Chem 1311 A D .G .V anD erveer/T.F.Block Fall,2001 Tentative Syllabus Date Lecture Reading 1 Problem s M A ugust20 Introduction O FB,sec.7-1 to 7-4 W A ugust22 Chem icalEquilibrium O FB,sec.7-5 to 7-7 O FB Ch 7,# 2,4,6,8, 12,18,20,27,30 F A ugust24 Chem icalEquilbirium O FB,sec.8-1 to 8-3 O FB Ch 7,# 40,42,48, 53,56,60,62 M A ugust27 Bronsted-Low ry Theory O FB Ch 8,# 4,6,10, W A ugust29 Bronsted-Low ry Theory O FB,sec.8-4 O FB Ch 8,# 16,20,22 F A ugust31 Hydrolysis O FB,sec.8-5 O FB Ch 8,# 26b,28, 32,38 M Septem ber3 LaborD ay W Septem ber5 Buffers O FB,sec.8-6 O FB Ch 8,# 46,50 F Septem ber7 Titrations O FB,sec.8-7 M Septem ber10 Polyprotic Acids W Septem ber12 Review F Septem ber14 EXAM I M Septem ber17 Exam Return and Review N orm an,Ch.5; O FB,sec.8-8 W Septem ber19 Lew isA cidsand Bases; A cid and Base A nhydrides O FB,sec.9-1 to 9-3 F Septem ber21 Solubility Product O FB,sec.9-4 M Septem ber24 EffectofpH on Solubility O FB,sec.9-5 W Septem ber26 Com plex Ionsand Solubility O FB,sec.20-1 and 20-2 F Septem ber28 Sym m etry and Structure O FB,sec.20-3 and 20-4 M O ctober1 CrystalStructure and Defects W O ctober3 Lattice Energy O FB,sec.21-2 and 21-3 F O ctober5 SilicateM inerals O FB,sec.21-4 and 21-5 M O ctober8 Ceram ics W O ctober10 Review F O ctober12 EXAM II M O ctober15 N o School N orm an,Ch.1 and 2; O FB,sec.18-1;W inter, Ch.3 W O ctober17 Exam Return and Review F O ctober19 M O Theory W inter,Ch.4; O FB,sec.3-6 and 18-2 O FB,Ch.18, #2, 4, 6. 8, 10,12,14,16,18 M O ctober22 V SEPR And Hybrid O rbitals W inter,Ch.5 and 6 O FB,Ch.18,#20,22, 24,26,31;W inter,sec. 4.4,#1(first6 only)and 2 W O ctober24 M O’sforPolyatom ic M olecules W inter,sec.5.7,#1 and 2;sec.6.8,#3 The Reading is for the next class Problems are for that day’s class. Problems for each week (MWF) are due the following Monday.
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The Reading is for the next class. Problems are for that day’s class. Problems for each week (MWF) are due the following Monday.
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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.