs� pOH= -log[OH-]� pKa = -log K

Relationships� KW = [H+][OH-]� -log KW = -log([H+][OH-])� -log KW = -log[H+]+ -log[OH-]� pKW = pH + pOH

� KW = 1.0 x10-14

� 14.00 = pH + pOH� [H+],[OH-],pH and pOH

Given any one of these we can find the other three.

Problems� If a solution has a [H+] of .0035M what

is the pH?

� [OH-]?

� pOH?

3

Problems� If a solution has a pOH of 9.28 what is

the [H+]?

� If a solution has a pH of 9.28 what is the [OH-]?

BasicAcidic Neutral

100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14

[H+]

0 1 3 5 7 9 11 13 14

pH

Basic 10010-110-310-510-710-910-1110-1310-14

[OH-]

013579111314

pOH

Calculating pH of Solutions� Always write down the major ions in

solution.� Remember these are equilibria.� Remember the chemistry.� Don’t try to memorize there is no one

way to do this.

Strong Acids� HCl, HBr, HI, HNO3, H2SO4, HClO4� Completely dissociated� [H+] = [HA]

� [OH-] is going to be small because of equilibrium

� 10-14 = [H+][OH-]� If [HA]< 10-7 water contributes H+

Weak Acids� Ka will be small.� It will be an equilibrium problem from

the start.� Determine whether most of the H+ will

come from the acid or the water.

� Compare Ka or Kw� Rest is just like last chapter.

Example� Calculate the pH of 1.6 M HCl(aq)

� Calculate the pH of 1.6 x 10-10 M HCl(aq)

� Calculate the pH of 2.0 M acetic acid HC2H3O2 with a Ka = 1.8 x10-5

� Calculate pOH, [OH-], [H+]

4

A mixture of Weak Acids� The process is the same.� Determine the major species.� The stronger acid will predominate.� Bigger Ka if concentrations are

comparable

� Calculate the pH of a mixture 1.20 M HF (Ka = 7.2 x 10-4) and 3.4 M HOC6H5(Ka = 1.6 x 10-10)

Percent dissociation� = amount dissociated x 100

initial concentration� For a weak acid percent dissociation

increases as acid becomes more dilute.� Calculate the % dissociation of 1.00 M

and .00100 M Acetic acid (Ka = 1.8 x 10-5)� As [HA]0 decreases [H+] decreases but %

dissociation increases.� Le Châtelier

The other way� What is the Ka of a weak acid that is

8.1 % dissociated as 0.100 M solution?

Bases� The OH- is a strong base. � Hydroxides of the alkali metals are

strong bases because they dissociate completely when dissolved.

� The hydroxides of alkaline earths Ca(OH)2 etc. are strong dibasic bases, but they don’t dissolve well in water.

� Used as antacids because [OH- ] can’t build up.

Bases without OH-

� Bases are proton acceptors.� NH3 + H2O NH4

+ + OH-

� It is the lone pair on nitrogen that accepts the proton.

� Many weak bases contain N� B(aq) + H2O(l) BH+(aq) + OH- (aq)

� Kb = [BH+][OH- ][B]

Strength of Bases

N:

� Hydroxides are strong.� Others are weak.� Smaller Kb weaker base.� Calculate the pH of a solution of 4.0 M

pyridine (Kb = 1.7 x 10-9)

5

Polyprotic acids� Always dissociate stepwise.� The first H+ comes of much easier than

the second.� Ka for the first step is much bigger than

Ka for the second.� Denoted Ka1, Ka2, Ka3

Polyprotic acid� H2CO3 H+ + HCO3

-

Ka1= 4.3 x 10-7

� HCO3- H+ + CO3

-2

Ka2= 4.3 x 10-10

� Base in first step is acid in second.� In calculations we can normally ignore

the second dissociation.

Calculate the Concentration� Of all the ions in a solution of 1.00 M

Arsenic acid H3AsO4�Ka1 = 5.0 x 10

-3

�Ka2 = 8.0 x 10-8

�Ka3 = 6.0 x 10-10

Sulfuric acid is special� In first step it is a strong acid.

� Ka2 = 1.2 x 10-2

� Calculate the concentrations in a 2.0 M solution of H2SO4

� Calculate the concentrations in a 2.0 x 10-3 M solution of H2SO4

Salts as acids and bases� Salts are ionic compounds.� Salts of the cation of strong bases and

the anion of strong acids are neutral.� for example NaCl, KNO3� There is no equilibrium for strong acids

and bases.� We ignore the reverse reaction.

Basic Salts� If the anion of a salt is the conjugate

base of a weak acid – solution is basic.� In an aqueous solution of NaF� The major species are Na+, F-, and H2O� F- + H2O HF + OH-

� Kb =[HF][OH-][F- ]

� For HF- the acid form- Ka = [H+][F-][HF]

6

Basic Salts� Ka x Kb = [HF][OH-] x [H+][F-]

[F- ] [HF]


[F- ] [HF]


[F- ] [HF]


[F- ] [HF]� Ka x Kb =[OH-] [H+]


[F- ] [HF]� Ka x Kb =[OH-] [H+]� Ka x Kb = KW

Ka tells us Kb

� The anion of a weak acid is a weak base.� Calculate the pH of a solution of 1.00 M

NaF. Ka of HF is 7.2 x 10-4

� H+ + F- HF� The F- ion competes with OH- for the H+

7

Acidic salts� A salt with the cation of a weak base and

the anion of a strong acid will be acidic.� The same development as bases leads

to

� Ka x Kb = KW� Calculate the pH of a solution of 0.40 M

NH4Cl (the Kb of NH3 1.8 x 10-5).� Other acidic salts are those of highly

charged metal ions.� More on this later.

Anion of weak acid, cation of weak base

� Ka > Kb acidic� Ka < Kb basic� Ka = Kb Neutral� NH4CN

– Ka for HCN is 6.2 x 10-10

– Kb for NH3 is 1.8 x 10-5

Structure and Acid base Properties

� Any molecule with an H in it is a potential acid.

� The stronger the X-H bond the less acidic (compare bond dissociation energies).

� The more polar the X-H bond the stronger the acid (use electronegativities).

� The more polar H-O-X bond -stronger acid.

Strength of oxyacids� The more oxygen hooked to the central

atom, the more acidic the hydrogen.� HClO4 > HClO3 > HClO2 > HClO� Remember that the H is attached to an

oxygen atom.� The oxygens are electronegative� Pull electrons away from hydrogen

Strength of oxyacids

Electron Density

Cl O H


Electron Density

Cl O HO

8


Cl O H

O

O

Electron Density


Cl O H

O

O

O

Electron Density

Hydrated metals� Highly charged metal

ions pull the electrons of surrounding water molecules toward them.

� Make it easier for H+

to come off.� Make solution acidic

Al+3 O

H

H

Acid-Base Properties of Oxides� Non-metal oxides dissolved in water

can make acids.� SO3 (g) + H2O(l) H2SO4(aq)� Ionic oxides dissolve in water to

produce bases. (metal oxides)� CaO(s) + H2O(l) Ca(OH)2(aq)� Hydroxides

Lewis Acids and Bases� Most general definition.� Acids are electron pair acceptors.� Bases are electron pair donors.

B F

F

F

:NH

H

H

Lewis Acids and Bases� Boron triflouride wants more electrons.

B F

F

F

:NH

H

H

9

Lewis Acids and Bases� Boron triflouride wants more electrons.� BF3 is Lewis base NH3 is a Lewis Acid.

BF

F

F

N

H

H

H

Lewis Acids and Bases

Al+3 ( )HH

O

Al ( )6

H

HO

+ 6

+3

1

Chapter 15

Applying equilibrium

The Common Ion Effect� When the salt with the anion of a weak

acid is added to that acid,� It reverses the dissociation of the acid.� Lowers the percent dissociation of the

acid.� The same principle applies to salts with

the cation of a weak base.� The calculations are the same as last

chapter.

Buffered solutions� A solution that resists a change in pH.� Either a weak acid and its salt or a weak

base and its salt.� We can make a buffer of any pH by

varying the concentrations of these solutions.

� Same calculations as before.� Calculate the pH of a solution that is .50 M

HAc and .25 M NaAc (Ka = 1.8 x 10-5)

Na+ is a spectator and the reaction we are worried about is

HAc H+ + Ac-

� Choose x to be small

x

� We can fill in the table

xx-x

0.50-x 0.25+x

Initial 0.50 M 0 0.25 M ∆

Final

� Do the math� Ka = 1.8 x 10-5

1.8 x 10-5 =x (0.25+x)

(0.50-x)� Assume x is small

=x (0.25)

(0.50)

x = 3.6 x 10-5

� Assumption is valid� pH = -log (3.6 x 10-5) = 4.44

HAc H+ + Ac-

x

xx-x

0.50-x 0.25+x

Initial 0.50 M 0 0.25 M ∆

Final

Adding a strong acid or base� Do the stoichiometry first.

–Use moles not molar� A strong base will grab protons from the

weak acid reducing [HA]0� A strong acid will add its proton to the anion

of the salt reducing [A-]0� Then do the equilibrium problem.� What is the pH of 1.0 L of the previous

solution when 0.010 mol of solid NaOH is added?

2

HAc H+ + Ac-

� In the initial mixture M x L = mol� 0.50 M HAc x 1.0 L = 0.50 mol HAc

� Adding 0.010 mol OH- will reduce the HAc and increase the Ac- by 0.010 mole

� Because it is in 1.0 L, we can convert it to molarity

0.50 mol 0.25 mol

� 0.25 M Ac- x 1.0 L = 0.25 mol Ac-

0.49 mol 0.26 mol

HAc H+ + Ac-

� In the initial mixture M x L = mol� 0.50 M HAc x 1.0 L = 0.50 mol HAc� 0.25 M Ac- x 1.0 L = 0.25 mol Ac-

� Adding 0.010 mol OH- will reduce the HAc and increase the Ac- by 0.010 mole

� Because it is in 1.0 L, we can convert it to molarity

0.50 mol 0.25 mol

0.49 M 0.26 M

HAc H+ + Ac-

� Fill in the table

0.50 mol 0.25 mol

0.49 M 0.26 M

HAc H+ + Ac-

x

xx-x

0.49-x 0.26+x

Initial 0.49 M 0 0.26 M ∆

Final

� Do the math� Ka = 1.8 x 10-5

1.8 x 10-5 =x (0.26+x)

(0.49-x)� Assume x is small

=x (0.26)

(0.49)

x = 3.4 x 10-5

� Assumption is valid� pH = -log (3.4 x 10-5) = 4.47

HAc H+ + Ac-

x

xx-x

0.49-x 0.26+x

Initial 0.49 M 0 0.26 M ∆

Final

Notice� If we had added 0.010 mol of NaOH to

1 L of water, the pH would have been.� 0.010 M OH-

� pOH = 2� pH = 12� But with a mixture of an acid and its

conjugate base the pH doesn’t change much

� Called a buffer.

General equation� Ka = [H+] [A-]

[HA]� so [H+] = Ka [HA]

[A-]� The [H+] depends on the ratio [HA]/[A-]� taking the negative log of both sides� pH = -log(Ka [HA]/[A-])� pH = -log(Ka)-log([HA]/[A-])� pH = pKa + log([A-]/[HA])

3

This is called the Henderson-Hasselbach equation

� pH = pKa + log([A-]/[HA])� pH = pKa + log(base/acid)� Works for an acid and its salt� Like HNO2 and NaNO2

� Or a base and its salt� Like NH3 and NH4Cl� But remember to change Kb to Ka

-4 0.25 MpH = -log(1.4 x 10 ) + log

0.75 M

� Calculate the pH of the following� 0.75 M lactic acid (HC3H5O3) and 0.25

M sodium lactate (Ka = 1.4 x 10-4)

�pH = 3.38

-10 0.25 MpH = -log(5.6 x 10 ) + log

0.40 M

� Calculate the pH of the following� 0.25 M NH3 and 0.40 M NH4Cl

(Kb = 1.8 x 10-5)� Ka = 1 x 10-14

1.8 x 10-5

� Ka = 5.6 x 10-10

� remember its the ratio base over acid

�pH = 9.05

Prove they’re buffers� What would the pH be if .020 mol of HCl

is added to 1.0 L of both of the preceding solutions.

� What would the pH be if 0.050 mol of solid NaOH is added to 1.0 L of each of the proceeding.

� Remember adding acids increases the acid side,

� Adding base increases the base side.

Prove they’re buffers� What would the pH be if .020 mol of HCl is

added to 1.0 L of preceding solutions.� 0.75 M lactic acid (HC3H5O3) and 0.25 M

sodium lactate (Ka = 1.4 x 10-4)HC3H5O3 H

+ + C3H5O3-

Initially 0.75 mol 0 0.25 molAfter acid 0.77 mol 0.23 mol

-4 0.23 MpH = -log(1.4 x 10 ) + log 3.33

0.77 M =

Compared to 3.38 before acid was added

Prove they’re buffers� What would the pH be if 0.050 mol of solid

NaOH is added to 1.0 L of the solutions.� 0.75 M lactic acid (HC3H5O3) and 0.25 M

sodium lactate (Ka = 1.4 x 10-4)HC3H5O3 H

+ + C3H5O3-


-4 0.30 MpH = -log(1.4 x 10 ) + log 3.48

0.70 M =


4

Prove they’re buffers� What would the pH be if .020 mol of HCl is

added to 1.0 L of preceding solutions.� 0.25 M NH3 and 0.40 M NH4Cl

Kb = 1.8 x 10-5 so Ka = 5.6 x 10-10

NH4+ H+ + NH3


-10 0.23 MpH = -log(5.6 x 10 ) + log 8.99

0.42 M =


Prove they’re buffers� What would the pH be if 0.050 mol of solid

NaOH is added to 1.0 L each solutions.� 0.25 M NH3 and 0.40 M NH4Cl

Kb = 1.8 x 10-5 so Ka = 5.6 x 10-10

NH4+ H+ + NH3


-10 0.30 MpH = -log(5.6 x 10 ) + log 9.18

0.35 M =


Buffer capacity� The pH of a buffered solution is

determined by the ratio [A-]/[HA].� As long as this doesn’t change much

the pH won’t change much.� The more concentrated these two are

the more H+ and OH- the solution will be able to absorb.

� Larger concentrations = bigger buffer capacity.

Buffer Capacity� Calculate the change in pH that occurs

when 0.020 mol of HCl(g) is added to 1.0 L of each of the following:

� 5.00 M HAc and 5.00 M NaAc� 0.050 M HAc and 0.050 M NaAc� Ka= 1.8x10-5

� pH = pKa

-5 5.00 MpH = -log(1.8 x 10 ) + log 4.74

5.00 M =


when 0.040 mol of HCl(g) is added to 1.0 L of 5.00 M HAc and 5.00 M NaAc

� Ka= 1.8x10-5

HAc H+ + Ac-


-5 4.96 MpH = -log(1.8 x 10 ) + log 4.74

5.04 M =



when 0.040 mol of HCl(g) is added to 1.0 L of 0.050 M HAc and 0.050 M NaAc

� Ka= 1.8x10-5

HAc H+ + Ac-

Initially 0.050 mol 0 0.050 molAfter acid 0.090 mol 0 0.010 mol

-5 0.010MpH = -log(1.8 x 10 ) + log 3.79

0.090 M =


5

Buffer capacity� The best buffers have a ratio

[A-]/[HA] = 1� This is most resistant to change� True when [A-] = [HA]� Makes pH = pKa (since log 1 = 0)

Titrations� Millimole (mmol) = 1/1000 mol� Molarity = mmol/mL = mol/L � Makes calculations easier because we

will rarely add liters of solution.� Adding a solution of known

concentration until the substance being tested is consumed.

� This is called the equivalence point.� Graph of pH vs. mL is a titration curve.

Titration Curves

pH

mL of Base added

7

� Strong acid with strong Base

� Equivalence at pH 7

pH

mL of Base added

>7

� Weak acid with strong Base� Equivalence at pH >7� When the acid is neutralized it makes a

weak base

7

pH

mL of acid added

7

� Strong base with strong acid� Equivalence at pH 7

6

pH

mL of acid added

<7

� Weak base with strong acid� Equivalence at pH <7� When the base is

neutralized it makes a weak acid

7

Strong acid with Strong Base� Do the stoichiometry.� mL x M = mmol� There is no equilibrium .� They both dissociate completely.

� The reaction is H+ + OH- → HOH� Use [H+] or [OH-] to figure pH or pOH� The titration of 50.0 mL of 0.200 M HNO3

with 0.100 M NaOH

Weak acid with Strong base� There is an equilibrium.� Do stoichiometry.

–Use moles� Determine major species� Then do equilibrium.� Titrate 50.0 mL of 0.10 M HF

(Ka = 7.2 x 10-4) with 0.10 M NaOH

Summary� Strong acid and base just stoichiometry.� Weak acid with 0 ml of base - Ka

� Weak acid before equivalence point–Stoichiometry first–Then Henderson-Hasselbach

� Weak acid at equivalence point- Kb-Calculate concentration

� Weak acid after equivalence - leftover strong base.

-Calculate concentration

Summary� Weak base before equivalence point.

–Stoichiometry first–Then Henderson-Hasselbach

� Weak base at equivalence point Ka. -Calculate concentration

� Weak base after equivalence – left over strong acid.

-Calculate concentration

Indicators� Weak acids that change color when they

become bases.� weak acid written HIn� Weak base� HIn H+ + In-

clear red� Equilibrium is controlled by pH� End point - when the indicator changes

color.� Try to match the equivalence point

7

Indicators� Since it is an equilibrium the color change

is gradual.� It is noticeable when the ratio of

[In-]/[HI] or [HI]/[In-] is 1/10� Since the Indicator is a weak acid, it has a

Ka.

� pH the indicator changes at is.� pH=pKa +log([In-]/[HI]) = pKa +log(1/10) � pH=pKa - 1 on the way up

Indicators� pH=pKa + log([HI]/[In-]) = pKa + log(10)� pH=pKa+1 on the way down� Choose the indicator with a pKa 1 more

than the pH at equivalence point if you are titrating with base.

� Choose the indicator with a pKa 1 less than the pH at equivalence point if you are titrating with acid.

Solubility Equilibria

Will it all dissolve, and if not, how much?

� All dissolving is an equilibrium.� If there is not much solid it will all

dissolve.� As more solid is added the solution will

become saturated.� Solid dissolved� The solid will precipitate as fast as it

dissolves .� Equilibrium

General equation� M+ stands for the cation (usually metal).� Nm- stands for the anion (a nonmetal).� MaNmb(s) aM+(aq) + bNm- (aq) � K = [M+]a[Nm-]b/[MaNmb]� But the concentration of a solid doesn’t

change.� Ksp = [M+]a[Nm-]b

� Called the solubility product for each compound.

Watch out� Solubility is not the same as solubility

product.� Solubility product is an equilibrium

constant.� it doesn’t change except with

temperature.� Solubility is an equilibrium position for

how much can dissolve.� A common ion can change this.

8

Calculating Ksp� The solubility of iron(II) oxalate FeC2O4

is 65.9 mg/L � The solubility of Li2CO3 is 5.48 g/L

Calculating Solubility� The solubility is determined by

equilibrium.� Its an equilibrium problem.� Watch the coefficients� Calculate the solubility of SrSO4, with a

Ksp of 3.2 x 10-7 in M and g/L.� Calculate the solubility of Ag2CrO4, with

a Ksp of 9.0 x 10-12 in M and g/L.

Relative solubilities� Ksp will only allow us to compare the

solubility of solids that fall apart into the same number of ions.

� The bigger the Ksp of those the more soluble.

� If they fall apart into different number of pieces you have to do the math.

Common Ion Effect� If we try to dissolve the solid in a

solution with either the cation or anion already present less will dissolve.

� Calculate the solubility of SrSO4, with a Ksp of 3.2 x 10-7 in M and g/L in a solution of 0.010 M Na2SO4.

� Calculate the solubility of SrSO4, with a Ksp of 3.2 x 10-7 in M and g/L in a solution of 0.010 M SrNO3.

pH and solubility� OH- can be a common ion.� More soluble in acid.� For other anions if they come from a

weak acid they are more soluble in a acidic solution than in water.

� CaC2O4 Ca+2 + C2O4-2

� H+ + C2O4-2 HC2O4

-

� Reduces [C2O4-2] in acidic solution.

Precipitation� Ion Product, Q =[M+]a[Nm-]b

� If Q>Ksp a precipitate forms.� If Q<Ksp No precipitate.� If Q = Ksp equilibrium.� A solution of 750.0 mL of 4.00 x 10-3M

Ce(NO3)3 is added to 300.0 mL of2.00 x 10-2M KIO3. Will Ce(IO3)3 (Ksp= 1.9 x 10-10M) precipitate and if so, what is the concentration of the ions?

9

Selective Precipitations� Used to separate mixtures of metal ions

in solutions.� Add anions that will only precipitate

certain metals at a time.� Used to purify mixtures.

� Often use H2S because in acidic solution Hg+2, Cd+2, Bi+3, Cu+2, Sn+4

will precipitate.

Selective Precipitation� Then add OH-solution [S-2] will increase

so more soluble sulfides will precipitate.� Co+2, Zn+2, Mn+2, Ni+2, Fe+2, Cr(OH)3,

Al(OH)3

Selective precipitation� Follow the steps � First with insoluble chlorides (Ag, Pb,

Ba)� Then sulfides in Acid.� Then sulfides in base.� Then insoluble carbonate (Ca, Ba, Mg)� Alkali metals and NH4

+ remain in solution.

Complex ion Equilibria� A charged ion surrounded by ligands.� Ligands are Lewis bases using their

lone pair to stabilize the charged metal ions.

� Common ligands are NH3, H2O, Cl-,CN-

� Coordination number is the number of attached ligands.

� Cu(NH3)42+ has a coordination # of 4

The addition of each ligandhas its own equilibrium

� Usually the ligand is in large excess.� And the individual K’s will be large so

we can treat them as if they go to completion.

� The complex ion will be the biggest ion in solution.

� Calculate the concentrations of Ag+, Ag(S2O3)-, and Ag(S2O3)-3 in a solution made by mixing 150.0 mL of 0.010 M AgNO3 with 200.0 mL of 5.00 M Na2S2O3

� Ag+ + S2O3-2 Ag(S2O3)-

K1=7.4 x 108

� Ag(S2O3)- + S2O3-2 Ag(S2O3)2

-3

K2=3.9 x 104

1

Page 1

Chapter 16

Spontaneity, entropy and free energy

Spontaneous�A reaction that will occur without

outside intervention.�We can’t determine how fast.�We need both thermodynamics and

kinetics to describe a reaction completely.

�Thermodynamics compares initial and final states.

�Kinetics describes pathway between.

Thermodynamics�1st Law- the energy of the universe is

constant.�Keeps track of thermodynamics doesn’t

correctly predict spontaneity.�Entropy (S)

– Number of ways things can be arranged– Looks like disorder or randomness

�2nd Law the entropy of the universe increases in any change

Entropy�Defined in terms of probability.�Substances take the arrangement that

is most likely.�The most likely is the most random.�Calculate the number of arrangements

for a system.

�2 possible arrangements

�50 % chance of finding the left empty


�25% chance of finding the left empty

�50 % chance of them being evenly dispersed

2

Page 2

�4 atoms



Gases�Gases completely fill their chamber

because there are many more ways to do that than to leave half empty.

�Ssolid <Sliquid <<Sgas� there are many more ways for the

molecules to be arranged as a liquid than a solid.

�Gases have a huge number of positions possible.

Entropy�Solutions form because there are many

more possible arrangements of dissolved pieces than if they stay separate.

�2nd Law

�∆Suniv = ∆Ssys + ∆Ssurr� If ∆Suniv is positive the process is

spontaneous.� If ∆Suniv is negative the process is

spontaneous in the opposite direction.

�For exothermic processes ∆Ssurr is positive.

�For endothermic processes ∆Ssurr is negative.

�Consider this processH2O(l)→ H2O(g)

�∆Ssys is positive�∆Ssurr is negative�∆Suniv depends on temperature.

Temperature and Spontaneity

�Entropy changes in the surroundings are determined by the heat flow.

�An exothermic process is favored because by giving up heat the entropy of the surroundings increases.

�The size of ∆Ssurr depends on temperature

�∆Ssurr = -∆H/T

∆Ssys

-∆H/T∆Ssurr ∆Suniv Spontaneous?

+ + +

- --

+ - ?

+- ?

Yes

No, Reverse

At Low temp.

At High temp.

3

Page 3

Gibb's Free Energy�G=H-TS�Never used this way.�∆G=∆H-T∆S at constant temperature�Divide by -T� -∆G/T = -∆H/T-∆S� -∆G/T = ∆Ssurr + ∆S � -∆G/T = ∆Suniv� If ∆G is negative at constant T and P,

the Process is spontaneous.

Let’s Check�For the reaction H2O(s) → H2O(l)�∆Sº = 22.1 J/K mol ∆Hº =6030 J/mol�Calculate ∆G at 10ºC and -10ºC�When does it become spontaneous?

�Look at the equation ∆G=∆H-T∆S�Spontaneity can be predicted from the

sign of ∆H and ∆S.

∆G=∆H-T∆S∆H∆S Spontaneous?

+ - At all Temperatures

+ + At high temperatures, “entropy driven”

- - At low temperatures, “enthalpy driven”

+- Not at any temperature,Reverse is spontaneous

Third Law of Thermo�The entropy of a pure crystal at 0 K is 0.�Gives us a starting point.�All others must be>0.�Standard Entropies Sº ( at 298 K and 1

atm) of substances are listed.

�Products - reactants to find ∆Sº (a state function).

�More complex molecules higher Sº.

Free Energy in Reactions�∆Gº = standard free energy change.�Free energy change that will occur if

reactants in their standard state turn to products in their standard state.

�Can’t be measured directly, can be calculated from other measurements.

�∆Gº=∆Hº-T∆Sº�Use Hess’s Law with known reactions.

Free Energy in Reactions�There are tables of ∆Gºf .�Products-reactants because it is a state

function.� The standard free energy of formation

for any element in its standard state is 0.�Remember- Spontaneity tells us nothing

about rate.

4

Page 4

Free energy and Pressure�∆G = ∆Gº +RTln(Q) where Q is the

reaction quotients (P of the products /P of the reactants).

�CO(g) + 2H2(g) → CH3OH(l)�Would the reaction be spontaneous at

25ºC with the H2 pressure of 5.0 atm and the CO pressure of 3.0 atm?

�∆Gºf CH3OH(l) = -166 kJ �∆Gºf CO(g) = -137 kJ ∆Gºf H2(g) = 0 kJ

How far?�∆G tells us spontaneity at current

conditions. When will it stop?� It will go to the lowest possible free

energy which may be an equilibrium.

�At equilibrium ∆G = 0, Q = K�∆Gº = -RTlnK

∆Gº K=0 =1<0 >1>0 <1

∆Gº = -RTlnK

At 1500°C for the reactionCO(g) + 2H2(g) → CH3OH(g)

the equilibrium constant isKp = 1.4 x 10-7. Is ∆H°at this temperature:

A. positive B. negativeC. zeroD. can not be determined

The standard free energy (∆Grxn0 for the

reaction N2(g) + 3H2(g) → 2NH3(g)

is -32.9 kJ. Calculate the equilibrium constant for this reaction at 25oC.

A. 13.3 B. 5.8 x 105

C. 2.5D. 4.0 x 10-6

E. 9.1 x 108

Temperature dependence of K

�∆Gº= -RTlnK = ∆Hº - T∆Sº

�A straight line of lnK vs 1/T

�With slope -∆Hº/R

Hº 1 Sºln(K) = - +

R T R

∆ ∆

5

Page 5

Free energy And Work�Free energy is that energy free to do

work.�The maximum amount of work possible

at a given temperature and pressure.

�∆E = q + w�Never really achieved because some of

the free energy is changed to heat during a change, so it can’t be used to do work.

�Can’t be 100% efficient

1

Page 1

Chapter 16

Spontaneity, entropy and free energy

Spontaneous�A reaction that will occur without

outside intervention.�We can’t determine how fast.�We need both thermodynamics and

kinetics to describe a reaction completely.

�Thermodynamics compares initial and final states.

�Kinetics describes pathway between.

Thermodynamics�1st Law- the energy of the universe is

constant.�Keeps track of thermodynamics doesn’t

correctly predict spontaneity.�Entropy (S)

– Number of ways things can be arranged– Looks like disorder or randomness

�2nd Law the entropy of the universe increases in any change

Entropy�Defined in terms of probability.�Substances take the arrangement that

is most likely.�The most likely is the most random.�Calculate the number of arrangements

for a system.


�50 % chance of finding the left empty




2

Page 2

�4 atoms



Gases�Gases completely fill their chamber

because there are many more ways to do that than to leave half empty.

�Ssolid <Sliquid <<Sgas� there are many more ways for the

molecules to be arranged as a liquid than a solid.

�Gases have a huge number of positions possible.

Entropy�Solutions form because there are many

more possible arrangements of dissolved pieces than if they stay separate.

�2nd Law

�∆Suniv = ∆Ssys + ∆Ssurr� If ∆Suniv is positive the process is

spontaneous.� If ∆Suniv is negative the process is

spontaneous in the opposite direction.

�For exothermic processes ∆Ssurr is positive.

�For endothermic processes ∆Ssurr is negative.

�Consider this processH2O(l)→ H2O(g)

�∆Ssys is positive�∆Ssurr is negative�∆Suniv depends on temperature.

Temperature and Spontaneity

�Entropy changes in the surroundings are determined by the heat flow.

�An exothermic process is favored because by giving up heat the entropy of the surroundings increases.

�The size of ∆Ssurr depends on temperature

�∆Ssurr = -∆H/T

∆Ssys

-∆H/T∆Ssurr ∆Suniv Spontaneous?

+ + +

- --

+ - ?

+- ?

Yes

No, Reverse

At Low temp.

At High temp.

3

Page 3

Gibb's Free Energy�G=H-TS�Never used this way.�∆G=∆H-T∆S at constant temperature�Divide by -T� -∆G/T = -∆H/T-∆S� -∆G/T = ∆Ssurr + ∆S � -∆G/T = ∆Suniv� If ∆G is negative at constant T and P,

the Process is spontaneous.

Let’s Check�For the reaction H2O(s) → H2O(l)�∆Sº = 22.1 J/K mol ∆Hº =6030 J/mol�Calculate ∆G at 10ºC and -10ºC�When does it become spontaneous?

�Look at the equation ∆G=∆H-T∆S�Spontaneity can be predicted from the

sign of ∆H and ∆S.

∆G=∆H-T∆S∆H∆S Spontaneous?

+ - At all Temperatures

+ + At high temperatures, “entropy driven”

- - At low temperatures, “enthalpy driven”

+- Not at any temperature,Reverse is spontaneous

Third Law of Thermo�The entropy of a pure crystal at 0 K is 0.�Gives us a starting point.�All others must be>0.�Standard Entropies Sº ( at 298 K and 1

atm) of substances are listed.

�Products - reactants to find ∆Sº (a state function).

�More complex molecules higher Sº.

Free Energy in Reactions�∆Gº = standard free energy change.�Free energy change that will occur if

reactants in their standard state turn to products in their standard state.

�Can’t be measured directly, can be calculated from other measurements.

�∆Gº=∆Hº-T∆Sº�Use Hess’s Law with known reactions.

Free Energy in Reactions�There are tables of ∆Gºf .�Products-reactants because it is a state

function.� The standard free energy of formation

for any element in its standard state is 0.�Remember- Spontaneity tells us nothing

about rate.

4

Page 4

Free energy and Pressure�∆G = ∆Gº +RTln(Q) where Q is the

reaction quotients (P of the products /P of the reactants).

�CO(g) + 2H2(g) → CH3OH(l)�Would the reaction be spontaneous at

25ºC with the H2 pressure of 5.0 atm and the CO pressure of 3.0 atm?

�∆Gºf CH3OH(l) = -166 kJ �∆Gºf CO(g) = -137 kJ ∆Gºf H2(g) = 0 kJ

How far?�∆G tells us spontaneity at current

conditions. When will it stop?� It will go to the lowest possible free

energy which may be an equilibrium.

�At equilibrium ∆G = 0, Q = K�∆Gº = -RTlnK

∆Gº K=0 =1<0 >1>0 <1

∆Gº = -RTlnK

At 1500°C for the reactionCO(g) + 2H2(g) → CH3OH(g)

the equilibrium constant isKp = 1.4 x 10-7. Is ∆H°at this temperature:

A. positive B. negativeC. zeroD. can not be determined

The standard free energy (∆Grxn0 for the

reaction N2(g) + 3H2(g) → 2NH3(g)

is -32.9 kJ. Calculate the equilibrium constant for this reaction at 25oC.

A. 13.3 B. 5.8 x 105

C. 2.5D. 4.0 x 10-6

E. 9.1 x 108

Temperature dependence of K

�∆Gº= -RTlnK = ∆Hº - T∆Sº

�A straight line of lnK vs 1/T

�With slope -∆Hº/R

Hº 1 Sºln(K) = - +

R T R

∆ ∆

5

Page 5

Free energy And Work�Free energy is that energy free to do

work.�The maximum amount of work possible

at a given temperature and pressure.

�∆E = q + w�Never really achieved because some of

the free energy is changed to heat during a change, so it can’t be used to do work.

�Can’t be 100% efficient

1

1

Chapter 4

Aqueous solutionsTypes of reactions

2

Parts of Solutions� Solution- homogeneous mixture.� Solute- what gets dissolved.� Solvent- what does the dissolving.� Soluble- Can be dissolved.

� Miscible- liquids dissolve in each other.

3

Aqueous solutions� Dissolved in water.� Water is a good solvent

because the molecules are polar.

� The oxygen atoms have a partial negative charge.

� The hydrogen atoms have a partial positive charge.

� The angle is 105ºC.

4

Hydration� The process of breaking the ions of

salts apart.

� Ions have charges and are attracted to the opposite charges on the water molecules.

5

How Ionic solids dissolve

H

H

H

HH

Click here for Animation

6

Solubility� How much of a substance will dissolve

in a given amount of water.

� Usually g/100 mL� Varies greatly, but if they do dissolve

the ions are separated,� and they can move around.� Water can also dissolve non-ionic

compounds if they have polar bonds.

2

7

Electrolytes� Electricity is moving charges.� The ions that are dissolved can move.� Solutions of ionic compounds can

conduct electricity.� Electrolytes.� Solutions are classified three ways.

8

Types of solutions� Strong electrolytes- completely

dissociate (fall apart into ions).

–Many ions- Conduct well.� Weak electrolytes- Partially fall apart

into ions.–Few ions -Conduct electricity slightly.

� Non-electrolytes- Don’t fall apart.–No ions- Don’t conduct.

9

Types of solutions� Acids- form H+ ions when dissolved.� Strong acids fall apart completely.� many ions� Memorize this list

H2SO4 HNO3 HCl HBr HI HClO4� Weak acids- don’ dissociate completely.� Bases - form OH- ions when dissolved.� Strong bases- many ions.� KOH NaOH

10

Measuring Solutions� Concentration- how much is dissolved.� Molarity = Moles of solute

Liters of solution� abbreviated M� 1 M = 1 mol solute / 1 liter solution� Calculate the molarity of a solution with

34.6 g of NaCl dissolved in 125 mL of solution.

11

Molarity� How many grams of HCl would be

required to make 50.0 mL of a 2.7 M solution?

� What would the concentration be if you used 27g of CaCl2 to make 500. mL of solution?

� What is the concentration of each ion?

12

Molarity� Calculate the concentration of a solution

made by dissolving 45.6 g of Fe2(SO4)3to 475 mL.

� What is the concentration of each ion?

3

13

Making solutions� Describe how to make 100.0 mL of a

1.0 M K2Cr2O4 solution.

� Describe how to make 250. mL of an 2.0 M copper (II) sulfate dihydrate solution.

14

Dilution

� Adding more solvent to a known solution.� The moles of solute stay the same.� moles = M x L� M1 V1 = M2 V2� moles = moles� Stock solution is a solution of known

concentration used to make more dilute solutions

15

Dilution

� What volume of a 1.7 M solutions is needed to make 250 mL of a 0.50 M solution?

� 18.5 mL of 2.3 M HCl is added to 250 mL of water. What is the concentration of the solution?

� 18.5 mL of 2.3 M HCl is diluted to 250 mL with water. What is the concentration of the solution?

16

Dilution� You have a 4.0 M stock solution.

Describe how to make 1.0L of a 0.75 M solution.

� 25 mL 0.67 M of H2SO4 is added to 35 mL of 0.40 M CaCl2 . What mass CaSO4 is formed?

17

Types of Reactions1 Precipitation reactions� When aqueous solutions of ionic

compounds are poured together a solid forms.

� A solid that forms from mixed solutions is a precipitate

� If you’re not a part of the solution, your part of the precipitate

18

Precipitation reactions� NaOH(aq) + FeCl3(aq) →

NaCl(aq) + Fe(OH)3(s)� is really� Na+(aq)+OH-(aq) + Fe+3 + Cl-(aq) →

Na+ (aq) + Cl- (aq) + Fe(OH)3(s)� So all that really happens is� OH-(aq) + Fe+3 → Fe(OH)3(s)� Double replacement reaction

4

19

Precipitation reaction� We can predict the products� Can only be certain by experimenting� The anion and cation switch partners� AgNO3(aq) + KCl(aq) →� Zn(NO3)2(aq) + BaCr2O7(aq) →� CdCl2(aq) + Na2S(aq) →

20

Precipitations Reactions� Only happen if one of the products is

insoluble

� Otherwise all the ions stay in solution-nothing has happened.

� Need to memorize the rules for solubility (pg 145)

21

Solubility Rules1 All nitrates are soluble2 Alkali metals ions and NH4

+ ions are soluble

3 Halides are soluble except Ag+, Pb+2, and Hg2

+2

4 Most sulfates are soluble, except Pb+2, Ba+2, Hg+2,and Ca+2

22

Solubility Rules5 Most hydroxides are slightly soluble

(insoluble) except NaOH and KOH

6 Sulfides, carbonates, chromates, and phosphates are insoluble

∗ Lower number rules supersede so Na2S is soluble

23

Three Types of Equations� Molecular Equation- written as whole

formulas, not the ions.� K2CrO4(aq) + Ba(NO3)2(aq) →� Complete Ionic equation show dissolved

electrolytes as the ions.� 2K+ + CrO4

-2 + Ba+2 + 2 NO3- →

BaCrO4(s) + 2K+ + 2 NO3-

� Spectator ions are those that don’t react.

24

Three Type of Equations� Net Ionic equations show only those

ions that react, not the spectator ions

� Ba+2 + CrO4-2 → BaCrO4(s)

� Write the three types of equations for the reactions when these solutions are mixed.

� iron (III) sulfate and potassium sulfide Lead (II) nitrate and sulfuric acid.

5

25

Stoichiometry of Precipitation

� Exactly the same, except you may have to figure out what the pieces are.

� What mass of solid is formed when 100.00 mL of 0.100 M Barium chloride is mixed with 100.00 mL of 0.100 M sodium hydroxide?

� What volume of 0.204 M HCl is needed to precipitate the silver from 50.ml of 0.0500 M silver nitrate solution ?

26

Types of Reactions2 Acid-Base� For our purposes an acid is a proton

donor.

� a base is a proton acceptor usually OH-

� What is the net ionic equation for the reaction of HCl(aq) and KOH(aq)?

� Acid + Base → salt + water� H+ + OH- → H2O

27

Acid - Base Reactions

� Often called a neutralization reaction Because the acid neutralizes the base.

� Often titrate to determine concentrations.� Solution of known concentration (titrant),� is added to the unknown (analyte),� until the equivalence point is reached

where enough titrant has been added to neutralize it.

28

Titration� Where the indicator changes color is the

endpoint.

� Not always at the equivalence point.� A 50.00 mL sample of aqueous

Ca(OH)2 requires 34.66 mL of 0.0980 M Nitric acid for neutralization. What is [Ca(OH)2 ]?

� # of H+ x MA x VA = # of OH- x MB x VB

29

Acid-Base Reaction� 75 mL of 0.25M HCl is mixed with 225

mL of 0.055 M Ba(OH)2 . What is the concentration of the excess H+ or OH- ?

30

Types of Reaction3 Oxidation-Reduction called Redox� Ionic compounds are formed through

the transfer of electrons.� An Oxidation-reduction reaction

involves the transfer of electrons.� We need a way of keeping track.

6

31

Oxidation States� A way of keeping track of the electrons.� Not necessarily true of what is in nature,

but it works.� need the rules for assigning

(memorize).1 The oxidation state of elements in their

standard states is zero.2 Oxidation state for monoatomic ions are

the same as their charge.

32

Oxidation states3 Oxygen is assigned an oxidation state of -

2 in its covalent compounds except as a peroxide.

4 In compounds with nonmetals hydrogen is assigned the oxidation state +1.

5 In its compounds fluorine is always –1.6 The sum of the oxidation states must be

zero in compounds or equal the charge of the ion.

33

Oxidation States� Assign the oxidation states to each

element in the following.

� CO2� NO3

-

� H2SO4� Fe2O3� Fe3O4

34

Oxidation-Reduction� Transfer electrons, so the oxidation

states change.

� Na + 2Cl2 → 2NaCl

� CH4 + 2O2 → CO2 + 2H2O� Oxidation is the loss of electrons.� Reduction is the gain of electrons.� OIL RIG� LEO GER

35

Oxidation-Reduction� Oxidation means an increase in

oxidation state - lose electrons.

� Reduction means a decrease in oxidation state - gain electrons.

� The substance that is oxidized is called the reducing agent.

� The substance that is reduced is called the oxidizing agent.

36

Redox Reactions

7

37

Agents� Oxidizing agent gets reduced.� Gains electrons.� More negative oxidation state.� Reducing agent gets oxidized.

� Loses electrons.� More positive oxidation state.

38

Identify the � Oxidizing agent� Reducing agent� Substance oxidized� Substance reduced� in the following reactions� Fe (s) + O2(g) → Fe2O3(s) � Fe2O3(s)+ 3 CO(g) → 2 Fe(l) + 3 CO2(g)� SO3

2- + H+ + MnO4- →SO4

2- + H2O + Mn2+

39

Half-Reactions� All redox reactions can be thought of as

happening in two halves.� One produces electrons - Oxidation half.� The other requires electrons - Reduction

half.� Write the half reactions for the following.

� Na + Cl2 →→→→ Na+ + Cl-

� SO32- + H+ + MnO4

- →→→→SO4

2- + H2O + Mn+2

40

Balancing Redox Equations� In aqueous solutions the key is the

number of electrons produced must be the same as those required.

� For reactions in acidic solution an 8 step procedure.

1 Write separate half reactions2 For each half reaction balance all

reactants except H and O3 Balance O using H2O

41

Acidic Solution

4 Balance H using H+

5 Balance charge using e-

6 Multiply equations to make electrons equal

7 Add equations and cancel identical species

8 Check that charges and elements are balanced.

42

Practice� The following reactions occur in aqueous

solution. Balance them

� MnO4- + Fe+2 → Mn+2 + Fe+3

� Cu + NO3- → Cu+2 + NO(g)

� Pb + PbO2 + SO4-2 → PbSO4

� Mn+2 + NaBiO3 → Bi+3 + MnO4-

8

43

Now for a tough one� Fe(CN)6

-4 + MnO4- →

Mn+2 + Fe+3 + CO2 + NO3-

44

Basic Solution� Do everything you would with acid, but

add one more step.� Add enough OH- to both sides to

neutralize the H+

� Makes water

� CrI3 + Cl2 → CrO42- + IO4

- + Cl-

� Fe(OH)2 + H2O2 → Fe(OH)-

� Cr(OH)3 + OCl- + OH- → CrO4

2- + Cl- + H2O

45

Redox Titrations

� Same as any other titration.� the permanganate ion is used often

because it is its own indicator. MnO4- is

purple, Mn+2 is colorless. When reaction solution remains clear, MnO4

-

is gone.� Chromate ion is also useful, but color

change, orangish yellow to green, is harder to detect.

46

Example� The iron content of iron ore can be

determined by titration with standard KMnO4 solution. The iron ore is dissolved in excess HCl, and the iron reduced to Fe+2 ions. This solution is then titrated with KMnO4 solution, producing Fe+3 and Mn+2 ions in acidic solution. If it requires 41.95 mL of 0.205 M KMnO4 to titrate a solution made with 0.6128 g of iron ore, what percent of the ore was iron?

47

Extra Credit� Nuclear Power� Write a paper that describes � 1. How does it work? � 2. What are the advantages?� 3. What are the disadvantages?� 4. Using your information to support your

conclusion, answer the question, “What role should nuclear power play in future energy generation for the United States?”

� 5-7 pages� Researched using MLA style with in text citations.� Due Oct. 26- no exceptions.

1

1

Chapter 5Chapter 5

The Gas LawsThe Gas Laws

2

PressurePressure�� Force per unit area.Force per unit area.�� Gas molecules fill container.Gas molecules fill container.�� Molecules move around and hit Molecules move around and hit

sides.sides.�� Collisions are the force.Collisions are the force.�� Container has the area.Container has the area.�� Measured with a barometer.Measured with a barometer.

3

BarometerBarometer�� The pressure of the The pressure of the

atmosphere at sea atmosphere at sea level will hold a level will hold a column of mercury column of mercury 760 mm Hg.760 mm Hg.

�� 1 atm = 760 mm Hg1 atm = 760 mm Hg

1 atm Pressure

760 mm Hg

Vacuum

4

ManometerManometer

Gas

h

�� Column of Column of mercury to mercury to measure measure pressure.pressure.

�� h is how much h is how much lower the lower the pressure is pressure is than outside. than outside.

5

ManometerManometer�� h is how much h is how much

higher the gas higher the gas pressure is pressure is than the than the atmosphere.atmosphere.h

Gas

6

Units of pressureUnits of pressure�� 1 atmosphere = 760 mm Hg1 atmosphere = 760 mm Hg�� 1 mm Hg = 1 torr1 mm Hg = 1 torr�� 1 atm = 101,325 Pascals = 101.325 kPa1 atm = 101,325 Pascals = 101.325 kPa�� Can make conversion factors from Can make conversion factors from

these.these.�� What is 724 mm Hg in kPa?What is 724 mm Hg in kPa?�� in torr?in torr?�� in atm?in atm?

2

7

The Gas LawsThe Gas Laws�� Boyle’s LawBoyle’s Law�� Pressure and volume are inversely Pressure and volume are inversely

related at constant temperature.related at constant temperature.�� PV= kPV= k�� As one goes up, the other goes As one goes up, the other goes

down.down.�� PP11VV11 = P= P22 VV22�� GraphicallyGraphically

8

V

P (at constant T)

9

V

1/P (at constant T)

Slope = k

10

PV

P (at constant T)

CO2

O2

22.

41

L a

tm

11

ExamplesExamples�� 20.5 L of nitrogen at 25ºC and 742 20.5 L of nitrogen at 25ºC and 742

torr are compressed to 9.8 atm at torr are compressed to 9.8 atm at constant T. What is the new volume?constant T. What is the new volume?

�� 30.6 mL of carbon dioxide at 740 torr 30.6 mL of carbon dioxide at 740 torr is expanded at constant temperature is expanded at constant temperature to 750 mL. What is the final pressure to 750 mL. What is the final pressure in kPa? in kPa?

12

Charles’ LawCharles’ Law�� Volume of a gas varies directly with Volume of a gas varies directly with

the absolute temperature at constant the absolute temperature at constant pressure.pressure.

�� V = kT (if T is in Kelvin)V = kT (if T is in Kelvin)

�� VV1 1 = V= V22

TT11 = T= T22

�� GraphicallyGraphically

3

13

V (

L)

T (ºC)

H2O

HeCH4

H2

-273.15ºC14

ExamplesExamples�� What would the final volume be if 247 What would the final volume be if 247

mL of gas at 22ºC is heated to 98ºC , mL of gas at 22ºC is heated to 98ºC , if the pressure is held constant?if the pressure is held constant?

15

ExamplesExamples�� At what temperature would 40.5 L of At what temperature would 40.5 L of

gas at 23.4ºC have a volume of 81.0 gas at 23.4ºC have a volume of 81.0 L at constant pressure? L at constant pressure?

16

Avogadro's LawAvogadro's Law�� Avagadro’sAvagadro’s�� At constant temperature and At constant temperature and

pressure, the volume of gas is pressure, the volume of gas is directly related to the number of directly related to the number of moles.moles.

�� V = k n (n is the number of moles)V = k n (n is the number of moles)

�� VV1 1 = V= V22

nn11 = n= n22

17

GayGay-- Lussac LawLussac Law�� At constant volume, pressure and At constant volume, pressure and

absolute temperature are directly absolute temperature are directly related.related.

�� P = k TP = k T

�� PP1 1 = P= P22

TT11 = T= T22

18

Combined Gas LawCombined Gas Law�� If the moles of gas remains constant, If the moles of gas remains constant,

use this formula and cancel out the use this formula and cancel out the other things that don’t change.other things that don’t change.

�� PP1 1 VV11 = P= P22 VV22

.. TT11 TT22

4

19

ExamplesExamples�� A deodorant can has a volume of 175 A deodorant can has a volume of 175

mL and a pressure of 3.8 atm at 22ºC. mL and a pressure of 3.8 atm at 22ºC. What would the pressure be if the What would the pressure be if the can was heated to 100.ºC?can was heated to 100.ºC?

�� What volume of gas could the can What volume of gas could the can release at 22ºC and 743 torr?release at 22ºC and 743 torr?

20

Ideal Gas LawIdeal Gas Law�� PV = nRTPV = nRT�� V = 22.42 L at 1 atm, 0ºC, n = 1 mole, V = 22.42 L at 1 atm, 0ºC, n = 1 mole,

what is R?what is R?�� R is the ideal gas constant.R is the ideal gas constant.�� R = 0.08206 L atm/ mol KR = 0.08206 L atm/ mol K�� Tells you about a gas is NOW.Tells you about a gas is NOW.�� The other laws tell you about a gas The other laws tell you about a gas

when it changes. when it changes.

21

Ideal Gas LawIdeal Gas Law�� An An equation of stateequation of state..�� Independent of how you end up Independent of how you end up

where you are at. where you are at. �� Does not depend on the path.Does not depend on the path.�� Given 3 you can determine the Given 3 you can determine the

fourth.fourth.�� An Empirical Equation An Empirical Equation -- based on based on

experimental evidence.experimental evidence.

22

Ideal Gas LawIdeal Gas Law�� A hypothetical substance A hypothetical substance -- the ideal the ideal

gasgas�� Think of it as a limit.Think of it as a limit.�� Gases only approach ideal behavior Gases only approach ideal behavior

at low pressure (< 1 atm) and high at low pressure (< 1 atm) and high temperature.temperature.

�� Use the laws anyway, unless told to Use the laws anyway, unless told to do otherwise.do otherwise.

�� They give good estimates.They give good estimates.

23

ExamplesExamples�� A 47.3 L container containing 1.62 mol of A 47.3 L container containing 1.62 mol of

He is heated until the pressure reaches He is heated until the pressure reaches 1.85 atm. What is the temperature?1.85 atm. What is the temperature?

�� Kr gas in a 18.5 L cylinder exerts a Kr gas in a 18.5 L cylinder exerts a pressure of 8.61 atm at 24.8ºC What is pressure of 8.61 atm at 24.8ºC What is the mass of Kr?the mass of Kr?

�� A sample of gas has a volume of 4.18 L A sample of gas has a volume of 4.18 L at 29ºC and 732 torr. What would its at 29ºC and 732 torr. What would its volume be at 24.8ºC and 756 torr?volume be at 24.8ºC and 756 torr?

24

Gas Density and Molar MassGas Density and Molar Mass�� D = m/VD = m/V�� Let Let MM stand for molar massstand for molar mass�� MM = m/n = m/n �� n= PV/RTn= PV/RT�� MM = m = m

PV/RTPV/RT�� MM = mRT = m RT = DRT= mRT = m RT = DRT

PV PV V PV P PP

5

25

Examples Examples �� What is the density of ammonia at What is the density of ammonia at

23ºC and 735 torr?23ºC and 735 torr?�� A compound has the empirical A compound has the empirical

formula CHCl. A 256 mL flask at formula CHCl. A 256 mL flask at 100.ºC and 750 torr contains .80 g of 100.ºC and 750 torr contains .80 g of the gaseous compound. What is the the gaseous compound. What is the molecular formula?molecular formula?

26

Gases and StoichiometryGases and Stoichiometry�� Reactions happen in molesReactions happen in moles�� At Standard Temperature and At Standard Temperature and

Pressure (STP, 0ºC and 1 atm) 1 Pressure (STP, 0ºC and 1 atm) 1 mole of gas occupies 22.42 L.mole of gas occupies 22.42 L.

�� If not at STP, use the ideal gas law to If not at STP, use the ideal gas law to calculate moles of reactant or calculate moles of reactant or volume of product.volume of product.

27

ExamplesExamples�� Mercury can be achieved by the Mercury can be achieved by the

following reactionfollowing reaction

What volume of oxygen gas can What volume of oxygen gas can be produced from 4.10 g of mercury be produced from 4.10 g of mercury (II) oxide at STP?(II) oxide at STP?

�� At 400.ºC and 740 torr?At 400.ºC and 740 torr?

→HgO Hg(l) + O (g) heat

2

28

ExamplesExamples�� Using the following reactionUsing the following reaction

calculate the mass of sodium hydrogen calculate the mass of sodium hydrogen carbonate necessary to produce 2.87 L carbonate necessary to produce 2.87 L of carbon dioxide at 25ºC and 2.00 atm.of carbon dioxide at 25ºC and 2.00 atm.

�� If 27 L of gas are produced at 26ºC and If 27 L of gas are produced at 26ºC and 745 torr when 2.6 L of HCl are added 745 torr when 2.6 L of HCl are added what is the concentration of HCl?what is the concentration of HCl?

NaCl(aq) + CO (g) +H O(l)2 2

NaHCO (s) + HCl 3 →

29

ExamplesExamples�� Consider the following reactionConsider the following reaction

What volume of NO at 1.0 atm and What volume of NO at 1.0 atm and 1000ºC can be produced from 10.0 L 1000ºC can be produced from 10.0 L of NHof NH33 and excess Oand excess O 22 at the same at the same temperature and pressure?temperature and pressure?

�� What volume of OWhat volume of O 22 measured at STP measured at STP will be consumed when 10.0 kg NHwill be consumed when 10.0 kg NH 33is reacted?is reacted?

4NH (g) + 5 O 4 NO(g)+ 6H O(g)3 22( )g →

30

ExamplesExamples

�� What mass of HWhat mass of H 22O will be produced O will be produced from 65.0 L of Ofrom 65.0 L of O 22 and 75.0 L of NHand 75.0 L of NH 33both measured at STP? both measured at STP?

�� What volume Of NO would be What volume Of NO would be produced?produced?

�� What mass of NO is produced from What mass of NO is produced from 500. L of NH500. L of NH 33 at 250.0ºC and 3.00 at 250.0ºC and 3.00 atm?atm?

4NH (g) + 5 O 4 NO(g)+ 6H O(g)3 22( )g →

6

31

Dalton’s LawDalton’s Law�� The total pressure in a container is The total pressure in a container is

the sum of the pressure each gas the sum of the pressure each gas would exert if it were alone in the would exert if it were alone in the container.container.

�� The total pressure is the sum of the The total pressure is the sum of the partial pressures.partial pressures.

�� PPTotalTotal = P= P11 + P+ P22 + P+ P33 + P+ P44 + P+ P55 ......�� For each P = nRT/VFor each P = nRT/V

32

Dalton's LawDalton's Law�� PPTotalTotal = n= n11RT + nRT + n22RT + nRT + n33RT +...RT +...

V V VV VV

�� In the same container R, T and V are In the same container R, T and V are the same.the same.

�� PPTotalTotal = (n= (n11+ n+ n22 + n+ n33+...)RT+...)RTVV

�� PPTotalTotal = (n= (nTotalTotal )RT)RTVV

33

The mole fractionThe mole fraction�� Ratio of moles of the substance to Ratio of moles of the substance to

the total moles.the total moles.

�� symbol is Greek letter chi symbol is Greek letter chi χχχχχχχχ�� χχχχχχχχ11111111 = n= n11 = P= P1 1

nnTotal Total PPTotalTotal

34

ExamplesExamples�� The partial pressure of nitrogen in air The partial pressure of nitrogen in air

is 592 torr. Air pressure is 752 torr, is 592 torr. Air pressure is 752 torr, what is the mole fraction of nitrogen?what is the mole fraction of nitrogen?

�� What is the partial pressure of What is the partial pressure of nitrogen if the container holding the nitrogen if the container holding the air is compressed to 5.25 atm?air is compressed to 5.25 atm?

35

ExamplesExamples

3.50 LO2

1.50 LN2

2.70 atm�� When these valves are opened, what is When these valves are opened, what is

each partial pressure and the total each partial pressure and the total pressure?pressure?

4.00 LCH4

4.58 atm 0.752 atm

36

Vapor PressureVapor Pressure�� Water evaporates!Water evaporates!�� When that water evaporates, the When that water evaporates, the

vapor has a pressure.vapor has a pressure.�� Gases are often collected over water Gases are often collected over water

so the vapor pressure of water must so the vapor pressure of water must be subtracted from the total pressure be subtracted from the total pressure to find the pressure of the gas.to find the pressure of the gas.

�� It must be given.It must be given.

7

37

ExampleExample�� NN22O can be produced by the O can be produced by the

following reactionfollowing reaction

what volume of Nwhat volume of N 22O collected over O collected over water at a total pressure of 94 kPa water at a total pressure of 94 kPa and 22ºC can be produced from 2.6 g and 22ºC can be produced from 2.6 g of NHof NH44NONO33? ( the vapor pressure of ? ( the vapor pressure of water at 22ºC is 21 torr)water at 22ºC is 21 torr)

NH NO NO (g) + 2H O4 heat

23 2( ) ( )s l →

38

Kinetic Molecular TheoryKinetic Molecular Theory�� Theory tells why the things happen.Theory tells why the things happen.�� explains why ideal gases behave the explains why ideal gases behave the

way they do.way they do.�� Assumptions that simplify the Assumptions that simplify the

theory, but don’t work in real gases.theory, but don’t work in real gases.11 The particles are so small we can The particles are so small we can

ignore their volume.ignore their volume.22 The particles are in constant motion The particles are in constant motion

and their collisions cause pressure. and their collisions cause pressure.

39

Kinetic Molecular TheoryKinetic Molecular Theory33 The particles do not affect each The particles do not affect each

other, neither attracting or repelling.other, neither attracting or repelling.44 The average kinetic energy is The average kinetic energy is

proportional to the Kelvin proportional to the Kelvin temperature.temperature.

�� Appendix 2 shows the derivation of Appendix 2 shows the derivation of the ideal gas law and the definition of the ideal gas law and the definition of temperature.temperature.

�� We need the formula KE = 1/2 mvWe need the formula KE = 1/2 mv 22

40

What it tells usWhat it tells us�� (KE)(KE)avgavg = 3/2 RT= 3/2 RT�� This the meaning of temperature.This the meaning of temperature.�� u is the particle velocity.u is the particle velocity.�� u is the average particle velocity.u is the average particle velocity.

�� u u 22 is the average of the squared is the average of the squared particle velocity.particle velocity.

�� the root mean square velocity is the root mean square velocity is

√√√√√√√√ u u 2 = 2 = uu rmsrms

41

Combine these two equationsCombine these two equations

�� For a mole of gas For a mole of gas �� NNA A is Avogadro's numberis Avogadro's number��

�� 2A

1 3N ( mu ) = RT

2 2

avg

3(KE) = RT

2

2avg A

1(KE) =N ( mu )

2

2

A

3RTu =

N m42

Combine these two equationsCombine these two equations��

�� m is kg for one particle, so Nm is kg for one particle, so N aam is kg m is kg for a mole of particles. We will call it for a mole of particles. We will call it MM

�� Where Where MM is the molar mass in is the molar mass in kgkg/mole, /mole, and R has the units 8.3145 J/Kmol.and R has the units 8.3145 J/Kmol.

�� The velocity will be in m/sThe velocity will be in m/s

2

A

3RTu = u

N rmsm=

3RT u = rms M

8

43

Example Example �� Calculate the root mean square Calculate the root mean square

velocity of carbon dioxide at 25ºC.velocity of carbon dioxide at 25ºC.�� Calculate the root mean square Calculate the root mean square

velocity of hydrogen at 25ºC.velocity of hydrogen at 25ºC.�� Calculate the root mean square Calculate the root mean square

velocity of chlorine at 250ºC.velocity of chlorine at 250ºC.

44

Range of velocitiesRange of velocities�� The average distance a molecule The average distance a molecule

travels before colliding with another travels before colliding with another is called the mean free path and is is called the mean free path and is small (near 10small (near 10 --77))

�� Temperature is an average. There are Temperature is an average. There are molecules of many speeds in the molecules of many speeds in the average.average.

�� Shown on a graph called a velocity Shown on a graph called a velocity distributiondistribution

45

nu

mb

er

of p

art

icle

s

Molecular Velocity

273 K2RTMv

223 2

evRT

Μ4 f(v)

−

2=

ππ

46

nu

mb

er

of p

art

icle

s

Molecular Velocity

273 K

1273 K

2RTMv

223 2

evRT

Μ4 f(v)

−

2=

ππ

47

VelocityVelocity�� Average increases as temperature Average increases as temperature

increases.increases.�� Spread increases as temperature Spread increases as temperature

increases.increases.

2RT

Mv2

2

3 2

evRT

Μ4 f(v)

−

2=

ππ

48

EffusionEffusion�� Passage of gas through a small hole, Passage of gas through a small hole,

into a vacuum.into a vacuum.�� The effusion rate measures how fast The effusion rate measures how fast

this happens.this happens.�� Graham’s Law the rate of effusion is Graham’s Law the rate of effusion is

inversely proportional to the square inversely proportional to the square root of the mass of its particles.root of the mass of its particles.

9

49

EffusionEffusion�� Passage of gas through a small hole, Passage of gas through a small hole,

into a vacuum.into a vacuum.�� The effusion rate measures how fast The effusion rate measures how fast

this happens.this happens.�� Graham’s Law the rate of effusion is Graham’s Law the rate of effusion is

inversely proportional to the square inversely proportional to the square root of the mass of its particles.root of the mass of its particles.

1

2

2 gasfor effusion of Rate

1 gasfor effusion of Rate

M

M=

50

DerivingDeriving�� The rate of effusion should be The rate of effusion should be

proportional to uproportional to u rmsrms

�� Effusion Rate 1 = uEffusion Rate 1 = u rms rms 11Effusion Rate 2 = uEffusion Rate 2 = u rms rms 22

51

DerivingDeriving�� The rate of effusion should be The rate of effusion should be

proportional to uproportional to u rmsrms

�� Effusion Rate 1 = uEffusion Rate 1 = u rms rms 11Effusion Rate 2 = uEffusion Rate 2 = u rms rms 22

effusion rate 1

effusion rate 2

u 1

u 2

3RT

M

3RT

M2

M

Mrms

rms

1= = = 2

1

52

DiffusionDiffusion�� The spreading of a gas through a The spreading of a gas through a

room.room.�� Slow considering molecules move at Slow considering molecules move at

100’s of meters per second.100’s of meters per second.�� Collisions with other molecules slow Collisions with other molecules slow

down diffusions.down diffusions.�� Best estimate is Graham’s Law.Best estimate is Graham’s Law.

53

ExamplesExamples�� Helium effuses through a porous Helium effuses through a porous

cylinder 3.20 time faster than a cylinder 3.20 time faster than a compound. What is it’s molar mass?compound. What is it’s molar mass?

�� If 0.00251 mol of NHIf 0.00251 mol of NH 33 effuse through a effuse through a hole in 2.47 min, how much HCl would hole in 2.47 min, how much HCl would effuse in the same time?effuse in the same time?

�� A sample of NA sample of N 22 effuses through a hole effuses through a hole in 38 seconds. what must be the in 38 seconds. what must be the molecular weight of gas that effuses in molecular weight of gas that effuses in 55 seconds under identical conditions? 55 seconds under identical conditions?

54

DiffusionDiffusion�� The spreading of a gas through a The spreading of a gas through a

room.room.�� Slow considering molecules move at Slow considering molecules move at

100’s of meters per second.100’s of meters per second.�� Collisions with other molecules slow Collisions with other molecules slow

down diffusions.down diffusions.�� Best estimate is Graham’s Law.Best estimate is Graham’s Law.

10

55

Real GasesReal Gases�� Real molecules do take up space and Real molecules do take up space and

they do interact with each other they do interact with each other (especially polar molecules).(especially polar molecules).

�� Need to add correction factors to the Need to add correction factors to the ideal gas law to account for these.ideal gas law to account for these.

56

Volume CorrectionVolume Correction�� The actual volume free to move in is The actual volume free to move in is

less because of particle size.less because of particle size.�� More molecules will have more effect.More molecules will have more effect.�� Bigger molecules have more effectBigger molecules have more effect�� Corrected volume V’ = V Corrected volume V’ = V -- nbnb�� b is a constant that differs for each gas.b is a constant that differs for each gas.

�� PP’’ = nRT= nRT(V(V--nb)nb)

57

Pressure correctionPressure correction�� Because the molecules are attracted Because the molecules are attracted

to each other, the pressure on the to each other, the pressure on the container will be less than ideal gascontainer will be less than ideal gas

�� Depends on the type of moleculeDepends on the type of molecule�� depends on the number of molecules depends on the number of molecules

per liter.per liter.�� since two molecules interact, the since two molecules interact, the

effect must be squared.effect must be squared.

58

Pressure correctionPressure correction�� Because the molecules are attracted Because the molecules are attracted

to each other, the pressure on the to each other, the pressure on the container will be less than idealcontainer will be less than ideal

�� depends on the number of molecules depends on the number of molecules per liter.per liter.

�� since two molecules interact, the since two molecules interact, the effect must be squared.effect must be squared. 2

observed V

nP'-a P

=

59

AltogetherAltogether��

�� Called the Van der Waal’s equation if Called the Van der Waal’s equation if rearrangedrearranged

�� Corrected Corrected Corrected Corrected Pressure Pressure VolumeVolume

( )P + an

V x V- nb nRTobs

=2

2

observed V

na-

nb-V

nRT P

=

60

Where does it come fromWhere does it come from�� a and b are determined by a and b are determined by

experiment.experiment.�� Different for each gas.Different for each gas.�� Look them upLook them up�� Bigger molecules have larger b.Bigger molecules have larger b.�� a depends on both size and polarity.a depends on both size and polarity.�� once given, plug and chug.once given, plug and chug.

11

61

ExampleExample�� Calculate the pressure exerted by Calculate the pressure exerted by

0.5000 mol Cl0.5000 mol Cl 22 in a 1.000 L container in a 1.000 L container at 25.0ºCat 25.0ºC

�� Using the ideal gas law.Using the ideal gas law.�� Van der Waal’s equationVan der Waal’s equation

––a = 6.49 atm La = 6.49 atm L 22 /mol/mol 22

––b = 0.0562 L/molb = 0.0562 L/mol

1

11

Chapter 6Chapter 6

EnergyEnergy

ThermodynamicsThermodynamics

22

Energy is...Energy is...�� The ability to do work.The ability to do work.

�� Conserved.Conserved.

�� made of heat and work.made of heat and work.

�� a state function.a state function.

�� independent of the path, or how you get independent of the path, or how you get from point A to B.from point A to B.

�� Work is a force acting over a distance.Work is a force acting over a distance.

�� Heat is energy transferred between Heat is energy transferred between objects because of temperature difference.objects because of temperature difference.

33

The universeThe universe�� is divided into two halves.is divided into two halves.

�� the system and the surroundings.the system and the surroundings.

�� The system is the part you are The system is the part you are concerned with.concerned with.

�� The surroundings are the rest.The surroundings are the rest.

�� Exothermic reactions release energy to Exothermic reactions release energy to the surroundings.the surroundings.

�� Endothermic reactions absorb energy Endothermic reactions absorb energy from the surroundings.from the surroundings.

44

CH + 2O CO + 2H O + Heat4 2 2 2→

CH + 2O 4 2

CO + 2 H O 2 2Po

ten

tia

l en

erg

y

Heat

55

N + O2 2 Po

ten

tia

l en

erg

y

Heat

2NO

N + O 2NO2 2 + heat →

66

DirectionDirection�� Every energy measurement has three Every energy measurement has three

parts.parts.

1.1. A unit ( Joules of calories).A unit ( Joules of calories).

2.2. A number how many.A number how many.

3.3. and a sign to tell direction.and a sign to tell direction.

�� negative negative -- exothermicexothermic

�� positivepositive-- endothermicendothermic

2

77

System

Surroundings

Energy

∆E <0

88

System

Surroundings

Energy

∆E >0

99

Same rules for heat and workSame rules for heat and work�� Heat given off is negative.Heat given off is negative.

�� Heat absorbed is positive.Heat absorbed is positive.

�� Work done by system on surroundings Work done by system on surroundings is positive.is positive.

�� Work done on system by surroundings Work done on system by surroundings is negative.is negative.

�� ThermodynamicsThermodynamics-- The study of energy The study of energy and the changes it undergoes.and the changes it undergoes.

1010

First Law of ThermodynamicsFirst Law of Thermodynamics�� The energy of the universe is constant.The energy of the universe is constant.

�� Law of conservation of energy.Law of conservation of energy.

�� q = heatq = heat

�� w = workw = work

�� ∆∆E = q + wE = q + w

�� Take the systems point of view to Take the systems point of view to decide signs.decide signs.

�� Energy is state functionEnergy is state function

�� Heat and work are notHeat and work are not

1111

What is work?What is work?�� Work is a force acting over a distance.Work is a force acting over a distance.

�� w= F x w= F x ∆∆dd

�� P = F/areaP = F/area

�� d = V/aread = V/area

�� w= (P x area) x w= (P x area) x ∆∆ (V/area)= P(V/area)= P∆∆VV

�� Work can be calculated by multiplying Work can be calculated by multiplying pressure by the change in volume at pressure by the change in volume at constant pressure.constant pressure.

�� units of liter x atm = Lunits of liter x atm = L--atmatm

1212

Work needs a signWork needs a sign�� If the volume of a gas increases, the If the volume of a gas increases, the

system has done work on the system has done work on the surroundings.surroundings.

�� work is negativework is negative

�� w = w = -- PP∆∆VV

�� Expanding work is negative.Expanding work is negative.

�� Contracting, surroundings do work on Contracting, surroundings do work on the system w is positive.the system w is positive.

�� 1 L atm = 101.3 J1 L atm = 101.3 J

3

1313

ExamplesExamples�� What amount of work is done when 15 What amount of work is done when 15

L of gas is expanded to 25 L at 2.4 atm L of gas is expanded to 25 L at 2.4 atm pressure?pressure?

�� If 2.36 J of heat are absorbed by the gas If 2.36 J of heat are absorbed by the gas above, what is the change in energy?above, what is the change in energy?

�� How much heat would it take to change How much heat would it take to change the gas without changing the internal the gas without changing the internal energy of the gas? energy of the gas?

1414

EnthalpyEnthalpy�� abbreviated Habbreviated H

�� H = E + PV (that’s the definition)H = E + PV (that’s the definition)

�� ∆∆H = H = ∆∆E + E + ∆∆PVPV

�� at constant pressure.at constant pressure.

�� ∆∆H = H = ∆∆E + PE + P∆∆VV

�� the heat at constant pressure qthe heat at constant pressure qpp can be can be calculated fromcalculated from

�� ∆∆E = qE = qpp + w = q+ w = qpp -- PP∆∆VV

�� qqpp = = ∆∆E + P E + P ∆∆V = V = ∆∆HH

1515

CalorimetryCalorimetry�� Measuring heat.Measuring heat.�� Use a calorimeter.Use a calorimeter.�� Two kindsTwo kinds�� Constant pressure calorimeter (called a Constant pressure calorimeter (called a

coffee cup calorimeter)coffee cup calorimeter)�� heat capacity for a material, C is heat capacity for a material, C is

calculated calculated �� C= heat absorbed/ C= heat absorbed/ ∆∆T = T = ∆∆H/ H/ ∆∆TT�� specific heat capacity = C/mass specific heat capacity = C/mass �� Q = Cm Q = Cm ∆∆TT

1616

CalorimetryCalorimetry�� molar heat capacity = C/molesmolar heat capacity = C/moles

�� heat = specific heat x m x heat = specific heat x m x ∆∆TT

�� heat = molar heat x moles x heat = molar heat x moles x ∆∆TT

�� Make the units work and you’ve done Make the units work and you’ve done the problem right.the problem right.

�� A coffee cup calorimeter measures A coffee cup calorimeter measures ∆∆H.H.

�� An insulated cup, full of water. An insulated cup, full of water.

�� The specific heat of water is 1 cal/gºCThe specific heat of water is 1 cal/gºC

�� Heat of reaction= Heat of reaction= ∆∆H = C x mass x H = C x mass x ∆∆TT

1717

ExamplesExamples�� The specific heat of graphite is 0.71 The specific heat of graphite is 0.71

J/gºC. Calculate the energy needed to J/gºC. Calculate the energy needed to raise the temperature of 75 kg of raise the temperature of 75 kg of graphite from 294 K to 348 K.graphite from 294 K to 348 K.

�� A 46.2 g sample of copper is heated to A 46.2 g sample of copper is heated to 95.4ºC and then placed in a calorimeter 95.4ºC and then placed in a calorimeter containing 75.0 g of water at 19.6ºC. The containing 75.0 g of water at 19.6ºC. The final temperature of both the water and final temperature of both the water and the copper is 21.8ºC. What is the specific the copper is 21.8ºC. What is the specific heat of copper?heat of copper?

1818

CalorimetryCalorimetry�� Constant volume calorimeter is called a Constant volume calorimeter is called a

bomb calorimeter.bomb calorimeter.

�� Material is put in a container with pure Material is put in a container with pure oxygen. Wires are used to start the oxygen. Wires are used to start the combustion. The container is put into a combustion. The container is put into a container of water.container of water.

�� The heat capacity of the calorimeter is The heat capacity of the calorimeter is known and/or tested.known and/or tested.

�� Since Since ∆∆V = 0, PV = 0, P∆∆V = 0, V = 0, ∆∆E = q E = q

4

1919

Bomb CalorimeterBomb Calorimeter

�� thermometerthermometer

�� stirrerstirrer

�� full of waterfull of water

�� ignition wireignition wire

�� Steel bombSteel bomb

�� samplesample

2020

2121

PropertiesProperties�� intensive properties not related to the intensive properties not related to the

amount of substance.amount of substance.

�� density, specific heat, temperature.density, specific heat, temperature.

�� Extensive property Extensive property -- does depend on does depend on the amount of stuff.the amount of stuff.

�� Heat capacity, mass, heat from a Heat capacity, mass, heat from a reaction.reaction.

2222

Hess’s LawHess’s Law�� Enthalpy is a state function.Enthalpy is a state function.

�� It is independent of the path.It is independent of the path.

�� We can add equations to come up with We can add equations to come up with the desired final product, and add the the desired final product, and add the ∆∆HH

�� Two rulesTwo rules

– If the reaction is reversed the sign of ∆H is changed

– If the reaction is multiplied, so is ∆H

2323

N2 2O2

68 kJ

NO2

O2 2NO

180 kJ

-112 kJ

H (

kJ)

2424

Standard EnthalpyStandard Enthalpy�� The enthalpy change for a reaction at The enthalpy change for a reaction at

standard conditions (25ºC, 1 atm , 1 M standard conditions (25ºC, 1 atm , 1 M solutions)solutions)

�� Symbol Symbol ∆∆HºHº

�� When using Hess’s Law, work by When using Hess’s Law, work by adding the equations up to make it look adding the equations up to make it look like the answer. like the answer.

�� The other parts will cancel out.The other parts will cancel out.

5

2525

H (g) + 1

2O (g) H (l) 2 2 2O→

C(s) + O (g) CO (g) 2 2→∆Hº= -394 kJ

∆Hº= -286 kJ

C H (g) + 5

2O (g) 2CO (g) + H O( ) 2 2 2 2 2→ l

ExampleExample�� GivenGiven

calculate calculate ∆∆Hº for this reactionHº for this reaction

∆Hº= -1300. kJ

2C(s) + H (g) C H (g) 2 2 2→2626

ExampleExample

O (g) + H (g) 2OH(g) 2 2 →O (g) 2O(g)2 →H (g) 2H(g)2 →

O(g) + H(g) OH(g) →

Given

Calculate ∆Hº for this reaction

∆Hº= +77.9kJ

∆Hº= +495 kJ

∆Hº= +435.9kJ

2727

Standard Enthalpies of FormationStandard Enthalpies of Formation�� Hess’s Law is much more useful if you Hess’s Law is much more useful if you

know lots of reactions.know lots of reactions.

�� Made a table of Made a table of standard heats of standard heats of formationformation. The amount of heat needed . The amount of heat needed to for 1 mole of a compound from its to for 1 mole of a compound from its elements in their standard states.elements in their standard states.

�� Standard states are 1 atm, 1M and 25ºCStandard states are 1 atm, 1M and 25ºC

�� For an element it is 0For an element it is 0

�� There is a table in Appendix 4 (pg A22)There is a table in Appendix 4 (pg A22)

2828

Standard Enthalpies of FormationStandard Enthalpies of Formation�� Need to be able to write the equations.Need to be able to write the equations.

�� What is the equation that would give What is the equation that would give you the heat of formation of NOyou the heat of formation of NO22 ??

�� ½N½N2 2 (g) + O(g) + O22 (g) (g) →→ NONO22 (g)(g)

�� Have to make Have to make one mole one mole to meet the to meet the definition.definition.

�� Write the equation that would give you Write the equation that would give you the heat of formation of methanol, the heat of formation of methanol, CHCH33OH.OH.

2929

Since we can manipulate the Since we can manipulate the equationsequations

�� We can use heats of formation to figure We can use heats of formation to figure out the heat of reaction.out the heat of reaction.

�� Lets do it with this equation.Lets do it with this equation.

�� CC22HH55OH +3OOH +3O22(g) (g) →→ 2CO2CO22 + 3H+ 3H22OO

�� which leads us to this rule.which leads us to this rule.

3030

Since we can manipulate the Since we can manipulate the equationsequations

�� We can use heats of formation to figure We can use heats of formation to figure out the heat of reaction.out the heat of reaction.

�� Lets do it with this equation.Lets do it with this equation.

�� CC22HH55OH +3OOH +3O22(g) (g) →→ 2CO2CO22 + 3H+ 3H22OO

�� which leads us to this rule.which leads us to this rule.

∑ ∑( H products) - ( H reactants) = Hfo

fo o∆ ∆ ∆

6

3131

�� What is the enthalpy change for the What is the enthalpy change for the reaction reaction 2C2C22HH66+ 5O+ 5O22(g) → 4CO(g) + 6H(g) → 4CO(g) + 6H22O(l)? O(l)? Heats of formation areHeats of formation are

--84.7kJ/mole for C84.7kJ/mole for C22HH66

--110.5 kJ/mole for CO(g)110.5 kJ/mole for CO(g)--241.8 kJ/mole for H241.8 kJ/mole for H22O(g)O(g)--285.8 kJ/mole for H285.8 kJ/mole for H22O(l).) O(l).)

1

1

Chapter 7

Atomic Structure

2

Light� Made up of electromagnetic radiation� Waves of electric and magnetic fields

at right angles to each other.

3

Parts of a wave

λWavelength

Frequency = number of cycles in one secondMeasured in hertz 1 hertz = 1 cycle/second

4

Frequency = ν

5

Kinds of EM waves � There are many � different λλλλ and νννν� Radio waves, microwaves, x rays and

gamma rays are all examples� Light is only the part our eyes can

detect

GammaRays

Radiowaves

6

The speed of light� in a vacuum is 2.998 x 10 8 m/s� = c� c = λνλνλνλν� What is the wavelength of light with a

frequency 5.89 x 10 5 Hz?� What is the frequency of blue light

with a wavelength of 484 nm?

2

7

In 1900� Matter and energy were seen as

different from each other in fundamental ways

� Matter was particles� Energy could come in waves, with

any frequency.� Max Planck found that the cooling of

hot objects couldn’t be explained by viewing energy as a wave.

8

Energy is Quantized� Planck found ∆∆∆∆E came in chunks with

size h νννν� ∆∆∆∆E = nhνννν� where n is an integer.� and h is Planck’s constant � h = 6.626 x 10 -34 J s� these packets of h νννν are called

quantum

9

Einsteinis next� Said electromagnetic radiation is

quantized in particles called photons

� Each photon has energy = h νννν = hc/λλλλ� Combine this with E = mc 2

� you get the apparent mass of a photon

� m = h / (λλλλc)

10

Which is it?� Is energy a wave like light, or a

particle?� Yes � Concept is called the Wave -Particle

duality.� What about the other way, is matter a

wave? � Yes

11

Matter as a wave� Using the velocity v instead of the

frequency νννν we get

� De Broglie’s equation λλλλ = h/mv� can calculate the wavelength of an

object

12

Examples� The laser light of a CD is 7.80 x 10 2 m.

What is the frequency of this light?� What is the energy of a photon of this

light?� What is the apparent mass of a

photon of this light?� What is the energy of a mole of these

photons?

3

13

What is the wavelength?� of an electron with a mass of

9.11 x 10-31 kg traveling at

1.0 x 107 m/s?

� Of a softball with a mass of 0.10 kg moving at 125 mi/hr?

14

How do they know?� When light passes through, or

reflects off, a series of thinly spaced lines, it creates a rainbow effect

� because the waves interfere with each other.

15

A wave moves toward a slit.

16

Comes out as a curve

17

with two holes

18

with two holes Two Curves

4

19

Two Curveswith two holes

Interfere with each other

20

Two Curveswith two holes

Interfere with each other

crests add up

21

Several waves

22

Several wavesSeveral Curves

23

Several wavesSeveral waves

Interference Pattern

Several Curves

24

What will an electron do?� It has mass, so it is matter.� A particle can only go through one

hole� A wave goes through both holes� Light shows interference patterns

5

Electron “gun”

Electron as Particle

Electron “gun”

Electron as wave

Which did it do?

� It made the diffraction pattern�The electron is a wave�Led to Schrödingers equation

28

What will an electron do?� An electron does go though both,

and makes an interference pattern.� It behaves like a wave.� Other matter has wavelengths too

short to notice.

Image

29

Spectrum� The range of frequencies present in

light.� White light has a continuous

spectrum.� All the colors are possible.� A rainbow.

30

Hydrogen spectrum� Emission spectrum because these

are the colors it gives off or emits� Called a line spectrum.� There are just a few discrete lines

showing

410 nm

434 nm

486 nm

656 nm

•Spectrum

6

31

What this means� Only certain energies are allowed for

the hydrogen atom.� Can only give off certain energies.� Use ∆∆∆∆E = hνννν = hc / λλλλ� Energy in the atom is quantized

32

Niels Bohr� Developed the quantum model of the

hydrogen atom.� He said the atom was like a solar

system� The electrons were attracted to the

nucleus because of opposite charges.

� Didn’t fall in to the nucleus because it was moving around

33

The Bohr Ring Atom� He didn’t know why but only certain

energies were allowed.� He called these allowed energies

energy levels.� Putting energy into the atom moved

the electron away from the nucleus� From ground state to excited state.� When it returns to ground state it

gives off light of a certain energy34

The Bohr Ring Atom

n = 3n = 4

n = 2n = 1

35

The Bohr Model� n is the energy level� for each energy level the energy is� Z is the nuclear charge, which is +1

for hydrogen.

� E = -2.178 x 10-18 J (Z2 / n2 )

� n = 1 is called the ground state

� when the electron is removed, n = ∞� E = 0

36

We are worried about the change � When the electron moves from one

energy level to another.

� ∆∆∆∆E = Efinal - Einitial

� ∆∆∆∆E = -2.178 x 10-18 J Z2 (1/ nf2 - 1/ ni

2)

7

37

Examples� Calculate the energy need to move an

electron from its to the third energy level.

� Calculate the energy released when an electron moves from n= 4 to n=2 in a hydrogen atom.

� Calculate the energy released when an electron moves from n= 5 to n=3 in a He+1 ion

38

When is it true?� Only for hydrogen atoms and other

monoelectronic species.� Why the negative sign?� To increase the energy of the

electron you make it further to the nucleus.

� the maximum energy an electron can have is zero, at an infinite distance.

39

The Bohr Model� Doesn’t work� only works for hydrogen atoms� electrons don’t move in circles� the quantization of energy is right,

but not because they are circling like planets.

40

The Quantum Mechanical Model� A totally new approach� De Broglie said matter could be like a

wave.� De Broglie said they were like

standing waves.� The vibrations of a stringed

instrument

41 42

What’s possible?� You can only have a standing wave if

you have complete waves.� There are only certain allowed waves.� In the atom there are certain allowed

waves called electrons.� 1925 Erwin Schroedinger described

the wave function of the electron� Much math, but what is important are

the solutions

8

43

Schrödinger’s Equation� The wave function is a F(x, y, z)� Actually F(r, θ,φ)� Solutions to the equation are called

orbitals.� These are not Bohr orbits.� Each solution is tied to a certain

energy � These are the energy levels

•Animation

44

There is a limit to what we can know

� We can’t know how the electron is moving or how it gets from one energy level to another.

� The Heisenberg Uncertainty Principle� There is a limit to how well we can

know both the position and the momentum of an object.

45

Mathematically� ∆∆∆∆x · ∆∆∆∆(mv) > h/4 ππππ� ∆∆∆∆x is the uncertainty in the position� ∆∆∆∆(mv) is the uncertainty in the

momentum.� the minimum uncertainty is h/4 ππππ

46

Examples� What is the uncertainty in the

position of an electron. mass 9.31 x 10-31 kg with an uncertainty in the speed of 0.100 m/s

� What is the uncertainty in the position of a baseball, mass 0.145 kg with an uncertainty in the speed of 0.100 m/s

47

What does the wave Function mean?

� nothing.� it is not possible to visually map it.� The square of the function is the

probability of finding an electron near a particular spot.

� best way to visualize it is by mapping the places where the electron is likely to be found.

48

Pro

bab

ility

Distance from nucleus

9

49

Su

m o

f a

ll P

rob

abili

ties

Distance from nucleus50

Defining the size� The nodal surface.� The size that encloses 90% to the

total electron probability.� NOT at a certain distance, but a most

likely distance.� For the first solution it is a a sphere.

51

Quantum Numbers� There are many solutions to

Schrödinger’s equation� Each solution can be described with

quantum numbers that describe some aspect of the solution.

� Principal quantum number (n) size and energy of an orbital

� Has integer values >0

52

Quantum numbers� Angular momentum quantum number l � shape of the orbital� integer values from 0 to n-1� l = 0 is called s� l = 1 is called p� l =2 is called d� l =3 is called f� l =4 is called g

53

S orbitals

54

P orbitals

10

55

P Orbitals

56

D orbitals

57

F orbitals

58

F orbitals

59

Quantum numbers� Magnetic quantum number (m l)

– integer values between - l and + l– tells direction in each shape

� Electron spin quantum number (m s) –Can have 2 values –either +1/2 or -1/2

60

1. A2. B3. C4. A5. B6. A7. B8. A9. A

11

61

Polyelectronic Atoms� More than one electron� three energy contributions� The kinetic energy of moving electrons� The potential energy of the attraction

between the nucleus and the electrons.

� The potential energy from repulsion of electrons

62

Polyelectronic atoms� Can’t solve Schrödinger’s equation

exactly� Difficulty is repulsion of other

electrons.� Solution is to treat each electron as if it

were effected by the net field of charge from the attraction of the nucleus and the repulsion of the electrons.

� Effective nuclear charge

63

+11

11 electrons

e-Zeff

Sodium Atom

+11 10 otherelectrons

e-

64

Effective Nuclear charge � Can be calculated from

E = -2.178 x 10-18 J (Zeff2 / n2 )

� and

� ∆∆∆∆E = -2.178 x 10-18 J Zeff2 (1/ nf

2 - 1/ ni2)

65

The Periodic Table � Developed independently by German

Julius Lothar Meyer and Russian Dmitri Mendeleev (1870”s)

� Didn’t know much about atom.� Put in columns by similar properties.� Predicted properties of missing

elements.

66

Aufbau Principle� Aufbau is German for building up� As the protons are added one by

one, the electrons fill up hydrogen-like orbitals.

� Fill up in order of energy

12

67

Incr

eas

ing

ener

gy

1s

2s

3s

4s

5s6s7s

2p

3p

4p

5p6p

3d

4d

5d

7p6d

4f

5f6f

Orbitals available to a Hydrogen atom

68

Incr

eas

ing

ener

gy

1s

2s

3s

4s

5s6s

7s

2p

3p

4p

5p

6p

3d

4d

5d

7p 6d

4f

5f

With more electrons, repulsion changes the energy of the orbitals.

69

Incr

eas

ing

ener

gy

1s

2s

3s

4s

5s6s

7s

2p

3p

4p

5p

6p

3d

4d

5d

7p 6d

4f

5f

He with 2 electrons

70

Incr

eas

ing

ener

gy

1s

2s

3s

4s

5s6s

7s

2p

3p

4p

5p

6p

3d

4d

5d

7p 6d

4f

5f

71

Details� Valence electrons- the electrons in

the outermost energy levels (not d).� Core electrons- the inner electrons� Hund’s Rule- The lowest energy

configuration for an atom is the one have the maximum number of unpaired electrons in the orbital.

� C 1s2 2s2 2p2

72

Fill from the bottom up following the arrows

1s2s 2p3s 3p 3d4s 4p 4d 4f

5s 5p 5d 5f6s 6p 6d 6f7s 7p 7d 7f

• 1s2

• 2• electrons

2s2

• 4

2p6 3s2

• 12

3p6 4s2

• 20

3d10 4p6

5s2

• 38

4d105p6 6s2

• 56

13

73

Details� Elements in the same column have

the same electron configuration.� Put in columns because of similar

properties.� Similar properties because of electron

configuration.� Noble gases have filled energy levels.� Transition metals are filling the d

orbitals74

The Shorthand� Write the symbol of the noble gas

before the element � Then the rest of the electrons.� Aluminum - full configuration� 1s22s22p63s23p1

� Ne is 1s 22s22p6

� so Al is [Ne] 3s 23p1

75

The Shorthand

Sn- 50 electrons

The noble gas before it is Kr

[ Kr ]

Takes care of 36

Next 5s 2

5s2Then 4d 10

4d10Finally 5p 2 5p2

[ Kr ] 5s 24d10 5p2

76

Exceptions� Ti = [Ar] 4s 2 3d2

� V = [Ar] 4s 2 3d3

� Cr = [Ar] 4s 1 3d5

� Mn = [Ar] 4s 2 3d5

� Half filled orbitals � Scientists aren’t certain why it

happens� same for Cu [Ar] 3d 10 4s1

77

More exceptions

� Lanthanum La: [Xe] 5d 1 6s2

� Cerium Ce: [Xe] 5d 1 4f16s2

� Promethium Pr: [Xe] 4f 3 6s2

� Gadolinium Gd: [Xe] 4f 7 5d1 6s2

� Lutetium Pr: [Xe] 4f 14 5d1 6s2

� We’ll just pretend that all except Cu and Cr follow the rules.

78

More Polyelectronic� We can use Z eff to predict properties,

if we determine it’s pattern on the periodic table.

� Can use the amount of energy it takes to remove an electron for this.

� Ionization Energy- The energy necessary to remove an electron from a gaseous atom.

14

79

Remember this� E = -2.18 x 10-18 J(Z2/n2)� was true for Bohr atom.� Can be derived from quantum

mechanical model as well� for a mole of electrons being removed � E =(6.02 x 1023/mol)2.18 x 10 -18 J(Z2/n2)� E= 1.13 x 106 J/mol(Z 2/n2)� E= 1310 kJ/mol(Z 2/n2)

80

Example � Calculate the ionization energy of B +4

81

Remember our simplified atom

+11

11 e-

Zeff

1 e-

82

This gives us� Ionization energy =

1310 kJ/mol(Z eff2/n2)

� So we can measure Z eff

� The ionization energy for a 1s electron from sodium is 1.39 x 10 5 kJ/mol .

� The ionization energy for a 3s electron from sodium is 4.95 x 10 2 kJ/mol .

� Demonstrates shielding

83

Shielding� Electrons on the higher energy levels

tend to be farther out.� Have to look through the other

electrons to see the nucleus.� They are less effected by the nucleus.� lower effective nuclear charge� If shielding were completely effective,

Zeff = 1� Why isn’t it?

84

Penetration� There are levels to the electron

distribution for each orbital

2s

15

85

Graphically

Penetration

2s

Ra

dia

l Pro

bab

ility

Distance from nucleus86

Graphically

Ra

dia

l Pro

bab

ility


3s

87

Ra

dia

l Pro

bab

ility


3p

88

Ra

dia

l Pro

bab

ility


3d

89

Ra

dia

l Pro

bab

ility


4s

3d

90

Penetration effect� The outer energy levels penetrate the

inner levels so the shielding of the core electrons is not totally effective.

� from most penetration to least penetration the order is

� ns > np > nd > nf (within the same energy level)

� This is what gives us our order of filling, electrons prefer s and p

16

91

How orbitals differ� The more positive the nucleus, the

smaller the orbital.� A sodium 1s orbital is the same

shape as a hydrogen 1s orbital, but it is smaller because the electron is more strongly attracted to the nucleus.

� The helium 1s is smaller as well� This provides for better shielding

92

Zef

f

1

2

4

5

1Atomic Number

93

Zef

f

1

2

4

5

1

If shielding is perfect Z= 1

Atomic Number94

Zef

f

1

2

4

5

1

No

shie

ldin

gZ

= Z ef

f

Atomic Number

95

Zef

f

1

2

4

5

16Atomic Number

96

Periodic Trends� Ionization energy the energy required

to remove an electron form a gaseous atom

� Highest energy electron removed first. � First ionization energy (I 1) is that

required to remove the first electron.� Second ionization energy (I 2) - the

second electron� etc. etc.

17

97

Trends in ionization energy� for Mg

• I1 = 735 kJ/mole• I2 = 1445 kJ/mole• I3 = 7730 kJ/mole

� The effective nuclear charge increases as you remove electrons.

� It takes much more energy to remove a core electron than a valence electron because there is less shielding

98

Explain this trend� For Al

• I1 = 580 kJ/mole• I2 = 1815 kJ/mole• I3 = 2740 kJ/mole• I4 = 11,600 kJ/mole

99

Across a Period� Generally from left to right, I 1

increases because � there is a greater nuclear charge with

the same shielding.� As you go down a group I 1 decreases

because electrons are further away and there is more shielding

100

It is not that simple� Zeff changes as you go across a

period, so will I 1� Half-filled and filled orbitals are

harder to remove electrons from� here’s what it looks like

101

Firs

t Ion

izat

ion

ener

gy

Atomic number 102

Firs

t Ion

izat

ion

ener

gy

Atomic number

18

103

Firs

t Ion

izat

ion

ener

gy

Atomic number 104

Atomic Size� First problem where do you start

measuring� The electron cloud doesn’t have a

definite edge.� They get around this by measuring

more than 1 atom at a time

105

Atomic Size

�Atomic Radius = half the distance between two nuclei of a diatomic molecule

}Radius

106

Trends in Atomic Size � Influenced by two factors� Shielding� More shielding is further away� Charge on nucleus� More charge pulls electrons in

closer

107

Group trends� As we go down a

group� Each atom has

another energy level

� So the atoms get bigger

HLi

Na

K

Rb

108

Periodic Trends� As you go across a period the radius

gets smaller.� Same energy level� More nuclear charge� Outermost electrons are closer

Na Mg Al Si P S Cl Ar

19

109

Overall

Atomic Number

Ato

mic

Rad

ius

(nm

)

H

Li

Ne

Ar

10

Na

K

Kr

Rb

110

Electron Affinity� The energy change associated with

adding an electron to a gaseous atom� High electron affinity gives you energy-� exothermic� More negative � Increase (more - ) from left to right

–greater nuclear charge.� Decrease as we go down a group

–More shielding

111

Ionic Size� Cations form by losing electrons� Cations are smaller than the atom

they come from� Metals form cations� Cations of representative elements

have noble gas configuration.

112

Ionic size� Anions form by gaining electrons� Anions are bigger than the atom they

come from� Nonmetals form anions� Anions of representative elements

have noble gas configuration.

113

Configuration of Ions� Ions always have noble gas

configuration� Na is 1s 22s22p63s1

� Forms a 1+ ion - 1s 22s22p6

� Same configuration as neon� Metals form ions with the

configuration of the noble gas before them - they lose electrons

114

Configuration of Ions� Non-metals form ions by gaining

electrons to achieve noble gas configuration.

� They end up with the configuration of the noble gas after them.

20

115

Group trends� Adding energy level� Ions get bigger as

you go downLi+1

Na+1

K+1

Rb+1

Cs+1

116

Periodic Trends� Across the period nuclear charge

increases so they get smaller.� Energy level changes between

anions and cations

Li+1

Be+2

B+3

C+4

N-3O-2 F-1

117

Size of Isoelectronic ions� Iso - same� Iso electronic ions have the same #

of electrons� Al+3 Mg+2 Na+1 Ne F-1 O-2 and N -3

� all have 10 electrons� all have the configuration 1s 22s22p6

118

Size of Isoelectronic ions� Positive ions have more protons so

they are smaller

Al+3

Mg+2

Na+1 Ne F-1 O-2 N-3

119

Electronegativity

120

Electronegativity� The tendency for an atom to attract

electrons to itself when it is chemically combined with another element.

� How “greedy”� Big electronegativity means it pulls

the electron toward itself.� Atoms with large negative electron

affinity have larger electronegativity.

21

121

Group Trend� The further down a group more

shielding� Less attracted (Z eff)� Low electronegativity.

122

Periodic Trend� Metals are at the left end� Low ionization energy- low effective

nuclear charge� Low electronegativity� At the right end are the nonmetals� More negative electron affinity� High electronegativity� Except noble gases

123

Ionization energy, electronegativity

Electron affinity INCREASE

124

Atomic size increases,

Ionic size increases

125

Parts of the Periodic Table

126

The information it hides� Know the special groups� It is the number and type of valence

electrons that determine an atom’s chemistry.

� You can get the electron configuration from it.

� Metals lose electrons have the lowest IE� Non metals- gain electrons most

negative electron affinities

22

127

The Alkali Metals� Doesn’t include hydrogen- it behaves

as a non-metal� decrease in IE� increase in radius� Decrease in density� decrease in melting point� Behave as reducing agents

128

Reducing ability� Lower IE< better reducing agents� Cs>Rb>K>Na>Li� works for solids, but not in aqueous

solutions.� In solution Li>K>Na� Why?� It’s the water -there is an energy

change associated with dissolving

129

Hydration Energy� Li+(g) → Li+(aq) is exothermic� for Li + -510 kJ/mol� for Na + -402 kJ/mol� for K + -314 kJ/mol� Li is so big because of it has a high

charge density, a lot of charge on a small atom.

� Li loses its electron more easily because of this in aqueous solutions

130

The reaction with water� Na and K react explosively with water� Li doesn’t.� Even though the reaction of Li has a

more negative ∆∆∆∆H than that of Na and K� Na and K melt� ∆∆∆∆H does not tell you speed of reaction� More in Chapter 12.

1

Chapter 8

Bonding

What is a Bond?� A force that holds atoms together.� Why?� We will look at it in terms of energy.� Bond energy- the energy required to

break a bond.� Why are compounds formed?� Because it gives the system the

lowest energy.

Ionic Bonding� An atom with a low ionization energy

reacts with an atom with high electron affinity.

� A metal and a non metal� The electron moves.� Opposite charges hold the atoms

together.

Coulomb's Law� E= 2.31 x 10-19 J · nm(Q1Q2)/r� Q is the charge.� r is the distance between the centers.� If charges are opposite, E is negative� exothermic� Same charge, positive E, requires

energy to bring them together.

What about covalent compounds?

� The electrons in each atom are attracted to the nucleus of the other.

� The electrons repel each other,� The nuclei repel each other.� The reach a distance with the lowest

possible energy.� The distance between is the bond length.

0

Ene

rgy

Internuclear Distance

2

0

Ene

rgy


0

Ene

rgy


0

Ene

rgy


0

Ene

rgy


Bond Length

0

Ene

rgy


Bond Energy

Covalent Bonding� Electrons are shared by atoms.� These are two extremes.� In between are polar covalent bonds.� The electrons are not shared evenly.� One end is slightly positive, the other

negative.� Indicated using small delta δ.

3

H - Fδ+ δ-

H - Fδ+ δ-

H - F

δ+δ-H - Fδ+

δ-

H -F

δ+δ-

H -F δ+δ-

H -Fδ+δ-

H - F

δ+δ-

H -F

δ+δ-

H - Fδ+ δ-

H - F

δ+δ-H - Fδ+

δ-

H -F

δ+δ-

H -F δ+δ-

H -Fδ+δ-

H - F

δ+δ-

H -F

δ+δ-

+- Electronegativity� The ability of an electron to attract

shared electrons to itself.� Pauling method� Imaginary molecule HX� Expected H-X energy =

H-H energy + X-X energy2

� ∆ = (H-X) actual - (H-X)expected

Electronegativity� ∆ is known for almost every element� Gives us relative electronegativities of

all elements.� Tends to increase left to right.� decreases as you go down a group.� Most noble gases aren’t discussed.� Difference in electronegativity between

atoms tells us how polar the bond is.

Electronegativitydifference

Bond Type

Zero

Intermediate

Large

Covalent

Polar Covalent

Ionic

Covalent C

haracterdecreases

Ionic Character increases

4

Dipole Moments� A molecule with a center of negative

charge and a center of positive charge is dipolar (two poles),

� or has a dipole moment.� Center of charge doesn’t have to be

on an atom.� Will line up in the presence of an

electric field.

H - Fδ+ δ-

H - F

δ+δ-H - Fδ+

δ-

H -F

δ+δ-

H -F δ+δ-

H -Fδ+δ-

H - F

δ+δ-

H -F

δ+δ-

+-

How It is drawn

H - Fδ+ δ-

Which Molecules Have Dipoles?

� Any two atom molecule with a polar bond.

� With three or more atoms there are two considerations.

1) There must be a polar bond.2) Geometry can’t cancel it out.

Geometry and polarity� Three shapes will cancel them out.� Linear

Geometry and polarity� Three shapes will cancel them out.� Planar triangles

120º

5

Geometry and polarity� Three shapes will cancel them out.� Tetrahedral

Geometry and polarity� Others don’t cancel� Bent

Geometry and polarity� Others don’t cancel� Trigonal Pyramidal

Ions� Atoms tend to react to form noble gas

configuration.� Metals lose electrons to form cations� Nonmetals can share electrons in

covalent bonds. –When two non-metals react.(more later)

� Or they can gain electrons to form anions.

Ionic Compounds� We mean the solid crystal.� Ions align themselves to maximize

attractions between opposite charges,� and to minimize repulsion between like

ions.� Can stabilize ions that would be unstable

as a gas.� React to achieve noble gas configuration

Size of ions� Ion size increases down a group.� Cations are smaller than the atoms

they came from.� Anions are larger.� across a row they get smaller, and

then suddenly larger.� First half are cations.� Second half are anions.

6

Periodic Trends� Across the period nuclear charge

increases so they get smaller.� Energy level changes between anions

and cations.

Li+1

Be+2

B+3

C+4

N-3O-2 F-1

Size of Isoelectronic ions� Positive ions have more protons so

they are smaller.

Al+3

Mg+2

Na+1 Ne F-1 O-2 N-3

Forming Ionic Compounds� Lattice energy - the energy associated

with making a solid ionic compound from its gaseous ions.

� M+(g) + X-(g) → MX(s)� This is the energy that “pays” for

making ionic compounds.� Energy is a state function so we can

get from reactants to products in a round about way.

Na(s) + ½F2(g) → NaF(s)� First sublime Na Na(s) → Na(g)

∆H = 109 kJ/mol� Ionize Na(g) Na(g) → Na+(g) + e-

∆H = 495 kJ/mol� Break F-F Bond ½F2(g) → F(g)

∆H = 77 kJ/mol� Add electron to F F(g) + e- → F-(g)

∆H = -328 kJ/mol

Na(s) + ½F2(g) → NaF(s)� Lattice energy

Na+(g) + F-(g) → NaF(s)∆H = -928 kJ/mol

Calculating Lattice Energy� Lattice Energy = k(Q1Q2 / r)� k is a constant that depends on the

structure of the crystal.� Q’s are charges.� r is internuclear distance.� Lattice energy is with smaller ions� Lattice energy is greater with more

highly charged ions.

7

Calculating Lattice Energy� This bigger lattice energy “pays” for

the extra ionization energy.� Also “pays” for unfavorable electron

affinity.� O2-(g) is unstable, but will form as part

of a crystal

Bonding

Partial Ionic Character�There are probably no totally ionic

bonds between individual atoms.�Calculate % ionic character.�Compare measured dipole of X-Y

bonds to the calculated dipole of X+Y-

the completely ionic case.�% dipole = Measured X-Y x 100

Calculated X+Y-

� In the gas phase.

% Io

nic

Cha

ract

er

Electronegativity difference

25%

50%

75%

How do we deal with it?� If bonds can’t be ionic, what are ionic

compounds?�And what about polyatomic ions?�An ionic compound will be defined as

any substance that conducts electricity when melted.

�Also use the generic term salt.

The Covalent Bond�The forces that causes a group of atoms

to behave as a unit.�Why?�Due to the tendency of atoms to achieve

the lowest energy state.� It takes 1652 kJ to dissociate a mole of

CH4 into its ions�Since each hydrogen is hooked to the

carbon, we get the average energy = 413 kJ/mol

8

�CH3Cl has 3 C-H, and 1 C - Cl� the C-Cl bond is 339 kJ/mol�The bond is a human invention.� It is a method of explaining the energy

change associated with forming molecules.

�Bonds don’t exist in nature, but are useful.

�We have a model of a bond.

What is a Model?�Explains how nature operates.�Derived from observations.� It simplifies them and categorizes the

information.�A model must be sensible, but it has

limitations.

Properties of a Model�A human inventions, not a blown up picture

of nature.�Models can be wrong, because they are

based on speculations and oversimplification.

�Become more complicated with age.�You must understand the assumptions in

the model, and look for weaknesses.�We learn more when the model is wrong

than when it is right.

Covalent Bond Energies�We made some simplifications in

describing the bond energy of CH4�Each C-H bond has a different energy.�CH4 → CH3 + H ∆H = 435 kJ/mol�CH3 → CH2 + H ∆H = 453 kJ/mol�CH2 → CH + H ∆H = 425 kJ/mol�CH→ C + H ∆H = 339 kJ/mol�Each bond is sensitive to its

environment.

Averages�There is a table of the averages of

different types of bonds pg. 365�single bond- one pair of electrons is

shared.�double bond- two pair of electrons are

shared.� triple bond- three pair of electrons are

shared.�More bonds, more bond energy, but

shorter bond length.

Using Bond Energies�We can estimate ∆H for a reaction.� It takes energy to break bonds, and end

up with atoms (+).�We get energy when we use atoms to

form bonds (-).� If we add up the energy it took to break

the bonds, and subtract the energy we get from forming the bonds we get the ∆H.

�Energy and Enthalpy are state functions.

9

Find the energy for this

2 CH2 = CHCH3

+

2NH3 O2+

→→→→ 2 CH2 = CHC ≡≡≡≡ N

+

6 H2O

C-H 413 kJ/molC=C 614kJ/molN-H 391 kJ/mol

O-H 467 kJ/molO=O 495 kJ/molC≡≡≡≡N 891 kJ/mol

C-C 347 kJ/mol

Localized Electron Model� Simple model, easily applied.� A molecule is composed of atoms that

are bound together by sharing pairs of electrons using the atomic orbitals of the bound atoms.

� Three Parts1) Valence electrons using Lewis structures2) Prediction of geometry using VSEPR3) Description of the types of orbitals

(Chapt 9)

Lewis Structure�Shows how the valence electrons are

arranged.�One dot for each valence electron.�A stable compound has all its atoms

with a noble gas configuration.�Hydrogen follows the duet rule.�The rest follow the octet rule.�Bonding pair is the one between the

symbols.

Rules�Sum the valence electrons.�Use a pair to form a bond between

each pair of atoms.�Arrange the rest to fulfill the octet rule

(except for H and the duet).�H2O�A line can be used instead of a pair.

Quiz Answers

1. D2. B3. A4. D5. C6. E7. D8. E9. E

A useful equation�(happy-have) / 2 = bonds

�CO2 C is central atom

�POCl3 P is central atom

�SO42- S is central atom

�SO32- S is central atom

�PO43- P is central atom

�SCl2 S is central atom

10

Exceptions to the octet�BH3�Be and B often do not achieve octet�Have less than an octet, for electron

deficient molecules.�SF6�Third row and larger elements can

exceed the octet�Use 3d orbitals?� I3

-

Exceptions to the octet�When we must exceed the octet, extra

electrons go on central atom.�(Happy – have)/2 won’t work

�ClF3

�XeO3� ICl4

-

�BeCl2

Resonance�Sometimes there is more than one valid

structure for an molecule or ion.�NO3

-

�Use double arrows to indicate it is the “average” of the structures.

� It doesn’t switch between them.�NO2

-

�Localized electron model is based on pairs of electrons, doesn’t deal with odd numbers.

Formal Charge�For molecules and polyatomic ions

that exceed the octet there are several different structures.

�Use charges on atoms to help decide which.

�Trying to use the oxidation numbers to put charges on atoms in molecules doesn’t work.

Formal Charge�The difference between the number of

valence electrons on the free atom and that assigned in the molecule or ion.

�We count half the electrons in each bond as “belonging” to the atom.

�SO4-2

�Molecules try to achieve as low a formal charge as possible.

�Negative formal charges should be on electronegative elements.

Examples�XeO3

�NO43-

�SO2Cl2

11

VSEPR�Lewis structures tell us how the atoms

are connected to each other.�They don’t tell us anything about

shape.�The shape of a molecule can greatly

affect its properties.�Valence Shell Electron Pair Repulsion

Theory allows us to predict geometry

VSEPR�Molecules take a shape that puts

electron pairs as far away from each other as possible.

�Have to draw the Lewis structure to determine electron pairs.

�bonding�nonbonding lone pair�Lone pair take more space.�Multiple bonds count as one pair.

VSEPR� The number of pairs determines

–bond angles–underlying structure

� The number of atoms determines –actual shape

VSEPRElectron

pairsBond

AnglesUnderlyingShape

2 180° Linear

3 120° Trigonal Planar

4 109.5° Tetrahedral

5 90°&120°

TrigonalBipyramidal

6 90° Octagonal

Actual shape

ElectronPairs

BondingPairs

Non-Bonding

Pairs Shape

2 2 0 linear

3 3 0 trigonal planar

3 2 1 bent4 4 0 tetrahedral4 3 1 trigonal pyramidal4 2 2 bent

Actual Shape

ElectronPairs

BondingPairs

Non-Bonding

Pairs Shape

5 5 0 trigonal bipyrimidal

5 4 1 See-saw

5 3 2 T-shaped5 2 3 linear

12

Actual Shape

ElectronPairs

BondingPairs

Non-Bonding

Pairs Shape

6 6 0 Octahedral

6 5 1 Square Pyramidal

6 4 2 Square Planar6 3 3 T-shaped6 2 1 linear

Examples� SiF4

� SeF4

� KrF4

� BF3

� PF3

� BrF3

No central atom� Can predict the geometry of each

angle.� build it piece by piece.

How well does it work?� Does an outstanding job for such a

simple model.� Predictions are almost always

accurate.� Like all simple models, it has

exceptions.� Doesn’t deal with odd electrons

Polar molecules� Must have polar bonds� Must not be symmetrical� Symmetrical shapes include

–Linear–Trigonal planar–Tetrahedral–Trigonal bipyrimidal–Octahedral–Square planar

1

1

Chapter 9

Orbitals and Covalent Bond

2

Molecular Orbitals� The overlap of atomic orbitals from

separate atoms makes molecular orbitals

� Each molecular orbital has room for two electrons

� Two types of MO

–Sigma ( σ ) between atoms–Pi ( π ) above and below atoms

3

Sigma bonding orbitals � From s orbitals on separate atoms

+ +

s orbital s orbital

+ ++ +

Sigma bondingmolecular orbital

4

Sigma bonding orbitals � From p orbitals on separate atoms

p orbital p orbital

Sigma bondingmolecular orbital

⊕ ⊕

⊕ ⊕⊕ ⊕

5

Pi bonding orbitals� p orbitals on separate atoms

⊕ ⊕ ⊕ ⊕⊕ ⊕⊕ ⊕

Pi bondingmolecular orbital

6

Sigma and pi bonds� All single bonds are sigma bonds� A double bond is one sigma and one pi

bond� A triple bond is one sigma and two pi

bonds.

2

7

Atomic Orbitals Don’t Work� to explain molecular geometry.� In methane, CH4 , the shape is

tetrahedral.� The valence electrons of carbon should

be two in s, and two in p.� the p orbitals would have to be at right

angles.� The atomic orbitals change when

making a molecule

8

Hybridization� We blend the s and p orbitals of the

valence electrons and end up with the tetrahedral geometry.

� We combine one s orbital and 3 p orbitals.

� sp3 hybridization has tetrahedral geometry.

9 10

11

In terms of energy

En

erg

y

2p

2s

Hybridization sp3

12

How we get to hybridization� We know the geometry from experiment.� We know the orbitals of the atom� hybridizing atomic orbitals can explain

the geometry.� So if the geometry requires a tetrahedral

shape, it is sp3 hybridized� This includes bent and trigonal pyramidal

molecules because one of the sp3 lobes holds the lone pair.

3

13

sp2 hybridization� C2H4� Double bond acts as one pair.� trigonal planar� Have to end up with three blended

orbitals.� Use one s and two p orbitals to make

sp2 orbitals.� Leaves one p orbital perpendicular.

14

15 16

In terms of energy

En

erg

y

2p

2s

sp2Hybridization

2p

17

Where is the P orbital?� Perpendicular� The overlap of

orbitals makes a sigma bond (σbond)

18

Two types of Bonds� Sigma bonds from overlap of orbitals.� Between the atoms.

� Pi bond (π bond) above and below atoms� Between adjacent p orbitals.� The two bonds of a

double bond.

4

19

CCH

H

H

H

20

sp2 hybridization� When three things come off atom.� trigonal planar� 120º

� One π bond, σ + lp =3

21

What about two� When two things come off.� One s and one p hybridize.� linear

22

sp hybridization� End up with two lobes 180º

apart.� p orbitals are at right

angles� Makes room for two π

bonds and two sigma bonds.

� A triple bond or two double bonds.

23

In terms of energy

En

erg

y

2p

2s

Hybridizationsp

2p

24

CO2

� C can make two σ and two π� O can make one σ and one π

CO O

5

25

N2

26

N2

27

Breaking the octet

� PCl5� The model predicts that we must use

the d orbitals.� dsp3 hybridization� There is some controversy about how

involved the d orbitals are.

28

dsp3

� Trigonalbipyrimidal

� can only σ bond.� can’t π bond.� basic shape for

five things.

29

PCl5

Can’t tell the hybridization of Cl

Assume sp3 to minimize repulsion of electron pairs.

30

d2sp3

� gets us to six things around

� Octahedral� Only σ bond

6

31

Molecular Orbital Model� Localized Model we have learned explains

much about bonding.� It doesn’t deal well with the ideal of

resonance, unpaired electrons, and bond energy.

� The MO model is a parallel of the atomic orbital, using quantum mechanics.

� Each MO can hold two electrons with opposite spins

� Square of wave function tells probability32

What do you get?

� Solve the equations for H2� HA HB

� get two orbitals

� MO2 = 1sA - 1sB

� MO1 = 1sA + 1sB

33

The Molecular Orbital Model• The molecular orbitals are centered on

a line through the nuclei–MO1 the greatest probability is

between the nuclei–MO2 it is on either side of the nuclei– this shape is called a sigma molecular

orbital

34

The Molecular Orbital Model• In the molecule only the molecular

orbitals exist, the atomic orbitals are gone• MO1 is lower in energy than the 1s

orbitals they came from.–This favors molecule formation–Called an bonding orbital

• MO2 is higher in energy–This goes against bonding–antibonding orbital

35

The Molecular Orbital Model

En

erg

y

MO2

MO1

1s1s

H2

36

The Molecular Orbital Model• We use labels to indicate shapes, and

whether the MO’s are bonding or antibonding.

–MO1 = σ1s–MO2 = σ1s* (* indicates antibonding)

• Can write them the same way as atomic orbitals

–H2 = σ1s2

7

37

The Molecular Orbital Model• Each MO can hold two electrons, but

they must have opposite spins• Orbitals are conserved.

• The number of molecular orbitals must equal the number atomic orbitals that are used to make them.

38

H2-

En

erg

y

σ1s

σ1s*

1s1s

39

Bond Order� The difference between the number of

bonding electrons and the number of antibonding electrons divided by two

Bond Order = # bonding-#antibonding

2

40

Only outer orbitals bond� The 1s orbital is much smaller than the

2s orbital� When only the 2s orbitals

are involved in bonding

� Don’t use the σ1s or σ1s*for Li2

� Li2 = (σ2s)2

� In order to participate in bonds the orbitals must overlap in space.

41

Bonding in Homonuclear Diatomic Molecules

� Need to use Homonuclear so that we know the relative energies.

� Li2-

� (σ2s)2 (σ2s*)1

� Be2

� (σ2s)2 (σ2s*)2

� What about the p orbitals? How do they form orbitals?

� Remember that orbitals must be conserved. 42

B2

8

43

B2

σ2p*

σ2p

π2p*

π2p44

Expected Energy Diagram

Ene

rgy

2s 2s

2p2p

σ2s

σ2p*

σ2p

σ2s*

π2p* π2p*π2pπ2p

45

B2

Ene

rgy

2s 2s

2p2p

46

B2� (σ2s)2(σ2s*)2 (σ2p)2

� Bond order = (4-2) / 2� Should be stable.� This assumes there is no interaction

between the s and p orbitals.� Hard to believe since they overlap� proof comes from magnetism.

47

Magnetism� Magnetism has to do with electrons.� Remember that spin is how an electron

reacts to a magnetic field� Paramagnetism attracted by a magnet.

–associated with unpaired electrons.� Diamagnetism repelled by a magnet.

–associated with paired electrons.� B2 is paramagnetic.

48

Magnetism� The energies of of the π2p and the σ2p

are reversed by p and s interacting

� The σ2s and the σ2s* are no longer equally spaced.

� Here’s what it looks like.

9

49

Correct energy diagram

2s 2s

2p2p

σ2s

σ2p*

σ2p

σ2s*

π2p* π2p*

π2pπ2p

50

B2

2s 2s

2p2p

σ2s

σ2p*

σ2p

σ2s*

π2p*

π2p

51

Patterns� As bond order increases, bond energy

increases.� As bond order increases, bond length

decreases.� Supports basis of MO model.� There is not a direct correlation of bond

order to bond energy.� O2 is known to be paramagnetic.� Movie.

52

Magnetism� Ferromagnetic strongly attracted� Paramagnetic weakly attracted

–Liquid Oxygen� Diamagnetic weakly repelled

–Graphite–Water Frog

53

Examples� C2� N2� O2� F2� P2

54

Heteronuclear Diatomic Species� Simple type has them in the same

energy level, so can use the orbitals we already know.

� Slight energy differences.� NO

10

55

NO

2s2s

2p2p

56

You try� NO+

� CN-

� What if they come from completely different orbitals and energy?

� HF� Simplify first by assuming that F only

uses one if its 2p orbitals.� F holds onto its electrons, so they have

low energy

57

1s

2p

σ∗

σ 58

Consequences� Paramagnetic� Since 2p is lower in energy, favored by

electrons.� Electrons spend time closer to fluorine.� Compatible with polarity and

electronegativity.

59

Names� sp orbitals are called the Localized

electron model

� σ and π Μolecular orbital model� Localized is good for geometry, doesn’t

deal well with resonance.

� seeing σ bonds as localized works well� It is the π bonds in the resonance

structures that can move.

60

π delocalized bonding� C6H6

H

H

H

HH

H

H

H

H

HH

H

11

61

C2H6

62

NO3-

1

Chapter 2Chapter 2

Atoms, Molecules, and Ions Atoms, Molecules, and Ions

HistoryHistory�� GreeksGreeks

�� Democritus and Leucippus Democritus and Leucippus -- atomosatomos

�� AristotleAristotle-- elementselements

�� AlchemyAlchemy

�� 1660 1660 -- Robert BoyleRobert Boyle-- experimental experimental definition of element.definition of element.

�� LavoisierLavoisier-- Father of modern chemistryFather of modern chemistry

�� He wrote the bookHe wrote the book-- used measurementused measurement

LawsLaws�� Conservation of MassConservation of Mass

�� Law of Definite ProportionLaw of Definite Proportion-- compounds compounds have a constant composition by mass.have a constant composition by mass.

�� They react in specific ratios by mass.They react in specific ratios by mass.

�� Multiple ProportionsMultiple Proportions-- When two elements When two elements form more than one compound, the ratios of form more than one compound, the ratios of the masses of the second element that the masses of the second element that combine with one gram of the first can be combine with one gram of the first can be reduced to small whole numbers.reduced to small whole numbers.

What?!What?!�� Water has 8 g of oxygen per g of hydrogen.Water has 8 g of oxygen per g of hydrogen.

�� Hydrogen peroxide has 16 g of oxygen per Hydrogen peroxide has 16 g of oxygen per g of hydrogen.g of hydrogen.

�� 16/8 = 2/116/8 = 2/1

�� Small whole number ratiosSmall whole number ratios

Example of Law Of Multiple Example of Law Of Multiple ProportionsProportions

�� Mercury has two oxides. One is 96.2 % Mercury has two oxides. One is 96.2 % mercury by mass, the other is 92.6 % mercury by mass, the other is 92.6 % mercury by mass.mercury by mass.

�� Show that these compounds follow the law Show that these compounds follow the law of multiple proportion.of multiple proportion.

�� Speculate on the formula of the two oxides.Speculate on the formula of the two oxides.

Your TurnYour Turn�� Nitrogen and oxygen form two compounds.Nitrogen and oxygen form two compounds.

Show that they follow the law of multiple Show that they follow the law of multiple proportionsproportions

Amount NAmount N Amount OAmount O

Compound ACompound A 1.206 g1.206 g 2.755 g2.755 g

Compound BCompound B 1.651g1.651g 4.714 g4.714 g

2

Dalton’s Atomic TheoryDalton’s Atomic Theory�� 1. Elements are made up of atoms1. Elements are made up of atoms

�� 2. Atoms of each element are identical. 2. Atoms of each element are identical. Atoms of different elements are different.Atoms of different elements are different.

�� 3. Compounds are formed when atoms 3. Compounds are formed when atoms combine. Each compound has a specific combine. Each compound has a specific number and kinds of atom.number and kinds of atom.

�� 4. Chemical reactions are rearrangement of 4. Chemical reactions are rearrangement of atoms. Atoms are not created or destroyed.atoms. Atoms are not created or destroyed.

�� GayGay--LussacLussac-- under the same conditions of under the same conditions of temperature and pressure, compounds temperature and pressure, compounds always react in whole number ratios by always react in whole number ratios by volume.volume.

�� AvagadroAvagadro-- interpreted that to mean interpreted that to mean

�� at the same temperature and pressure, equal at the same temperature and pressure, equal volumes of gas contain the same number of volumes of gas contain the same number of particlesparticles

�� (called Avagadro’s hypothesis)(called Avagadro’s hypothesis)

A Helpful ObservationA Helpful Observation

Experiments to determine what Experiments to determine what an atom wasan atom was

�� J. J. ThomsonJ. J. Thomson-- used Cathode ray tubesused Cathode ray tubes

Thomson’s ExperimentThomson’s Experiment

Voltage source

+-


Voltage source

+-


Voltage source

+-

3

�� Passing an electric current makes a beam Passing an electric current makes a beam appear to move from the negative to the appear to move from the negative to the positive endpositive end


Voltage source

+-



Voltage source

+-



Voltage source

+-



Voltage source

+-

Voltage source


�� By adding an electric field By adding an electric field

Voltage source



+

-

4

Voltage source



+

-

Voltage source



+

-

Voltage source



+

-

Voltage source



+

-

Voltage source


�� By adding an electric field he found that the By adding an electric field he found that the moving pieces were negative moving pieces were negative

+

-

Thomsom’s ModelThomsom’s Model�� Found the electronFound the electron

�� Couldn’t find Couldn’t find positive (for a while) positive (for a while)

�� Said the atom was Said the atom was like plum puddinglike plum pudding

�� A bunch of positive A bunch of positive stuff, with the stuff, with the electrons able to be electrons able to be removed removed

5

Millikan’s ExperimentMillikan’s Experiment

Atomizer

Microscope

-

+

Oil


Oil

Atomizer

Microscope

-

+

Oil droplets


X-rays

X-rays give some drops a charge by knocking offelectrons


+


They put an electric charge on the plates

++

--


Some drops would hover

++

--

6


+

+ + + + + + +

- - - - - - -


Measure the drop and find volume from 4/3πr3

Find mass from M = D x V

++

--


From the mass of the drop and the charge on the plates, he calculated the charge on an electron

++

--

RadioactivityRadioactivity�� Discovered by accidentDiscovered by accident

�� BequerelBequerel

�� Three types Three types

–– alphaalpha-- helium nucleus (+2 charge, large helium nucleus (+2 charge, large mass)mass)

–– betabeta-- high speed electronhigh speed electron

–– gammagamma-- high energy lighthigh energy light

Rutherford’s ExperimentRutherford’s Experiment�� Used uranium to produce alpha particlesUsed uranium to produce alpha particles

�� Aimed alpha particles at gold foil by Aimed alpha particles at gold foil by drilling hole in lead blockdrilling hole in lead block

�� Since the mass is evenly distributed in Since the mass is evenly distributed in gold atoms alpha particles should go gold atoms alpha particles should go straight through.straight through.

�� Used gold foil because it could be made Used gold foil because it could be made atoms thinatoms thin

Lead block

Uranium

Gold Foil

Florescent Screen

7

What he expected Because

Because, he thought the mass was evenly distributed in the atom

What he got

How he explained it

+

�� Atom is mostly emptyAtom is mostly empty

�� Small dense,Small dense,positive piecepositive piece

at centerat center

�� Alpha particles Alpha particles are deflected byare deflected byit if they get closeit if they get closeenoughenough

+

8

Modern ViewModern View�� The atom is mostly The atom is mostly

empty spaceempty space

�� Two regionsTwo regions

�� NucleusNucleus-- protons and protons and neutronsneutrons

�� Electron cloudElectron cloud-- region region where you have a where you have a chance of finding an chance of finding an electronelectron

SubSub--atomic Particlesatomic Particles�� Z Z -- atomic number = number of protons atomic number = number of protons

determines type of atomdetermines type of atom

�� A A -- mass number = number of protons + mass number = number of protons + neutronsneutrons

�� Number of protons = number of electrons if Number of protons = number of electrons if neutralneutral

SymbolsSymbols

XA

Z

Na23

11

Chemical BondsChemical Bonds�� The forces that hold atoms togetherThe forces that hold atoms together

�� Covalent bonding Covalent bonding -- sharing electronssharing electrons

�� makes moleculesmakes molecules

�� Chemical formulaChemical formula-- the number and type of the number and type of atoms in a molecule atoms in a molecule

�� CC22HH66 -- 2 carbon atoms, 6 hydrogen atoms, 2 carbon atoms, 6 hydrogen atoms,

�� Structural formula shows the connections, Structural formula shows the connections, but not necessarily the shape.but not necessarily the shape.

H

H

H H

H

HC C

�� Structural FormulaStructural Formula�� There are also other model that attempt to There are also other model that attempt to

show three dimensional shapeshow three dimensional shape

�� Ball and stick (see the models in room)Ball and stick (see the models in room)

�� Space Filling Space Filling

9

IonsIons�� Atoms or groups of atoms with a chargeAtoms or groups of atoms with a charge

�� CationsCations-- positive ions positive ions -- get by losing get by losing electrons(s)electrons(s)

�� AnionsAnions-- negative ions negative ions -- get by gaining get by gaining electron(s)electron(s)

�� Ionic bondingIonic bonding-- held together by the opposite held together by the opposite chargescharges

�� Ionic solids are called saltsIonic solids are called salts

Polyatomic Ions Polyatomic Ions �� Groups of atoms that have a chargeGroups of atoms that have a charge

�� Yes, you have to memorize them.Yes, you have to memorize them.

�� List on page 65List on page 65

Periodic TablePeriodic Table MetalsMetals�� ConductorsConductors

�� Lose electronsLose electrons

�� Malleable and ductileMalleable and ductile

NonmetalsNonmetals�� BrittleBrittle

�� Gain electronsGain electrons

�� Covalent bondsCovalent bonds

SemiSemi--metals or Metalloidsmetals or Metalloids

10

Alkali Metals Alkaline Earth Metals

Halogens Transition metals

Noble Gases Inner Transition Metals

11

+1+2 -1-2-3Naming compoundsNaming compounds

�� Two typesTwo types

�� IonicIonic -- metal and non metal or polyatomicsmetal and non metal or polyatomics

�� CovalentCovalent-- we will just learn the rules for 2 we will just learn the rules for 2 nonnon--metalsmetals

Ionic compoundsIonic compounds�� If the cation is monoatomicIf the cation is monoatomic-- Name the Name the

metal (cation) just write the name.metal (cation) just write the name.

�� If the cation is polyatomicIf the cation is polyatomic-- name itname it

�� If the anion is monoatomicIf the anion is monoatomic-- name it but name it but change the ending to change the ending to --ideide

�� If the anion is poly atomicIf the anion is poly atomic-- just name itjust name it

�� practicepractice

Covalent compoundsCovalent compounds�� Two words, with prefixesTwo words, with prefixes

�� Prefixes tell you how many.Prefixes tell you how many.

�� mono, di, tri, tetra, penta, hexa, septa, nona, mono, di, tri, tetra, penta, hexa, septa, nona, decadeca

�� First element whole name with the First element whole name with the appropriate prefix, except monoappropriate prefix, except mono

�� Second element, Second element, --ide ide ending with ending with appropriate prefixappropriate prefix

�� PracticePractice

More NamingMore Naming

Ionic compoundsIonic compounds�� If the cation is monoatomicIf the cation is monoatomic-- Name the Name the

metal (cation) just write the name.metal (cation) just write the name.

�� If the cation is polyatomicIf the cation is polyatomic-- name itname it

�� If the anion is monoatomicIf the anion is monoatomic-- name it but name it but change the ending to change the ending to --ideide

�� If the anion is poly atomicIf the anion is poly atomic-- just name itjust name it

�� practicepractice

12

Ionic CompoundsIonic Compounds�� Have to know what ions they formHave to know what ions they form

�� off table, polyatomic, or figure it outoff table, polyatomic, or figure it out

�� CaSCaS

�� KK22SS

�� AlPOAlPO44

�� KK22SOSO44

�� FeSFeS

�� CoICoI33

Ionic CompoundsIonic Compounds�� FeFe22(C(C22OO44))

�� MgOMgO

�� MnOMnO

�� KMnOKMnO44

�� NHNH44NONO33

�� HgHg22ClCl22�� CrCr22OO33

Ionic CompoundsIonic Compounds�� KClOKClO44�� NaClONaClO33�� YBrOYBrO22�� Cr(ClO)Cr(ClO)66

Naming Covalent CompoundsNaming Covalent Compounds�� Two words, with prefixesTwo words, with prefixes

�� Prefixes tell you how many.Prefixes tell you how many.

�� mono, di, tri, tetra, penta, hexa, septa, nona, mono, di, tri, tetra, penta, hexa, septa, nona, decadeca

�� First element whole name with the First element whole name with the appropriate prefix, except monoappropriate prefix, except mono

�� Second element, Second element, --ide ide ending with ending with appropriate prefixappropriate prefix

�� PracticePractice

�� COCO22�� CO CO

�� CClCCl44

�� NN22OO44

�� XeFXeF66�� NN44OO44�� PP22OO1010

Naming Covalent CompoundsNaming Covalent Compounds Writing FormulasWriting Formulas�� Two sets of rules, ionic and covalentTwo sets of rules, ionic and covalent

�� To decide which to use, decide what the To decide which to use, decide what the first word is.first word is.

�� If is a metal or polyatomic use ionic.If is a metal or polyatomic use ionic.

�� If it is a nonIf it is a non--metal use covalentmetal use covalent

13

Ionic FormulasIonic Formulas�� Charges must add up to zeroCharges must add up to zero

�� get charges from table, name of metal ion, get charges from table, name of metal ion, or memorized from the listor memorized from the list

�� use parenthesis to indicate multiple use parenthesis to indicate multiple polyatomicspolyatomics

Ionic FormulasIonic Formulas�� Sodium nitrideSodium nitride

�� sodiumsodium-- Na is always +1Na is always +1

�� nitride nitride -- ide tells you it comes from the tableide tells you it comes from the table

�� nitride is Nnitride is N--33





�� doesn’t add up to zerodoesn’t add up to zero

Na+1 N-3





�� doesn’t add up to zerodoesn’t add up to zero

�� Need 3 NaNeed 3 Na

Na+1 N-3 Na3N

Ionic CompoundsIonic Compounds�� Sodium sulfiteSodium sulfite

�� calcium iodidecalcium iodide

�� Lead (II) oxide Lead (II) oxide

�� Lead (IV) oxideLead (IV) oxide

�� Mercury (I) sulfideMercury (I) sulfide

�� Barium chromateBarium chromate

�� Aluminum hydrogen sulfateAluminum hydrogen sulfate

�� Cerium (IV) nitriteCerium (IV) nitrite

Covalent compoundsCovalent compounds�� The name tells you how to write the The name tells you how to write the

formulaformula

�� duhduh

�� Sulfur dioxideSulfur dioxide

�� diflourine monoxidediflourine monoxide

�� nitrogen trichloridenitrogen trichloride

�� diphosphorus pentoxidediphosphorus pentoxide

14

More Names and formulasMore Names and formulas

AcidsAcids�� Substances that produce HSubstances that produce H++ ions when ions when

dissolved in waterdissolved in water

�� All acids begin with HAll acids begin with H

�� Two types of acids Two types of acids

�� OxyacidsOxyacids

�� non oxyacidsnon oxyacids

Naming acidsNaming acids�� If the formula has oxygen in itIf the formula has oxygen in it

�� write the name of the anion, but change write the name of the anion, but change

–– ate to ate to --ic acidic acid

–– ite to ite to --ous acidous acid

�� Watch out for sulfWatch out for sulfururic and sulfic and sulfururousous

�� HH22CrOCrO44

�� HMnOHMnO44�� HNOHNO22

Naming acidsNaming acids�� If the acid doesn’t have oxygenIf the acid doesn’t have oxygen

�� add the prefix hydroadd the prefix hydro--

�� change the suffix change the suffix --ide to ide to --ic acidic acid

�� HClHCl

�� HH22SS

�� HCNHCN

Formulas for acidsFormulas for acids�� Backwards from namesBackwards from names

�� If it has hydroIf it has hydro-- in the name it has no oxygenin the name it has no oxygen

�� anion ends in anion ends in --ideide

�� No hydro, anion ends in No hydro, anion ends in --ate or ate or --iteite

�� Write anion and add enough H to balance Write anion and add enough H to balance the charges.the charges.

Formulas for acidsFormulas for acids�� hydrofluoric acidhydrofluoric acid

�� dichromic aciddichromic acid

�� carbonic acidcarbonic acid

�� hydrophosphoric acidhydrophosphoric acid

�� hypofluorous acidhypofluorous acid

�� perchloric acidperchloric acid

�� phosphorous acid phosphorous acid

15

HydratesHydrates�� Some salts trap water crystals when they Some salts trap water crystals when they

form crystalsform crystals

�� these are hydrates.these are hydrates.

�� Both the name and the formula needs to Both the name and the formula needs to indicate how many water molecules are indicate how many water molecules are trappedtrapped

�� In the name we add the word hydrate with a In the name we add the word hydrate with a prefix that tells us how many water prefix that tells us how many water molecules molecules

HydratesHydrates�� In the formula you put a dot and then write In the formula you put a dot and then write

the number of molecules.the number of molecules.

�� Calcium chloride dihydrate = CaClCalcium chloride dihydrate = CaCl22•2Η•2Η22ΟΟ�� Chromium (III) nitrate hexahydrate = Chromium (III) nitrate hexahydrate =

Cr(NOCr(NO33))33•• 6H6H22O O

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