transition state help
TRANSCRIPT
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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
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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!
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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
\
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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
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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
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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
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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 !
!
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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)
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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
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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
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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
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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 !!$!! ´§
g ¹¹ º
¸©©ª
¨
By inserting this in the
result of constraint 1 and
assuming close lying states
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IC T
IC-13/44 Lecture-3 02-10-2003
Partition Functions
Utilizing this in constraint 2
2
2
2
2
2
2
2
2
2
2
2
0
2
2
1
2
2
2
1
22
222
P
P
T
P
T
P
P
T
I
I I
PI P
B
B
B B
B
k
ci
i
k
i
i
ii B k
hml k
h
ml k k
hml k
d icei
q
e
P T k
B B
i
!!$!!!´§
§
g
Thus
T 12 !P I.e. temperature is just a Lagrange multiplyer
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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
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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
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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!
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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
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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
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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 ???
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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
N
!
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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
N
!
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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
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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
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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:
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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
! § §
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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.
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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
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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??
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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
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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
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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?
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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
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IC T
IC-33/44 Lecture-3 02-10-2003
Transition State Theory
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
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IC T
IC-34/44 Lecture-3 02-10-2003
Transition State Theory
? A ? A ? A Rq
q
e
Rq
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
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IC T
IC-35/44 Lecture-3 02-10-2003
Transition State Theory
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:
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IC T
IC-36/44 Lecture-3 02-10-2003
Transition State Theory
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
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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:
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IC T
IC-38/44 Lecture-3 02-10-2003
Transition State Theory on Surfaces
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
N
N K
3
2#
## Y!!
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IC T
IC-39/44 Lecture-3 02-10-2003
Transition State Theory on Surfaces
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
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IC T
IC-40/44 Lecture-3 02-10-2003
Transition State Theory on Surfaces
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?
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IC T
IC-41/44 Lecture-3 02-10-2003
Transition State Theory on Surfaces
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
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!!
(
(
(
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IC T
IC-42/44 Lecture-3 02-10-2003
Transition State Theory on Surfaces
)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
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IC T
IC-43/44 Lecture-3 02-10-2003
Transition State Theory on Surfaces
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
qT S
(
(
(
!!
!!
0
0
2
2
2
#0"
2
2
#0
22
#
20
||
)(
R
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IC T
IC-44/44 Lecture-3 02-10-2003
Transition State Theory on Surfaces
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
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
qqT S
##
0 )( !
Notice that if the precursor is sufficiently loose S0(T)=1.
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IC T
IC-45/44 Lecture-3 02-10-2003
Transition State Theory on Surfaces
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 !
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IC T
IC-46/44 Lecture-3 02-10-2003
Transition State Theory on Surfaces
)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
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IC T
IC-47/44 Lecture-3 02-10-2003
Transition State Theory on Surfaces
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
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
q
h
T k
K k dt
d
UUUU
R
R
UUR UUU
|!
!!
#'
#
#
#
#
#
#
B A
AB
B A
AB
q K
### !!
UU
U
#
m AB B A
p AB AB #
Reactions between surface species:
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IC T
IC-48/44 Lecture-3 02-10-2003
Transition State Theory on Surfaces
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'#' (
(
!!!
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IC T
IC-49/44 Lecture-3 02-10-2003
Transition State Theory on Surfaces
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.
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IC T
IC-50/44 Lecture-3 02-10-2003
Transition State Theory on Surfaces
#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
hT k
M
N K k
dt
d
U
U
R UR U
U
(
!
$
!!!
#0'
#'
# ##
#
AB
T k
E
B
a
e UR
!
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IC T
IC-51/44 Lecture-3 02-10-2003
Transition State Theory on Surfaces
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?
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IC T
Transition State Theory on Surfaces
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.