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
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.)
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
27 Oct 97 Chemical Equilibrium 4
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)
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
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
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
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
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
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
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)
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
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)
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)
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) :
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
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
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
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
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
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
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
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
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
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?
27 Oct 97 Chemical Equilibrium 26
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
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
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?
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]
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
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
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.
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
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
27 Oct 97 Chemical Equilibrium 35
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
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
27 Oct 97 Chemical Equilibrium 37
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?
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)
27 Oct 97 Chemical Equilibrium 39
EQUILIBRIUM AND EXTERNAL EFFECTS
• Add catalyst ---> no change in K• A catalyst only affects the RATE of approach
to equilibrium.
Catalytic exhaust system
27 Oct 97 Chemical Equilibrium 40
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)