che 452 lecture 15 transition state theory 1. conventional transition state theory (ctst) 2 tst ...
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ChE 452 Lecture 15 Transition State Theory
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Conventional Transition State Theory (CTST)
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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
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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
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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
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(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
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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
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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
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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
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Use TST to calculate the rate of the reaction.
F H HF H2 (7.C.1)
Data
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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:
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(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
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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.
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1
2g B
tp
2πm k Tq =
h
Consequently
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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:
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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:
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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)
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Next: Calculate The Ratio Of The Rotational Partition Functions
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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.
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(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)
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(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
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-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:
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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
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(7.C.22)
Putting This All Together, Allows One To Calculate A Pre-exponential
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(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:
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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.
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Collision Theory Continued
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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
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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
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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?
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