Bonding: General
Concepts
A computer representation of K3C60, a
superconducting substance formed by reacting
potassium with buckminster fullerine (C60)
Source: Photo Researchers, Inc.
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13a–2
Two forms of carbon; graphite and diamond.
Source: Grant Hellman
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13a–3
Quartz grows in beautiful, regular crystals
SiO2
Vs
CO2
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13a–4
Figure 13.1: (a) The interaction of two hydrogen atoms
(b) Energy profile as a function of the distance
between the nuclei of the hydrogen atoms.
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13a–5
Figure 13.1: (a) The interaction of two hydrogen atoms
(b) Energy profile as a function of the distance
between the nuclei of the hydrogen atoms.
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13a–6
WHERE DO THE ELECTRONS GO?
• Are they shared equally?
• Are they more on one atom than the other?
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13a–7
WHERE DO THE ELECTRONS GO?
• Are they shared equally?
• Are they more on one atom than the other?
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13a–8
WHERE DO THE ELECTRONS GO?
• Are they shared equally?
• Are they more on one atom than the other?
ANSWER:
It depends who is pulling harder
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13a–9
WHERE DO THE ELECTRONS GO?
• Are they shared equally?
• Are they more on one atom than the other?
ANSWER:
It depends who is pulling harder
(“Electro negativity”)
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13a–10
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13a–11
The HCL molecule
has a dipole moment
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13a–12
Pauling and his electronegativity
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13a–13
Pauling and his electronegativity
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13a–14
Figure 13.3: The Pauling electronegativity
values as updated by A.L. Allred in 1961.
Expect A-B bond energy to be average of A-A and B-B
If it is not electro negativities must be different
(Stronger bond – more ionic – larger EN difference)
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13a–15
Figure 13.3: The Pauling electronegativity values
as updated by A.L. Allred in 1961. (cont’d)
Arbitrarily set F as 4
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13a–16
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13a–17
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13a–18
The HCL molecule
has a dipole moment
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13a–19
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13a–20
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13a–21
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13a–22
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13a–23
models
Now lets consider
more than two atoms
in a molecule
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13a–24
Linear molecules
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13a–25
Planar molecules
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13a–26
Tetrahedral molecules
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13a–27
MODELS
• These are only models
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13a–28
MODELS
• These are only models
• But…. Models are very useful for describing
properties.
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13a–29
MODELS
• These are only models
• But…. Models are very useful for describing
properties.
Newton:
particles
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13a–30
MODELS
• These are only models
• But…. Models are very useful for describing
properties.
Newton:
particles
Huygens:
waves
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13a–31
models
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13a–32
Ball-and-stick model of a protein segment
illustrating the alpha helix.
Source: Photo Researchers, Inc.
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13a–33
The concept of individual bonds makes it much
easier to deal with complex molecules such as DNA.
Source: Photo Researchers, Inc.
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13a–34
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13a–35
Polarity of Molecules
Dipole Moments of Polyatomic Molecules
Example: in CO2, each C-O dipole is canceled because
the molecule is linear. In H2O, the H-O dipoles do not
cancel because the molecule is bent.
O
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C
O
13a–36
Skeletal Structure
• Hydrogen atoms are always terminal
atoms.
• Central atoms are generally those with the
lowest electronegativity.
• Carbon atoms are always central atoms.
• Generally structures are compact and
symmetrical.
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13a–37
Skeletal Structure
• Identify central and terminal atoms in the
molecule C2H6O (ethyl alcohol or ethanol).
NOTE: Terminal atoms
are all bonded to only
one other atoms.
Central atoms are
bonded to two or more
other atoms
Now where
do the
electrons go?
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H H
H C C O H
H
H
13a–38
Writing Lewis Structures
• All the valence e- of atoms must appear.
• Usually, the e- are paired.
• Usually, each atom requires an octet.
– H only requires 2 e-.
• Multiple bonds may be needed.
– Readily formed by C, N, O, S, and P.
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13a–39
Lewis Structures
Draw Lewis structures for:
CH4:



H N H

H
H

H C
H

H
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
or
or
or
or



H O
H


NH3:


H2O:
H 
F

HF:


H F


H O
H


H N H
H
H
H C H
H
Exceptions to the Octet Rule
• Molecules with an odd number of electrons.
• Molecules in which an atom has less than an octet
of electrons.
• Molecules in which an atom
has more than an octet of
electrons.
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Resonance Forms
• Lewis structures that differ only in the
placement of electrons are resonance forms.
For O3:




O
O
O
=





 O O =O




• Experimentally, it is found that both bonds
are 0.128 nm long.
• The Lewis structure of O3 must show both
resonance forms.
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SAMPLE PROBLEM 10.4
O
Writing Resonance Structures
PROBLEM:
Write resonance structures for the nitrate ion, NO3-.
SOLUTION:
Nitrate has 1(5) + 3(6) + 1 = 24 valence e-
O
O
O
N
N
N
O
O
O
O
O
N does not have an
octet; a pair of ewill move in to form
a double bond.
O
O
O
O
N
N
N
O
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O
O
O
O
13a–45
Draw the resonance hybrid structure of NO3-
O
O
O
N
N
O
O
O
-1
-1
N
O
O
O
The three resonance structures
O
Hybrid structure.
Averaging leads to partial
N-O double bond character
(solid + dotted lines)
-1
N
O
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O
13a–46
Resonance Forms
C 2H 6O
20 ve’s
H
H H
H


 H C O C H
H C C O
H


H H
H
H
Methyl ether
Ethyl alcohol

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Benzene
Fredrich August von Kekule (German chemist) said that he
discovered the ring-shaped chemical structure of benzene
because of a strange, reptilian dream he had in 1865: "I turned
my chair to the fire and dozed. Again the atoms were
gamboling before my eyes. ... My mental eye... could not
distinguish larger structures, of manifold conformation; long
rows, sometimes more closely fitted together; all twining and
twisting in snakelike motion. But look! What was that? One of
the snakes had seized hold of its own tail, and the form
whirled mockingly before my eyes. As if by a flash of lighting
I awoke... " ( From "Creativity, Beyond the Myth of Genius" by
Robert Weisberg published by W. H. Freeman 1992
.) Although some scholars now believe that Kekule's dream
was a hoax to avoid sharing credit for the discovery of the
hexagonal shape of benzene, it still makes a wonderful story.
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13a–48
Resonance Forms
30 ve’s
C 6H 6
H
H
C
C
H
C
C
H
C
C
H
H

H
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H
C
C
H
C
C
H
C
C
H
H
Nonequivalent Forms
The molecule N2O has three resonance forms:




N
NO


N
=N=O





N N O

Which one contributes most to the overall
electronic structure?
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Formal Charge: Selecting the Best Resonance Structure
An atom “owns” all of its nonbonding electrons and half of its bonding electrons.
Formal charge of atom =
# valence e- - (# unshared electrons + 1/2 # shared electrons)
B
For OA
# valence
e-
O
# nonbonding
# bonding
e-
# valence e- = 6
O
=6
e-
For OC
=4
= 4 X 1/2 = 2
A
For OB
Formal charge = 0
O
# nonbonding e- = 6
C
# valence
# bonding e- = 2 X 1/2 = 1
Formal charge = -1
e-
=6
# nonbonding e- = 2
# bonding e- = 6 X 1/2 = 3
Formal charge = +1
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13a–51
Formal Charges
The formal charge on an atom assumes that the
bonding electrons are shared equally.
Formal charge = # valence electrons
- # nonbonding electrons
- ½ # bonding electrons
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
0

N
=N=O


0 +1 -1
N N O





N
N
O


-1 +1

-2 +1 +1

Formal
charges
in N2O:
Formal Charges
The most favored resonance form is the one that
has:
• The smallest formal charges.
• No formal charges with the same sign on
adjacent atoms.
• Negative formal charges on the most
electronegative atoms.
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Formal Charges
The formal charge on an atom assumes that the
bonding electrons are shared equally.
Formal charge = # valence electrons
- # nonbonding electrons
- ½ # bonding electrons

0

N
=N=O


0 +1 -1
N N O





N
N
O


-1 +1

-2 +1 +1

Formal
charges
in N2O:
Most favorable
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Resonance (continued)
EXAMPLE: NCO- has 3 possible resonance forms -
N C
O
N C
A
N C
O
B
O
C
formal charges
-2
0
N C
+1
O
-1
0
N C
0
O
0
0
N C
-1
O
Forms B and C have negative formal charges on N and O; this makes them
more important than form A.
Form C has a negative charge on O which is the more electronegative
element, therefore C contributes the most to the resonance hybrid.
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13a–55
Electron-Deficient Atoms
• Molecules containing Be and B can have less
than an octet for these atoms.

F = Be = 
F


0

0
0

F
Be
F





• BeF2:
+1 -2 +1
Most
Favorable
• The small Be atom can’t accomodate a
large formal charge.
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Electron-Deficient Atoms
+1
-1

BF3:
0


F
=B F




F
B
F




F
B F


F

0 F







0

0

0


0 F


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F
B= F




F


Most
Favorable

Electron-Rich Atoms
0
An atom that has a
d subshell in the
valence electron shell
E
can accomodate
more than an octet
of electrons.
Nitrogen can’t
accommodate more
than 8 electrons.
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4s
3s
 2s
3p
   2p
N in NF3
 1s
3d
Electron-Rich Atoms
An atom that
0
has a d subshell
4s
3d



3p
in the valence
 3s
electron shell
E
   2p
can accomodate
 2s
more than an
octet of electrons.
P can accommodate
P in PF5
more than 8 electrons
 1s
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Electron-Rich Atoms
PF5




F

F



F

P

F



F


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Exceptions to the Octet Rule
• Expanded octets.
P
Cl
F
S
P
Cl
Cl
Cl
F
F
F
F
••
Cl
••
••
Cl
F
••
Cl
••
••
Cl
••
••
••
••
••
••
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13a–61
Molecular Shapes
•Lewis structures give atomic connectivity: they tell us
which atoms are physically connected to which.
•The shape of a molecule is determined by its bond
angles.
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13a–62
Molecular Shapes
In order to predict molecular shape, we assume the
valence electrons repel each other. Therefore, the
molecule adopts whichever 3D geometry minimizes this
repulsion.
We call this process Valence Shell Electron Pair
Repulsion (VSEPR) theory.
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13a–63
The VSEPR Model
Predicting Molecular Geometries
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13a–64
The VSEPR Model
Predicting Molecular Geometries
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13a–65
Molecular Shapes
experimentally we find all Cl-C-Cl bond angles are 109.5.
Therefore, the molecule cannot be planar.
All Cl atoms are located at the vertices of a tetrahedron with the C at its center.
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13a–66
The VSEPR Model
Predicting Molecular Geometries
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13a–67
The VSEPR Model
Predicting Molecular Geometries
To determine the electron pair geometry:
•draw the Lewis structure
•count the total number of electron pairs around the central
atom
•arrange the electron pairs in one of the above geometries to
minimize e--e- repulsion
•multiple bounds count as one bonding pair
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13a–68
The VSEPR Model
Predicting Molecular Geometries
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13a–69
The VSEPR Model
Predicting Molecular Geometries
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13a–70
AB2
Linear
AB3
Trigonal planar
AB5
Trigonal bipyramidal
Molecular Shapes
AB2E
Angular or Bent
AB4
Tetrahedral
AB4E
Irregular tetrahedral
(see saw)
AB6
Octahedral
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AB3E
Trigonal
pyramidal
AB3E2
T-shaped
AB5E
Square pyramidal
AB2E2
Angular
or Bent
AB2E3
Linear
AB4E2
Square planar
13a–71
The VSEPR Model
The valence electrons in a molecule are the bonding
pairs of electrons as well as the lone pairs.
There are 11 shapes that are important to us:
Number of atoms, formula
Shapes
(3 atoms, AB2)
(4 atoms, AB3)
(5 atoms, AB4)
(6 atoms, AB5)
(7 atoms, AB6)
linear or bent
trigonal planar, trigonal bipyramidal,
or T-shaped
tetrahedral, square planar, or see-saw
trigonal bipyramidal or square pyramidal
octahedral
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13a–72
The VSEPR Model
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13a–73
The VSEPR Model
The Effect of Nonbonding Electrons and
Multiple Bonds on Bond Angles
By experiment, the H-X-H bond angle decreases on
moving from C to N to O:
H
H C H
H
109.5O
H N H
H
107O
O
H
H
104.5O
Since electrons in a bond are attracted by two nuclei, they do
not repel as much as lone pairs.
Therefore, the bond angle decreases as the number of lone
pairs increase.
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13a–74
The VSEPR Model
The Effect of Nonbonding Electrons and
Multiple Bonds on Bond Angles
Similarly, electrons in multiple bonds repel more than
electrons in single bonds.
Cl
111.4o
Cl
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C O
124.3o
13a–75
The VSEPR Model
Molecules with Expanded Valence Shells
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13a–76
The VSEPR Model
Molecules with Expanded Valence Shells
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13a–77
The VSEPR Model
Molecules with More than One Central Atom
In acetic acid, CH3COOH, there are three central
atoms.
We assign the geometry about each central atom
separately.
H
O
H
C
C
O
3
4
H
H
Number of electron domains
Electron-domain geometry
Predicted bond angles
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4
Tetrahedral
Trigonal
planar
Tetrahedral
109.5o
120o
109.5o
13a–78
The VSEPR Model
Molecules with More than One Central Atom
In acetic acid, CH3COOH, there are three central
atoms.
We assign the geometry about each central atom
separately.
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13a–79
Figure 13.6: The carbon dioxide molecule
• Nonpolar Molecules
– Dipole moments are symmetrical and cancel out.
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13a–82
Determining Molecular Polarity
• Nonpolar Molecules
– Dipole moments are symmetrical and cancel out.
F
BF3
B
F
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Copyright © Houghton Mifflin Company. All rights reserved.
F
13a–83
Determining Molecular Polarity
• Polar Molecules
– Dipole moments are asymmetrical and don’t
cancel .
O
H2O
H
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Copyright © Houghton Mifflin Company. All rights reserved.
H
net
dipole
moment
13a–84
Determining Molecular Polarity
polar molecules have...
– asymmetrical shape (lone pairs) or
– asymmetrical atoms
H
CHCl3
Cl
Cl
Cl
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Copyright © Houghton Mifflin Company. All rights reserved.
net
dipole
moment
13a–85
Figure 13.5: (a) The structure and charge distribution
of the ammonia molecule. (b) The dipole moment of the
ammonia molecule oriented in an electric field.
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13a–86
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13a–87
Sulfur has a partial
positive charge
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13a–88
Hydrogen atoms have a partial positive charge
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13a–89
Hydrogen atoms and a small partial negative
charge on the carbon
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13a–90
Polar Bonds
..
F
N
O
Cl
H
H
Polar
H
F
Polar
H
B
Polar
Cl
F
F
Cl
Polar
F
Cl
F
H
Cl
F
H
F
Nonpolar
H
C
Xe
F
F
Nonpolar
C
Cl
Cl
Nonpolar
H
H
Polar
A molecule has a zero dipole moment when bond dipoles cancel one another.
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13a–91
Covalent Bond Energy
• Covalent bond energy is measured by the
energy required to break the bond.
• The bond enthalpy, D(X-Y) is the average H
for breaking one mole of X-Y bonds in the gas
phase:
C + O


C O





D(C-O) = H
= 358 kJ
• When one mole of X-Y bonds is formed, the
enthalpy change is -D(X-Y).
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Bond Enthalpies and Bond Lengths
As bond order increases, the bond enthalpy
increases and the bond length decreases.
D(C-C) = 348 kJ
0.154 nm
D(C=C) = 614 kJ
0.134 nm
D(CC) = 839 kJ
0.120 nm
D(C-O) = 358 kJ
0.143 nm
D(C=O) = 799 kJ
0.123 nm
D(CO) = 1072 kJ
0.113 nm
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13a–94
Bond Enthalpies and Hrxn
Consider the reaction of H2 and O2 to form H2O:


H H + H H + O
=O




 H O
H + H O
H


H for breaking bonds = 2 D(H–H) + D(O=O)
H for forming bonds = 4-D(O–H)
Hrxn = 2 D(H–H) + D(O=O) - 4 D(O–H)
H =  D(bonds broken) -  D(bonds formed)
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Bond Enthalpies and Hrxn
Estimate H for the combustion of CH4:
H


H C H + 2 O
=O





H
 O=C=O
+2 H O
H



H = 4 D(C–H) + 2 D(O=O)
- 2 D(C=O) - 4 D(O–H)
= [ 4(413) + 2(495) - 2(799) - 4(463) ] kJ
= -808 kJ
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Comparing fuels
• Natural gas:
CH4 + 2O2 → CO2 + 2 H2O
ΔH=-808 kJ/mol
• Coal:
C + O2 → CO2
ΔH=-393.5 kJ/mol
• Oil:
C20H42 + 30½O2 → 20CO2 + 21 H2O
ΔH=-13315 kJ/mol
ΔH=-666 kJ/mol.CO2
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13a–97
Comparing fuels
Production of 1 GigaJoule of energy releases:
• Natural gas:
109 J x 0.044 kg/mol ÷ 808,000 J/mol =
54.5 kg CO2
• Coal:
112 kg CO2
• Oil:
66 kg CO2
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13a–98
N3 high energy density
3
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2
13a–99
ATP energy
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13a–100
ATP energy
Repulsion weakens these bonds
Resonance!
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13a–101