AP Notes Chapter 15Principles of Reactivity:Principles of Reactivity:

Chemical KineticsChemical Kinetics

Chemical KineticsChemical Kinetics The study of how fast chemical The study of how fast chemical

reactions occur. reactions occur. How fast species are appearing and How fast species are appearing and

disappearing.disappearing.

It has nothing to do with the extent to which It has nothing to do with the extent to which

a reaction will occur – that is thermodynamics.a reaction will occur – that is thermodynamics.

There is a difference between the There is a difference between the probability that molecules will react probability that molecules will react when they meet each other and once when they meet each other and once they do react how stable is the product.they do react how stable is the product.

First, probability of reaction has to do First, probability of reaction has to do with overcoming an energy barrier or with overcoming an energy barrier or activation energy.activation energy.

Second, has to do with the Second, has to do with the differences between reactants and differences between reactants and products.products.

Thermodynamics:Thermodynamics: Spontaneity deals with how likely a Spontaneity deals with how likely a

reaction will take place.reaction will take place.

Spontaneous reactions are reactions Spontaneous reactions are reactions that will happen - but we can’t tell that will happen - but we can’t tell how fast.how fast.

Diamond will spontaneously turn to Diamond will spontaneously turn to graphite – eventually.graphite – eventually.

Factors Affecting Rates

1. Surface area Powders

2. Temperature Dye

3. Pressure4. Catalyst /Inhibitor5. Concentration H202 & I2

Reaction Rate

The change in concentration of a reactant or product per unit time

For a reaction A B

Average Rate = Change in Moles of B Change in time

Rate = [B] = - [A] t t

Reaction RateReaction Rate Rate = Rate = Conc. of A at tConc. of A at t22 -Conc. of A at -Conc. of A at

tt11 tt22- t- t11

Rate =Rate =[A][A] tt Change in concentration per unit timeChange in concentration per unit time Changes in Pressure Changes in Pressure to changes in to changes in

concentration concentration

NN22 + 3H + 3H22 2NH 2NH33

As the reaction progresses the As the reaction progresses the concentration Hconcentration H22 goes down goes down

Concentration

Time

[H[H22]]

NN22 + 3H + 3H22 2NH 2NH33 As the reaction progresses the As the reaction progresses the

concentration Nconcentration N22 goes down 1/3 as fast goes down 1/3 as fastConcentration

Time

[H[H22]]

[N[N22]]

NN22 + 3H + 3H22 2NH 2NH33 As the reaction progresses the As the reaction progresses the

concentration NHconcentration NH33 goes up. goes up.Concentration

Time

[H[H22]]

[N[N22]]

[NH[NH33]]

Calculating RatesCalculating Rates Average ratesAverage rates are taken over long are taken over long

intervalsintervals

Instantaneous rates are determined by are determined by finding the slope of a line tangent to finding the slope of a line tangent to the curve at any given point because the curve at any given point because the rate can change over timethe rate can change over time

Derivative and Integrated Rates..

Average slope methodAverage slope method

Concentration

Time

[H[H22]]

tt

Instantaneous slope method.Instantaneous slope method.

Concentration

Time

[H[H22]]

tt

Defining RateDefining Rate We can define rate in terms of the We can define rate in terms of the

disappearance of the reactant or in disappearance of the reactant or in terms of the rate of appearance of the terms of the rate of appearance of the product.product.

In our exampleIn our example

NN22 + 3H + 3H22 2NH 2NH33

--[N[N22] ] = = - - [H[H22] ] = = [NH[NH33] ] t t 3 3 t t 2 2 t t

RATE LAWMathematical model that contains specific information about what reactants determine the rate and how the rate is controlled

Rate LawsRate Laws Reactions are reversible.Reactions are reversible. As products accumulate they can As products accumulate they can

begin to turn back into reactants.begin to turn back into reactants. Early on the rate will depend on only Early on the rate will depend on only

the amount of reactants present.the amount of reactants present. We want to measure the reactants as We want to measure the reactants as

soon as they are mixedsoon as they are mixed.. This is called the This is called the Initial rate methodInitial rate method..

Two key pointsTwo key points The concentration of the products do The concentration of the products do

not appear in the rate law because not appear in the rate law because this is an initial rate.this is an initial rate.

The order must be determined The order must be determined experimentally,experimentally,

can’tcan’t be obtained from the equation be obtained from the equation

Rate LawsRate Laws

You will find that the rate will only You will find that the rate will only depend on the concentration of the depend on the concentration of the reactants.reactants.

Rate = Rate = kk[NO[NO22]]nn

This is called a This is called a rate law expressionrate law expression.. kk is called the rate constant. is called the rate constant. n is the order of the reactant -usually n is the order of the reactant -usually

a positive integer.a positive integer.

2 NO2 NO2 2 22 NO + ONO + O22

The rate of appearance of OThe rate of appearance of O22 can be said to can be said to be.be.

Rate' = Rate' = [O[O22]] = = [NO[NO22] ] = k'[NO = k'[NO22]] tt 2 2 tt

Because there are 2 NOBecause there are 2 NO22 for each O for each O22

Rate = 2 x Rate'Rate = 2 x Rate' So k[NOSo k[NO22]]

nn = 2 x k'[NO= 2 x k'[NO22]]nn

So k = 2 x k' So k = 2 x k'

2 NO2 NO2 2 22 NO + ONO + O22

Types of Rate LawsTypes of Rate Laws Differential Rate law - describes how Differential Rate law - describes how

rate depends on concentration.rate depends on concentration. Integrated Rate Law - Describes how Integrated Rate Law - Describes how

concentration depends on time.concentration depends on time. For each type of differential rate law For each type of differential rate law

there is an integrated rate law and there is an integrated rate law and vice versa.vice versa.

Rate laws can help us better Rate laws can help us better understand reaction mechanisms.understand reaction mechanisms.

Determining Rate LawsDetermining Rate Laws The first step is to determine the form of The first step is to determine the form of

the rate law (especially its order).the rate law (especially its order). Must be determined from experimental Must be determined from experimental

data.data. For this reactionFor this reaction 2 2

NN22OO55 (aq) 4NO (aq) 4NO22 (aq) + O (aq) + O22(g)(g)

The reverse reaction won’t play a roleThe reverse reaction won’t play a role

[N[N22OO55] (mol/L) ] (mol/L) Time (s) Time (s)

1.001.00 00

0.880.88 200200

0.780.78 400400

0.690.69 600600

0.610.61 800800

0.540.54 10001000

0.480.48 12001200

0.430.43 14001400

0.380.38 16001600

0.340.34 18001800

0.300.30 20002000

Now graph Now graph the datathe data

0

0.2

0.4

0.6

0.8

1

1.2

0 200

400

600

800

1000

1200

1400

1600

1800

2000 To find rate we have to find the slope at two pointsTo find rate we have to find the slope at two points

We will use the tangent method.We will use the tangent method.

0

0.2

0.4

0.6

0.8

1

1.2

0 200

400

600

800

1000

1200

1400

1600

1800

2000At .90 M the rate is

(.98 - .76) = 0.22 =- 5.5x 10 -4 (0-400) -400

0

0.2

0.4

0.6

0.8

1

1.2

0 200

400

600

800

1000

1200

1400

1600

1800

2000

At .40 M the rate is

(.52 - .31) = 0.22 =- 2.7 x 10 -4 (1000-1800) -800

Since the rate at twice the Since the rate at twice the concentration is twice as fast the concentration is twice as fast the rate law must be..rate law must be..

Rate = -Rate = -[N[N22OO55]] = k[N = k[N22OO55]]11 = = k[Nk[N22OO55]] t t

We say this reaction is first order in We say this reaction is first order in NN22OO55

The only way to determine order is The only way to determine order is to run the experiment.to run the experiment.

The method of Initial The method of Initial RatesRates

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

The initial concentrations of the The initial concentrations of the reactants are varied.reactants are varied.

The reaction rate is measured bust The reaction rate is measured bust after the reactants are mixed.after the reactants are mixed.

Eliminates the effect of the reverse Eliminates the effect of the reverse reaction.reaction.

Integrated Rate LawIntegrated Rate Law Expresses the reaction concentration Expresses the reaction concentration

as a function of time.as a function of time. Form of the equation depends on the Form of the equation depends on the

order of the rate law (differential).order of the rate law (differential). Rate = Rate = [A][A]nn

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

although negative and fractional although negative and fractional existexist

General Rate LawaA + bB cCRate -[A] or -[B]r = k[A]m[B]n

Where: m = order with respect to A n = order wrt B and n + m = order of reaction

Notice we are primarily discussing the reactants

For example: aA + bB For example: aA + bB cC + dD cC + dD Zero Order: r = kZero Order: r = k First Order in A: r = k [A]First Order in A: r = k [A] First Order in B: r = k [B]First Order in B: r = k [B] Second Order in A: r = k [A]Second Order in A: r = k [A]22

Second Order in B: r = k [B]Second Order in B: r = k [B]22

First Order in A, first Order in B, for a First Order in A, first Order in B, for a second order overall: r = k [A][B]second order overall: r = k [A][B]

Initial Rate Method to establish the exact rate law

R0 = k[A0]m[B0]n

but experimental data must provide orders

Use ratio method to solve for m and n

R0 = k[A0]m[B0]n

Exp Rate [A0] [B0] 1 0.05 0.50 0.20 2 0.05 0.75 0.20 3 0.10 0.50 0.40

R0 = k[A0]m[B0]n

075.0

50.01

20.075.005.0

20.050.005.0:

2 Exp

1

mk

kExpm

nm

nm

140.0

20.0

2

1

40.050.010.0

20.050.005.0:

3 Exp

1

nk

kExpn

nm

nm

Thus, rate = k[A]0[B]1 or rate = k[B]

Where k is the rate constant , with appropriate units.

rate = k[B]

Using Exp 1:

1-s 25.0

20.005.0

k

L

molk

sL

mol

INTEGRATED RATE LAWS

Each “order” has a separate “law” based on the integration of the rate expression.

You NEED TO KNOW the rate expressions& half-life formulas for 0, 1st, and 2nd

order reactions.

Third order and fractional order reactions exist, but are rare.

Nuclear reactions are typically 1st order, therefore, use the first-order rate law and half-life to solve problems of this type

aAaABB Zero Zero OrderOrder

First OrderFirst Order Second Second OrderOrder

Rate LawRate Law r=k[A]r=k[A]oo r=k[A]r=k[A]11 r=k[A]r=k[A]22

d[A] = d[A] = k[A]mk[A]m

y=mx +by=mx +b

[A]=-kt +[A]=-kt +[A][A]oo

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

1 1 = kt + = kt + 1 1

[A] [A][A] [A]oo

Plot for a Plot for a linear linear graphgraph

[A] vs t[A] vs t ln[A] vs tln[A] vs t 1 1 vs t vs t

ln[A]ln[A]

slope of slope of linear plotlinear plot

-k-k -k-k kk

half-lifehalf-life

t1/2 whent1/2 whentt1/21/2= = [A][A]oo

2k2k

tt1/21/2= = 0.6930.693

kk

tt1/21/2= = 1 1

k[A]k[A]oo

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

Also applies to enzyme chemistry.Also applies to enzyme chemistry.

Zero Order Rate LawZero Order Rate Law

Time

Concentration

Time

Concentration

A]/t

t

k =

A]

Zero Order Rate LawZero Order Rate Law Rate = k[A]Rate = k[A]00 = k = k Rate does not change with Rate does not change with

concentration.concentration. Integrated [A] = -kt + [A]Integrated [A] = -kt + [A]00 When [A] = [A]When [A] = [A]00 /2 t = t /2 t = t1/2 1/2

tt1/2 = 1/2 = [A][A]00 /2k /2k

For a zero-order reaction, the rate is independent of the concentration.

Expressed in terms of calculus, a zero - order reaction would be expressed as:

dtkor

kBrate

dB dt

dB- 0

tB

BdtkdB

00

t

[B] m = -k

00 0 BktBortkBB

Time for the concentration to become one-half of its initial value

0

021

t

into substitute

B if

BkB

then

B

k2

Bt

Bt kB

0

0021

21

21

Half-life 0-Order Reaction

First OrderFirst Order

For the reaction 2NFor the reaction 2N22OO55 4NO 4NO22 + O + O22

We found the Rate = k[We found the Rate = k[NN22OO55]]11

If concentration doubles rate doubles.If concentration doubles rate doubles. If we integrate this equation with If we integrate this equation with

respect to time we get the Integrated respect to time we get the Integrated Rate LawRate Law

ln[Nln[N22OO55] = - kt + ln[N] = - kt + ln[N22OO55]]00

ln is the natural logln is the natural log [N[N22OO55]]00 is the initial concentration. is the initial concentration.

General form Rate = General form Rate = [A] / [A] / t = k[A]t = k[A] ln[A] = - kt + ln[A]ln[A] = - kt + ln[A]00

In the form y = mx + bIn the form y = mx + b y = ln[A]y = ln[A] m = -km = -k x = tx = t b = ln[A]b = ln[A]00

A graph of ln[A] vs time is a straight A graph of ln[A] vs time is a straight line.line.

First OrderFirst Order

By getting the straight line you can By getting the straight line you can prove it is first orderprove it is first order

Often expressed in a ratio Often expressed in a ratio

First OrderFirst Order

By getting the straight line you can By getting the straight line you can prove it is first orderprove it is first order

Often expressed in a ratio Often expressed in a ratio lnA

A = kt0

First OrderFirst Order

The time required to reach half The time required to reach half the original concentration.the original concentration.

If the reaction is first orderIf the reaction is first order [A] = [A][A] = [A]00/2 when t = t/2 when t = t1/21/2

First OrderFirst OrderNuclear Decay and Half lifeNuclear Decay and Half life

• The time required to reach half the original concentration.

• If the reaction is first order

• [A] = [A]0/2 when t = t1/2

ln

A

A = kt0

01 2

2

ln(2) = ktln(2) = kt1/21/2

tt1/21/2 = 0.693/k = 0.693/k The time to reach half the The time to reach half the

original concentration does not original concentration does not depend on the starting depend on the starting concentration.concentration.

An easy way to find kAn easy way to find k

dtkA

dAorkA

dt

dA- 1

For a 1st-order reaction, the rate is dependent of the concentration. Expressed in terms of calculus, a 1st - order reaction would be expressed as:

tA

Adtk

A

dA00

t

ln [

A]

m = -k

0ln0

tk

A

A

0Aln kt - A ln

kor

693.0

k

2 ln t k t2ln

21

21

21 tln

tln

0

021

0

kA

A

kA

A

Half-life 1st -order reactionk

693.0 t

21

Second OrderSecond Order Rate = -Rate = -[A] / [A] / t = k[A]t = k[A]22 integrated rate law integrated rate law 1/[A] = kt + 1/[A]1/[A] = kt + 1/[A]00 y= 1/[A] y= 1/[A] m = km = k x= tx= t b = 1/[A]b = 1/[A]00 A straight line if 1/[A] vs t is graphedA straight line if 1/[A] vs t is graphed Knowing k and [A]Knowing k and [A]0 0 you can you can

calculate [A] at any time tcalculate [A] at any time t

Second Order Half LifeSecond Order Half Life

[A] = [A][A] = [A]00 /2 at t = t /2 at t = t1/21/2 1

20

2[ ]A = kt +

1

[A]10

22[ [A]

- 1

A] = kt

0 01

tk[A]1 =

1

02

1

[A] = kt

01 2

dt B

dB

22 korkB

dt

dB

For a 2nd-order reaction, the rate is dependent of the concentration. Expressed in terms of calculus, a 2nd - order reaction would be expressed as:

t

0

B

B 2 dtkB

dB0

t

1/[B]

m = k

0B k

1t

21 Half-life for 2nd - order

0021

0

B

1t k

B

1

B

1t k

B

1

21

00

1

B

1 0

11

Bktortk

BB

General Rate LawaA + bB + cC dD

p

n

m

k[C] t

- rate

k[B] t

B- rate

k[A] t

A- rate

rateorC

rateor

rateor

A + B + C products

Exp [A] [B] [C] InitialRate

1 1.25 1.25 1.25 8.7

2 2.50 1.25 1.25 17.4

3 1.25 3.02 1.25 50.8

4 1.25 3.02 3.75 457

5 3.01 1.00 1.15 ??

Exp [A] [B] [C] InitialRate

1 1.25 1.25 1.25 8.7

2 2.50 1.25 1.25 17.4

•If [B] and [C] are held constant, then any change in rate is caused by [A].

•[A] doubles from 1.25 to 2.50 the rate double from 8.7 to 17.4

therefore order for A is 1st order

Exp [A] [B] [C] InitialRate

3 1.25 3.02 1.25 50.84 1.25 3.02 3.75 457

•If [A] and [B] are held constant, then any change in rate is caused by [C].•[C] doubles from 1.25 to 3.75 the rate quadruples or a squaring of the change from 50.8 to 475•therefore order for C is 2nd order

Exp [A] [B] [C] InitialRate

1 1.25 1.25 1.25 8.73 1.25 3.02 1.25 50.8

•If [A] and [C] are held constant, then any change in rate is caused by [B].•[B] doubles from 1.25 to 3.02 the rate quadruples or a squaring of the change from 8.7 to 50.8•therefore order for B is 2nd order

aA + bB + cC dDA + B + C products

rate = k[A]1 [B]2 [C]2

if we make substitutions with data fromexperiment 1 to solve for k

8.7 = k[1.25]1 [1.25]2 [1.25]2

k=2.85

Exp [A] [B] [C] InitialRate

5 3.01 1.00 1.15 ??

rate = 2.85[3.01]1 [1.00]2 [1.15]2

rate =11.35

For the reactionFor the reaction

BrOBrO33-- + 5 Br + 5 Br-- + 6H + 6H++ 3Br 3Br22 + 3 H + 3 H22OO

The general form of the Rate Law is The general form of the Rate Law is

Rate = k[BrORate = k[BrO33--]]nn[Br[Br--]]mm[H[H++]]pp

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

Another Complex Reaction

BrOBrO33-- + 5 Br + 5 Br-- + 6H + 6H++ 3Br 3Br22 + 3 H + 3 H22OO

For this reaction we found the rate law For this reaction we found the rate law to beto be

Rate = k[BrORate = k[BrO33--][Br][Br--][H][H++]]22

To investigate this reaction rate we To investigate this reaction rate we need to control the conditions need to control the conditions

Rate = k[BrORate = k[BrO33--][Br][Br--][H][H++]]22

We set up the experiment so that We set up the experiment so that two of the reactants are in large two of the reactants are in large excess.excess.

[BrO[BrO33--]]00= 1.0 x 10= 1.0 x 10-3-3 M M

[Br[Br--]]0 0 = 1.0 M= 1.0 M [H[H++]]0 0 = 1.0 M= 1.0 M As the reaction proceeds [BrOAs the reaction proceeds [BrO33

--] ]

changes noticably changes noticably [Br[Br--] and [H] and [H++] don’t] don’t

This rate law can be rewrittenThis rate law can be rewritten Rate = k[BrORate = k[BrO33

--][Br][Br--]]00[H[H++]]00

22

Rate = k[BrRate = k[Br--]]00[H[H++]]00

22[BrO[BrO33--]]

Rate = k’[BrORate = k’[BrO33--]]

This is called a pseudo first order rate This is called a pseudo first order rate law.law.

k =k = k’ k’

[Br[Br--]]00[H[H++]]00

22

Rate = k[BrORate = k[BrO33--][Br][Br--][H][H++]]22

Initial concentrations (M)

Rate (M/s)

BrOBrO33-- BrBr-- HH++

0.100.10 0.100.10 0.100.10 8.0 x 108.0 x 10--

44

0.200.20 0.100.10 0.100.10 1.6 x 101.6 x 10--

33

0.200.20 0.200.20 0.100.10 3.2 x 103.2 x 10--

33

0.100.10 0.100.10 0.200.20 3.2 x 103.2 x 10--

33

Now we have to see how the rate Now we have to see how the rate changes with concentrationchanges with concentration

Reaction Reaction MechanismsMechanisms

The series of steps that actually The series of steps that actually occur in a chemical reaction.occur in a chemical reaction.

Kinetics can tell us something about Kinetics can tell us something about the mechanismthe mechanism

A balanced equation does not tell us A balanced equation does not tell us how the reactants become products.how the reactants become products.

2NO2NO22 + F + F22 2NO2NO22FF Rate = k[NORate = k[NO22][F][F22]]

The proposed mechanism isThe proposed mechanism is NONO22 + F + F22 NONO22F + F F + F (slow)(slow) F + NOF + NO22 NONO22FF (fast) (fast)

F is called an F is called an intermediate intermediate It is formed It is formed then consumed in the reactionthen consumed in the reaction

Reaction Reaction MechanismsMechanisms

Each of the two reactions is called an Each of the two reactions is called an elementary stepelementary step . .

The rate for a reaction can be written The rate for a reaction can be written from its from its molecularitymolecularity . .

Molecularity is the number of pieces Molecularity is the number of pieces that must come together.that must come together.

Reaction MechanismsReaction Mechanisms

Mechanism must pass two tests to be accepted

1. Sum of mechanism steps produces the net equation.

2. Rate laws of all rate determining steps must be combined to produce experimental rate law.

UnimolecularUnimolecular step involves one molecule - step involves one molecule - Rate is rirst order.Rate is rirst order.

BimolecularBimolecular step - requires two molecules - step - requires two molecules - Rate is second orderRate is second order

TermolecularTermolecular step- requires three molecules step- requires three molecules - Rate is third order - Rate is third order

TermolecularTermolecular steps are almost never heard steps are almost never heard of because the chances of three molecules of because the chances of three molecules coming into contact at the same time are coming into contact at the same time are miniscule.miniscule.

A A products products Rate = k[A]Rate = k[A] A+A productsA+A products Rate= k[A]Rate= k[A]22

2A 2A products products Rate= k[A]Rate= k[A]22

A+B productsA+B products Rate= k[A][B]Rate= k[A][B] A+A+B Products A+A+B Products Rate= k[A]Rate= k[A]22[B][B] 2A+B Products 2A+B Products Rate= k[A]Rate= k[A]22[B][B] A+B+C Products A+B+C Products Rate= k[A][B][C]Rate= k[A][B][C]

How to get rid of How to get rid of intermediatesintermediates

Use the reactions that form themUse the reactions that form them If the reactions are fast and If the reactions are fast and

irreversible - the concentration of irreversible - the concentration of the intermediate is based on the intermediate is based on stoichiometry.stoichiometry.

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

Formed in reversible Formed in reversible reactionsreactions

2 NO + O2 NO + O22 2 NO 2 NO22

MechanismMechanism 2 NO 2 NO NN22OO22 (fast)(fast) NN22OO22 + O + O2 2 2 NO 2 NO2 2 (slow)(slow)

rate = krate = k22[N[N22OO22][O][O22]] kk11[NO][NO]22 = k = k-1-1[N[N22OO22]] rate = krate = k22 (k (k11/ k/ k-1-1)[NO])[NO]22[O[O22]=k[NO]]=k[NO]22[O[O22]]

Formed in fast reactionsFormed in fast reactions

2 IBr 2 IBr II22+ Br+ Br22 MechanismMechanism IBrIBr I + Br I + Br (fast)(fast) IBr + Br I + BrIBr + Br I + Br22 (slow)(slow) I + II + I II22 (fast)(fast) Rate = k[IBr][Br] but [Br]= [IBr]Rate = k[IBr][Br] but [Br]= [IBr] Rate = k[IBr][IBr] = k[IBr]Rate = k[IBr][IBr] = k[IBr]22

Collision theoryCollision theory

Molecules must collide to react.Molecules must collide to react. Concentration affects rates because Concentration affects rates because

collisions are more likely.collisions are more likely. Must collide hard enough.Must collide hard enough. Temperature and rate are related.Temperature and rate are related. Only a small number of collisions Only a small number of collisions

produce reactions.produce reactions.

Potential Energy

Reaction Coordinate

Reactants

Products

Potential Energy

Reaction Coordinate

Reactants

Products

Activation Energy Ea

Potential Energy

Reaction Coordinate

Reactants

Products

Activated complex

Potential Energy

Reaction Coordinate

Reactants

ProductsE}

Potential Energy

Reaction Coordinate

2BrNO

2NO + Br

Br---NO

Br---NO

2

Transition State

A + B --> A-B* --> AB

A + B

A-B*

AB

HEa(cat)

Reaction Coordinate

En

erg

y

Misconception about catalysts

Catalysts do not take part in reactions

TermsTerms

Activation energy - the minimum Activation energy - the minimum energy needed to make a reaction energy needed to make a reaction happen.happen.

Activated Complex or Transition Activated Complex or Transition State - The arrangement of atoms at State - The arrangement of atoms at the top of the energy barrier.the top of the energy barrier.

ArrheniusArrhenius

Said the reaction rate should Said the reaction rate should increase with temperature.increase with temperature.

At high temperature more molecules At high temperature more molecules have the energy required to get over have the energy required to get over the barrier.the barrier.

The number of collisions with the The number of collisions with the necessary energy increases necessary energy increases exponentially.exponentially.

Arrhenius EquationArrhenius EquationNumber of collisions with the required Number of collisions with the required

kk == Ae ^(-EAe ^(-Eaa/RT)/RT) k = rate constantk = rate constant A = total collisions A = total collisions ≈≈ frequency factor frequency factor e is Euler’s number (opposite of ln)e is Euler’s number (opposite of ln) EEa a = activation energy= activation energy R = Universal gas constantR = Universal gas constant T = temperature in KelvinT = temperature in Kelvin

Linearized Arrhenius EquationLinearized Arrhenius Equation

lnk = lnA – (Elnk = lnA – (Eaa/RT)/RT)

Using the linear equation we can find Using the linear equation we can find a number of thingsa number of things

Plot lnk vs 1/T the slope will give you Plot lnk vs 1/T the slope will give you the activation energythe activation energy

If we know Ea and k at one If we know Ea and k at one temperature then you can find k at a temperature then you can find k at a different temperaturedifferent temperature

ln ln k2k2 = = EEaa 1 1 - - 1 1 k1 R T1 T2k1 R T1 T2

ProblemsProblems

Observed rate is less than the Observed rate is less than the number of collisions that have the number of collisions that have the minimum energy.minimum energy.

Due to Molecular orientationDue to Molecular orientation written into equation as p the written into equation as p the steric steric

factor.factor.

ON

Br

ON

Br

ON

Br

ON

Br

O N Br ONBr ONBr

O NBr

O N BrONBr No Reaction

Arrhenius EquationArrhenius Equation

k = zpek = zpe-E-Eaa/RT /RT = Ae= Ae-E-Eaa/RT/RT

A is called the frequency factor = zpA is called the frequency factor = zp

ln k = -(Eln k = -(Eaa/R)(1/T) + ln A/R)(1/T) + ln A

Another line !!!!Another line !!!!

ln k vs t is a straight lineln k vs t is a straight line

Mechanisms and rates Mechanisms and rates There is an activation energy for There is an activation energy for

each elementary step.each elementary step. Activation energy determines k.Activation energy determines k. k = Aek = Ae- (E- (Eaa/RT) /RT)

k determines ratek determines rate Slowest step (rate determining) must Slowest step (rate determining) must

have the highest activation energy.have the highest activation energy.

This reaction takes place in three This reaction takes place in three stepssteps

Ea

First step is fastFirst step is fast

Low activation energyLow activation energy

Second step is slowHigh activation energy

Ea

Ea

Third step is fastLow activation energy

Second step is rate determining

Intermediates are present

Activated Complexes or Transition States

Possible Mechanismsa. CO + NO2 CO2 + NO

b. 2NO2 N2O4

N2O4 + 2CO 2CO2 + 2NO (slow)

422

2

42

22

ON''kNO'k

ON''k)eq everser(rate

NO'k)eq forward(rate

CO NOkrate 2

242 CO ON'''k)slo(rate

222 CONOkrate

c. 2NO2 NO3 + NO (slow)

NO3 + CO NO2 + CO2 (fast)

22NOkrate

d. 2NO2 2NO + O2 (slow)

2 CO + O2 2CO2 (fast)

22NOkrate

Transition State Theory

transition state = activated complex

A + B ---> A--B* ---> AB

A + B --> A-B* --> AB

A + B

A-B*

AB

H

Ea

Reaction Coordinate

En

erg

yactivated complex

activation energy

Heat of reaction

A + B

A-B*

AB

HEa

Reaction Coordinate

En

erg

y

A + B --> A-B* --> AB

In General

rate(R) [A]

R = k[A]k = rate constant

If one examines the relationship between the rate constant, k, and temperature, we

find that ln k 1/T

1/T

ln k

m = -Ea/R

AlnT

1

R

E- k ln a

Arrhenius Equation

RT

Ea

Aek

A = frequency factorR = 8.314 J/mol.K

AlnT

1

R

E- k ln

AlnT

1

R

E- k ln

2

a2

1

a1

12

a

2

1

T

1

T

1

R

E

k

k ln

CatalystsCatalysts Speed up a reaction without being Speed up a reaction without being

used up in the reaction.used up in the reaction. Enzymes are biological catalysts.Enzymes are biological catalysts. Homogenous CatalystsHomogenous Catalysts are in the are in the

same phase as the reactants.same phase as the reactants. Heterogeneous CatalystsHeterogeneous Catalysts are in a are in a

different phase as the reactants.different phase as the reactants.

How Catalysts WorkHow Catalysts Work Catalysts allow reactions to Catalysts allow reactions to

proceed by a different mechanism proceed by a different mechanism - a new pathway.- a new pathway.

New pathway has a lower New pathway has a lower activation energy.activation energy.

More molecules will have this More molecules will have this activation energy.activation energy.

Do not change Do not change EE

Pt surface

HH

HH

HH

HH

Hydrogen bonds to Hydrogen bonds to surface of metal.surface of metal.

Break H-H bondsBreak H-H bonds

Heterogenous CatalystsHeterogenous Catalysts

Pt surface

HH

HH

Heterogenous CatalystsHeterogenous Catalysts

C HH C

HH

Pt surface

HH

HH

Heterogenous CatalystsHeterogenous Catalysts

C HH C

HH

The double bond breaks and bonds The double bond breaks and bonds to the catalyst.to the catalyst.

Pt surface

HH

HH

Heterogenous CatalystsHeterogenous Catalysts

C HH C

HH

The hydrogen atoms bond with the The hydrogen atoms bond with the carboncarbon

Pt surface

H

Heterogenous CatalystsHeterogenous Catalysts

C HH C

HH

H HH

Homogenous CatalystsHomogenous Catalysts Chlorofluorocarbons catalyze the Chlorofluorocarbons catalyze the

decomposition of ozone.decomposition of ozone. Enzymes regulating the body Enzymes regulating the body

processes. (Protein catalysts)processes. (Protein catalysts)

Catalysts and rateCatalysts and rate Catalysts will speed up a reaction but Catalysts will speed up a reaction but

only to a certain point.only to a certain point. Past a certain point adding more Past a certain point adding more

reactants won’t change the rate.reactants won’t change the rate. Zero OrderZero Order

Catalysts and rate.Catalysts and rate.

Concentration of reactants

Rate

Rate increases until the active Rate increases until the active sites of catalyst are filled.sites of catalyst are filled.

Then rate is independent of Then rate is independent of concentrationconcentration