the reading is for the next class. problems are for that day’s class. problems for each week (mwf)...
TRANSCRIPT
Chem 1311 A D. G. VanDerveer/T. F. Block Fall, 2001 Tentative Syllabus
Date Lecture Reading1 Problems
M August 20 Introduction OFB, sec. 7-1 to 7-4 W August 22 Chemical Equilibrium OFB, sec. 7-5 to 7-7 OFB Ch 7, # 2, 4, 6, 8,
12, 18, 20, 27, 30 F August 24 Chemical Equilbirium OFB, sec. 8-1 to 8-3 OFB Ch 7, # 40, 42, 48,
53, 56, 60, 62 M August 27 Bronsted-Lowry Theory OFB Ch 8, # 4, 6, 10, W August 29 Bronsted-Lowry Theory OFB, sec. 8-4 OFB Ch 8, # 16, 20, 22 F August 31 Hydrolysis OFB, sec. 8-5 OFB Ch 8, # 26b, 28,
32, 38 M September 3 Labor Day W September 5 Buffers OFB, sec. 8-6 OFB Ch 8, # 46, 50 F September 7 Titrations OFB, sec. 8-7 M September 10 Polyprotic Acids
W September 12 Review F September 14 EXAM I
M September 17 Exam Return and Review
Norman, Ch. 5; OFB, sec. 8-8
W September 19 Lewis Acids and Bases; Acid and Base Anhydrides
OFB, sec. 9-1 to 9-3
F September 21 Solubility Product OFB, sec. 9-4 M September 24 Effect of pH on
Solubility OFB, sec. 9-5
W September 26 Complex Ions and Solubility
OFB, sec. 20-1 and 20-2
F September 28 Symmetry and Structure OFB, sec. 20-3 and 20-4 M October 1 Crystal Structure and
Defects
W October 3 Lattice Energy OFB, sec. 21-2 and 21-3 F October 5 Silicate Minerals OFB, sec. 21-4 and 21-5 M October 8 Ceramics W October 10 Review F October 12 EXAM II M October 15 No School Norman, Ch. 1 and 2;
OFB, sec. 18-1; Winter, Ch. 3
W October 17 Exam Return and Review
F October 19 MO Theory Winter, Ch. 4; OFB, sec. 3-6 and 18-2
OFB, Ch. 18, #2, 4, 6. 8, 10, 12, 14, 16, 18
M October 22 VSEPR And Hybrid Orbitals
Winter, Ch. 5 and 6 OFB, Ch. 18, #20, 22, 24, 26, 31;Winter, sec. 4.4, #1(first 6 only) and 2
W October 24 MO’s for Polyatomic Molecules
Winter, sec. 5.7, #1 and 2; sec. 6.8, #3
The Reading isfor the next class.
Problems are forthat day’s class.
Problems for eachweek (MWF) aredue the followingMonday.
Reaction quotient
PCPD
PAPBa b
c d
Q = or[C]c[D]d
[A]a[B]b
Reaction quotient
PCPD
PAPBa b
c d
Q =
PA , PB, etc. not at equilibrium
Reaction quotient
PCPD
PAPBa b
c d
Q =
Q will go to the value of K as
the partial pressures go to
equilibrium
Q vs K can
predict where
the reaction is
in respect to
equilibrium.
PCPD
PAPBa b
c d
K =
PCPD
PAPBa b
c d
Q =
Q < K
aA + bB cC + dD
PCPD
PAPBa b
c d
K =
PCPD
PAPBa b
c d
Q =
Q < K
aA + bB cC + dD
Too much reactant,not enough product.
PCPD
PAPBa b
c d
K =
PCPD
PAPBa b
c d
Q =
Q < K
Q > K
aA + bB cC + dD
aA + bB cC + dD
PCPD
PAPBa b
c d
K =
PCPD
PAPBa b
c d
Q =
Q < K
Q > K
aA + bB cC + dD
aA + bB cC + dD
Too much product, not enoughreactant.
PCPD
PAPBa b
c d
K =
Q =Q > K
aA + bB cC + dD
aA + bB cC + dD
Q < K
Q < K
Large vs c dPCPD
c dPCPD
a bPAPB
a bPAPB
Q =c dPCPD
a bPAPB
Exercise page 291
P4 2 P2
K = 1.39 @ 400oC
1.40 mol P4
1.25 mol P2
Volume = 25.0 L
P4 2 P2
Q = (PP )2
2
PP4
P4 2 P2
Q = (PP )2
2
PP4
P =nRT
V
T = 673 K
V = 25.0 L
R = 0.0820578 L atm mol-1 K-1
P4 2 P2
Q = (PP )2
2
PP4
PP =nRT
V4
=(1.4)(0.0820578)(673)
25.0= 3.09 atm
P4 2 P2
Q = (PP )2
2
PP4
PP =nRT
V2
=(1.25)(0.0820578)(673)
25.0= 2.76 atm
PP =nRT
V4
=(1.4)(0.0820578)(673)
25.0= 3.09 atm
P4 2 P2
Q = (PP )2
2
PP4PP =
2
2.76 atm
PP =4
3.09 atm
=(2.76)2
3.09=
P4 2 P2
(PP )2
2
PP4PP =
2
2.76 atm
PP =4
3.09 atm
=(2.76)2
3.09=
2.46Q =
Q =
K = 1.39
P4 2 P2
(PP )2
2
PP4PP =
2
2.76 atm
PP =4
3.09 atm
=(2.76)2
3.09=
2.46Q =
Q =
K = 1.39
P4 2 P2
Converting between partial
pressures and concentrations.
Converting between partial
pressures and concentrations.
Concentration = moles
V
Converting between partial
pressures and concentrations.
Concentration = moles
V
PV = nRT
Converting between partial
pressures and concentrations.
Concentration = moles
V
For gas ‘A’ [A] =nA
V
PV = nRT
Converting between partial
pressures and concentrations.
Concentration = moles
V
For gas ‘A’ [A] =nA
V=
PA
RT
PV = nRT
Converting between partial
pressures and concentrations.
Concentration = moles
V
For gas ‘A’ [A] =nA
V=
PA
RT
PA = RT[A]PV = nRT
PA = RT[A]
2 NO2(g)N2O4(g)
PA = RT[A]
2 NO2(g)N2O4(g)
Pref = 1 atm
PA = RT[A]
2 NO2(g)N2O4(g)
Pref = 1 atm
PN O2 4
/Pref
(PNO2
/Pref)2= K
PA = RT[A]
2 NO2(g)N2O4(g)
Pref = 1 atm
4
/Pref
(PNO2
/Pref)2= K
PN O2
PN O2
4
= RT[N2O4]
PA = RT[A]
2 NO2(g)N2O4(g)
Pref = 1 atm
4
/Pref
(PNO2
/Pref)2= K
PN O2
PN O2
4
= RT[N2O4]
PNO2
= RT[NO2]
2 NO2(g)N2O4(g)
4
/Pref
(PNO2
/Pref)2= K
PN O2
PN O24
= RT[N2O4]
PNO2
= RT[NO2]
[N2O4](RT/Pref)
[NO2]2(RT/Pref)=
[N2O4]
[NO2]2
xRT
Pref
)(-1
K =
2 NO2(g)N2O4(g)
[N2O4](RT/Pref)
[NO2]2(RT/Pref)= xK =
[N2O4]
[NO2]2
[N2O4]
[NO2]2
= K RT
Pref
)(
RT
Pref
)(-1
[N2O4]
[NO2]2= K RT
Pref
)(
aA + bB cC + dD
c
= K[C]c[D]dPCPD
PAPBa b
d
[A]a[B]b
= K ( RTPref
)a+b-c-d
?
[C]c[D]d
[A]a[B]b
= K ( RTPref
)a+b-c-d
Exercise page 297
CH4 + H2O CO + 3 H2
K = 0.172
[H2]=[CO]=[H2O]= 0.00642 mol L-1
[C]c[D]d
[A]a[B]b
= K ( RTPref
)a+b-c-d
CH4 + H2O CO + 3 H2
K = 0.172
[H2]=[CO]=[H2O]= 0.00642 mol L-1
(0.00642)4
[CH4](0.00642)= 0.172 ( RT
Pref
)-2
[CH4]= 0.172 ( RT
Pref
)-2
CH4 + H2O CO + 3 H2
[CH4] =
K(RT)-2 = 3.153 x 10-5
(0.00642)3
(0.00642)3
3.153 x 10-5= 8.39 x 10-2 mol L-1
Le Chatelier’s Principle
Le Chatelier’s Principle
A system in equilibrium that is
subjected to a stress reacts in a
way that counteracts the stress.
Le Chatelier’s Principle
Any system in chemical equilibrium, as a result
of the variation in one of the factors determining
the equilibrium, undergoes a change such that, if
this change had occurred by itself, it would have
introduced a variation of the factor considered
in the opposite direction.
Le Chatelier’s Principle
Any system in chemical equilibrium, as a result
of the variation in one of the factors determining
the equilibrium, undergoes a change such that, if
this change had occurred by itself, it would have
introduced a variation of the factor considered
in the opposite direction.
Le Chatelier’s Principle
A system in equilibrium that is
subjected to a stress reacts in a
way that counteracts the stress.
Stress = a factor affecting equilibrium
Stress = a factor affecting equilibrium
K =[C]c[D]d
[A]a[B]b
Anything causing a change in the concentration (or partial pressure) ofa reactant or product is a stress.
Factors affecting equilibrium:
Factors affecting equilibrium:
temperature
Factors affecting equilibrium:
temperature
pressure
Factors affecting equilibrium:
temperature
pressure
volume
Factors affecting equilibrium:
temperature
pressure
removal or addition of product
volume
Factors affecting equilibrium:
temperature
pressure
removal or addition of reactant
volume
removal or addition of product
Removal or addition of a reactant
or product.
Removal or addition of a reactant
or product.
Increases or decreases concentration.
PA[A]
PB [B]
Removal or addition of a reactant
or product.
Increases or decreases concentration.
K =[B]
[A]
I2 + H2 2 HI
I2 + H2 2 HI
PHI= 3.009 atm
PI2
= 0.4756 atm
PH2
= 0.2056 atm
I2 + H2 2 HI
= 0.4756 atm
= 0.2056 atm
= 3.009 atm
PHI
PHI
( )2
PI2
PI2
( ) PH2PH
2
( )= K
I2 + H2 2 HI
= 0.4756 atm
= 0.2056 atm
= 3.009 atm
PHI
PHI
( )2
PI2
PI2
( ) PH2PH
2
( )= K
(3.009)2
(0.4756)(0.2056)=
I2 + H2 2 HI
= 0.4756 atm
= 0.2056 atm
= 3.009 atm
PHI
PHI
( )2
PI2
PI2
( ) PH2PH
2
( )= K
(3.009)2
(0.4756)(0.2056)= 92.60
I2 + H2 2 HI
Increase PI2
to 2.00 atm.
I2 + H2 2 HIK = 92.60
Start P 2.000 0.2056 3.009
P
2nd eq P
Increase PI2
to 2.000 atm
I2 + H2 2 HIK = 92.60
Start P 2.000 0.2056 3.009
P
2nd eq P
Increase PI2
to 2.000 atm
-x -x +2x
I2 + H2 2 HIK = 92.60
Start P 2.000 0.2056 3.009
P
2nd eq P
Increase PI2
to 2.000 atm
-x -x +2x
2.000-x 0.2056-x 3.009+2x
I2 + H2 2 HIK = 92.60
Start P 2.000 0.2056 3.009
P
2nd eq P
Increase PI2
to 2.000 atm
-x -x +2x
2.000-x 0.2056-x 3.009+2x
92.6 =(3.009+2x)2
(2.000-x)(0.2056-x)=
92.6 =(3.009+2x)2
(2.000-x)(0.2056-x)=
92.6 =(3.009+2x)2
(2.000-x)(0.2056-x)=
(9.054 + 12.036x + 4x2)
(0.4112 - 2.2056x + x2)= 92.6
92.6 =(3.009+2x)2
(2.000-x)(0.2056-x)=
(9.054 + 12.036x + 4x2)= 92.6
9.054 + 12.036x + 4x2 = 92.6 (0.4112 - 2.2056x + x2)
(0.4112 - 2.2056x + x2)
92.6 =(3.009+2x)2
(2.000-x)(0.2056-x)=
(9.054 + 12.036x + 4x2)= 92.6
(0.4112 - 2.2056x + x2)
(0.4112 - 2.2056x + x2)
9.054 + 12.036x + 4x2 = 38.08 - 204.24x + 92.6x2
9.054 + 12.036x + 4x2 = 92.6
9.054 + 12.036x + 4x2 = 38.08 - 204.24x + 92.6x2
88.60x2 - 216.276x + 29.023 = 0
x =-b b2 - 4 ac
2a
a b c
=
9.054 + 12.036x + 4x2 = 38.08 - 204.24x + 92.6x2
88.60x2 - 216.276x + 29.023 = 0
x =-b b2 - 4 ac
2a
a b c
= 0.142 or 2.30
Start P 2.000 0.2056 3.009
P
2nd eq P
-x -x +2x
2.000-x 0.2056-x 3.009+2x
0.142 or 2.30x =
I2 + H2 2 HI
Start P 2.000 0.2056 3.009
P
2nd eq P
-x -x +2x
2.000-x 0.2056-x 3.009+2x
0.142 or 2.30x =
I2 + H2 2 HI
x = 2.30 would give a negative pressure for I2
Start P 2.000 0.2056 3.009
P
2nd eq P
-0.142 -0.142 0.284
1.858 0.0636 3.293
0.142 or 2.30x =
I2 + H2 2 HI
Calculate K with new partial
pressures.
Calculate K with new partial
pressures.
(3.293)2
(1.858)(0.0636)=
Calculate K with new partial
pressures.
(3.293)2
(1.858)(0.0636)= 91.77
K given = 92.60
Changing volume of system
Changing volume of system
V 1
P
Changing volume of system
V 1
P
If V is reduced, P increases.
PV = nRT
Changing volume of system
V 1
P
If V is reduced, P increases.
Le Chatlier’s principle requires that the
equilibrium shift so that P decreases.
2 NO2(g)N2O4(g)
2 NO2(g)N2O4(g)
An increase in pressure should
favor an increase in the product,
N2O4.
2 NO2(g)N2O4(g)
An increase in pressure should
favor an increase in the product,
N2O4.
Each N2O4 produced means a loss of two NO2 molecules, a net loss of onemolecule and a lower pressure.
2 NO2(g)N2O4(g)
A decrease in pressure favors a shift to
NO2.
Raising or lowering the temperature
Raising or lowering the temperature
Some reactions liberate heat to
form products.
Raising or lowering the temperature
Some reactions liberate heat to
form products.
This is an exothermic reaction.
Raising or lowering the temperature
This is an endothermic reaction.
Other reactions absorb heat to
produce products.
Raising the temperature of an
endothermic reaction will favor
the formation of more product.
R P
Raising the temperature of an
exothermic reaction will favor
the formation of more reactant.
R P
NaOH(s) + H2O(l) Na+(aq) + OH-
(aq) + H2O(l)
NaOH(s) + H2O(l) Na+(aq) + OH-
(aq) + H2O(l)
The solution of solid sodium hydroxide
into water is exothermic.
K = [P]
[R]
K = [P]
[R]
If a reaction is exothermic and the
temperature is raised, K will decrease.
K = [P]
[R]
If a reaction is endothermic and the
temperature is raised, K will increase.
Driving reactions to completion
Driving reactions to completion
Completion = 100% yield of product
Cl-(aq) + Ag+
(aq) AgCl(s)
Cl-(aq) + Ag+
(aq) AgCl(s)
AgCl precipitates from the
solution.
Cl-(aq) + Ag+
(aq) AgCl(s)
AgCl precipitates from the
solution.
As the AgCl precipitates,
product is removed from solution.
Cl-(aq) + Ag+
(aq) AgCl(s)
N2 + 3 H2 2 NH3
All gases
N2 + 3 H2 2 NH3
All gases
exothermic
N2 + 3 H2 2 NH3
All gases
exothermic cool
N2 + 3 H2 2 NH3
All gases
exothermic cool
N2 + 3 H2 2 NH3
Although a lower temperature
favors more NH3 formed, the
lower temperature also leads to
a very slow reaction.
N2 + 3 H2 2 NH3
An increase in pressure should favor product.
N2 + 3 H2 2 NH3
An increase in pressure should favor product.
N2 + 3 H2 2 NH3
Ultimate solution: react at high
Temperature to speed up reaction,
cool until NH3 becomes liquid.
Remove from reaction vessel and repeat.
(time)
CONCENTRATION
Heterogeneous equilibrium
Heterogeneous equilibrium
Involves at least two phases.
Heterogeneous equilibrium
Involves at least two phases.
What is the concentration of
a pure liquid or a pure solid?
Concentrations are not a valid
way to define a pure liquid
or solid.
Concentrations are not a valid
way to define a pure liquid
or solid.
Moles water
Liters solvent= ?
The concentration of a pure liquid
or solid is defined as 1.
Law of Mass Action
Law of Mass Action
1. Gases enter equilibrium expressionsas partial pressures in atmospheres.
Law of Mass Action
1. Gases enter equilibrium expressionsas partial pressures in atmospheres.
2. Dissolved species enter as concentrations in mol L-1.
Law of Mass Action1. Gases enter equilibrium expressionsas partial pressures in atmospheres.
2. Dissolved species enter as concentrations in mol L-1.
3. Pure solids and liquids are represented by 1at equilibrium , a dilute solvent is 1.
Law of Mass Action
1. Gases enter equilibrium expressionsas partial pressures in atmospheres.
2. Dissolved species enter as concentrations in mol L-1.
3. Pure solids and liquids are represented by 1at equilibrium , a dilute solvent is 1.
4. Partial pressures or concentrations ofproducts appear in the numerator, reactants inthe denominator. Each is raised to the powerof its coefficient.
hemoglobin
hemoglobinFe
heme Fe
Feheme + O2 Feheme O2
Feheme + O2 Feheme O2
4 heme Fe sites to bond O2
4 heme Fe sites to bond O2
Increase PO more Feheme O2
Feheme + O2 Feheme O2
2
4 heme Fe sites to bond O2
Feheme O2
Feheme + O2 Feheme O2
Increase PO by changing local
atmosphere.
2
Increase PO more2
4 heme Fe sites to bond O2
Decrease PO less Feheme O2
Feheme + O2 Feheme O2
2
4 heme Fe sites to bond O2
Decrease PO less Feheme O2
Feheme + O2 Feheme O2
2
Decrease PO2
by going to
higher altitude.
Feheme + CO Feheme CO
Feheme + O2 Feheme O2
Feheme + CO Feheme CO
Feheme + O2 Feheme O2
PO2
= 0.20 atm
Affinity for CO over O2 factor of 200+.
Feheme + CO Feheme CO
Feheme + O2 Feheme O2
PO2
= 0.20 atm
Affinity for CO over O2 factor of 200+.
CO 1 x 103 ppm < 1% PCO < 0.01 atm
Feheme + CO Feheme CO
Feheme + O2 Feheme O2
PO2
= 0.20 atm
CO 1 x 103 ppm < 1% PCO < 0.01 atm
Kheme-CO >> Kheme-O2