Chemical Bonding and
Molecular Structure
(Chapter 9)
• Ionic vs. covalent bonding
• Molecular orbitals and the covalent bond (Ch. 10)
• Valence electron Lewis dot structures
octet vs. non-octet
resonance structures
formal charges
• VSEPR - predicting shapes of molecules
• Bond properties
bond order, bond strength
polarity, electronegativity
27 Oct 97
Chemical Equilibrium
1
Bond Polarity
+
H
-
••
Cl ••
••
HCl is POLAR because it has a positive end
and a negative end (partly ionic).
Polarity arises because Cl has a greater share
of the bonding electrons than H.
Calculated charge by CAChe:
H (red)
is +ve (+0.20 e-)
Cl (yellow) is -ve (-0.20 e-).
(See PARTCHRG folder in MODELS.)
27 Oct 97
Chemical Equilibrium
2
+
Bond Polarity (2)
H
-
••
Cl ••
••
• Due to the bond polarity, the H—Cl bond
energy is GREATER than expected for a
“pure” covalent bond.
BOND
ENERGY
“pure” bond
339 kJ/mol calculated
real bond
432 kJ/mol measured
Difference
92 kJ/mol.
This difference is the contribution of IONIC bonding
It is proportional to the difference in
ELECTRONEGATIVITY, c.
27 Oct 97
Chemical Equilibrium
3
Electronegativity, c
c is a measure of the ability of an atom in a
molecule to attract electrons to itself.
Concept proposed by
Linus Pauling (1901-94)
Nobel prizes:
Chemistry (54), Peace (63)
See p. 425; 008vd3.mov (CD)
27 Oct 97
Chemical Equilibrium
4
4
N
3.5
Cl
C
3
2.5
F
O
H
Si
2
P
S
Electronegativity, c
Figure 9.7
1.5
1
0.5
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14 15
16
17
18
• F has maximum c.
• Atom with lowest c is the center atom in most
molecules.
• Relative values of c determines BOND
POLARITY (and point of attack on a molecule).
27 Oct 97
Chemical Equilibrium
5
Bond Polarity
Which bond is more polar ? (has larger bond DIPOLE)
O—H
O—F
c H
c(A) - c(B) 3.5 - 2.1
3.5 - 4.0
0.5
Dc
1.4
Dc(O-H) > Dc(O-F)
Therefore OH is more polar than OF
2.1
O F
3.5 4.0
Also note that polarity is “reversed.”
O—H
- +
27 Oct 97
O—F
+ -
Chemical Equilibrium
6
Molecular Polarity
• Molecules such as HCl and H2O are POLAR
• They have a DIPOLE MOMENT.
• Polar molecules turn to align their dipole with
an electric field.
POSITIVE
• A molecule will be polar
ONLY if
a) it contains polar bonds AND
b) the molecule is NOT “symmetric”
H—Cl


NEGATIVE
Symmetric molecules
27 Oct 97
Chemical Equilibrium
7
Molecular Polarity: H2O
••
O
••
H
H
polar
O
H
H
+
Water is polar because:
a) O-H bond is polar
b) water is non-symmetric
The dipole associated with polar H2O
is the basis for absorption of microwaves
used in cooking with a microwave oven
27 Oct 97
Chemical Equilibrium
8
Molecular Polarity in
NON-symmetric molecules
F
B +ve
F -ve
B
F
B
F
F
B—F bonds are polar
molecule is symmetric
BF3 is NOT polar
27 Oct 97
Atom Chg. c
H
F
B
H
F
+ve
+ve
-ve
2.0
2.1
4.0
B—F bonds are polar
molecule is NOT symmetric
HBF2 is polar
Chemical Equilibrium
9
Fluorine-substituted Ethylene: C2H2F2
C—F bonds are MUCH more polar than C—H bonds.
Dc(C-F) = 1.5, Dc(C-H) = 0.4
CIS isomer
• both C—F bonds on same side
 molecule is POLAR.
TRANS isomer
• both C—F bonds on opposite side
 molecule is NOT POLAR.
27 Oct 97
Chemical Equilibrium
10
CHEMICAL EQUILIBRIUM
Chapter 16
• equilibrium vs. completed reactions
• equilibrium constant expressions
• Reaction quotient
• computing positions of equilibria: examples
• Le Chatelier’s principle - effect on equilibria of:
• addition of reactant or product
• pressure
• temperature
YOU ARE NOT RESPONSIBLE for section 16.7 (relation to kinetics)
27 Oct 97
Chemical Equilibrium
11
Properties of an Equilibrium
Equilibrium systems are
• DYNAMIC (in constant motion)
• REVERSIBLE
• can be approached from either
direction
Co(H2O)6Cl2 (aq)
Co(H2O)6Cl2 (aq) + 2 H2O
Pink to blue
Co(H2O)6Cl2 ---> Co(H2O)4Cl2 + 2 H2O
16_CoCl2.mov
(16z01vd1.mov)
Blue to pink
Co(H2O)4Cl2 + 2 H2O ---> Co(H2O)6Cl2
27 Oct 97
Chemical Equilibrium
12
Chemical Equilibrium
Fe3+ + SCN-
FeCl3 (aq)
NaSCN(aq)
FeSCN2+
FeSCN (aq)
• After a period of time, the concentrations of
reactants and products are constant.
• The forward and reverse reactions continue after
equilibrium is attained.
16_FeSCN.mov
16m03an1.mov
27 Oct 97
Chemical Equilibrium
13
Chemical
Equilibria
CaCO3(s) + H2O(l) + CO2(g)
Ca2+(aq) + 2 HCO3-(aq)
At a given T and pressure of CO2,
[Ca2+] and [HCO3-] can be found from the
EQUILIBRIUM CONSTANT.
27 Oct 97
Chemical Equilibrium
14
THE EQUILIBRIUM CONSTANT
For any type of chemical equilibrium of the type
aA + bB
cC + dD
the following is a CONSTANT (at a given T) :
conc. of products
K =
[C]c [D]d
[A]a [B]b
equilibrium constant
conc. of reactants
If K is known, then we can predict
concentrations of products or reactants.
27 Oct 97
Chemical Equilibrium
15
Determining K
2 NOCl(g)
2 NO(g) + Cl2(g)
Place 2.00 mol of NOCl is a 1.00 L flask. At
equilibrium you find 0.66 mol/L of NO.
Calculate K.
Solution
1. Set up a table of concentrations:
[NOCl]
[NO]
[Cl2]
Before
2.00
0
0
Change
-0.66
+0.66
+0.33
Equilibrium
1.34
0.66
0.33
27 Oct 97
Chemical Equilibrium
16
Calculate K from equil. [ ]
2 NOCl(g)
Before
Change
Equilibrium
2 NO(g) + Cl2(g)
[NOCl]
[NO]
2.00
0
-0.66
+0.66
1.34
0.66
K
[Cl2]
0
+0.33
0.33
2
[NO] [Cl2 ]
[NOCl]2
2
K=
27 Oct 97
(0.66) (0.33)
(1.34)2
Chemical Equilibrium
= 0.080
17
Writing and Manipulating
Equilibrium Expressions
S
O
Solids and liquids NEVER appear
in equilibrium expressions.
S(s) + O2(g)
NH3(aq) + H2O(liq)
SO2(g)
O
[SO2 ]
K
[O2 ]
NH4+(aq) + OH-(aq)
[NH4+ ][OH- ]
K
[NH3 ]
27 Oct 97
Chemical Equilibrium
18
Manipulating K: adding reactions
Adding equations for reactions
S(s) + O2(g)
SO2(g) + 1/2 O2(g)
SO2(g)
SO3(g)
K1 = [SO2] / [O2]
K2 =
[SO3]
[SO2][O2]1/2
NET EQUATION
S(s) + 3/2 O2(g)
SO3(g)
ADD REACTIONS  MULTIPLY K
27 Oct 97
Chemical Equilibrium
Ktot =
[SO3]
[O2]3/2
Ktot = K1 x K2
19
Manipulating K: Reverse reactions
Changing direction
S(s) + O2(g)
SO2(g)
SO2(g)
S(s) + O2(g)
K
[SO2 ]
[O2 ]
[O2 ]
Knew 
[SO2 ]
[O2 ]
1
Knew 
=
[SO2 ]
Kold
27 Oct 97
Chemical Equilibrium
20
Chemistry of Sulfur
Elemental S : stable form is S8 (s)
sources:
desulfurizing natural gas
roasting metal sulfides
Oxides of S :
SO2 (g) and SO3 (g) - significant in atmospheric pollution
Industrially:
Oxides generated as needed; ‘stored’ as the hydrate
SO3 (g) + H2O (l)  H2SO4 (aq)
Sulfuric acid is HIGHEST VOLUME
chemical (fertilizers, refining, manufacturing)
27 Oct 97
Chemical Equilibrium
21
Manipulating K : Kp for gas rxns
Concentration Units
We have been writing K in terms of mol/L.
These are designated by Kc
But with gases, P = (n/V)•RT = conc • RT
P is proportional to concentration, so we can write K in
terms of PARTIAL PRESSURES.
These constants are called Kp.
Kc and Kp have DIFFERENT VALUES
(unless same number of species on both sides of equation)
27 Oct 97
Chemical Equilibrium
22
The Meaning of K
1. Can tell if a reaction is
product-favored or reactant-favored.
2 H2(g) + O2(g)
Kp =
P(H2O)2
2 H2O (g)
= 1.5 x 1080 K >> 1
P(H2)2 P(O2)
Concentration of products is much greater
than that of reactants at equilibrium.
The reaction is strongly product-favored.
27 Oct 97
Chemical Equilibrium
23
Meaning of K: AgCl rxn
AgCl(s)
Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 1.8 x 10-5 K << 1
Conc. of products is much less
than that of reactants at equilibrium.
This reaction is strongly reactant-favored.
What about the reverse reaction ?
Ag+(aq) + Cl-(aq)
AgCl(s)
Krev = Kc-1 = 5.6x104. It is strongly product-favored.
27 Oct 97
Chemical Equilibrium
24
Meaning of K : butane isomerization
2. Can tell if a reaction is at equilibrium.
If not, which way it moves to approach equilibrium.
n-butane
H H
CH3 —C —C —CH3
H H
[iso]
K =
[n]
iso-butane
H
CH3—C—CH3
= 2.5
CH3
If [iso] = 0.35 M and [n] = 0.25 M, is the system at equilibrium?
If not, which way does the rxn “shift” to approach equilibrium?
27 Oct 97
Chemical Equilibrium
25
Q - the reaction quotient
All reacting chemical systems can be characterized by their
REACTION QUOTIENT, Q.
Q has the same form as K,
. . . but uses existing concentrations
If Q = K, then system is at equilibrium.
0.35
[iso]
For
Q=
=
= 1.40
n-Butane
iso-Butane
[n]
0.25
Q = 1.4 which is LESS THAN K =2.5
Reaction is NOT at equilibrium. To reach EQUILIBRIUM
[Iso] must INCREASE and [n] must DECREASE.
27 Oct 97
Chemical Equilibrium
26
Typical EQUILIBRIUM Calculations
2 general types:
a. Given set of concentrations, Calculate Q
is system at equilibrium ? compare to K
IF:
Q > K or Q/K > 1
 REACTANTS
Q/K
1
Q < K or Q/K < 1
 PRODUCTS
Q
Q=K
27 Oct 97
Q=K
Chemical Equilibrium
at EQUILIBRIUM
27
Examples of equilibrium questions
b. From an initial non-equilibrium condition,
what are the concentrations at equilibrium?
H2(g) + I2(g)
2 HI(g)
Place 1.00 mol each of H2 and I2 in a 1.00 L
flask. Calculate equilibrium concentrations.
27 Oct 97
Chemical Equilibrium
28
H2(g) + I2(g)
2 HI(g)
2
[HI]
Kc =
[H2 ][I2 ]
Step 1.
Kc = 55.3
= 55.3
Set up table to define EQUILIBRIUM concentrations
in terms of initial concentrations and a change variable
Initial
[H2]
[I2]
[HI]
1.00
1.00
0
DEFINE x = [H2] consumed to get to equilibrium.
Change
-x
At equilibrium 1.00-x
27 Oct 97
-x
+2x
1.00-x
2x
Chemical Equilibrium
29
H2(g) + I2(g)
2 HI(g)
Kc = 55.3
Step 1 Define equilibrium condition in terms of initial condition
and a change variable
[H2]
At equilibrium 1.00-x
[I2]
1.00-x
[HI]
2x
Step 2
Put equilibrium concentrations into Kc expression.
Kc =
27 Oct 97
2
[2x]
[1.00 - x][1.00 - x]
Chemical Equilibrium
= 55.3
30
H2(g) + I2(g)
2 HI(g)
Kc = 55.3
Step 3. Solve for x. 55.3 = (2x)2/(1-x)2
In this case, take square root of both sides.
2x
7.44=
1.00 - x
Solution gives: x = 0.79
Therefore, at equilibrium
[H2] = [I2] = 1.00 - x = 0.21 M
[HI] = 2x = 1.58 M
27 Oct 97
Chemical Equilibrium
31
EQUILIBRIUM AND EXTERNAL EFFECTS
• The position of equilibrium is changed when
there is a change in:
– pressure
– changes in concentration
– temperature
• The outcome is governed by
LE CHATELIER’S PRINCIPLE
“...if a system at equilibrium is disturbed, the
system tends to shift its equilibrium position
to counter the effect of the disturbance.”
27 Oct 97
Chemical Equilibrium
Henri Le Chatelier
1850-1936
- Studied mining
engineering
- specialized in glass
and ceramics.
32
Shifts in EQUILIBRIUM : Concentration
• If concentration of one species changes,
concentrations of other species CHANGES
to keep the value of K the same (at constant T)
• no change in K - only position of equilibrium changes.
ADDING PRODUCTS
- equilibrium shifts to REACTANTS
ADDING REACTANTS
- equilibrium shifts to PRODUCTS
- GAS-FORMING; PRECIPITATION
REMOVING PRODUCTS
- often used to DRIVE REACTION TO COMPLETION
27 Oct 97
Chemical Equilibrium
33
Effect of changed [ ] on an equilibrium
K =
n-Butane
Isobutane
[iso]
[n]
= 2.5
INITIALLY: [n] = 0.50 M
[iso] = 1.25 M
CHANGE: ADD +1.50 M n-butane
What happens ?
Solution
A. Calculate Q with extra 1.50 M n-butane.
16_butane.mov
Q = [iso] / [n] = 1.25 / (0.50 + 1.50) = 0.63(16m13an1.mov)
Q < K . Therefore, reaction shifts to PRODUCT
27 Oct 97
Chemical Equilibrium
34
Butane/Isobutane
Solution
B. Solve for NEW EQUILIBRIUM
- set up concentration table
[n-butane]
[isobutane]
Initial
0.50 + 1.50
1.25
Change
-x
+x
Equilibrium 2.00 - x
1.25 + x
B
A
[isobutane]
1.25 + x
K = 2.50 =

[butane]
2.00 - x
x = 1.07 M. At new equilibrium position,
[n-butane] = 0.93 M [isobutane] = 2.32 M.
Equilibrium has shifted toward isobutane.
27 Oct 97
Chemical Equilibrium
35
Effect of Pressure (gas equilibrium)
N2O4(g)
2 NO2(g)
[NO2 ]2
Kc =
= 0.0059 at 298 K
[N2O4 ]
Increase P in the system
by reducing the volume.
Increasing P shifts equilibrium to side
with fewer molecules (to try to reduce P).
Here, reaction shifts LEFT
PN2O4 increases
16_NO2.mov
(16m14an1.mov)
PNO2 decreases
See Ass#2 - question #6
27 Oct 97
Chemical Equilibrium
36
EQUILIBRIUM AND EXTERNAL EFFECTS
• Temperature change  change in
• Consider the fizz in a soft drink
CO2(g) + H2O(liq)
CO2(aq) + heat
LOWER T
K
Kc = [CO2(aq)]/[CO2(g)]
HIGHER T
• Change T: New equilib. position? New value of K?
• Increase T
Equilibrium shifts left: [CO2(g)]  [CO2 (aq)] 
K decreases as T goes up.
•Decrease T
[CO2 (aq)] increases and [CO2(g)] decreases.
K increases as T goes down
27 Oct 97
Chemical Equilibrium
37
Temperature Effects on
Chemical Equilibrium
N2O4 + heat
(colorless)
2 NO2
(brown)
[NO2 ]2 K = 0.00077 at 273 K
c
Kc 
[N2O 4 ] K = 0.00590 at 298 K
c
DHorxn = + 57.2 kJ
Increasing T changes K so as to
shift equilibrium in
ENDOTHERMIC direction
27 Oct 97
Chemical Equilibrium
16_NO2RX.mov
(16m14an1.mov)
38
EQUILIBRIUM AND EXTERNAL EFFECTS
Catalytic exhaust system
• Add catalyst ---> no change in
• A catalyst only affects the RATE of
approach to equilibrium.
27 Oct 97
Chemical Equilibrium
K
39
CHEMICAL EQUILIBRIUM
Chapter 16
• equilibrium vs. completed reactions
• equilibrium constant expressions
• Reaction quotient
• computing positions of equilibria: examples
• Le Chatelier’s principle - effect on equilibria of:
• addition of reactant or product
• pressure
• temperature
YOU ARE NOT RESPONSIBLE for section 16.7 (relation to kinetics)
27 Oct 97
Chemical Equilibrium
40