1 kinetics — the study of reaction rates and their relation to the way the reaction proceeds,...

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• KINETICSKINETICS — the study of — the study of REACTION REACTION RATESRATES and their relation to the way the and their relation to the way the reaction proceeds, i.e., its reaction proceeds, i.e., its MECHANISMMECHANISM..

• KINETICSKINETICS — the study of — the study of REACTION REACTION RATESRATES and their relation to the way the and their relation to the way the reaction proceeds, i.e., its reaction proceeds, i.e., its MECHANISMMECHANISM..

Chemical KineticsChemical KineticsChapter 15Chapter 15

Chemical KineticsChemical KineticsChapter 15Chapter 15

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Reaction MechanismsReaction MechanismsReaction MechanismsReaction MechanismsThe sequence of events at the molecular The sequence of events at the molecular

level that control the speed and level that control the speed and outcome of a reaction.outcome of a reaction.

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

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• change in concentration of a reactant change in concentration of a reactant or product with timeor product with time

• initial rateinitial rate• average rateaverage rate• instantaneous rateinstantaneous rate

Reaction Rates Reaction Rates Reaction Rates Reaction Rates

4Concentration/Time Concentration/Time Relations Relations

Concentration/Time Concentration/Time Relations Relations

Need to know what conc. of reactant is as Need to know what conc. of reactant is as function of time. Consider FIRST ORDER function of time. Consider FIRST ORDER REACTIONSREACTIONS

For 1st order reactions, the rate law isFor 1st order reactions, the rate law is- (- ( [A] / [A] / time) = k [A] time) = k [A]

Need to know what conc. of reactant is as Need to know what conc. of reactant is as function of time. Consider FIRST ORDER function of time. Consider FIRST ORDER REACTIONSREACTIONS

For 1st order reactions, the rate law isFor 1st order reactions, the rate law is- (- ( [A] / [A] / time) = k [A] time) = k [A]

CisplatinCisplatin

5Concentration/Time Concentration/Time Relations Relations

Concentration/Time Concentration/Time Relations Relations

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

CisplatinCisplatin

[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 =fraction remaining after time

t has elapsed.t has elapsed.

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

t has elapsed.t has elapsed.

Called the Called the integrated first-order rate lawintegrated first-order rate law..

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Half-LifeHalf-LifeSection 15.4 and Screen 15.8Section 15.4 and Screen 15.8


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

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

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Half-LifeHalf-LifeHalf-LifeHalf-Life

• Reaction after 654 Reaction after 654 min, 1 half-life.min, 1 half-life.

• 1/2 of the reactant 1/2 of the reactant remains.remains.

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• Reaction after Reaction after 1306 min, or 2 1306 min, or 2 half-lives.half-lives.


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• Reaction after 3 Reaction after 3 half-lives, or 1962 half-lives, or 1962 min.min.


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SolutionSolution

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

Therefore, ln (1/2) = - k • tTherefore, ln (1/2) = - k • t1/21/2

- 0.693 = - k • t- 0.693 = - k • t1/21/2

tt1/21/2 = 0.693 / k = 0.693 / kSo, for sugar, So, for sugar,

tt1/21/2 = 0.693 / k = 2100 sec = = 0.693 / k = 2100 sec = 35 min35 min

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

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SolutionSolution2 hr and 20 min = 4 half-lives2 hr and 20 min = 4 half-livesHalf-life Half-life Time ElapsedTime Elapsed Mass LeftMass Left1st1st 35 min35 min 5.00 g5.00 g2nd2nd 7070 2.50 g2.50 g3rd3rd 105105 1.25 g1.25 g4th4th 140140 0.625 g0.625 g

Rate = k[sugar] and k = 3.3 x 10Rate = k[sugar] and k = 3.3 x 10-4-4 sec sec-1-1. Half-life . Half-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?left after 2 hr and 20 min?

12Half-LifeHalf-LifeSection 15.4 and Screen 15.8Section 15.4 and Screen 15.8


SolutionSolution

ln [A] / [A]ln [A] / [A]00 = -kt = -kt

[A] = ?[A] = ? [A][A]00 = 1.50 mg = 1.50 mg t = 49.2 mgt = 49.2 mg

Need k, so we calc k from: k = 0.693 / tNeed k, so we calc k from: k = 0.693 / t1/21/2

Obtain k = 0.0564 yObtain k = 0.0564 y-1-1

Now ln [A] / [A]Now ln [A] / [A]00 = -kt = - (0.0564 y = -kt = - (0.0564 y-1-1) • (49.2 y) ) • (49.2 y)

= - 2.77= - 2.77

Take antilog: [A] / [A]Take antilog: [A] / [A]00 = e = e-2.77-2.77 = 0.0627 = 0.0627

0.0627 is the 0.0627 is the fraction remainingfraction remaining!!

Start with 1.50 mg of tritium, how much is left after 49.2 Start with 1.50 mg of tritium, how much is left after 49.2 years? tyears? t1/21/2 = 12.3 years = 12.3 years

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• ConcentrationsConcentrations and and physical physical statestate of reactants and of reactants and products products

• TemperatureTemperature

• CatalystsCatalysts

Factors Affecting Rates Factors Affecting Rates Section 15.2Section 15.2

Factors Affecting Rates Factors Affecting Rates Section 15.2Section 15.2

14Collision TheoryCollision TheoryCollision TheoryCollision Theory

Reactions require Reactions require

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

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

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





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

single single ELEMENTARYELEMENTARY step. step.

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Collision TheoryCollision Theory explains effects explains effects Of Conc. & Temp on Rates!Of Conc. & Temp on Rates!

• Molecules must collide

• Molecules must collide with enough energy

• Molecules must collide with the right orientation

16Collision TheoryCollision TheoryCollision TheoryCollision Theory









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Concentrations and Rates Concentrations and Rates Concentrations and Rates Concentrations and Rates

To postulate a reaction To postulate a reaction mechanism, we studymechanism, we study

•• reaction ratereaction rate and and

•• its its concentration dependenceconcentration dependence

To postulate a reaction To postulate a reaction mechanism, we studymechanism, we study

•• reaction ratereaction rate and and

•• its its concentration dependenceconcentration dependence

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Arrhenius Equation

k = A e k = A e –Ea/Rt–Ea/Rt

A = frequency of collisions with correct geometry.

e e –Ea/Rt–Ea/Rt = fraction of molecules

with mimimum energy for the reaction

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Arrhenius Equation

k = A e k = A e –Ea/RT–Ea/RT

ln k = ln A - (Ea/RT)ln k = ln A - (Ea/RT)

ln k = ln A - Ea/R ( 1/T)ln k = ln A - Ea/R ( 1/T)

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Arrhenius Equation

As Temperature increases, the As Temperature increases, the fraction of fraction of molecules with sufficient activation molecules with sufficient activation energy energy increases.increases.

Temp (K) e -Ea/Rt

298 9.7 x 10-8

400 5.9 x 10-6

600 3.3 x 10-4

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MECHANISMSMECHANISMS

Sections 15.5 and 15.6Sections 15.5 and 15.6

MECHANISMSMECHANISMS

Sections 15.5 and 15.6Sections 15.5 and 15.6

How are reactants converted to products at How are reactants converted to products at the the molecular levelmolecular level??

RATE LAW ----> RATE LAW ----> MECHANISMMECHANISM

experiment ---->experiment ----> theorytheory

How are reactants converted to products at How are reactants converted to products at the the molecular levelmolecular level??

RATE LAW ----> RATE LAW ----> MECHANISMMECHANISM

experiment ---->experiment ----> theorytheory

22MECHANISMSMECHANISMSMECHANISMSMECHANISMS

For exampleFor example

Rate = k [trans-2-butene]Rate = k [trans-2-butene]

Conversion requires twisting around the C=C Conversion requires twisting around the C=C bond.bond.

For exampleFor example

Rate = k [trans-2-butene]Rate = k [trans-2-butene]

Conversion requires twisting around the C=C Conversion requires twisting around the C=C bond.bond.

H3C

C C

CH3

H H

H3C

C C

H

H CH3trans-2-butene cis-2-butene

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Energy involved in conversion of trans to cis Energy involved in conversion of trans to cis butenebutene

Energy involved in conversion of trans to cis Energy involved in conversion of trans to cis butenebutene

trans

cis

energy ActivatedComplex

26.9 kJ/mol

30.2 kJ/mol3.9 kJ/mol

233 kJ 229 kJ

MECHANISMSMECHANISMSMECHANISMSMECHANISMS

See Figure 15.15See Figure 15.15


TRANSITION STATETRANSITION STATE

ACTIVATION ACTIVATION ENERGY, EENERGY, Eaa = =

energy req’d to form energy req’d to form activated complex.activated complex.

Here EHere Eaa = 233 kJ/mol = 233 kJ/mol

TRANSITION STATETRANSITION STATE

ACTIVATION ACTIVATION ENERGY, EENERGY, Eaa = =

energy req’d to form energy req’d to form activated complex.activated complex.

Here EHere Eaa = 233 kJ/mol = 233 kJ/mol

trans

cis


26.9 kJ/mol


233 kJ 229 kJ

trans

cis


26.9 kJ/mol


233 kJ 229 kJ

25Activation EnergyActivation EnergyActivation EnergyActivation Energy

Molecules are moving…..but how many of them Molecules are moving…..but how many of them have enough Energy to go to product? have enough Energy to go to product?

What does increasing T do?What does increasing T do?

A flask full of trans-butene is stable because A flask full of trans-butene is stable because only a tiny fraction of trans molecules have only a tiny fraction of trans molecules have enough energy to convert to cis.enough energy to convert to cis.

In general, In general, differences in activation energydifferences in activation energy are the reason reactions vary from fast to are the reason reactions vary from fast to slow.slow.

Molecules are moving…..but how many of them Molecules are moving…..but how many of them have enough Energy to go to product? have enough Energy to go to product?

What does increasing T do?What does increasing T do?

A flask full of trans-butene is stable because A flask full of trans-butene is stable because only a tiny fraction of trans molecules have only a tiny fraction of trans molecules have enough energy to convert to cis.enough energy to convert to cis.

In general, In general, differences in activation energydifferences in activation energy are the reason reactions vary from fast to are the reason reactions vary from fast to slow.slow.


1.1. Why is reaction observed to be 1st Why is reaction observed to be 1st order?order?

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

2.2. Why is the reaction faster at higher Why is the reaction faster at higher temperature?temperature?

Fraction of molecules with sufficient Fraction of molecules with sufficient activation energy increases with T.activation energy increases with T.

1.1. Why is reaction observed to be 1st Why is reaction observed to be 1st order?order?

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

2.2. Why is the reaction faster at higher Why is the reaction faster at higher temperature?temperature?

Fraction of molecules with sufficient Fraction of molecules with sufficient activation energy increases with T.activation energy increases with T.

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Reaction of Reaction of trans --> cistrans --> cisis is UNIMOLECULARUNIMOLECULAR- only one reactant is - only one reactant is involved.involved.

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

Reaction of Reaction of trans --> cistrans --> cisis is UNIMOLECULARUNIMOLECULAR- only one reactant is - only one reactant is involved.involved.

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

MECHANISMSMECHANISMSMECHANISMSMECHANISMS

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BIMOLECULARBIMOLECULAR — two — two different molecules different molecules must collide must collide

--> --> productsproducts

BIMOLECULARBIMOLECULAR — two — two different molecules different molecules must collide must collide

--> --> productsproducts

A bimolecular reactionA bimolecular reaction

Exo- or endothermic? Exo- or endothermic?


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

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

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

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

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Relationship to Reaction order

• Molecularity of an Elementary Step and its order are the same.

• Not necessarily true for overall reaction, just for elementary steps!

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Rate Equations again

A unimolecular Rate = k [A]

A + B Bimolecular Rate = k [A][B]

A + A Bimolecular Rate = k [A]2

2 A + B

Termolecular Rate = k [A]2[B]


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

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

Step 1 —HOOH + IStep 1 —HOOH + I-- --> HOI + OH --> HOI + OH--

Step 2 —HOI + IStep 2 —HOI + I-- --> I --> I22 + OH + OH--

Step 3 —2 OHStep 3 —2 OH- - + 2 H + 2 H++ --> 2 H --> 2 H22OO

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


Step 1 —HOOH + IStep 1 —HOOH + I-- --> HOI + OH --> HOI + OH--

Step 2 —HOI + IStep 2 —HOI + I-- --> I --> I22 + OH + OH--

Step 3 —2 OHStep 3 —2 OH- - + 2 H + 2 H++ --> 2 H --> 2 H22OO


Step 1Step 1 is is bimolecularbimolecular and involves I and involves I-- and HOOH. and HOOH. Therefore, this predicts the rate law should beTherefore, this predicts the rate law should 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.

Step 1Step 1 is is bimolecularbimolecular and involves I and involves I-- and HOOH. and HOOH. Therefore, this predicts the rate law should beTherefore, this predicts the rate law should 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 — slowStep 1 — slow HOOH + IHOOH + I-- --> HOI + OH --> HOI + OH--

Step 2 — fastStep 2 — fast HOI + IHOI + I-- --> I --> I22 + OH + OH--

Step 3 — fastStep 3 — fast 2 OH2 OH- - + 2 H + 2 H++ --> 2 H --> 2 H22OO

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Exercise 15.12

• 2 NO N2O2

• N2O2 + H2 N2O + H2O

• N2O + H2 N2 + H2O

• Molecularity? Rate Eqns? Sum of Steps?

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

RATE DETERMINING STEPRATE DETERMINING STEP

Rate can be no faster than Rate can be no faster than RDSRDS!!

38CATALYSISCATALYSISCATALYSISCATALYSIS

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

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

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Catalysts in Industry

• Petroleum refining

• Industrial production of chemicals, pharmaceuticals

• Environmental controls

Heterogeneous vs. Homogeneous


In auto exhaust systems — Pt, NiOIn auto exhaust systems — Pt, NiO

2 CO + O2 CO + O22 ---> 2 CO ---> 2 CO22

2 NO ---> N2 NO ---> N22 + O + O22

In auto exhaust systems — Pt, NiOIn auto exhaust systems — Pt, NiO

2 CO + O2 CO + O22 ---> 2 CO ---> 2 CO22

2 NO ---> N2 NO ---> N22 + O + O22


2.2. Polymers: Polymers: HH22C=CHC=CH22 ---> polyethylene ---> polyethylene

3.3. Acetic acid: Acetic acid:

CHCH33OH + CO --> CHOH + CO --> CH33COCO22HH

4. 4. Enzymes — biological catalystsEnzymes — biological catalysts

2.2. Polymers: Polymers: HH22C=CHC=CH22 ---> polyethylene ---> polyethylene

3.3. Acetic acid: Acetic acid:

CHCH33OH + CO --> CHOH + CO --> CH33COCO22HH

4. 4. Enzymes — biological catalystsEnzymes — biological catalysts


Catalysis and activation energyCatalysis and activation energy

Catalysis and activation energyCatalysis and activation energy

Uncatalyzed reactionUncatalyzed reaction

Catalyzed reactionCatalyzed reaction

MnOMnO22 catalyzes catalyzes

decomposition of Hdecomposition of H22OO22

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

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MnOMnO22 catalyzes decomposition of H catalyzes decomposition of H22OO22

Figure 15.18

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Iodine-Catalyzed Isomerization Iodine-Catalyzed Isomerization of cis-2-Buteneof cis-2-Butene

Figure 15.19Figure 15.19

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Cis-2-butene trans-2-butene

• 1) I2 2 I•

• 2) I • + cis-2-butene I-cis-2-butene

• 3) I-cis-2-butene I-trans-2-butene

• 4) I-trans-2-butene I • + trans-2-butene

• 5) I • + I • I2

Rate = k [cis-2-butene][I2]1/2

One I2 gets broken, one I • used, but regenerated. In the end, the two I can recombine. No net consumption of I2!

1 kinetics — the study of reaction rates and their relation to the way the reaction proceeds,...

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