Chemical Bonding and Molecular Structure (Chapter 9)
description
Transcript of Chemical Bonding and Molecular Structure (Chapter 9)
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27 Oct 97 Chemical Equilibrium 1
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-octetresonance structuresformal charges
• VSEPR - predicting shapes of molecules• Bond properties
bond order, bond strengthpolarity, electronegativity
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27 Oct 97 Chemical Equilibrium 2
Bond 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.
Cl
-+
•••H••
••
Calculated charge by CAChe:
H (red) is +ve (+0.20 e-)
Cl (yellow) is -ve (-0.20 e-).
(See PARTCHRG folder in MODELS.)
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27 Oct 97 Chemical Equilibrium 3
• Due to the bond polarity, the H—Cl bond energy is GREATER than expected for a “pure” covalent bond.
Cl
-+
•••H••
••
Bond Polarity (2)
BOND ENERGY
“pure” bond 339 kJ/mol calculated
real bond 432 kJ/mol measured
ELECTRONEGATIVITY, .
Difference 92 kJ/mol.
This difference is the contribution of IONIC bondingIt is proportional to the difference in
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Electronegativity,
is a measure of the ability of an atom in a molecule to attract electrons to itself.
Concept proposed byLinus Pauling (1901-94)Nobel prizes:Chemistry (54), Peace (63)
See p. 425; 008vd3.mov (CD)
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27 Oct 97 Chemical Equilibrium 5
• F has maximum .
• Atom with lowest is the center atom in most molecules.
• Relative values of determines BOND POLARITY (and point of attack on a molecule).
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 180
0.5
1
1.5
2
2.5
3
3.5
4
H
FCl
CN
O
SP
Si
Electronegativity,
Figure 9.7
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27 Oct 97 Chemical Equilibrium 6
Bond Polarity
(A) - (B) 3.5 - 2.1
1.4
+ -+-O—FO—H
H 2.1O F3.5 4.0
Also note that polarity is “reversed.”
Which bond is more polar ? (has larger bond DIPOLE)
O—H O—F
3.5 - 4.0 0.5
(O-H) > (O-F) Therefore OH is more polar than OF
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27 Oct 97 Chemical Equilibrium 7
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
NEGATIVE
H—Cl
POSITIVE
NEGATIVE
H—Cl • A molecule will be polar
ONLY if
a) it contains polar bonds ANDb) the molecule is NOT “symmetric”
Symmetric molecules
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27 Oct 97 Chemical Equilibrium 8
H
HH H
O
••
••
O
+polar
Molecular Polarity: H2O
Water is polar because:
a) O-H bond is polarb) water is non-symmetric
The dipole associated with polar H2O is the basis for absorption of microwaves used in cooking with a microwave oven
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27 Oct 97 Chemical Equilibrium 9
B—F bonds are polarmolecule is NOT symmetric
B—F bonds are polarmolecule is symmetric
Molecular Polarity in NON-symmetric molecules
F
F FB
B +ve F -ve
H
F FB
Atom Chg. B +ve 2.0H +ve 2.1F -ve 4.0
BF3 is NOT polar HBF2 is polar
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27 Oct 97 Chemical Equilibrium 10
Fluorine-substituted Ethylene: C2H2F2
CIS isomer
• both C—F bonds on same side
molecule is POLAR.
C—F bonds are MUCH more polar than C—H bonds.
TRANS isomer
• both C—F bonds on opposite side
molecule is NOT POLAR.
(C-F) = 1.5, (C-H) = 0.4
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27 Oct 97 Chemical Equilibrium 11
CHEMICAL EQUILIBRIUMChapter 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)
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27 Oct 97 Chemical Equilibrium 12
16_CoCl2.mov(16z01vd1.mov)
Properties of an Equilibrium
Equilibrium systems are• DYNAMIC (in constant motion)• REVERSIBLE • can be approached from either
direction
Pink to blueCo(H2O)6Cl2 ---> Co(H2O)4Cl2 + 2 H2O
Blue to pinkCo(H2O)4Cl2 + 2 H2O ---> Co(H2O)6Cl2
Co(H2O)6Cl2 (aq) Co(H2O)6Cl2 (aq) + 2 H2O
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27 Oct 97 Chemical Equilibrium 13
Chemical Equilibrium
• After a period of time, the concentrations of reactants and products are constant.
• The forward and reverse reactions continue after equilibrium is attained.
Fe3+ + SCN- FeSCN2+
16_FeSCN.mov16m03an1.mov
FeCl3 (aq) NaSCN(aq)
FeSCN (aq)
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27 Oct 97 Chemical Equilibrium 14
CaCO3(s) + H2O(l) + CO2(g)
Chemical Equilibria
At a given T and pressure of CO2,
[Ca2+] and [HCO3-] can be found from the
EQUILIBRIUM CONSTANT.
Ca2+(aq) + 2 HCO3-(aq)
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27 Oct 97 Chemical Equilibrium 15
For any type of chemical equilibrium of the type
THE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANT
K =[C]c [D]d
[A]a [B]b
conc. of products
conc. of reactantsequilibrium constant
If K is known, then we can predict concentrations of products or reactants.
a A + b B c C + d D
the following is a CONSTANT (at a given T) :
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27 Oct 97 Chemical Equilibrium 16
Determining 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]
2 NOCl(g) 2 NO(g) + Cl2(g)
Before 2.00 0 0Change -0.66 +0.66 +0.33Equilibrium 1.34 0.66 0.33
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27 Oct 97 Chemical Equilibrium 17
K [NO]2[Cl2 ]
[NOCl]2
Calculate 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
K = (0.66)2(0.33)
(1.34)2 = 0.080
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27 Oct 97 Chemical Equilibrium 18
Writing and ManipulatingEquilibrium Expressions
Solids and liquids NEVER appear in equilibrium expressions.
K [SO2 ][O2 ]
S(s) + O2(g) SO2(g)
K [NH4
+][OH- ][NH3 ]
NH3(aq) + H2O(liq) NH4+(aq) + OH-(aq)
S
O O
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27 Oct 97 Chemical Equilibrium 19
Ktot = K1 x K2
S(s) + 3/2 O2(g) SO3(g)
Adding equations for reactions
S(s) + O2(g) SO2(g)
Manipulating K: adding reactions
NET EQUATION
SO2(g) + 1/2 O2(g) SO3(g) [SO3]
[SO2][O2]1/2K2 =
K1 = [SO2] / [O2]
[SO3]
[O2]3/2Ktot =
ADD REACTIONS MULTIPLY K
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27 Oct 97 Chemical Equilibrium 20
Changing directionK
[SO2 ]
[O2 ]
Knew [O2 ][SO2 ]
= 1
Kold
Knew [O2 ][SO2 ]
S(s) + O2(g) SO2(g)
SO2(g) S(s) + O2(g)
Manipulating K: Reverse reactions
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27 Oct 97 Chemical Equilibrium 21
Chemistry of SulfurElemental S : stable form is S8 (s)
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)
sources: desulfurizing natural gas
roasting metal sulfides
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27 Oct 97 Chemical Equilibrium 22
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)
Manipulating K : Kp for gas rxns
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27 Oct 97 Chemical Equilibrium 23
Concentration of products is much greater than that of reactants at equilibrium.
The Meaning of K1. Can tell if a reaction is
product-favored or reactant-favored.
K =p P(H2O)2
P(H2)2 P(O2) = 1.5 x 1080
2 H2(g) + O2(g) 2 H2O (g)
The reaction is strongly product-favored.
K >> 1
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27 Oct 97 Chemical Equilibrium 24
What about the reverse reaction ?
This reaction is strongly reactant-favored.
Conc. of products is much less
than that of reactants at equilibrium.
Meaning of K: AgCl rxn
Kc = K << 1
Ag+(aq) + Cl-(aq) AgCl(s)
Krev = Kc-1 = 5.6x104. It is strongly product-favored.
AgCl(s) Ag+(aq) + Cl-(aq)
[Ag+] [Cl-] = 1.8 x 10-5
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27 Oct 97 Chemical Equilibrium 25
CH3 —C —C —CH3
H
H
H
H
n-butane
HCH3—C—CH3
iso-butane
CH3
2. Can tell if a reaction is at equilibrium. If not, which way it moves to approach equilibrium.
Meaning of K : butane isomerization
K = [iso]
[n] = 2.5
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?
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All reacting chemical systems can be characterized by their REACTION QUOTIENT, Q.
Q = =0.350.25
= 1.40
Q - the reaction quotient
If Q = K, then system is at equilibrium.
Reaction is NOT at equilibrium.
[Iso] must INCREASE and [n] must DECREASE.
Q = 1.4 which is LESS THAN K =2.5
To reach EQUILIBRIUM
[iso][n]
Q has the same form as K, . . . but uses existing concentrations
For n-Butane iso-Butane
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27 Oct 97 Chemical Equilibrium 27
Q/K
Q
Typical EQUILIBRIUM Calculations2 general types: a. Given set of concentrations, is system at equilibrium ?
Calculate Q compare to K
1
Q = K
IF:Q > K or Q/K > 1 REACTANTS
Q < K or Q/K < 1 PRODUCTS
Q=K at EQUILIBRIUM
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27 Oct 97 Chemical Equilibrium 28
Examples of equilibrium questions
Place 1.00 mol each of H2 and I2 in a 1.00 L flask. Calculate equilibrium concentrations.
H2(g) + I2(g) 2 HI(g)
b. From an initial non-equilibrium condition, what are the concentrations at equilibrium?
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27 Oct 97 Chemical Equilibrium 29
H2(g) + I2(g) 2 HI(g) Kc = 55.3
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
Kc = [HI]2
[H2 ][I2 ] = 55.3
Step 1. Set up table to define EQUILIBRIUM concentrationsin terms of initial concentrations and a change variable
[H2] [I2] [HI]
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27 Oct 97 Chemical Equilibrium 30
Step 2 Put equilibrium concentrations into Kc expression.
Kc = [2x]2
[1.00 - x][1.00 - x] = 55.3
Step 1 Define equilibrium condition in terms of initial condition and a change variable
[H2] [I2] [HI]
At equilibrium 1.00-x 1.00-x 2x
H2(g) + I2(g) 2 HI(g) Kc = 55.3
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27 Oct 97 Chemical Equilibrium 31
[H2] = [I2] = 1.00 - x = 0.21 M
[HI] = 2x = 1.58 M
Step 3. Solve for x. 55.3 = (2x)2/(1-x)2
In this case, take square root of both sides.
7.44=2x
1.00 x-
H2(g) + I2(g) 2 HI(g) Kc = 55.3
Solution gives: x = 0.79Therefore, at equilibrium
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27 Oct 97 Chemical Equilibrium 32
“...if a system at equilibrium is disturbed, the
system tends to shift its equilibrium position
to counter the effect of the disturbance.”
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 PRINCIPLEHenri Le Chatelier
1850-1936
- Studied mining engineering
- specialized in glass and ceramics.
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27 Oct 97 Chemical Equilibrium 33
• If concentration of one species changes, concentrations of other species CHANGESto keep the value of K the same (at constant T)
• no change in K - only position of equilibrium changes.
Shifts in EQUILIBRIUM : Concentration
ADDING REACTANTS- equilibrium shifts to PRODUCTS
ADDING PRODUCTS- equilibrium shifts to REACTANTS
REMOVING PRODUCTS - often used to DRIVE REACTION TO COMPLETION
- GAS-FORMING; PRECIPITATION
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27 Oct 97 Chemical Equilibrium 34
Solution
A. Calculate Q with extra 1.50 M n-butane.
INITIALLY: [n] = 0.50 M [iso] = 1.25 M
CHANGE: ADD +1.50 M n-butane What happens ?What happens ?
Q < K . Therefore, reaction shifts to PRODUCT
Q = [iso] / [n] = 1.25 / (0.50 + 1.50) = 0.63
n-Butane Isobutane
Effect of changed [ ] on an equilibrium
16_butane.mov(16m13an1.mov)
K = [iso]
[n] = 2.5
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Solution
B. Solve for NEW EQUILIBRIUM - set up concentration table [n-butane] [isobutane]Initial 0.50 + 1.50 1.25Change - x + xEquilibrium 2.00 - x 1.25 + x
Butane/Isobutane
K = 2.50 = [isobutane]
[butane]
1.25 + x2.00 - x
x = 1.07 M. At new equilibrium position, [n-butane] = 0.93 M [isobutane] = 2.32 M.
Equilibrium has shifted toward isobutane.
AB
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27 Oct 97 Chemical Equilibrium 36
Effect of Pressure (gas equilibrium)
Increase P in the system by reducing the volume.
Kc = [NO2 ]2
[N2O4 ] = 0.0059 at 298 K
NN22OO44(g) (g) 2 NO 2 NO22(g)(g)
16_NO2.mov(16m14an1.mov)
Increasing P shifts equilibrium to side with fewer molecules (to try to reduce P). Here, reaction shifts LEFT PN2O4 increases
See Ass#2 - question #6
PNO2 decreases
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EQUILIBRIUM AND EXTERNAL EFFECTS
• Temperature change change in K• Consider the fizz in a soft drinkCO2(g) + H2O(liq) CO2(aq) + heat
• Increase TEquilibrium 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
Kc = [CO2(aq)]/[CO2(g)]HIGHER T
LOWER T
• Change T: New equilib. position? New value of K?
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27 Oct 97 Chemical Equilibrium 38
Temperature Effects on Chemical Equilibrium
Kc = 0.00077 at 273 K
Kc = 0.00590 at 298 KKc
[NO2 ]2
[N2O4 ]
N2O4 + heat 2 NO2 (colorless) (brown)
Horxn = + 57.2 kJ
Increasing T changes K so as to shift equilibrium in ENDOTHERMIC direction
16_NO2RX.mov(16m14an1.mov)
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EQUILIBRIUM AND EXTERNAL EFFECTS
• Add catalyst ---> no change in K• A catalyst only affects the RATE of approach
to equilibrium.
Catalytic exhaust system
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CHEMICAL EQUILIBRIUMChapter 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)