Chapter 14 Chemical Kinetics Chemical Kinetics. Kinetics The study of reaction rates – the speed...

Chapter 14Chapter 14

Chemical KineticsChemical Kinetics

KineticsKinetics• The The study of reaction rates study of reaction rates – the – the

speed at which reactants are converted speed at which reactants are converted to products.to products.

• Explosions happen in seconds- very fast.Explosions happen in seconds- very fast.• Coal (carbon) will turn into diamond Coal (carbon) will turn into diamond

eventually.eventually.• Spontaneous reactions are reactions that Spontaneous reactions are reactions that

will happen will happen but we can’t tell how fast. but we can’t tell how fast.• Kinetics is very important because it Kinetics is very important because it

allows us to manipulate and control allows us to manipulate and control reactionsreactions..

Reactions occur when . . .Reactions occur when . . .

• Molecules Molecules find each otherfind each other• Effective collisions happen Effective collisions happen

between moleculesbetween molecules• Frequency of collisions is Frequency of collisions is

highhigh

Factors that affect Reaction Rate:Factors that affect Reaction Rate:

1. Temperature1. Temperature• Hot = fast Cold = slowHot = fast Cold = slow2. Concentration of Reactants2. Concentration of Reactants• more concentrated = fastermore concentrated = faster3. Physical conditions of reactants.3. Physical conditions of reactants.• more surface area = fastermore surface area = faster4. Presence of a Catalyst –4. Presence of a Catalyst –

• increases rate without being used upincreases rate without being used up

Reaction RateReaction Rate

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

Rates decrease with timeRates decrease with time

Typically rates are expressed as positive Typically rates are expressed as positive valuesvalues

Instantaneous rate can be determined Instantaneous rate can be determined by finding the slop of a line tangent to a by finding the slop of a line tangent to a point representing a particular timepoint representing a particular time

Reaction Rate - “speed of the reaction”Reaction Rate - “speed of the reaction”

• Rate = Rate = Conc. of A at tConc. of A at t22 Conc. of A at t Conc. of A at t11

tt2 2 t t11

• Rate =Rate =[A][A] tt

• Units: Change in concentration per Units: Change in concentration per timetime

• For this reactionFor this reaction

• NN22 + 3H + 3H22 2NH 2NH33

• As the reaction progresses the concentration As the reaction progresses the concentration of Hof H22 goes down goes down NN22 + 3H + 3H22 2NH 2NH33

Concentration

Time

[H[H22]]

• As the reaction progresses the concentration NAs the reaction progresses the concentration N22

goes down 1/3 as fast. Why? goes down 1/3 as fast. Why? NN22 + 3H + 3H22

2NH2NH33

Concentration

Time

[H[H22]]

[N[N22]]

• As the reaction progresses the concentration As the reaction progresses the concentration NHNH33 goes up. goes up. NN22 + 3H + 3H22 2NH 2NH33

Concentration

Time

[H[H22]]

[N[N22]]

[NH[NH33]]

Calculating RatesCalculating Rates1. Average rates are taken over long 1. Average rates are taken over long

intervalsintervals

2. Instantaneous rates are determined 2. Instantaneous rates are determined by finding the slope of a line tangent by finding the slope of a line tangent to the curve at any given point to the curve at any given point because the rate can change over because the rate can change over time (Finding the time (Finding the Derivative.)Derivative.)

• Average slope methodAverage slope method

Concentration

Time

[H[H22]]

tt

• Instantaneous slope method.Instantaneous slope method.

Concentration

Time

[H[H22]]

tt

Find the Instantaneous rate of Find the Instantaneous rate of disappearance:disappearance:

In order to find the instantaneous rate In order to find the instantaneous rate of disappearance at time 0, a tangent of disappearance at time 0, a tangent line was drawn and the following line was drawn and the following information was obtained: the information was obtained: the concentration of the unknown changed concentration of the unknown changed from 0.100 M to 0.060 M from 0 seconds from 0.100 M to 0.060 M from 0 seconds to 200 seconds. What is the to 200 seconds. What is the instantaneous rate of disappearance at instantaneous rate of disappearance at time 0?time 0?

2.0 x 102.0 x 10-4-4 M/s M/s

Reaction Rates and Reaction Rates and StoichiometryStoichiometry

• Remember, the rate of disappearance is Remember, the rate of disappearance is a a slope and the rate of appearance is a slope and the rate of appearance is a + slope+ slope

• For general equation:For general equation:

aaA + A + bbB B ccC + C + ddDD

General rate equation: General rate equation: Rate = -Rate = -11[A] [A] = = -1-1[B] [B] = = 11[C] [C] = = 1∆[D1∆[D] ]

a a t t bb t t cc t t dd∆t∆t

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 terms of the rate of appearance of the product.the product.• In our exampleIn our example

NN22 + 3H + 3H22 2NH 2NH3 3

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

PracticePracticeHow is the rate of disappearance of How is the rate of disappearance of ozone related to the rate of ozone related to the rate of appearance of oxygen in the appearance of oxygen in the following equation: 2Ofollowing equation: 2O33 3O 3O22??

If the rate of appearance of OIf the rate of appearance of O22, is , is 6.0x106.0x10-5-5 M/s at a particular instant, M/s at a particular instant, what is the value of the rate of what is the value of the rate of disappearance of Odisappearance of O33 at this same at this same time?time?

4.0x104.0x10-5-5 M/s M/s

The decomposition of NThe decomposition of N22OO55 proceeds according to the proceeds according to the following equation:following equation:

2N2N22OO55 4NO 4NO22 + O + O22

If the rate of decomposition of NIf the rate of decomposition of N22OO55 at a particular instant in a reaction at a particular instant in a reaction vessel is 4.2x10vessel is 4.2x10-7-7 M/s, what is the M/s, what is the rate of appearance of NOrate of appearance of NO22? O? O22??

NONO22 = 8.4x10 = 8.4x10-7-7 M/s M/s

OO22 = 2.1 x 10 = 2.1 x 10-7-7 M/s M/s

Rate LawsRate Laws

To study the rate of a reaction and To study the rate of a reaction and determine the rate law, a determine the rate law, a series of series of experimentsexperiments must be done varying the must be done varying the initial concentrations of reactants.initial concentrations of reactants.

This is called This is called the Initial rate methodthe Initial rate method..

The products do not appear in the rate The products do not appear in the rate law because we are only concerned law because we are only concerned with the very start of the experiment with the very start of the experiment where conc. of reactants are carefully where conc. of reactants are carefully chosen.chosen.

The Rate LawThe Rate Law• The The rate lawrate law expresses the relationship of the expresses the relationship of the

rate of a reaction to the rate constant and the rate of a reaction to the rate constant and the concentrations of the reactants raised to some concentrations of the reactants raised to some powers.powers.

• aaA + A + bbB B ccC + C + ddDD• Rate = Rate = kk [A] [A]xx[B][B]y y } The rate law expression} The rate law expression• kk is theis the rate constant rate constant (temperature dependent)(temperature dependent)• UnitsUnits ofof k k vary vary

• xx and and yy are integers, usually 0, 1 or 2 are integers, usually 0, 1 or 2• reaction is reaction is xxthth order in A, reaction is order in A, reaction is yythth order in order in

BB• reaction is (x +y)reaction is (x +y)thth order overall order overall

FF22 ( (gg) + 2ClO) + 2ClO22 ( (gg) 2FClO) 2FClO22 ( (g)g)

rate = k [F2]x [ClO2]y

Initial [ ]

Double [F2] with [ClO2] constant: rate doublesQuadruple [ClO2] with [F2] constant: rate quadruples

Therefore, rate law is: rate = k [F2]1 [ClO2]1

We say, “first order with respect to F2, first order with respect to ClO2, and 2nd order overall”

(All readingsrecorded atsame temp.)

1. Rate laws are 1. Rate laws are alwaysalways determined determined experimentally.experimentally.

2. Reaction order is 2. Reaction order is alwaysalways defined in terms of defined in terms of reactant (not product) concentrations.reactant (not product) concentrations.

3. 3. The order of a reactant The order of a reactant is is notnot related to the related to the stoichiometric coefficient of the reactant in stoichiometric coefficient of the reactant in the balanced chemical equation.the balanced chemical equation.

Rate Laws - Key PointsRate Laws - Key Points

F2 (g) + 2ClO2 (g) 2FClO2 (g)

rate = k [F2]1 [ClO2]1

4. Once you obtain the Rate Law and the value of k, you can calculate the rate of reaction for any set of concentrations.

Try this. A + B ---> C + DTry this. A + B ---> C + D

Run # Initial [A] Initial [B] Initial Rate (vRun # Initial [A] Initial [B] Initial Rate (v00) ) ([A] ([A]00) ([B]) ([B]00))

11 1.00 M 1.00 M 1.00 M 1.00 M 1.25 x 10 1.25 x 10-2-2 M/s M/s 22 1.00 M 1.00 M 2.00 M 2.00 M 2.5 x 10 2.5 x 10-2-2 M/s M/s 33 2.00 M 2.00 M 2.00 M 2.00 M 2.5 x 10 2.5 x 10-2-2 M/s M/s

What is the order with respect to A?

What is the order with respect to B?

What is the overall order of the reaction?

Write the rate law expression

0

1

1

Rate = k [A]0 [B]1

Try this. 2NO Try this. 2NO (g)(g) + Cl + Cl22 (g)(g) ---> 2NOCl ---> 2NOCl

[NO[NO(g)(g)] (M) [Cl] (M) [Cl2(g)2(g)] (M) Initial Rate(Ms] (M) Initial Rate(Ms-1-1))  

0.2500.250   0.250 0.250   1.43 x 10 1.43 x 10-6-6 0.2500.250   0.500 0.500   2.86 x 10 2.86 x 10-6-6 0.5000.500   0.500 0.500   1.14 x 10 1.14 x 10-5-5

What is the order with respect to Cl2?

What is the order with respect to NO?

What is the overall order of the reaction?

Write the rate law expression

Calculate the value of k

1

2

3

Rate = k [NO]2 [Cl2]1

9.15 x 10-5 1/M2•s

Units of Units of kk1st Order overall units: 1/s1st Order overall units: 1/s

2nd Order overall units: 1/M2nd Order overall units: 1/M••ss

3rd Order overall units: 1/ M3rd Order overall units: 1/ M22••ss

What the Order means:•Zero order means doubling or tripling the [reactant]o has NO effect on the rate.•First order means doubling the [reactant]o doubles the rate, tripling the [reactant]o triples the rate, etc…•Second order means doubling the [reactant]o quadruples the rate (raises to a power of 2), etc. . .

Initial concentrations (M)Rate (M/s)

[BrO[BrO33--] [Br[Br--] [H[H++]

0.100.10 0.100.10 0.100.10 8.0 x 10 8.0 x 10-4-4

0.200.20 0.100.10 0.100.10 1.6 x 10 1.6 x 10-3-3

0.200.20 0.200.20 0.100.10 3.2 x 10 3.2 x 10-3-3

0.100.10 0.100.10 0.200.20 3.2 x 10 3.2 x 10-3-3Determine the rate law Determine the rate law expressionexpression

Calculate the value of Calculate the value of kk

Try this: BrO3- + 5 Br- + 6H+ ----> 3Br2 + 3 H2O

Rate = k [BrO3]1 [Br-]1 [H+]2

fourth order overall

8.0 1/M3•s

Determine the Rate Law ExpressionDetermine the Rate Law Expression

Determine the value for kDetermine the value for k

[NH[NH44++ ] ] [NO[NO22

--]] Initial Rate (M/s)Initial Rate (M/s)

0.100 M0.100 M 0.0050 M0.0050 M 1.35 x 101.35 x 10-7-7

0.100 M0.100 M 0.010 M0.010 M 2.70 x 102.70 x 10-7-7

0.200 M0.200 M 0.010 M0.010 M 5.40 x 105.40 x 10-7-7

NHNH44++ + NO + NO22

-- N N22 + 2H + 2H22OO

Types of Rate LawsTypes of Rate Laws

Differential Rate LawDifferential Rate Law – describes how – describes how rate depends on concentrationrate depends on concentration

Rate = Rate = kk [A] [A]xx[B][B]yy

Integrated Rate LawIntegrated Rate Law –– describes how describes how concentration depends on timeconcentration depends on time

For each type of differential rate law, For each type of differential rate law, there is an integrated rate law and vise there is an integrated rate law and vise versa.versa.

We can use these laws to help us better We can use these laws to help us better understand reaction mechanismsunderstand reaction mechanisms

Integrated Rate Law Equations allow us to:Integrated Rate Law Equations allow us to:

1. Determine the concentration of 1. Determine the concentration of reactant remaining at any time after reactant remaining at any time after the reaction has started.the reaction has started.2. Determine the time required for a 2. Determine the time required for a fraction of a sample to react.fraction of a sample to react.3. Determine the time for a reactant 3. Determine the time for a reactant concentration to fall to a certain level.concentration to fall to a certain level.4. Verify whether a reaction is first 4. Verify whether a reaction is first order and determine the rate constant.order and determine the rate constant.

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).

• A Changing Rate = A Changing Rate = [A][A]nn tt

• We will only work with n = 0, 1, and We will only work with n = 0, 1, and 22

First OrderFirst Order• For the reaction 2NFor the reaction 2N22OO55 4NO 4NO22 + O + O22

• Rate = k[Rate = k[NN22OO55]]11

• If concentration doubles the rate If concentration doubles the rate doubles.doubles.

• If we integrate this equation with respect If we integrate this equation with respect to time we get the Integrated Rate Lawto time we get the Integrated Rate 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] ; m = -k ; x = t ;b = ln[A]y = ln[A] ; m = -k ; x = t ;b = ln[A]00

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

First OrderFirst Order

• A graph of ln[A] vs time is a A graph of ln[A] vs time is a straight linestraight line..• By getting the straight line you can prove By getting the straight line you can prove

it is first orderit is first order

First OrderFirst Order

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

• Rate = k[NRate = k[N22OO55]]1 1

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

First OrderFirst Order

Try thisTry this• The reaction 2A B is first order in A with a The reaction 2A B is first order in A with a

rate constant of 2.8 x 10rate constant of 2.8 x 10-2-2 s s-1-1 at 80 at 8000C. How long will C. How long will it take for A to decrease from 0.88 it take for A to decrease from 0.88 MM to 0.14 to 0.14 M M ??

ln[A] = - ln[A] = - kkt + ln[A]t + ln[A]00

ln[A] - ln[A]ln[A] - ln[A]00 = - = - kktt

ln[0.14 M] - ln[0.88M]0 = - (2.8 x 10-2 s-1) t- 1.97 - 0.128 = - (2.8 x 10-2 s-1) t

- 1.84 = - (2.8 x 10-2 s-1) t

66 s = t

Second Order (Overall)Second Order (Overall)• Rate = -Rate = -[A] / [A] / t = k[A]t = k[A]22

• integrated rate law: 1/[A] = kt + 1/[A]integrated rate law: 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 time is A straight line if 1/[A] vs time is graphedgraphed

• Knowing k and [A]Knowing k and [A]0 0 you can calculate you can calculate [A] at any time t[A] at any time t

Second Order Second Order •Looking at the data alone, you cannot tell if 1st or 2nd order.•To tell the difference you must graph the data, or look at the units of k.

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

• In the form y = mx + bIn the form y = mx + b• 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.

Time

Concentration

Zero Order

Time

Concentration

A]/t

t

k =

A]

Zero Order

Zero OrderZero Order

Half LifeHalf Life• The The half-lifehalf-life, , tt½½,, is the time required for the is the time required for the

concentration of a reactant to decrease to half concentration of a reactant to decrease to half of its initial concentration.of its initial concentration.

• tt½½ = = tt when [A] = [A] when [A] = [A]00/2/2

Half LifeHalf Life• The time required to reach half The time required to reach half

the original concentration.the original concentration.• If the reaction is If the reaction is firstfirst order order• tt½½ = = tt when [A] = [A] when [A] = [A]00/2/2

• If the reaction is If the reaction is 2nd2nd order: order:• tt½½ = = tt when [A] = [A] when [A] = [A]00/2/2

• If the reaction is If the reaction is zerozero order: order:

tt½½ = = tt when [A] = [A] when [A] = [A]00/2/2 t½ =

[A]0

2k

t½ =1

k[A]0

ln[A]0

[A]0/2

k=t½

ln 2k

=.693k

=

Try ThisTry This• What is the half-life of NWhat is the half-life of N22OO55 if it if it

decomposes with a rate constant of 5.7 decomposes with a rate constant of 5.7 x 10x 10-4-4 s s-1-1??

• Which order is it? How do you know?Which order is it? How do you know?

t½Ln 2k

=0.693

5.7 x 10-4 s-1= = 1200 s = 20 minutes

Summary of Rate LawsSummary of Rate Laws

OrderOrder Rate LawRate Law Integrated Integrated Rate Law Rate Law EquationEquation

Half-LifeHalf-Life

Reaction RatesReaction Rates

So, WHY do some reactions go So, WHY do some reactions go fast and others are slow?fast and others are slow?

Reaction Mechanisms Collision Theoryand Activation Energy

Reaction MechanismsReaction Mechanisms• The series of steps that actually The series of steps that actually

occur in a chemical reaction.occur in a chemical reaction.• Each step is called an elementary Each step is called an elementary

step.step.• 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 2NO2NO22F F

• Rate = k[NORate = k[NO22] [F] [F22] ] (determined experimentally)(determined experimentally)

• 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

• The slow step is theThe slow step is the rate rate determiningdetermining step!step!

Reaction Mechanisms -- ExampleReaction Mechanisms -- Example

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

• MolecularityMolecularity is the number of pieces that is the number of pieces that must collide to produce the reaction must collide to produce the reaction indicated by the step.indicated by the step.

• MolecularityMolecularity is the number of molecules is the number of molecules which must collide effectively for reaction which must collide effectively for reaction to occur.to occur.

Reaction MechanismsReaction Mechanisms

• UnimolecularUnimolecular step involves one step involves one molecule - Rate is first order.molecule - Rate is first order.

• Bimolecular Bimolecular step - requires two step - requires two molecules - Rate is second ordermolecules - Rate is second order

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

• TermolecularTermolecular steps are almost never steps are almost never heard of because the chances of heard of because the chances of three molecules colliding effectively three molecules colliding effectively at the same time are miniscule.at the same time are 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= Rate=

k[A]k[A]22[B][B]• 2A+B Products 2A+B Products Rate= Rate=

k[A]k[A]22[B][B]• A+B+C Products; Rate= k[A][B][C]A+B+C Products; Rate= k[A][B][C]

Rate Laws for Elementary Steps (these ARE based on the stoichiometry)

Going back to our example:Going back to our example:

• 2NO2NO22 + F + F22 2NO2NO22FF (overall reaction)(overall reaction)

• Rate = k[NORate = k[NO22] [F] [F22]] (determined experimentally)(determined experimentally)

• The proposed mechanism consists of The proposed mechanism consists of • 2 elementary steps:2 elementary steps:• NONO22 + F + F22 NONO22F + F F + F (slow)(slow)• F + NOF + NO22 NONO22FF (fast) (fast) • F is called an intermediate - It is formed then F is called an intermediate - It is formed then

consumed in the reaction.consumed in the reaction.• The The slow bimolecular elementary step slow bimolecular elementary step

determines the rate of the reactiondetermines the rate of the reaction and and sets the rate law as: sets the rate law as:

2nd order overall.A + B P 2nd order overall.A + B P rate = k[A] [B]rate = k[A] [B]

Multistep ExampleMultistep Example

It has been proposed that the It has been proposed that the conversion of ozone into O2 proceeds conversion of ozone into O2 proceeds via two elementary steps:via two elementary steps:– OO33 O O22 + O + O

– OO33 + O + O 2O 2O22

Describe the Molecularity of each stepDescribe the Molecularity of each step

Write the equation for the overall Write the equation for the overall reaction.reaction.

Identify the intermediate(s)Identify the intermediate(s)

For the Equation: For the Equation: Mo(CO)Mo(CO)66 + P(CH + P(CH33))33 Mo(CO) Mo(CO)55P(CHP(CH33))33 + CO + CO

The proposed mechanism is:The proposed mechanism is:– Mo(CO)Mo(CO)66 Mo(CO) Mo(CO)55 + CO + CO

– Mo(CO)Mo(CO)55 + P(CH + P(CH33))33 Mo(CO) Mo(CO)55P(CHP(CH33))33

Is the proposed mechanism consistent Is the proposed mechanism consistent with the overall reaction?with the overall reaction?

Identify the intermediate(s)Identify the intermediate(s)

Consider the following reaction: Consider the following reaction:

2NO+ Br2NO+ Br22 2NOBr 2NOBr

Write the rate law for the Write the rate law for the reaction assuming it involves a reaction assuming it involves a single step.single step.

Is a single-step mechanism likely Is a single-step mechanism likely for this reaction? Explain.for this reaction? Explain.

In summaryIn summary• Reaction mechanisms describe what really Reaction mechanisms describe what really

happens during a reaction: how the happens during a reaction: how the molecules collide and how many are molecules collide and how many are involved.involved.

• There is a bit of guessing as to what the There is a bit of guessing as to what the mechanism might be.mechanism might be.

• Chemists try to match a proposed Chemists try to match a proposed mechanism to an actual experimentally mechanism to an actual experimentally determined rate law.determined rate law.

• The rate of the reaction is the slowest The rate of the reaction is the slowest elementary step of the mechanism. The elementary step of the mechanism. The overall rate order is set by this step.overall rate order is set by this step.

What about the role of temperature?What about the role of temperature?

Temperature, Rate, and Temperature, Rate, and

Effective CollisionsEffective Collisions

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.• In order to react, colliding In order to react, colliding

molecules must have a total KE molecules must have a total KE equal or greater than some equal or greater than some minimum value.minimum value.

ON

Br

ON

Br

ON

Br

ON

Br

O N Br ONBr ONBr

O NBr

O N BrONBr No Reaction

2 NOBr 2 NO + Br2

No reaction

an effective collision

Potential Energy

Reaction Coordinate

Reactants

Products

Energy Diagram

Potential Energy

Reaction Coordinate

Reactants

Products

Activation Energy Ea

Energy Diagram

Potential Energy

Reaction Coordinate

Reactants

Products

Activated complex

Energy Diagram

Potential Energy

Reaction Coordinate

Reactants

ProductsE}

Energy Diagram

Potential Energy

Reaction Coordinate

2BrNO

2NO + Br

Br---NO

Br---NO

2

Transition State

TermsTerms• Activation energy - the minimum Activation energy - the minimum

kinetic energy needed to make a kinetic energy needed to make a reaction happen.reaction 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.

A + B C + DA + B C + D

Exothermic Endothermic

The activation energy (Ea) is the minimum amount of energy required to initiate a chemical reaction.

ExothermicExothermic and the Reverse and the Reverse EndothermicEndothermic

(a)Activation energy (Ea) for the forward reaction

(a)Activation energy (Ea) for the reverse reaction

(c) ∆H

50 kJ/mol50 kJ/mol 300 kJ/mol300 kJ/mol

150 kJ/mol150 kJ/mol 100 kJ/mol100 kJ/mol

-100 kJ/mol-100 kJ/mol +200 kJ/mol+200 kJ/mol

Exothermic Endothermic

ArrheniusArrhenius• Said that molecules must possess a Said that molecules must possess a

certain minimum energy in order to react.certain minimum energy in order to react.• Said the reaction rate should increase with Said the reaction rate should increase with

temperature.temperature.• At high temperature more molecules have At high temperature more molecules have

the energy required to get over the the energy required to get over the barrier.barrier.

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

Arrhenius EquationArrhenius Equation• Number of collisions with the required Number of collisions with the required

energy,energy,

• worked into an equation:worked into an equation:

• kk = Ae = Ae-E-Eaa/RT/RT

• kk = rate constant = rate constant• A = frequency factor (total collisions)A = frequency factor (total collisions)• e is Euler’s number (opposite of ln)e is Euler’s number (opposite of ln)• EEa a = activation energy= activation energy• R = ideal gas law constantR = ideal gas law constant• T is temperature in KelvinT is temperature in Kelvin

To find the activation energy:To find the activation energy:

• Rearrange Arrhenius Equation by Rearrange Arrhenius Equation by taking the natural log of each side taking the natural log of each side

ln ln kk == Ea Ea + ln + ln AA RTRT

Activation Energy and RatesActivation Energy and Rates

The final sagaThe final saga

Mechanisms and rates Mechanisms and rates • There is an activation energy (hill to There is an activation energy (hill to

get over) for each elementary step.get over) for each elementary step.• Activation energy determines Activation energy determines kk..

• kk = Ae = Ae- (E- (Eaa/RT) /RT)

• kk determines rate 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 stepsThis reaction takes place in three steps

E

Reaction Progress

1

Ea

First step is fastFirst step is fast

Low activation energyLow activation energy

Second step is slowHigh activation energy

2

Ea

3

Ea

Third step is fastLow activation energy

Second step is rate determining

Intermediates are present

Activated Complexes or Transition States

Catalysts - How do they work?Catalysts - How do they work?• 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 proceed by Catalysts allow reactions to proceed by a different mechanism - a new a different mechanism - a new pathway.pathway.

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

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

• Do not change Do not change E (∆H) of the reaction.E (∆H) of the reaction.

A A catalystcatalyst is a substance that increases the rate of a is a substance that increases the rate of a chemical reaction without itself being consumed. chemical reaction without itself being consumed.

No catalyst With a catalyst No catalyst With a catalyst

uncatalyzed catalyzed

ratecatalyzed > rateuncatalyzedEa , k

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

Heterogeneous CatalystsHeterogeneous Catalysts

H2 + C2H4 -----> C2H6

Pt surface

HH

HH

Heterogeneous CatalystsHeterogeneous Catalysts

C HH C

HH

H2 + C2H4 -----> C2H6

Pt surface

HH

HH

Heterogeneous CatalystsHeterogeneous Catalysts

C HH C

HH

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

H2 + C2H4 -----> C2H6

Pt surface

HH

HH

Heterogeneous CatalystsHeterogeneous Catalysts

C HH C

HH

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

H2 + C2H4 -----> C2H6

Pt surface

H

Heterogeneous CatalystsHeterogeneous Catalysts

C HH C

HH

H HH

H2 + C2H4 -----> C2H6

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 RateCatalysts 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