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IC

T

IC-1/44 Lecture-3 02-10-2003

Reaction Rate Theory

(E

reaction coordinate

E

+

+k

A B AB



IC

T

IC-2/44 Lecture-3 02-10-2003

Svante Arrhenius1859 - 1927

Nobel Prize 1903

k = v e -Eact /RT

Eact

reaction parameter

E

The Arrhenius Equation

+

+k

A B AB

r = = k [A][B]d[AB]

dt

Empirical!



IC

T

IC-3/44 Lecture-3 02-10-2003

Transition State Theory

To determine the rate we must know the concentration on top

of the barrier.

The Chemical Equilibrium is given by the chemical potential

of the reactant and the product. That we know how to calculate.

The relative concentration between a reactant and product in a

Chemical reaction is given by the Chemical Equilibrium

DC B A DC

k

k

B A YYYY p

n

0!!! §i

ii D D B B A Ad

dG QY QY QY QY QY

\



IC

T

IC-4/44 Lecture-3 02-10-2003

The Chemical Equilibrium

The chemical potential for the reactant and the product can be

determined if we know their Partition Functions Q.

)(ln

)!

(ln)(ln

i

i

i

i

N

i

i

ii

N

q

kT N

N

q

kT N

Q

kT

i

}¹¹¹¹¹

º

¸

©©©©©

ª

¨

!¹¹ º

¸

©©ª

¨

! H

H

H

H Q

Here Qi is the partition function for the gas i and qi the

partition function for the gas molecule i

Let us assume that we know qi then

¡

¢

C i

B

B

A

A

D

D

C

C

i i

i qqqqqYYYYY

¹¹ º

¸©©ª

¨¹¹

º

¸©©ª

¨¹¹

º

¸©©ª

¨¹¹

º

¸©©ª

¨¹¹

º

¸©©ª

¨ 1



IC

T

IC-5/44 Lecture-3 02-10-2003

The Chemical Equilibrium

If we assume an ideal gaskT

V p N ii !

and normalize the pressure with p0=1 bar

B A

D£

B A D£

B A

D£

p

p

p

p

p

p

p

p

p

kT

V

q

V

q

V q

V q

T K

B A

D

B A

D

YY

YY

YYYY

YY

YY

¹¹ º

¸©©ª

¨¹¹ º

¸©©ª

¨

¹¹ º

¸©©ª

¨¹¹ º

¸©©ª

¨

!¹¹ º

¸©©ª

¨

¹ º

¸©ª

¨¹ º

¸©ª

¨

¹ º ¸©ª̈¹ º ¸©ª̈!

00

00

0

)(

We obtain the important result that the Equilibrium Constant K(T)

is given by the Partitions Functions of the reactants and products

Thus we can determine the concentration of a product on top of the

barrier if we know the relevant Partion Functions



IC

T

IC-6/44 Lecture-3 02-10-2003

Partition Functions

Obviously are Partition Functions relevant. We shall here dealwith the Canonical Partition Function in which N, V, and T are

fixed.

Remember, that although we talk of a partition function for an

individual molecule we always should keep in mind that this only

applicable for a large ensample of molecules, i.e. statistics

§g

!

|0

/

i

T k

i Bie g q

I

§g

!

!

0

/

/

i

T k

T k

i

Bi

Bi

e

e P

I

I

Consider a system with i energy levels with energy ei and

degeneration gi

Where Pi is the probability for finding the system in state i



IC T

IC-7/44 Lecture-3 02-10-2003

Ludwig Boltzmann(1844-1906)

S = k ln (W)

P e

e

i

RT

RT

i

i

i

!

!

g

§

I

I

/

/

0Boltzmann Statistics:

The high temperature/diluted limit of

R eal statistical thermodynamics

There is some really interesting Physics here!!

)ln(W k S Bt ot !

!

i

i

W !

!



IC T

IC-8/44 Lecture-3 02-10-2003

Partition Functions

Why does the partition function look like this?Lets see if we can rationalize the expression:

Let us consider a system of N particles, which can be distributed

on i states with each the energy ei and Ni particles. It is assumed

the system is very dilute. I.e. many more available states than particles.

§§§ !!!!i

i

i

i

i

ii

i N

N N N

N

N 1;;

N E P N E t ot

i

iii

i

it ot !!! §§ I I I ;

Constraint 1

Constraint 2

Requirement: The Entropy should be maximized (Ludwig Boltzmann)



IC T

IC-9/44 Lecture-3 02-10-2003

Partition Functions

Problem: Optimize the entropy and fulfill the two constraints at thesame time. USE LAGRANGE UNDERTERMINED MULTIPLIERS

)(ln i

i

i Bt ot k

N

S S §!! Where we have utilized

N N N N } )(ln)!(lnStirling approximation:

Only valid for huge N

)()1()()( _

21 I I PP !! §§ i

i

i

i

iii S S P f

0

)()(

21

!

¹ º

¸©ª

¨

!§§

i

i

i

i

i

ii

i

i

P

P P P S

P

P f

H

I H

H

H

0 0



IC T

IC-10/44 Lecture-3 02-10-2003

Partition Functions

Result: The Entropy Maximized when

i B B k k

i ee P I

PP21 1 ¹¹

º

¸©©ª

¨

!

§§§¹¹

º

¸©©ª

¨¹¹

º

¸©©ª

¨

!!!i

k k

i

k k

i

i

i B B

i B B eeee P

I PP

I PP 2121 11

1

If we now utilize the first constraints: §§§ !!!!i

i

i

i

i

ii

i N N P N N

N N P 1;;

§¹¹

º

¸©©ª

¨

!i

k k i

B B eeI 21 1

Which reminds us of q the partition function



IC T

IC-11/44 Lecture-3 02-10-2003

Partition Functions

The second constraint: N

E P N E t ot

i

iii

i

it ot !!! §§ I I I ;

We have to relate the average energy to some thermodynamical data

2T k B!I

Now if we wants to perform the

sum above we need to have an

analytical expression for the

energy in state i

Le g t of ox

Am

l i t

e

article i a ox

0

0 L

i=2, I! 2

/8mL2

i=1, I! 2

/8mL2

i=3, I! 2

/8mL2

]ibsi (iTx/L)This can be found by considering

a particle confined in a box



IC T

IC-12/44 Lecture-3 02-10-2003

Partition Functions

m H

2

22

!J! H Ö H Ö

H

iii H m

H =!=

! I Ö,2

Ö22

J

Le g t of ox

Am

l i

t e

article i a ox

0

0 L

i=2, I!h2

/8mL2

i=1, I!h2

/8mL2

i=3, I!h2

/8mL2

]ibsi (iTx/L)

ciml

hi

i

2

2

22

8 !!I

2

2

2

20

1 2

2

12

221

P

T

P

TP

I PP

h

ml k

c

k d ieeeq B Bk

ci

i

k k B

i B B !!$!! ´§

g ¹¹ º

¸©©ª

¨

By inserting this in the

result of constraint 1 and

assuming close lying states



IC T

IC-13/44 Lecture-3 02-10-2003

Partition Functions

Utilizing this in constraint 2

2

2

2

2

2

2

2

2

2

2

2

0

2

2

1

2

2

2

1

22

222

P

P

T

P

T

P

P

T

I

I I

PI P

B

B

B B

B

k

ci

i

k

i

i

ii B k

hml k

h

ml k k

hml k

d icei

q

e

P T k

B B

i

!!$!!!´§

§

g

Thus

T 12 !P I.e. temperature is just a Lagrange multiplyer



IC T

IC-14/44 Lecture-3 02-10-2003

Partition Functions

Since constraint 1 gave §¹¹

º ¸©©

ª¨ ¤

!i

k k i

B B eeI PP 21 1

T

12 !P

Since constraint 2 gave

and the entropy is max for i B B k k

i ee P I

PP21 1 ¹¹

º

¸©©ª

¨

!

q

e

e

e P

T k

i

T k

T k

i

B

i

B

i

B

i I

I

I

!!

§

Thus the form of the partition

function comes as a result of

ma imizing the entropy with

2 constraints



IC T

IC-15/44 Lecture-3 02-10-2003

Translational Partition Functions

As we have assumed the system to be a particle capable of movingin one dimension we have determined the one-dimensional

partition function for translational motion in a box of length l

hmT k l q B

tran s T2!

Now what happens when we have several degrees of freedom?

If the different degrees of freedom are independent the Hamiltoniancan be written as a sum of Hamiltonians for each degree of freedom

H t ot = H 1+ H 2+«.

Discuss the validity of this: When does this not work? Give examples



IC T

IC-16/44 Lecture-3 02-10-2003

Translational Partition Functions

If the hamiltonian can be written as a sum the different coordinates

are indrependant and

cba

l

T k

k

T k

i

T k

ik l

T k

ik l

T k

H H H

j

T k

E

j

T k

H

qqqeeee

eeeq

B

l

B

k

B

i

B

c

l b

k a

i

B

cba

B

j

B

!!!

!!!

§§§§§§§

FI I I I I I )(

)(

Thus for translational motion in 3. Dimensions.

qtrans3D = qtrans

x qtransy qtrans

z=3

2/33 )2(

h

T k mV q B D

tr an s

T!



IC T

IC-17/44 Lecture-3 02-10-2003

Partition Functions

It is now possible to understand we the Maxwell-Boltzmandistribution comes from

d i

h

mT k l

ed i

q

edp p f P

B

T k T k

x x

i

i

B

i

B

i

´´´§g

g

g

g

!!!!00 2

)(1T

I I

x

B

T k m

p

B

T k m

pi

x x dpmT k

ed i p

mT k

edp p f

B

x

B

´´´g

g

g

g

g

!!TT 2

22

)(2

0

2

222

x

B

T mk

p

x x dpT mk

edp p f

B

x

T2)(

2

2

!

)()()(),,( z y x z y x p f p f p f p p p f !

T k mv

B

Bev

T k

mv f

2/2

2/3

2

2

4)(

¹¹

º

¸©©

ª

¨!

T

T



IC T

IC-18/44 Lecture-3 02-10-2003

Maxwell-Boltzmann

distribution of velocities

Average:

500 ± 1500 m/s at 300 K

T k mv

B

BevT k

mv f 2/2

2/32

24)(

¹¹ º ¸

©©ª¨!

TT

2/12/1

8;

8¹¹ º

¸©©ª

¨!¹¹

º

¸©©ª

¨!

QTTT k

um

T k v B B



IC T

IC-19/44 Lecture-3 02-10-2003

Partition Functions

Similarly can we separate the internal motions of a molecule inPart involving vibrations, rotation and nuclei motion, and

electronic motion i.e. for a molecule we have

nucl elecvibr ot tran s qqqqqq !

Now we create a system of many molecules N that are in principle

independent and as they are indistinguishable we get an overall

partition function Q

! N

q N

!

What if they were distinguishable ???



IC T

IC-20/44 Lecture-3 02-10-2003

Partition FunctionsWhat was the advantage of having the Partition Function?

T V

B N

QT k

,

)ln(¹ º

¸©ª

¨!

H H Q

V N

BT

QT k E

,

2 )ln(¹

º

¸©

ª

¨!H

H

T N

B

V

QT k p

,

)ln(¹

º

¸©

ª

¨!

H

H

N V B QT k T

S )(lnH H !

B A

D¥

B A D¥

B A

D¥

p

p

p

p

p p

p p

p

kT

V

q

V

q

V q

V q

T K

B A

D

B A

D

YY

YY

YYYY

YY

YY

¹¹ º

¸©©ª

¨¹¹

º

¸©©ª

¨

¹¹ º ¸©©

ª¨¹¹

º ¸©©

ª¨

!¹¹ º

¸©©ª

¨

¹ º

¸©ª

¨¹

º

¸©ª

¨

¹ º ¸©

ª¨¹

º ¸©

ª¨

!

00

00

0

)(

! N

qQ

N

!



IC T

IC-21/44 Lecture-3 02-10-2003

Partition Functions

Similarly can we separate the internal motions of a molecule inPart involving vibrations, rotation and nuclei motion, and

electronic motion i.e. for a mulecule we have

nucl elecvibr ot tran s qqqqqq !

Now we create a system of many molecules N that are in principle

independent and as they are indistinguishable we get an overall

partition function Q

! N

qQ

N

!



IC T

IC-22/44 Lecture-3 02-10-2003

The Vibrational Partition Function

R I hii )( 2

1

!Consider a harmonic potentialR I R I hhi ii 2

1'!!

T k h

T k h

i

T k hi

vib B

B

B

e

eeq

/

/

0

/)(

1

21

21

R

R R

g

!

!! §

T k hi

T k hi

vib B

B

eeq

/0

/

1

1R

R

g

!

!! §

T k h for h

T k q B B

classvib } R R

',

If there are several normal modes:

!

iT k h

T k h

vib Bi

Bi

e

eq

/

/

1

21

R

R



IC T

IC-23/44 Lecture-3 02-10-2003

The Rotational (Nuclear) Partition Function

I

h j j j 2

2

8

)1(

TI

!

2r I Q!21

111

mm!

Q

B

B

T k I

h j j

T k I

h j j

j

r ot

k I

hT for

h

T k I

e j jd

e jq

B

B

2

2

2

2

8

)1(

0

8

)1(

0

8

81

))1((1

)12(1

2

2

2

2

T

T

W

W

W

T

T

"!

}

!

g

g

!

´

§

Notice: Is not valid for H

2 WHY

? TR H2=85K, TRCO=3K



IC T

IC-24/44 Lecture-3 02-10-2003

The Rotational (Nuclear) Partition Function

The Symmetri factor:

This has strong impact

on the rotational energy

levels. Results in fx

Ortho- and para-hydrogen

Molecular symmetry W Types of molecules

C1, Ci, and

Cs

1 CO, CHFClBr, meso-

tartraric acid, and CH3OH

C2, C

2v, and

C2h

2 H2, H

2O

2, H

2O, and trans-

dichloroethylene

C3v and C3h 3 NH3, and planar B(OH)3

C B A B

r ot I I I h

T k q T

T

W

2/3

2

281¹¹

º

¸©©

ª

¨!

For a non-linear molecule:



IC T

IC-25/44 Lecture-3 02-10-2003

Effect of bosons and fermions

If two fermions (half intergral spin) are interchanges the total wavefunction must be anti symmetric i.e. change sign.

Consider Hydrogen each nuclei spin is I=1/2

From two spin particles we can form 2 nuclear wave function:

and which are (I+1)(2I+1)=3 and I(I+1)=1degenerate respectively

s ymmetric

nuclear = Antis ymmetricnuclear =

Since the rotation wave function has the symmetry

is it easily seen that if the nuclear function is even must j be odd and

visa versa

( 1) J J

Rot = w

2 2

2 2

( 1) ( 1)

8 8

, (2 1) (2 1) ( 1)(2 1) (2 1) B B

j j h j j h

I k T I k T

r ot nucl

J even J odd

q I I J e I I J eT T

g g

! § §2 2

2 2

( 1) ( 1)

8 8

, (2 1) 3 (2 1)

B B

j j h j j h

I k T I k T

r ot nucl J even J odd q

J

eJ

e

T T

g g

! § §



IC T

IC-26/44 Lecture-3 02-10-2003

Ortho and Para Hydrogen

This means that our hydrogen comes in two forms: OrthoH

ydrogenWhich has odd J and Para Hydrogen which has even J incl. 0

Notice there is 3 times as much Ortho than Para, but Para has the

lowest energy a low temperature.

If liquidH

ydrogen should ever be a fuel we shall see advertisements

Absolute Ortho free Hydrogen for longer mileages

Hydroprod Inc.



IC T

IC-27/44 Lecture-3 02-10-2003

Liquid Hydrogen

This has severe consequences for manufacturing Liq H2 !!

The ortho-para exchange is slow but will eventually happen so if we

have made liq. hydrogen without this exchange being in equilibrium

we have build a heating source into our liq. H2

as ¾ of the H2

will

End in J=1 instead of 0.

2 2

1 2 2

( 1)2 2 85 170 1, 4

8 8 j B

j j h hk R k J

I I I

T T!

! ! ! } !

2

252,87

1

30,9 / 1,06 /

4

T

H va p J H k J mol Inter nal energ y k J mol I !

!! ! !o

i.e. 11% loss due to the internal conversion of Ortho into Para hydrogen



IC T

IC-28/44 Lecture-3 02-10-2003

Dista ce (ar . its)

0 2 4 6 8

P o t e t i al E

e r g y (

0

1

2

-De

-D0

hR /2

2XX2

The Electronic Partition Function

...)(

10

10 ! T k

e

T k

eel B B eeT q

I I

[[

200

YI

h D D e !!

Does usually not contribute

exceptions are NO and fx. H

atoms which will be twice

degenerate due to spin

What about He, Ne, Ar etc??



IC T

IC-29/44 Lecture-3 02-10-2003

translation vibration rotation

qm k T

htr an s

B/l

( )/

21 2T

qe

vib h k T B!

1

1R /

qIk T

hr ot

B!8

2

2

T

Huge for any

reasonable si e of lH2: 1.8*1010m-1 at 500 K

CO: 6.8*1010m-1 at 500 K

Cl2: 1.1*1011m-1 at 500 K

usually equals 1

unless vibrations ave

very low frequency

H2: 1.000 at 500 K

CO: 1.002 at 500 K

Cl2: 1.250 at 500 K

large:H2: 2.9 at 500 K

CO: 180 at 500 K

Cl2: 710 at 500 K

artition f unctions of a iatomic molecule (per egree of freedom)

Partition Functions Summary

W



IC T

IC-30/44 Lecture-3 02-10-2003

Partition Functions Example

Knowing the degrees of internal coordinates and their energydistribution calculate the amount of molecules dissociated into

atoms a different temperatures.

T(K) K H2(T) p

H/p0 K N2(T) p N/p0 K O2(T) pO/p0

298 5.81*10-72 2.41*10-36 6.35*10-

160

2.52*10-80 6.13*10-81 7.83*10-41

1000 5.24*10-18 2.29 *10-9 2.55*10-43 5.05*10-22 4.12*10-19 6.42*10-10

2000 3.13*10-6 1.76*10-3 2.23*10-18 1.80*10-9 1.22*10-5 3.49*10-3

3000 1.77*10-3 1.72*10-1 1.01*10-9 3.18*10-5 5.04*10-1 5.01*10-1

We see why we cannot make ammonia in the gas phase but O radicals

may make NO at elevated temperatures



IC T

IC-31/44 Lecture-3 02-10-2003

Surface Collisions

( t

T mk

p

m

T k

T k

pv

T k

p

dveT k

mvT k

p

dvv f vdvvV v f t A

r

B

B

B

x

B

x

T k

mv

B

x

B

x x x x x x sur f coll

B

x

TT

T

V V

22

2

)()()(1

2

0

00

2

!!!

!

!(

!

g

gg

´

´´

V tAv x !(

Consider a box with volume V

T k

p

B

! V

What are

the

numbers?



IC T

IC-32/44 Lecture-3 02-10-2003

Surface Collisions

Density

E n e r g y ( k J / m ol )

0

10

20

30

40

50

Reaction Coordinate

(E

(H

vx > vmin

vx < vmin

T=300K

T=600K

T=1000K

T k E

T k u

B

u

T k u

B

u

B

B

B

e

dueuT k

dueu

T k

duuu f

duuu f

P

/

0

2/3

2/3

2/3

2/3

0

2

min

2

min

24

2

4

)(

)(

(

g

g

g

g

!

¹¹ º

¸©©ª

¨

¹¹

º

¸©©

ª

¨

!

!

´

´

´

´

Q

Q

T

QT

T

QT

How many are successful in reacting?Simple Maxwell-Boltzman distribution



IC T

IC-33/44 Lecture-3 02-10-2003


Consider the following reaction:

? A ? A ? A ? A Rqq R K R

t d P d

#

##

d

!!! R R R

How?

We assume that R and R # are in Equilibrium

P R R ppn #

(E

q

q#

R

P

R #

T V

B

N

QT k

,

)ln(¹

º

¸©

ª

¨!

H

H Q

#

R R

Q Q !? A

? A

##

#

q

q

R

R K

d!!

q`#

R is a frequency or trial factor in the

reaction coordinate



IC T

IC-34/44 Lecture-3 02-10-2003


? A ? A ? A Rq

q

e

Rq

qq

t d

P d

T k

h

B

##

#

1

1R R R R

!d!

By splitting the partition function in the transition state

? A ? A ? A Rk R

q

q

h

T k

t d

P d T S T

B !!#

q

q

h

T k k B

TST

#

|

T k h for h

T k

q B

B

}d R R R

Assuming YhT k B "" xe xs}

s 1



IC T

IC-35/44 Lecture-3 02-10-2003


T k E T k E BT S T

B B eeqq

hT k k //

#

0 (( !! R

P R R ppn #

(E

q

q#

R

P

R #

q`#=q`#vq

#0e

-(E/kT

q`#

Which basically is the Arrhenius form

If q0# ~ q R ~1x1013s-1

Relation to Thermodynamics

# K h

T k k B

T S T !RT H RS B RT G B

T S T eeh

T k e

h

T k k

/// #

0#

0#

0 ((( !!

The partition function q# can conveniently be split further:



IC T

IC-36/44 Lecture-3 02-10-2003


r ti r i t

R

P

R#

r ti r i t

R

P

R#

Loose TST:

q# >> q

Tight TST:

q# << q

1013 < R < 10

17s

-110

9 < R < 1013

s-1

( (

(S# g tiv (S# sitiv

r ti r i t

R

P

R#

r ti r i t

R

P

R#

Loose TST:

q# >> q

Tight TST:

q# << q

1013 < R < 10

17s

-110

9 < R < 1013

s-1

( (

(S# g tiv (S# sitiv

Think of

some e amples

Temperature

dependence of

prefactor



IC T

IC-37/44 Lecture-3 02-10-2003

Transition State Theory on Surfaces

An atom adsorbs into a 2-dim mobile state, we have Ng gas atoms,

M sites on the surface, and N# atoms in the transition state

mobile A A #m

p mobil mobie A A # A M N !0

g

A

N K N dt

d N ##*

R R !!# Q Q ! g M

N A A

!U

ATST A

B

g A A pk pT k

V

M

K

M

N K

M d t

d N

d t

d |!!!

##

R R U

Indirect adsorption of atoms:



IC T

IC-38/44 Lecture-3 02-10-2003


Now what is K # ?

N QT k B H H Q )ln(!

# Q Q ! g

!;!#

#

#

#

N

q

Q N

q

Q

N

g

N

g

g a s

g

!!

D

tr an s

D

tr an s g qqqqq 2#

#

3 ; Y!!

D

tran s

D

tran s

g q

qq

N

N K

3

2#

## Y!!



IC T

IC-39/44 Lecture-3 02-10-2003


T k m N T k m M

A

T k V

hT k mV

hT k m Ah M

T k

T k

V

q

q

M

q

T k

V

M

K k

B B

B B

B B

B

D

tran s

D

tran s

B

T S T

T T

T

T

R R Y

2

1

2

/2

/2

0

32/3

2

3

2##

!!

$

!!

ATST A

B

g A A pk pT k

V M

K M

N K

Md t d N

d t d |!!!

##

R R U

T k m N

p

d t

d

B

A A

T

U

20

* !

T k m

pr

d t

d N

Ad t

d

A

M

d t

d N

B

A sur f acecoll

A A A

T

UU

2

1

.

***0

!

!!!

This corres ponds t o the

collision on a sur f ace since

the at oms are still f ree t omove in two dimen sion s



IC T

IC-40/44 Lecture-3 02-10-2003


Direct adsorption of atoms:

immobile A A #

p immobileimmobile A A #

20

1

a A

M N !!

#)()!(!

! #

##

# N q

N M N

M Q

d

d!

# Q Q ! g

g g q

q N M

N

N K

#

#

## )( d!!

g q

qM K

#

#*

#)( UU !

M is total number

of sites

M´ is number of

free sites

)1()( *#* AUUUU !$Why?



IC T

IC-41/44 Lecture-3 02-10-2003


T mk N

T S

T mk N

T S

eT mk

hh

T k

T mk N

N

eT mk

hh

T k

T mk

eT mk V

hh

T k

h

V

q

q

h

V

T k

V

q

qq

T k

V

q

q

M

N M

T k

V

M

K k

B

A

B

T k

h E

B

D

B

B

T k

h E

B

D

B

B

T k

h E

B

D

B

D

tran s

vib D

B

D

tran s

vib D

B g B

T S T

B

vib D

B

vib D

B

vib D

T

U

T

U

T

R

T

U

T

R

T

U

T

R U

U

UUR

R R

R

R

R

Y

2

)()1(

2

)(

)2(2

)2(2

)2(

)(

)(

0

0

0

0*

)(2

2

2

0

*0

)(2

2

2*

)(

2/3

3

2

2*

3

#

2*

3

#

2

#

#*

##

#

2

2

2

!|

¹¹ º

¸©©ª

¨

!

¹¹ º

¸©©ª

¨

!

¹¹ º ¸©©

ª¨

$

$

!

d!!

(

(

(



IC T

IC-42/44 Lecture-3 02-10-2003


)1(2

)(0

0 A

B

A AT S T

A

T mk N pT S pk

dt d U

T U !!

D

unit cell tr an s

vib D

D

tr an s

vib D

D

tr an s

vib D

q

q

M

q

q

q

qM T S

2

#

2

2

#

2

2

#

20 )(

!!!

2

2

2

22 )2()2(

h

T mk a

h

T mk

M

A

M

qq B B

D

tr an s D

unit cell tr an s

TT!!!

09.010*0.4

)2()(

3

2

2

)(

2

2

2

2

2

#

20

_ 2

2

!¹¹ º

¸

©©ª

¨

!

¹¹ º ¸©©

ª¨

!!

(

T k

h

D

B

T k

h E

B

D

B

D

unit cell tr an s

vib D

B

vib D

B

vib D

eh

T k

eT mk a

hh

T k

q

qT S

R

R

Y

T

Y Notice adsorptionalways result in loss of

entropy

There may also be steric

hindrance leading to

reduced S



IC T

IC-43/44 Lecture-3 02-10-2003


What happens in the regime between direct and indirect adsorption?

))2

cos()2

cos(2(2

1),( 0

a

y

a

xV y xV

T T !

The atoms breaks free of the site and start to diffuse around in

2

22

2

)2(2

h

T mk ae

h

T k BT k

h

D

B B

vib D TY

R

p¹¹ º

¸©©ª

¨

Eventually

T k

E

T k

h E

D

unit cell tran s

vib D

T k

E

D

unit cell tran s

vib D D

unit cell tran s

vib D

B

act

B

B

eS e

q

q

eq

qq

qT S

(

(

(

!!

!!

0

0

2

2

2

#0"

2

2

#0

22

#

20

||

)(

R



IC T

IC-44/44 Lecture-3 02-10-2003


Indirect adsorption of molecules:

AT S T A

B

g A A pk pT k

V

M

K

M

N K

dt M

d N

dt

d |!!!

##

R R U

gas g q

q

N

N

K

#

#

!!

g a s

vib

g a s

r ot

vibr ot

B

g a s

vib

g a s

r ot

D

tr an s

vibr ot

D

tr an s

B

TST

qq

qq

T mk N

qqq

qqq

hM

V

TM k

VK k

##

0

3

##2#

2

1

T

R

!

!!

)(2

0

0

T S T mk N

p pk

dt

d

B

A AT S T

A

T

U!!

gas

vib

gas

r ot

vibr ot

qq

qqT S

##

0 )( !

Notice that if the precursor is sufficiently loose S0(T)=1.



IC T

IC-45/44 Lecture-3 02-10-2003


Direct adsorption of molecules:

)1()(2

10

0

3

#'

*

#

A

B

g a s

vib

g a s

r ot

D

tr an s B

TST

T S T mk N

qqq

q

h

V

T k

V

M

K k

UT

UR

!

!!

#)()!(!

! #

##

# N q

N M N

M Q

d

d!

# Q Q ! g

g g q

q N M

N

N K

#

#

## )( d!! g q

qM K

#

#*

#)( UU !



IC T

IC-46/44 Lecture-3 02-10-2003


)1()(20

0UT

U!! T S T mk N

p pk dt

d

B

A AT S T

A

gas

vib

gas

r ot

D

unit cell tran s

gas

vib

gas

r ot

D

tran s qqq

q

qqq

MqT S

2

#'

2

#'

0 )(

!!

T k

E

tran s f rusr ot f rusvib

T k

E

B B eqqqeqq

(

(

!! 0#0#0#0#'#'

-vibration rustrated

otation x

rustrated

otation y

rustrated

Translation y

rustrated

Translation x

eaction

coordinate-vibration rustrated

otation x

rustrated

otation y

rustrated

Translation y

rustrated

Translation x

eaction

coordinate



IC T

IC-47/44 Lecture-3 02-10-2003


T k

E

T k

E

B A

AB B

B A

AB B B

AB B A

B

AB B A

ek eqq

q

h

T k

qq

q

h

T k k

##

##

0

#0'#' (

(

!!!

B A B A

B A

AB

AB

B AB

B A AB B AT S T AB

k

qq

q

h

T k

K k dt

d

UUUU

R

R

UUR UUU

|!

!!

#'

#

#

#

#

#

#

B A

AB

B A

AB

qq

q K

### !!

UU

U

#

m AB B A

p AB AB #

Reactions between surface species:



IC T

IC-48/44 Lecture-3 02-10-2003


m # AB AB

The reverse process:

p B A AB #

*UUU

AB AB k

dt

d

!

T k

E

T k

E

AB

AB B

AB

AB B B

AB AB

B

AB AB

ek eq

q

h

T k

q

q

h

T k k

##

##

0

#0'#' (

(

!!!



IC T

IC-49/44 Lecture-3 02-10-2003


0*

#

!! UUUUU

AB B A AB k k

dt

d

Considering both processes and equilibrium:

T k

E E

B A

ABeq

b

AB B A AB AB

eqq

qk k K

## ((

!|

Reaction Coordinate

A*+B*

AB#**

AB*+*

(EA+B-AB#

(H

(EAB-AB#

T k

H

k

S

eqb B ee K

(

(

!

Notice how the K eq is alone determined

from initial and final state partition functions.



IC T

IC-50/44 Lecture-3 02-10-2003


#m AB AB

Desorption:

p AB AB #

AB

AB

AB

AB

AB AB

AB AB

AB q

q K

q

q

M

N #

#

###

# !!!!U

UUU

AB

T k

E

AB

AB B

AB

AB

AB B

AB AB AB AB ABT S T

AB

Be

q

q

h

T k

qq

hT k

M

N K k

dt

d

U

U

R UR U

U

(

!

$

!!!

#0'

#'

# ##

#

AB

T k

E

B

a

e UR

!



IC T

IC-51/44 Lecture-3 02-10-2003


ABT k

E

AB B

a

ed t

d UR U

!

AB

AB B Ba

q

q

h

T k eT k E E

#0'

; !(! R

Adsorbed Transition Desorbed Preexponential

state state state f actor

b 1015

s-1

b 1013

s-1

b 1014-16

s-1

b 1013 s-1

mobile

immobile

immobile

mobile

System Prefactor s-1 Ea

kJ/mol

CO/Co(0001) 1015 118

CO/Ni(111) 1015 130

CO/Ni(111) 1017 155

CO/Ni(111) 1015 126

CO/Ni(100) 1014 130

CO/Cu(100) 10

14

67CO/Ru (001) 1016 160

CO/Rh(111) 1014 134

How?



IC T


If the details of the transition state can be determined can the rate

over the barrier be calculated.

Details of the transition state are difficult to access:Low concentration

Short lifetime.

Often determined by ̀ `First Principle´´ calculations, but are onlyaccurate to say 0.1 eV or 10 kJ/mol.

transition state help

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