physical chemistry- kinetics

8/10/2019 Physical Chemistry- Kinetics

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Techniques to Study Kinetics (come up with mechanisms)

ultimately - conc as f(time)

1. pressure changes

- at least one component must be gas

2. spectroscopy- use Beer's law A ac

- only one species must absorb at particular l

3. Electrochemical methods- can monitor change in pH or conductivity of solution

4. Other more sophisticated methods


3/66

dt

didt

idn

dni /V

dt i

V

ddt

dcidt

i

V

d

dt

dci

dt

Assuming closed systems:

aA + bB cC + dD

moles of i : ni= ni(t) = nio+ i

i= stoichiometric coef. - unitless - rcts, + pdts

dni=id noi= constant = extent of rxn = mol

const. V

For const. V: rate ui= = rate of species i


4/66

dtNH

dc

dtH

dc

dtN

dc

dtd

V3

212

3121

32

1

23

1

2

2NHHNdt

Ndc

uuu

32

3

23

2

2NHNHdt

Hdc

uuu

dcNH3

dtu

NH3 2u

N2

2

3uH2

Example: N2 + 3H2= 2NH3

rate of disappearance N2=

rate of disappearance H2=

rate of formation NH3=

u(t) = rate of rxn is related to concentrations of various chemical

species present at time t. This relationship called rate law.

u= rate of reaction =


5/66

]][[][

22 IBrkdt

BrId

]/[]['1

][2/1][][

2

22

BrHBrk

HBrk

dt

HBrd

which components of rxn are kinetically active, i.e. a

change in its concentration will alter the rate of rxn. or

included in rate lawRate law often looks like following:

u= k[A]a[B]b k = rate constant = f(T) c

Br2+ I2 = 2BrI

Br2+ H2 = 2HBr

must be determined experimentally, cannot be deduced

from stoichiometry of rxn.Example: rate law


6/66

In general: [A]a[B]b[D]d

a = order with respect to species A

b = order with respect to species B

d = order with respect to species D

(a+b+d) = overall order of rxn

a can be fraction, negative

For 1st rxn: order w.r.t. Br2and I2is 1, overall order =2

rate law happens to reflect rxn stoichiometry

For 2nd rxn: concept of order does not apply

rate law is complicated and rxn occurs by multistep process


7/66

If concentration of any kinetically active species is

nearly constant it can be incorporated into k and we get

pseudo nth order rxn. C = catalyst for instance:

CBAkdt

Ad

][

(pseudo 2nd order)

can use to determine the order (isolation method)

if [C]~ constant

BAkdt

Ad'

][

(3rd order)


8/66

to find rate of reaction at any time, measure change of

concentration with time at constant T

rate at any time = slope of curve at that time

easier to compare integrated rate expression

time

Co

equilibriumConc.


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dt

Ad ][

kdtA

Ad

][][

tA

oA dtkA

Ad

0

][

][ ][][

ln[A]

[A]o kt

First Order Reactions

A products or A + B products

= k[A]

Plot ln[A] vs t slope = -k

u(t) =

[A] = [A]oe

-kt


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21][

2/][ln ktAA

o

o 2

121ln kt

kt

2ln

21

Half-life: time required for half of the specified

reactant to disappear

t = t1/2when [A] = [A]o/2

For first order reaction:

.)(21 concft


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Second Order Reactions

A products or A + B products

kdtA

Ad

2][][

tA

oA dtkA

Ad

0

][

][2][

][

1

[A

]

1

A o

kt

Plot 1/[A] vs t slope = k

dt

Ad ][ = k[A]2u(t) =

kt

o

AA 1

][

1

kt

oAA 1

][

1


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oAkt

1

21

Half-life for Second Order Reaction

t = t1/2when [A] = [A]o/2

.)(2

1 concinitialft

2

1

1

][

2kt

oo AA

oA

kt

][

1

21


13/66

need to integrate but we have a problem

need to know how [B] varies with [A]

[A] = [A]o- /V and [B] = [B]o- /V

dt

Bd

dt

Ad ][][

= k[A][B]u(t) =

Second Order Reactions

kdtBAAd

][

/V = [A]o-[A] so [B] = [B]o-[A]o+[A]

kdt

AABA

Ad

oo

][


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method of partial fractions: for [B]o [A]o

AAB

q

A

p

AABA oooo

1

1 = p([B]o-[A]o+[A] ) + q[A] = p([B]o-[A]o) +(p+q)[A]

1= p([B]o-[A]o) 0 = p+q

oo ABp

1

oo ABq

1

A

oA AoAoBA

1

A

oA

oooo

A

oA

o AABAB

Ad

A

o

BA

Ad


15/66

oo

ooo

ooo

oo BABA

ABkt

BAAABA

ABln1ln1

ktB

AAB

ABA

A

AB ooo

ooooo

ln1

ln1


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Second Order Autocatalytic Reaction

AB

[A] = [A]o- /V and [B] = [B]o+ /V

dt

Bd

dt

Ad ][][ = k[A][B]u(t) =

kdtBA

Ad ][

/V = [A]o-[A] so [B] = [B]o+ [A]o- [A]

kdt

AABA

Ad

oo

][


17/66

method of partial fractions: for [B]o [A]o

AAB

q

A

p

AABA oooo

1

1 = p([B]o+[A]o-[A] ) +q[A] = p([B]o+[A]o) - (p-q)[A]

1= p([B]o+ [A]o) 0 = p-q p = q

oo ABp

1

A

oA AoAoBA

1

A

o

A

A

o

A oAoB

o

AoB

Ad

oA

o

BA

Ad


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Determining the Rate Law

1. Take c vs. t , guess n, plot as discussed

1st order ln c vs. t straight line n=12nd order 1/c vs. t straight line n = 2

- perform experiment at const. T where [B] and [D] are in excess

ba DBAkdt

Ad

DBAkdt

Adlogloglogloglog ba

2. Isolation method:

aAkdt

Ad ' where k' includes [B] and [D]

- can determine k' and a

- do for other components as well


21/66

2 . Method of Initial Rates

- measure rates at initial concentrations

- run experiment at series of initial concentration

rates plot log ovs. log [A]o slope = a

3. Half-life Method

nth order: =k[A]n for n > 2

Determining rate law: need 2 different [A]o at const. T (k is same)

solve for n knowing [A]o(1), t (1), [A]o(2), t(2) from experiments

rate law must be of form = k[A]n

dt

Ad ][

1

o

o

1-n

o

1-n

o

1-n

o

1)-(n

1-n

o

1)-(n

21

21

1[A]

2[A]

1[A]

2[A]

21)[A]-k(n1)-(2

11)[A]-k(n

1)-(2

2

1

n

o

o

At

At

1-no

1)-(n

21

1)[A]-k(n

1)-(2t

au oo Ak'

dt

Ad ][


22/66

A = B k1, k-1AB k1B A k-1

Reversible Reactions

- usually study reactions far from equilibrium where

reverse reaction is not important

For forward rxn:

[A] = [A]o- 1/V + 2/V and

[B] = [B]o+ 1/V - 2/V = [B]o+[A]o-[A] assume [B]o= 0

For reverse rxn:

For both rxns:

][][][

1

Akdt

Bd

dt

Ad

][][][

1 Bkdt

Bd

dt

Ad

][][][][

11 BkAkdt

Bd

dt

Ad

= -k1[A] + k-1([B]o+[A]o-[A]) = -(k1+k-1)[A] + k-1[A]odt

Ad ][


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tA

oA

dtAd

01-1o1-

)[A]k(k-[A]k

tAk A

oAo

|A)k(kln)k(k

11-11

1-1

tAk

Ak

o

o)k(k 1-1

1

1-11 A)k(kln

Solve for [A]

)k(k

)k(k

)k(k

)k(k

1

1-1

1

1-1

1

1

1-1

1

1-1

o

oo

Ak

t

ekAkteAkA


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Ea= activation energy f(T) - energy needed for rct to proceed to products

A = preexponential factor or frequency factor

Effect of Temperature on Reaction Rate

- most common relationship is exponential one

- reaction rate increases by a factor of 2 or 3 for each 10orise in T

expt'ly we see: plot of ln k vs. 1/T is linear

Arrhenius eqn. - empirical relationship2

ln

RT

E

dT

kd a

integrate ARTEk a lnln RT

Ea

Aek

or

will talk more about significance of A and Ealater

A has units of k

1st order k = s-1

2nd order k = M-1s-1

The higher Ea, the stronger the T dependence of k

Ea= 0 means rate independent of T

Ea< 0 means rate as T


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Elementary Reactions

- rxns that occur in a single step - only involve 1 or 2

molecules- most rxns do not take place in a single collision but have

mechanisms that involve several elementary reactions

molecularity of rxn - # of particles involved in simple collisionalprocess for elementary rxn

- theoretical concept, whereas order is empirical

unimolecular - one that takes place w/o collision

- spontaneous disruption or transformation

- radioactive disintegration

- absorption of radiation to give energy

bimolecular - 2 molecules collide


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tktko ekek

kkAC 21 12

21

11

Can calculate time at which [B] = max by setting 0dt

Bd

time

Conc.

[A]o

[A]

[B]

[C]

eqm.


30/66

Parallel 1st Order Reactions

AU

A

VAW

AkAkkkAkAkAk

dt

Ad 321321

ktAA

o

][][ln [A] = [A]oe-kt

ktoeAkAk

dt

Ud 11

t

kt

o

U

U

dteAkUd

o 0

1 o1kt-

o1 A

k

keA

k

k-U-[U] o

k k


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products are in a constant ratio to each other

dependent on time, [A]oand k's.

okt-1 Ae-1k

kU[U] o okt-2 Ae-1

k

kV[V] o

okt-3 Ae-1k

kW[W] o If [U]o= [V]o= [W]o= 0

13

k

k

U

[W]

12

k

k

U

[V]and

time

Conc.

[A]o

[A]

[V]

[U]

eqm.

[W]


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steady state approximation - assumes that the conc. of

all rxn intermediates (B) remain constant thruout exp't.

Steady State Approximation

A = B k1, k-1B C k2

- usually intermediates are very reactive & thus are

present in small conc.

BkAk

dt

Ad11

BkBkAk

dt

Bd211

Bk

dt

Cd2


33/66

0

dt

Bd

BkBkAk

dt

Bd211

21

1

kk

AkB

2121

kk

Akk

dt

Cd

dt

Cd

kk

Akk

kk

kAk

kk

AkkAk

dt

Ad

21

21

21

11

21

111 1

t

kk

kk

oeAA 2121


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when k2


35/66

3 main rxn types:

1. initiation - unimolecular or bimolecular (can be

photoinitiated)

2. chain propagation - often bimolecular - one

intermediate and one substrate - intermediate is

product

3. termination - converts chain propagating

intermediates to stable molecules4. inhibition - rxn where product is destroyed, i.e. is rct

(intermediate is destroyed & no new one produced)

Chain Reactions

-rxn where intermediates are produced which generate

more intermediates thus propagating the rxn.


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C2H6 2CH3. k1 initiation

CH3.+ C2H6 CH4+ C2H5

.k2 initiation

C2H5.

C2H4+ H

.

k3 propagationH.+ C2H6 H2+ C2H5. k4 propagation

H.+ C2H5 C2H6 k5 termination

methane is a byproduct

Example: Ethane decomposition

C2H6C2H4+ H2


37/66

dt

HCd 62

dt

Hd

dt

CHd 3

dt

HCd 52

k3[C2H5] -k4[H][C2H6] -k5[C2H5][H] =0

[CH3] = 2k1/k22k1[C2H6] - k2[CH3][C2H6] = 0

2k1[C2H6]=2k5[C2H5][H]

-k1[C2H6] - k2[CH3][C2H6] -k4[H][C2H6] +k5[C2H5][H]

k2[CH3][C2H6]-k3[C2H5]+k4[H][C2H6]-k5[C2H5][H]=0

1.

2.

3.

1+3 k2[CH3][C2H6]-2k5[C2H5][H]=0 Plug in for [CH3]

][][ 525624

523

HCkHCk

HCkH

From (1)


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5253525624621

5

62152

][][

2

2

HCkk

HCkHCkHCk

Hk

HCkHC

rearrange k3k5[C2H5]2 - k1k5[C2H6][C2H5] - k1k4[C2H6]2 = 0

][2

][4][][62

53

2

625431

2

62516251

52 HCkkk

HCkkkkHCkkHCkkHC

53

2

5431

2

5151

2

4

kk

kkkkkkkkk

'

][][ 625624

623 kHCkkHCk

HCkkH

]Hk[Ck'k]Hk[Ck]H[Ck

]H][CHk[Ckk-]H[C

k

2kk-]H[C-k 625

625624

62623462

2

12621

62

dt

HCd

=k[C2H6] first order

dt

HCd 62

Fl E t d St d St t A i ti


39/66

Fluorescence Expt and Steady State Approximation

tris(2,2-bipyridine) ruthenium +2= R+2

R+2 + h R+2* absorption (rate: Iabs)

R+2* R+2 fluorescence

R+2* + Fe+3 R+3+ Fe+2 quenching (rate constant:kq)

03*2*2

*2

FeRkRkIdt

Rdqsabs

3

*2

FekkI

Rqs

abs

Intensity of fluorescence a[R+2*]

I

Io intensity when no quencher present ([Fe+3]=0)= intensity when quencher present

sqs

qsabs

sabso

k

Fekk

FekkI

kI

I

I ][

][/

/ 3

3

][1 3 Fe

k

k

I

I

s

qo


40/66

I

Io

sqs

qsabs

sabso

k

Fekk

FekkI

kI

I

I ][

][/

/ 3

3

intensity when no quencher present ([Fe+3

]=0)=intensity when quencher present

][1 3 Fekk

II

s

qo


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Lindemann-Hinshelwood Mechanism

1. A + A A* + A k1 activation

2. A* + A A + A k-1 deactivation

3. A* P______ k2 spontaneous decay- unimolecular

A P k overall rxn

Unimolecular Reaction

(Lindemann-Hinshelwood Mechanism)

- unimolecular rxns seen as complicated rxn involving

multistep mechanism- collisions are bimolecular, how can a rxn be first order

- from exptl data, it has been seen that rate laws are often

1storder at high concentrations, but 2ndorder at low

concentrationslook at Lindemann-HinshelwoodMechanism tp help explain this


42/66

step 1: excitation due to collision, A's translational

kinetic E or with an inert molecule N2or Ar

step 2: loss of energy due to collision of A* with A

step 3: truly unimolecular rxn.assume steady state approximation

0[A*]k-A*][k-][Ak

*21-

2

1 Adt

Ad

0kAk-

][Ak*

21-

2

1

A 21-

2

212

k][k

][Akk[A*]k

Adt

Pd

1. low P: fewer collisions, A* has time to decompose k2>>k-1[A]uB = k1[A]

2 second order

2. high P - more collisions [A] large k-1[A]>>k2first order (3rd step rate determining)

neither 1st or 2nd order2 limiting cases:

1

21

k

Akk

dt

Pdu


43/66

heterogeneous catalyst - catalyst in different phase,

i.e. solid in gaseous rxn.

Homogeneous Catalysis (same phase)

catalyst - changes rate of desired reaction- changes mechanism to one with lower Ea

enzymes - biological catalysts - read about Michaelis-

Menten rxn.

homogeneous catalyst - catalyst in same phase as rxn mixture

G Ph C l i


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Gas Phase Catalysis

NO2+ SO2 NO + SO3 bimolecular

NO2oxidizes SO2in low Eabimolecular process

we spoke of autocatalysis - product accelerates rxn.

2SO2+ O22SO3 slow

termolecular rxn needs to take place in order to occur

2NO + O22NO2 fast

termolecular rxn with NO well known

In general: uncatalyzed A + B P slow uP=k[A][B]

catalyzed A + C X fast if k2>>k1

X + B P + C fast uP=k1[A][C]

catalyzed rxns often found expt'ly to be 1st order w.r.t. [catalyst]

C = catalystk1

k2


45/66

Michaelis-Menten mechanism

-enzyme catalyzed reaction

Substrate = S products = P enzyme = E ES = intermediate

E + S = ES P + E ESkdt

Pd2

k2

k-1

k1

0][][ 211 ESkESkSEkdtESdSteady-state approximation

21

1

kk

SEkES

[E]o= [E] + [ES]

[S]o= [S] + [ES]negligible

only a little enzyme is added so [S] >> [E]


46/66

21

1

kk

SESEkES oo

ESkk

Sk

kk

SEkES ooo

21

1

21

1

21

1

21

11kk

SEk

kk

SkES ooo

o

oo

Skkk

kk

kk

SEkES

121

21

21

1

ooo

Skkk

SEk

ES121

1

ESk

dt

Pd2

o

o

oo EkSkkk

ESkk

121

21 Since [S] = [S]o~ const.

oM

o

SK

Skk

2

1

21

k

kkK

M

Michaelis const. mol/L

oMoo

SK

ESk

dt

Pd

2

Rate varies linearly with enzyme conc.


47/66

o

Ekdt

Pd][

rate of formation of product 1storder w.r.t. enzyme not so with substrate

oM

o

SK

Skk

2

max2

][ u oEk

dt

Pd0 order in S: so much substrate, rate is constant

- rate of formation of products is maximum

- maximum velocity of enzymolysis

- S forces eqm. to right so only dependent on P formation

when KM>> [S]o [S]olow reaction rate a[S]o and [E]o

M

oo

K

ESk

dt

Pd2

when [S]o>> KMk= k2


48/66

Th t i l d l f d ibi h h


49/66

2. Kinetic molecular theory

- calculate # of molecular collisions per unit volume per time and fraction of

those collisions whose energy exceeds some threshold energy

- atoms treated as rigid spheres - if collision betw. 2 spheres has enough

energy will react to form products

- does not deal with how large threshold energy must be or effectiveness of a

collision as a function of relative molecular orientation

Theoretcial models for describing how chem rxns occur

1. Activated Complex Theory or Transition State Theory

- whether or not rxn occurs determined by movement of atomicnuclei on a multidimensional potential energy surface

- valleys or holes = energy minima = stable molecular species

- heights or saddle points or cols = energy barrier to rxn

- rate = concentration of activated species x frequency with which

they pass over addle point

- need statistical mechanics to determine [activated species]


50/66

Kinetic Model of Gases

3 assumptions

1. gas consists of molecules of mass m in ceaseless

random motion

2. size of molecules negligible dmol


51/66

Look at volume swept out by sphere of radius d

where d = (dA+dB)/2

d

dA

dB

dBdB

dA= diam of molecule A

dB= diam of molecule B

assume center of one A molecule is at center of cylinder

when the center of a B molecule is within this volume, A & B collide

a collision occurs when the distance betw. centers becomes assmall as d

B molecule at rest, A molecule moves thru volume

d 2 3/


52/66

Volume swept out per unit time = relcd2 = m3/s

218

Tkc Brel

BA

BA

mm

mm

relcrelative mean speedmean speed with which one molecule

approaches another

= reduced mass

kB=R/NA= 1.38 x 10-23J/K

s = d2= collision cross-section = m2

If there are (nBNA/V) B molecules per unit volume then the # of

collisions of one A molecule with the B molecules in this V is given

by the collision frequency, z

nB= moles BNA= Avogadros #

s

molecules

Tk

Pc

V

Nncz

B

relAB

rel

ss

If th i t b t ( N /V) A l l i thi l th


53/66

If there is not one, but (nANA/V) A molecules in this volume then

the # of collisions betw. A and B molecules is given by the

collision frequency, ZAB

sm

moleculesBANTkVNn

VNncZ A

BAAABrelAB 3

222

1

8

ss

22

21

2

8

2

1AN

m

TkZ A

A

BAA

s

For a single gas molecule A = B

22

2

A

A

A m

m

m

# of collisions of A with other A molecules

in this volume. The is included so as

not to count each collision twice

3410AAZ collisoins/m3s for typical gas at std. conditions

H l t t t l ?


54/66

How can we relate to rate laws?

A + B P u= k2[A][B]

Only a fraction of collisions successful, f

- energy must exceed a minimum value E > Ea

rate of rxn. = fZAB=

dt

Ad

NBAN

Tk

f AAB

22

1

8

s

BAN

Tkf

dt

AdA

B2

1

8

s

Experimentally we found: BAkdt

Ad2

AB NTk

fk2

1

2

8

s

Ea


55/66

According to ArrheniusRT

a

Aek

2

Ea= activation energyA = preexponential factor

There exists a threshold energy below which rexn will

not proceed. For most rxns E >>kBT

The fraction of molecules with E > Ethresholdis given by

Boltzmann distribution

0

1

1

dEeTk

dEeTk

TkE

B

E

TkE

B

B

th

B

Eanyinmoleculesthatprob.

EEinmoleculesthatprob. th

E


56/66

Tk

ETkE

TkE

E

TkE

B

thB

th

B

th

B

ee

e

e

10

0

f = fraction of molecules with E > Eth=RT

NETk

EAth

B

th

ee

RTNE

AB

AB AtheN

TkkN

Tkfk

/2

1

2

21

2

88

s

s

= Ae-Ea/RT

?

AB NTk

A2

1

8

s

Ea= E

thN

A

Arrhenius assumes A not f(T) here

shows slight dependence

l Ekd


57/66

Arrhenius found 2ln

RT

E

dT

kd a If A ~ constant

A

BAth NkTRTNEk s

21

8lnln21ln

22

2

1ln

RT

E

TRT

NE

dT

kd aAth

Ea= EthNA+ RT if Ea>> RT can neglect

Then Ea= EthNA

AB NTk

A2

1

8

s

Arrhenius assumes A not f(T) here

shows slight dependence

Steric Effects


58/66

Steric Effects

- experimental values of A often much less than theoretical

- orientation of molecules may be a factor in rxn

CH3I (g) + Rb(g)

Rb CH

H

ICH

H

I Rb

(not good) (good)

Instead of suse s* reactive cross-section s* = sp

p = steric factor p


59/66

Activated Complex Theory

- look at reaction in terms of potential energy of rcts and

pdts & progress of rxn (reaction coordinate)

- formulate reaction rate in terms of properties of rcts &

activated complex for that reaction

- apply statistical mechanics to reactants & activated

complexes

Partition function for molecule:

- probability that molecule will exist in any one of its

possible energy states i aq (q is unitless)

i

TkE Bieq /

q gives indication of the average # of states that are

thermally accessible to a molecule at temperature T

C it titi f ti i t f titi f ti f


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q = qvqrqtqe

CBA

qom= standard molar partition function

DEo= difference in ground state of

products & reactants

= Eo(C) - Eo(A)Eo(B)

Can write partition function in terms of partition functions of

vibration, rotation, translation, electronic

BA

CC

eNqNq

Nq

K RTE

AmBAmA

AmC o

D

/

,,

,

//

/

D i ti f R t E ti ith A ti t d C l Th


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Derivation of Rate Equation with Activated Complex Theory

Rate of rxn. = # of activated complexes passing over top of

potential energy barrier per second

= x average frequency with which complex moves to

product side,

PCBA

Ck

bimolecular unimolecular decay

C

A+B

P

rxn coordinate

PotentialE

assume rcts in eqm withC

dtdPCk u

transition state

transition statecertain energetic configuration of maximum

potential energy


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BACC

K

RT

PC

barP 1where

P

RTBAKC BA

PRTKk

dtPd

BAk

dt

Pd2Experimentally

P

RTKkk2

k = frequency of vibration of= transmission coef. = success of vibration to

push complex thru transition state ~ 0.5 - 1

C

RTE

mBmA

mCARTE

AmBAmA

AmC oo eqq

qNe

NqNq

NqK

/

,,

,/

,,

,

//

/D

D

C Complex moves to products when a vibration


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C Co p e o es o p oduc s e a b a opushes it thru transition state

qv 1 eh / kBT 1Look at partition function of vibration

If hkBT

1 then

eh / kBT 1 h

kBT1

2

h

kBT

2

...1 h

kBT

uh

Tkq Bv Then CvC qqq

where

etrC qqqq

qC

kBT

hqC

Then write K

kBT

hK

where

RTE

mBmA

mCA oeqq

qNK

/

,,

, D

has one vibrational mode missing

and

RT k T RT


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k2kK

RT

P

kBT

h

RT

PK

KPRThTkk B2 Eyring eqn.

Thermodynamics of Activated Complex Theory

- look at Gibbs Energy, Enthalpy & Entropy of activation

D KRTG ln eqm const without vibrational mode of C#

RTGeK /D

RTGB eP

RT

h

Tkk /2

D

STHG DDD RTHRS eBek //2 DD

P

RT

h

TkB B Assume ~1where

RTTk


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RTH

RST

hPRkk B

DD 22 lnlnln

2

2 2ln

RT

H

TdT

kd D

2

2ln

RT

E

dT

kd a Arrhenius found

RTHEa 2D

RTERS aeeBek /2/

2

D

2/ eBeA RSD

RTEaAek /

2

experimentally

preexpoential factor

RTHRS eBek //2 DD P

RT

h

TkB B


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D 2ln/ln 22B

AReTRkhAPRS B

RTEH a 2D

STHG DDD

PRT

hTkB B A = preexponential factor

For bimolecular collision 2 molecules form complex 0DS

physical chemistry- kinetics

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