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ChE 452 Lecture 15 Transition State Theory
1
Conventional Transition State Theory (CTST)
2
Reaction Cordinate
En
erg
yReactants
Products
Barrier
A‡
Figure 7.5 Polanyi’s picture ofexcited molecules.
TST Model motion over a barrier Use stat mech to estimate key terms
Motion Over PE Surface
3
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.
1 1.5 2 2.5 3
1
1.5
2
2.5
3
C-H Distance (Angstroms)
*
H-H
DIS
TAN
CE
(A
NG
ST
RO
MS
)C-H Distance
X
X
Y
Y
transition stateEnergy
H-H
Dis
tanc
e
transition state
Approximate Derivation Of TST For A+BCAB+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
Combining
reaction rate = K0 KEQU[A][BC]
or rate constant=K0 KEQU
5
]BC][A[
]ABC[K
†
EQU =
From Statistical Mechanics
6
(7.37)
where Kequ, the equilibrium constant for the production of the hot reactive molecules is: E† is the average energy needed to traverse
the barrier, qABC
†
is the partition function for the hot molecules, qA and qBC are the partition function for A and BC, B is Boltzman’s constant and T is temperature.
+T
ABC B
E†-k T
equA BC
qK = e
q q
Bk
Combining
7
Equation (7.38) is exact but we will need an expression for K0. We can get it from collision theory.
(7.38)
here kABC 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.
+T
ABC B
E†-k T
A BC 0A BC
qk =K e
q q
Next: Estimate K0 From Collision Theory
8
First, let us define a new partition coefficient q+, by:
(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.
tBCA
ABC†
Tq
Combining Equation (7.38) And (7.39) Yields:
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(7.40)
+
B
E+ -k Tt T
A BC 0 A BCA BC
qk =(K q ) e
q q
Key Approximation
10
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 qq
q qexp - E / TA BC A
tBC
T
A BC
‡B
‡
K0
(7.41)
††
B
†E† -
k Tt TA BC 0 A BC
A BC
qk =(K q ) e
q q
Derivation Continued
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We want TST to go to collision theory when qv’s are all one. After pages of algebra we obtain:
Substituting equation (7.42) into equation (7.41) yields:
††
B
†E† -
k TB TA BC
p A BC
k T qk = e
h q q
†† B
0 A BCp
k TK q =
h
(7.42)
(7.43)
Example 7.C A True Transition State Theory Calculation
12
Use TST to calculate the rate of the reaction.
F H HF H2 (7.C.1)
Data
13
Exact Used for transition state calculations
F H2
rHF 1.34Å 1.602Å
rHH 0.801Å 0.756Å 0.7417Å
H-H stretch about 3750cm-1 4007cm-1 4395.2cm-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 19 AMU
2 AMU
I 5.48AMU-Å2 7.09AMU-Å2 0.275AMU-Å2
ge 4 4 4 1
Transition State Reactants
Solution
According to transition state theory:
14
(7.C.2)
††
2 B
2
2
†E†T -
F H k TBF H
p H F
qk Tk = e
h q q
Solution Continued
15
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.,:
kT
hl
q
q q
q
q q
q
q q
q
q qeF H
B
P
‡‡
H Ftrans
‡
H Fvibration
‡
H Frotation
‡
H Felect
E T2
2 2 2 2
T‡
B
/
(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.
Next: Substitute Expressions From Tables 6.5
According to Table 6.5:
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
1
2g B
tp
2πm k Tq =
h
Consequently
17
q
q q
m T
h
m T
h
m T
h
‡
H Ftrans
‡ B
P2
F B
P2
H B
P2
2 2
2
2 2
3 2
3 2 3 2
/
/ /
(7.C.5) where mF,
2Hm and m‡ are the masses of
fluorine, H2 and the transition state.
Performing The Algebra
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q
q q
m
m m
h
T
‡
F Htrans
‡
F H
P2
B2 2
3 2 3 2
2
/ /
(7.C.6)
Next: Calculate The Last Term In Equation (7.C.6)
Rearranging the last term shows:
Plugging in the numbers yields:
Doing the arithmetic yields:
19
h
T
300K
T
h
(300 K)P2
B
P2
B2 2
3 2 3 2 3 2
/ / /
h
T
300K
T
kg mÅ
m
AMU
kg
2 1.381 10 kg m / sec - K K
P2
B
2
-23 2 22
6 626 1010
166 10
300
3 2 3 234
210 2
27
3 2
/ /
/
. / sec.
h
T
300K
TÅ AMUP
2
B
3
21024
3 2 3 23 2
/ //.
(7.C.8)
(7.C.9)
(7.C.7)
Solution Continued Combining Equations (7.C.6) And (7.C.9) Yields:
20
2/33
2/32/3
HF
‡
transHF
‡
AMUÅ024.1K300
M
M
q
22
(7.C.10) Setting T = 300K M‡ = 21AMU, MF = 19AMU,
2HM = 2AMU yields:
q
q q
AMU
U AMUÅ AMU Å
‡
F Htrans
3 3
2
21
21024 0 42
3 23 2
//. .
(7.C.11)
23
23
Next: Calculate The Ratio Of The Rotational Partition Functions
21
The fluorine atom does not rotate so:
q
q q
q
q
‡
H Frot
‡
Hrot2 2 2
(7.C.12)
According to equation (6.5)
q8K T I
hr
B
P3
(7.C.13)
where B is Boltzmann’s constant, T is temperature, hp is Plank’s constant and I is the moment of inertia of the molecule.
Combining (7.C.12) And (7.C.13) Yields
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q
q
8 T I / h
8 T I / h
I
I
‡
Hrot
B‡
P2
B H P2
‡
H2 2 2
(7.C.14)
Substituting in the adjusted value of I‡ and IH2
from Table 7.C.1 yields:
q
q
I
I
7.011AMU Å
0.275AMU Å
‡
Hrot
‡
H
2
22 2
258.
(7.C.15)
Next: Calculate The Vibrational Partition Functions.
23
(7.C.16)
v
p B
1q =
1-exp -h υ/k T
First Get An Expression For The Term In Exponential In Equation
(7.C.16)
24
(7.C.17)
-1pp
-1B B
h 1cmh υ 300K υ=
k T T 1cm k 300K
Substitution In Values At hp And kB From The Appendix Yields
25
-1pp
-1B B
h 1cmh υ 300K υ=
k T T 1cm k 300K
p -3-1
B
h υ υ 300K=4.78×10
k T 1cm T
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:
7.C.19
7.C.18
Substituting
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Table 7.C.2 The vibrational partition function. Mode hP/BT qv qHH‡ 4395.2 cm-1 21. 1.0
(qHHH2) 4007 cm-1 19.2 1.0
qBend‡ 379.9 cm-1 1.82 1.19
The vibrational partition function ratio equals:
q
q
q q q
q
1 1.19 1.19
11.42
‡
Hvib
HH‡
Bend‡
Bend‡
H H H2 2
(7.C.20)
Next: Calculate The Ratio Of The Partition Functions For The Electronic
StateOnly consider the ground electronic
state:
27
q
q q
g
g g 1 4
‡
H Felec
e‡
e H e F2 2
4
1
(7.C.21)
Finally: Calculate kBT/hP
B
P
23 2
30 2
T
h
1.381 10 Kg M / sec - mole K K
6.626 10 Kg M / sec
I
300 K
T
300K
3006 05 1012. / sec
28
(7.C.22)
Putting This All Together, Allows One To Calculate A Pre-exponential
29
(7.C.23)
Plugging in the numbers:
k 1T
h
q
q q
q
q q
q
q q
q
q qo‡ B
P
‡
H Ftrans
‡
H Frot
‡
H Fvib
‡
H Felec2 2 2 2
k 6.65 10 / molecule sec 0.42Å 25.8 1.42 1 2.05 10 Å / molecule seco12 3 14 3 2
(7.C.24)
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:
30
k 6.65 10 / molecule sec 0.42Å 18.9 1.4 1 1.56 10 Å / molecule seco12 3 14 3 2
q
q
I
I
5.48 AMU Å
0.275 AMU Å
‡
Hrot
‡
Hrot
2
22 2
19 9.
(7.C.25)
ko becomes:
(7.C.26)
Comparison To Collision Theory
In collision theory, one considers the translations and rotations, but not the vibrations., i.e.,:
k lT
h
q
q q
q
q q0‡ B
p
‡
H Ftrans
‡
H Frot2 2
(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
32
If we assume a collision diameter of 2.3 (i.e., the sum of the Van der Wall radii) we obtain:
I r = 2.31Å
2AMU 19AMU
21 AMU9.57Å AMU‡
F H
2
FH2 2
2 2
(7.C.28)
Plugging into equation 7.C.25 using the results above:
k 6.65 10 / mole sec 0.42Å9.57Å AMU
0.275Å AMU1.9 10 Å / mole seco
12 32
2
14 3
2
(7.C.29)
oA
Å
Comparison Of Results
33
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
2.05 1014 Å3/mole-sec
ko Transition state theory with exact transition state geometry
1.65 1014 Å3/mole-sec
ko Collision theory 1.9 1014 Å3/mole-sec ko Experiment 2.3 1014 Å3/mole-sec
Summary: Transition State Theory Makes Two Corrections To Collision
Theory
34
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.
Question
What did you learn new in this lecture?
35
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