Title: Chemical Bonding and Molecular Structure (Chapter 9)
1Chemical 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
2Bond Polarity
- 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.)
3Bond Polarity (2)
- Due to the bond polarity, the HCl 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.
4Electronegativity, 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)
5Electronegativity, c
Figure 9.7
- 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).
6Bond Polarity
Which bond is more polar ? (has larger bond
DIPOLE) OH OF
c H 2.1 O F 3.5 4.0
- c(A) - c(B) 3.5 - 2.1
- Dc 1.4
3.5 - 4.0 0.5
??(O-H) gt ??(O-F) Therefore OH is more polar
than OF
Also note that polarity is reversed.
7Molecular 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.
- A molecule will be polar
- ONLY if
- a) it contains polar bonds AND
- b) the molecule is NOT symmetric
8Molecular Polarity H2O
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
9Molecular Polarity in NON-symmetric molecules
BF bonds are polar molecule is NOT
symmetric
- BF bonds are polar
- molecule is symmetric
-
Atom Chg. ? B ve 2.0 H ve
2.1 F -ve 4.0
B ve F -ve
10Fluorine-substituted Ethylene C2H2F2
CF bonds are MUCH more polar than CH bonds.
??(C-F) 1.5, ??(C-H) 0.4
- CIS isomer
- both CF bonds on same side
- ? molecule is POLAR.
- TRANS isomer
- both CF bonds on opposite side
- ? molecule is NOT POLAR.
11CHEMICAL EQUILIBRIUMChapter 16
- equilibrium vs. completed reactions
- equilibrium constant expressions
- Reaction quotient
- computing positions of equilibria examples
- Le Chateliers principle - effect on equilibria
of - addition of reactant or product
- pressure
- temperature
YOU ARE NOT RESPONSIBLE for section 16.7
(relation to kinetics)
12Properties of an Equilibrium
- Equilibrium systems are
- DYNAMIC (in constant motion)
- REVERSIBLE
- can be approached from either direction
16_CoCl2.mov (16z01vd1.mov)
Pink to blue Co(H2O)6Cl2 ---gt Co(H2O)4Cl2 2
H2O
Blue to pink Co(H2O)4Cl2 2 H2O ---gt Co(H2O)6Cl2
13Chemical Equilibrium
- 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
14Chemical Equilibria
Ca2(aq) 2 HCO3-(aq)
At a given T and pressure of CO2, Ca2 and
HCO3- can be found from the EQUILIBRIUM
CONSTANT.
15THE EQUILIBRIUM CONSTANT
- For any type of chemical equilibrium of the type
the following is a CONSTANT (at a given T)
If K is known, then we can predict
concentrations of products or reactants.
16Determining K
- 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 Equ
ilibrium 1.34 0.66 0.33
17Calculate K from equil.
- 2 NOCl(g) 2 NO(g) Cl2(g)
- NOCl NO Cl2
- Before 2.00 0 0
- Change -0.66 0.66 0.33
- Equilibrium 1.34 0.66 0.33
18Writing and ManipulatingEquilibrium Expressions
- Solids and liquids NEVER appear in equilibrium
expressions.
19Manipulating K adding reactions
- Adding equations for reactions
K1 SO2 / O2
NET EQUATION
Ktot
Ktot K1 x K2
ADD REACTIONS ? MULTIPLY K
20Manipulating K Reverse reactions
21Chemistry 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)
22Manipulating 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)
23The Meaning of K
- 1. Can tell if a reaction is
- product-favored or reactant-favored.
1.5 x 1080
K gtgt 1
Concentration of products is much greater than
that of reactants at equilibrium.
The reaction is strongly product-favored.
24Meaning of K AgCl rxn
Kc
K ltlt 1
Ag Cl-
1.8 x 10-5
Conc. of products is much less than that of
reactants at equilibrium.
This reaction is strongly reactant-favored.
What about the reverse reaction ?
Krev Kc-1 5.6x104. It is strongly
product-favored.
25Meaning of K butane isomerization
- 2. Can tell if a reaction is at equilibrium.
- If not, which way it moves to approach
equilibrium.
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?
26Q - 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.
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.
27Typical EQUILIBRIUM Calculations
2 general types a. Given set of
concentrations, is system at
equilibrium ?
Calculate Q compare to K
IF Q gt K or Q/K gt 1 ?
REACTANTS Q lt K or Q/K lt 1 ?
PRODUCTS
QK at EQUILIBRIUM
28Examples of equilibrium questions
b. From an initial non-equilibrium condition,
what are the concentrations at equilibrium?
Place 1.00 mol each of H2 and I2 in a 1.00 L
flask. Calculate equilibrium concentrations.
29H2(g) I2(g) 2 HI(g) Kc 55.3
Step 1. Set up table to define EQUILIBRIUM
concentrations in terms of initial concentrations
and a change variable
H2 I2 HI
- Initial 1.00 1.00 0
- DEFINE x H2 consumed to get to equilibrium.
- Change -x -x 2x
- At equilibrium 1.00-x 1.00-x 2x
30H2(g) I2(g) 2 HI(g) Kc 55.3
Step 1 Define equilibrium condition in terms of
initial condition and a change variable H2 I
2 HI At equilibrium 1.00-x 1.00-x 2x
- Step 2
- Put equilibrium concentrations into Kc expression.
31H2(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.
Solution gives x 0.79 Therefore, at
equilibrium
H2 I2 1.00 - x 0.21 M HI 2x
1.58 M
32EQUILIBRIUM 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 CHATELIERS PRINCIPLE
Henri Le Chatelier 1850-1936 - Studied mining
engineering - specialized in glass and ceramics.
...if a system at equilibrium is disturbed, the
system tends to shift its equilibrium position to
counter the effect of the disturbance.
33Shifts 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
34Effect of changed on an equilibrium
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 (16m13an1.mov)
Q iso / n 1.25 / (0.50 1.50) 0.63
Q lt K . Therefore, reaction shifts to PRODUCT
35Butane/Isobutane
B. Solve for NEW EQUILIBRIUM - set up
concentration table n-butane isobutane Ini
tial 0.50 1.50 1.25 Change - x
x Equilibrium 2.00 - x 1.25 x
x 1.07 M. At new equilibrium position,
n-butane 0.93 M isobutane 2.32 M.
Equilibrium has shifted toward isobutane.
36Effect of Pressure (gas equilibrium)
- 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 Ass2 - question 6
37EQUILIBRIUM AND EXTERNAL EFFECTS
- Temperature change ? change in K
- Consider the fizz in a soft drink
Kc CO2(aq)/CO2(g)
- 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
38Temperature Effects on Chemical Equilibrium
- Kc 0.00077 at 273 K
- Kc 0.00590 at 298 K
?Horxn 57.2 kJ
Increasing T changes K so as to shift equilibrium
in ENDOTHERMIC direction
16_NO2RX.mov (16m14an1.mov)
39EQUILIBRIUM AND EXTERNAL EFFECTS
Catalytic exhaust system
- Add catalyst ---gt no change in K
- A catalyst only affects the RATE of approach to
equilibrium.
40CHEMICAL EQUILIBRIUMChapter 16
- equilibrium vs. completed reactions
- equilibrium constant expressions
- Reaction quotient
- computing positions of equilibria examples
- Le Chateliers principle - effect on equilibria
of - addition of reactant or product
- pressure
- temperature
YOU ARE NOT RESPONSIBLE for section 16.7
(relation to kinetics)