the reading is for the next class. problems are for that day’s class. problems for each week (mwf)...

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

PAPBa b

Q = or[C]c[D]d

[A]a[B]b

Reaction quotient

PAPBa b

PA , PB, etc. not at equilibrium

Reaction quotient

PAPBa b

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.

PAPBa b

aA + bB cC + dD

PAPBa b

aA + bB cC + dD

Too much reactant,not enough product.

PAPBa b

aA + bB cC + dD

PAPBa b

aA + bB cC + dD

Too much product, not enoughreactant.

PAPBa b

Q =Q > K

aA + bB cC + dD

Large vs c dPCPD

c dPCPD

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

P4 2 P2

Q = (PP )2

P =nRT

T = 673 K

V = 25.0 L

R = 0.0820578 L atm mol-1 K-1

P4 2 P2

Q = (PP )2

PP =nRT

=(1.4)(0.0820578)(673)

25.0= 3.09 atm

P4 2 P2

Q = (PP )2

PP =nRT

=(1.25)(0.0820578)(673)

25.0= 2.76 atm

PP =nRT

=(1.4)(0.0820578)(673)

25.0= 3.09 atm

P4 2 P2

Q = (PP )2

PP4PP =

2.76 atm

3.09 atm

=(2.76)2

P4 2 P2

(PP )2

PP4PP =

2.76 atm

3.09 atm

=(2.76)2

2.46Q =

K = 1.39

P4 2 P2

(PP )2

PP4PP =

2.76 atm

3.09 atm

=(2.76)2

2.46Q =

K = 1.39

P4 2 P2

Converting between partial

pressures and concentrations.

Concentration = moles

PV = nRT

For gas ‘A’ [A] =nA

PV = nRT

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)2= K

PA = RT[A]

2 NO2(g)N2O4(g)

Pref = 1 atm

/Pref)2= K

= RT[N2O4]

PA = RT[A]

2 NO2(g)N2O4(g)

Pref = 1 atm

/Pref)2= K

= RT[N2O4]

= RT[NO2]

2 NO2(g)N2O4(g)

/Pref)2= K

PN O24

= RT[N2O4]

= RT[NO2]

[N2O4](RT/Pref)

[NO2]2(RT/Pref)=

[N2O4]

[NO2]2

2 NO2(g)N2O4(g)

[N2O4](RT/Pref)

[NO2]2(RT/Pref)= xK =

[N2O4]

[NO2]2

[N2O4]

[NO2]2

= K RT

[N2O4]

[NO2]2= K RT

aA + bB cC + dD

= K[C]c[D]dPCPD

PAPBa b

[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

[CH4]= 0.172 ( RT

CH4 + H2O CO + 3 H2

[CH4] =

K(RT)-2 = 3.153 x 10-5

(0.00642)3

3.153 x 10-5= 8.39 x 10-2 mol L-1

Le Chatelier’s Principle

A system in equilibrium that is

subjected to a stress reacts in a

way that counteracts the stress.

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.

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.

A system in equilibrium that is

subjected to a stress reacts in a

way that counteracts the stress.

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:

temperature

pressure

temperature

pressure

volume

temperature

pressure

removal or addition of product

volume

temperature

pressure

removal or addition of reactant

volume

removal or addition of product

Removal or addition of a reactant

or product.

Increases or decreases concentration.

PB [B]

or product.

Increases or decreases concentration.

K =[B]

I2 + H2 2 HI

PHI= 3.009 atm

= 0.4756 atm

= 0.2056 atm

I2 + H2 2 HI

= 0.4756 atm

= 0.2056 atm

= 3.009 atm

( ) PH2PH

( )= K

I2 + H2 2 HI

= 0.4756 atm

= 0.2056 atm

= 3.009 atm

( ) PH2PH

( )= K

(3.009)2

(0.4756)(0.2056)=

I2 + H2 2 HI

= 0.4756 atm

= 0.2056 atm

= 3.009 atm

( ) PH2PH

( )= 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

2nd eq P

Increase PI2

to 2.000 atm

I2 + H2 2 HIK = 92.60

Start P 2.000 0.2056 3.009

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

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

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)

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

9.054 + 12.036x + 4x2 = 38.08 - 204.24x + 92.6x2

88.60x2 - 216.276x + 29.023 = 0

x =-b b2 - 4 ac

= 0.142 or 2.30

Start P 2.000 0.2056 3.009

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

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

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.

(3.293)2

(1.858)(0.0636)=

pressures.

(3.293)2

(1.858)(0.0636)= 91.77

K given = 92.60

Changing volume of system

If V is reduced, P increases.

PV = nRT

Changing volume of system

If V is reduced, P increases.

Le Chatlier’s principle requires that the

equilibrium shift so that P decreases.

2 NO2(g)N2O4(g)

An increase in pressure should

favor an increase in the product,

2 NO2(g)N2O4(g)

An increase in pressure should

favor an increase in the product,

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

Raising or lowering the temperature

Some reactions liberate heat to

form products.

Some reactions liberate heat to

form products.

This is an exothermic reaction.

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.

Raising the temperature of an

exothermic reaction will favor

the formation of more reactant.

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]

If a reaction is exothermic and the

temperature is raised, K will decrease.

K = [P]

If a reaction is endothermic and the

temperature is raised, K will increase.

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

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

1. Gases enter equilibrium expressionsas partial pressures in atmospheres.

Law of Mass Action

2. Dissolved species enter as concentrations in mol L-1.

Law of Mass Action1. Gases enter equilibrium expressionsas partial pressures in atmospheres.

3. Pure solids and liquids are represented by 1at equilibrium , a dilute solvent is 1.

Law of Mass Action

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

4 heme Fe sites to bond O2

Increase PO more Feheme O2

Feheme O2

Increase PO by changing local

atmosphere.

Increase PO more2

Decrease PO less Feheme O2

Decrease PO2

by going to

higher altitude.

Feheme + CO Feheme CO

= 0.20 atm

Affinity for CO over O2 factor of 200+.

= 0.20 atm

Affinity for CO over O2 factor of 200+.

CO 1 x 103 ppm < 1% PCO < 0.01 atm

= 0.20 atm

CO 1 x 103 ppm < 1% PCO < 0.01 atm

Kheme-CO >> Kheme-O2

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