Transcript Sulfur Dioxide, SO2 - McMaster University
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
+
H
Bond Polarity
-
•• 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
•
Due to the bond polarity, the H—Cl bond energy is GREATER than expected for a “pure” covalent bond.
BOND “pure” bond ENERGY 339 kJ/mol calculated real bond
Difference
432 kJ/mol measured
92 kJ/mol. This difference is the contribution of IONIC bonding It is proportional to the difference in
ELECTRONEGATIVITY,
c
.
-
•• Cl •• •••
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 3.5
3 2.5
2 1.5
1 0.5
0
• • •
H C N O F Si P S Cl
Electronegativity, Figure 9.7
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).
c 27 Oct 97 Chemical Equilibrium 5
Bond Polarity
Which bond is more polar ? (has larger bond DIPOLE) O —H O —F
c (A) Dc c (B) 3.5 - 2.1
3.5 - 4.0
1.4
Dc
(O-H) >
Dc
(O-F) 0.5
Therefore OH is more polar than OF
c
H 2.1
O F 3.5 4.0
Also note that polarity is “reversed.” O—H
+
O—F +
-
27 Oct 97 Chemical Equilibrium 6
Molecular Polarity
• • • •
Molecules such as HCl and H They have a 2 O are DIPOLE MOMENT . POLAR Polar molecules turn to align their dipole with an electric field.
POSITIVE A molecule will be polar H—Cl
NEGATIVE ONLY if a) it contains polar bonds AND b) the molecule is NOT “symmetric” Symmetric molecules
27 Oct 97 Chemical Equilibrium 7
•• O •• H polar H Molecular Polarity: H 2 O H O + H Water is polar because: a) O-H bond is polar b) water is non-symmetric
The dipole associated with polar H 2 O is the basis for absorption of microwaves used in cooking with a microwave oven 27 Oct 97 Chemical Equilibrium 8
F F B
Molecular Polarity in NON-symmetric molecules
F B +ve F -ve F H B F Atom Chg.
c
B +ve 2.0
H +ve 2.1
F -ve 4.0
B—F bonds are polar molecule is symmetric BF 3 is NOT polar B—F bonds are polar molecule is NOT symmetric HBF 2 is polar
27 Oct 97 Chemical Equilibrium 9
Fluorine-substituted Ethylene: C 2 H 2 F 2 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 .
27 Oct 97 TRANS isomer • both C—F bonds on opposite side molecule is NOT POLAR .
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(H 2 O) 6 Cl 2
(aq)
Co(H 2 O) 6 Cl 2
(aq) + 2 H
2 O Pink to blue Co(H 2 O) 6 Cl 2 ---> Co(H 2 O) 4 Cl 2 + 2 H 2 O Blue to pink Co(H 2 O) 4 Cl 2 + 2 H 2 O ---> Co(H 2 O) 6 Cl 2 16_CoCl2.mov
(16z01vd1.mov) 27 Oct 97 Chemical Equilibrium 12
FeCl 3 (aq) NaSCN(aq)
Chemical Equilibrium
Fe 3+ + SCN FeSCN 2+
• •
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
CaCO 3
(s) + H
2
O(l) + CO
2
(g)
Ca 2+
(aq)
+ 2
HCO 3 -
(aq)
At a given T and pressure of CO 2 , [Ca 2+ ] and [HCO 3 ] 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 a A + b B c C + d D
the following is a CONSTANT (at a given T) : conc. of products [C]c [D]d K = [A]a [B]b conc. of reactants equilibrium constant 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) + Cl 2 (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: Before Change Equilibrium [NOCl] 2.00
-0.66
1.34
[NO] 0 +0.66
0.66
[Cl 2 ] 0 +0.33
0.33
27 Oct 97 Chemical Equilibrium 16
Calculate K from equil. [ ]
2 NOCl(g) Before Change Equilibrium 2 NO(g) + Cl 2 (g) [NOCl] [NO] 2.00
0 -0.66
1.34
+0.66
0.66
K
[NO] 2 [Cl 2 ] [NOCl] 2 K = (0.66) 2 (0.33) (1.34) 2 = 0.080
27 Oct 97 Chemical Equilibrium
[Cl 2 ] 0 +0.33
0.33
17
Writing and Manipulating Equilibrium Expressions
Solids and liquids NEVER appear in equilibrium expressions.
S(s) + O 2 (g) SO 2 (g) O S O K
[SO 2 ] [O 2 ] NH 3 (aq) + H 2 O(liq) NH 4 + (aq) + OH (aq)
27 Oct 97
K
[NH 4 + ][OH ] [NH 3 ]
Chemical Equilibrium 18
Manipulating K: adding reactions
Adding equations for reactions
SO 2 S(s) + O 2 (g) + 1/2 O (g) SO 2 (g) SO 2 3 (g) (g) K K 1 = [SO 2 ] / [O 2 ] 2 = [SO 3 ] [SO 2 ][O 2 ] 1/2 NET EQUATION S(s) + 3/2 O 2 (g) SO 3 (g) ADD REACTIONS
MULTIPLY K [SO 3 ] K tot = [O 2 ] 3/2 K tot = K 1 x K 2
27 Oct 97 Chemical Equilibrium 19
Manipulating K: Reverse reactions
Changing direction S(s) + O 2 (g) SO 2 (g) SO 2 (g) S(s) + O 2 (g) K
[SO 2 ] [O 2 ] K new
[O 2 ] [SO 2 ] K new
[O 2 ] [SO 2 ] = 1 K old
27 Oct 97 Chemical Equilibrium 20
Chemistry of Sulfur
Elemental S : stable form is S 8 (s) sources: desulfurizing natural gas roasting metal sulfides Oxides of S : SO 2 (g) and SO 3 (g) - significant in atmospheric pollution Industrially: Oxides generated as needed; ‘stored’ as the hydrate SO 3
(g) + H
2
O (l)
H 2 SO 4
(aq)
Sulfuric acid is HIGHEST VOLUME chemical (fertilizers, refining, manufacturing)
27 Oct 97 Chemical Equilibrium 21
Manipulating K : K p for gas rxns Concentration Units We have been writing K in terms of mol/L. These are designated by
K
c 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
K
p . K c and K p 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 H 2 (g) + O 2 (g) 2 H 2 O (g) K p = P (H 2 O) 2 P (H 2 ) 2 P (O 2 ) = 1.5 x 10 80 K >> 1 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)
K
c
= [Ag
+
] [Cl
-
] = 1.8 x 10
-5 K << 1 Conc. of products is
much less
than that o f reactants at equilibrium. This reaction is strongly
reactant-favored
.
What about the reverse reaction ?
Ag + (aq) + Cl (aq) AgCl(s) K rev = K c -1 = 5.6x10
4 . 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 CH 3 —C —C —CH 3 H H [iso] K = [n] = 2.5
iso-butane CH 3 H —C—CH CH 3 3 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
For n-Butane iso-Butane Q = [iso] [n] = 0.25
= 1.40
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, is system at equilibrium ? Calculate Q compare to K Q/K 1
27 Oct 97
IF: Q > K or Q/K > 1
REACTANTS Q < K or Q/K < 1
PRODUCTS Q = K Q
Q=K at EQUILIBRIUM
Chemical Equilibrium 27
Examples of equilibrium questions
b. From an initial non-equilibrium condition, what are the concentrations at equilibrium?
H
2
(g) + I
2
(g) 2 HI(g) Place 1.00 mol each of H
2
and I
2
in a 1.00 L flask. Calculate equilibrium concentrations.
27 Oct 97 Chemical Equilibrium 28
H 2 (g) + I 2 (g) 2 HI(g) K c = 55.3
K c = [HI] 2 [H 2 ][I 2 ] = 55.3
Step 1. Set up table to define EQUILIBRIUM concentrations in terms of initial concentrations and a change variable Initial [H 2 ] 1.00
[I 2 ] 1.00
[HI] 0 DEFINE x = [H 2 ] consumed to get to equilibrium.
Change -x -x +2x At equilibrium 1.00-x
27 Oct 97
1.00-x
Chemical Equilibrium
2x
29
H 2 (g) + I 2 (g) 2 HI(g) K c = 55.3
Step 1 Define equilibrium condition in terms of initial condition and a change variable
At equilibrium [H 2 ] 1.00-x [I 2 ] 1.00-x [HI] 2x
Step 2
Put equilibrium concentrations into K c expression.
K c = [2x] 2 [1.00 - x][1.00 - x] = 55.3
27 Oct 97 Chemical Equilibrium 30
H 2 (g) + I 2 (g) 2 HI(g) K c = 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= Solution gives: x = 0.79
Therefore, at equilibrium [H 2 ] = [I 2 ] = 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.” Henri Le Chatelier 1850-1936 - Studied mining engineering - specialized in glass and ceramics.
27 Oct 97 Chemical Equilibrium 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 REMOVING PRODUCTS - GAS-FORMING; PRECIPITATION - often used to DRIVE REACTION TO COMPLETION
27 Oct 97 Chemical Equilibrium 33
Effect of changed [ ] on an equilibrium n-Butane Isobutane
K = [iso] [n] = 2.5
INITIALLY: [n] = 0.50 M [iso] = 1.25 M CHANGE: ADD +1.50 M n-butane Solution A. Calculate Q with extra 1.50 M n-butane.
What happens ?
16_butane.mov
Q = [iso] / [n] = 1.25 / (0.50 + 1.50) = 0.63
Q < K . Therefore, reaction shifts to PRODUCT
27 Oct 97 Chemical Equilibrium 34
Butane/Isobutane
Solution
A B. Solve for NEW EQUILIBRIUM - set up concentration table [n-butane] [isobutane] Initial Change 0.50 + 1.50
- x 1.25
+ x Equilibrium 2.00 - x
K = 2.50 = [isobutane] [butane]
1.25 + x
1.25 + x 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 B
Effect of Pressure (gas equilibrium) N 2 O 4 (g) 2 NO 2 (g)
K c = [NO 2 ] 2 [N 2 O 4 ] = 0.0059 at 298 K
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 P N 2 O 4 increases
16_NO2.mov
(16m14an1.mov) P NO 2 decreases See Ass#2 - question #6 27 Oct 97 Chemical Equilibrium 36
• • EQUILIBRIUM AND EXTERNAL EFFECTS
Temperature change
change in K
Consider the fizz in a soft drink
CO 2 (g) + H 2 O(liq)
LOWER T
CO 2 (aq) + heat
K c = [CO 2 (aq)]/[CO 2 (g)] HIGHER T
• Change T: New equilib. position? New value of K?
•
Increase T
Equilibrium shifts left: [CO 2 (g)] K decreases as T goes up .
•
Decrease T
[CO 2 (aq)] [CO 2 (aq)] increases and [CO 2 (g)] decreases.
K increases as T goes down 27 Oct 97 Chemical Equilibrium 37
Temperature Effects on Chemical Equilibrium
N 2 O 4 + heat (colorless) 2 NO 2 (brown) K c
[NO 2 ] [N 2 O 4 ] 2
K
c
= 0.00077 at 273 K K
c
= 0.00590 at 298 K
D
H o rxn = + 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 38
EQUILIBRIUM AND EXTERNAL EFFECTS
Catalytic exhaust system
• •
Add catalyst
---> no change in K
A catalyst only affects the RATE of approach to equilibrium.
27 Oct 97 Chemical Equilibrium 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