1 © 2009 Brooks/Cole - Cengage Chemical Kinetics Chapter 12 H 2 O 2 decomposition in an insect H 2...

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© 2009 Brooks/Cole - Cengage

Chemical KineticsChapter 12

H2O2 decomposition in an insect

H2O2 decomposition catalyzed by MnO2

2


Objectives• Understand rates of reaction and the

conditions affecting rates.• Derive a rate of reaction, rate constant,

and reaction order from experimental data

• Use integrated rate laws• Discuss collision theory and the role of

activation energy in a reaction• Discuss reaction mechanisms and their

effect on rate law

3


• We can use thermodynamics to tell if a reaction is product- or reactant-favored.

• But this gives us no info on HOW FAST reaction goes from reactants to products.

• KINETICS — the study of REACTION RATES and their relation to the way the reaction proceeds, i.e., its MECHANISM.

Chemical Kinetics

4


Reaction Mechanisms• A reaction mechanism is a

sequence of events at the molecular level that control the speed and outcome of a reaction.

• The reaction mechanism is our goal!

5


• Reaction rate = change in concentration of a reactant or product with time.

• Three “types” of rates –initial rate–average rate–instantaneous rate

Reaction Rates Section 12.1

6


Factors Affecting Reaction Rates• Physical State of the Reactants

– Gas, liquid or solid – how molecules are able to interact with each other

– Solids react faster when surface area is greater so fine powders react faster than big chunks

• Concentration of Reactants– As the concentration of reactants increases, so does the

likelihood that reactant molecules will collide.

• Temperature– At higher temperatures, reactant molecules have more kinetic

energy, move faster, and collide more often and with greater energy.

• Catalysts– Speed up reaction by changing mechanism.– Catalysts don’t get used up themselves

7


Concentrations & Rates

0.3 M HCl 6 M HCl

Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g)

8


• Physical state of reactants

Factors Affecting Rates

9


Catalysts: catalyzed decomp of H2O2

2 H2O2 2 H2O + O2


10


• Temperature


Bleach at 54 ˚C Bleach at 22 ˚C

11


Determining a Reaction Rate

Blue dye is oxidized with bleach.

Its concentration decreases with time.

The rate — the change in dye conc with time — can be determined from the plot.

Blue dye is oxidized with bleach.

Its concentration decreases with time.

The rate — the change in dye conc with time — can be determined from the plot.

Dy

e C

on

c

Time

12


Determining a Reaction Rate

13


Reaction Rates

The average rate of the reaction over each interval is the change in concentration divided by the change in time:

C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)

Average Rate, M/s

14


Reaction Rates

• Note that the average rate decreases as the reaction proceeds.

• This is because as the reaction goes forward, there are fewer collisions between reactant molecules.


15


Reaction Rates

• A plot of concentration vs. time for this reaction yields a curve like this.

• The slope of a line tangent to the curve at any point is the instantaneous rate at that time.


16


Reaction Rates

• The reaction slows down with time because the concentration of the reactants decreases.


[A] Change in molarity

t Change in time

17


Reaction Rates and Stoichiometry

• In this reaction, the ratio of C4H9Cl to C4H9OH is 1:1.

• Thus, the rate of disappearance of C4H9Cl is the same as the rate of appearance of C4H9OH.


Rxn Rate =

-[C4H9Cl]t

=[C4H9OH]

t

18


Reaction Rates

In this reaction, the concentration of nitrogen dioxide, NO2, was measured at various times, t.

2NO2(g) 2NO(g) + O2(g)

[C4H9Cl] M

19


Reaction Rates and Stoichiometry• What if the ratio is not 1:1?

2NO2(g) 2NO(g) + O2(g)

• 2NO can be made from 2NO2 consumed, but only 1 O2 is produced.

Read as: the rate of consumption of NO2 is the same as the rate of production of NO. This is because their coefficients are the same.

20


Reaction RatesBut since the coefficient for oxygen is ½ of the other two, it’s rate is of production rate is half as fast. Or 2 x rate O2 = rate of NO

21


Reaction Rate and Stoichiometry summary:• For the reaction


we know • In general for

aA + bB cC + dD

Rxn

Reaction Rates

22


Reaction Rates Practice1. What is the relative rate of disappearance of

the reactants and relative rate of appearance of the products for each reaction?

A. 2O3(g) 3O2(g)

B. 2HOF(g) 2HF(g) + O2(g)

2. In the synthesis of ammonia, if Δ[H2]/Δt =-4.5 x 10-4 mol/L•min, what is rate with respect to NH3? wrt N2?

N2(g) + 3H2(g) 2NH3(g)

23


Concentrations & Rates

Rate of reaction is proportional to [NO2]

We express this as a RATE LAW

Rate of reaction = k [NO2]n

where k = rate constant

k is independent of conc. but increases with T

n is the order of the reactant

2NO2(g) 2NO(g) + O2(g)

24


Concentrations, Rates, & Rate Laws

In general, for

a A + b B x X with a catalyst C

Rate = k [A]m[B]n[C]p

The exponents m, n, and p:

• sum of m, n and p are the reaction order

• can be 0, 1, 2 or fractions like 3/2

• must be determined by experiment! They are not simply related to stoichiometry!

25


Ch. 12.3 - Interpreting Rate Laws Determining Rate by inspection (quick method):

Rate = k [A]m[B]n[C]p

• If m = 1, rxn. is 1st order in A

Rate = k [A]1

If [A] doubles, then rate goes up by factor of _2_ • If m = 2, rxn. is 2nd order in A.

Rate = k [A]2

Doubling [A] increases rate by ___4_____• If m = 0, rxn. is zero order.

Rate = k [A]0

If [A] doubles, rate _Stays the same_

26


The method of Initial Rates

• This method requires that a reaction be run several times.

• The initial concentrations of the reactants are varied.

• The reaction rate is measured just after the reactants are mixed.

• Eliminates the effect of the reverse reaction.

27


Deriving Rate Laws

Expt. [CH3CHO] Disappear of CH3CHO(mol/L) (mol/L•sec)

1 0.10 0.020

2 0.20 0.081

3 0.30 0.182

4 0.40 0.318

Derive rate law and k for CH3CHO(g) CH4(g) + CO(g)

from experimental data for rate of disappearance of CH3CHO

28


Deriving Rate Laws

Determination of Rate by inspection:

Here the rate goes up by __4___ when initial conc. doubles. Therefore, we say this reaction is ________2nd_________ order.

So, the Rate of rxn = k [CH3CHO]2

Now determine the value of k. Use expt. #3 data—

0.182 mol/L•s = k (0.30 mol/L)2

k = 2.0 (L / mol•s)Using k you can calc. rate at other values of [CH3CHO] at same T.

29


Deriving Rate LawsDetermination of Rate by rigorous math:Rate = k[A]n

Note: For easier math, put the faster rate in the numerator.

4=2nn=2

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Determination of Rate Law Practice

For the reaction

BrO3- + 5Br- + 6H+ 3Br2 + 3H2O

• The general form of the Rate Law is Rate = k[BrO3

-]n[Br-]m[H+]p

• We use experimental data to determine the values of n, m, and p

• We will choose trials that vary only one of the concentrations to determine the order based upon that reactant.

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Initial concentrations (M)

Rate (M/s)BrO3- Br- H+

0.10 0.10 0.10 0.8 x 10-3

0.20 0.10 0.10 1.6 x 10-3

0.20 0.20 0.10 3.2 x 10-3

0.10 0.10 0.20 3.2 x 10-3

1) Determine the rate law wrt each reactant

2) Determine the rate constant, k (with units)

3) Determine the overall order of reaction

Determination of Rate Law Practice

32


Ch. 12.4 – Integrated Rate Law (Conc vs time)

What is concentration of reactant as function of time?

Consider FIRST ORDER REACTIONS

The rate law is

And the units of k are sec-1.

𝐑𝐚𝐭𝐞=−𝚫 [ 𝑨 ]Δ𝒕𝒊𝒎𝒆

=𝒌 [𝑨 ]

33


Integrated Rate Law

• Expresses the reaction concentration as a function of time.

• Form of the equation depends on the order of the rate law (differential).

• Changes Rate = D[A]n Dt

• We will only work with n=0, 1, and 2

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Concentration/Time Relations

Integrating - (∆ [A] / ∆ time) = k [A], we get

ln is natural logarithm

[A]0 at time = 0

[A] / [A]0 =fraction remaining after time t has elapsed.

[A] / [A]0 =fraction remaining after time t has elapsed.

Called the integrated first-order rate law.

ln [A]

[A]0 = - kt

35


Concentration/Time Relations

Sucrose decomposes to simpler sugars

Rate of disappearance of sucrose = k [sucrose]

Glucose

If k = 0.21 hr-1

and [sucrose] = 0.010 M

How long to drop 90% (to 0.0010 M)?

36


Concentration/Time Relations Rate of disappear of sucrose = k [sucrose], k = 0.21 hr-1. If initial [sucrose] = 0.010 M, how long to drop 90% or to 0.0010 M?

Use the first order integrated rate law

ln (0.100) = - 2.3 = - (0.21 hr-1)(time)

time = 11 hours

ln

0.0010

0.010

= - (0.21 h-1) t

37


Using the Integrated Rate Law

The integrated rate law suggests a way to tell the order based on experiment.

2 N2O5(g) 4 NO2(g) + O2(g)

Time (min) [N2O5] (M) ln [N2O5]

0 1.00 01.0 0.705 -0.352.0 0.497 -0.705.0 0.173 -1.75

Rate = k [N2O5]

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2 N2O5(g) 4 NO2(g) + O2(g) Rate = k [N2O5]

[N2O5] vs. time

time

1

0

0 5

[N2O5] vs. time

time

1

0

0 5

l n [N2O5] vs. time

time

0

-2

0 5

l n [N2O5] vs. time

time

0

-2

0 5

Data of conc. vs. time plot do not fit straight line.

Plot of ln [N2O5] vs. time is a straight line!

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All 1st order reactions have straight line plot for ln [A] vs. time.

ln [N2O5] vs. time

time

0

-2

0 5

ln [N2O5] vs. time

time

0

-2

0 5

Plot of ln [N2O5] vs. time is a straight line! Eqn. for straight line: y = mx + b

Plot of ln [N2O5] vs. time is a straight line! Eqn. for straight line: y = mx + b ln[N2O5 ] = -kt + ln[N2O5]0

conc at rate const conc attime t =-slope time=0

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Half-Life

HALF-LIFE is the time it takes for 1/2 a sample is disappear.

For 1st order reactions, the concept of HALF-LIFE is especially useful.

See Active Figure 15.9

41


• Reaction is 1st order decomposition of H2O2.

Half-Life

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Half-Life

• Reaction after 1 half-life.

• 1/2 of the reactant has been consumed and 1/2 remains.

43


Half-Life

• After 2 half-lives 1/4 of the reactant remains.

44


Half-Life

• A 3 half-lives 1/8 of the reactant remains.

45


Half-Life

• After 4 half-lives 1/16 of the reactant remains.

46


Half-Life

Sugar is fermented in a 1st order process (using an

enzyme as a catalyst).

sugar + enzyme products

Rate of disappear of sugar = k[sugar]

k = 3.3 x 10-4 sec-1

What is the half-life of this reaction?

47


Half-Life

Solution

[A] / [A]0 = fraction remaining

when t = t1/2 then fraction remaining = _______

Therefore, ln (1/2) = - k · t1/2

- 0.693 = - k · t1/2

t1/2 = 0.693 / kSo, for sugar,

t1/2 = 0.693 / k = 2100 sec = 35 min

Rate = k[sugar] and k = 3.3 x 10-4 sec-1. What is the half-life of this reaction?

NOTE: For a first-order process, the half-life does not depend on [A]0.

48


Half-Life

Solution2 hr and 20 min = 4 half-livesHalf-life Time Elapsed Mass Left1st 35 min 2.50 g2nd 70 1.25 g3rd 105 0.625 g4th 140 0.313 g

Rate = k[sugar] and k = 3.3 x 10-4 sec-1. Half-life is 35 min. Start with 5.00 g sugar. How much is left after 2 hr and 20 min (140 min)?

49


Second Order• Rate = -Δ[A]/Δt = k[A]2 • integrated rate law • 1/[A] = kt + 1/[A]0 • y= 1/[A] m = k• x= t b = 1/[A]0 • A straight line if 1/[A] vs t is plotted• Knowing k and [A]0 you can calculate [A] at

any time t

50


Determining rxn order

The decomposition of NO2 at 300°C is described by the equation

NO2 (g) NO (g) + 1/2 O2 (g)

and yields these data:

Time (s) [NO2], M

0.0 0.01000

50.0 0.00787

100.0 0.00649

200.0 0.00481

300.0 0.00380

51


Graphing ln [NO2] vs. t yields:

Time (s) [NO2], M ln [NO2]

0.0 0.01000 -4.610

50.0 0.00787 -4.845

100.0 0.00649 -5.038

200.0 0.00481 -5.337

300.0 0.00380 -5.573

• The plot is not a straight line, so the process is not first-order in [A].

Determining rxn order

Does not fit:

52


Second-Order Processes

A graph of 1/[NO2] vs. t gives this plot.

Time (s) [NO2], M 1/[NO2]

0.0 0.01000 100

50.0 0.00787 127

100.0 0.00649 154

200.0 0.00481 208

300.0 0.00380 263

• This is a straight line. Therefore, the process is second-order in [NO2].

53


Zero Order Rate Law• Rate = k[A]0 = k• Rate does not change with

concentration.• Integrated [A] = -kt + [A]0 • When [A] = [A]0/2 then t = t1/2

• t1/2 = [A]0 /2k

54


• Most often when reaction happens on a surface because the surface area stays constant.

• Also applies to enzyme chemistry.

Zero Order Rate Law

55


Time

Concentration

[D A]/Dt

Dt

k =

[D A]

56


Rate Laws SummaryZero Order First Order Second Order

Rate Law Rate = k Rate = k[A] Rate = k[A]2

Integrated Rate Law

[A] = -kt + [A]0

ln[A] = -kt + ln[A]0

Plot that produces a straight line

[A] versus t ln[A] versus t

Relationship of rate constant to slope of straight line

Slope = -k Slope = -k Slope = k

Half-Life

1

[ ]versus t

A

0

1 1

[ ] [ ]kt

A A

01/ 2

[ ]

2

At

k 1/ 2

0.693t

k 1/ 2

0

1

[ ]t

k A

57


Ch. 12.5 - Reaction Mechanisms

Ch. 12.5 - Reaction Mechanisms

Mechanism: how reactants are converted to products at the molecular level.

RATE LAW MECHANISMexperiment theory

58


Reaction Mechanisms

• The molecularity of a process tells how many molecules are involved in the process.

• The rate law for an elementary step is written directly from that step.

59


Reaction Mechanisms• If a reaction is an elementary step,

then we know its rate law.• All of the molecules involved in an

elementary step are defined in the rate law.

• The full molecularity has to be included.

• The sum of the elementary steps in a mechanism must give the overall balanced equation for the reaction.

60


Molecularity Example• Given the reactions and their rate

laws below, which is the only one that can be an elementary step?

A) 2A P Rate = k[A]B) A + B P Rate = k[A][B]C) A + 2B P Rate = k[A]2

D) A + B + C P Rate = k[A][C]

61


Mechanisms

Most rxns. involve a sequence of elementary steps.

2 I- + H2O2 + 2 H+ I2 + 2 H2O

Rate = k [I-] [H2O2]NOTE1. Rate law comes from experiment2. Order and stoichiometric coefficients not

necessarily the same!3. Rate law reflects all chemistry

down to and including the slowest step in multistep reaction.

62


Mechanisms

Proposed Mechanism

Step 1 — slow HOOH + I- HOI + OH-

Step 2 — fast HOI + I- I2 + OH-

Step 3 — fast 2 OH- + 2 H+ 2 H2O

Rate of the reaction controlled by slow step —

RATE DETERMINING STEP, rds.

Rate can be no faster than rds!

Most rxns. involve a sequence of elementary steps. 2 I- + H2O2 + 2 H+ I2 + 2 H2O

Rate = k [I-] [H2O2]

63


Mechanisms

Elementary Step 1 is bimolecular and involves I- and HOOH. Therefore, this predicts the rate law should be

Rate [I-] [H2O2] — as observed!!

The species HOI and OH- are reaction intermediates.

More on intermediates coming up.

2 I- + H2O2 + 2 H+ I2 + 2 H2O

Rate = k [I-] [H2O2]

Step 1 — slow HOOH + I- HOI + OH-

Step 2 — fast HOI + I- I2 + OH-

Step 3 — fast 2 OH- + 2 H+ 2 H2O

64


Rate Laws and Mechanisms

NO2 + CO reaction: Rate = k[NO2]2

Single step

Two possible mechanisms

Two steps: step 1

Two steps: step 2

65


Slow Initial Step

• The rate law for this reaction is found experimentally to be

Rate = k [NO2]2

• CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration.

• This suggests the reaction occurs in two steps.

NO2 (g) + CO (g) NO (g) + CO2 (g)

66


Slow Initial Step• A proposed mechanism for this reaction is

Step 1: NO2 + NO2 NO3 + NO (slow)

Step 2: NO3 + CO NO2 + CO2 (fast)

• The NO3 intermediate is consumed in the second step.

• As CO is not involved in the slow, rate-determining step, it

does not appear in the rate law.

67


Fast Initial Step

• The rate law for this reaction is found (experimentally) to be

• Because termolecular (aka trimolecular) processes are rare, this rate law suggests a two-step mechanism.

67

68


Fast Initial Step• A proposed mechanism is

Step 1 is an equilibrium- it includes the forward and reverse reactions.

69


Fast Initial Step

• The rate of the overall reaction depends upon the rate of the slow step.

• The rate law for that step would be

• But how can we find [NOBr2]?

70


Intermediates• [NOBr2] is an intermediate, an unstable

molecule of low, unknown concentration

• Intermediates can’t appear in the rate law.

• Remember: only reactants (and products, rarely) can appear in a rate law.

• Slow step determines the rate and the rate law.

71


Intermediates• Use the reactions that form them• If the reactions are fast and

irreversible the concentration of the intermediate is based on stoichiometry.

• If it is formed by a reversible reaction set the rates equal to each other.

72


Fast Initial Step

• NOBr2 can react two ways:– With NO to form NOBr– By decomposition to reform NO and Br2

• The reactants and products of the first step are in equilibrium with each other.

• Therefore,

Ratef = Rater

73


Fast Initial Step

• Because Ratef = Rater ,

k1 [NO] [Br2] = k−1 [NOBr2]

Solving for [NOBr2] gives us

k1/k-1 [NO] [Br2] = [NOBr2]

74


Fast Initial Step

Substituting this expression for [NOBr2] in the rate law for the rate-determining step gives

74

75


Ch. 12.6 – A Model for Kinetics

Molecules need a minimum amount of energy to react.

Visualized as an energy barrier - activation energy, Ea.

Reaction coordinate diagram

76


MECHANISMS& Activation Energy

Conversion of cis to trans-2-butene requires

twisting around the C=C bond.

Rate = k [trans-2-butene]

77


MECHANISMS

Activation energy barrier

Cis TransTransition state

78


Energy involved in conversion of trans to cis butene

transcis

energy ActivatedComplex

+262 kJ -266 kJ

MECHANISMS

4 kJ/mol

79


Collision TheoryCollision Theory

Reactions require

(a) activation energy and

(b) correct orientation.

O3(g) + NO(g) O2(g) + NO2(g)

2. Activation energy and geometry1. Activation energy

80


Mechanisms

O3 + NO reaction occurs in a single ELEMENTARY step. Most others involve a sequence of elementary steps.

Adding elementary steps gives NET reaction.

81


Mechanisms

• Reaction passes thru a TRANSITION STATE where there is an

activated complex that has sufficient energy to become a product.

ACTIVATION ENERGY, Ea

= energy req’d to form activated complex.

Here Ea = 262 kJ/mol

82


Also note that trans-butene is MORE STABLE than cis-butene by about 4 kJ/mol.

Therefore, cis trans is EXOTHERMIC

This is the connection between thermo-dynamics and kinetics.

MECHANISMS

83


Effect of TemperatureEffect of Temperature

• Reactions generally occur slower at lower T.

Iodine clock reaction.

H2O2 + 2 I- + 2 H+ 2 H2O + I2

Room temperature

In ice at 0 oC

84


Activation Energy and Temperature

Reactions are faster at higher T because a larger

fraction of reactant molecules have enough energy to

convert to product molecules.

In general, differences in activation energy cause reactions to vary from fast to slow.

In general, differences in activation energy cause reactions to vary from fast to slow.

85


Mechanisms

1. Why is trans-butene cis-butene reaction observed to be 1st order?

As [trans] doubles, number of molecules with enough E also doubles.

2. Why is the trans cis reaction faster at higher temperature?

Fraction of molecules with sufficient activation energy increases with T.

86


More About Activation EnergyMore About Activation Energy

k Ae-Ea/RT

ln k = - (EaR

)(1T

) + ln A

Arrhenius equation —Arrhenius equation —

Rate constant

Temp (K)

8.31 x 10-3 kJ/K•molActivation energy

Frequency factor

Frequency factor related to frequency of collisions with correct geometry.

Plot ln k vs. 1/T straight line. slope = -Ea/R

87


Arrhenius Equation Example

• Calculate the Ea from the data below.

• T(⁰C) k(s-1)• 189.7 2.52 X 10-5

• 198.9 5.25 X 10-5

• 230.3 6.30 X 10-4

• 251.2 3.16 X 10-3

88



• 1st step change T(⁰C) to T(K), find 1/T and change k to ln k.

T(K) 1/T ln(k)462.9 0.002160 -10.589472.1 0.002118 -9.855503.5 0.001986 -7.370524.4 0.001907 -5.757

89


0.00185 0.0019 0.00195 0.002 0.00205 0.0021 0.00215 0.0022

-11

-10

-9

-8

-7

-6

-5

-4

f(x) = − 19079.7028647974 x + 30.5823130270649R² = 0.999461034136358

Arrhenius Plot

1/T

ln k

90



2nd Step find the slope (If not using graphing calculator or excel)

Slope = (-5.757 – (-9.855))/(0.001907 – 0.002118) = 4.098/(-0.000211) = -19422 K3rd Step Multiply the slope by -8.314 J/mol K

to get Ea

= -19422 s-1 * -8.314 J/mol K = 161472 J = 161 kJ

91


Determining the Activation Energy• If we do not have a lot of data, then we recognize

122

1

2121

22

11

11ln

lnlnlnln

lnln and lnln

TTRE

kk

ARTE

ARTE

kk

ARTE

kARTE

k

a

aa

aa

More on the Arrhenius Equation

92


Example Problem

• The Ea of a certain reaction is 65.7 kJ/mol. How many times faster is the rxn at 50⁰C than at 0⁰C?

93


Ch. 12.7 - CATALYSISCh. 12.7 - CATALYSIS

Catalysts speed up reactions by altering the mechanism to lower the activation energy barrier.

Dr. James Cusumano, Catalytica Inc.

What is a catalyst?

Catalysts and society

Catalysts and the environment

PLAY MOVIE

PLAY MOVIE

PLAY MOVIE

94


CATALYSIS

In auto exhaust systems — Pt, NiO

2 CO + O2 2 CO2

2 NO N2 + O2

95


CATALYSIS

2. Polymers: H2C=CH2 polyethylene

3. Acetic acid:

CH3OH + CO CH3CO2H

4. Enzymes — biological catalysts

96


CATALYSISCatalysis and activation energy

Uncatalyzed reaction

Catalyzed reaction

MnO2 catalyzes decomposition of H2O2

2 H2O2 2 H2O + O2

97


Iodine-Catalyzed Isomerization of cis-2-

Butene


Butene

See Figure 15.15

98



Butene


Butene

99


AP Exam Practice• 2010 AP Exam #3b-d• 2009B AP Exam #2• 2009 AP exam #3d-e• 2008 AP Exam #3d-f• 2008B AP Exam #2a-g• 2005B AP Exam #3• 2005 AP Exam #3• 2003 AP Exam #3a-c

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