Chemical Kinetics - Georgia Southern University...

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CHAPTER 15

Chemical Kinetics

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�� We can use We can use thermodynamicsthermodynamics to tell if a reaction is to tell if a reaction is productproduct-- or reactantor reactant--favored.favored.

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

�� KINETICSKINETICS —— the study of the study of REACTION RATESREACTION RATES and and their relation to the way the reaction proceeds, their relation to the way the reaction proceeds, i.e., its i.e., its MECHANISMMECHANISM..

�� The reaction mechanism is our goal!The reaction mechanism is our goal!

Chemical KineticsChemical Kinetics

The Rate of a Reaction

� Kinetics is the study of rates of chemical

reactions and the mechanisms by which they

occur.

� The reaction rate is the increase in concentration

of a product per unit time or decrease in

concentration of a reactant per unit time.

� A reaction mechanism is the series of molecular

steps by which a reaction occurs.

4

Reaction MechanismsReaction Mechanisms

The sequence of events at the molecular level that control the The sequence of events at the molecular level that control the

speed and outcome of a reaction.speed and outcome of a reaction.

Br from biomass burning destroys stratospheric ozone. Br from biomass burning destroys stratospheric ozone.

(See R.J. Cicerone, (See R.J. Cicerone, ScienceScience, volume 263, page 1243, 1994.), volume 263, page 1243, 1994.)

Step 1:Step 1: Br + OBr + O33 ------> > BrOBrO + O+ O22

Step 2:Step 2: ClCl + O+ O33 ------> > ClOClO + O+ O22

Step 3:Step 3: BrOBrO + + ClOClO + light + light ------> Br + > Br + ClCl + O+ O22

NET: NET: 2 O2 O33 ------> 3 O> 3 O22

The Rate of a Reaction

� Thermodynamics determines if a reaction can occur.

� Kinetics determines how quickly a reaction occurs.

� Some reactions that are thermodynamically feasible are kinetically so slow as to be imperceptible.

( ) ( )

( ) ( ) ( )

OUSINSTANTANE

kJ -79=G OHOH+H

SLOW VERY

kJ 396G COO C

o

2982

-

aq

+

aq

o

298g2g2diamond

∆→

−=∆→+

l

The Rate of Reaction� Consider the

hypothetical reaction,

A(g) + B(g) → C(g) + D(g)

� equimolar amounts of

reactants, A and B, will

be consumed while

products, C and D, will

be formed as indicated

in this graph:

0

0.2

0.4

0.6

0.8

1

1.2

0 50 100

150

200

250

300

350

Time

Concentrations of

Reactants & Products

[A] & [B]

[C] & [D]

[ ] [ ] [ ] [ ]t d

D+

t c

C+or

t b

B-

t a

A-= Rate

∆∆

=∆∆

∆∆

=∆∆

2

7

�� Reaction rate = change in concentration of a reactant or productReaction rate = change in concentration of a reactant or product

with time.with time.

�� Three Three ““typestypes”” of rates of rates

�� initial rateinitial rate

�� average rateaverage rate

�� instantaneous rateinstantaneous rate

Reaction Rates Reaction Rates

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Determining a Reaction RateDetermining a Reaction Rate

Blue dye is oxidized with Blue dye is oxidized with

bleach. bleach.

Its concentration decreases Its concentration decreases

with time.with time.

The rate The rate —— the change in dye the change in dye

concconc with time with time —— can be can be

determined from the plot.determined from the plot.

Slope of tangent line to initial Slope of tangent line to initial

part of curve.part of curve.

Dy

e

Dy

e C

on

cC

on

c

Time

9

Determining a Reaction RateDetermining a Reaction Rate Rate Law� Example: The rate of a simple one-step, elementary reaction is

directly proportional to the concentration of the reacting substance.

� [A] is the concentration of A in molarity, k is the rate constant.

� Example: proportional to the molar concentration of the reactant raised to the power of the number of molecules involved in the reaction.

[ ] [ ]Ak = Rateor ARate

C + BA (g)(g)(g)

∝

→

( ) ( ) ( )

[ ] [ ]22

ggg

Xk = Rateor XRate

Z+ YX 2

∝

→

11

Concentrations, Rates, & Concentrations, Rates, & Rate LawsRate Laws

In general, for In general, for

a A + b B a A + b B ----> x X> x X

Rate = k [Rate = k [A]A]mm[B][B]nn

K is the rate constantK is the rate constant

The exponents m, n, and p The exponents m, n, and p

•• are the are the reaction orderreaction order

•• can be 0, 1, 2 or fractionscan be 0, 1, 2 or fractions

•• must be determined by experimentmust be determined by experiment

THEY ARE NOT BALANCED EQUATION THEY ARE NOT BALANCED EQUATION COEFFICIENTSCOEFFICIENTS 12

Interpreting Rate Interpreting Rate Laws Laws

Rate = k [Rate = k [A]A]mm[B][B]nn

�� If m = 1, If m = 1, rxnrxn. is 1st order in A. is 1st order in A

Rate = k [A]Rate = k [A]11

If [A] doubles, then rate goes up by factor of __ If [A] doubles, then rate goes up by factor of __

�� If m = 2, If m = 2, rxnrxn. is 2nd order in A.. is 2nd order in A.


Doubling [A] increases rate by ________Doubling [A] increases rate by ________

�� If m = 0, If m = 0, rxnrxn. is zero order.. is zero order.


If [A] doubles, rate ________If [A] doubles, rate ________

3

The Rate of Reaction – Rate Law

� Rate Law Expressions must be determined experimentally.

� The rate law cannot be determined from the final

balanced chemical equation, only from simple, one step

mechanistic equations

�The order of a reaction can be expressed in terms of

either: each reactant in the reaction or the overall reaction.

�Order for the overall reaction is the sum of the orders for each

reactant in the reaction.

�For example: ( ) ( ) ( )

[ ]

overall.order first and

ONin order first isreaction This

ONk= Rate

O + NO4ON 2

52

52

g2g2g52 →

The Rate of Reaction� Example: ( ) ( ) ( )

[ ]

overallorder third and ,Oin order first

NO,in order second isreaction This

isreaction This Ok[NO]=Rate

NO 2O+NO 2

2

2

2

g2g2g →

( ) ( ) ( )

[ ]2ggg

Ak = Rate

CBA 2 +→�Example:

�Because it is a second order rate-law

expression:

�If the [A] is doubled the rate of the reaction

will increase by a factor of ?????

�If the [A] is halved the rate of the reaction

will decrease by a factor of ??????

Factors That Affect Reaction

Rates

� There are several factors that can influence the

rate of a reaction:

1. The nature of the reactants.

2. The concentration of the reactants.

3. The temperature of the reaction.

4. The presence of a catalyst.

� We will look at each factor individually.

Nature of Reactants� This is a very broad category that encompasses the different

reacting properties of substances.

� For example sodium reacts with water explosively at room

temperature to liberate hydrogen and form sodium hydroxide.

( ) ( ) ( ) ( )

burns. and ignites H The

reaction. rapid and violent a is This

HNaOH 2OH 2Na 2

2

g2aq2s +→+l

( ) ( ) reaction No OH Mg 2s →+l

•The reaction of magnesium with water at room temperature is so

slow that that the evolution of hydrogen is not perceptible to the

human eye.

Concentrations of Reactants: The

Rate-Law Expression

� This is a simplified representation of the effect of

different numbers of molecules in the same

volume.

� The increase in the molecule numbers is indicative of

an increase in concentration.

A(g) + B (g) →→→→ Products

A B

A B

A B

B

A B

A B

A B

A B

4 different possible

A-B collisions


A-B collisions


A-B collisions18

Deriving Rate LawsDeriving Rate Laws

ExptExpt. . [CH[CH33CHO]CHO] Disappear of CHDisappear of CH33CHOCHO

(mol/L)(mol/L) (mol/L(mol/L••sec)sec)

11 0.100.10 0.0200.020

22 0.200.20 0.0810.081

33 0.300.30 0.1820.182

44 0.400.40 0.3180.318

Derive rate law and k for Derive rate law and k for CHCH33CHO(g) CHO(g) ----> CH> CH44(g) + CO(g)(g) + CO(g)

from experimental data for rate of disappearance from experimental data for rate of disappearance of CHof CH33CHOCHO

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Deriving Rate LawsDeriving Rate Laws

Rate of Rate of rxnrxn = ??= ??

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


Rate-Law Expression� Example: The following rate data were obtained at 25oC

for the following reaction. What are the rate-law expression and the specific rate-constant for this reaction?

2 A(g) + B(g) → 3 C(g)

2.0 x 10-40.200.103

4.0 x 10-40.300.202

2.0 x 10-40.100.101

Initial rate of

formation of C

(M/s)

Initial [B]

(M)

Initial [A]

(M)

Experiment

Number

[ ] [ ]yxBAk=Rate


Rate-Law Expression2 A(g) + B(g) + 2 C(g) → 3 D(g) + 2 E(g)

2.0 x 10-40.100.100.201

1.8 x 10-30.400.300.604

2.0 x 10-40.300.100.203

6.0 x 10-40.200.300.202

Initial rate

of

formation

of D (M/s)

Initial

[C]

(M)

Initial [B]

(M)

Initial [A]

(M)

Experimen

t


Rate-Law Expression

� Example : consider a chemical reaction between

compounds A and B that is first order with respect to A,

first order with respect to B, and second order overall.

From the information given below, fill in the blanks.

?0.403.2 x 10-23

0.050?1.6 x 10-22

0.0500.204.0 x 10-31

Initial [B]

(M)

Initial [A]

(M)

Initial Rate

(M/s)Experiment

Concentration vs. Time: The Integrated Rate Equation

� The integrated rate equation relates time and concentration for

chemical and nuclear reactions.

� From the integrated rate equation we can predict the amount

of product that is produced in a given amount of time.

� First order reactions: 1st order in the reactant and 1st order

overall.

A → products

This is a common reaction type for many chemical reactions and all simple radioactive decays.

�Two examples of this type are:

2 N2O5(g) → 2 N2O4(g) + O2(g)

238U → 234Th + 4He 24

Concentration/Time Concentration/Time RelationsRelations

What is concentration of reactant as function of What is concentration of reactant as function of

time? time?

Consider Consider FIRST ORDER REACTIONSFIRST ORDER REACTIONS

The rate law isThe rate law is

Rate ==== -∆∆∆∆[A]

∆∆∆∆time = k [A]

5

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Concentration/Time RelationsConcentration/Time Relations

Integrating Integrating -- ((∆∆ [A] / [A] / ∆∆

time) = k [A], we gettime) = k [A], we get[A] = - k tln[A]o

naturallogarithm [A] at time = 0

[A] = - k tln[A]o

naturallogarithm [A] at time = 0

[A] / [A][A] / [A]00 =fraction remaining after time t =fraction remaining after time t

has elapsed.has elapsed.

Called the Called the integrated firstintegrated first--order rate laworder rate law..

oAktA ]ln[]ln[ +−=26

Concentration/Time RelationsConcentration/Time Relations

Sucrose decomposes to simpler sugarsSucrose decomposes to simpler sugars

Rate of disappearance of sucrose = k Rate of disappearance of sucrose = k

[sucrose][sucrose]

GlucoseGlucose

If k = 0.21 hrIf k = 0.21 hr--11

and [sucrose] = 0.010 Mand [sucrose] = 0.010 M

How long to drop 90% How long to drop 90%

(to 0.0010 M)?(to 0.0010 M)?

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Using the Integrated Rate LawUsing the Integrated Rate Law

The integrated rate law suggests a way to tell the order based oThe integrated rate law suggests a way to tell the order based on n

experiment. experiment.

2 N2 N22OO55(g) (g) ------> 4 NO> 4 NO22(g) + O(g) + O22(g)(g)

Time (min)Time (min) [N[N22OO55]]00 (M) (M) lnln [N[N22OO55]]00

00 1.001.00 00

1.01.0 0.7050.705 --0.350.35

2.02.0 0.4970.497 --0.700.70

5.05.0 0.1730.173 --1.751.75

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2 N2 N22OO55(g) (g) ------> 4 NO> 4 NO22(g) + O(g) + O22(g)(g)

[N2 O5 ] vs . t i me

t i me

1

0

0 5

l n [N2 O5 ] vs. t i me

t i me

0

-2

0 5

Data of conc. vs. Data of conc. vs.

time plot do not fit time plot do not fit

straight line.straight line.

Plot of Plot of lnln [N[N22OO55] vs. ] vs.

time is a straight time is a straight

line!line!

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All 1st order reactions have straight line plot for All 1st order reactions have straight line plot for lnln [A] [A]

vs. time. vs. time.

(2nd order gives straight line for plot of 1/[A] vs. time)(2nd order gives straight line for plot of 1/[A] vs. time)

l n [N2 O5 ] vs. t i me

t i me

0

-2

0 5

Plot of Plot of lnln [N[N22OO55] vs. time ] vs. time

is a straight line! is a straight line!

EqnEqn. for straight line: . for straight line:

y = y = mxmx + b + b

ln [N 2O5] = - kt + ln [N 2O5]o

conc at time t

rate const = slope

conc at time = 0

Rate = k [NRate = k [N22OO55]]

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HalfHalf--LifeLifeSection 15.4 & Screen 15.8Section 15.4 & Screen 15.8

HALFHALF--LIFELIFE is is

the time it the time it

takes for 1/2 a takes for 1/2 a

sample is sample is

disappear.disappear.

For 1st order For 1st order

reactions, the reactions, the

concept of concept of

HALFHALF--LIFE LIFE

is especially is especially

useful.useful.Active Figure 15.9

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31

�� Reaction is 1st order Reaction is 1st order

decomposition of decomposition of

HH22OO22..

HalfHalf--LifeLife

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HalfHalf--LifeLife

�� Reaction after 1 Reaction after 1

halfhalf--life.life.

�� 1/2 of the 1/2 of the

reactant has been reactant has been

consumed and consumed and

1/2 remains.1/2 remains.

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HalfHalf--LifeLife

�� After 2 halfAfter 2 half--lives lives

1/4 of the reactant 1/4 of the reactant

remains.remains.

34

HalfHalf--LifeLife

�� A 3 halfA 3 half--lives 1/8 of lives 1/8 of

the reactant the reactant

remains.remains.

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HalfHalf--LifeLife

�� After 4 halfAfter 4 half--lives lives

1/16 of the reactant 1/16 of the reactant

remains.remains.

Concentration vs. Time: The

Integrated Rate Equation

� Solve the first order integrated rate equation for t.

� Define the half-life, t1/2, of a reactant as the time required

for half of the reactant to be consumed, or the time at

which [A]=1/2[A]0.

[ ][ ]

−=

oA

Aln

k

1t

7



� At time t = t1/2, the expression becomes:

[ ][ ]

( )

k

693.0t

5.0lnk

1t

A

A2/1ln

k

1t

1/2

1/2

0

01/2

=

−=

−=

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HalfHalf--LifeLife

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

an enzyme as a catalyst).an enzyme as a catalyst).

sugar + enzyme sugar + enzyme ----> products> products

Rate of disappear of sugar = k[sugar]Rate of disappear of sugar = k[sugar]

k = 3.3 x 10k = 3.3 x 10--44 secsec--11

What is the What is the halfhalf--lifelife of this reaction?of this reaction?

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Rate = k[sugar] and k = 3.3 x 10Rate = k[sugar] and k = 3.3 x 10--44 secsec--11. Half. Half--life life is 35 min. Start with 5.00 g sugar. How much is is 35 min. Start with 5.00 g sugar. How much is left after 2 hr and 20 min (140 min)?left after 2 hr and 20 min (140 min)?

40


Radioactive decay is a first order process. Radioactive decay is a first order process.

Tritium Tritium ------> electron + helium> electron + helium33HH 00

--11ee 33HeHe

tt1/21/2 = 12.3 years= 12.3 years

If you have 1.50 mg of tritium, how much is left If you have 1.50 mg of tritium, how much is left

after 49.2 years? after 49.2 years?

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HalfHalf--Lives of Radioactive ElementsLives of Radioactive Elements

Rate of decay of radioactive isotopes given in Rate of decay of radioactive isotopes given in

terms of 1/2terms of 1/2--life. life.

238238U U ----> > 234234Th + HeTh + He 4.5 x 104.5 x 1099 yy

1414C C ----> > 1414N + betaN + beta 5730 y5730 y

131131I I ----> > 131131Xe + betaXe + beta 8.05 d8.05 d

Element 106 Element 106 -- seaborgiumseaborgium263263Sg Sg 0.9 s0.9 s



� Example: Cyclopropane, an anesthetic,

decomposes to propene according to the

following equation.

The reaction is first order in cyclopropane with k = 9.2

s-1 at 10000C. Calculate the half life of cyclopropane

at 10000C.

CH2

CH2

CH2

CH2

CH3

CH

(g)(g)

8



� Example: Cyclopropane, an anesthetic,

decomposes to propene according to the

following equation.

The reaction is first order in cyclopropane with k = 9.2

s-1 at 10000C. Calculate the half life of cyclopropane

at 10000C.

CH2

CH2

CH2

CH2

CH3

CH

(g)(g)



� Example: Refer to previous. How much of a 3.0

g sample of cyclopropane remains after 0.50

seconds?

� The integrated rate laws can be used for any unit that

represents moles or concentration.

� In this example we will use grams rather than mol/L.



� For reactions that are second order with respect to a

particular reactant and second order overall, the rate

equation is:

� Where:

[A]0= mol/L of A at time t=0. [A] = mol/L of A at time t.

k = specific rate constant. t = time elapsed since

beginning of reaction.

[ ] [ ]k t

A

1

A

1

0

=−

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Properties of ReactionsProperties of Reactions

Reaction Mechanisms and the

Rate-Law Expression

� Use the experimental rate-law to postulate a

mechanism.

� The slowest step in a reaction mechanism is the

rate determining step.

� Consider the iodide ion catalyzed decomposition

of hydrogen peroxide to water and oxygen.

( ) ( ) ( )g22

I

22 O + OH 2 OH 2-

ll→

Reaction Mechanisms and the

Rate-Law Expression

� This reaction is known to be first order in H2O2 , first

order in I- , and second order overall.

� The mechanism for this reaction is thought to be:

[ ][ ]-22

2222

-

2222

-

2

--

22

IOHk=R law rate alExperiment

O+OH 2OH 2reaction Overall

I+O+OHOH+ IO stepFast

OH+IOI+OH step Slow

→

→

→

UNIMOLECULAR UNIMOLECULAR -- only one reactant is involved.only one reactant is involved.

BIMOLECULARBIMOLECULAR —— two different molecules must collide two different molecules must collide ----> products> products

9

49

Collision Theory of

Reaction Rates

� Three basic events must occur for a reaction to

occur the atoms, molecules or ions must:

1. Collide.

2. Collide with enough energy to break and form bonds.

3. Collide with the proper orientation for a reaction to

occur.

50

Collision Theory of

Reaction Rates

� One method to increase the number of collisions and the

energy necessary to break and reform bonds is to heat the

molecules.

� As an example, look at the reaction of methane and

oxygen:

� We must start the reaction with a match.

� This provides the initial energy necessary to break the first few

bonds.

� Afterwards the reaction is self-sustaining.

kJ 891 OH CO O CH (g)22(g)2(g)4(g) ++→+

51

Collision Theory of

Reaction Rates

� Illustrate the proper orientation of molecules that

is necessary for this reaction.

X2(g) + Y2(g) →2 XY(g)

� Some possible ineffective collisions are :

X

X

Y YY

Y

X X X X Y Y

52

Collision Theory of

Reaction Rates

� An example of an effective collision is:

X Y

X Y

X Y

X Y

X Y

+

X Y

53

Collision TheoryCollision Theory

Reactions require Reactions require

(a) activation energy and (a) activation energy and

(b) correct geometry. (b) correct geometry.

OO33(g) + NO(g) (g) + NO(g) ------> O> O22(g) + NO(g) + NO22(g)(g)

2. Activation energy 2. Activation energy

and geometryand geometry1. Activation energy 1. Activation energy

54

MechanismsMechanisms

OO33 + NO reaction occurs in a single + NO reaction occurs in a single

ELEMENTARYELEMENTARY step. Most others involve a step. Most others involve a

sequence of elementary steps.sequence of elementary steps.

Adding elementary steps gives NET reaction.Adding elementary steps gives NET reaction.

10

55


Most Most rxnsrxns. involve a sequence of elementary . involve a sequence of elementary

steps.steps.

2 I2 I-- + H+ H22OO22 + 2 H+ 2 H++ ------> I> I22 + 2 H+ 2 H22OO

Rate = k [IRate = k [I--] [H] [H22OO22]]

NOTENOTE

1.1. Rate law comes from experimentRate law comes from experiment

2.2. Order and Order and stoichiometricstoichiometric coefficients not coefficients not necessarily the same!necessarily the same!

3.3. Rate law reflects all chemistry down Rate law reflects all chemistry down to and including the slowest step in to and including the slowest step in multistepmultistep reaction.reaction.

56


Proposed MechanismProposed Mechanism

Step 1 Step 1 —— slowslow HOOH + IHOOH + I-- ----> HOI + OH> HOI + OH--

Step 2 Step 2 —— fastfast HOI + IHOI + I-- ----> I> I22 + OH+ OH--

Step 3 Step 3 —— fastfast 2 OH2 OH-- + 2 H+ 2 H++ ----> 2 H> 2 H22OO

Rate of the reaction controlled by slow step Rate of the reaction controlled by slow step ——

RATE DETERMINING STEPRATE DETERMINING STEP, , rdsrds..

Rate can be no faster than Rate can be no faster than rdsrds!!

Most Most rxnsrxns. involve a sequence of elementary steps.. involve a sequence of elementary steps.

2 I2 I-- + H+ H22OO22 + 2 H+ 2 H++ ------> I> I22 + 2 H+ 2 H22OO


57


Elementary Step 1Elementary Step 1 is is bimolecularbimolecular and involves Iand involves I--

and HOOH. Therefore, this predicts the rate law and HOOH. Therefore, this predicts the rate law

should beshould be

Rate Rate ∝∝ [I[I--] [H] [H22OO22] ] —— as observed!!as observed!!

The species HOI and OHThe species HOI and OH-- are are reaction reaction intermediates.intermediates.

2 I2 I-- + H+ H22OO22 + 2 H+ 2 H++ ------> I> I22 + 2 H+ 2 H22OO


Step 1 Step 1 —— slowslow HOOH + IHOOH + I-- ----> HOI + OH> HOI + OH--

Step 2 Step 2 —— fastfast HOI + IHOI + I-- ----> I> I22 + OH+ OH--

Step 3 Step 3 —— fastfast 2 OH2 OH-- + 2 H+ 2 H++ ----> 2 H> 2 H22OO

58

Rate Laws and Rate Laws and MechanismsMechanisms

NO2 + CO reaction:

Rate = k[NO2]2

Single step

Two possible Two possible

mechanismsmechanismsTwo steps: step 1

Two steps: step 2

59

Ozone Decomposition Ozone Decomposition over Antarcticaover Antarctica

2 O2 O33 (g) (g) ------> 3 O> 3 O22 (g)(g) 60

Ozone Decomposition Ozone Decomposition MechanismMechanism

Proposed mechanismProposed mechanism

Step 1: fast, equilibriumStep 1: fast, equilibrium

O3 (g) + O2 (g) + O (g)

Step 2: slow O3 (g) + O (g) ---> 2O2 (g)

2 O2 O33 (g) (g) ------> 3 O> 3 O22 (g)(g)

Rate = k [O3]

2

[O2 ]

11

Transition State Theory

� Transition state theory postulates that reactants

form a high energy intermediate, the transition

state, which then falls apart into the products.

� For a reaction to occur, the reactants must acquire

sufficient energy to form the transition state.

� This energy is called the activation energy or Ea.

� Look at a mechanical analog for activation energy


∆Epot = mg∆h

Cross section

of mountain

Boulder

Eactivation

∆h

h2

h1

Epot=mgh2

Epot=mgh1

Height


Potential

Energy

Reaction Coordinate

X2 + Y2

2 XY

Eactivation - a kinetic quantity

∆E ≈∆H

a thermodynamic

quantity

Representation of a chemical reaction.


� The relationship between the activation energy for

forward and reverse reactions is

� Forward reaction = Ea

� Reverse reaction = Ea + ∆E

� difference = ∆E

65

Activation Energy and TemperatureActivation Energy and Temperature

Reactions are Reactions are faster at higher Tfaster at higher T because a larger fraction of because a larger fraction of

reactant molecules have enough energy to convert to product reactant molecules have enough energy to convert to product

molecules.molecules.

In general, In general,

differences in differences in activation energyactivation energycause reactions to cause reactions to

vary from fast to vary from fast to

slow.slow.

66

More About Activation EnergyMore About Activation Energy

k = Ae-E

a/RT

ln k = - (Ea

R)(

1

T) + ln A

Arrhenius equation Arrhenius equation ——

Rate Rate

constantconstant

Temp (K)Temp (K)

8.31 x 108.31 x 10--33 kJ/KkJ/K••molmolActivation Activation

energyenergyFrequency factorFrequency factor

Frequency factor related to frequency of collisions

with correct geometry.

Plot Plot lnln k vs. 1/T k vs. 1/T

------> >

straight line. straight line.

slope = slope = --EEaa/R/R

12

Temperature:

The Arrhenius Equation

� If the Arrhenius equation is written for two temperatures,

T2 and T1 with T2 >T1.

ln k ln A -E

RT

and

ln k ln A -E

RT

1a

1

2a

2

=

=

Temperature:


1. Subtract one equation from the other.

ln k k A - ln A -E

RT

E

RT

ln k kE

RT-

E

RT

2 1a

2

a

1

2 1a

1

a

2

− = − −

− =

ln ln

ln

Temperature:


2. Rearrange and solve for ln k2/k1.

ln k

k

E

R T T

or

ln k

k

E

R

T - T

T T

2

1

a

1 2

2

1

a 2 1

2 1

= −

=

1 1

Temperature:


� Consider the rate of a reaction for which Ea=50 kJ/mol, at 20oC (293 K) and at 30oC (303 K). � How much do the two rates differ?

Temperature:


� For reactions that have an Ea≈50 kJ/mol, the rate approximatelydoubles for a 100C rise in temperature, near room temperature.

� Consider:

2 ICl(g) + H2(g) → I2(g) + 2 HCl(g)

� The rate-law expression is known to be R=k[ICl][H2].

At 230 C, k = 0.163 s

At 240 C, k = 0.348 s

k approximately doubles

0 -1 -1

0 -1 -1

M

M

Catalysts� Catalysts change reaction rates by providing an

alternative reaction pathway with a different activation

energy.

� Homogeneous catalysts exist in same phase as the

reactants.

� Heterogeneous catalysts exist in different phases than the

reactants.

13

Catalysts

� Examples of commercial catalyst systems include:

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

systemconverter catalytic Automobile

ONNO 2

CO 2O+CO 2

OH 18CO16O 25+HC

g2g2

Pt and NiO

g

g2

Pt and NiO

g2g

g2g2

Pt and NiO

g2g188

+ →

→

+ →

74

CATALYSISCATALYSIS

Catalysis and activation energyCatalysis and activation energy

UncatalyzedUncatalyzed reactionreaction

Catalyzed reactionCatalyzed reaction

MnOMnO22 catalyzes catalyzes

decomposition of Hdecomposition of H22OO22

2 H2 H22OO22 ------> 2 H> 2 H22O + OO + O22

Chemical Kinetics - Georgia Southern University...

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