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Kinetics

Reaction rate:

how fast to you make products

Thermodynamics:

what happens eventually

H2 + O2 gives H2O + heat

Kinetics

Should I care

Concentrations as a function of time

Road Map from React to Products

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KineticsRate law - concentrations related to their changes- differential

integral form - conc as function of time [A](t)

Rate law from reaction mechanism:path from Reac to Prod

Reaction Mechanism- Sequence of Elementary Steps

A+B –> C; C+D–> E overall A+B+D–>E

Rate determining step Steady state approximation

compareA+B vs C+D something about C

Collision theory TEMPERATURE Arrhenius A factor and Eact

Reaction profile - Activation energy MB Distribution as a pdf

Auto catalytic and cyclic reactions:Some things are worth doingover and over

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Rate constant k(T,P) - ArrheniusRate laws - mechanistic interpretation

Microscopic theory of chemicalreaction rates

Reaction pathsMaxwell-Boltzman Distribution andactivation energy

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Rate Laws

Instantaneous Rate laws - relationbetween concentrations and theirrate of change (reaction rate) -differential equationsIntegrated rates laws -concentrations as function of times

Instantaneous Rate law –––> Integrated rate law

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Rate Law

relates Rate to instantaneous concentrations

aA + bB cC + dD

Expectation the more A and B the more C producedin a given time ie

Rate is proportional to concentration

To a power!

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NO2 + CO NO + CO2

Measuringreaction

rates

One of severalchoices. Do Ineed to measureall of them?

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Reaction Rates

[A](

t)

t(s)

x

Rate = d A(t) dt

B+C A + D

What is therelation between

rate andstoichiometry

Re-phrasing

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aA + bB cC + dD

1 d [C](t) c dt

1 d [D](t) d dt

- 1 d [A](t) a dt

- 1 d [B](t) b dt

What does overall reaction tell you about rate law

Stiochiometry and Rates

reaction mechanism

Stiochiometry gives ratios of A, B, C, DNot

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Rate Law

Relates rate to instantaneous concentrations

aA + bB cC + dD

Rate(t) = k {[A](t)}na {[B](t)}

nb{[C](t)}nc{[D](t)}

nd

Order of reaction = na+ nb+ nc+ nd

Rate constant-not a function

of time butk(T,P)

-1 d A(t) = rate(t) a dt

So rate law is a diff. eq.

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First Order Rate LawUnimolecular Decay

A products

Examples:Photodissociation•O2+ hv ––> O2

* ––> O + O absorbhigh frequency photons•O3+ hv ––> O3

*––> O2+O absorb uv

ONLY ACTIVATED STATEDECAYS

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First Order Rate LawUnimolecular Decay

-d [A](t) = k [A](t) dt

A products

- 1 d[A](t) = k dt[A](t)

Divide by [A]

Differential to Integrated rate law

dln(x)=dx/x

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• -d ln [A](t) = k dt

• Integrate from (t=0 A(0)) to (t=t, A(t))

ln(A(t))- ln (A(0))=-kt

• A(t)/A(0) = exp(-kt) lnB-lnA=lnB/A

• Half life : t for which A( t) = A(0)/2

0.5*A(0)/A(0) = exp(-k t)

ln2/k= t

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[A]=c

d ln c = (1/c)dc-lnc

t

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Second Order Rate Law

-d [A](t) = 2k [A](t)2

dt

- 1 d[A](t) = 2k dt[A](t)2

2A products

Note: convenient toinclude 1/2 otherwiserate is for loosing 2A’s

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[ ]( ) ( )

( )

( ) ( )

( ) ( )( ) ( )

( ) ( ) / ( )

( )

( )

( )

( )

A t c t

dc

c tkdt

dc

ck dt

c c t ckt

c c t

c c t

ktc c c t

c

c tt

c

c t

=

-=

-=

= - = =-

= -

Ú Ú2 20

0

0

2 2

1 1 10

200

2 0 0 1

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Half Life

•2ktc(0)=c(0)/c(t)-1

•2kt1/2 c(0)=c(0)/c(0)/2-1=1

•t1/2=1/(2kc(0))

Concentration dependent

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Half lifedependson c0

Half life:

-1/(c0/2)+1/c0=-2kt

1/ c0=2kt

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2000 Words

-lnc 1/c

t t

First order second order

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Chemical Deceit

Initial Conditions can make thingsappear other than they are

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A+B products

c0A c0B initialconcentrations

x= amount/V of A reacted

-d[A]/dt= k [A](t) [B](t)(dx/dt) =k(c0A-x)(c0B-x)

dx/ [(c0A-x)(c0B-x )]=kdt

c0A-x= concentration of A at time t = [A](t)

c0B-x= concentration of B at time t = [B](t)

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1/ [(c0A-x)(c0B-x )] = C/ [c0A-x]+D/ [c0B-x ]

-dx / [c0A-x]+dx/ [c0B-x ] =(c0A- c0B)kdt = Kdt

dln[c0A-x] - dln [c0B-x ] = Kdt

(c0A-x)/(c0B-x )=eKt for c0Aπ c0B

Effective rate constant depends concentration

1= D [c0A-x]+C [c0B-x ] multiply by [(c0A-x)(c0B-x )]

1= D c0A+C c0B 0= D x+Cx difference of these two equations

is the previous line (underdetermined

So that C=-D and 1/(c0A- c0B)=D

OK only if initial concentrations are distinct

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Plotln (c0A-x)/(c0B-x )=Kt

Two cases: A or B in excess

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A in excessln (c0A-x)/(c0B-x )=Kt

-ln(c0B-x )=Kt- ln (c0A-x) ~ Kt- ln (c0A)

-ln[B]

t

Looks like unimolecular decayof B

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Complex Rate Laws

Building blocks are

Elementary Reactions

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Elementary Reaction

aA + bB––––> cC + dD

-(1/a)d/dt[A](t)=k [A](t)na [B](t)

nb[C](t)nc[D](t)

nd

na=a nb=b nc=0 nd=0

Definition:Stoichiometry gives rate law

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Beyond Forward Reactions

Overall rate of single reaction isforward rate - backward rate

A+B ––> C+D

as soon as C and D exist

C+D ––> A+B

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Kinetics and EquilibriumDetailed Balance

Detailed balance

at equilibrium the forward and backward rates of eachelementary reaction are balanced

A + B C + D kf [A][B] = kb[C][D]K= kf / kb

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Reaction Mechanism

Sequence of Elementary Reactions

Elementary Reaction

DEFINITION

Stoichiometry gives rate law

nA + mB+cC––––> products

d/dt[Prod] = k [A]n [B]m[C]c

D(12)

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Complex mechanism

Observed reaction: 2A+2B––> C+D

Elementary Steps:

A+A<–––> F consider forward and Back

F+B––––>G consider forward

G+B––––>C+D consider forward

SUM

2A+2B––––>C+D

Why someforward onlysome both

REACTIONPATH

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Rules• Rate law for individual steps unaffected by

existence other steps

• Concentration of any species is the ‘net’ ofall reactions in which it is involved;

Net=sources - sinks

!d/dt[C] = k3[G][B]

!d/dt[F]=k1[A]2 – k-1[F]–k2[F][B]

A+A<–––> FF+B––––>G

G+B––––>C+D

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RATE LAW is deduced from

!d/dt[C] = k3[G][B]

But how to get [G]?

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Determining Rate Laws

Rate determining step (prior equilibrium)

Steady state approximation

(Reactive intermediate)

Simplifying assumptions

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Rate determining step

Slow step in a reaction mechanism

Rate of reaction is rate of this step

Prior steps reach equilibrium

Subsequent step(s) fast/irrelevant

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Building a reaction

2NO+O2 ––> 2NO2

Built from

NO+NO <–––> N2O2 in equilibrium

N2O2 + O2 ––> 2NO2

rate determining

Note sum of elementary reactions is the overall reaction

Rate=1/2d/dt[NO2] = k2[N2O2][O2]

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2NO+O2 2NO2

Three bodycollisions rare

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Classic Example

H2 + I2 2HI rate=k[ H2][ I2]

But

hv + I2 2I dramatically increased rate!!!!

•M + I2 I + I + M fast K 1 = [I]2/[I2]!

•I + H2 H2I fast K 2 = [H2I] /[H2][I]

•H2I + I 2HI slow rate= [H2I] ][I]

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rate= k[H2I] ][I] K 2 = [H2I]

/[H2][I]

K 2 [H2][I]= [H2I]

rate= kK2 [H2][I] [I]

= kK2 [H2][I]2 K1 = [I]2/[I2]

=kK2K1[I2] [H2]

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Steady State Approximation

No rate determining step

Applies to (an) intermediate I d [I](t) = sources - sinks = 0 dt

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Lindemann Activated ComplexMechanism: Collisional Activation

Observation: A in an inert gas ( buffer gas) decomposes

rate constant is unimolecular(sum ofexponents=1) for ‘ normal’ gas pressure

for low gas pressure rate is bimolecular( sumof exponents =2)

EXPLAIN

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Steady- State Approximation andLindemann Activated Complex

Mechanism: Collisional Activation

A+M A* + M

A* products

NET: A ––––> prod

d [A*](t) = 0 dt

k1[A][M] -k-1 [A*][M]-k2[A*] = 0

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k1[A][M} / {k-1 [M]+k2}=[A*]

rate=k2[A*]=k2k1[A][M]/{k-1[M]+k2}

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Steady- State Approximation and Lindemann Activated ComplexMechansim

N2O5 ––> 2NO2 + 1/2 O2

N2O5 +M <––> N2O5*+M

N2O5* ––> 2NO2 + 1/2 O2

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M

=

N2O5

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EnzymeCatalysis MichaelisMenton Eq

Finish start

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Testing a Mechanism

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Verifying Mechanisms

Test predicted rate law

Chemical methods- isotopicsubstitution

12C18O ( no 18O in NO)

NO2 + CO ––> NO + CO2

D13

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Collisiontheory

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Microscopic description ofA+A -----> products

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Rate equation from collision theory

# of Collisions

Start with 1 particle - sweeping out a volume

collisions = volume times density

volume = x section velocity (–> per unit time)

velocity = MB-velocity

total collisions mult by N particles / volume

/2 since A collides with A-A

Units: [A] = density/N0

d[A]/dt = -twice the collision frequency

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Collision Theory

V x density(almost)=# of collisions

u from MB

Area x velocity= volume/time

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Collision frequency=collisions/time= Z1

{[ pd2<u>]} density÷2 = Z1 =

4d2r÷(pRT/M)

particles in vol swept out

÷(8RT/(Mp)) = <u>

Vol swept out/time

Average velocity

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N such molecules so

Total number of collisions/time= 1/2 N Z1

Total number of collisions/time/vol =1/2 N/V Z1 = 1/2r Z1

=1/2r4d2r÷(pRT/M)= ZAA

Total collisionrate/vol

N Z1 /Vx1/2

Notational caveat

kN0 = R

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(1/V)dN/dt= -2ZAA

[ ] = Molar density = r/N0

d[A]/dt= -2ZAA/N0

Rate= (-1/2) d[A]/dt= ZAA /N0 =

2d2r2 (N0 /N02)÷(pRT/M)=

[A]2 [2d2N0÷(pRT/M)]=k [A]2

One collision eliminates two As A+A ––––––> products

k = [2d2N0÷÷÷÷(ppppRT/M)]

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Does every collision cause a reaction?

11 more

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Do all k have dramatic temperaturedependence

CH3 + CH3 no

But in general yes

Why is CH3 + CH3 different?

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Origin of Activation Energy

Marcelin-1915 not all collisions cause reaction

collision partners must have sufficient

energy to react

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Maxwell-Boltzmannand reaction profile

Average is not everything

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Maxwell-Boltzmann Distribution

Fraction of molecules withenergy > Eact decreases like

exp(-Eact/kT)

<u> = average speed =

÷(8RT/(Mp))

where g u f u du g( ) ( )Ú =

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k = [2d2N0÷÷÷÷(ppppRT/M)]exp(-Eact/RT)

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PROBABILITYDENSITY(distribution) FUNCTION

f(u)du is the probability of having u (velocity) between uand u+du

f u du Normalization( ) -�

�

Ú = 1

f

u

u+duu

f(u) ≥ 0

g u f u du g( ) ( )Ú =

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Coming soon another pdf

Y*(x) Y(x)

* is complex conjugate

Y(x) is the wave function obtained from the

solution of Schrödinger Equation

Expectation - average - value

Y Y* ( ) ( ) ( )x g x x dx gÚ =

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Rate Constants Temperature Dependence

Arrhenius-1889 k(T)=A exp(-Eact/kT)Eact -activation energy-is independent of TA is the Arrhenius or prefactor. It has as asmall temperature dependence.What is the microscopic origin of Eact and thetemperature dependence of A.

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Origin of Eact

(10 +cl)

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Is the theorycomplete yet?

k= 2d2N0÷(pRT/M) exp (-Eact/kT)

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Not all directions ofapproach areequivalent

O–– N–– O + C––O –––> O–––N + O––C––O

O–– N–– O + O–– C ––––>

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k = P

2d2N0÷÷÷÷(ppppRT/M)

exp (-Eact/RT)

steric factor(P)

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Two pieces missing from collisiontheory

k=P 2d2N0÷(pRT/M) exp (-Eact/RT)

steric factor(P)

exp (-Eact/kT)Activation

steriochemistry

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Reaction sequences cangive deviations fromArrhenius behaviour

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Negative Activation Energy

What is the T-dependence of k?

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expE

xo/R

T

TT

exp-

E/R

T

T E RT E RT E RT

T E RT E RT E RT

= = � - = = �

= � = - = =

0 0

0 1 1

/ exp( / ) exp( / )

/ exp( / ) exp( / )

Can the RHS occur

Rate (k) ~ exp(-E/RT)increasesexponentiallywith T

Rate (k) ~ exp(E/RT)decreasesexponentiallywith T

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Negative Activation Energies

2NO(g) + O2(g) 2NO2 (g)

NO + NO <––––> N2O2 fast

N2O2 + O2 ––––––> 2NO2 slow

Rate = k2K1[NO]2[O2]

K1 = k1/k-1 will use Arrhenius k to evaluate K

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As T increases k2

increases but K1

decreases

K=Afexp-E af /RT

A bexp-E ab/RT =exp-(E af -E ab)/RT =expExo/RT

Ex0

expE

xo/R

T

TT

exp-

E/R

T

- + = >E E EaF aB xo 0

keff = k2K1k E E RTeff a x~ exp[ ( ) / ]`- -2 0

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Surface

Catalysis

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Autocatalytic and Oscillatoryreactions

Rate= k[A][P] A P

[A]=c0A-x [P]=c0P+x

-d[A]/dt = dx/dt=

k(c0A-x ) (c0P+x)

Usual trick

dx/((c0A-x ) (c0P+x))=kdt

A decreases with time butP increases

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Algebra

1/ [(c0A-x)(c0P+x )] = C/ [c0A-x]+D/ [c0P+x ]

1= D [c0A-x]+C [c0P+x ]

1= D c0A+C c0P 0= -D x+Cx

So that C=D and 1/(c0A+ c0P)=D

+dx / [c0A-x]+dx/ [c0P+x ] =(c0A+ c0P)kdt = adt

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dln(c0A -x)=-1/(c0A -x) dln(c0P +x)=1/(c0P +x)

C0X =[X]0

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Key point:rate starts slow and increases with time

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Auto catalytic reactions can beused as the building block of

cyclic reactions

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Autocatalytic(AC)-cyclic reaction(CR)

AC––––> CR

Far from equilibrium

‘intermediates’ key

Cant use steady stateapproxmation

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Lotka-VolterraModel Auto catalytic reaction

Steady state condition [A] constant

NOT steady state approximation

feedback

Startwith little

X and y

A–––––––––>B

First order-Not AC

AC

AC

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Lotka-Volterra

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d A

dtk A X A X X A Xa

[ ][ ][ ] ( )= - + Æ Æ2

r k Y X Y X Y X Yb b= - + Æ Æ[ ][ ] ( )2

X

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