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101 chem
Instructor: Dr.
Course# and Name: Chem -101, General Chemistry
Semester Credit Hours: 2+1
Email Address:
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GENERAL CHEMISTRY (CHEM 101)
Recommended Books
1- Chemistry, The central Science 11E or 12E
BROWN, LEMAY, BURSTEN, MURPHY,
WOODWARD
Pearson International Edition
2
2- Chemistry: The Molecular Nature of
Matter, 6E
Jespersen/Brady/Hyslop
Wiley Edition
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Chp. 1
Atomic Structure and
Periodic Table
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Complete Symbols of Element
Contain the symbol of the element, the
mass number and the atomic number.
4
X Mass
number
Atomic
number Subscript →
Superscript →
Symbols
a) number of protons
b)number of neutrons
c) number of electrons
d)Atomic number
e) Mass Number
Br 80
35
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Periodic Table
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The horizontal rows of the periodic table are
called PERIODS.
For a certain element, the number of occupied
shells indicated the number of period
The Periodic Table of Element
8
The vertical columns of the periodic table are called
GROUPS, or FAMILIES. For a certain element, the number of outer most
shell’s electrons indicated the number of group
The elements in any group of the periodic table have
similar physical and chemical properties!
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Periodic Table
Group 3 – 7 A
Group 1 - 2A
For a certain element, the number of occupied shells
indicated the number of period
For a certain element, the number of outer most shell’s
electrons indicated the number of group
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Na
2, 8, 1 or 1s2, 2s2, 2p6, 3s1
Na in Group 1 and period 3
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Cl
2, 8, 7 or 1s2, 2s2, 2p6, 3s2, 3p5
Cl in Group 7 and period 3
31
Ga
2, 8, 18, 3 or 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p1
Ga in Group 3 and period 4
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General Properties of Periodic Table
The periodic trends of the following properties will
be studied here.
Atomic radius
Metallic and Non-metallic character
Ionization (energy) potential
Electron affinity
Electronegativity
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Atomic Radius
It is the distance from the centre of the
nucleus to the outermost shell of an atom.
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Trends in atomic radius
down a group
It is fairly obvious that the
atoms get bigger as you go
down groups. Due to adding
extra layers of electrons.
Trends in atomic radius
across periods
atomic radius decreases with
increase in atomic number
(in a period left to right. Due
to Electrons are pulled close
to the nucleus by the
increased Zeff
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Atomic Radius
Metallic character is the tendency of an element to
lose electrons and form positive ions (cations). For
e.g., alkali metals are the most electropositive
elements.
It is also known as electropositivity.
Metallic and Non-metallic Character
The tendency of an element to accept electrons to
form an anion is called its non-metallic or
electronegative character. For e.g., chlorine,
oxygen and phosphorous show greater
electronegative or non-metallic character."
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Trends across the period
Metallic character of elements decreases as we move
to the right. So, the elements at the left have metallic
character while those to the right have a non-metallic
character.
Trends across the group
As we move down the group the number of shells
increases. This causes the atomic size increases.
The electrons of the outermost shell experience less
nuclear attraction and so can lose electrons easily thus
showing increased metallic character.
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The amount of energy required to remove the most loosely
bound electron from an isolated gaseous atom is called
ionization energy (IE).
It is measured in the units of electron volts (eV) per atom
or kilo joules per mole of atoms (kJ mol-1).
Ionization Energy (IE)
Variation along a period
The ionization energy increases with increasing atomic
number in a period.
Variation down a group
The ionization energy gradually decreases in moving from
top to bottom in a group.
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Electron Affinity (EA) Electron affinity is the amount of energy released when
an electron is added to an isolated gaseous atom.
Electron affinity is the ability of an atom to hold an additional electron.
If the atom has more tendency to accept an electron then the energy
released will be large and consequently the electron affinity will be
high. Just like a strong magnet
Variation along a period The size of an atom decreases thus, Hence the electron affinity
increases in a period from left to right
Variation down a group As we move down a group the atomic size and nuclear size increases.
Consequently the electron affinity decreases.
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Electronegativity This the relative tendency of an atom in a molecule to
attract a shared pair of electrons towards itself.
Variation along a period
As the nuclear charge increases from going left to right in a
period because the electrons enter the same shell, the
shielding is less effective. Thus the increased nuclear charge
attract the shared pair of electrons more strongly resulting in
higher electronegativity from going left to right in a period.
Variation down the group
Electronegativity decreases down the group because the
atomic size increases. The larger the size of the atom the
lesser the tendency to attract the shared pair of electrons.
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Periodic Properties of the elements
Electron affinity
Ionization energy
Atomic radius
Atomic properties and the periodic table—a summary
Atomic Properties
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Chp. 2
Chemical Bonding
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Chemical Bonds
Attractive forces that hold atoms together in complex
substances
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•The tendency of an atom to take part in chemical combination is
determined by the number of valence electrons (electrons in the
outermost shell of an atom).
•The atoms acquire the stable noble gas configuration of having eight
electrons in the outermost shell (called octet rule) by mutual sharing or
by transfer of one or more electrons.
•The valency (number of electrons an atom loses, gains or mutually
shares to attain noble gas configuration) of an element is either equal to
the number of valence electrons or equal to 8 minus the number of
valence electrons.
The type of chemical bond developed between the two combining
atoms depends upon the way these atoms acquire a stable noble gas
configuration
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Octet Rule = atoms tend to gain, lose or share electrons so as to
have 8 electrons
C would like to
N would like to
O would like to
Gain 4 electrons
Gain 3 electrons
Gain 2 electrons 24
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Lewis electron dot symbols
The valence electrons are then written as
dots or (small cross marks) around the
symbol of the atom. They are spread in a
pair on four sides of the symbol.
In case of ions the charge is shown
with the symbol.
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Lewis Symbols • Electron bookkeeping method
• Way to keep track of e–’s
• Write chemical symbol surrounded by dots for each e–
H
Li
Na
Be
Mg
B
Al
C
Si
He
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Lewis Symbols
N O F Ne
P S Cl Ar
For the representative elements
Group # = # valence e–’s
He
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Give the dot structure for the following atoms:
1) Na 2) K 3) Al 4) B 5) N 6) P
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Learning Check
A. X would be the electron dot formula for
1) Na 2) K 3) Al
B. X would be the electron dot formula for
1) B 2) N 3) P
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Why are electrons important?
Elements have different electron configurations
– different electron configurations mean different levels
of bonding
– Chemical bonds: an attempt to fill electron shells
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1. Ionic Bonds When the chemical bond occurs by the complete transfer of
electron(s) from the atom(s) of metal to atoms of nonmetal is
called Ionic (electrovalent) bond.
2. Covalent Bonds When the shared electrons are contributed by the two
combining atoms (nonmetal with nonmetal), the bond formed
is called Covalent bond.
3. Metallic Bonds Metal atoms bonded to several other metal atoms
Types of Chemical Bonds
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Ionic Bond
• Between atoms of metals and nonmetals with very
different electronegativity
• Bond formed by transfer of electrons
• Produce charged ions all states.
• Conductors in a liquid state only and have high
melting point.
• Examples; NaCl, CaCl2, K2O
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Formation of Ions from Metals
Ionic compounds result when metals react with
nonmetals
Metals lose electrons to match the number of valence
electrons of their nearest noble gas
Positive ions form when the number of electrons are
less than the number of protons
Group 1 metals ion 1+
Group 2 metals ion 2+
• Group 13 metals ion 3+
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Formation of Sodium Ion
Sodium atom Sodium ion
Na – e Na +
2-8-1 2-8 ( = Ne)
11 p+ 11 p+
11 e- 10 e-
0 1+
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Formation of Magnesium Ion
Magnesium atom Magnesium ion
Mg – 2e Mg2+
2-8-2 2-8 (=Ne)
12 p+ 12 p+
12 e- 10 e-
0 2+
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Some Typical Ions with Positive
Charges (Cations)
Group 1 Group 2 Group 13
H+ Mg2+ Al3+
Li+ Ca2+
Na+ Sr2+
K+ Ba2+
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Learning Check
A. Number of valence electrons in aluminum
1) 1 e- 2) 2 e- 3) 3 e-
B. Change in electrons for octet
1) lose 3e- 2) gain 3 e- 3) gain 5 e-
C. Ionic charge of aluminum
1) 3- 2) 5- 3) 3+
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Solution
A. Number of valence electrons in aluminum
3) 3 e-
B. Change in electrons for octet
1) lose 3e-
C. Ionic charge of aluminum
3) 3+
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Learning Check
Give the ionic charge for each of the following:
A. 12 p+ and 10 e-
1) 0 2) 2+ 3) 2-
B. 50p+ and 46 e-
1) 2+ 2) 4+ 3) 4-
C. 15 p+ and 18e-
2) 3+ 2) 3- 3) 5-
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Ions from Non-metal Ions
• In ionic compounds, nonmetals in groups
5, 6, and 7 gain electrons from metals
• Nonmetal add electrons to achieve the
octet arrangement
• Nonmetal ionic charge: 3-, 2-, or 1-
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Fluoride Ion
unpaired electron octet
1 -
: F + e : F :
2-7 2-8 (= Ne)
9 p+ 9 p+
9 e- 10 e-
0 1 -
ionic charge
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Ions of Representative Elements Can use periodic table to predict ion charges
When we use North American numbering of groups:
Cation positive charge = group #
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Ionic Bonding
Na Cl The metal (sodium) tends to lose its one electron
from the outer level.
The nonmetal (chlorine) needs to gain one more
to fill its outer level, and will accept the one
electron that sodium is going to lose.
Formation of Sodium Chloride
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Ionic Bonding
Na+ Cl -
Note: Remember that NO DOTS are
now shown for the cation!
Formation of Sodium Chloride
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1). Ionic bond – electron from Na is transferred to
Cl, this causes a charge imbalance in each atom.
The Na becomes (Na+) and the Cl becomes (Cl-),
charged particles or ions. 46
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Formation of Magnesium fluoride
The Ionic Bond
[Ne] 3s2 1s22s22p5 1s22s22p6
[Ne] [Ne]
Mg2+ F - + F Mg F -
1s22s22p6 [Ne] 3s0
[Ne]
Mg2+ + 2e- Mg
2e- + F F - 2 2
MgF2 + 2F Mg
2
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Determining Ionic Formulas
“Cross-cross” rule
– Make magnitude of charge on one ion into subscript
for other
– When doing this, make sure that subscripts are
reduced to lowest whole number.
Ex. What is the formula of ionic compound
formed between aluminum and oxygen ions?
Al2O3 Al3+ O2–
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Determining Ionic Formulas
Ex. Formula of ionic compound formed when
magnesium reacts with oxygen
– Mg is group 2A
• Forms +2 ion or Mg2+
– O is group 6A
• Forms –2 ion or O2–
– To get electrically neutral particle need
• 1:1 ratio of Mg2+ and O2–
– Formula: MgO
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Your Turn! Which of the following is the correct formula for
the formula unit composed of potassium and
oxygen ions?
A.KO
B.KO2
C.K2O
D.P2O3
E.K2O2
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Your Turn! Which of the following is the correct formula for
the formula unit composed of Fe3+ and sulfide ions?
A.FeS
B.Fe3S2
C.FeS3
D.Fe2S3
E.Fe4S6
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Covalent Bond • Between nonmetallic elements of similar
electronegativity.
• Formed by sharing electron pairs
• Stable non-ionizing particles, they are not
conductors at any state
• Examples; O2, CO2, C2H6, H2O, SiC
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A covalent bond is a chemical bond in which two
or more electrons are shared by two atoms.
Why should two atoms share electrons?
F F +
7e- 7e-
F F
8e- 8e-
F F
F F
Lewis structure of F2
lone pairs lone pairs
lone pairs lone pairs
single covalent bond
single covalent bond
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8e-
H H O + + O H H O H H or
2e- 2e-
Lewis structure of water
Double bond – two atoms share two pairs of electrons
single covalent bonds
O C O or O C O
8e- 8e- 8e- double bonds
double bonds
Triple bond – two atoms share three pairs of electrons
N N
8e- 8e-
N N
triple bond triple bond
or
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Electronegativity • The ability of atoms in a molecule to attract
electrons to itself.
• On the periodic chart, electronegativity increases
as you go…
– …from left to right across a row.
– …from the bottom to the top of a column.
.
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Types of covalent bonds
When two atoms bond their DIFFERENCE in
electronegativity determines the covalent bond
type.
1- Polar Covalent Bonds
2- Non Polar Covalent Bonds
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Polar Covalent Bonds 1- Occurs between metals and non metals
2-The greater the difference in electronegativity,
3- Electrons are unequallly shared
4- The more polar is the bond.
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Occurs between non metals
When electrons are shared equally
Has almost no electronegativity difference (0.0 to 0.4).
Examples:
Electronegativity
Atoms Difference Type of Bond
N-N 3.0 - 3.0 = 0.0 Nonpolar covalent
Cl-Br 3.0 - 2.8 = 0.2 Nonpolar covalent
H-Si 2.1 - 1.8 = 0.3 Nonpolar covalent
Non-polar Covalent Bonds
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Covalent bonds; Two atoms share one or more pairs of outer-
shell electrons. Oxygen Atom Oxygen Atom
Oxygen Molecule (O2)
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Your Turn! Which species is most likely covalently bonded?
A. CsCl
B. NaF
C. CaF2
D. CO
E. MgBr2
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Lewis Structures
Diatomic Gases:
• H and Halogens
H2
• H· + ·H H:H or HH
• Each H has 2 e–’s through sharing
• Can write shared pair of e–’s as line ()
• : =
• covalent bond
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Lewis Structures Diatomic Gases:
F2
• Each F has complete octet
• Only need to form one bond to complete octet
• Pairs of e–’s not included in covalent bond are
called Lone Pairs
• Same for rest of Halogens: Cl2, Br2, I2
F F+ F F F F
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Lewis Structures
Diatomic Gases:
HF
• Same for HCl, HBr, HI.
• Molecules are diatomics as need only 1e– to
complete octet
• Separate molecules
– Gas in most cases because very weak intermolecular
forces
H F+ H F H F
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Your Turn! How many electrons are required to complete the
octet around nitrogen, when it forms N2 ?
A. 2
B. 3
C. 1
D. 4
E. 6
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Lewis Structures • Many nonmetals form more than 1 covalent bond
C
H
CH H
H H
CH H
H
NH H
H
NH H
H
O H
H
O H
H
ON
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Double Bonds
• 2 pairs of e–’s shared between 2 atoms
Ex. CO2
Triple bond
• 3 pairs of e–’s shared between 2 atoms
Ex. N2
O OC C OO C OO
N N N N N N
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Your Turn! Which species is most likely to have multiple bonds
?
A. CO
B. H2O
C. PH3
D. BF3
E. CH4
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Metallic Bond
• Formed between atoms of metallic elements
• Electron cloud around atoms
• Good conductors at all states, lustrous, different
melting points
• Examples; Na, Fe, Al, Au, Co
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Intermolecular
Attractions and the
Properties of Liquids
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Intermolecular Forces (IMF)
• Important differences between gases,
solids, and liquids:
– Gases
• Expand to fill their container
– Liquids
• Retain volume, but not shape
– Solids
• Retain volume and shape
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Inter vs. Intra-Molecular Forces • Intramolecular forces
– Covalent bonds within molecule
– Strong
• Intermolecular forces
– Attraction forces between molecules
– Weak
Cl H Cl H
Covalent Bond (strong) Intermolecular attraction (weak)
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Intermolecular Forces
The attractions between molecules are not
nearly as strong as the intramolecular
attractions that hold compounds together.
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Intermolecular Forces
Intermolecular forces are attractive forces between molecules.
Intramolecular forces hold atoms together in a molecule.
Intermolecular vs Intramolecular
Generally, intermolecular forces are
much weaker than intramolecular forces.
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Types of intermolecular Forces
• Dipole-dipole forces
• Hydrogen bonding
• London dispersion forces
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Dipole-Dipole Forces
1- Attractive forces between polar molecules
Orientation of Polar Molecules in a Solid
2- Electrostatic interaction between the oppositely charged regions
of polar molecule and negative charged regions in another
molecules (dipoles).
Exampes: HCl, CO
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Ex: Which of the following molecules have dipole
interactions?
A) F2
B) CH4
C) H2O
D) CH3Cl
E) NH3
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Hydrogen bonds
What are they?
A special case of permanent dipole-dipole
interactions
They are stronger than dipole –dipole interaction.
Molecules with hydrogen bonds have higher boiling
points than molecules that don’t.
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Hydrogen bonds
What do you need?
A hydrogen atom covalently bonded to an
electronegative atom … N, O or F.
If only one of these conditions is met, you don’t
get hydrogen bonding.
A lone pair of electrons on the electronegative
atom.
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Hydrogen bonds
Give me an example!
This does not have any hydrogen bonds. Carbon is
not very electronegative, and it has no lone pairs of
electrons in methane.
methane, CH4 …
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Hydrogen bonds
Give me a real example!
This does have hydrogen bonds.
Nitrogen is very electronegative, and it has one
lone pair of electrons in ammonia.
ammonia, NH3 …
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Hydrogen bonds
Give me another example!
This has not one, but two hydrogen bonds.
Oxygen is very electronegative, and it has two
lone pairs of electrons in water.
water, H2O …
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Examples Hydrogen bonds
Remember, you need:
A hydrogen atom covalently bonded to an
electronegative atom … N, O or F.
If only one of these conditions is met,
you don’t get hydrogen bonding.
A lone pair of electrons on the electronegative
atom.
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Ex: Which of the following molecules can have
hydrogen bonding?
A) F2
B) CH4
C) H2O
D) CH3Cl
E) NH3
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London Dispersion Forces 1-They are the weakest kind of intermolecular attraction
2- occur between non polar molecules molecules.
3-They are thought to be caused by the motion of electrons.
This is because they are a temporary attractive force that results when the electrons in two adjacent atoms occupy a position that make the atoms form temporary dipoles.
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Dispersion Forces
• Dispersion Forces are also known as
• London forces and van
der Waals forces.
• They were named the London forces in honor of the German physicist; Fritz London who studied these forces!!
Johannes van Der Waals (1837-
1923)
Fritz London (1900-1954)
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Factors Affecting dispersion or
London Forces
The strength of dispersion forces tends to increase
with increased
1-molecular weight.
2- Larger atoms have larger electron clouds, which
are easier to polarize.
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S
Ex. What type(s) of intermolecular forces exist between
each of the following molecules?
HBr HBr is a polar molecule: dipole-dipole forces. There are also
dispersion forces between HBr molecules.
CH4
CH4 is nonpolar: dispersion forces.
SO2
SO2 is a polar molecule: dipole-dipole forces. There are also
dispersion forces between SO2 molecules.
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Intermolecular Forces and the Properties
of Liquids
• In summary, intermolecular forces play a large
role in many of the physical properties of liquids
and gases. These include:
• Vapor pressure
• Boiling point
• Surface tension
• Viscosity
• Solubility
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Table 12.1 Boiling Points of Halogens and Noble Gases
Larger molecules have stronger London forces and thus
higher boiling points.
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Vapor Pressure
Vapor pressures of liquids decreases with
decreases the intermolecular forces
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Propane, C3H8
BP –42.1oC
Hexane, C6H14
BP 68.7oC
More sites (marked with *) along its chain where attraction
to other molecules can occur
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Boiling Point
Your Turn!
Which species has a higher boiling point, Cl2 or
HCl; F2 or HF ?
A. HCl; F2
B. Cl2; F2
C. HCl; HF
D. Cl2 ; HF
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Surface Tension • Liquids containing molecules
with strong intermolecular
forces have high surface
tension
– Allows us to fill glass above rim
• Gives surface rounded appearance
• Surface acts as “skin” that lets
water pile up
• Surface resists expansion and
pushes back
Surface Tension
as IMF
Surface Tension
as IMF
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Viscosity • Resistance of a liquid
to flow is called
viscosity.
• It is related to the
ease with which
molecules can move
past each other.
• Viscosity increases
with stronger
intermolecular forces
and decreases with
higher temperature.
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Effect of Intermolecular Forces on Viscosity
Acetone
• Polar molecule
– Dipole-dipole and
– London forces
Ethylene glycol
• Polar molecule
– Hydrogen-bonding
– Dipole-dipole and
– London forces
Which is more viscous??
98
Your Turn! For each pair given, which is more viscose ?
• CH3CH2CH2CH2OH, CH3CH2CH2CHO;
• C6H14, C12H26; NH3(l ), PH3(l )
A. CH3CH2CH2CH2OH; C6H14; NH3(l )
B. CH3CH2CH2CH2OH; C12H26; NH3(l )
C. CH3CH2CH2CHO; C6H14; PH3(l )
D. CH3CH2CH2CHO; C12H26; NH3(l )
E. CH3CH2CH2CH2OH; C12H26; PH3(l )
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Solubility • “Like dissolves like”
–To dissolve polar substance, use polar
solvent
–To dissolve nonpolar substance, use
nonpolar solvent
• Compare relative polarity of two substances
–Similar polarity means greater ability to
interact with each other
–Differing polarity means that they don’t
interact; move past each other 100
Factors Affecting Solubility
• Chemists use the axiom “like dissolves like."
– Polar substances tend to dissolve in polar
solvents.
– Nonpolar substances tend to dissolve in
nonpolar solvents.
101
Factors Affecting Solubility
The more similar the intermolecular
attractions, the more likely one substance
is to be soluble in another.
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Factors Affecting Solubility
Glucose (which has hydrogen bonding) is very soluble in water, while cyclohexane
(which only has dispersion forces) is not.
103
Factors Affecting Solubility
• Vitamin A is soluble in nonpolar compounds
(like fats).
• Vitamin C is soluble in water.
104
Temperature
Generally, the
solubility of solid
solutes in liquid
solvents
increases with
increasing
temperature.
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Temperature
• The opposite is true
of gases.
– Carbonated soft
drinks are more
“bubbly” if stored
in the refrigerator.
– Warm lakes have
less O2 dissolved
in them than cool
lakes.
106
Your Turn!
Ex: Which of the following are not expected
to be soluble in water?
A.HF
B.CH4
C.CH3OH
D.All are soluble
107
Chp. 3
The Mole
and Stoichiometry
Chemistry: The Molecular Nature of Matter, 6E
Jespersen/Brady/Hyslop
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The Mole • Number of atoms in exactly 12 grams of 12C
atoms
How many atoms in 1 mole of 12C ?
– Based on experimental evidence
1 mole of 12C = 6.022 × 1023 atoms = 12.011 g
Avogadro’s number = NA
– Number of atoms, molecules or particles in one
mole
• 1 mole of X = 6.022 × 1023 units of X
• 1 mole Xe = 6.022×1023 Xe atoms
• 1 mole NO2 = 6.022×1023 NO2 molecules 109
The Mole The amount of substance that contains as many atoms as there
are in exactly 12 grams of pure 12C. That is 6.022 x 1023 atoms
1 mole of particles contains 6.022 x1023 particles (Avogadro’s
number )
For example
One mole of hydrogen atoms = 6.023 x 1023 atoms of
hydrogen
One mole of hydrogen molecules = 6.023 x 1023 molecule of
hydrogen
One mole of electrons = 6.023 x 1023 electrons
One mole of sodium ions (Na+) = 6.023 x 1023 Na+ ions
110
Avogadro’s Number
• NA = 6.02 x 1023
1 mole of 12C has a mass of 12 g. 111
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Molecular & Formula Weight
Formula Mass is sometimes used in a more general sense to include Molecular Mass, but its formal definition refers to the sum of the atomic weights of the atoms in ionic bonded compounds
112
Molecular Mass (also referred to Molecular Weight (MW)) is the sum of the atomic weights of all atoms in a covalently bonded molecule – organic compounds, oxides, etc.
Molar Mass A substance’s molar mass (molecular weight) is the mass
in grams of one mole of the compound.
C=12 , O=16 amu
CO2 = 44.01 grams / mole
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 atom of 12C = 12.00 amu
1 mole 12C atoms = 12.00 g 12C
For any element atomic mass (amu) = molar mass (grams)
113
Ex: What is the molar mass of caffeine, C8H10N4O2?
C = 12.0107 g/mol H = 1.00794 g/mol
N = 14.0067 g/mol O = 15.9994 g/mol
12.0107 g/mol C x 8 mol C/mol C8H10N4O2 = 96.0856 g/mol C8H10N4O2
1.00794 g/mol H x 10 mol H/mol C8H10N4O2 = 10.0794 g/mol C8H10N4O2
14.0067 g/mol N x 4 mol N/mol C8H10N4O2 = 56.0268 g/mol C8H10N4O2
15.9994 g/mol O x 2 mol O/mol C8H10N4O2 = 31.9988 g/mol C8H10N4O2
Add total the elemental masses to get the molecular mass of caffeine
96.0856 + 10.0794 + 56.0268 + 31.9988 =
194.1906 g/mol C8H10N4O2
114
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Moles of Compounds Atoms
– Atomic Mass
• Mass of atom (from periodic table)
• 1 mole of atoms = gram atomic mass
= 6.022×1023 atoms
Molecules
– Molecular Mass
• Sum of atomic masses of all atoms in compound’s
formula
1 mole of molecule X = gram molecular mass of X
= 6.022 × 1023 molecules
115
Moles of Compounds Ionic compounds
– Formula Mass
• Sum of atomic masses of all atoms in ionic compound’s
formula
1 mole ionic compound X = gram formula mass of X
= 6.022 × 1023 formula units
General
Molar mass (MM)
• Mass of 1 mole of substance (element, molecule, or ionic
compound) under consideration
1 mol of X = gram molar mass of X
= 6.022 × 1023 formula units
116
Ex: How many H atoms are in 72.5 g of C3H8O?
1 mol C3H8O molecules = 8 mol H atoms
1 mol H = 6.023 x 1023 atoms H
1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O
x 8 mol H atoms x 6.023 x 1023 H atoms =
= 5.82 x 1024 atoms H
72.5 g C3H8O
60 g C3H8O
117
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Ex: How many moles of iron (Fe) are in 15.34 g
Fe?
• What do we know?
1 mol Fe = 55.85 g Fe
• What do we want to determine?
15.34 g Fe = ? Mol Fe
• Set up ratio so that what you want is on top & what you
start with is on the bottom
118
Fe g 55.85
Fe mol 1Fe g 15.34 = 0.2747 mole Fe
Start End
Ex: If we need 0.168 mole Ca3(PO4)2 for an
experiment, how many grams do we need to weigh
out?
• Calculate MM of Ca3(PO4)2
3 × mass Ca = 3 × 40.08 g = 120.24 g
2 × mass P = 2 × 30.97 g = 61.94 g
8 × mass O = 8 × 16.00 g = 128.00 g
1 mole Ca3(PO4)2 = 310.18 g Ca3(PO4)2
• What do we want to determine?
0.168 g Ca3(PO4)2 = ? Mol Ca3(PO4)2
119
Start End
• Set up ratio so that what you want is on the top &
what you start with is on the bottom
120
= 52.11 g Ca3(PO4)2
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Your Turn! Ex: How many moles of CO2 are there in 10.0 g?
A.1.00 mol
B.0.0227 mol
C.4.401 mol
D.44.01 mol
E.0.227 mol
121
= 0.227 mol CO2
2
22
CO g 44.01
CO mol 1 CO g 0.10
Molar mass of CO2
1 × 12.01 g = 12.01 g C
2 × 16.00 g = 32.00 g O
1 mol CO2 = 44.01 g CO2
Your Turn! Ex: How many grams of platinum (Pt) are in 0.475
mole Pt?
A.195 g
B.0.0108 g
C.0.000513 g
D.0.00243 g
E.92.7 g
Pt mol 1
Pt g 195.08Pt mol 475.0
= 92.7 g Pt
Molar mass of Pt = 195.08 g/mol
122
Percent Composition Percent composition by mass:
The mass of one element in a compound divided by
the mass of the entire compound
Steps to determine percentage composition:
1. Calculate the mass of each individual element
in the compound
2. Add up all the masses of each element to get
the total mass of compound
3. Divide the mass of each individual element
with the total mass of compound
123
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% element = x 100 (number of atoms)(atomic weight)
(FW of the compound)
Percent Composition
The Percent of C in C2H6
%C = (2)(12.0 amu)
(30.0 amu)
= 24.0 amu x 100
30.0 amu
= 80.0 %
124
Percent composition of an element in a compound
n x molar mass of element molar mass of compound
x 100%
n is the number of moles of the element in 1 mole of the
compound
C2H6O
%C = 2 x (12.01 g)
46.07 g x 100% = 52.14%
%H = 6 x (1.008 g)
46.07 g x 100% = 13.13%
%O = 1 x (16.00 g)
46.07 g x 100% = 34.73%
52.14% + 13.13% + 34.73% = 100.00%
Percent composition
=
125
Types of Formulas •Empirical Formula: (E.F.)
•Simplest whole-number ratio of atoms in a
compound where “empirical” means derived from
experiment
•Molecular Formula: (M.F.)
• Chemical formula of a compound that expresses the
actual number of atoms present in one molecule.
-The molecular formula will either be exactly the
same or some multiple of the empirical formula!
H2O E.F same as M.F 126
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Empirical Formula Ex: The compound para-aminobenzoic acid is composed of
carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and
oxygen (23.33%). Find the empirical formula of PABA.
Assuming 100.00 g of para-aminobenzoic acid,
C: 61.31 g x = 5.105 mol C
H: 5.14 g x = 5.09 mol H
N: 10.21 g x = 0.7288 mol N
O: 23.33 g x = 1.456 mol O
1 mol
12.01 g
1 mol
14.01 g
1 mol
1.01 g
1 mol
16.00 g
127
Calculate the mole ratio by dividing by the smallest number of
moles:
C: = 7.005 7
H: = 6.984 7
N: = 1.000
O: = 2.001 2
5.105 mol
0.7288 mol
5.09 mol
0.7288 mol
0.7288 mol
0.7288 mol
1.458 mol
0.7288 mol
These are the subscripts for the empirical formula:
C7H7NO2
128
Molecular Formula
Molecular formula of a compound that
expresses the actual number of atoms present
in one molecule.
The molecular formula will either be exactly
the same or some multiple of the empirical
formula
129
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Determining Molecular Formula
1. Follow the same steps for determining
empirical formula.
2. For the Molecular Formula, you need to be
given the molar mass of the compound. Find
the molar mass of the empirical formula.
3. Divide the molar mass of the compound by
the molar mass of the empirical formula to get
the factor with which to multiply each
subscript in the empirical formula
130
Ex: A compound contains 75% carbon and 25% hydrogen.
Determine its empirical formula. The molecular mass of this
compound is 16 amu. Determine its molecular formula also.
The atomic masses are: C = 12 amu, H = 1 amu.
So, The empirical formulae of the compound = C1H4 or CH4
The empirical formula mass = (1 x 12) + (4 x 1) = 12 + 4 = 16 amu
Molecular mass (given) = 16 amu
Therefore, Molecular formula = 1 x Empirical formula = 1 x CH4 = CH4
Molecular mass
Empirical formula mass
16 amu
16 amu
131
= So, N =
Chemical Equation
CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g)
Reactants appear on the left side of the equation
Products appear on the right side of the equation 132
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Chemical Equation
CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g)
The states of the reactants and products are written in
parentheses to the right of each compound 133
Chemical Equation
CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g)
Coefficients are inserted to balance the equation
134
Subscripts and Coefficients
•Subscripts tell the number of atoms of each element in a
molecule
•Coefficients tell the number of molecules.
135
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Balancing Equation
Balancing by Inspection GUIDELINES
1. Count the number of elements on both sides
of the equation
2. Change the coefficients (NEVER the
subscripts) to get the same number of elements
on both sides of the equation
136
Balancing Equation
A representation of a chemical reaction:
C2H5OH + O2 → CO2 + H2O
reactants products
Unbalanced !
C2H5OH + 3O2 → 2CO2 + 3H2O
The equation is balanced.
1 mole of ethanol reacts with 3 moles of oxygen
to produce
2 moles of carbon dioxide and 3 moles of water 137
138
Stoichiometry
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Ex: Methanol burns in air according to the equation
2CH3OH + 3O2 2CO2 + 4H2O
If 209 g of methanol are used up in the combustion, what
mass of water is produced?
= 235 g H2O
2CH3OH + 3O2 2CO2 + 4H2O
2 mol 4 mol
2x(12+4+16) g 4x(2+16) g
209 g m
209 x 4(2+16)
m =
2(12+4+16)
139
Ex: Without using a calculator, arrange the
following samples in order of increasing
numbers of carbon atoms:
12 g 12C, 1 mol C2H2, 9 x 1023 molecules of
CO2
12 g 12C (6 x 1023 C atoms) < 9 x 1023 CO2
molecules (9 x 1023 C atoms) < 1 mol C2H2 (12
x 1023 C atoms).
140
Ex: What is the mass in grams of 1.000 mol of glucose,
C6H12O6?
C6H12O6 has a molar mass of 180.0 g/mol
Ex: Calculate the number of moles of glucose (C6H12O6) in
5.380 g of C6H12O6
141
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Ex: Calculate the mass, in grams, of 0.433 mol of calcium
nitrate
Ex: How many glucose molecules are in 5.23 g of C6H12O6?
(b) How many oxygen atoms are in this sample?
142
Ex: Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and
54.50% O by mass. What is the empirical formula of ascorbic
acid?
143
Ex: Calculate the percentage of carbon, hydrogen,
and oxygen (by mass) in C12H22O11
144
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49
Ex: Calculate the number of H atoms in 0.350 mol of C6H12O6
145
The percentage by mass of
phosphorus in Na3PO4 is:
a. 44.0%
b. 11.7%
c. 26.7%
d. 18.9%
146
Your Turn! Calculate the number of moles of calcium in 2.53 moles of
Ca3(PO4)2
A.2.53 mol Ca
B.0.432 mol Ca
C.3.00 mol Ca
D.7.59 mol Ca
E.0.843 mol Ca
2.53 moles of Ca3(PO4)2 = ? mol Ca
3 mol Ca 1 mol Ca3(PO4)2
243243
)PO(Ca mol 1
Ca mol 3 )PO(Ca mol 53.2
= 7.59 mol Ca 147
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Ex: How many g of iron are required to use up all of 25.6 g
of oxygen atoms (O) to form Fe2O3?
A.59.6 g
B.29.8 g
C.89.4 g
D.134 g
E.52.4 g
148
Fe mol 1
Fe g 55.845
O mol 3
Fe mol 2
O g 16.0
O mol 1O g 25.6
mass O mol O mol Fe mass Fe
25.6 g O ? g Fe
3 mol O 2 mol Fe
= 59.6 g Fe
Ex: Balance the equation
__C3H8(g) + __O2(g) __CO2(g) + __H2O(ℓ)
Assume 1 in front of C3H8
3C 1C 3
8H 2H 4
1C3H8(g) + __O2(g) 3CO2(g) + 4H2O(ℓ)
2O 5 =10 O = (3 2) + 4 = 10
8H H = 2 4 = 8
1C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(ℓ)
149
___KClO3(s) → ___KCl(s) +___ O2(g) 2 2 3
__ Ba(OH)2(aq) +__ Na2SO4(aq) → __ BaSO4(s) + __ NaOH(aq) 1 1 2 1
__H3PO4(aq) + __ Ba(OH)2(aq) → __Ba3(PO4)2(s) + __H2O(ℓ) 3 1 6 2
150
Ex: Balance each of the following equations.
What are the coefficients in front of each compound?
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151
Chp. 4
States of Matters
State of Matters
How many state of matter found in
nature? Matter exists on earth in three physical states :
-Gases -Liquid -Solids
-Example water
-In the solid state H2O is known as ice .
-In the liquid state is called water.
-In the gases state is known as steam or water
Vapour.
152
The various points of distinction
between the three states
153
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Solid Liquid Gas
Melt Evaporate
Condense Freeze
154
The Gaseous State
155
Characteristics of Gases
•Unlike liquids and solids, gases
–expand to fill their containers;
–are highly compressible;
–have extremely low densities.
• Gas volume changes greatly with pressure
• Gas volume changes greatly with temperature
• Gas volume is a function of the amount of gas
156
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Introduction to Pressure
Pressure = Force
Area
Barometer
Units of Pressure
1 pascal (Pa) = 1 N/m2 = 1 kg/m.s2
1 atm = 760 mmHg = 760 torr
1 atm = 101,325 Pa(~105) = 101.325 kPa
Mercury Barometer At sea level,
h = 760 mm Hg 157
Example: a) Convert 0.357 atm to torr. 1 atm = 760 torr 0.357 atm = ? torr
760 x 0.357 P= ____________ = 271 torr 1
b) Convert 0.066 torr to atm c) Convert 147.2 kPa to torr
158
The Gas Laws
Four variables define the state of a gas
1)Temperature T
2) Pressure P
3) Volume V
4) Number of moles n
159
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Boyle’s Law
The volume of a fixed quantity of gas at
constant temperature is inversely
proportional to the pressure.
160
Decrease the volume Increase the pressure
P and V are inversely proportional
Constant temperature
Constant amount of gas 161
A plot of V versus P results in a curve
A plot of V versus 1/P will be a straight line
P a 1/V P = k (1/V )
P x V = constant
P1 x V1 = P2 x V2 This means a plot of P versus
1/V will be a straight line 162
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Ex: A sample of chlorine gas occupies a volume of 946 mL
at a pressure of 726 mmHg. What is the pressure of the gas
(in mmHg) if the volume is reduced at constant temperature
to 154 mL?
P1 x V1 = P2 x V2
P1 = 726 mmHg
V1 = 946 mL
P2 = ?
V2 = 154 mL
P2 = P1 x V1
V2
726 mmHg x 946 mL 154 mL
= = 4460 mmHg
163
Charles’ Law
The volume of a fixed
amount of gas at
constant pressure is
directly proportional
to its absolute
temperature i.e., V = k T
A plot of V versus T will be a straight line 164
The volume occupied by any sample of gas at constant
pressure is directly proportional to its absolute temperature
V a T
V = constant x T
V1/T1 = V2/T2
Note:
Temperature must be
in ____________
T (K) = t (°C) + 273.15 165
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Ex: A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of
1.54 L if the pressure remains constant?
V1 = 3.20 L
T1 = 398.15 K
V2 = 1.54 L
T2 = ?
T2 = V2 x T1
V1
1.54 L x 398.15 K 3.20 L
= = 192 K
V1/T1 = V2/T2
T2 = 192 – 273 = -81 0C 166
Gay-Lussac’s Law
V= constant, n= Constant
The pressure of a fixed amount of gas at constant
volume is directly proportional to its absolute
temperature
P / T = K
P1T2 = P2 T1
Constant V Constant n
167
Combined Gas Law
In the event that all three
parameters, P, V, and T,
are changing, their
combined relationship is
defined as follows
assuming the mass of
the gas (number of
moles) is constant.
•For Fixed Amount of Gas
• n = constant
• P . V = k
• V / T = k
• P / T = k
•P.V / T = k
•P1V1 / T1 = P2V2 / T2
168
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Ex: The pressure of 4.0 L of Nitrogen in a flexible container is
decreased to one-half its original pressure, and its absolute
temperature is increased to double the original temperature. The
new volume is now
a. 2.0 L b. 4.0 L c. 8.0 L d. 16.0 L e. 32.0 L
169
Avogadro’s Law
The volume of a gas at constant temperature and
pressure is directly proportional to the number of
moles of the gas
Mathematically, this means V = k.n
170
– The volume of one mole of gas is called the: molar gas volume, Vm
– Volumes of gases are often compared at standard temperature and pressure (STP), chosen to be
0 °C (273.15 ° K) and 1 atm pressure – At STP, the molar volume, Vm, that is, the volume occupied by one mole of any gas, is 22.4 L/mol
Avogadro’s Law
171
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V a number of moles (n)
V = constant x n
V = k.n
V1/n1 = V2/n2
172
Ex: A sample of Fluorine gas has a volume of 5.80 L at
150.0 °C and 10.5 atm of pressure. How many moles of
Fluorine gas are present?
First, use the combined empirical gas law to determine
the volume at STP. Then, use Avagadro’s law to
determine the number of moles
173
Ideal Gas Equation
174
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Ex: Ammonia burns in oxygen to form nitric oxide (NO) and
water vapor. How many volumes of NO are obtained from
one volume of ammonia at the same temperature and
pressure?
175
176
The Ideal Gas Law
The numerical value of “R” can be derived using
Avogadro’s law, which states that one mole of any
gas at STP will occupy 22.4 liters
177
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The Ideal Gas Law
The ideal gas equation is usually expressed in the
following form
178
Ex: Argon is an inert gas used in light bulbs to retard the
vaporization of the filament. A certain light bulb containing
argon at 1.20 atm and 18 °C is heated to 85 °C at constant
volume. What is the final pressure of argon in the light bulb
(in atm)?
179
Ex: A steel tank has a volume of 438 L and is filled with
0.885 kg of O2. Calculate the pressure of oxygen in the tank
at 21oC
180
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Ex: A 50-L cylinder of nitrogen, N2, has a pressure of 17.1 atm
at 23 °C. What is the mass (g) of nitrogen in the cylinder?
181
Molecular Mass
Molecular Weight Determination
182
Density of Gas
183
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Ex: What is the density of Methane gas (natural gas),
CH4, at 125 °C and 3.50 atm?
184
Gas Stoichiometry Ex: Suppose you heat 0.0100 mol of Potassium Chlorate,
KClO3, in a test tube. How many liters of Oxygen can you
produce at 298 K and 1.02 atm?
185
Gas Stoichiometry
Ex: What is the volume of CO2 produced at 37°C and 1.00
atm when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)
g C6H12O6 mol C6H12O6 mol CO2 V CO2
186
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Ex: Calcium carbonate, CaCO3(s), decomposes upon heating
to give CaO(s) and CO2(g). A sample of CaCO3 is
decomposed, and the carbon dioxide is collected in a 250 mL
flask. After the decomposition is complete, the gas has a
pressure of 1.3 atm at a temperature of 31 °C. How many
moles of CO2 gas were generated?
187
Ex: The gas pressure in an aerosol can is 1.5 atm at 25 °C.
Assuming that the gas inside obeys the ideal-gas equation,
what would the pressure be if the can were heated to 450 °C?
188
Ex: An inflated balloon has a volume of 6.0 L at sea level (1.0
atm) and is allowed to ascend in altitude until the pressure is
0.45 atm. During ascent the temperature of the gas falls from
22 °C to –21 °C. Calculate the volume of the balloon at its
final altitude.
189
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Ex: What is the density of carbon tetrachloride
(CCl4) vapor at 714 torr and 125 °C?
190
191
192
Chp. 5
Redox Reactions
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65
193
Oxidation–Reduction Reactions
Involves 2 processes:
Oxidation = Loss of Electrons (LEO)
Na Na+ + e Oxidation Half-Reaction
Reduction = Gain of electrons (GER)
Cl2 + 2e 2Cl Reduction Half-Reaction
Net reaction:
2Na + Cl2 2Na+ + 2Cl
– Oxidation & reduction always occur together
– Can't have one without the other
Oxidation Reduction Reaction Oxidizing Agent
• Substance that accepts e's
– Accepts e's from another substance
– Substance that is reduced
– Cl2 + 2e 2Cl–
Reducing Agent
• Substance that donates e's
– Releases e's to another substance
– Substance that is oxidized
– Na Na+ + e–
194
Your Turn! Which species functions as the oxidizing agent in the
following oxidation-reduction reaction?
Zn(s) + Pt2+(aq) Pt(s) + Zn2+(aq)
A.Pt(s)
B.Zn2+(aq)
C.Pt2+(aq)
D.Zn(s)
E.None of these, as this is not a redox reaction.
195
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Rules for Assigning Oxidation Numbers
1. Oxidation numbers must add up to charge on
molecule, formula unit or ion.
2. Atoms of free elements have oxidation numbers of
zero.
3. Metals in Groups 1A, 2A, and Al have +1, +2, and
+3 oxidation numbers, respectively.
4. H & F in compounds have +1 & –1 oxidation
numbers, respectively.
5. Oxygen has –2 oxidation number.
6. Group 7A elements have –1 oxidation number.
196
Rules for Assigning Oxidation Numbers
7. Group 6A elements have –2 oxidation number.
8. Group 5A elements have –3 oxidation number.
9. When there is a conflict between 2 of these rules or
ambiguity in assigning an oxidation number, apply
rule with lower oxidation number & ignore
conflicting rule.
Oxidation State
– Used interchangeably with oxidation number
– Indicates charge on monatomic ions
– Iron (III) means +3 oxidation state of Fe or Fe3+
197
Ex: Assigning Oxidation Number
1. Li2O
Li (2 atoms) × (+1) = +2 (Rule 3)
O (1 atom) × (–2) = –2 (Rule 5)
sum = 0 (Rule 1)
+2 –2 = 0 so the charges are balanced to zero
2. CO2
C (1 atom) × (x) = x
O (2 atoms) × (–2) = –4 (Rule 5)
sum = 0 (Rule 1)
x 4 = 0 or x = +4
C is in +4 oxidation state
198
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199
Learning Check
Assign oxidation numbers to all atoms:
Ex. ClO4
O (4 atoms) × (–2) = –8
Cl (1 atom) × (–1) = –1
(molecular ion) sum ≠ –1 (violates Rule 1)
Rule 5 for O comes before Rule 6 for halogens
O (4 atoms) × (–2) = –8
Cl (1 atom) × (x) = x
sum = –1 (Rule 1)
–8 + x = –1 or x = 8 –1
So x = +7; Cl is oxidation state +7
Learning Check Ex: Assign Oxidation States To All Atoms:
• MgCr2O7
Mg =+2; O = –2; and Cr = x (unknown)
+2 + 2x + {7 × (–2)} = 0
2x – 12 = 0 x = +3
Cr is oxidation # of +3
• KMnO4
K =+1; O = – 2; so Mn = x
+1 + x + {4 × (–2)} = 0
x – 7 = 0 x = +7
Mn is oxidation # of +7
200
Your Turn! Ex: What is the oxidation number of each atom in
H3PO4?
A. H = –1; P = +5; O = –2
B. H = 0; P = +3; O = –2
C. H = +1; P = +7; O = –2
D. H = +1; P = +1; O = –1
E. H = +1; P = +5; O = –2
201
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Redefine Oxidation-Reduction
in Terms of Oxidation Number • A redox reaction occurs when there is a change
in oxidation number.
Oxidation
– Increase in oxidation number
– e loss
Reduction
– Decrease in oxidation number
– e gain
202
Using Oxidation Numbers to
Recognize Redox Reactions • Sometimes literal electron transfer:
Cu: oxidation number decreases by 2
reduction
Zn: oxidation number increases by 2
oxidation
203
+ ++2 +20 0
increase oxidation
decrease reduction
Cu2+ Zn Zn2+ Cu
Balancing Redox Reactions Some Redox reactions are simple:
Ex. Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq)
Break into half-reactions
Zn(s) Zn2+(aq) + 2e oxidation
LEO
Reducing agent
Cu2+(aq) + 2e Cu(s) reduction
GER
Oxidizing agent 204
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Example Zn(s) Zn2+(aq) + 2e oxidation
Cu2+(aq) + 2e Cu(s) reduction
• Each half-reaction is balanced for atoms
– Same # atoms of each type on each side
• Each half-reaction is balanced for charge
– Same sum of charges on each side
• Add both equations algebraically, canceling e’s
• NEVER have e's in net ionic equation
Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq)
205
Redox in Aqueous Solution Ex. Mix solutions of K2Cr2O7 & FeSO4
– Dichromate ion, Cr2O72–, oxidizes Fe2+ to Fe3+
– Cr2O72– is reduced to form Cr3+
– Acidity of mixture decreases as H+ reacts with oxygen
to form water
Skeletal Eqn. Cr2O72– + Fe2+ Cr3+ + Fe3+
206
Ox. # Cr = +6 Fe = +2 Cr = +3 Fe = +3
Balance Ion-Electron Method –
Acidic Solution
1. Divide equation into 2 half-reactions
2. Balance atoms other than H & O
3. Balance O by adding H2O to side that needs O
4. Balance H by adding H+ to side that needs H
5. Balance net charge by adding e–
6. Make e– gain equal e– loss; then add half-reactions
7. Cancel anything that is the same on both sides
207
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Ion Electron Method
Ex. Balance in Acidic Solution
Cr2O72– + Fe2+ Cr3+ + Fe3+
1. Break into half-reactions
Cr2O72 Cr3+
Fe2+ Fe3+
2. Balance atoms other than H & O
Cr2O72 2Cr3+
– Put in 2 coefficient to balance Cr
Fe2+ Fe3+
– Fe already balanced 208
Ex. Ion-Electron Method in Acid
3. Balance O by adding H2O to the side that needs
O.
Cr2O72 2Cr3+
• Right side has 7 O atoms
• Left side has none
• Add 7 H2O to left side
Fe2+ Fe3+
• No O to balance
209
+ 7 H2O
Ex. Ion-Electron Method in Acid
4. Balance H by adding H+ to side that needs H
Cr2O72 2Cr3+ + 7H2O
• Left side has 14 H atoms
• Right side has none
• Add 14 H+ to right side
Fe2+ Fe3+
• No H to balance
210
14H+ +
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Ex. Ion-Electron Method in Acid
5. Balance net charge by adding electrons.
14H+ + Cr2O72 2Cr3+ + 7H2O
– 6 electrons must be added to reactant side
Fe2+ Fe3+
– 1 electron must be added to product side
• Now both half-reactions balanced for mass &
charge
211
6e +
+ e
Net Charge = 2(+3)+7(0) = 6
Net Charge = 14(+1) (–2) = 12
Ex. Ion-Electron Method in Acid 6. Make e– gain equal e– loss; then add half-
reactions
6e + 14H+ + Cr2O72– 2Cr3+ + 7H2O
Fe2+ Fe3+ + e
7. Cancel anything that's the same on both sides
6[ ]
212
6e + 6Fe2+ + 14H+ + Cr2O7
2 6Fe3+ + 2Cr3+
+ 7H2O + 6e
6Fe2+ + 14H+ + Cr2O7
2 6Fe3+ + 2Cr3+
+ 7H2O
Ex. Ion-Electron Method in Base Returning to our example of Cr2O7
2 & Fe2+
8. Add to both sides of equation the same number of
OH– as there are H+.
9. Combine H+ and OH– to form H2O.
10. Cancel any H2O that you can
213
6Fe2++ 14H+
+ Cr2O72
6Fe3+ + 2Cr3+
+ 7H2O + 14 OH– + 14 OH–
6Fe2+ + 14H2O + Cr2O7
2 6Fe3+ + 2Cr3+
+ 7H2O + 14OH
7
6Fe2+ + 7H2O + Cr2O7
2 6Fe3+ + 2Cr3+
+ 14OH
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Your Turn! Which of the following is a correctly balanced
reduction half-reaction?
A.Fe3+ + e– Fe°
B.2Fe + 6HNO3 2Fe(NO3)3 + 3H2
C.Mn2+ + 4H2O MnO4– + 8H+ + 5e–
D.2O2– O2 + 4e–
E.Mg2+ + 2e– Mg°
214
Ex: Ion-Electron Method Balance the following equation in acidic solution:
MnO4– + HSO3
– Mn2+ + SO42
1. Break it into half-reactions
MnO4– Mn2+
HSO3– SO4
2–
2. Balance atoms other than H & O
MnO4 Mn2+
– Balanced for Mn
HSO3 SO4
2
– Balanced for S
215
Ex: Ion-Electron Method
3. Add H2O to balance O
MnO4 Mn2+
HSO3 SO4
2
4. Add H+ to balance H
MnO4 Mn2+ + 4H2O
H2O + HSO3 SO4
2
216
+ 4H2O
H2O +
8H+ +
+ 3H+
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Ex: Ion-Electron Method
5. Balance net charge by adding e–.
8H+ + MnO4 Mn2+ + 4H2O
8(+1) + (–1) = +7 +2 + 0 = +2
Add 5 e– to reactant side
H2O + HSO3 SO4
2 + 3H+
0 + (–1) = –1 –2 + 3(+1) = +1
Add 2 e– to product side
5e– +
+ 2 e–
217
Ex: Ion-Electron Method
6. Make e– gain equal e– loss
5e– + 8H+ + MnO4 Mn2+ + 4H2O
H2O + HSO3 SO4
2 + 3H+ + 2e–
– Must multiply Mn half-reaction by 2
– Must multiply S half-reaction by 5
– Now have 10 e– on each side
218
2[ ]
5[ ]
10e– + 16H+ + 2MnO4
+ 5H2O + 5HSO3
2Mn2+ + 8H2O +
5SO42 + 15H+ + 10e
Ex: Ion-Electron Method
6. Then add the two half-reactions
10e– + 16H+ + 2MnO4 2Mn2+ + 8H2O
5H2O + 5HSO3 5SO4
2 + 15H+ + 10e–
7. Cancel anything that is the same on both sides.
Balanced in acid. 219
3 1
H+ + 2MnO4
+ 5HSO3
2Mn2+ + 3H2O + 5SO4
2
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220
Balance each equation in Acid & Base using the Ion
Electron Method.
MnO4– + C2O4
2– MnO2 + CO32–
Acid: 2MnO4– + 3C2O4
2– + 2H2O 2MnO2 + 4H+ + 6CO32–
ClO– + VO3– ClO3
– + V(OH)3
Acid: ClO– + 2H2O + 2VO3– + 2H+ ClO3
–+ 2V(OH)3
Your Turn!
Oxidation–Reduction in Biological
Systems In biological systems, oxidation may involve
the loss of H
the gain of O
In biological systems, reduction may involve
the gain of H
the loss of O
221
Oxidation–Reduction in Biological
Systems
• Plants represent one of the most basic
examples of biological oxidation and
reduction. The chemical conversion of carbon
dioxide and water into sugar (glucose) and
oxygen is a light-driven reduction process:
6CO2 + 6H2O → C6H12O6 + 6O2
• The light-driven reduction of CO2
222
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Aerobic respiration • Aerobic respiration may be represented by the general
equation
C6H12O6 + 6O2 → 6CO2 + 6H2O
• About 3000 kJ mol-1 of energy is released. Burning
glucose in air would release this amount of energy in one
go. However, it is not as simple as this in aerobic
respiration. Aerobic respiration is a series of enzyme-
controlled reactions that release the energy stored up in
carbohydrates and lipids during photosynthesis and make
it available to living organisms
223
Chp. 6
Chemical
Equilibrium
224
Reversible Reaction
Reversible reactions : Reaction in which entire amount of
the reactants is not converted into products is termed
as reversible reaction.
Characteristics of reversible reactions
(a) These reactions can be started from either side,
(b) These reactions are never complete,
(c) These reactions have a tendency to attain a state of
equilibrium
Example of reversible reactions
Salt hydrolysis, e.g., FeCl3 + 3H2O ⇄ Fe(OH)3 + 3HCl
225
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Irreversible Reaction
• Irreversible reactions : Reaction in which entire amount of the
reactants is converted into products is termed as irreversible reaction.
Characteristics of irreversible reactions
(a) These reactions proceed only in one direction (forward
direction),
(b) These reactions can proceed to completion,
(d) The arrow ( ) is placed between reactants and products,
Example of irreversible reactions
Neutralisation between strong acid and strong base e.g.,
NaOH + HCl → NaCl + H2O + 13.7 kcal
226
What is Equilibrium?
227
This is not Equilibrium?
228
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The Concept of Equilibrium
• As a system approaches
equilibrium, both the
forward and reverse
reactions are occurring.
• At equilibrium, the
forward and reverse
reactions are proceeding at
the same rate.
229
A System at Equilibrium
Once equilibrium is
achieved, the amount
of each reactant and
product remains
constant.
230
Expression of Equilibrium Constant
Forward reaction:
Backward reaction:
Reverse reaction:
Rate Law Rate law
231
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The Equilibrium Constant
At equilibrium
Rearranging gives:
232
The Equilibrium Constant
The ratio of the rate constants is a constant (as long as T is constant).
The expression becomes
233
The Equilibrium Constant
To generalize, the reaction:
Has the equilibrium expression:
234
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Homogeneous Equilibrium The equilibrium constant always has the same value (provided you
don't change the temperature), irrespective of the amounts of A, B, C
and D you started with. It is also unaffected by a change in pressure
or whether or not you are using a catalyst
235
Equilibrium-Constant Expressions
Write the equilibrium expression Kc for the following:
236
Heterogeneous Equilibrium The equilibrium established if steam is in contact with red hot
carbon. Here we have gases in contact with a solid. Everything is
exactly the same as before in the equilibrium constant expression,
except that you leave out the solid carbon.
Writing an expression for Kc for a heterogeneous equilibrium.
The important difference this time is that you don't include any
term for a solid in the equilibrium expression 237
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Equilibrium Constant Using Partial Pressures
• Consider the following reaction in gaseous phase:
wA + xB ⇄ yC + zD
The equilibrium constant of gases are expressed in terms of
partial pressures, instead of its concentrations and denoted
by Kp
• Kp =
Example:
2 SO2(g) + O2(g) ⇄ 2 SO3(g)
y z
C D
w x
A B
(P ) (P )
(P ) (P )
O2
2
SO2
2
SO3
PP
PKP
238
Relationship Between Kc and Kp
• Consider a general reaction:
• wA + xB ⇄ yC + zD
• Kp =
• PA = [A]RT; PB = [B]RT; PC = [C]RT; PD = [D]RT
• Kp =
• Kp = Kc(RT)∆n; (where ∆n = (y+z)-(w+x))
y z
C D
w x
A B
(P ) (P )
(P ) (P )
y z y z(y+z)-(w+x)
w x w x
([C]RT) ([D]RT) [C] [D](RT)
([A]RT) ([B]RT) [A] [B]
Dn = number of moles of gaseous products – number of moles of
gaseous reactants in chemical equation
239
Relationship between Kc and Kp
For other reactions:
1.2NO2(g) ⇄ N2O4(g); Kp = Kc(RT)-1
2.H2(g) + I2(g) ⇄ 2 HI(g); Kp = Kc
3.N2(g) + 3H2(g) ⇄ 2 NH3(g); Kp = Kc(RT)-2
240
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241
Interpreting KC
• Large K (K>>1)
– Means product rich mixture
– Reaction goes far toward
completion
Ex.
2SO2(g) + O2(g) 2SO3(g)
Kc = 7.0 1025 at 25 ° C
1
100.7
][O][SO
][SO 25
22
2
23
cK
242
Interpreting KC
• Small K (K<<1)
– Means reactant rich mixture
– Only very small amounts of
product formed
Ex.
H2(g) + Br2(g) 2HBr(g)
Kc = 1.4 10–21 at 25 °C
1
104.1
]][Br[H
[HBr] 21
22
2 cK
243
Interpreting KC
• K 1
– Means product and
reactant concentrations
close to equal
– Reaction goes only ~
halfway
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244
• Size of K gives measure of how reaction
proceeds
• K >> 1 [products] >> [reactants]
• K = 1 [products] = [reactants]
• K << 1 [products] << [reactants]
245
Ex: Consider the reaction of
2NO2(g) N2O4(g)
If Kp = 0.480 at 25°C, does the reaction favor product or reactant?
K is small (K < 1)
Reaction favors reactant
Since K is close to 1, significant amounts of
both reactant and product are present
Le Châtelier's Principle
• Le Châtelier's principle states that,
when a system at equilibrium is subjected to a change, the system will response by shifting in the direction that tends to minimize the change while reaching a new equilibrium position.
246
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247
Factors Affect Equilibrium According
Le Châtelier’s Principle
1. Concentration
2. Pressure and volume
3. Temperature
4. Catalysts
5. Adding inert gas to system at constant volume
Effect of Change in Concentration
2SO2(g) + O2(g) → 2SO3(g)
Kc = 2.4 x 10-3 at 700 oC
• Which direction will the reaction move if 0.125 moles of O2 is added to an equilibrium mixture ?
the equilibrum shift Towards the products
Adding or removing concentration change equilibrium concentration and not affect on KC or KP values (remain constant)
248
Shift Due to Pressure Change
• For the reaction: N2(g) + 3H2(g) ⇄ 2NH3(g),
• Increasing p, the equilibrium shift to lower
moles (shift to right) and KC or KP remain
constant
• decreasing p, the equilibrium shift to higher
moles (shift to left) and KC or KP remain
constant
249
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Shift Due to Pressure Change
For the reaction: CH4(g) + H2O(g) ⇄ CO(g) + 3H2(g),
– Compression will increase pressure; equilibrium
shifts left, reducing the number of molecules and
pressure.
– Expansion causes pressure to decrease; equilibrium
shifts right, increasing the number of molecules
pressure.
– This type of reactions favor low pressure for product
formation.
250
Shift Due to Pressure Change
• For the reaction: CO(g) + H2O(g) ⇄ CO2(g) + H2(g),
Increasing or decreasing P ont affect equilibrium
position or Kc or Kp due to the number of
gaseous mole of reactant = number of mole of
products).
251
Shift Due to Volume Change
• For the reaction: N2(g) + 3H2(g) ⇄ 2NH3(g),
• Increasing V of reaction container, the equilibrium
shift to higher moles (shift to left) and KC or KP
remain constant
• decreasing P, the equilibrium shift to lower moles
(shift to right) and KC or KP remain constant
252
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Shift due to Temperature Change
• Consider the following reactions:
(1) N2(g) + 3H2(g) ⇄ 2NH3(g); DHo = -92 kJ
(2) CH4(g) + H2O(g) ⇄ CO(g) + 3H2(g); DHo = 206 kJ
– For exothermic reactions, like reaction (1), raising the
temperature causes a shift to the left;
– For endothermic reactions, like reaction (2),
increasing the temperature causes a shift to the right;
– Exothermic reactions favor low temperature, while
endothermic reactions favor high temperature for
products formation. 253
254
Shift due to Temperature Change
• T shifts reaction in direction that produces
endothermic (heat absorbing) change
• T shifts reaction in direction that produces
exothermic (heat releasing) change
• Changes in T change value of mass action
expression at equilibrium, so K changed
– K depends on T
– T of exothermic reaction makes K smaller
• More heat (product) forces equilibrium to reactants
– T of endothermic reaction makes K larger
• More heat (reactant) forces equilibrium to products
255
Effect of Catalysts
• Catalyst lowers Ea for both forward and reverse reaction
• Change in Ea affects rates kr and kf equally
• Catalysts have no effect on equilibrium
• Equilibrium position and
KC or KP remain constant
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256
Effect of Adding Inert Gas
Inert gas
– One that does not react with components of reaction
Ex. Argon, Helium, Neon, usually N2
• Adding inert gas to reaction at fixed V (n and T),
P of all reactants and products
• Since it doesn’t react with anything
– No change in concentrations of reactants or products
– No net effect on reaction
Chp. 7
Chemistry: The Molecular Nature of Matter, 6E
Jespersen/Brady/Hyslop
Acids and Bases
257
Arrhenius Acids and Bases Acid produces H3O
+ in water
Base gives OH–
Acid-base neutralization
– Acid and base combine to produce water and a salt.
Ex. HCl(aq) + NaOH(aq) H2O + NaCl(aq)
H3O+(aq) + Cl–(aq) + Na+(aq) + OH–(aq)
2H2O + Cl–(aq) + Na+(aq)
• Many reactions resemble this without forming
H3O+ or OH– in solution
258
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Brønsted-Lowry Definition • Acid = proton donor
• Base = proton acceptor
• Allows for gas phase acid-base reactions
Ex. HCl + H2O H3O+ + Cl–
– HCl = acid
• Donates H+
– Water = base
• Accepts H+ 259
Conjugate Acid-Base Pair • Species that differ by H+
Ex. HCl + H2O H3O+ + Cl–
• HCl = acid
• Water = base
• H3O+
– Conjugate acid of H2O
• Cl–
– Conjugate base of HCl
260
Formic Acid is Bronsted Acid
• Formic acid (HCHO2) is a weak acid
• Must consider equilibrium
– HCHO2(aq) + H2O CHO2–(aq) + H3O
+(aq)
• Focus on forward reaction
261
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Formate Ion is Bronsted Base • Now consider reverse reaction
• Hydronium ion transfers H+ to CHO2
262
H3O+ + CHO2
HCHO2 + H2O
conjugate pair
conjugate pair
acid base acid base
Learning Check
263
• Ex: Identify the Conjugate Partner for Each
Cl–
NH4+
C2H3O2–
HCN
F–
Learning Check
• Ex: Write a reaction that shows that HCO3– is a
Brønsted acid when reacted with OH–
• HCO3–(aq) + OH–(aq)
• Write a reaction that shows that HCO3– is a
Brønsted base when reacted with H3O+(aq)
• HCO3–(aq) + H3O
+(aq)
264
H2CO3(aq) + H2O(ℓ)
H2O(ℓ) + CO32–(aq)
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Your Turn! Ex: In the following reaction, identify the acid/base
conjugate pair.
(CH3)2NH + H2SO4 → (CH3)2NH2+ + HSO4
-
A. (CH3)2NH / H2SO4 ; (CH3)2NH2+ / HSO4
-
B. (CH3)2NH / (CH3)2NH2+ ; H2SO4 / HSO4
-
C. H2SO4 / HSO4- ; (CH3)2NH2
+ / (CH3)2NH
D. H2SO4 / (CH3)2NH ; (CH3)2NH2+ / HSO4
-
265
Amphoteric Substances • Can act as either acid or base
– Also called amphiprotic
– Can be either molecules or ions
Ex. hydrogen carbonate ion:
– Acid:
HCO3–(aq) + OH–(aq) CO3
2–(aq) + H2O(ℓ)
– Base:
HCO3–(aq) + H3O
+(aq) H2CO3(aq) + H2O(ℓ)
2H2O(ℓ) + CO2(g)
266
Your Turn! Ex: Which of the following can act as an
amphoteric substance?
A. CH3COOH
B. HCl
C. NO2-
D. HPO42-
267
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Strengths of Acids and Bases
Strength of Acid
– Measure of its ability to transfer H+
– Strong acids • React completely with water Ex. HCl and HNO3
– Weak acids • Less than completely ionized Ex. CH3COOH and CHOOH
Strength of Base classified in similar fashion: – Strong bases
• React completely with water Ex. Oxide ion (O2) and OH
– Weak bases • Undergo incomplete reactions
Ex. NH3 and NRH2 (NH2CH3, methylamine)
268
Strong vs. Weak Acid
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 508 269
Reactions of Strong Acids and Bases
In water
• Strongest acid = hydronium ion, H3O+
– If more powerful H+ donor added to H2O
– Reacts with H2O to produce H3O+
Similarly,
• Strongest base is hydroxide ion (OH)
– More powerful H+ acceptors
– React with H2O to produce OH
270
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In General • Stronger acids and bases tend to react with each
other to produce their weaker conjugates
– Stronger Brønsted acid has weaker conjugate
base
– Weaker Brønsted acid has stronger conjugate
base
• Can be applied to binary acids (acids made from
hydrogen and one other element)
271
Autoionization of Water • Self Ionization of Water
• Pure water undergoes auto-ionization to produce
Hydronium and Hydroxide ions
2 H2O(l) H3O+(aq) + OH-(aq)
• The extent of this process is described by an auto-
ionization (ion-product) constant, Kw
• The concentrations of Hydronium and Hydroxide
ions in any aqueous solution must obey the auto-
ionization equilibrium
+ - -14w 3K = [H O ][OH ] = 1.0 10
272
273
• In aqueous solution,
• Product of [H+] and [OH] equals Kw
• [H+] and [OH] may not actually equal each other
Solution Classification
Autoionization of water
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Learning Check
EX: In a sample of blood at 25 °C, [H+] = 4.6 108
M. Find [OH] and determine if the solution is acidic,
basic or neutral.
•So 2.2 107 M > 4.6 108 M
•[OH] > [H3O+]
•Solution slightly basic 274
14w 101OHH ]][[K
7
8
14
10221064
1001
HOH
.
.
.
][][ wK
The pH Concept • With most weak acids and bases [x] small
• Must compare values and exponents
• Easier to compare if you take base 10 logarithm
of each side
• Define
• Get back to concentration by
• In general
– Can adapt to many values
275
]log[ HpHpH10H ][
XpX log
]log[ OHpOH
0014p ww .log KK
Redefine Acidic, Basic and Neutral
Solutions in terms of pH!
• As pH , [H+] ; pOH , and [OH]
• As pH , [H+] ; pOH , and [OH]
276
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Your Turn! Ex: Kw increases with increasing temperature. At
50 oC, Kw = 5.476 x 10-14. What is the pH of a
neutral solution at 50 oC ?
A. 7.00
B. 6.63
C. 7.37
D. 15.3
277
Learning Check Ex: What are [H+] and [OH] of pH = 3.00 solution?
• [H+] = 103.00 = 1.0 103 M
• [OH] = = 1.0 1011 M
Ex: What are [H+] and [OH] of pH = 4.00 solution?
• pH = 4.00 [H+] = 1.0 104 M
• [OH] = = 1.0 1010 M
• Or pH 4.00 solution has 10 times less H+ than pH 3.00
solution
4
14
1001
1001
.
.
278
3
14
1001
1001
.
.
Sample pH Calculations Ex: Calculate pH and pOH of blood in Ex. 1.
• We found [H+] = 4.6 108 M
[OH] = 2.2 x 107 M
pH = log(4.6 x 108) = 7.34
pOH = log(2.2 x 107) = 6.66
14.00 = pKw
Or
pOH = 14.00 pH = 14.00 – 7.34 = 6.66
279
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Ex: What is the pH of NaOH solution at 25 °C in
which the OH concentration is 0.0026 M?
[OH] = 0.0026 M
pOH = log(0.0026) = 2.59
pH = 14.00 – pOH
= 14.00 – 2.59
= 11.41
280
Your Turn! Ex: A sample of fresh pressed apple juice has a pH
of 3.76. Calculate [H+].
A.7.6 x 103 M
B.3.76 M
C.10.24 M
D.5.9 x 109 M
E.1.7 x 104 M
281
H10]H[ p
= 103.76 = 1.7 x 104 M
Learning Check EX: What is the [H3O
+] and pH of a solution that has [OH–] = 3.2 × 10–3 M?
• [H3O+][OH-] = 1 x 10-14
• [H3O+] = 1 x 10-14/3.2 x 10-3 =3.1 x 10-12 M
• pH = -log [H3O+] = -log(3.1 x 10-12)= 11.50
282
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Your Turn!
283
Ex: What is the [OH–] and pH of a solution that has
[H3O+] = 2.3 × 10–5 M?
Learning Check
284
Ex: What is the pOH and the [H3O+] of a solution
that has a pH of 2.33?
[H3O+]= 4.7×10–3
pOH = 11.67
Your Turn!
285
Ex: What is the pH and the [H3O+] of a solution that
has a pOH of 1.89?
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pH of Dilute Solutions of Strong
Bases
• 1 mole OH for every 1 mole B
• [OH] = [B] for strong bases
• 2 mole OH for every 1 mole B
• [OH] = 2*[B] for strong bases
286
Learning Check Ex: Calculate the pH of 0.011 M Ca(OH)2.
Ca(OH)2(s) + H2O Ca2+(aq) + 2 OH(aq)
• [OH] = 2*[Ca(OH)2] = 2*0.011M = 0.022M
• pOH = – log (0.022) = 1.66
• pH = 14.00 – pOH
• = 14.00 – 1.66 = 12.34
• What is this in the [H+] of the solution?
• [H+] = 1012.34 = 4.6 x 1013 M
287
Learning Check
Ex: What is the pH of 0.1M HCl?
• Assume 100% dissociation
HCl(aq) + H2O(ℓ) H+(aq) + OH(aq)
I 0.1 N/A 0 0
C -0.1 -0.1 0.1 0.1
End 0 N/A 0.1 0.1
288
pH = –log(0.1) = 1
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Learning Check
Ex: What is the pH of 0.5M Ca(OH)2?
• Assume 100% dissociation
Ca(OH)2 (aq) Ca2+ (aq) + 2 OH– (aq)
I 0.5 0 0
C -0.5 +0.5 +0.52
E 0 0.5 1.0
289
pOH = -log(1.0) = 0
pH = 14.00 – pOH = 14.00 – 0 = 14
Chp. 8
Organic Chemistry
Chemistry: The Molecular Nature of Matter, 6E
Jespersen/Brady/Hyslop
290
What Organic Chemistry
• Chemistry of the compounds present
in living organisms.
• They all contain carbon.
• Organic Chemistry is the Chemistry
of Carbon.
291
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Living
things
Carbohydrates /
Proteins / Fats /
Vitamins / Antibiotics
Natural Sources of Organic Compounds
A variety of organic products
obtained from living things
292
The Unique Nature of Carbon CATENATION is the binding of an element to itself through covalent bonds to
form chain or ring molecules.
Carbon forms chains and rings, with single, double and triple covalent bonds, because it is
able to FORM STRONG COVALENT BONDS WITH OTHER CARBON ATOMS
Carbon forms a vast number of carbon compounds because of the strength of
the C-C covalent bond. Other Group IV elements can do it but their chemistry
is limited due to the weaker bond strength.
BOND ATOMIC RADIUS BOND ENTHALPY
C-C 0.077 nm +348 kJmol-1
Si-Si 0.117 nm +176 kJmol-1
The larger the atoms, the weaker the bond. Shielding due to filled inner orbitals
and greater distance from the nucleus means that the shared electron pair is held
less strongly.
293
CHAINS AND RINGS
CARBON ATOMS CAN BE ARRANGED IN
STRAIGHT CHAINS
BRANCHED CHAINS
and RINGS
The Unique Nature of Carbon
You can also get a combination of rings and chains
MULTIPLE BONDING AND SUBSTITUENTS
CARBON-CARBON COVALENT BONDS CAN BE SINGLE, DOUBLE OR
TRIPLE
294
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DIFFERENT ATOMS / GROUPS OF ATOMS CAN BE PLACED ON THE
CARBONS
The basic atom is HYDROGEN but groups containing OXYGEN, NITROGEN, HALOGENS
and SULPHUR are very common.
CARBON SKELETON FUNCTIONAL CARBON SKELETON FUNCTIONAL
GROUP GROUP
The chemistry of an organic compound is determined by its FUNCTIONAL GROUP
The Unique Nature of Carbon
295
Classification of Hydrocarbons
Section 14.2
296
Aliphatic Hydrocarbons
• Aliphatic hydrocarbons are hydrocarbons having
no benzene rings.
• Aliphatic hydrocarbons can be divided into four
major divisions:
– Alkanes
– Cycloalkanes
– Alkenes
– Alkynes
Section 14.3
297
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Alkanes • Alkanes are hydrocarbons that contain only single
bonds.
• Alkanes are said to be saturated hydrocarbons
– Because their hydrogen content is at a maximum.
• Alkane general formula CnH2n + 2
• The names of alkanes all end in “-ane.”
• Methane butane are gases
• Pentane C17H36 are liquids
• C18H38 and higher are solids
Section 14.3
298
Straight Chained Alkanes
299
Alkyl Groups Alkane type groups added to parent chain are
known as alkyl groups. Consist of alkane, minus
one H atom. Name always ends in -yl
Example
CH4 : now remove one H which yields –CH3
• Naming of –CH3
• Start with parent name, which is methane
• Drop –ane and add –yl
• So methane becomes methyl group
300
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Hydrocarbon Nomenclature Rules for naming alkanes
Established by IUPAC
1. Name ends in “-ane”
2. Complete name uses that of parent
compound with constituent groups added.
3. Parent is longest continuous carbon chain.
4. Name of longest chain based on the number
of carbons.
5. Carbon atoms are numbered starting at the
end that gives the lowest number for the
first branch.
301
6. Aryl groups names are prefixed to parent
name.
7. Multiple aryl groups on a parent are
numbered and named alphabetically.
8. When there are multiple identical groups
add di, tri, tetra to the aryl name.
9. If multiple, identical aryl groups are
attached to the same carbon repeat the
carbon number.
302
Hydrocarbon Nomenclature
Example • The longest continuous chain of C atoms is
five
• Therefore this compound is a pentane
derivative with an attached methyl group
– Start numbering from end nearest the
substituent
– The methyl group is in the #2 position
• The compound’s name is 2-methylpentane.
303
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102
Example What is the name of the compound shown?
1. The longest carbon chain (parent) is
four. Parent name is butane.
2. Start numbering from the left to get the
smallest number for the attached group.
304
Example
3. The attached alkyl group is a methyl group.
– Thus, the correct name is:
• 2-methylbutane
– What is the name of the following
compound?
305
Example • The parent chain contains five carbons.
• Thus, the parent name is pentane.
• Number from the left to obtain the smallest
number for the first alkyl group.
• The alkyl groups are at the 2 and 3 positions.
• The 2 and 3 positions each contain a methyl
group. 306
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Example
• Thus, the correct name is:
• 2,3-dimethylpentane
• Let’s consider an alkane with two substituents
on the same carbon.
307
Example
• The parent chain is six carbons long.
• The lowest correct numbering of positions is
shown below.
• There are methyl and ethyl groups attached to
carbon 3.
308
Example
• The correct name is:
3-ethyl-3-methylhexane
309
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Your Turn!
What is the correct name for the molecule shown
below?
A. 3-butylpentane
B. 1,1-diethylpentane
C. 3-ethylheptane
D. 5-ethylheptane
310
Your Turn!
What is the name of the compound shown
below?
A. 3-methyl-3-methyloctane
B. 3,3-dimethyloctane
C. 2-ethyl-2-methylheptane
D. 6,6-dimethyloctane 311
Chemical Properties of Alkanes
• Alkanes are relatively unreactive
• Not reactive in conc. NaOH or H2SO4 at room
temperature.
• React with hot HNO3
• Will react with Cl2 and Br2 to form
halogenated hydrocarbons.
• Examples are CH3Cl, CH2Cl2 and CHCl3
• Can crack molecules like ethane under
controlled conditions to form CH2CH2
• Will react with O2 to form CO2, CO, and H2O 312
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Alkenes and Alkynes
• Alkenes contain one or more double bonds
– General form: CnH2n
• Alkynes contain one or more triple bonds
– General form: CnH2n-2
• Non-polar compounds are not water soluble
• Examples:
313
Alkenes and Alkynes
• Nomenclature
– The parent chain must contain the multiple
bond even if it is a smaller chain length than
one without a multiple bond
– Number from end that gives the lowest
number to the first carbon of the multiple
bond
– The number is given as -x- and placed just
before the –ene or –yne of the parent name.
For example, but-2-ene. The double bond
starts on carbon 2 of the chain. 314
Alkene Examples
• Start numbering from the left to get the lowest
number for the first carbon with the double
bond
• The parent is heptene and the correct naming
including the double bond location would be
hep-2-ene
315
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Example
• The parent chain is four carbons
• 2,3-dimethylbut-2-ene
• We would not name this 2-methyl-3-methylbut-2-
ene
316
Naming Polyenes
• How do we name compounds such as the
following?
• This compound contains two double bonds and
is known as a diene
• We want the lowest number for the first carbon
of each of the double bonds
• Start numbering from the right
317
Naming Polyenes
• The correct name would be hex-1,3-diene
• Three double bonds would be a triene
hex-1,3,5-triene
318
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Cyclic Alkenes
• Number ring to obtain lowest number for first
carbon of the double bond
319
Cyclic Alkenes
• Correct name is 1,6-dimethylcyclohex-1-ene
• Other ring examples
320
Your Turn! What is the correct name for the compound
shown
below?
A. 1,4-dimethylcyclopent-1-ene
B. 1,3-dimethylcyclopent-1-ene
C. 1-methyl-4-methylcyclopent-1-ene
D. 1,3-dimethylcyclo-1-pentene
321
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Your Turn! What is the correct structure for 3,3-dimethylpro-1-
ene?
A.
B.
C.
D.
322
Geometric Isomers
• Groups cannot freely rotate about a double
bond
• Therefore, it is possible to have geometric
isomers
• Examples:
323
Reactions of Alkene
• Alkenes readily add across the double bond
• Examples of an addition reaction:
– CH2CH2 + H2 CH3CH3
hydrogenation
– CH2CH2 + HCl → CH3CH2Cl
– CH2CH2 + H2O → CH3CH2OH
– CH CH + Cl → CH ClCH Cl
Pt
324
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Aromatic Hydrocarbons
• The most common aromatic compound is
benzene and its derivatives
• Representation of bonding
• Delocalized π bonds create unique stability,
called resonance stabilization. The circle in the
ring represents delocalization.
325
Aromatic Hydrocarbons
• Aromatic hydrocarbons contain one or
more benzene ring.
• Benzene (C6H6) is the most important
aromatic hydrocarbon.
• It is a clear, colorless liquid with a distinct
odor, and is a carcinogen (cancer-causing
agent.)
• Traditional Lewis Structure
Section 14.2
326
Other Aromatic Hydrocarbons
• Toluene is used in modeling glue. Naphthalene is use
in mothballs, and Phenanthrene are used in the
synthesis of dyes, explosives, and drugs.
Section 14.2
327
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When Other Atoms are Substituted for
the H’s in the Benzene Ring A vast array of other compounds can be produced
Section 14.2
328
Naming for Benzene Derivatives
• Draw the structural formula for 1,3-dibromobenzene.
• First, Draw a benzene ring.
Section 14.2
329
Drawing Structures for Benzene
Derivatives
• Second, attach a bromine atom (“bromo”) to
the carbon atom at the ring position you
choose to be number 1.
Section 14.2
330
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Drawing Structures for Benzene
Derivatives
• Third, attach a second (“di”) bromine atom to ring
position 3 (you may number either clockwise or
counterclockwise from carbon 1) and you have the
answer.
• 1,3-dibromobenzene
Section 14.2
331
Drawing Structures for Benzene
Derivatives
• Draw the structural formula for 1-chloro-2-fluorobenzene.
1. Draw a benzene ring.
3. Attach a fluorine atom to
ring position 2 and you have
the answer.
• 1-chloro-2-fluorobenzene
Section 14.2
Cl
F
2. Attach a chlorine atom (“chloro”) to the carbon atom at the ring position you choose to
be number 1.
332
Reactions of Aromatics
• Substitution reactions maintain benzene’s
resonance structure.
• Addition reactions, like those of alkenes,
destroy resonance structure
• Substitution reaction:
333
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Addition Reaction
• Notice that you have reduced the double
bonding in the ring and altered the resonance
stabilization of the ring
334
Learning Check What product would form if benzene reacted with
nitric acid using an appropriate catalyst?
• Sulfuric acid is the catalyst
• A substitution reaction occurs
335
Your Turn! Which product is most likely formed when
sulfuric acid reacts with benzene?
A.
B.
C.
D.
336
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Organic Compounds Containing Oxygen
Important functional groups:
337
Alcohols and Ethers • Common alcohols: names end in -ol
– CH3OH methanol
– CH3CH2OH ethanol
– CH3CH2CH2OH propan-1-ol
• If the –OH group was attached to the
central carbon then the alcohol would be
propan-2-ol
– Alcohols form hydrogen bonds, causing
their boiling points to be higher than
predicted. 338
Alcohols and Ethers
• Primary alcohols:
• Secondary alcohols:
• Tertiary alcohols: 339
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Alcohols and Ethers
• Common ethers:
– CH3OCH3 dimethyl ether
– CH3CH2OCH2CH3 diethyl ether
– CH3OCH2CH3 methyl ethyl ether
– No hydrogen bonding occurs, thus, boiling
points are lower than corresponding
alcohols
– Like alkanes, ethers are not very reactive 340
Reactions of Alcohols
• Alcohols can undergo oxidation to form a
variety of products. Oxidation removes an H
atom from the alcoholic carbon as well as the
H on the –OH group.
• Primary alcohols can be oxidized to aldehydes
and carboxylic acids
341
Reactions of Alcohols • Aldehydes are more readily oxidized than
alcohols
• Secondary alcohols can be oxidized to ketones
342
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Reactions of Alcohols
• Ketones are not further oxidized
• Tertiary alcohols have no H atom on the
alcoholic carbon and thus, do not undergo
oxidation
• Alcohols undergo elimination reactions in the
presence of concentrated H2SO4 forming water
and alkenes
• -OH group readily accepts a proton from
sulfuric acid
343
Elimination Reaction
• Dehydration of an alcohol
• During the reaction a very unstable
carbocation is formed. This ion eliminates a
proton to form the alkene.
344
Substitution Reactions of Alcohols
• Using heat and concentrated HBr, HI, or HCl,
a halogen will replace the –OH group
• A proton adds to the –OH forming –OH2+
• Water leaves and the halogen ion attaches to
the carbon site where the –OH was attached
345
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Aldehydes and Ketones
• Naming Aldehydes
– Parent name ends in –al, replacing –e in the
alkane name
– The aldehyde group is always at the end of a
chain and numbering starts with that end of the
chain 346
Naming Aldehydes
• Number from the Aldehyde end
• Do not use -1- for Aldehyde:
• 3-methylpropan-1-al, or 3-methyl-1-propanal
would be wrong 347
Learning Check What is the name of the following aldehyde?
4-ethylhexanal
348
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Naming Ketones
• Parent name ends in –one
• Parent chain must contain carbonyl group
• Numbering so carbonyl carbon has lowest
possible number
4-ethylheptan-3-one
NOT: 4-ethylheptan-5-one
349
Your Turn! What is the correct name for the aldehyde shown
below?
A. 2,4-dimethylpentanal
B. 2,4-dimethyl-1-pentanal
C. 2-methyl-4-methylpropanal
D. 2,4-dimethyl-5-pentanal 350
Your Turn! - Solution
• Aldehydes are numbered from the aldehyde
end of the molecule
• There are two identical groups, (methyl) so we
use –di in the naming
351
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Your Turn!
What is the correct name for the ketone shown
below?
A. 4-methyl-3-ethylhexan-2-one
B. 4-ethyl-3-methylhexan-5-one
C. 3-ethyl-4-methylhexan-2-one
D. 3,4-diethylpentan-2-one 352
Your Turn! - Solution • Number to give lowest number to keto group
so you start from the right
• Alkyl groups are ordered alphabetically so
ethyl comes before methyl 353
Reactions of Aldehydes and Ketones
• Aldehydes and ketones add hydrogen across
the C=O bond
• Process is hydrogenation or reduction
354
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Carboxylic Acids and Esters
355
Naming Carboxlic Acids
• Name ends in –oic, replacing –e in the parent
name
• Numbering begins with carboxyl group
• -COOH or -CO2H is the condensed form
• CH3COOH is ethanoic acid (acetic acid)
356
Naming Carboxylic Acids
• Benzoic acid
• Propanoic acid
357
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Naming Ethers
• Name begins with alkyl group attached to the –O
• Name of parent acid is separate from the alkyl
group name and –oic is replaced with –ate
• Ethyl propanate
358
Learning Check
• What is the name of the following ester?
• Alkyl group is propyl
• Number, starting with
the ester carbon
• Propyl 4-methylpentanate
359
Your Turn!
What is the correct name for the product when 3-
methylbutan-1-ol is completely oxidized?
A. 3-methylbutanoic acid
B. 2-methyl-1-butanoic acid
C. 2-methlybutan-1-oic acid
D. 3-methylbutan-1-oic acid
360
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Reactions of Carboxylic Acids
• The –COOH is weakly acidic and therefore
reacts with base
• RCOOH + OH- → RCOO- + H2O
361
Formation of Esters • Esters give fruits their characteristic odor
ethyl pentanoate
362
Saponification • Strong base reacts with an ester to form alcohol
and the ester’s anion forms pentanoate ion
363
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Your Turn!
Name the ester formed when methanol reacts
with
hexanoic acid.
A. 1-methyl hexanoate
B. methylhexanoate
C. methyl hexanoate
D. methyl hexan-1-oate
364
Organic Derivatives of Ammonia
• Amines are derived from ammonia with one or
more H atoms replaced with organic groups
• Like ammonia, amines are weakly basic
• Amines react with acids
365
Acid Property of Protonated Amines
• Ethylmethylammonium ion is the conjugate
acid of ethylmethylamine
pKa = 10.76 pKb= 3.24
366
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Amides
• General form
• Where (H)R indicates either an H atom or an R
group attached
• Naming
– The name of the parent acid is amended
dropping the –oic ending and replacing it
with -amide 367
Amides
• Propanamide
• 4-ethylhexamide
• These are examples
of simple amides
368
Synthesis of Simple Amides • An organic acid reacts with aqueous NH3 to
form an amide
2-methylpropanoic acid yields 2-methylpropanamide
369
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Amide Reactions • Amides can be hydrolyzed back to their acid
form producing ammonia in the process
370
Amide Reactions • Urea, an amide, ultimately hydrolyzes to NH3,
CO2 and water
• Carbonic acid is formed, which then
decomposes to carbon dioxide and water
• The overall reaction is:
371
Basicity of Amides • Amides are not basic like amines
• The lone pair on the N atom is delocalized and
thus not readily available for donation to a
proton
• Amides are neutral in an acid-base sense
372
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Your Turn!
What is the correct name for the molecule shown
below?
A. 4,5-dimethylhexanamide
B. 2,3-dimethyl-6-hexanamide
C. 4-methyl-5-methylhexanamide
D. 4-isopropyl-4-methylpropanamide
373
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