AP Chapter 3_Combine
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Transcript of 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
Balancing equationsBalancing equations
CH4 + O2 → CO2 + 2 H2OReactantsReactants ProductsProducts
C1 1
O2 3
H4 2 4
4
4
Balancing equationsBalancing equations
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
3
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
3
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.
7
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.
4
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.
6
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
Filled Molecular Orbitals
Empty Molecular Orbitals
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
7
Filled Molecular Orbitals
Empty Molecular Orbitals
1s
2s
2p
3s
3p
Magnesium Atoms
The 3s and 3p orbitals overlap and form
molecular orbitals.
Filled Molecular Orbitals
Empty Molecular Orbitals
1s
2s
2p
3s
3p
Magnesium Atoms
Electrons in these energy levels can
travel freely throughout the crystal.
Filled Molecular Orbitals
Empty Molecular Orbitals
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.
9
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.
10
�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
11
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
12
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
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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
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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
Heterogenous Catalysts
C HH C
HH
Pt surface
HH
HH
Heterogenous Catalysts
C HH C
HH
� The double bond breaks and bonds to the catalyst.
Pt surface
HH
HH
Heterogenous Catalysts
C HH C
HH
� The hydrogen atoms bond with the carbon
Pt surfaceH
Heterogenous Catalysts
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
2NOCl 2NO + Cl2Initial 1.20 0.45 0.87
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
2NOCl 2NO + Cl2Initial 1.20 0.45 0.87
Change X-.45
Final X
� Now we can determine final concentrations
� Add
2NOCl 2NO + Cl2Initial 1.20 0.45 0.87
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 -
2NOCl 2NO + Cl2Initial 1.20 0.45 0.87
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
2NOCl 2NO + Cl2Initial 1.20 0.45 0.87
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!!!
2NOCl 2NO + Cl2Initial 1.20 0.45 0.87
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]
Basic Salts� Ka x Kb = [HF][OH-] x [H+][F-]
[F- ] [HF]
Basic Salts� Ka x Kb = [HF][OH-] x [H+][F-]
[F- ] [HF]
Basic Salts� Ka x Kb = [HF][OH-] x [H+][F-]
[F- ] [HF]� Ka x Kb =[OH-] [H+]
Basic Salts� Ka x Kb = [HF][OH-] x [H+][F-]
[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
Strength of oxyacids
Electron Density
Cl O HO
8
Strength of oxyacids
Cl O H
O
O
Electron Density
Strength of oxyacids
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-
Initially 0.75 mol 0 0.25 molAfter acid 0.70 mol 0.30 mol
-4 0.30 MpH = -log(1.4 x 10 ) + log 3.48
0.70 M =
Compared to 3.38 before acid was added
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
Initially 0.40 mol 0 0.25 molAfter acid 0.42 mol 0.23 mol
-10 0.23 MpH = -log(5.6 x 10 ) + log 8.99
0.42 M =
Compared to 9.05 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 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
Initially 0.40 mol 0 0.25 molAfter acid 0.35 mol 0.30 mol
-10 0.30 MpH = -log(5.6 x 10 ) + log 9.18
0.35 M =
Compared to 9.05 before acid was added
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 =
Buffer Capacity� Calculate the change in pH that occurs
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-
Initially 5.00 mol 0 5.00 molAfter acid 5.04 mol 4.96 mol
-5 4.96 MpH = -log(1.8 x 10 ) + log 4.74
5.04 M =
Compared to 4.74 before acid was added
Buffer Capacity� Calculate the change in pH that occurs
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 =
Compared to 4.74 before acid was added
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
�4 possible arrangements
�25% chance of finding the left empty
�50 % chance of them being evenly dispersed
2
Page 2
�4 atoms
�8% chance of finding the left empty
�50 % chance of them being evenly dispersed
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.
�2 possible arrangements
�50 % chance of finding the left empty
�4 possible arrangements
�25% chance of finding the left empty
�50 % chance of them being evenly dispersed
2
Page 2
�4 atoms
�8% chance of finding the left empty
�50 % chance of them being evenly dispersed
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
Distance from nucleus
3s
87
Ra
dia
l Pro
bab
ility
Distance from nucleus
3p
88
Ra
dia
l Pro
bab
ility
Distance from nucleus
3d
89
Ra
dia
l Pro
bab
ility
Distance from nucleus
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
Internuclear Distance
0
Ene
rgy
Internuclear Distance
0
Ene
rgy
Internuclear Distance
0
Ene
rgy
Internuclear Distance
Bond Length
0
Ene
rgy
Internuclear Distance
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
+-
Thomson’s ExperimentThomson’s Experiment
Voltage source
+-
Thomson’s ExperimentThomson’s Experiment
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
Thomson’s ExperimentThomson’s Experiment
Voltage source
+-
�� 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
Thomson’s ExperimentThomson’s Experiment
Voltage source
+-
�� 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
Thomson’s ExperimentThomson’s Experiment
Voltage source
+-
�� 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
Thomson’s ExperimentThomson’s Experiment
Voltage source
+-
Voltage source
Thomson’s ExperimentThomson’s Experiment
�� By adding an electric field By adding an electric field
Voltage source
Thomson’s ExperimentThomson’s Experiment
�� By adding an electric field By adding an electric field
+
-
4
Voltage source
Thomson’s ExperimentThomson’s Experiment
�� By adding an electric field By adding an electric field
+
-
Voltage source
Thomson’s ExperimentThomson’s Experiment
�� By adding an electric field By adding an electric field
+
-
Voltage source
Thomson’s ExperimentThomson’s Experiment
�� By adding an electric field By adding an electric field
+
-
Voltage source
Thomson’s ExperimentThomson’s Experiment
�� By adding an electric field By adding an electric field
+
-
Voltage source
Thomson’s ExperimentThomson’s Experiment
�� 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
Millikan’s ExperimentMillikan’s Experiment
Oil
Atomizer
Microscope
-
+
Oil droplets
Millikan’s ExperimentMillikan’s Experiment
X-rays
X-rays give some drops a charge by knocking offelectrons
Millikan’s ExperimentMillikan’s Experiment
+
Millikan’s ExperimentMillikan’s Experiment
They put an electric charge on the plates
++
--
Millikan’s ExperimentMillikan’s Experiment
Some drops would hover
++
--
6
Millikan’s ExperimentMillikan’s Experiment
+
+ + + + + + +
- - - - - - -
Millikan’s ExperimentMillikan’s Experiment
Measure the drop and find volume from 4/3πr3
Find mass from M = D x V
++
--
Millikan’s ExperimentMillikan’s Experiment
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
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
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
�� 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