ChE 452 Lecture 22
Transition State Theory
1
Energy
Conventional Transition State Theory
(CTST)
A
‡
Barrier
Reactants
Products
Reaction Cordinate
Figure 7.5 Polanyi’s picture of
excited molecules.
TST
 Model motion over a barrier
 Use stat mech to estimate key terms
2
Motion Over PE Surface
transition
state
Energy
X
2.5
transition
state
2
*
H-H DISTANCE (ANGSTROMS)
3
C-H Dis
ta
nce
HH
X
Dis
tan
ce
Y
1.5
1
1
1.5
2
2.5
Y
3
C-H Distance (Angstroms)
Figure 7.6 A potential energy surface for the reaction H + CH3OH 
H2 + CH2OH from the calculations of Blowers and Masel. The lines in
the figure are contours of constant energy. The lines are spaced 5
kcal/mole apart.
3
Approximate Derivation Of TST
For A+BCAB+C
Assume Arrhenius’ Model
 Two populations of A-BC complexes



Cold A-BC complexes
Hot A-BC complexes that are in right
configuration and have enough energy to
react (A-BC could be far apart).
Equilibrium between the 2 populations
4
Derivation Continued

Assume reaction rate=K0 [ABC†]
Where K0 is the rate constant for reaction of the hot
molecules

From equilibrium
†
[ABC ]
K EQU =
[A][BC]

Combining
reaction rate = K0 KEQU[A][BC]
or rate constant=K0 KEQU
5
From Statistical Mechanics
K equ =
q
†
ABC
q A q BC
e
E T+
k BT
(7.37)
where Kequ, the equilibrium constant for the
production of the hot reactive molecules is:
E † is the average energy needed to traverse
†
q
the barrier, ABC is the partition function for
the hot molecules, qA and qBC are the
partition function for A and BC, kBB is
Boltzman’s constant and T is temperature.
6
Combining
k A BC =K 0
q
†
ABC
q A q BC
e
E T+
k BT
(7.38)
here kABC is the rate constant for reaction (7.38);
kB is Boltzman’s constant; T is the absolute temperature; K0 is the rate
constant; qA is the microcanonical partition function per unit volume of
the reactant A; qBC is the microcanonical partition function per unit
volume for the reactant BC;
is the average energy of the hot molecules and, is the average
partition function of the molecules which react.
Equation (7.38) is exact but we will need an expression
for K0. We can get it from collision theory.
7
Next: Estimate K0 From Collision
Theory
First, let us define a new partition coefficient q+,
by:
†
q

q T  t ABC
q A BC
(7.39)
In equation (7.39) is the partition function for
the translation of A toward BC and q+ is the
partition function for all of the other modes of
the reacting A-B-C complex.
8
Combining Equation (7.38) And
(7.39) Yields:
k A BC =(K 0 q
t
A  BC
+
T
q
)
e
q A q BC
E+
k BT
(7.40)
9
Key Approximation
+
Eyring proposed that one could replace q in
equation (7.5) with q ‡T , the partition function
+
at the transition state and E . The average
energy of the molecules with react with E‡,
the energy at the transition state. The result
is:
k
k AABCBC=(K
 q

‡
†
†
T
q
)

q q
qT
t
t
‡
K 0q A0 BC
exp - E
A BC
q A q BC A BC
(7.41)
†
†
-
E
k BT
e/ B T
10
Derivation Continued
We want TST to go to collision theory when qv’s are all
one. After pages of algebra we obtain:
K 0q
†
†
A  BC
 k BT 
=

 h 
 p 
(7.42)
Substituting equation (7.42) into equation (7.41)
yields:
†
†
T
 k BT  q
k A  BC = 
e

 h q q
 p  A BC
†
†
-
E
k BT
(7.43)
11
Example 7.C A True Transition State Theory
Calculation
Use TST to calculate the rate of the reaction.
F  H 2  HF  H
(7.C.1)
12
Data
Transition State
Reactants
Exact
Used for transition
state calculations
rHF
1.34Å
1.602Å
rHH
0.801Å
0.756Å
0.7417Å
4007cm-1
4395.2cm-1
H-H stretch about 3750cm-1
FH2 Bend
?
397.9 cm-1
FH2 Bend
?
397.9 cm-1
Curvature
barrier
?
310 cm-1
E‡
5.6 kcal/mole
1.7 kcal/mole
M
21 AMU
21 AMU
I
5.48AMU-Å2
7.09AMU-Å2
ge
4
4
F
19
AMU
H2
2 AMU
0.275AMU-Å2
4
1
13
Solution
According to transition state theory:
†
†T
F H2
 k BT  q
k F H 2 = 
e

 h q q
 p  H2 F
†
†
-
E
k BT
(7.C.2)
14
Solution Continued
It is useful to divide up the partition
functions in equation (7.C.2) into the
contributions from the translation, vibration,
rotation and electronic modes, i.e.,:
k F H 2
‡
kBBT ‡  q 


l 

hP
q
q
 H2 F 
‡
 q‡ 
 q‡ 
 q ‡ 
 ET /  B T






e
q q 
q q 
 q q 
 H 2 F  vibration  H 2 F  rotation  H 2 F elect
trans
(7.C.3)
where l‡ is an extra factor of 2 that arises
because there are two equivalent transition
states, one with the fluorine attacking one
hydrogen, and the other with one fluorine
attaching the other hydrogen.
15
Next: Substitute Expressions
From Tables 6.5
According to Table 6.5:
2πm k T 

q=
g
t
1
2
B
hp
7.C.4
where qt is the translational partition function for a
single translational mode of a molecule, m is the
mass of the molecule, kB is Boltzmann’s constant, T
is temperature, and hP is Plank’s constant. For our
particular reaction, the fluorine can translate in
three directions; the H2 can translate in three
directions; the transition state can translate in three
directions.
16
Consequently
 2 m‡ kBBT 


2
hP


3/ 2
 q‡ 



3/ 2
3/ 2
 qH qF 
kBBT   2 m H 2 kBBT 
 2  trans  2 m F 

 

2
2
hP
hP

 

(7.C.5)
where mF, m and m‡ are the masses of
fluorine, H2 and the transition state.
H2
17
Performing The Algebra
 q 
 m‡ 



 
q q 
 F H 2  trans  mF mH 2 
‡
3/ 2




 2  kBBT
2
hP
3/ 2
(7.C.6)
18
Next: Calculate The Last Term
In Equation (7.C.6)
Rearranging the last term shows:
 h 2P 


 2  kBB T
3/ 2
 300K 


 T 
3/ 2


h 2P



 2  kBB (300 K) 
3/ 2
(7.C.7)
Plugging in the numbers yields:

 10 Å  
AMU
 6.626  10  kgm / sec 
 
 m   166
.  10
 h

 300K  

 
 
 T 
 2  k T 
2 1.381  10  kgm / sec - K300 K


 34
2
P
3/ 2
2
2
10
2
3/ 2
-23
BB

2
(7.C.8)
2




 27
kg  




3/ 2
Doing the arithmetic yields:
 h 2P 


 2  kBB T
3/ 2
 300K 


 T 
3/ 2
(7.C.9)
1024
. Å 3 AMU 3/ 2
19
Solution Continued Combining Equations
(7.C.6) And (7.C.9) Yields:
 q

q q

‡
F
H2





M


  M
‡
trans
F
  300 K 
 
   

3/ 2
H2
3/ 2
1.024 Å AMU 3 2
3
3/ 2
(7.C.10)
Setting T = 300K M‡ = 21AMU, MF =
19AMU, M H 2 = 2AMU yields:
3/ 2
 q‡ 


21AMU



1.024Å 3AMU 33/22  0.42Å

q q 
 F H 2  trans  U2AMU 
(7.C.11)
20
Next: Calculate The Ratio Of The
Rotational Partition Functions
The fluorine atom does not rotate so:
 q‡ 
 q‡ 




q q 


 H 2 F2  rot  q H 2  rot
(7.C.12)
According to equation (6.5)
8πk
8K
B T I
B TI
qrr= 3 3
hP hP
(7.C.13)
(7.C.13)
where kBBis Boltzmann’s constant, T is
h p is Plank’s constant and I is
temperature, hp
the moment of inertia of the molecule.
21
Combining (7.C.12) And (7.C.13)
Yields
‡
 q‡ 
 8 k B T I‡ / h2 
I
B
P 

 

2


q 
 H 2  rot  8 kBBT I H 2 / h P  I H 2
(7.C.14)
‡
Substituting in the adjusted value of I and
I H 2 from Table 7.C.1 yields:
‡
2
 q‡ 
I
7.011AMU

Å

 

 25.8
2
 qH 
 2  rot I H 2 0.275AMU  Å
(7.C.15)
22
Next: Calculate The Vibrational Partition
Functions.
1
qv =
1-exp  -h p υ/k BT 
(7.C.16)
23
First Get An Expression For The Term
In Exponential In Equation (7.C.16)
h p υ  300K  υ  h p 1cm 
=

-1 
k BT  T  1cm  k B  300K 
-1
(7.C.17)
24
Substitution In Values At hp And kB From
The Appendix Yields
-1
h p υ  300K  υ  h p 1cm 
=


k BT  T  1cm -1  k B  300K 
7.C.18
Note that we actually used hpc/Na and kB/Na in
equation 7.C.16, and not hp where Na is
Avogadro’s number and c is the speed of light in
order to get the units right
Doing the arithmetic in equation 7.C.18 yields:
hpυ
υ  300K 
-3 
=4.78×10 

-1 
k BT
 1cm  T 
7.C.19
25
Substituting
Table 7.C.2 The vibrational partition function.
Mode
‡
HH
q
(q HH ) H 2
q ‡Bend

4395.2 cm-1
4007 cm-1
379.9 cm-1
hP/kBBT
21.
19.2
1.82
qv
1.0
1.0
1.19
The vibrational partition function ratio equals:
‡
‡
‡
 q‡ 
q
q
q

  HH Bend Bend  11.191.19  1.42
q 
1
q H  H  H 2
 H 2  vib
(7.C.20)
26
Next: Calculate The Ratio Of The Partition
Functions For The Electronic State
Only consider the ground electronic state:
‡
 q‡ 
ge
4




1
qH qF 
 2  elec g e  H 2 g e  F 1  4
(7.C.21)
27
Finally: Calculate kBT/hP
k
 BBT
hP
1.381  10 Kg  M


 6.626  10
23
/ sec - mole  K 300 K  I 
 T 
12

6
.
05

10
/
sec





 30
2




300K
300
K
Kg  M / sec
2
(7.C.22)
28
Putting This All Together, Allows One To
Calculate A Pre-exponential
‡ 
‡ 
‡ 
‡ 




q
q
q
q
‡  kBBT  


 
 

ko  1 

 qH qF   qH qF   qH qF 
 h P   q H 2 q F 
 2  rot  2  vib  2  elec
trans
(7.C.23)
Plugging in the numbers:
k o  26.65  1012 / molecule  sec0.42Å 3 25.81.421  2.05  1014 Å 3 / molecule  sec
(7.C.24)
29
Note: Calculation Used A Fitted
Geometry
If one uses the actual transition state geometry,
the only thing that changes significantly is the
rotational term. One obtains:




 q‡ 
 I‡ 
5.48 AMU  Å 2

 
 
 19.9
2
q 
I 
 H 2  rot  H 2  rot 0.275 AMU  Å
(7.C.25)
ko becomes:
k o  26.65  10 12 / molecule  sec0.42Å 3 18.91.41  1.56  1014 Å 3 / molecule  sec
(7.C.26)
30
Comparison To Collision Theory
In collision theory, one considers the
translations and rotations, but not the
vibrations., i.e.,:
‡


q


kBB T 
‡ 


k 0  l 
 h p   q H 2 q F 
 q‡ 


q q 
 H 2 F  rot
trans
(7.C.27)
in equation (7.C.26), the rotational partition
function should be calculated at the collision
diameter and not the transition state
geometry.
31
Collision Theory Continued
If we assume a collision diameter of 2.3 Å

(i.e., the sum of the Van der Wall radii) we
obtain:

I ‡  rF H 2

2
2  2AMU 19AMU  
2
o
 FH 2 = 2.31Å  

9.57Å
AMU

A


21 AMU

(7.C.28)
Plugging into equation 7.C.25 using the
results above:
 9.57Å 2 AMU 
14
3
k o  26.65  10 / mole  sec0.42Å 
  1.9  10 Å / mole  sec
2
 0.275Å AMU 
12
3
(7.C.29)
32
Comparison Of Results
Table 7.C.3 A comparison of the preexponential calculated
by transition state theory and collision theory to the
experimental value.
ko Transition state theory with
adjusted transition state geometry
ko Transition state theory with exact
transition state geometry
ko Collision theory
ko Experiment
2.05 1014 Å3/mole-sec
1.65 1014 Å3/mole-sec
1.9 1014 Å3/mole-sec
2.3 1014 Å3/mole-sec
33
Summary: Transition State Theory Makes
Two Corrections To Collision Theory
1.Transition state theory uses the transition
state diameter rather than the collision
diameter in the calculation.
2.Transition state theory multiplies by two
extra terms: the ratio of the vibrational
partition function, and the electronic
partition function for the transition state,
and the reactants.
34
Question

What did you learn new in this lecture?
35