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SAT II CHEM PREP PPT
Mrs. Gupta
Modified from Mark Rosengarten’s Powerpoint
Setup of the SAT II Chem Exam
• 85 total questions, 1 hour (about 42 s/question)
• - All multiple choice, 1/4th point taken off for every incorrect
answer
• - if you can narrow down to two choices, then guess
otherwise leave blank
• - scoring scale from 200-800
What to Bring to the Exam
• #2 pencil, eraser
• No calculators allowed (brush up on your basic math skills)
• Your brain. Please don’t leave it at home.:)
How To Prepare
• DO NOT CRAM. Get your studying done with by the night
before. Get a good night’s sleep and have breakfast the
morning of the exam.
• Actively participate in any and all review classes and
activities offered by your teacher.
Matter
1) Properties of Phases
2) Types of Matter
3) Phase Changes
Properties of Phases
• Solids: Crystal lattice (regular geometric pattern), vibration
motion only
• Liquids: particles flow past each other but are still attracted
to each other.
• Gases: particles are small and far apart, they travel in a
straight line until they hit something, they bounce off without
losing any energy, they are so far apart from each other that
they have effectively no attractive forces and their speed is
directly proportional to the Kelvin temperature (KineticMolecular Theory, Ideal Gas Theory)
Solids
The positive and
negative ions
alternate in the
ionic crystal lattice
of NaCl.
Liquids
When heated, the ions move
faster and eventually
separate from each other to
form a liquid. The ions are
loosely held together by the
oppositely charged ions, but
the ions are moving too fast
for the crystal lattice to stay
together.
Gases
Since all gas molecules spread out
the same way, equal volumes of
gas under equal conditions of
temperature and pressure will
contain equal numbers of
molecules of gas. 22.4 L of any
gas at STP (1.00 atm and 273K)
will contain one mole
(6.02 X 1023) gas molecules.
Since there is space between gas
molecules, gases are affected by
changes in pressure.
Types of Matter
• Substances (Homogeneous)
– Elements (cannot be decomposed by chemical
change): Al, Ne, O, Br, H
– Compounds (can be decomposed by chemical
change): NaCl, Cu(ClO3)2, KBr, H2O, C2H6
• Mixtures
– Homogeneous: Solutions (solvent + solute)
– Heterogeneous: soil, Italian dressing, etc.
Elements
• A sample of lead atoms (Pb). All
atoms in the sample consist of lead,
so the substance is homogeneous.
• A sample of chlorine atoms (Cl). All
atoms in the sample consist of
chlorine, so the substance is
homogeneous.
Compounds
• Lead has two charges listed, +2
and +4. This is a sample of lead
(II) chloride (PbCl2). Two or more
elements bonded in a wholenumber ratio is a COMPOUND.
• This compound is formed from
the +4 version of lead. This is
lead (IV) chloride (PbCl4). Notice
how both samples of lead
compounds have consistent
composition throughout?
Compounds are homogeneous!
Mixtures
• A mixture of lead atoms and
chlorine atoms. They exist in no
particular ratio and are not
chemically combined with each
other. They can be separated by
physical means.
• A mixture of PbCl2 and PbCl4
formula units. Again, they are in
no particular ratio to each other
and can be separated without
chemical change.
The Atom
1) Nucleons
2) Isotopes
3) Natural Radioactivity
4) Half-Life
5) Nuclear Power
6) Electron Configuation
7) Development of the Atomic Model
Nucleons
• Protons: +1 each, determines identity of element, mass of 1
amu, determined using atomic number, nuclear charge
• Neutrons: no charge, determines identity of isotope of an
element, 1 amu, determined using mass number - atomic
number (amu = atomic mass unit)
• 3216S and 3316S are both isotopes of S
• S-32 has 16 protons and 16 neutrons
• S-33 has 16 protons and 17 neutrons
• All atoms of S have a nuclear charge of +16 due to the 16
protons.
Isotopes
• Atoms of the same element MUST contain the same number
of protons.
• Atoms of the same element can vary in their numbers of
neutrons, therefore many different atomic masses can exist
for any one element. These are called isotopes.
• The atomic mass on the Periodic Table is the weight-average
atomic mass, taking into account the different isotope masses
and their relative abundance.
• Rounding off the atomic mass on the Periodic Table will tell
you what the most common isotope of that element is.
Weight-Average Atomic Mass
•
WAM = ((% A of A/100) X Mass of A) + ((% A of B/100) X Mass of B) + …
• What is the WAM of an element if its isotope masses and
abundances are:
– X-200: Mass = 200.0 amu, % abundance = 20.0 %
– X-204: Mass = 204.0 amu, % abundance = 80.0%
– amu = atomic mass unit (1.66 × 10-27 kilograms/amu)
Most Common Isotope
• The weight-average atomic mass of Zinc is
65.39 amu. What is the most common isotope
of Zinc? Zn-65!
• What are the most common isotopes of: C, H
and O?
• FACT: one atomic mass unit (1.66 × 10-27
kilograms) is defined as 1/12 of the mass of an
atom of C-12.
Natural Radioactivity
•
•
•
•
•
Alpha Decay
Beta Decay
Positron Decay
Gamma Decay
Charges of Decay Particles
• Natural decay starts with a parent
nuclide that ejects a decay particle to
form a daughter nuclide which is more
stable than the parent nuclide was.
Alpha Decay
• The nucleus ejects two protons and two neutrons. The
atomic mass decreases by 4, the atomic number
decreases by 2.
• 23892U 
Beta Decay
• A neutron decays into a proton and an electron. The
electron is ejected from the nucleus as a beta particle.
The atomic mass remains the same, but the atomic
number increases by 1.
• 146C 
Positron Decay
• A proton is converted into a neutron and a positron. The
positron is ejected by the nucleus. The mass remains the
same, but the atomic number decreases by 1.
• 5326Fe 
Gamma Decay
• The nucleus has energy levels just like electrons, but the
involve a lot more energy. When the nucleus becomes
more stable, a gamma ray may be released. This is a
photon of high-energy light, and has no mass or charge.
The atomic mass and number do not change with gamma.
Gamma may occur by itself, or in conjunction with any
other decay type.
Charges of Decay Particles
Half-Life
• Half life is the time it takes for half of
the nuclei in a radioactive sample to
undergo decay.
• Problem Types:
– Going forwards in time
– Going backwards in time
– Radioactive Dating
Going Forwards in Time
• How many grams of a 10.0 gram sample of I-131 (half-life
of 8 days) will remain in 24 days?
• #HL = t/T = 24/8 = 3
• Cut 10.0g in half 3 times: 5.00, 2.50, 1.25g
Going Backwards in Time
• How many grams of a 10.0 gram sample of I-131 (half-life
of 8 days) would there have been 24 days ago?
• #HL = t/T = 24/8 = 3
• Double 10.0g 3 times: 20.0, 40.0, 80.0 g
Radioactive Dating
• A sample of an ancient scroll contains 50% of the original
steady-state concentration of C-14. How old is the scroll?
• 50% = 1 HL
• 1 HL X 5730 y/HL = 5730y
Nuclear Power
•
•
•
•
Artificial Transmutation
Particle Accelerators
Nuclear Fission
Nuclear Fusion
Artificial Transmutation
•
40
20Ca + _____ ----->
•
96
2 H -----> 1 n + _____
Mo
+
42
1
0
40 K
19
+ 11H
• Nuclide + Bullet --> New Element + Fragment(s)
• The masses and atomic numbers must add
up to be the same on both sides of the
arrow.
Nuclear Fission
 9236Kr + 14156Ba + 3 10n + energy
• The three neutrons given off can be reabsorbed by other U235 nuclei to continue fission as a chain reaction
•
235 U
92
+
1 n
0
• A tiny bit of mass is lost (mass defect) and converted into a
huge amount of energy.
Chain Reaction
Nuclear Fusion
• 21H + 21H  42He + energy
• Two small, positively-charged nuclei smash
together at high temperatures and pressures
to form one larger nucleus.
• A small bit of mass is destroyed and
converted into a huge amount of energy,
more than even fission.
Electron Configuration
•
•
•
•
•
•
•
Basic Configuration
Valence Electrons
Electron-Dot (Lewis Dot) Diagrams
Excited vs. Ground State
Rules for Electron Filling
Para and Diamagnetic
Lewis Structures and Hybridization
Basic Configuration
• The number of electrons is determined from the atomic
number.
• Look up the basic configuration below the atomic number on
the periodic table. (PEL: principal energy level = shell)
• He: 2 (2 e- in the 1st PEL)
• Na: 2-8-1 (2 e- in the 1st PEL, 8 in the 2nd and 1 in the 3rd)
• Br: 2-8-18-7 (2 e- in the 1st PEL, 8 in the 2nd, 18 in the 3rd and
7 in the 4th)
Valence Electrons
• The valence electrons are responsible for all chemical
bonding.
• The valence electrons are the electrons in the outermost PEL
(shell).
• He: 2 (2 valence electrons)
• Na: 2-8-1 (1 valence electron)
• Br: 2-8-18-7 (7 valence electrons)
• The maximum number of valence electrons an atom can have
is EIGHT, called a STABLE OCTET.
Electron-Dot Diagrams
• The number of dots equals the number of valence electrons.
• The number of unpaired valence electrons in a nonmetal tells
you how many covalent bonds that atom can form with other
nonmetals or how many electrons it wants to gain from
metals to form an ion.
• The number of valence electrons in a metal tells you how
many electrons the metal will lose to nonmetals to form an
ion. Caution: May not work with transition metals.
• EXAMPLE DOT DIAGRAMS
(c) 2006, Mark Rosengarten
Example Dot Diagrams
Carbon can also have this dot diagram, which it
has when it forms organic compounds.
Excited vs. Ground State
• Configurations on the Periodic Table are ground state
configurations.
• If electrons are given energy, they rise to higher energy levels
(excited state).
• If the total number of electrons matches in the configuration,
but the configuration doesn’t match, the atom is in the
excited state.
• Na (ground, on table): 2-8-1
• Example of excited states: 2-7-2, 2-8-0-1, 2-6-3
Ways to Represent Electron Configuration
1.Expanded Electron Configuration
2.Condensed Electron Configurations
3.Orbital Notation
4.Electron Dot Structure
Write the above four electron configurations for Zinc, Zinc ion and
Cu ion.
Electron Configuration of Ions
• -Group configurations: s block ns1-2, p block
ns2 np 1-6, d block ns0-2 (n-1) d 1-10, f block ns 02 (n-1) d 1 (n-2) f 1-14
• - Remember that outmost electrons are lost
first (which means that it will always be s or p
electrons lost, never d or f). Ex. Sc+ or Sc3+
electron configuration would be:
Rules for Electron Filling
• - Afbau’s Principle: Electrons tend to occupy
the lowest energy orbitals first.
• - Hund’s Rule: Pairing of e in the degenerate
orbitals does not take place till every orbital
has one e.
• - Pauli’s Exclusion Principle: No two electrons
can have all four same quantum numbers.
Diamagnetism, Paramagnetism
• Diamagnetism: does not show magnetic
properties in external magnetic field. No
unpaired electrons.
• Paramagnetism: shows magnetic properties
in external magnetic field. Has unpaired
electrons.
• Best way to predict dia or paramagnetism is
by drawing orbital diagrams.
Writing Lewis Structures
• Lewis structures are used to
depict bonding pairs and
lone pairs of electron in the
molecule.
Step 1
• Total number of valence
electrons in the system: Sum
the number of valence
electrons on all the atoms .
Add the total negative
charge if you have an anion.
Subtract the charge if you
have a cation.
Example: CO32Step 2
• Number of electrons if each
atom is to be happy: Atoms
in our example will need 8 e
(octet rule) or 2 e (
hydrogen). So, for the ex.
•
Step 3
Step 3
• Calculate number of bonds in the system:
Covalent bonds are made by sharing of e.
You need 32 and you have 24. You are 8 e
deficient. If you make 4 bonds ( with 2 e per
bond) , you will make up the deficiency.
Therefore,
•
# of bonds= ( e in step 2- e in step 1)/2 =(3224)/2= 4 bonds
Step 4
• Draw the structure: The central atom is C (
usually the atom with least electro
negativity will be in the center). The oxygens
surround it . Because there are four bonds
and only three atoms, there will be one
double bond.
Step 5
• Double check your answer by counting total
number of electrons.
© 2009, Prentice-Hall, Inc.
Formal Charges: Writing Lewis
Structures
• Then assign formal charges.
– For each atom, count the electrons in lone pairs and half
the electrons it shares with other atoms.
– Subtract that from the number of valence electrons for that
atom: the difference is its formal charge.
© 2009, Prentice-Hall, Inc.
Writing Lewis Structures
• The best Lewis structure…
– …is the one with the fewest charges.
– …puts a negative charge on the most
electronegative atom.
© 2009, Prentice-Hall, Inc.
Resonance
• One Lewis structure
cannot accurately
depict a molecule
like ozone.
• We use multiple
structures,
resonance
structures, to
describe the
molecule.
© 2009, Prentice-Hall, Inc.
Resonance
Just as green is a synthesis of
blue and yellow…
…ozone is a synthesis of
these two resonance
structures.
© 2009, Prentice-Hall, Inc.
Molecular Shapes
• The shape of a molecule
plays an important role
in its reactivity.
• By noting the number of
bonding and nonbonding
electron pairs we can
easily predict the shape
of the molecule.
© 2009, Prentice-Hall, Inc.
What Determines the Shape of a
Molecule?
• Simply put, electron
pairs, whether they be
bonding or
nonbonding, repel each
other.
• By assuming the
electron pairs are
placed as far as
possible from each
other, we can predict
the shape of the
molecule.
© 2009, Prentice-Hall, Inc.
Electron Domains
• The central atom in
this molecule, A, has
four electron
domains.
• We can refer to the
electron pairs as electron
domains.
• In a double or triple bond,
all electrons shared
between those two atoms
are on the same side of the
central atom; therefore,
they count as one electron
domain.
© 2009, Prentice-Hall, Inc.
Valence Shell Electron Pair Repulsion
Theory (VSEPR)
“The best
arrangement of a
given number of
electron domains is
the one that minimizes
the repulsions among
them.”
© 2009, Prentice-Hall, Inc.
Electron-Domain
Geometries
These are the
electron-domain
geometries for two
through six electron
domains around a
central atom.
© 2009, Prentice-Hall, Inc.
Hybridization
• Refers to mixing of
orbitals.
• Atomic orbitals of
central atom
undergo change to
accommodate
incoming atoms.
• Hybridization could
be sp, sp2, sp3,
sp3d and sp3d2.
• How do you tell the
hybridization on
the central atom?
9.1
–
9.2:
V.S.E.P.R.
Valence-shell electron-pair repulsion theory
– Because e- pairs repel, molecular shape adjusts so the valence epairs are as far apart as possible around the central atom.
– Electron domains: areas of valence e- density around the central
atom; result in different molecular shapes
• Includes bonding e- pairs and nonbonding e- pairs
• A single, double, or triple bond counts as one domain
Summary of LmABn (Tables 9.1 - 9.3):
L = lone or non-bonding pairs
A = central atom
B = bonded atoms
Bond angles notation used here:
< xº
<< xº
means ~2-3º less than predicted
means ~4-6º less than predicted
Tables 9.1 - 9.3
# of edomains
&
# and type
of hybrid
orbitals
2
Two sp
hybrid
orbitals
e- domain
geometry
Formula
&
Molecular
geometry
X
B
A
|X
A
|B
Linear
AB2
Linear
Predicted
bond
angle(s)
180º
Example
(Lewis
structure with
molecular
shape)
BeF2
CO2
B
B
120º
A
|B
BF3
3
X
Three sp2
hybrid
orbitals
X
A
|X
AB3
Trigonal planar
B
Cl2CO
:
A
|B
Trigonal
planar
Cl-C-Cl
<< 120º
LAB2
Bent
< 120º
NO21-
Example: CH4
H
|
H—C—H
|
H
109.5º
109.5º
109.5º
109.5º
Molecular shape =
tetrahedral
Bond angle = 109.5º
B
B
A
B
X
X
4
A
X
Four sp3
hybrid
orbitals
B
AB4
Tetrahedral
B
:
X
or
109.5º
CH4
< 109.5º
A
B
X
X
LAB3 B
Trigonal
pyramidal
Ex: NH3 =
107º
NH3
B
A
X
<<109.5º
:
A
X
:
Tetrahedral
L2AB2
B
Bent
Ex: H2O =
104.5º
H2O
PCl5
:
:
: :
: :
:Cl: :Cl:
\ /
:Cl—P—Cl:
|
:Cl:
:
90º
90º
120º
90º
120º
Molecular shape =
trigonal bipyramidal
Bond angles
equatorial = 120º
axial = 90º
B
B
A
B
|
B
5
X
X
X
B
AB5
Trigonal
bipyramidal
Axial = 90º
PCl5
|
X
X
:
Five
sp3d
hybrid
orbitals
A
Equatorial
= 120º
Equatorial
< 120º
B-A-B
B
Trigonal
bipyramidal
B
LAB4
Seesaw
Axial
< 90º
SF4
B
:
A
B
|
:
B
5
X
X
Five
sp3d
hybrid
orbitals
A
X
L2AB3
T-shaped
|
X
X
ClF3
B
:
A
|
:
Trigonal
bipyramidal
Axial
<< 90º
B
L3AB2
Linear
:
Axial =
180º
XeF2
X
X
X
B
B
X
A
A
|
|
X
X
6
B
B
B
90º
AB6
Octahedral
or
Six sp3d2
hybrid
orbitals
B
SF6
B
B
X
X
X
X
A
|
B
A
X
X
Octahedral
B
|
..
B
LAB5
Square
pyramidal
< 90º
BrF5
B
..
B
or
A
6
Six sp3d2
hybrid
orbitals
B
..
B
|
..
B
A
B
90º
|
B
..
B
L2AB4
Square planar
..
B
..
..
A
B
XeF4
|
..
..
or
B
B
A
|
..
L3AB3
T-shaped
B
<90º
B
KrCl31-
What Is Light?
• Light is formed when electrons drop from the
excited state to the ground state.
• The lines on a bright-line spectrum come from
specific energy level drops and are unique to
each element.
• Ex. Emission and Absorption Spectra ( line
spectra)
EXAMPLE SPECTRUM
This is the bright-line spectrum of hydrogen. The top
numbers represent the energy level transition change that produces the
light with that color and the bottom number is the
wavelength of the light (in nanometers, or 10-9 m).
No other element has the same bright-line spectrum as
hydrogen, so these spectra can be used to identify
elements or mixtures of elements.
Development of the Atomic Model
•
•
•
•
Thompson Model
Rutherford Gold Foil Experiment and Model
Bohr Model
Quantum-Mechanical Model
Thompson Model
• The atom is a positively charged diffuse mass
with negatively charged electrons stuck in it.
Rutherford Model
• The atom is made of a small, dense, positively charged
nucleus with electrons at a distance, the vast majority of the
volume of the atom is empty space.
Alpha particles shot
at a thin sheet of gold
foil: most go through
(empty space). Some
deflect or bounce off
(small + charged
nucleus).
Bohr Model
• Electrons orbit around the nucleus in energy levels (shells).
Atomic bright-line spectra was the clue.
Quantum-Mechanical Model
• Electron energy levels are wave functions.
• Electrons are found in orbitals, regions of space where an
electron is most likely to be found.
• You can’t know both where the electron is and where it is
going at the same time.
• Electrons buzz around the nucleus like gnats buzzing around
your head.
Orbital Quantum Numbers
Symbol
Name
n
Principle
Q.N.
l
Description
Energy level
(i.e. Bohr’s
theory)
Meaning
Shell number
n = 1, 2, 3, 4,
5, 6, 7
Subshell
number
l = 0, 1, 2, 3
General
Angular
probability l = 0 means
Momentum plot (“shape” “s”
Q.N.
of the
l = 1 means
orbitals)
“p”
l = 2 means
“d”
Equations
n = 1, 2, 3,
…
l = 0, 1, 2,
…, n – 1
Ex: If n = 1,
l can only be
0; if n = 2, l
can be 0 or
1.
Symbol
ml
ms
Name
Magnetic
Q.N.
Spin Q.N.
Description
Meaning
3-D orientation
of the orbital
s has 1
p has 3
d has 5
f has 7
Spin of the
electron
Parallel or
antiparallel
to field
Equations
ml = -l, -l +1, …,
0, l, …, +l
There are
(2l + 1) values.
ms = +½ or
-½
* s, p, d, and f come from the words sharp, principal, diffuse, and fundamental.
Permissible Quantum Numbers
(4, 1, 2, +½)
Not permissible; if l = 1, ml = 1, 0, or –1 (p
orbitals only have 3 subshells)
(5, 2, 0, 0)
Not permissible; ms = +½ or –½
(2, 2, 1, +½)
Not permissible; if n = 2, l = 0 or 1 (there is no
2d orbital)
76
Phase Changes
•
•
•
•
Phase Change Types
Phase Change Diagrams
Heat of Phase Change
Evaporation
Phase Change Types
Phase Change Diagrams
AB: Solid Phase
BC: Melting (S + L)
CD: Liquid Phase
DE: Boiling (L + G)
EF: Gas Phase
Notice how temperature remains constant during a phase change? That’s because
the PE is changing, not the KE.
Heat of Phase Change
• How many joules would it take to melt 100. g of H2O (s) at
0oC?
• q=mHf = (100. g)(334 J/g) = 33400 J
• How many joules would it take to boil 100. g of H2O (l) at
100oC?
• q=mHv = (100.g)(2260 J/g) = 226000 J
Evaporation
• When the surface molecules of a gas travel upwards at a
great enough speed to escape.
• The pressure a vapor exerts when sealed in a container at
equilibrium is called vapor pressure, and can be found on
Table H.
• When the liquid is heated, its vapor pressure increases.
• When the liquid’s vapor pressure equals the pressure
exerted on it by the outside atmosphere, the liquid can
boil.
• If the pressure exerted on a liquid increases, the boiling
point of the liquid increases (pressure cooker). If the
pressure decreases, the boiling point of the liquid
decreases (special cooking directions for high elevations).
Reference Table H: Vapor Pressure of
Four Liquids
(c) 2006, Mark Rosengarten
Phase diagrams: CO2
 Lines: 2 phases exist in
equilibrium
 Triple point: all 3 phases exist
together in equilibrium (X on
graph)
 Critical point, or critical
temperature & pressure: highest
T and P at which a liquid can exist
(Z on graph)
Temp (ºC)
 For most substances, inc P will cause a gas to condense (or deposit), a liquid to freeze,
and a solid to become more dense (to a limit.)
83
Phase diagrams: H2O
• For H2O, inc P
will cause ice
to melt.
84
The Periodic Table
•
•
•
•
•
•
Metals
Nonmetals
Metalloids
Chemistry of Groups
Electronegativity
Ionization Energy
Metals
• Have luster, are malleable and ductile, good
conductors of heat and electricity
• Lose electrons to nonmetal atoms to form positively
charged ions in ionic bonds
• Large atomic radii compared to nonmetal atoms
• Low electronegativity and ionization energy
• Left side of the periodic table (except H)
Nonmetals
• Are dull and brittle, poor conductors
• Gain electrons from metal atoms to form negatively
charged ions in ionic bonds
• Share unpaired valence electrons with other
nonmetal atoms to form covalent bonds and
molecules
• Small atomic radii compared to metal atoms
• High electronegativity and ionization energy
• Right side of the periodic table (except Group 18)
Metalloids
• Found lying on the jagged line between metals and
nonmetals flatly touching the line (except Al and Po).
• Share properties of metals and nonmetals (Si is shiny like a
metal, brittle like a nonmetal and is a semiconductor).
Chemistry of Groups
•
•
•
•
•
Group 1: Alkali Metals
Group 2: Alkaline Earth Metals
Groups 3-11: Transition Elements
Group 17: Halogens
Group 18: Noble Gases
• Diatomic Molecules
Group 1: Alkali Metals
• Most active metals, only found in compounds in
nature
• React violently with water to form hydrogen gas
and a strong base: 2 Na (s) + H2O (l)  2 NaOH
(aq) + H2 (g)
• 1 valence electron
• Form +1 ion by losing that valence electron
• Form oxides like Na2O, Li2O, K2O
Group 2: Alkaline Earth Metals
• Very active metals, only found in compounds in
nature
• React strongly with water to form hydrogen gas
and a base:
– Ca (s) + 2 H2O (l)  Ca(OH)2 (aq) + H2 (g)
• 2 valence electrons
• Form +2 ion by losing those valence electrons
• Form oxides like CaO, MgO, BaO
Groups 3-11: Transition Metals
• Many can form different possible charges of ions
• If there is more than one ion listed, give the charge as a
Roman numeral after the name
• Cu+1 = copper (I) Cu+2 = copper (II)
• Compounds containing these metals can be colored.
Group 17: Halogens
• Most reactive nonmetals
• React violently with metal atoms to form halide
compounds: 2 Na + Cl2  2 NaCl
• Only found in compounds in nature
• Have 7 valence electrons
• Gain 1 valence electron from a metal to form -1
ions
• Share 1 valence electron with another nonmetal
atom to form one covalent bond.
Group 18: Noble Gases
• Are completely nonreactive since they have eight
valence electrons, making a stable octet.
• Kr and Xe can be forced, in the laboratory, to give
up some valence electrons to react with fluorine.
• Since noble gases do not naturally bond to any
other elements, one atom of noble gas is
considered to be a molecule of noble gas. This is
called a monatomic molecule. Ne represents an
atom of Ne and a molecule of Ne.
Diatomic Molecules
• Br, I, N, Cl, H, O and F are so reactive that they exist in a more
chemically stable state when they covalently bond with
another atom of their own element to make two-atom, or
diatomic molecules.
• Br2, I2, N2, Cl2, H2, O2 and F2
• The decomposition of water: 2 H2O  2 H2 + O2
Electronegativity
•
•
•
•
An atom’s attraction to electrons in a chemical bond.
F has the highest, at 4.0
Fr has the lowest, at 0.7
If two atoms that are different in EN (END) from each other by
1.7 or more collide and bond (like a metal atom and a
nonmetal atom), the one with the higher electronegativity will
pull the valence electrons away from the atom with the lower
electronegativity to form a (-) ion. The atom that was stripped
of its valence electrons forms a (+) ion.
• If the two atoms have an END of less than 1.7, they will share
their unpaired valence electrons…covalent bond!
Ionization Energy
• The energy required to remove the most loosely held
valence electron from an atom in the gas phase.
• High electronegativity means high ionization energy
because if an atom is more attracted to electrons, it will
take more energy to remove those electrons.
• Metals have low ionization energy. They lose electrons
easily to form (+) charged ions.
• Nonmetals have high ionization energy but high
electronegativity. They gain electrons easily to form (-)
charged ions when reacted with metals, or share unpaired
valence electrons with other nonmetal atoms.
Ions
• Ions are charged particles formed by the gain or loss of
electrons.
– Metals lose electrons (oxidation) to form (+) charged
cations.
– Nonmetals gain electrons (reduction) to form (-) charged
anions.
• Atoms will gain or lose electrons in such a way that they end
up with 8 valence electrons (stable octet).
– The exceptions to this are H, Li, Be and B, which are not
large enough to support 8 valence electrons. They must
be satisfied with 2 (Li, Be, B) or 0 (H).
Metal Ions (Cations)
• Na: 2-8-1
• Na+1: 2-8
• Ca: 2-8-8-2
• Ca+2: 2-8-8
• Al: 2-8-3
• Al+3: 2-8
Note that when the atom loses its
valence electron, the next lower
PEL becomes the valence PEL.
Notice how the dot diagrams for
metal ions lack dots! Place
brackets around the element
symbol and put the charge on the
upper right outside!
Nonmetal Ions (Anions)
• F: 2-7
• F-1: 2-8
• O: 2-6
• O-2: 2-8
• N: 2-5
• N-3: 2-8
Note how the ions all have 8 valence
electrons. Also note the gained electrons
as red dots. Nonmetal ion dot diagrams
show 8 dots, with brackets around the dot
diagram and the charge of the ion written
to the upper right side outside the brackets.
Chemical Bonding
•Intermolecular Bonding: Ionic, Covalent, Metallic and Covalent
Network Bonds
•Intermolecular Bonding: H bond, dipole-dipole interaction, LDFs
Ionic Bonding
• If two atoms that are different in EN (END) from each other by
1.7 or more collide and bond (like a metal atom and a
nonmetal atom), the one with the higher electronegativity will
pull the valence electrons away from the atom with the lower
electronegativity to form a (-) ion. The atom that was stripped
of its valence electrons forms a (+) ion.
• The oppositely charged ions attract to form the bond. It is a
surface bond that can be broken by melting or dissolving in
water.
• Ionic bonding forms ionic crystal lattices, not molecules.
Example of Ionic Bonding
Covalent Bonding
• If two nonmetal atoms have an END of 1.7 or less, they
will share their unpaired valence electrons to form a
covalent bond.
• A particle made of covalently bonded nonmetal atoms is
called a molecule.
• If the END is between 0 and 0.4, the sharing of electrons is
equal, so there are no charged ends. This is NONPOLAR
covalent bonding.
• If the END is between 0.5 and 1.7, the sharing of electrons
is unequal. The atom with the higher EN will be d- and the
one with the lower EN will be d+ charged. This is a POLAR
covalent bonding. (d means “partial”)
Examples of Covalent Bonding
Sigma
Sigma (s) bond:
and Pi bonds
• Covalent bond that results from axial overlap of orbitals between atoms in
a molecule
• Lie directly on internuclear axis
• “Single” bonds, could form between s-s orbital or s-p orbital or p-p orbital
by axial overlapping
Ex: F2
Pi (p) bond:
• Covalent bond that results from side-by-side overlap of orbitals between
atoms in a molecule.
• Are “above & below” and “left & right” of the internuclear axis and
therefore have less total orbital overlap, so they are weaker than s bonds.
Forms between two p orbitals (py or pz)
• Make up the 2nd and 3rd bonds in double & triple bonds.
Ex: O2
N2
Metallic Bonding
• Metal atoms of the same element bond with each other by
sharing valence electrons that they lose to each other.
• This is a lot like an atomic game of “hot potato”, where metal
kernals (the atom inside the valence electrons) sit in a crystal
lattice, passing valence electrons back and forth between
each other).
• Since electrons can be forced to travel in a certain direction
within the metal, metals are very good at conducting
electricity in all phases.
Types of Compounds
• Ionic: made of metal and nonmetal ions. Form an ionic
crystal lattice when in the solid phase. Ions separate
when melted or dissolved in water, allowing electrical
conduction. Examples: NaCl, K2O, CaBr2
• Molecular: made of nonmetal atoms bonded to form a
distinct particle called a molecule. Bonds do not break
upon melting or dissolving, so molecular substances do
not conduct electricity. EXCEPTION: Acids [H+A- (aq)]
ionize in water to form H3O+ and A-, so they do conduct.
• Network: made up of nonmetal atoms bonded in a
seemingly endless matrix of covalent bonds with no
distinguishable molecules. Very high m.p., don’t conduct.
Ionic Compounds
Ionic Crystal Structure, then adding heat (or dissolving in water) to break
up the crystal into a liquid composed of free-moving ions.
(c) 2006, Mark Rosengarten
Molecular Compounds
(c) 2006, Mark Rosengarten
Network Solids
Network solids are made of nonmetal atoms covalently bonded together to
form large crystal lattices. No individual molecules can be distinguished.
Examples include C (diamond) and SiO2 (quartz). Corundum (Al2O3) also forms
these, even though Al is considered a metal. Network solids are among the
hardest materials known. They have extremely high melting points and do not
conduct electricity.
Intermolecular Forces
The attractions between molecules are not nearly
as strong as the intramolecular attractions that
hold compounds together.
© 2009, Prentice-Hall, Inc.
Intermolecular Forces
They are, however, strong enough to control
physical properties such as boiling and melting
points, vapor pressures, and viscosities.
© 2009, Prentice-Hall, Inc.
Intermolecular Forces
These intermolecular forces as a group are
referred to as van der Waals forces.
© 2009, Prentice-Hall, Inc.
van der Waals Forces
• Dipole-dipole interactions
• Hydrogen bonding
• London dispersion forces
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Ion-Dipole Interactions
• Ion-dipole interactions (a fourth type of force),
are important in solutions of ions.
• The strength of these forces are what make it
possible for ionic substances to dissolve in polar
solvents.
© 2009, Prentice-Hall, Inc.
Dipole-Dipole Interactions
• Molecules that have
permanent dipoles are
attracted to each other.
– The positive end of one is
attracted to the negative
end of the other and viceversa.
– These forces are only
important when the
molecules are close to each
other.
© 2009, Prentice-Hall, Inc.
Dipole-Dipole Interactions
The more polar the molecule, the higher is its
boiling point.
© 2009, Prentice-Hall, Inc.
London Dispersion Forces
While the electrons in the 1s orbital of helium
would repel each other (and, therefore, tend to
stay far away from each other), it does happen
that they occasionally wind up on the same side
of the atom.
© 2009, Prentice-Hall, Inc.
London Dispersion Forces
At that instant, then, the helium atom is polar,
with an excess of electrons on the left side and a
shortage on the right side.
© 2009, Prentice-Hall, Inc.
London Dispersion Forces
Another helium nearby, then, would have a dipole
induced in it, as the electrons on the left side of
helium atom 2 repel the electrons in the cloud on
helium atom 1.
© 2009, Prentice-Hall, Inc.
London Dispersion Forces
London dispersion forces, or dispersion forces, are
attractions between an instantaneous dipole and
an induced dipole.
© 2009, Prentice-Hall, Inc.
London Dispersion Forces
• These forces are present in all molecules,
whether they are polar or nonpolar.
• The tendency of an electron cloud to distort in
this way is called polarizability.
© 2009, Prentice-Hall, Inc.
Factors Affecting London Forces
• The shape of the molecule affects
the strength of dispersion forces:
long, skinny molecules (like npentane tend to have stronger
dispersion forces than short, fat
ones (like neopentane).
• This is due to the increased surface
area in n-pentane.
© 2009, Prentice-Hall, Inc.
Factors Affecting London Forces
• The strength of dispersion forces tends to
increase with increased molecular weight.
• Larger atoms have larger electron clouds which
are easier to polarize.
© 2009, Prentice-Hall, Inc.
Which Have a Greater Effect?
Dipole-Dipole Interactions or Dispersion Forces
• If two molecules are of comparable size and
shape, dipole-dipole interactions will likely the
dominating force.
• If one molecule is much larger than another,
dispersion forces will likely determine its
physical properties.
© 2009, Prentice-Hall, Inc.
How Do We Explain This?
• The nonpolar series
(SnH4 to CH4) follow
the expected trend.
• The polar series
follows the trend from
H2Te through H2S, but
water is quite an
anomaly.
© 2009, Prentice-Hall, Inc.
Hydrogen Bonding
• The dipole-dipole interactions
experienced when H is bonded to N,
O, or F are unusually strong.
• We call these interactions hydrogen
bonds.
© 2009, Prentice-Hall, Inc.
Hydrogen Bonding
• Hydrogen bonding arises
in part from the high
electronegativity of
nitrogen, oxygen, and
fluorine.
Also, when hydrogen is bonded to one of those very
electronegative elements, the hydrogen nucleus is
exposed.
© 2009, Prentice-Hall, Inc.
Attractive Forces
• Molecules have partially charged ends. The d+ end of one
molecule attracts to the d- end of another molecule.
• Ions are charged (+) or (-). Positively charged ions attract
other to form ionic bonds, a type of attractive force.
• Since partially charged ends result in weaker attractions than
fully charged ends, ionic compounds generally have much
higher melting points than molecular compounds.
• Determining Polarity of Molecules
• Hydrogen Bond Attractions
Determining Polarity of
Molecules
-----------------------------------------------------------------------------
(c) 2006, Mark Rosengarten
Hydrogen Bond
Attractions
A hydrogen bond attraction is a very strong
attractive force between the H end of one
polar molecule and the N, O or F end of
another polar molecule. This attraction is
so strong that water is a liquid at a
temperature where most compounds that
are much heavier than water (like
propane, C3H8) are gases. This also gives
water its surface tension and its ability to
form a meniscus in a narrow glass tube.
Summarizing Intermolecular Forces
© 2009, Prentice-Hall, Inc.
Compounds
1) Types of Compounds
2) Formula Writing
3) Formula Naming
4) Empirical Formulas
5) Molecular Formulas
6) Types of Chemical Reactions
7) Balancing Chemical Reactions
8) Attractive Forces
Formula Writing
• The charge of the (+) ion and the charge of the (-) ion must
cancel out to make the formula. Use subscripts to indicate
how many atoms of each element there are in the compound,
no subscript if there is only one atom of that element.
• Na+1 and Cl-1 = NaCl
• Ca+2 and Br-1 = CaBr2
• Al+3 and O-2 = Al2O3
• Zn+2 and PO4-3 = Zn3(PO4)2
• Try these problems!
Formulas to Write
•
•
•
•
•
•
•
•
Ba+2 and N-3
NH4+1 and SO4-2
Li+1 and S-2
Cu+2 and NO3-1
Al+3 and CO3-2
Fe+3 and Cl-1
Pb+4 and O-2
Pb+2 and O-2
Formula Naming
• Compounds are named from the elements or
polyatomic ions that form them.
• KCl = potassium chloride
• Na2SO4 = sodium sulfate
• (NH4)2S = ammonium sulfide
• AgNO3 = silver nitrate
• Notice all the metals listed here only have one
charge listed? So what do you do if a metal has
more than one charge listed? Take a peek!
The Stock System
• CrCl2 = chromium (II) chloride
• CrCl3 = chromium (III) chloride
• CrCl6 = chromium (VI) chloride
Try
Co(NO3)2 and
Co(NO3)3
• FeO = iron (II) oxide
MnS = manganese (II) sulfide
• Fe2O3 = iron (III) oxide
MnS2 = manganese (IV) sulfide
• The Roman numeral is the charge of the metal ion!
Math of Chemistry
1) Formula Mass
2) Percent Composition
3) Mole Problems
4) Gas Laws
5) Neutralization
6) Concentration
7) Significant Figures and Rounding
8) Metric Conversions
9) Calorimetry
Formula Mass
• Gram Formula Mass = sum of atomic masses of all elements in
the compound
• Round given atomic masses to the nearest tenth
• H2O: (2 X 1.0) + (1 X 16.0) = 18.0 grams/mole
• Na2SO4: (2 X 23.0)+(1 X 32.1)+(4 X 16.0) = 142.1 g/mole
• Now you try:
– BaBr2
– CaSO4
– Al2(CO3)3
Percent Composition
What is the % composition, by mass,
of each element in SiO2?
%Si =
(28.1/60.1) X 100 = 46.8%
%O = (2 X 16.0 = 32.0), (32.0/60.1) X 100 = 53.2%
The mass of part is the number of atoms of that element in the compound. The
mass of whole is the formula mass of the compound. Don’t forget to take atomic
mass to the nearest tenth! This is a problem for you to try.
Practice Percent
Composition Problem
• What is the percent by mass of each element in Li2SO4?
Mole Problems
• Grams <=> Moles
• Molecular Formula
• Stoichiometry
Grams <=> Moles
• How many grams will 3.00 moles of NaOH (40.0 g/mol)
weigh?
• 3.00 moles X 40.0 g/mol = 120. g
• How many moles of NaOH (40.0 g/mol) are represented
by 10.0 grams?
• (10.0 g) / (40.0 g/mol) = 0.250 mol
Molecular Formula
•
Molecular Formula = (Molecular Mass/Empirical Mass) X Empirical Formula
• What is the molecular formula of a compound with an
empirical formula of CH2 and a molecular mass of 70.0
grams/mole?
• 1) Find the Empirical Formula Mass: CH2 = 14.0
• 2) Divide the MM/EM: 70.0/14.0 = 5
• 3) Multiply the molecular formula by the result:
5 (CH2) = C5H10
Stoichiometry
•
Moles of Target = Moles of Given X (Coefficent of Target/Coefficient of given)
• Given the balanced equation N2 + 3 H2  2 NH3, How
many moles of H2 need to be completely reacted with N2
to yield 20.0 moles of NH3?
• 20.0 moles NH3 X (3 H2 / 2 NH3) = 30.0 moles H2
Limiting Reactant
• controls the amount of product formed.
CO(g) + 2H2 (g)  Ch3OH
a. If 500 mol of CO react with 750 mol of H2,
which is the limiting reactant? H
2
1.Use either given amount to calculate required
amount of other.
2.Compare calculated amount to amount given
b. How many moles of excess reactant remain
unchanged?
125 mol CO
147
Percent yield= (actual yield/ theoretical yield)*100
• Theoretical yield is the maximum amount of
product that can be produced from a given amount
of reactant
• Actual yield is the measured amount of a product
obtained from a reaction
Theoretical yield= 117.5 g SnF2
Actual yield = 113. 4g SnF2
Percent yield = 113.4 g SnF2 *100
117.5 g SnF2
148
Determining empirical formula from combustion data
When a compound containing C,H and O undergoes combustion, it forms CO2 and
H2O. Then from the mass of CO2 and H2O, we can calculate the mass of C and Hand
then find the mass of O by subtracting the sum of masses of C and H from total g
present of that substance. From the mass of C,H and O, we can calculate the moles
of C,H and O.Then the smallest whole number ratios of these moles will give the
empirical formula.
Ex. A 0.6349 g sample of the unknown produced 1.603 g of CO2 and 0.2810 g of
H2O. Determine the empirical formula of the compound. Ans. C7H6O2
Empirical Formulas
• Ionic formulas: represent the simplest whole number mole
ratio of elements in a compound.
• Ca3N2 means a 3:2 ratio of Ca ions to N ions in the compound.
• Many molecular formulas can be simplified to empirical
formulas
– Ethane (C2H6) can be simplified to CH3. This is the
empirical formula…the ratio of C to H in the molecule.
• All ionic compounds have empirical formulas.
Molecular Formulas
• The count of the actual number of atoms of each element
in a molecule.
• H2O: a molecule made of two H atoms and one O atom
covalently bonded together.
• C2H6O: A molecule made of two C atoms, six H atoms and
one O atom covalently bonded together.
• Molecular formulas are whole-number multiples of
empirical formulas:
– H2O = 1 X (H2O)
– C8H16 = 8 X (CH2)
• Calculating Molecular Formulas
Calculating Empirical Formulas
One can calculate the empirical formula from the
percent composition.
© 2009, Prentice-Hall,
Calculating Empirical Formulas
The compound para-aminobenzoic acid (you may have seen it listed as PABA on
your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%),
nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.
© 2009, Prentice-Hall, Inc.
Calculating Empirical Formulas
Assuming 100.00 g of para-aminobenzoic acid,
C:
61.31 g x
H:
5.14 g x
N:
10.21 g x
5.105 mol C
1=mol
12.01 g
= 5.09 mol H
1 mol
1.01
g
= 0.7288
mol N
23.33 g x
1 mol
= 1.456
14.01
g mol O
O:
1 mol
16.00 g
© 2009, Prentice-Hall, Inc.
Calculating Empirical Formulas
Calculate the mole ratio by dividing by the smallest number of moles:
C:
H:
5.105 mol
0.7288 mol
5.09 mol
0.7288 mol
N:
O:
= 7.005  7
= 6.984  7
= 1.000
0.7288 mol
0.7288 mol
= 2.001  2
1.458 mol
0.7288 mol
© 2009, Prentice-Hall, Inc.
Calculating Empirical Formulas
These are the subscripts for the empirical formula:
C7H7NO2
© 2009, Prentice-Hall, Inc.
Combustion Analysis
• Compounds containing C, H and O are routinely analyzed
through combustion in a chamber like this.
– C is determined from the mass of CO2 produced.
– H is determined from the mass of H2O produced.
– O is determined by difference after the C and H have been
determined.
© 2009, Prentice-Hall,
Stoichiometric Calculations
The coefficients in the balanced equation give the
ratio of moles of reactants and products.
© 2009, Prentice-Hall,
Stoichiometric Calculations
Starting with the mass
of Substance A you
can use the ratio of
the coefficients of A
and B to calculate the
mass of Substance B
formed (if it’s a
product) or used (if
it’s a reactant).
© 2009, Prentice-Hall,
Stoichiometric Calculations
C6H12O6 + 6 O2  6 CO2 + 6 H2O
Starting with 1.00 g of C6H12O6…
we calculate the moles of C6H12O6…
use the coefficients to find the moles of H2O…
and then turn the moles of water to grams.
© 2009, Prentice-Hall,
Molecular Formula
Actual ratio of atoms in a compound.
Ex. H2O, C6H12O6
To determine the molecular formula, divide the molar mass by empirical formula
mass. This will give the number of empirical formula units (n) in actual molecule.
n= Molar Mass/ Empirical Formula Mass
Ex. Determine the empirical and molecular formula of each of the following:
1.Ethylene glycol, the substance used as antifreeze has 38.70 % C, 9.70 % H
and 51.60 % O , mm= 62.10 g
2.Caffeine, a stimulant in coffee has the following percent composition:
49.50 % C, 5.15% H, 28.90 % N and 16.50 % O , molar mass= 195.00g
Types of Chemical Reactions
• Redox Reactions: driven by the loss (oxidation) and gain
(reduction) of electrons. Any species that does not
change charge is called the spectator ion.
– Synthesis
– Decomposition
– Single Replacement
• Ion Exchange Reaction: driven by the formation of an
insoluble precipitate. The ions that remain dissolved
throughout are the spectator ions.
– Double Replacement
Synthesis
• Two elements combine to form a compound
• 2 Na + O2  Na2O
• Same reaction, with charges added in:
– 2 Na0 + O20  Na2+1O-2
• Na0 is oxidized (loses electrons), is the reducing agent
• O20 is reduced (gains electrons), is the oxidizing agent
• Electrons are transferred from the Na0 to the O20.
• No spectator ions, there are only two elements here.
Decomposition
• A compound breaks down into its original elements.
• Na2O  2 Na + O2
• Same reaction, with charges added in:
– Na2+1O-2  2 Na0 + O20
• O-2 is oxidized (loses electrons), is the reducing agent
• Na+1 is reduced (gains electrons), is the oxidizing agent
• Electrons are transferred from the O-2 to the Na+1.
• No spectator ions, there are only two elements here.
Single Replacement
• An element replaces the same type of element in a
compound.
• Ca + 2 KCl  CaCl2 + 2 K
• Same reaction, with charges added in:
– Ca0 + 2 K+1Cl-1  Ca+2Cl2-1 + 2 K0
• Ca0 is oxidized (loses electrons), is the reducing agent
• K+1 is reduced (gains electrons), is the oxidizing agent
• Electrons are transferred from the Ca0 to the K+1.
• Cl-1 is the spectator ion, since it’s charge doesn’t change.
Double Replacement
• The (+) ion of one compound bonds to the (-) ion of another
compound to make an insoluble precipitate. The compounds
must both be dissolved in water to break the ionic bonds first.
• NaCl (aq) + AgNO3 (aq)  NaNO3 (aq) + AgCl (s)
• The Cl-1 and Ag+1 come together to make the insoluble
precipitate, which looks like snow in the test tube.
• No species change charge, so this is not a redox reaction.
• Since the Na+1 and NO3-1 ions remain dissolved throughout
the reaction, they are the spectator ions.
• How do identify the precipitate?
Identifying the Precipitate
• The precipitate is the compound that is insoluble. AgCl is
a precipitate because Cl- is a halide. Halides are soluble,
except when combined with Ag+ and others.
Balancing Chemical Reactions
•
•
•
•
Balance one element or ion at a time
Use a pencil
Use coefficients only, never change formulas
Revise if necessary
• The coefficient multiplies everything in the formula by that
amount
– 2 Ca(NO3)2 means that you have 2 Ca, 4 N and 12 O.
• Examples for you to try!
Reactions to Balance
• ___NaCl  ___Na + ___Cl2
• ___Al + ___O2  ___Al2O3
• ___SO3  ___SO2 + ___O2
• ___Ca + ___HNO3  ___Ca(NO3)2 + ___H2
• __FeCl3 + __Pb(NO3)2  __Fe(NO3)3 + __PbCl2
Writing Net Ionic Equations
• Cancel all the spectator ions.
• Dissociate all dissociable ionic compounds
(refer to solubility rules)
• All gases and liquids NEVER dissociate.
• Write the net ionic equation.
Gases
Manometers: measure P of a gas
1. Closed-end: difference in Hg levels (Dh) shows P of
gas in container compared to a vacuum
2.
http://www.chm.davidson.edu/ChemistryApplets/GasLaws/Pressure.html
closed
172
Open-end:
• Difference in 2.
Hg levels
(Dh) shows P of gas in
container compared to Patm
173
Gas Laws
• Make a data table to put the numbers so you can
eliminate the words.
• Make sure that any Celsius temperatures are converted to
Kelvin (add 273).
• Rearrange the equation before substituting in numbers. If
you are trying to solve for T2, get it out of the denominator
first by cross-multiplying.
• If one of the variables is constant, then eliminate it.
• Try these problems!
Gas Law Problem 1
• A 2.00 L sample of N2 gas at
STP is compressed to 4.00
atm at constant temperature. What is the new
volume of the gas?
• V 2 = P 1V 1 / P 2
• = (1.00 atm)(2.00 L) / (4.00
atm)
• = 0.500 L
Gas Law Problem 2
• To what temperature must a 3.000 L sample of O2 gas at 300.0
K be heated to raise the volume to 10.00 L?
• T2 = V2T1/V1
• = (10.00 L)(300.0 K) / (3.000 L) = 1000. K
Gas Law Problem 3
• A 3.00 L sample of NH3 gas at 100.0 kPa is cooled from 500.0 K
to 300.0 K and its pressure is reduced to 80.0 kPa. What is the
new volume of the gas?
• V2 = P1V1T2 / P2T1
• = (100.0 kPa)(3.00 L)(300. K) / (80.0 kPa)(500. K)
• = 2.25 L
Gay Lussac’s Law of Combining Volumes
When measured at the same
temperature and pressure, the ratio of
the volumes of reacting gases are small
whole numbers.
N2
+
3 H2
→
2 NH3
1 volume + 3 volumes → 2 volumes
V
VNH
H 2 3 32
==
VNH 113
2
Gay Lussac’s Law of Combining Volumes
When measured at the same
temperature and pressure, the ratio of
the volumes of reacting gases are small
whole numbers.
Avogadro’s Law
Equal volumes of different gases at the
same temperature and pressure contain
the same number of molecules.
Mole-Mass-Volume Relationships
• Volume of one mole of any gas at STP =
22.4 L.
• 22.4 L at STP is known as the molar
volume of any gas.
atmospheres
nRT
nT
VV =a= nRT
PV
P
P
Determination of Density
Using the Ideal Gas Equation
• Density = mass/volume
gRT
M=
PV
D = MP/ RT
Mole fraction (X):
• Ratio of moles of one component to the total moles in
the mixture (dimensionless, similar to a %)
Ex: What are the mole fractions of H2 and He in the previous example?
X H2
184
0.60

 0.29
2.10
X He
1.50

 0.714
2.10
Collecting Gases “over Water”
• When a gas is bubbled through water, the vapor
pressure of the water (partial pressure of the water)
must be subtracted from the pressure of the collected
gas:
PT = Pgas + PH2O
∴ Pgas = PT – PH2O

See Appendix B for vapor pressures of water at
different temperatures.
185
Graham’s Law of Effusion
• The rates of effusion of gases at the same
temperature and pressure are inversely
proportional to the square roots of their molar
masses.
• Rate of effusion of A = MB
Rate of effusion of B
MA
Neutralization
• 10.0 mL of 0.20 M HCl is neutralized by 40.0 mL of NaOH.
What is the concentration of the NaOH?
•
#H MaVa = #OH MbVb, so Mb = #H MaVa / #OH Vb
• = (1)(0.20 M)(10.0 mL) / (1) (40.0 mL) = 0.050 M
• How many mL of 2.00 M H2SO4 are needed to completely
neutralize 30.0 mL of 0.500 M KOH?
Concentration
•
•
•
•
Molarity
Parts per Million
Percent by Mass
Percent by Volume
Molarity
• What is the molarity of a 500.0 mL solution of NaOH (FM =
40.0) with 60.0 g of NaOH (aq)?
– Convert g to moles and mL to L first!
– M = moles / L = 1.50 moles / 0.5000 L = 3.00 M
• How many grams of NaOH does it take to make 2.0 L of a
0.100 M solution of NaOH (aq)?
– Moles = M X L = 0.100 M X 2.0 L = 0.200 moles
– Convert moles to grams: 0.200 moles X 40.0 g/mol = 8.00 g
Parts Per Million
• 100.0 grams of water is evaporated and analyzed for lead.
0.00010 grams of lead ions are found. What is the
concentration of the lead, in parts per million?
• ppm = (0.00010 g) / (100.0 g) X 1 000 000 = 1.0 ppm
• If the legal limit for lead in the water is 3.0 ppm, then the water
sample is within the legal limits (it’s OK!)
Percent by Mass
• A 50.0 gram sample of a solution is evaporated and found
to contain 0.100 grams of sodium chloride. What is the
percent by mass of sodium chloride in the solution?
• % Comp = (0.100 g) / (50.0 g) X 100 = 0.200%
Percent By Volume
• Substitute “volume” for “mass” in the above equation.
• What is the percent by volume of hexane if 20.0 mL of
hexane are dissolved in benzene to a total volume of 80.0
mL?
• % Comp = (20.0 mL) / (80.0 mL) X100 = 25.0%
Colligative Properties
•
•
•
•
•
Vapor Pressure Lowering
B.P. Elevation DTf= m. kf (or D Tb= m. kb)
F.P. Depression
Osmotic Pressure
Colligative properties depend upon # of
particles (ions, atoms, molecule= particle)
• Which will have lowest B.P. 1M NaCl, 1 M
C6H12O6 or 1M Na3PO4?
How many Sig Figs?
• Start counting sig figs at the first non-zero.
• All digits except place-holding zeroes are sig figs.
Measurement
# of Sig Figs
Measurement
# of Sig Figs
0.115 cm
3
234 cm
3
0.00034 cm
2
67000 cm
2
0.00304 cm
3
_
45000 cm
4
0.0560 cm
3
560. cm
3
0.00070700 cm
5
560.00 cm
5
What Precision?
• A number’s precision is determined by the furthest (smallest)
place the number is recorded to.
•
•
•
•
•
•
6000 mL : thousands place
6000. mL : ones place
6000.0 mL : tenths place
5.30 mL : hundredths place
8.7 mL : tenths place
23.740 mL : thousandths place
Rounding with addition and
subtraction
• Answers are rounded to the least precise place.
1) 4.732 cm
16.8
cm
+ 0.781 cm
---------22.313 cm
22.3 cm
2)
17.440 mL
3.895 mL
+ 16.77 mL
-------------38.105 mL
38.11 mL
3)
32.0
MW
+ 0.0059 MW
--------------32.0059 MW
32.0 MW
Rounding with multiplication
and division
• Answers are rounded to the fewest number of significant
figures.
1)
37.66 KW
x 2.2 h
---------82.852 KWh
83 KWh
2)
14.922 cm
x 2.0 cm
----------2
29.844 cm
2
30. cm
3) 98.11 kg
x 200 m
---------19 622 kgm
20 000 kgm
Metric Conversions
• Determine how many powers of ten
difference there are between the two
units (no prefix = 100) and create a
conversion factor. Multiply or divide
the given by the conversion factor.
How many kg are in 38.2 cg?
(38.2 cg) /(100000 cg/kg) = 0.000382 km
How many mL in 0.988 dL?
(0.988 dg) X (100 mL/dL) = 98.8 mL
Calorimetry
• This equation can be used to determine any of the
variables here. You will not have to solve for C, since we
will always assume that the energy transfer is being
absorbed by or released by a measured quantity of water,
whose specific heat is given above.
• Solving for q
• Solving for m
• Solving for DT
Solving for q
• How many joules are absorbed by 100.0 grams of water in a
calorimeter if the temperature of the water increases from
20.0oC to 50.0oC?
• q = mCDT = (100.0 g)(4.18 J/goC)(30.0oC) = 12500 J
Solving for m
• A sample of water in a calorimeter cup increases from 25oC to
50.oC by the addition of 500.0 joules of energy. What is the
mass of water in the calorimeter cup?
• q = mCDT, so m = q / CDT = (500.0 J) / (4.18 J/goC)(25oC) = 4.8 g
Solving for DT
• If a 50.0 gram sample of water in a calorimeter cup absorbs
1000.0 joules of energy, how much will the temperature rise
by?
• q = mCDT, so DT = q / mC = (1000.0 J)/(50.0 g)(4.18 J/goC) = 4.8oC
• If the water started at 20.0oC, what will the final temperature
be?
– Since the water ABSORBS the energy, its temperature will INCREASE by
the DT: 20.0oC + 4.8oC = 24.8oC
Reaction Rate
• Reactions happen when reacting particles collide with
sufficient energy (activation energy) and at the proper angle.
• Anything that makes more collisions in a given time will make
the reaction rate increase.
– Increasing temperature
– Increasing concentration (pressure for gases)
– Increasing surface area (solids)
• Adding a catalyst makes a reaction go faster by removing
steps from the mechanism and lowering the activation energy
without getting used up in the process.
Heat of Reaction
• Reactions either absorb PE (endothermic, +DH) or release PE
(exothermic, -DH)
Exothermic, PEKE, Temp
Endothermic, KEPE, Temp
Rewriting the equation with heat included:
4 Al(s) + 3 O2(g)  2 Al2O3(s) + 3351 kJ
N2(g) + O2(g) +182.6 kJ  2 NO(g)
5.3: Enthalpy, H
• Since most reactions occur in containers open to the air,
w is often negligible. If a reaction produces a gas, the gas
must do work to expand against the atmosphere. This
mechanical work of expansion is called PV (pressurevolume) work.
• Enthalpy (H): change in the heat content (qp) of a
reaction at constant pressure
H = E + PV
DH = DE + PDV
(at constant P)
DH = (qp + w) + (-w)
DH = qp
205
• Sign conventions
DH > 0
Heat is gained from surroundings
+ DH in endothermic reaction
DH < 0
Heat is released to surroundings
- DH in exothermic reaction
206
5.4: Enthalpy of Reaction (DHrxn)
•
Also called heat of reaction:
1. Enthalpy is an extensive property (depends on
amounts of reactants involved).
Ex: CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (l)
DHrxn = - 890. kJ
– Combustion of 1 mol CH4 produces 890. kJ
… of 2 mol CH4 → (2)(-890. kJ) = -1780 kJ
What is the DH of the combustion of 100. g CH4?
207
2. DHreaction = - DHreverse reaction
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (l)
DH = - 890. kJ
CO2 (g) + 2 H2O (l)  CH4 (g) + 2 O2 (g)
DH = +890. kJ
208
5.6: Hess’ Law
•
If a rxn is carried out in a series of steps,
DHrxn =  (DHsteps) = DH1 + DH2 + DH3 + …
Germain Hess
(1802-1850)
Ex. What is DHrxn of the combustion of propane?
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
3 C (s) + 4 H2 (g)  C3 H8 (g)
C3H8 (g)  3 C (s) + 4 H2 (g)
DH1 = -103.85 kJ
DH1 = +103.85 kJ
3[ C (s) + O2 (g)  CO2 (g)
]
DH2 = -393.5
3( kJ
)
4[ H2 (g) + ½ O2 (g)  H2O (l)
DH] 3 = -285.8
4( kJ
)
DHrxn = 103.85 + 3(- 393.5) + 4(- 285.8) = - 2219.8 kJ
209
5.7: Enthalpy of Formation (DHf)
• Formation: a reaction that describes a substance
formed from its elements
Ex: 2 N2 (g) + 4 H2 (g) + 3 O2 (g)  2
NH4NO3 (s)
• Standard enthalpy of formation (DHf): forms 1 mole of
compound from its elements in their standard state (at
298 K)
2 C (graphite) + 3 H2 (g) + ½ O2 (g) 
C2H5OH (l)
DHf = - 277.7 kJ
– DHf of the most stable form of any element equals
zero.
H2, N2 , O2 , F2 , Cl2 (g)
Br2 (l), Hg (l)
C (graphite), P4 (s, white), S8 (s), I2 (s)
210
Hess’ Law (again)
Ex. Combustion of propane:
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Given:
Compound
C3H8 (g)
CO2 (g)
H2O (l)
H2O (g)
DHrxn (kJ/mol)
-103.85
-393.5
-285.8
-241.82
DHrxn = [3(- 393.5) + 4(- 285.8)] – [1(-103.85) + 5(0)]
= - 2219.8 kJ
211
5.5: Calorimetry
• Measurement of heat flow
• Heat capacity, C: amount of heat required to raise T of an
object by 1 K
q = C DT
• Specific heat (or specific heat capacity, c): heat capacity
of 1 g of a substance
q = m c DT
Ex: How much energy is required to heat 40.0 g of iron (c
= 0.45 J/(g K) from 0.0ºC to 100.0ºC?
q = m c DT = (40.0 g)(0.45 J/(g K))(100.0 – 0.0 ºC)
= 1800 J
212
Potential Energy Diagrams
• Steps of a reactions:
– Reactants have a certain amount of PE stored in their
bonds (Heat of Reactants)
– The reactants are given enough energy to collide and
react (Activation Energy)
– The resulting intermediate has the highest energy that
the reaction can make (Heat of Activated Complex)
– The activated complex breaks down and forms the
products, which have a certain amount of PE stored in
their bonds (Heat of Products)
– Hproducts - Hreactants = DH
EXAMPLES
Making a PE Diagram
• X axis: Reaction Coordinate (time, no units)
• Y axis: PE (kJ)
• Three lines representing energy (Hreactants, Hactivated complex,
Hproducts)
• Two arrows representing energy changes:
– From Hreactants to Hactivated complex: Activation Energy
– From Hreactants to Hproducts : DH
• ENDOTHERMIC PE DIAGRAM
• EXOTHERMIC PE DIAGRAM
Endothermic PE Diagram
If a catalyst is added?
Endothermic with Catalyst
The red line represents the catalyzed reaction.
Exothermic PE Diagram
What does it look like with a catalyst?
Exothermic with a Catalyst
The red line represents the catalyzed reaction. Lower A.E. and faster
reaction time!
19.1: Spontaneous Processes
• Reversible reaction: can proceed forward and backward
along same path (equilibrium is possible)
Ex: H2O freezing & melting at 0ºC
• Irreversible reaction: cannot proceed forward and
backward along same path
Ex: ice melting at room temperature
Spontaneous reaction: an irreversible reaction that
occurs without outside intervention
Ex: Gases expand to fill a container, ice melts at room
temperature (even though endothermic), salts
dissolve in water
219
Entropy
Entropy (S): a measure of molecular randomness or
disorder
– S is a state function: DS = Sfinal - Sinitial
+ DS = more randomness
- DS = less randomness
–
For a reversible process that occurs at constant
T:
q rev
DSsystem 
T
–
Units: J/K
220
Examples of spontaneous reactions:
Particles are more evenly distributed
• Gases expand to fill a container:
Particles are no longer in an ordered crystal
•lattice
Ice melts at room temperature:
Ions are not locked in crystal lattice
• Salts dissolve in water:
221
19.3: 3rd Law of Thermodynamics
•
The entropy of a crystalline solid at 0 K is 0.
How to predict DS:
• Sgas > Sliquid > Ssolid
• Smore gas molecules > Sfewer gas molecules
• Shigh T > Slow T
Ex: Predict the sign of DS for the following:
+, solid to gas
1. CaCO3 (s) → CaO (s) + CO2 (g)
2. N2 (g) + 3 H2 (g) → 2 NH3 (g)
-, fewer moles produced
3. N2 (g) + O2 (g) → 2 NO (g)
?
222
19.5: Gibbs free energy, G
• Represents combination of two
forces that drive a reaction:
Josiah Willard Gibbs
DH (enthalpy) and DS (disorder)
(1839-1903)
• Units: kJ/mol
DG = DH - TDS
DG° = DH° - TDS°
(absolute T)



DG“free
 energy”
nGf products
- DGmG
Called
because
represents
f reactants
maximum useful work that can be done by the
system on its surroundings in a spontaneous
reaction. (See p. 708 for more details.)
223
Determining Spontaneity of a Reaction
If DG is: reaction is spontaneous (proceeds in
the forward direction
• Positive
Forward reaction is non-spontaneous; the
reverse reaction is spontaneous
• Zero
The system is at equilibrium
224
19.6: Free Energy & Temperature
• DG depends on enthalpy, entropy, and
temperature:
DG = DH - TDS
DH
-
DS
+
+
-
-
-
+
+
DG and reaction outcome
Always (-); spontaneous at all T
2 O3 (g) → 3 O2 (g)
Always +; non-spontaneous at all T
3 O2 (g) → 2 O3 (g)
Spontaneous at low T; non-spontaneous at high T
H2O (l) → H2O (s)
Spontaneous only at high T ; non-spontaneous at
low T
H2O (s) → H2O (l)
225
Solubility Curves
• Solubility: the maximum quantity of solute that can be
dissolved in a given quantity of solvent at a given temperature
to make a saturated solution.
• Saturated: a solution containing the maximum quantity of
solute that the solvent can hold. The limit of solubility.
• Supersaturated: the solution is holding more than it can
theoretically hold OR there is excess solute which precipitates
out. True supersaturation is rare.
• Unsaturated: There are still solvent molecules available to
dissolve more solute, so more can dissolve.
• How ionic solutes dissolve in water: polar water molecules
attach to the ions and tear them off the crystal.
Solubility
Solubility: go to the temperature and up to
the desired line, then across to the Y-axis.
This is how many g of solute are needed to
make a saturated solution of that solute in
100g of H2O at that particular temperature.
At 40oC, the solubility of KNO3 in 100g of
water is 64 g. In 200g of water, double that
amount. In 50g of water, cut it in half.
Supersaturated
If 120 g of NaNO3 are added to 100g of
water at 30oC:
1) The solution would be
SUPERSATURATED, because there is more
solute dissolved than the solubility allows
2) The extra 25g would precipitate out
3) If you heated the solution up by 24oC (to
54oC), the excess solute would dissolve.
Unsaturated
If 80 g of KNO3 are added to 100g of water
at 60oC:
1) The solution would be UNSATURATED,
because there is less solute dissolved than
the solubility allows
2) 26g more can be added to make a
saturated solution
3) If you cooled the solution down by 12oC
(to 48oC), the solution would become
saturated
How Ionic Solutes Dissolve in Water
Water solvent molecules attach to the
ions (H end to the Cl-, O end to the Na+)
Water solvent holds the ions apart and keeps
the ions from coming back together
Formulas, Naming and
Properties of Acids
• Arrhenius Definition of Acids: molecules that dissolve in water
to produce H3O+ (hydronium) as the only positively charged
ion in solution.
• HCl (g) + H2O (l)  H3O+ (aq) + Cl• Properties of Acids
• Naming of Acids
• Formula Writing of Acids
Properties of Acids
• Acids react with metals above H2 on Table J to
form H2(g) and a salt.
• Acids have a pH of less than 7.
• Dilute solutions of acids taste sour.
• Acids turn phenolphthalein CLEAR, litmus RED
and bromthymol blue YELLOW.
• Acids neutralize bases.
• Acids are formed when acid anhydrides (NO2,
SO2, CO2) react with water for form acids. This is
how acid rain forms from auto and industrial
emissions.
Naming of Acids
• Binary Acids (H+ and a nonmetal)
– hydro (nonmetal) -ide + ic acid
• HCl (aq) = hydrochloric acid
• Ternary Acids (H+ and a polyatomic ion)
– (polyatomic ion) -ate +ic acid
• HNO3 (aq) = nitric acid
– (polyatomic ion) -ide +ic acid
• HCN (aq) = cyanic acid
– (polyatomic ion) -ite +ous acid
• HNO2 (aq) = nitrous acid
Formula Writing of Acids
• Acids formulas get written like any other. Write the H+1
first, then figure out what the negative ion is based on the
name. Cancel out the charges to write the formula. Don’t
forget the (aq) after it…it’s only an acid if it’s in water!
• Hydrosulfuric acid: H+1 and S-2 = H2S (aq)
• Carbonic acid: H+1 and CO3-2 = H2CO3 (aq)
• Chlorous acid: H+1 and ClO2-1 = HClO2 (aq)
• Hydrobromic acid: H+1 and Br-1 = HBr (aq)
• Hydronitric acid:
• Hypochlorous acid:
• Perchloric acid:
Formulas, Naming and
Properties of Bases
• Arrhenius Definition of Bases: ionic compounds that dissolve
in water to produce OH- (hydroxide) as the only negatively
charged ion in solution.
• NaOH (s)  Na+1 (aq) + OH-1 (aq)
• Properties of Bases
• Naming of Bases
• Formula Writing of Bases
Properties of Bases
• Bases react with fats to form soap and glycerol. This process
is called saponification.
• Bases have a pH of more than 7.
• Dilute solutions of bases taste bitter.
• Bases turn phenolphthalein PINK, litmus BLUE and
bromthymol blue BLUE.
• Bases neutralize acids.
• Bases are formed when alkali metals or alkaline earth metals
react with water. The words “alkali” and “alkaline” mean
“basic”, as opposed to “acidic”.
Naming of Bases
• Bases are named like any ionic
compound, the name of the metal
ion first (with a Roman numeral if
necessary) followed by “hydroxide”.
Fe(OH)2 (aq) = iron (II) hydroxide
Fe(OH)3 (aq) = iron (III) hydroxide
Al(OH)3 (aq) = aluminum hydroxide
NH3 (aq) is the same thing as NH4OH:
NH3 + H2O  NH4OH
Also called ammonium hydroxide.
Formula Writing of Bases
• Formula writing of bases is the same as for any ionic formula
writing. The charges of the ions have to cancel out.
• Calcium hydroxide = Ca+2 and OH-1 = Ca(OH)2 (aq)
• Potassium hydroxide = K+1 and OH-1 = KOH (aq)
• Lead (II) hydroxide = Pb+2 and OH-1 = Pb(OH)2 (aq)
• Lead (IV) hydroxide = Pb+4 and OH-1 = Pb(OH)4 (aq)
• Lithium hydroxide =
• Copper (II) hydroxide =
• Magnesium hydroxide =
Neutralization
• H+1 + OH-1  HOH
• Acid + Base  Water + Salt (double replacement)
• HCl (aq) + NaOH (aq)  HOH (l) + NaCl (aq)
• H2SO4 (aq) + KOH (aq)  2 HOH (l) + K2SO4 (aq)
• HBr (aq) + LiOH (aq) 
• H2CrO4 (aq) + NaOH (aq) 
• HNO3 (aq) + Ca(OH)2 (aq) 
• H3PO4 (aq) + Mg(OH)2 (aq) 
pH
• A change of 1 in pH is a tenfold increase in acid or base
strength.
• A pH of 4 is 10 times more acidic than a pH of 5.
• A pH of 12 is 100 times more basic than a pH of 10.
16.2: Dissociation of Water
• Autoionization of water:
H2O (l) ↔ H+ (aq) + OH- (aq)


[ H ][OH ]
Kc 
[ H 2O ]


-14
 [ H ][OH ]  1.0 x 10  K w
KW = ion-product constant for water

H3O+ (aq) or H+ (aq) = hydronium
1000 g 1 mol

 55.6 M
[H 2 O (l)] 
1 L 18.0 g
241
Indicators
At a pH of 2:
Methyl Orange = red
Bromthymol Blue = yellow
Phenolphthalein = colorless
Litmus = red
Bromcresol Green = yellow
Thymol Blue = yellow
Methyl orange is red at a pH of
3.2 and below and yellow at a pH
of 4.4 and higher. In between the
two numbers, it is an
intermediate color that is not
listed on this table.
Alternate Theories
• Arrhenius Theory: acids and bases must be in aqueous
solution.
• Alternate Theory: Not necessarily so!
– Acid: proton (H+1) donor…gives up H+1 in a reaction.
– Base: proton (H+1) acceptor…gains H+1 in a reaction.
• HNO3 + H2O  H3O+1 + NO3-1
– Since HNO3 lost an H+1 during the reaction, it is an acid.
– Since H2O gained the H+1 that HNO3 lost, it is a base.
16.11: Lewis Acids & Bases
• Lewis acid: “e- pair acceptor”
– Brønsted-Lowry acid = H+ donor
– Arrhenius acid = produces H+
• Lewis base: “e- pair donor”
– B-L base = H+ acceptor
– Arrhenius base = produces OH-
Ex:
NH3 +
BF3
Lewis base Lewis acid
6 CN- +
Fe3+
Lewis base Lewis acid
Gilbert N. Lewis
(1875 – 1946)
→
NH3BF3
Lewis salt
→
Fe(CN)63Coordination compound
244
15.1: Chemical Equilibrium
• Occurs when opposing reactions are proceeding
at the same rate
– Forward rate = reverse rate of reaction
Ex:
• Vapor pressure: rate of vaporization = rate of condensation
• Saturated solution: rate of dissociation = rate of
crystallization
• Expressing concentrations:
– Gases: partial pressures, PX
– Solutes in liquids: molarity, [X]
245
Reversible Reactions and Rate
Reaction Rate
Forward rate
Equilibrium is established:
Forward rate = Backward rate
Backward rate
Time
When equilibrium is achieved:
[A] ≠ [B]
and
kf/kr = Keq
246
•
15.2:
Law
of
Mass
Action
Derived from rate laws by Guldberg and
Waage (1864)
– For a balanced chemical reaction
in equilibrium:
aA+bB↔cC+dD
– Equilibrium constant expression (Keq):
c
c
d
[C] [D]
Kc 
a
b
[A] [B]
Cato Guldberg Peter Waage
(1836-1902) (1833-1900)
or
d
(PC ) (PD )
Kp 
a
b
(PA ) (PB )
Keq is strictly based on stoichiometry of the reaction (is independent of the mechanism).
Units: Keq is considered dimensionless (no units)
247
Relating Kc and Kp
• Convert [A] into PA:
n
P
M 
V RT
PA  [ A]RT
(PC ) c (PD ) d ([C]RT) c ([D]RT) d [C]c [D]d (RT) cd
Kp 

a
b 
a
b
a
b
a b
(PA ) (PB )
([A]RT) ([B]RT)
[A] [B] (RT)
K p  K c (RT)
(c d) - (a  b)
 K c (RT)
Dn
where Dn =
= change in coefficents of products – reactants (gases only!)
= (c+d) - (a+b)
248
Magnitude of Keq
• Since Keq a [products]/[reactants], the magnitude of Keq
predicts which reaction direction is favored:
– If Keq > 1
then [products] > [reactants]
and equilibrium “lies to the right”
– If Keq < 1
then [products] < [reactants]
and equilibrium “lies to the left”
249
Relationship Between Q and K
• Reaction Quotient (Q): The particular ratio of
concentration terms that we write for a particular
reaction is called reaction quotient.
• For a reaction, A B, Q= [B]/[A]
• At equilibrium, Q= K
Reaction Direction: Comparing Q and K
• Q<K, reaction proceeds to right, until equilibrium
is achieved (or Q=K)
• Q>K, reaction proceeds to left, until Q=K
250
Value of K
For the
For the
For the
reference
reverse rxn, reaction,
rxn, A>B, B >A,
2A > 2B
For the rxn,
A > C
C > B
K= K(ref)2
K (overall)=
K1 X K2
K(ref)=
[B]/[A]
K= 1/K(ref)
251
15.3: Types of Equilibria
• Homogeneous: all components in same phase
1 g or aq)3
2
(usually
N2 (g) + H2 (g)c ↔ NH
d 3 (g)
(PC ) (PD )
KP 
a
b
(PA ) (PB )
KP 
(PNH3 )
1
Fritz Haber
(1868 – 1934)
2
(PN 2 ) (PH 2 )
3
252
• Heterogeneous: different phases
CaCO3 (s) ↔ CaO (s) + CO2 (g)
Definition:
What we use:
K eq 
[CaO] (PCO 2 )
[CaCO3 ]
K p  PCO 2

Concentrations of pure solids and pure liquids are not included in Keq expression
because their concentrations do not vary, and are “already included” in Keq (see p. 548).

Even though the concentrations of the solids or liquids do not appear in the
equilibrium expression, the substances must be present to achieve equilibrium.
253
15.4: Calculating Equilibrium Constants
Steps to use “ICE” table:
1. “I” = Tabulate known initial and equilibrium
concentrations of all species in equilibrium
expression
2. “C” = Determine the concentration change for
the species where initial and equilibrium are
known
• Use stoichiometry to calculate
concentration changes for all other species
involved in equilibrium
3. “E” = Calculate the equilibrium concentrations
254
• Ex: Enough ammonia is dissolved in 5.00 L of water
at 25ºC to produce a solution that is 0.0124 M
ammonia. The solution is then allowed to come to
equilibrium. Analysis of the equilibrium mixture
shows that [OH1-] is 4.64 x 10-4 M. Calculate Keq at
25ºC for the reaction:
NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1- (aq)
255
NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1- (aq)
NH3 (aq)
Initial
0.0124 M
Change
-x
Equilibrium
0.0119 M
H2O
(l)
X
X
X
NH41+ (aq)
OH1- (aq)
0M
0M
+x
4.64 x 10-4 M
+x
4.64 x 10-4 M
x = 4.64 x 10-4 M
1
4
[NH ][OH1- ] (4.64 10-4 ) 2
Kc 

 1.81x 10-5
[NH3 ]
0.0119
256
Equilibrium
When the rate of the forward reaction equals the rate of the reverse reaction.
(c) 2006, Mark Rosengarten
Examples of Equilibrium
• Solution Equilibrium: when a solution is saturated, the rate of
dissolving equals the rate of precipitating.
– NaCl (s)  Na+1 (aq) + Cl-1 (aq)
• Vapor-Liquid Equilibrium: when a liquid is trapped with air in
a container, the liquid evaporates until the rate of evaporation
equals the rate of condensation.
– H2O (l)  H2O (g)
• Phase equilibrium: At the melting point, the rate of solid
turning to liquid equals the rate of liquid turning back to solid.
– H2O (s)  H2O (l)
Le Châtelier’s Principle
• If a system at equilibrium is stressed, the equilibrium will
shift in a direction that relieves that stress.
• A stress is a factor that affects reaction rate. Since catalysts
affect both reaction rates equally, catalysts have no effect on a
system already at equilibrium.
• Equilibrium will shift AWAY from what is added
• Equilibrium will shift TOWARDS what is removed.
• This is because the shift will even out the change in reaction
rate and bring the system back to equilibrium
»NEXT
Steps to Relieving Stress
• 1) Equilibrium is subjected to a STRESS.
• 2) System SHIFTS towards what is removed from the system or
away from what is added.
• The shift results in a CHANGE OF CONCENTRATION for both
the products and the reactants.
– If the shift is towards the products, the concentration of
the products will increase and the concentration of the
reactants will decrease.
– If the shift is towards the reactants, the concentration of
the reactants will increase and the concentration of the
products will decrease.
» NEXT
Examples
• For the reaction N2(g) + 3H2(g)  2 NH3(g) + heat
– Adding N2 will cause the equilibrium to shift RIGHT, resulting in an
increase in the concentration of NH3 and a decrease in the
concentration of N2 and H2.
– Removing H2 will cause a shift to the LEFT, resulting in a decrease
in the concentration of NH3 and an increase in the concentration
of N2 and H2.
– Increasing the temperature will cause a shift to the LEFT, same
results as the one above.
– Decreasing the pressure will cause a shift to the LEFT, because
there is more gas on the left side, and making more gas will bring
the pressure back up to its equilibrium amount.
– Adding a catalyst will have no effect, so no shift will happen.
•
•
•
•
Oxidation Numbers
Rules for Assigning Oxidation States
The oxidation state of an atom in an uncombined element is 0.
The oxidation state of a monatomic ion is the same as its charge.
Oxygen is assigned an oxidation state of –2 in most of its covalent
compounds. Important exception: peroxides (compounds
containing the O2 2- group), in which each oxygen is assigned an
oxidation state of –1)
• In its covalent compounds with nonmetals, hydrogen is assigned
an oxidation state of +1
• For a compound, sum total of ON s is zero.
• For an ionic species (like a polyatomic ion), the sum of the
oxidation states must equal the overall charge on that ion.
16.6: Weak Acids
• Weak acids partially ionize in water
(equilibrium is somewhere between ions and
molecules). HA (aq) ↔ A- (aq) + H+ (aq)


[ H ][ A ]
Ka 
[ HA]eq
Ka = acid-dissociation constant in water
 Weak acids generally have Ka < 10-3
 See Appendix D for full listing of Ka values
263
Dissociation Constants
• For a generalized acid dissociation,
A- (aq) + H3O+ (aq)
HA (aq) + H2O (l)
the equilibrium expression would be
Kc =
[H3O+] [A-]
[HA]
• This equilibrium constant is called the aciddissociation constant, Ka.
© 2009, Prentice-Hall, Inc.
Dissociation Constants
The greater the value of Ka, the stronger is the acid.
© 2009, Prentice-Hall, Inc.
Calculating Ka from the pH
The pH of a 0.10 M solution of formic acid, HCOOH,
at 25C is 2.38. Calculate Ka for formic acid at this
temperature.
We know that
Ka =
[H3O+] [COO-]
[HCOOH]
© 2009, Prentice-Hall, Inc.
Calculating Ka from the pH
The pH of a 0.10 M solution of formic acid, HCOOH,
at 25C is 2.38. Calculate Ka for formic acid at this
temperature.
To calculate Ka, we need the equilibrium concentrations
of all three things.
We can find [H3O+], which is the same as [HCOO-], from
the pH.
© 2009, Prentice-Hall, Inc.
Calculating Ka from the pH
pH = -log [H3O+]
2.38 = -log [H3O+]
-2.38 = log [H3O+]
10-2.38 = 10log [H3O+] = [H3O+]
4.2  10-3 = [H3O+] = [HCOO-]
© 2009, Prentice-Hall, Inc.
Calculating Ka from pH
Now we can set up a table…
[HCOOH], M
[H3O+], M
[HCOO-], M
Initially
0.10
0
0
Change
- 4.2  10-3
+ 4.2  10-3
+ 4.2  10-3
0.10 - 4.2  10-3
= 0.0958 = 0.10
4.2  10-3
4.2  10-3
At Equilibrium
© 2009, Prentice-Hall, Inc.
Calculating Ka from pH
Ka =
[4.2  10-3] [4.2  10-3]
[0.10]
= 1.8  10-4
© 2009, Prentice-Hall, Inc.
Calculating Percent Ionization
[H3O+]eq

100
[HA]initial
• Percent Ionization =
• In this example
[H3O+]eq = 4.2  10-3 M
[HCOOH]initial = 0.10 M
4.2  10-3
Percent Ionization =
 100
0.10
= 4.2%
© 2009, Prentice-Hall, Inc.
Calculating pH from Ka
Calculate the pH of a 0.30 M solution of acetic
acid, HC2H3O2, at 25C.
HC2H3O2 (aq) + H2O (l)
H3O+ (aq) + C2H3O2- (aq)
Ka for acetic acid at 25C is 1.8  10-5.
© 2009, Prentice-Hall, Inc.
Calculating pH from Ka
The equilibrium constant expression is
Ka =
[H3O+] [C2H3O2-]
[HC2H3O2]
© 2009, Prentice-Hall, Inc.
Calculating pH from Ka
We next set up a table…
[C2H3O2], M
[H3O+], M
[C2H3O2-], M
Initially
0.30
0
0
Change
-x
+x
+x
0.30 - x  0.30
x
x
At Equilibrium
We are assuming that x will be very small compared to 0.30 and can,
therefore, be ignored.
© 2009, Prentice-Hall, Inc.
Calculating pH from Ka
Now,
1.8  10-5 =
(x)2
(0.30)
(1.8  10-5) (0.30) = x2
5.4  10-6 = x2
2.3  10-3 = x
© 2009, Prentice-Hall, Inc.
Calculating pH from Ka
pH = -log [H3O+]
pH = -log (2.3  10-3)
pH = 2.64
© 2009, Prentice-Hall, Inc.
Polyprotic Acids…
…have more than one acidic proton
If the difference between the Ka for the first
dissociation and subsequent Ka values is 103 or
more, the pH generally depends only on the first
dissociation.
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Weak Bases
Bases react with water to produce hydroxide ion.
© 2009, Prentice-Hall, Inc.
Weak Bases
The equilibrium constant expression for this
reaction is
[HB] [OH-]
Kb =
[B-]
where Kb is the base-dissociation constant.
© 2009, Prentice-Hall, Inc.
Weak Bases
Kb can be used to find [OH-] and, through it, pH.
© 2009, Prentice-Hall, Inc.
pH of Basic Solutions
What is the pH of a 0.15 M solution of NH3?
NH3 (aq) + H2O (l)
Kb =
NH4+ (aq) + OH- (aq)
[NH4+] [OH-]
= 1.8  10-5
[NH3]
© 2009, Prentice-Hall, Inc.
pH of Basic Solutions
Tabulate the data.
Initially
At Equilibrium
[NH3], M
[NH4+], M
[OH-], M
0.15
0
0
0.15 - x  0.15
x
x
© 2009, Prentice-Hall, Inc.
pH of Basic Solutions
1.8  10-5 =
(x)2
(0.15)
(1.8  10-5) (0.15) = x2
2.7  10-6 = x2
1.6  10-3 = x2
© 2009, Prentice-Hall, Inc.
pH of Basic Solutions
Therefore,
[OH-] = 1.6  10-3 M
pOH = -log (1.6  10-3)
pOH = 2.80
pH = 14.00 - 2.80
pH = 11.20
© 2009, Prentice-Hall, Inc.
Ka and Kb
Ka and Kb are related in this way:
Ka  Kb = Kw
Therefore, if you know one of them, you can
calculate the other.
© 2009, Prentice-Hall, Inc.
Factors Affecting Acid Strength
• The more polar the H-X bond and/or the weaker the HX bond, the more acidic the compound.
• So acidity increases from left to right across a row and
from top to bottom down a group.
© 2009, Prentice-Hall, Inc.
Factors Affecting Acid Strength
In oxyacids, in which an
-OH is bonded to
another atom, Y, the
more electronegative Y
is, the more acidic the
acid.
© 2009, Prentice-Hall, Inc.
Factors Affecting Acid Strength
For a series of oxyacids, acidity increases with the
number of oxygens.
© 2009, Prentice-Hall, Inc.
Factors Affecting Acid Strength
Resonance in the conjugate bases of carboxylic
acids stabilizes the base and makes the conjugate
acid more acidic.
© 2009, Prentice-Hall, Inc.
16.9: Salt Solutions as Acids & Bases
• Hydrolysis: acid/base reaction of ion with
water to produce H+ or OH– Anion (A-) = a conjugate base
A- (aq) + H2O (l) ↔ HA (aq) + OH- (aq)
– Cation (B+) = a conjugate acid
B+ (aq) + H2O (l) ↔ BOH (aq) + H+ (aq)
290
17.1: Common Ion Effect
• Addition of a “common ion”: solubility of
solids decrease because of Le Châtelier’s
principle.
Ex:
AgCl (s) ↔ Ag+ (aq) + Cl- (aq)
– Addition of Cl- shifts equilibrium toward solid
291
17.4:
Solubility
Equilibria
• Dissolving & precipitating of salts
– Solubility rules discussed earlier are generalized
qualitative observations of quantitative experiments.
Ex:
PbCl2 (s) ↔ Pb2+ (aq) + 2 Cl- (aq)
Ksp = [Pb2+][Cl-]2 = 1.6 x 10-5
Ksp = solubility-product constant (found in App. D)
– Recall that both aqueous ions and solid must be present
in solution to achieve equilibrium
– Changes in pH will affect the solubility of salts
composed of a weak acid or weak base ion.
292
Solubility Products
Consider the equilibrium that exists in a
saturated solution of BaSO4 in water:
BaSO4(s)
Ba2+(aq) + SO42−(aq)
© 2009, Prentice-Hall, Inc.
Solubility Products
The equilibrium constant expression for this
equilibrium is
Ksp = [Ba2+] [SO42−]
where the equilibrium constant, Ksp, is called the
solubility product.
© 2009, Prentice-Hall, Inc.
Solubility Products
• Ksp is not the same as solubility.
• Solubility is generally expressed as the mass of
solute dissolved in 1 L (g/L) or 100 mL (g/mL) of
solution, or in mol/L (M).
© 2009, Prentice-Hall, Inc.
Calculating Ksp from solubility
1. Calculate Ksp for Ag2CrO4, if its solubility is 0.022 g/L.
(Ans: 6.6 X 10^-5)
Calculating Solubility given Ksp
2. Ksp for MgF2 is 6.4 X 10^-9 at 250C. Calculate its solubility in
mol/L and g/L. (Ans: 1.2 X 10^-3 M, 7.3 X 10^-2 g/L)
Factors Affecting Solubility
• The Common-Ion Effect
– If one of the ions in a solution equilibrium is
already dissolved in the solution, the
equilibrium will shift to the left and the
solubility of the salt will decrease.
BaSO4(s)
Ba2+(aq) + SO42−(aq)
© 2009, Prentice-Hall, Inc.
Factors Affecting Solubility
• pH
– If a substance has a
basic anion, it will be
more soluble in an
acidic solution.
– Substances with acidic
cations are more
soluble in basic
solutions.
© 2009, Prentice-Hall, Inc.
Factors Affecting Solubility
• Amphoterism
– Amphoteric metal oxides
and hydroxides are
soluble in strong acid or
base, because they can
act either as acids or
bases.
– Examples of such cations
are Al3+, Zn2+, and Sn2+.
© 2009, Prentice-Hall, Inc.
•
17.2:
Buffers:
Solutions that resist drastic changes in pH upon
additions of small amounts of acid or base.
– Consist of a weak acid and its conjugate base
(usually in salt form)
Ex: acetic acid and sodium acetate:
HC2H3O2 + NaC2H3O2
– Or consist of a weak base and its conjugate acid
(usually in salt form)
Ex: ammonia and ammonium chloride:
NH3 + NH4Cl
301
Buffers
• Buffers are solutions of
a weak conjugate acidbase pair.
• They are particularly
resistant to pH
changes, even when
strong acid or base is
added.
© 2009, Prentice-Hall, Inc.
Buffers
If a small amount of hydroxide is added to an equimolar
solution of HF in NaF, for example, the HF reacts with the
OH− to make F− and water.
© 2009, Prentice-Hall, Inc.
Buffers
Similarly, if acid is added, the F− reacts with it to form HF
and water.
© 2009, Prentice-Hall, Inc.
Titration
In this technique a
known concentration of
base (or acid) is slowly
added to a solution of
acid (or base).
© 2009, Prentice-Hall, Inc.
Titration
A pH meter or
indicators are used to
determine when the
solution has reached
the equivalence point,
at which the
stoichiometric amount
of acid equals that of
base.
© 2009, Prentice-Hall, Inc.
Titration of a Strong Acid with a Strong
Base
From the start of the
titration to near the
equivalence point, the
pH goes up slowly.
© 2009, Prentice-Hall, Inc.
Titration of a Strong Acid with a Strong
Base
Just before (and after)
the equivalence point,
the pH increases
rapidly.
© 2009, Prentice-Hall, Inc.
Titration of a Strong Acid with a Strong
Base
At the equivalence
point, moles acid =
moles base, and the
solution contains only
water and the salt
from the cation of the
base and the anion of
the acid.
© 2009, Prentice-Hall, Inc.
Titration of a Strong Acid with a Strong
Base
As more base is
added, the increase in
pH again levels off.
© 2009, Prentice-Hall, Inc.
Titration of a Weak Acid with a Strong Base
• Unlike in the previous
case, the conjugate base
of the acid affects the
pH when it is formed.
• At the equivalence point
the pH is >7.
• Phenolphthalein is
commonly used as an
indicator in these
titrations.
© 2009, Prentice-Hall, Inc.
Practice Problem on Titration:
If 7.3 mL of 1.25 M HNO3 is required to neutralize 25.00 mL
of a potassium
hydroxide solution, what is the molarity of the potassium
hydroxide?
0.044 M KOH
Titration of a Weak Base and Strong Acid
14
pH 7
0
0
30
Volume of HCl added (mL)
- Half Equivalence Point , pH= pKa
- pka or pkb of weak acid or base in a buffer should be clsoe to
the desired pH of the buffer solution.
313
Redox:
Reduction occurs when an atom gains one or more
electrons.
Ex:
Oxidation occurs when an atom or ion loses one or
more electrons.
Ex:
LEO goes GER
Copper metal reacts with silver nitrate to form silver
metal and copper nitrate:
Cu + 2 Ag(NO3)  2 Ag + Cu(NO3)2.
Identifying OX, RD, SI
Species
• Ca0 + 2 H+1Cl-1  Ca+2Cl-12 + H20
• Oxidation = loss of electrons. The species becomes more
positive in charge. For example, Ca0  Ca+2, so Ca0 is the
species that is oxidized.
• Reduction = gain of electrons. The species becomes more
negative in charge. For example, H+1  H0, so the H+1 is
the species that is reduced.
• Spectator Ion = no change in charge. The species does not
gain or lose any electrons. For example, Cl-1  Cl-1, so the
Cl-1 is the spectator ion.
Oxidizing Agent and Reducing Agent:
Oxidizing agent gets reduced itself and reducing
agent gets oxidized itself, so a strong oxidizing agent
should have a great tendency to accept e and a strong
reducing agent should be willing to lose e easily. What
are strong oxidizing agents- metals or non metals?
Why?
Which is the strongest oxidizing agent and which is the
strongest reducing agent?
Agents
• Ca0 + 2 H+1Cl-1  Ca+2Cl-12 + H20
• Since Ca0 is being oxidized and H+1 is being reduced, the
electrons must be going from the Ca0 to the H+1.
• Since Ca0 would not lose electrons (be oxidized) if H+1 weren’t
there to gain them, H+1 is the cause, or agent, of Ca0’s
oxidation. H+1 is the oxidizing agent.
• Since H+1 would not gain electrons (be reduced) if Ca0 weren’t
there to lose them, Ca0 is the cause, or agent, of H+1’s
reduction. Ca0 is the reducing agent.
Steps for Balancing a Redox Reaction:
Half Reaction Method
In half reaction method, oxidation and reduction halfreactions are written and balanced separately before
combining them into a balanced redox reaction. It is
a good method for balancing redox reactions
because this method can be used both for reactions
carried out in acidic and basic medium .
Steps for Balancing Redox Reaction Using
Half Reaction Method IN ACIDIC MEDIUM:
Step 1: Write unbalanced equation in ionic form.
Step 2: Write separate half reactions for the oxidation and
reduction processes. (Use Oxidation Numbers for identifying
oxidation and reduction reactions)
Step 3: Balance atoms in the half reactions
•First, balance all atoms except H and O
•Balance O by adding H2O
•Balance H by adding H+
Step 4: Balance Charges on each half reaction, by adding
electrons.
Step 5: Multiply each half reaction by an appropriate number to
make the number of electrons equal in both half reactions.
Step 6: Add two half reactions and simplify where possible by
canceling species appearing in both sides.
Step 7: Check equation for same number of atoms and charges on
both sides.
Writing Half-Reactions
• Ca0 + 2 H+1Cl-1  Ca+2Cl-12 + H20
• Oxidation: Ca0  Ca+2 + 2e• Reduction: 2H+1 + 2e-  H20
The two electrons lost by Ca0
are gained by the two H+1
(each H+1 picks up an
electron).
PRACTICE SOME!
Practice Half-Reactions
• Don’t forget to determine the charge of each species first!
• 4 Li + O2  2 Li2O
• Oxidation Half-Reaction:
• Reduction Half-Reaction:
• Zn + Na2SO4  ZnSO4 + 2 Na
• Oxidation Half-Reaction:
• Reduction Half-Reaction:
Steps for Balancing Redox Reaction Using
Half Reaction Method IN BASIC MEDIUM:
For balancing redox reactions in basic solutions, all the
steps are the same as acidic medium balancing,
except you add one more step to it. The H+ ions can
then be “neutralized” by adding an equal number of
OH- ions to both sides of the equation. Ex.
Standard Cell Potential
Just as the water tends to flow from a higher level to a lower
level, electrons also move from a higher “potential” to a
lower potential. This potential difference is called the
electromotive force (EMF) of cell and is written as Ecell.
The standard for measuring the cell potentials is called a
SHE (Standard Hydrogen Electrode).
Description of SHE (Standard Hydrogen
Electrode)
Reaction 2H+(aq, 1M)+ 2e - H2(g, 101kPa) E0= 0.00 V
Standard Reduction Potentials
Many different half cells can be paired with the
SHE and the standard reduction potentials for
each half cell is obtained. Check the table for
values of reduction potential for various
substances:
Would substances with high reduction potential
be strong oxidizing agents or strong reducing
agents? Why?
Activity Series
• For metals, the higher up the chart the
element is, the more likely it is to be
oxidized. This is because metals like to
lose electrons, and the more active a
metallic element is, the more easily it can
lose them.
• For nonmetals, the higher up the chart the
element is, the more likely it is to be
reduced. This is because nonmetals like to
gain electrons, and the more active a
nonmetallic element is, the more easily it
can gain them.
Metal Activity
3 K0 + Fe+3Cl-13
REACTION
Fe0 + 3 K+1Cl-1
NO REACTION
• Metallic elements start out with a charge
of ZERO, so they can only be oxidized to
form (+) ions.
• The higher of two metals MUST undergo
oxidation in the reaction, or no reaction
will happen.
• The reaction 3 K + FeCl3  3 KCl + Fe
WILL happen, because K is being
oxidized, and that is what Table J says
should happen.
• The reaction Fe + 3 KCl  FeCl3 + 3 K will
NOT happen.
Voltaic Cells (Galvanic Cells)
A voltaic cell converts chemical energy from a
spontaneous redox reaction into electrical energy.
Ex: Cu and Zn voltaic cell (More positive reduction
potential is the cathode)
Key Words:
•Cathode
•Anode
•Salt Bridge
How a Voltaic Cell Works: An Ox, Red Cat
Representing Electrochemical Cells
Voltaic Cells
• Produce electrical current using a spontaneous redox reaction
• Used to make batteries!
• Materials needed: two beakers, piece of the metals (anode, electrode and cathode + electrode), solution of each metal,
porous material (salt bridge), solution of a salt that does not
contain either metal in the reaction, wire and a load to make
use of the generated current!
• Use Reference Table J to determine the metals to use
– Higher = (-) anode (lower reduction potential)
– Lower = (+) cathode (higher reduction potential)
Making Voltaic Cells
Electrolytic Cells
• Use electricity to force a nonspontaneous redox reaction
to take place.
• Uses for Electrolytic Cells:
– Decomposition of Alkali Metal Compounds
– Decomposition of Water into Hydrogen and Oxygen
– Electroplating
• Differences between Voltaic and Electrolytic Cells:
– ANODE:
Voltaic (-)
Electrolytic (+)
– CATHODE:
Voltaic (+)
Electrolytic (-)
– Voltaic: 2 half-cells, a salt bridge and a load
– Electrolytic: 1 cell, no salt bridge, IS the load
Decomposing Alkali
Metal Compounds
2 NaCl  2 Na + Cl2
The Na+1 is reduced at the (-)
cathode, picking up an efrom the battery
The Cl-1 is oxidized at the (+)
anode, the e- being pulled off
by the battery (DC)
Decomposing Water
2 H2O  2 H2 + O2
The H+ is reduced at the (-)
cathode, yielding H2 (g), which
is trapped in the tube.
The O-2 is oxidized at the (+)
anode, yielding O2 (g), which is
trapped in the tube.
Electroplating
The Ag0 is oxidized to Ag+1 when the
(+) end of the battery strips its
electrons off.
The Ag+1 migrates through the
solution towards the (-) charged
cathode (ring), where it picks up an
electron from the battery and forms
Ag0, which coats on to the ring.
Spontaneity of Redox Reactions:
E0 = E0red( reduction process-cathode) – E0red (oxidation process-anode)
A positive value of E0 indicates a spontaneous process
and a negative value of E0 indicates a nonspontaneous
value.
Steps for Predicting Spontaneity of Redox Reactions
•First write the reaction as oxidation and reduction half
reactions.
•Then plug standard reduction potential values in the
equation given above.
•Check for the spontaneity by a positive or a negative value
of E0
Ex:
Hydrocarbons
• Molecules made of Hydrogen and Carbon
• Carbon forms four bonds, hydrogen forms one bond
• Hydrocarbons come in three different homologous series:
– Alkanes (single bond between C’s, saturated)
– Alkenes (1 double bond between 2 C’s, unsaturated)
– Alkynes (1 triple bond between 2 C’s, unsaturated)
• These are called aliphatic, or open-chain, hydrocarbons.
• Count the number of carbons and add the appropriate suffix!
Alkanes
•
•
•
•
•
CH4 = methane
C2H6 = ethane
C3H8 = propane
C4H10 = butane
C5H12 = pentane
• To find the number of hydrogens,
double the number of carbons and add
2.
Methane
Meth-: one carbon
-ane: alkane
The simplest organic
molecule, also known as
natural gas!
Ethane
Eth-: two carbons
-ane: alkane
Propane
Prop-: three carbons
-ane: alkane
Also known as “cylinder gas”, usually stored under pressure and used for gas grills
and stoves. It’s also very handy as a fuel for Bunsen burners!
Butane
But-: four carbons
-ane: alkane
Liquefies with moderate pressure, useful for gas lighters. You have probably lit
your gas grill with a grill lighter fueled with butane!
Pentane
Pent-: five carbons
-ane: alkane
Your Turn!!!
Draw Hexane:
Draw Heptane:
Alkenes
•
•
•
•
C2H4 = Ethene
C3H6 = Propene
C4H8 = Butene
C5H10 = Pentene
• To find the number of hydrogens,
double the number of carbons.
Ethene
Two carbons, double bonded. Notice
how each carbon has four bonds? Two
to the other carbon and two to
hydrogen atoms.
Also called “ethylene”, is used for the production of polyethylene, which is an
extensively used plastic. Look for the “PE”, “HDPE” (#2 recycling) or “LDPE” (#4
recycling) on your plastic bags and containers!
Propene
Three carbons, two of them double
bonded. Notice how each carbon has
four bonds?
If you flipped this molecule so that the double bond was on the right side of the
molecule instead of the left, it would still be the same molecule. This is true of all
alkenes.
Used to make polypropylene (PP, recycling #5), used for dishwasher safe
containers and indoor/outdoor carpeting!
Butene
This is 1-butene, because the double bond is
between the 1st and 2nd carbon from the end.
The number 1 represents the lowest numbered
carbon the double bond is touching.
This is 2-butene. The double bond is between
the 2nd and 3rd carbon from the end. Always
count from the end the double bond is closest
to.
ISOMERS: Molecules that share the same molecular formula, but have different
structural formulas.
Pentene
This is 1-pentene. The double bond is on the first
carbon from the end.
This is 2-pentene. The double bond is on the
second carbon from the end.
This is not another isomer of pentene. This is also
2-pentene, just that the double bond is closer to
the right end.
Alkynes
4 C2H2 = Ethyne
4 C3H4 = Propyne
4 C4H6 = Butyne
4 C5H8 = Pentyne
4
To find the number of hydrogens, double the number
of carbons and subtract 2.
Ethyne
Now, try to draw propyne! Any isomers? Let’s see!
Also known as “acetylene”, used by miners by dripping water on CaC2 to light
up mining helmets. The “carbide lamps” were attached to miner’s helmets
by a clip and had a large reflective mirror that magnified the acetylene
flame.
Used for welding and cutting applications, as ethyne burns at temperatures
over 3000oC!
Propyne
This is propyne! Nope! No isomers.
OK, now draw butyne. If there are any isomers, draw them too.
Butyne
Well, here’s 1-butyne!
And here’s 2-butyne!
Is there a 3-butyne? Nope! That would be 1-butyne. With four carbons, the
double bond can only be between the 1st and 2nd carbon, or between the 2nd
and 3rd carbons.
Now, try pentyne!
Pentyne
1-pentyne
2-pentyne
Now, draw all of the possible isomers for hexyne!
Naming: Check
this link out
http://www.chem
.ucalgary.ca/cours
es/351/orgnom/ja
vanom/nomenclat
ure2StartPage.ht
m
Isomers
Isomers are compounds that have same molecular formula (same
number of atoms) but a different structure.
There are three types of isomers:
1. Structural Isomers: Same number of atoms, arranged
differently.
2. Geometric Isomers (Cis- trans-): Happens in = or triple bonded
compounds since these are inflexible bonds. Ex.
3. Optical Isomers ( D- and L-): Need a central atom that is
“Chiral” (all four groups attached to it are different). These are
non super imposable mirror images. Usually this central atom
is C.
Substituted Hydrocarbons
• Hydrocarbon chains can have three kinds of “dinglydanglies” attached to the chain. If the dingly-dangly is
made of anything other than hydrogen and carbon, the
molecule ceases to be a hydrocarbon and becomes
another type of organic molecule.
– Alkyl groups
– Halide groups
– Other functional groups
• To name a hydrocarbon with an attached group,
determine which carbon (use lowest possible number
value) the group is attached to. Use di- for 2 groups, trifor three.
Alkyl Groups
Halide Groups
Organic Families
• Each family has a functional group to identify it.
– Alcohol (R-OH, hydroxyl group)
– Organic Acid (R-COOH, primary carboxyl group)
– Aldehyde (R-CHO, primary carbonyl group)
– Ketone (R1-CO-R2, secondary carbonyl group)
– Ether (R1-O-R2)
– Ester (R1-COO-R2, carboxyl group in the middle)
– Amine (R-NH2, amine group)
– Amide (R-CONH2, amide group)
• These molecules are alkanes with functional groups attached.
The name is based on the alkane name.
Alcohol
On to DI and TRIHYDROXY ALCOHOLS
Di and Trihydroxy Alcohols
Positioning of
Functional Group
PRIMARY (1o): the functional group is bonded to a
carbon that is on the end of the chain.
SECONDARY (2o): The functional group is bonded to a
carbon in the middle of the chain.
TERTIARY (3o): The functional group is bonded to a
carbon that is itself directly bonded to three other
carbons.
Organic Acid
These are weak acids. The H on the right side is the one that ionized in water
to form H3O+. The -COOH (carboxyl) functional group is always on a PRIMARY
carbon.
Can be formed from the oxidation of primary alcohols using a KMnO4 catalyst.
Aldehyde
Aldehydes have the CO (carbonyl) groups ALWAYS on a PRIMARY carbon. This is
the only structural difference between aldehydes and ketones.
Formed by the oxidation of primary alcohols with a catalyst. Propanal is formed
from the oxidation of 1-propanol using pyridinium chlorochromate (PCC)
catalyst.*
Ketone
Ketones have the CO (carbonyl) groups ALWAYS on a SECONDARY carbon. This is
the only structural difference between ketones and aldehydes.
Can be formed from the dehydration of secondary alcohols with a catalyst.
Propanone is formed from the oxidation of 2-propanol using KMnO4 or PCC
catalyst.*
Ether
Ethers are made of two alkyl groups surrounding one oxygen atom. The ether is
named for the alkyl groups on “ether” side of the oxygen. If a three-carbon alkyl
group and a four-carbon alkyl group are on either side, the name would be propyl
butyl ether. Made with an etherfication reaction.
Ester
Esters are named for the alcohol and organic acid that reacted by esterification
to form the ester. If the alcohol was 1-propanol and the acid was hexanoic acid,
the name of the ester would be propyl hexanoate. Esters contain a COO
(carboxyl) group in the middle of the molecule, which differentiates them from
organic acids.
Amine
- Component of amino acids, and therefore proteins, RNA and DNA…life itself!
- Essentially ammonia (NH3) with the hydrogens replaced by one or more
hydrocarbon chains, hence the name “amine”!
Amide
Synthetic Polyamides: nylon, kevlar
Natural Polyamide: silk!
For more information on polymers, go here.
Organic Reactions
•
•
•
•
•
Combustion
Fermentation
Substitution
Addition
Dehydration Synthesis
– Etherification
– Esterification
• Saponification
• Polymerization
Combustion
• Happens when an organic molecule reacts with oxygen
gas to form carbon dioxide and water vapor. Also known
as “burning”.
Substitution
• Alkane + Halogen  Alkyl Halide + Hydrogen Halide
• The halogen atoms substitute for any of the hydrogen
atoms in the alkane. This happens one atom at a time.
The halide generally replaces an H on the end of the
molecule.
C2H6 + Cl2  C2H5Cl + HCl
The second Cl can then substitute for another H:
C2H5Cl + HCl  C2H4Cl2 + H2
Addition
• Alkene + Halogen  Alkyl Halide
• The double bond is broken, and the halogen adds at either
side of where the double bond was. One isomer possible.
(c) 2006, Mark Rosengarten
Etherification*
• Alcohol + Alcohol  Ether + Water
• A dehydrating agent (H2SO4) removes H from one alcohol’s
OH and removes the OH from the other. The two
molecules join where there H and OH were removed.
Note: dimethyl ether and diethyl ether are also produced from this reaction, but
can be separated out.
Esterification
• Organic Acid + Alcohol  Ester + Water
• A dehydrating agent (H2SO4) removes H from the organic acid and
removes the OH from the alcohol. The two molecules join where
there H and OH were removed.
Saponification
The process of making soap from glycerol esters (fats).
Glycerol ester + 3 NaOH  soap + glycerol
Glyceryl stearate + 3 NaOH  sodium stearate + glycerol
The sodium stearate is the soap! It emulsifies grease…surrounds globules with its
nonpolar ends, creating micelles with - charge that water can then wash away.
Hard water replaces Na+ with Ca+2 and/or other low solubility ions, which forms a
precipitate called “soap scum”.
Water softeners remove these hardening ions from your tap water, allowing the
soap to dissolve normally.
Polymerization
• A polymer is a very long-chain molecule made up of many
monomers (unit molecules) joined together.
• The polymer is named for the monomer that made it.
– Polystyrene is made of styrene monomer
– Polybutadiene is made of butadiene monomer
• Addition Polymers
• Condensation Polymers
• Rubber
Addition Polymers
Joining monomers together by breaking double bonds
Polyvinyl chloride (PVC): vinyl siding, PVC pipes, etc.
Vinyl chloride
polyvinyl chloride

n C2H3Cl
-(-C2H3Cl-)-n
Polytetrafluoroethene (PTFE, teflon):
TFE
n C2 F4
PTFE

-(-C2F4-)-n
Condensation Polymers
Condensation polymerization is just dehydration synthesis, except instead of
making one molecule of ether or ester, you make a monster molecule of
polyether or polyester.
Rubber
The process of toughing rubber by cross-linking the polymer strands with sulfur
is called...
THE END
(c) 2006, Mark Rosengarten