Chemical Kinetics

CHAPTER 14

Chemistry: The Molecular Nature of Matter, 6th editionBy Jesperson, Brady, & Hyslop

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CHAPTER 14 Chemical Kinetics

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Learning Objectives: Factors Affecting Reaction Rate:

o Concentrationo Stateo Surface Areao Temperatureo Catalyst

Collision Theory of Reactions and Effective Collisions Determining Reaction Order and Rate Law from Data Integrated Rate Laws Rate Law Concentration vs Rate Integrated Rate Law Concentration vs Time Units of Rate Constant and Overall Reaction Order Half Life vs Rate Constant (1st Order) Arrhenius Equation Mechanisms and Rate Laws Catalysts

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CHAPTER 14 Chemical Kinetics

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Lecture Road Map:

① Factors that affect reaction rates

② Measuring rates of reactions

③ Rate Laws

④ Collision Theory

⑤ Transition State Theory & Activation Energies

⑥ Mechanisms

⑦ Catalysts

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Mechanisms of Reactions

CHAPTER 14 Chemical Kinetics

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Mechanisms Overall vs Individual Steps

Sometimes rate law has simple form

– N2O5 NO2 + NO3

– NO2 + NO3 N2O5

But others are complex

– H2 + Br2 2 HBr

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Mechanisms Overall vs Individual Steps

Some reactions occur in a single step, as written

Others involve a sequence of stepsoReaction Mechanism

oEntire sequence of stepso Elementary Process

oEach individual step in mechanismoSingle step that occurs as written

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Mechanisms Overall vs Individual Steps

o Exponents in rate law for elementary process are equal to coefficients of reactants in balanced chemical equation for that elementary process

o Rate laws for elementary processes are directly related to stoichiometry

o Number of molecules that participate in elementary process defines molecularity of step

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Mechanisms Unimolecular Process

o Only one molecule as reactanto H3C—NC H3C—CNo Rate = k[CH3NC]

o 1st order overallo As number of molecules increases, number that rearrange

in given time interval increases proportionally

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Mechanisms Bimolecular Process

o Elementary step with two reactantso NO(g) + O3(g) NO2(g) + O2(g)

o Rate = k[NO][O3]

o 2nd order overallo From collision theory:

o If [A] doubles, number of collisions between A and B will double

o If [B] doubles, number of collisions between A and B will double

o Thus, process is 1st order in A, 1st order in B, and 2nd order overall

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Mechanisms Termolecular Process

o Elementary reaction with three molecules o Extremely rare

o Why? o Very low probability that three molecules

will collide simultaneouslyo 3rd order overall

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Mechanisms Elementary Processes

Molecularity

Elementary Step Rate Law

Unimolecular

A products Rate = k[A]

Bimolecular A + A products Rate = k[A]2

Bimolecular A + B products Rate = k[A][B]Significance of elementary steps: o If we know that reaction is elementary stepo Then we know its rate law

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Mechanisms Multi-step Mechanisms

o Contains two or more steps to yield net reactiono Elementary processes in multi-step mechanism must always

add up to give chemical equation of overall processo Any mechanism we propose must be consistent with

experimentally observed rate law

o Intermediate = species which are formed in one step and used up in subsequent stepso Species which are neither reactant nor product in overall

reactiono Mechanisms may involve one or more intermediates

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Mechanisms Example

The net reaction is:

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

The proposed mechanism is:

NO2(g) + NO2(g) NO3(g) + NO(g)

NO3(g) + CO(g) NO2(g) + CO2(g)

2NO2(g) + NO3(g) + CO(g) NO2(g) + NO3(g) + NO(g) + CO2(g)

or

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

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Mechanisms Rate Determining Step

o If process follows sequence of steps, slow step determines rate = rate determining step.

o Think of an assembly lineo Fast earlier steps may cause intermediates to pile upo Fast later steps may have to wait for slower initial steps

o Rate-determining step governs rate law for overall reaction

o Can only measure rate up to rate determining step

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Mechanisms Example: Rate Determining Step

(CH3)3CCl(aq) + OH–(aq) (CH3)3COH(aq) + Cl–(aq)

chlorotrimethylmethane trimethylmethanol

o Observed rate = k[(CH3)3CCl]

o If reaction was elementaryo Rate would depend on both reactantso Frequency of collisions depends on both concentrations

o Mechanism is more complex than single stepo What is mechanism?

o Evidence that it is a two step process

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Mechanisms Rate Determining Step as Initial Step

Step 1: (CH3)3CCl(aq) (CH3)3C+(aq) + Cl–(aq) (slow)

Step 2: (CH3)3C+(aq) + OH–(aq) (CH3)3COH(aq) (fast)

o Two steps each at different rateso Each step in multiple step mechanism is elementary process,

soo Has its own rate constant and its own rate law

o Hence only for each step can we write rate law directlyo Observed rate law says that step 1 is very slow compared to

step 2o In this case step 1 is rate determiningo Overall rate = k1[(CH3)3CCl]

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Mechanisms Mechanisms with Fast Initial Step

1st step involves fast, reversible reactionEx. Decomposition of ozone (No catalysts)

Net reaction: 2O3(g) 3O2(g)

Proposed mechanism:

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

O(g) + O3(g) 2O2(g) (slow)

]O[]O[

Rate Observed2

23k

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Mechanisms Is the Mechanism Rate Law Consistent?

o Rate of formation of O2 = Rate of reaction 2 = k2[O][O3]

o But O is intermediateo Need rate law in terms of reactants and products

o and possibly catalystso Rate (forward) = kf[O3]o Rate (reverse) = kr[O2][O]o When step 1 comes to equilibrium

o Rate (forward) = Rate (reverse) o kf[O3] = kr[O2][O]

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Mechanisms Is the Mechanism Rate Law Consistent?

o Solving this for intermediate O gives:

o Substitution into rate law for step 2 gives:

o Rate of reaction 2 = k2[O][O3] =

o where

o This is observed rate lawo Yes, mechanism consistent

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GroupProblem

The reaction mechanism that has been proposed for the decomposition of H2O2 is

1. H2O2 + I– → H2O + IO– (slow)

2. H2O2 + IO– → H2O + O2 + I– (fast)

What is the expected rate law?

First step is slow so the rate determining step defines the rate law

rate=k [H2O2][I–]

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GroupProblem

The reaction: A + 3B → D + F was studied and the following mechanism was finally determined:

1. A + B C (fast)

2. C + B → D + E (slow)

3. E + B → F (very fast)

What is the expected rate law?

Rate Step 2=k2[C][B] Rate forward = kf[A][B]Rate reverse = kr[C]kf[A][B] = kr[C][C]= kf[A][B]/kr Rate = kobs[A][B]2

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Catalysts

CHAPTER 14 Chemical Kinetics

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

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Catalyst Definition

o Substance that changes rate of chemical reaction

without itself being used up

o Speeds up reaction, but not consumed by reaction

o Appears in mechanism, but not in overall reaction

o Does not undergo permanent chemical change

o Regenerated at end of reaction mechanism

o May appear in rate law

o May be heterogeneous or homogeneous

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Catalyst Activation Energy

o By providing alternate mechanismo One with lower Ea

o Because Ea lower, more reactants and collisions have minimum KE, so reaction proceeds faster

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Catalyst Activation Energy

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Catalyst Homogeneous Catalyst

• Same phase as reactants

Consider : S(g) + O2(g) + H2O(g) H2SO4(g)

S(g) + O2(g) SO2(g)

NO2(g) + SO2(g) NO(g) + SO3(g) Catalytic pathway

SO3(g) + H2O(g) H2SO4(g)

NO(g) + ½O2(g) NO2(g) Regeneration of catalystNet: S(g) + O2(g) + H2O(g) H2SO4(g) • What is Catalyst?

– Reactant (used up) in early step– Product (regenerated) in later step

• Which are Intermediates?

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Catalyst Heterogeneous Catalyst

o Exists in separate phase from reactantso Usually a solido Many industrial catalysts are heterogeneouso Reaction takes place on solid catalyst

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

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Catalyst Heterogeneous Catalyst

H2 and N2 approach Fe catalyst

H2 and N2 bind to Fe& bonds break

N—H bonds forming

N—H bonds forming

NH3 formation complete

NH3 dissociates

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Enzymes: Superoxide Dismutase

Miller, Anne-Frances. “Fe Superoxide Dismutase” Handbook of Metalloproteins. John Wiley & Sons, Ltd, Chinchester, 2001Rodrigues, J. V; Abreu, I. A.; Cabelli, D; Teixeira, M. Biochemistry 2006, 45, 9266-9278.

Catalyst