chapter 8 bonding and molecular structure · chapter 8 bonding and molecular structure ... chemical...
TRANSCRIPT
Jeffrey Mack
California State University,
Sacramento
Chapter 8 Bonding and Molecular Structure
• Things we must consider:
• What holds the atoms in a molecule or ionic
compound together?
• Why are atoms in molecules often distributed
at strange angles?
• Why are molecules three dimensional?
• Can we predict the structure of a compound?
• How does structure relate to chemical and
physical properties?
Chemical Bonding
Chemical bond: attractive force holding two or more atoms
together
Covalent bond: a sharing electrons between the atoms.
non–metal / non–metal bonds.
Ionic bond: electrostatic in nature, transfer of
electrons from a metal to a nonmetal.
(cation + anion)
Metallic bond: attractive force holding pure metals
together.
Cations in a “sea” of electrons.
Chemical Bonding
Ionic Bonding: complete transfer of 1 or more electrons from one atom to another.
(metals and non-metals)
Covalent Bonding: valence electrons shared
between atoms.
(non-metals and non-
metals)
Most bonds are somewhere in between.
Two Extreme Forms of Connecting or Bonding Atoms
Ionic Bonding results when an electron or electrons are
transferred from one atom to another.
The transfer results in each attaining an octet or Noble gas
electron configuration.
Na + Cl
3s1 3s23p5
Na+ Cl
e–
2s22p6
3s23p6
Na+ Cl-
[Ne] [Ar]
Noble gas electron
configurations
Ionic Bonding
• There are many examples of compounds
having covalent bonds, including the gases in
our atmosphere (O2, N2, H2O, and CO2),
common fuels (CH4), and most of the
compounds in your body.
• Covalent bonding is also responsible for the
atom-atom connections in polyatomic ions.
Covalent Bonding
A covalent bond results from a overlap of
valence orbitals on neighboring atoms.
Valence Electrons and Lewis Symbols for Atoms:
Covalent Bond Formation
• The American chemist Gilbert Newton
Lewis (1875–1946) introduced a useful
way to represent the valence shell
electrons of an atom.
• The element’s symbol represents the
nucleus including the core electrons.
• Up to four valence electrons, indicated
by dots, are placed one at a time
around the symbol.
• If any valence electrons remain, they
are paired with ones already there or
shared electrons up to a total of eight.
G. N. Lewis
1875 - 1946
Covalent Bonding & Lewis Structures
Notice that the group numbers of the main group elements represents the number of valence electrons (dots) for each element.
Lewis Dot Symbols
• Electrons are divided between core and valence electrons
B 1s2 2s2 2p1
• Core = [He] , valence = 2s2 2p1
Br [Ar] 3d10 4s2 4p5
Core = [Ar] 3d10 , valence = 4s2 4p5
Valence Electrons
• Covalent bonding, a bond results when one or more
electron pairs are shared between two atoms.
• The electron pair bond between the two atoms of an H2
molecule is represented by a pair of dots or a line.
• The representation of a molecule in this fashion is
called a Lewis electron dot structure or just a Lewis
structure in honor of G. N. Lewis.
H–H H
H-atom: H2 molecule:
H H
Lewis Dot Structures & the Octet Rule
• The number of unpaired valence electrons gives a general
indication as to the number of bonds an atom will likely form.
• Hydrogen has only one electron and therefore can only
make one covalent bond.
• Gr 7A has only one unpaired electron, so it generally forms
one bond.
• Gr 6A had two unpaired electrons, thus the likelihood of two
bonds. ... and so on.
Lewis Dot Structures & the Octet Rule
F
When other covalent species form, there are
additional electron pairs that do not participate in
bonding.
These are called “lone pairs” (lp)
& three lone pairs (octet)
H + H F
one bonding pair
hydrogen fluoride: HF
“single bond”
H– F
Lewis Dot Structures & the Octet Rule
• No. of valence electrons of an atom = Group number
• For Groups 1A-4A, no. of bond pairs = group number
• For Groups 5A -7A, BP’s = 8 - Grp. No.
• Hydrogen can only form one bond and never has lone
pairs.
• Boron often forms only 3 bonds.
• Elements in the 3rd period and beyond can exceed the
octet rule.
• Some stable molecules can form with an odd number of
electrons.
The Octet Rule
N N : :
Triple bond!
Example: N2
Each nitrogen atom needs 8 electrons to
complete an octet…
But each nitrogen has only 5 valence electrons!
As a result, the electrons must be shared.
N
N
There are 10
electrons
available
N2 needs 16
electrons (28) There are 10 valence electrons present
Therefore 6 electrons must be shared.
The remaining 4 electrons are the lone pairs
1. In general, for a poly atomic molecule the atom that
is the lowest and to the left on the periodic table
will be in the center. (lowest Electron Affinity)
2.Determine the total number of valence electrons in
the molecule or ion.
• In a neutral molecule, this number will be the sum
of the valence electrons for each atom.
• For an anion, add the number of electrons equal to
the negative charge.
• For a cation, subtract the number of electrons equal
to the positive charge.
Drawing Lewis Electron Dot Structures
3.Place one pair of electrons between each pair of
bonded atoms to form single bonds.
4.Use the remaining pairs to form more bonds or add
in as lone pairs to complete the octet of each atom.
5. If the central atom has fewer than eight electrons at
this point, change one or more of the lone pairs on
the terminal atoms into a bonding pair between the
central and terminal atom to form a multiple bond.
• Hydrogen atoms will never have lone pairs or
multiple bonds!
Drawing Lewis Electron Dot Structures
Ammonia, NH3
1. Decide on the central atom; Central atom is
generally the atom of lowest affinity for electrons &
never hydrogen.
Therefore, N is the central atom.
2. Count valence electrons
H = 1 and N = 5
Total = (3 x 1) + 5
= 8 electrons / 4 pairs
Building a Lewis Dot Structure
3. Form a single bonds between the central atom and each surrounding atom.
H H
H
N
H ••
H
H
N 4. The remaining electrons form
LONE PAIRS to complete octet as
needed.
3 BOND PAIRS and 1 LONE PAIR.
Note that N has a share in 4 pairs (8 electrons), while
H shares 1 pair.
Building a Lewis Dot Structure
Step 1. Central atom = S
Step 2. Count valence electrons:
S = 6
3 x O = 3 x 6 = 18
Negative charge = 2
TOTAL = 26 e- or 13 pairs
Step 3. Form bonds
10 pairs of electrons are
now left.
Sulfite Ion, SO32−
4. The remaining pairs become lone pairs, first on outside atoms and then on central atom.
O O
O
S ••
•
•
••
••
•• ••
••
•
•
•
•
•
•
Each atom is surrounded by an octet
of electrons.
Sulfite Ion, SO32−
1. Central atom = _______
2. Valence electrons = __ or __ pairs
3. Form bonds.
4. Place lone pairs on outer atoms.
This leaves 6 pairs.
Carbon Dioxide, CO2
4. Place lone pairs on outer atoms.
The second bonding pair forms a double bond.
5. So that C has an octet, we shall form
DOUBLE BONDS between C and O.
Carbon Dioxide, CO2
C H O
4 + 2 × 1 + 6 = 12 electrons available
Carbon is the least electronegative,
hydrogen can form only one bond.
C
Formaldehyde, CH2O
Formaldehyde, CH2O
C H O
4 + 2 × 1 + 6 = 12 electrons available
Carbon is the least electronegative,
hydrogen can form only one bond.
C H H
O
C H O
4 + 2 × 1 + 6 = 12 electrons available
Carbon is the least electronegative,
hydrogen can form only one bond.
C H H
O
Formaldehyde, CH2O
C H O
4 + 2 × 1 + 6 = 12 electrons available
Carbon is the least electronegative,
hydrogen can form only one bond.
C H H
O
Formaldehyde, CH2O
C H O
4 + 2 × 1 + 6 = 12 electrons available
Carbon is the least electronegative,
hydrogen can form only one bond.
C H H
O
Formaldehyde, CH2O
C
O H H
: :
Additional
Stuctures: C H H
O
Formaldehyde, CH2O
C
O H H
: :
Additional
Stuctures: C H H
O
Formaldehyde, CH2O
C O H H
: :
C
O H H
: :
Additional
Stuctures: C H H
O
Formaldehyde, CH2O
correct structure
Not this one…
C O H H
: :
or this one…
C
O H H
: :
Additional
Stuctures: C H H
O
Formaldehyde, CH2O
correct structure
Not this one…
C O H H
: :
or this one…
nature likes symmetry!
C
O H H
: :
Additional
Stuctures: C H H
O
Formaldehyde, CH2O
correct structure
Not this one…
C O H H
: :
or this one…
nature likes symmetry!
You will see why the first structure is favored by formal
chargers later.
Double and even triple bonds are commonly observed for C, N, P, O, and S
H2CO
C2F4
SO3
Double & Triple Bonds
Lewis Structures of Common Oxoacids & Their Anions
Molecules and ions having the same number of valence electrons and the same Lewis structures are said to be isoelectronic
Common Isoelectric Molecules & Ions
• The formal charge is the charge that would reside on an
atom in a molecule or polyatomic ion if we assume that all
bonding electrons are shared equally.
• The formal charge for an atom in a molecule or ion is
calculated based on the Lewis structure of the molecule or
ion, using:
• NVE = the number of valence electrons in the uncombined
atom (and equal to its group number in the periodic table).
• LPE = number of lone pair electrons on an atom.
• BE = number of bonding electrons around an atom.
FC = NVE [LPE + ½ BE]
Atom Formal Charges in Covalent Molecules & Ions
Hydrogen Fluoride: H F
H F Number of valence
electrons:
Number of bonding
electrons:
Number of lone pair
electrons:
Formal charge:
1 7
2 2
0 6
2FC(H) 1 0
2
0
2FC(F) 7 6
2
H F
0
0 0
FC = NVE [LPE + ½ BE]
Calculating the Formal Charge on Each Atom in a Covalent Molecule
When the sum of the formal charges on the atoms in
a molecule equals the expected overall charge on the
molecule, the Lewis structure is valid.
HF is expected to be a neutral compound, and the
formal charges validate this Lewis dot structure.
H F 0 0 + = 0
Calculating the Formal Charge on Each Atom in a Covalent Molecule
• •
•
• O O C • •
•
•
O-atom: FC = 6 – [4 + ½ 4] = 0
C-atom: FC = 4 – [0 + ½ 8] = 0
0 + 0 + 0 = 0
Carbon Dioxide, CO2
[ :C N: ]–
FC(C)
–1
FC(N)
0
The negative charge resides on the carbon atom.
As expected the overall charge is 1
64 2
2
65 2
2
Use Formal Charges to Predict Which Atom Carries the Negative Charge on CN−
C
O H H
: :
C H H
O
C O H H
: :
0 0 0
0 0 0 +2
-2
not favorable!
0 0
-1 +1
not favorable!
This one is
most
favorable!
Generally, the structure with
the lowest formal charges on
each atom or negative
charges on the atoms with
the highest EA’s are often
most favored.
CH2O Formal Charges
Consider the nitrate anion: The double bond does not
have to be on the vertical O-atom. There are three
equivalent structures that can be drawn.
We see that the double bond moves around or
“resonates” between the structures indicating that
the molecule exhibits what is called “Resonance” .
N
O
O O :
:
: :
:
:
:
:
–
: N
O
O O :
: :
: : :
– :
: N
O
O O : :
:
: :
–
Resonance
All three of the Lewis structures are equivalent
Resonance Structures.
Resonance stabilizes the energy (lowers) by
distributing electron density over the entire molecule.
The measured fHo is more exothermic than the
predicted value.
N
O
O O :
:
: :
:
:
:
:
–
: N
O
O O :
: :
: : :
– :
: N
O
O O : :
:
: :
–
Resonance
Identifying Resonance: Compounds that exhibit
resonance will have fractional bond orders.
: N
O
O O :
: :
:
: :
–
# of A-B bondsBond Order =
# of A-B links
4 N O bondsBO 1.333
3 N O links
Resonance
Certain elements can violate the octet rule. Boron
may form stable compounds with only 6 valence
electrons.
Compounds in the 3rd period and beyond can have
more than 8 electrons.
BF3 SF4
Exceptions to the Octet Rule
There are three classes of exceptions to the
octet rule:
1. Molecules with an odd number of electrons.
2. Molecules in which one atom has less than
an octet
3. Molecules in which one atom has more than
an octet.
Exceptions to the Octet Rule
Examples: Generally molecules such as ClO2, NO,
and NO2 have an odd number of electrons.
NO (nitrogen monoxide) Available electrons = 11
There must be an odd electron!
Molecular Orbital Theory will explain why NO is
stable later.
N O N O
115.5...?
2
or
Odd Numbers of Electrons
• Relatively rare.
• Molecules with less than an octet are typical for
compounds of Groups 1A, 2A, and 3A
Most typical example is BH3.
Available Electrons = 3 + 3(1) = 6 B
H
H H
There are only 6 electrons on the central atom.
Hydrogen may only form one bond.
The formal charges (all zero) support the drawn
structure.
Incomplete Octet
• Atoms from the 3rd period onwards can
accommodate more than an octet.
• Beyond the third period, the d-orbitals are low
enough in energy to participate in bonding and
accept the extra electron density.
Example: BrF3
Available Electrons = 7 + 3(7) = 28
Each fluorine atom can only share one electron.
What does this mean??
Central Atoms that Exceed the Octet Rule
Bromine has 7 valence electrons (4s24p5)
Fluorine also has 7 (2s22p5)
BrF3
Br has the lowest electronegativity so it will be the
central atom in the molecule.
Each F shares
one electron to
complete an
octet.
There are 10
total electrons in
the Br valence Br F F
Available Electrons = 48
Sulfur has 6 valence electrons, fluorine has seven.
If each F shares one electron with one of the six
from S, F completes its octet.
This will yield a compound with
S in the center having 6 F’s
bonded.
12 valence electrons around
the S
S
F F
F F
F
F
SF6
Compounds in Which an Atom Has More Than Eight Valence Electrons
Lewis structures tell us how atoms are connected in a
molecule:
bonds (bp) & lone pairs (lp) etc…
The 3–D shape of a molecule is however, determined
by its bond angles.
The Lewis structure
suggests that the H–C–
H bond angles are 90o &
that the molecule is flat.
H
H
C H H
90o
Molecular Shapes
Experimentally however, the H–C–H bond angles are
found to be 109.5
If this is the case, the molecule cannot be planar…
The methane molecule must be 3-D since the sum of
the angles is greater than 360 degrees!
H
H
C H H 90° 90°
90° 90°
90° + 90° + 90° +90° = 360°
109.5 + 109.5 + 109.5 + 109.5 =
438.0°
Molecular Shapes
To accommodate the 109.5o bond angles, the atoms
adopt a new 3-D geometry to minimize the repulsion
between the atoms:
• The molecular geometry of CH4 is said to be
“tetrahedral”
• The H-atoms fit at the corners of a regular
tetrahedron shape with the carbon at the center.
H
H
C H H C H H
H
H
becomes…
109.5o
Molecular Shapes
• In order to predict molecular shape, we assume the
valence electrons of each atom in the molecule
repel one another.
• When this occurs, the molecule adopts a 3D
geometry that minimizes this repulsion where:
This process is known as:
Valence Shell Electron Pair Repulsion (VSEPR)
theory.
bp-bp lp-lp > lp-bp >
Molecular Shapes
Valence Shell Electron Pair Repulsion
A molecule can be described in terms of the
distribution of the bonding atoms about the
central atom:
Molecular Geometry (MG)
A molecule can be described in terms of the
distribution of the bonding pair electrons (bp)
and lone pair electrons (lp) about the central
atom:
Electron Pair Geometry (EPG)
Molecular Shapes
Central Atoms Surrounded Only by Single-Bond Pairs
2 pairs
of
electrons
3 pairs
of
electrons
4 pairs
of
electrons
5 pairs
of
electrons
6 pairs
of
electrons
Electron Pair Geometries
G e ome tr y E x a m pl e
N o. of e - P ai r s Ar o u n d Ce nt r a l Atom
G e ome tr y E x a m pl e
N o. of e - P ai r s Ar o u n d Ce nt r a l Atom
180°
l i n e ar 2 F—Be—F
G e ome tr y E x a m pl e
N o. of e - P ai r s Ar o u n d Ce nt r a l Atom
180°
l i n e ar 2 F—Be—F
120°
planar trigonal
F F B
F
3
G e ome tr y E x a m pl e
N o. of e - P ai r s Ar o u n d Ce nt r a l Atom
180°
l i n e ar 2 F—Be—F
120°
planar trigonal
F F B
F
3
H
H H
H
tetrahedral
109°
C 4
Electron Pair Geometries
4 bonding pairs of electrons about the central atom, no lone
pairs on the central atom:
4 groups of electrons (all bonding)
Electron Pair Geometry (EPG) :
Tetrahedral
Molecular Geometry (MG):
Tetrahedral
When there are no lone pairs of
electrons in a molecule, the molecular
and electronic geometries are the same!
Central Atoms with Single-Bond Pairs & Lone Pairs
4 total pairs of electrons about the central atom, three bonding,
one lone pair on the central atom:
Electron Pair Geometry (EPG) :
Tetrahedral
Molecular Geometry (MG):
Trigonal Pyramidal
The lone pair forces the bonds to
reduce below the normal 109.5°.
Central Atoms with Single-Bond Pairs & Lone Pairs
Four groups of electrons, 2 bonding, 2 lone pairs on the central
atom:
Electron Pair Geometry (EPG) :
Tetrahedral
Molecular Geometry (MG):
Bent
The lone pairs force the bonds to
reduce below the normal 109.5°.
Central Atoms with Single-Bond Pairs & Lone Pairs
BF3
Molecular and Electron Pair geometries are
described as trigonal planar.
12 pairs of electrons available, F
forms only 1 bond
The B atom is surrounded by only
3 electron pairs.
Bond angles are 120°
•
• F
B
••
•
•
Compounds with Less Than an Octet
All based on trigonal
bipyramidal shape
Molecular Geometries for 5 Electron Pairs
P has 5 valence electrons
Each Cl-atom needs one electron to complete the
octet
P : : . P
Cl
Cl
Cl
Cl
Cl
Molecular Geometry : Trigonal
Bipyramidal
Electron Pair Geometry: Trigonal
Bipyramidal
Example: PCl5
S has 6 valence electrons
Each F-atom needs one electron to complete the
octet
S : : : S
F
F
:
F
F
Molecular Geometry : See-Saw
Electron Pair Geometry: Trigonal
Bipyramidal
Example: SF4
All are based on the 8-
sided octahedron
Molecular Geometries for 6 Electron Pairs
6 electron pairs
F
F
F F
F F
O c t a h e d r o n
9 0°
9 0°
S
Compounds with 5 or 6 Pairs Around the Central Atom
Three groups of electrons, all bonding,
no lone pairs on the central atom:
The lone pair of electrons forces the molecule into a
bent molecular geometry.
S
:
SO2
3 groups of electrons (2 bonds, 1 lp)
Electron Pair Geometry (EPG) Trigonal Planar
Molecular Geometry (MG): bent
Example: SO2
Recall that electronegativity () measures the relative
tendency for an atom to polarize a bond. It follows Zeff on
the periodic table.
Bond Polarity & Electronegativity
As the difference in Electronegativity increases
(), so does the ionic character of the bond.
Increasing covalent character
Bond Polarity & Electronegativity
• A difference in electronegativity of 0 indicates
that the bond is purely covalent. There is equal
sharing of electrons.
• A difference of electronegativity between 0.1 to
0.3 indicates that the bond is non-polar
covalent.
• A difference of electronegativity between 0.4 to
1.7 indicates that the bond is polar covalent.
• A difference of electronegativity that is greater
than 1.8 indicates that the bond is ionic.
Bond Polarity & Electronegativity
The polarity of the individual bonds in a molecule will
determine the overall polarity of a molecule.
All homonuclear
diatomic molecules
a non-polar.
I I :
iodine (I2)
Notice the symmetry of the
molecule:
• When divided, the top and
bottom as well as the left and
right are mirror images of one
another.
• One also knows the molecule
is non polar because the
bond is non polar. = 2.5 2.5 = 0
Overall Polarity in Molecules
H F
• Fluorine has a larger
electroegativity value
than hydrogen.
• This means that the
electrons in the bond
are skewed toward
the F-atom.
• The electrons shift
toward the F-atom This polarizes the molecule
more positive end
(+)
more negative end
()
Example: HF
The greater the difference in electronegativity () the
more polar the covalent bond.
H vs. F
4.0 – 2.2 = 1.8
C vs. F
4.0 – 2.5 = 1.5
O vs. F
4.0 – 3.5 = 0.5
So in terms of polarity,
H-F > C-F > O-F
Electronegativity: H vs. F
When a molecule possesses a net dipole moment, it
is polar.
H2O
The molecule
is therefore
polar.
The individual O-H
bond dipoles result
in a “net” or
overall
dipole moment for
the molecule.
Bond Polarity & Molecular Polarity
Notice also that the molecule’s symmetry can be broken along
either O-H bond axis:
This side…
does not look like this side!
Whenever there exists a line or plane of asymmetry,
the molecule is Polar!
Molecular Shape & Molecular Polarity
CO2
non–polar
Symmetry across all
bond axes.
Molecular Shape & Molecular Polarity
Dipole Moments & Molecular Polarity
• The order of a bond is the number of bonding
electron pairs shared by two atoms in a
molecule
• Bond orders may be 1, 2, and 3, as well as
fractional values.
• Bond strength increases with bond order.
• Bond length is inversely proportional to bond
order.
• Bond strength also increases with the difference
in electronegativity between two covalently
bonded atoms.
Bond Properties: Order, Length, Energy
Covalent bond strength increases with increasing
example: HCl bond is stronger than the HBr bond
Covalent bond strength increases with increasing
bond order.
example: O=O bond in stronger than O–O bond
triple > double > single
Bond Order: 3 2 1
Bond length decreases with increasing bond order
(Strength)
example: O=O bond is shorter than O–O bond
Therefore shorter!
Bond Properties: Order, Length, Energy
Double bond Single bond
Triple
bond
Acrylonitrile
Bond Order: # of Bonds Between a Pair of Atoms
Fractional bond orders occur in molecules with resonance structures.
Consider NO2-
The N—O bond order = 1.5
O=NO
: :
: :
:
Fractional Bond Order
• Bond length is the distance between the nuclei of two bonded atoms.
Bond Length
Bond length depends on size of bonded atoms.
H—F
H—Cl
H—I
Bond distances measured
in Angstrom units where
1Å = 10-10 m.
Bond Length
Bond length depends on bond order.
Bond distances measured
in Angstrom units where
1Å = 10-10 m.
Bond Length
Bond order is proportional to two important bond properties:
(a) bond strength
(b) bond length
745 kJ
414 kJ
110 pm
123 pm
Bond Length & Bond Strength
BOND Bond dissociation enthalpy (kJ/mol)
H—H 436
C—C 346
C=C 602
CC 835
NN 945
The GREATER the number of bonds (bond order)
the HIGHER the bond strength and the SHORTER
the bond.
Bond Strength
The energy required to break a covalent bond is
called the bond dissociation enthalpy, D.
For the Cl2 molecule, D(Cl–Cl) is given by H for the
reaction:
Cl2(g) 2Cl(g)
When more than one bond is broken:
CH4(g) C(g) + 4H(g) H = 1660 kJ
the bond enthalpy is a fraction of H for the
atomization reaction:
D(C–H) = ¼H = ¼(1660 kJ) = 415 kJ
Strengths of Covalent Bonds
Estimate the energy of the reaction
H—H(g) + Cl—Cl(g) 2 H—Cl(g)
Net energy = ∆rH =
= energy required to break bonds
energy evolved when bonds are made
H—H = 436 kJ/mol
Cl—Cl = 242 kJ/mol
H—Cl = 432 kJ/mol
Using Bond Dissociation Enthalpies
It takes energy to break bonds. (endothermic)
Energy is releases when bonds are made.
(exothermic)
If more energy is released than is required to break
the reactant bonds, then the over all reaction is
exothermic.
If more energy is required to break bonds than is
released when the product bonds form, then the over
all reaction is endothermic.
rH = (bonds broken) (bonds formed)
Energy Supplied ( H>0)
Energy Released ( H<0)Molecule (g) Molecular fragments (g)
Estimate the rH for the following reaction using
average bond dissociation enthalpies.
CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
Estimate the rH for the following reaction using
average bond dissociation enthalpies.
First draw out the Lewis structures to determine
the number and types of bonds
CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
Estimate the rH for the following reaction using
average bond dissociation enthalpies.
First draw out the Lewis structures to determine
the number and types of bonds
C
H
H
H H + 2 O=O O=C=O O H
H + 2
CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
Estimate the rH for the following reaction using
average bond dissociation enthalpies.
First draw out the Lewis structures to determine
the number and types of bonds
C
H
H
H H + 2 O=O O=C=O O H
H + 2
Second, count up the bonds of each type:
CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
Estimate the rH for the following reaction using
average bond dissociation enthalpies.
CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
First draw out the Lewis structures to determine
the number and types of bonds
C
H
H
H H + 2 O=O O=C=O O H
H + 2
Second, count up the bonds of each type:
4 (C–H) + 2 (O=O) 2 (C=O) + 4 (O–H)
Now sum up the bonds broken and subtract the
bonds formed.
4 (C–H) + 2 (O=O) 2 (C=O) + 4 (O–H) – [ ] rH =
Now sum up the bonds broken and subtract the
bonds formed.
4 (C–H) + 2 (O=O) 2 (C=O) + 4 (O–H) – [ ] rH =
rH = {4 (413 kJ/mol) + 2 (495 kJ/mol)}
– { 2 (745 kJ/mol + 4 (463 kJ/mol) }
Now sum up the bonds broken and subtract the
bonds formed.
4 (C–H) + 2 (O=O) 2 (C=O) + 4 (O–H) – [ ] rH =
rH = {4 (413 kJ/mol) + 2 (495 kJ/mol)}
– { 2 (745 kJ/mol + 4 (463 kJ/mol) }
= –718 kJ/mol
Now sum up the bonds broken and subtract the
bonds formed.
4 (C–H) + 2 (O=O) 2 (C=O) + 4 (O–H) – [ ] rH =
rH = {4 (413 kJ/mol) + 2 (495 kJ/mol)}
– { 2 (745 kJ/mol + 4 (463 kJ/mol) }
= –718 kJ/mol
By way of comparison, rH from fHo = –890 kJ/mol
890 ( 718)% error = 100 19.3%
890