ChE 551 Lecture 19 Transition State Theory Revisited 1.
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Transcript of ChE 551 Lecture 19 Transition State Theory Revisited 1.
ChE 551 Lecture 19 Transition State Theory Revisited
1
Last Time We Discussed Advanced Collision Theory:
Method Use molecular dynamics to
simulate the collisions Integrate using statistical
mechanics
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r r (v , E , , b , , v )
D v E , , b , , v dv dE d db d dvA BC A BC A BC BC A BC BC
A BC BC A BC BC BC A BC BC A BC BC BC
R
R R
(8.20)
Summary:
Key finding – we need energy and momentum to be correct to get reaction to happen.
Many molecules which are hot enough do not make it.
Today how does that affect transition state theory?
3
Next: Review Arrhenius’ Model vs Transition State Theory
Arrhenius’ model: Consider two populations of molecules:
1. Hot molecules in the right configuration to react (i.e. molecules moving toward each other with the right velocity, impact parameter etc to react).
2. Molecules not hot enough or not moving together in the right configuration.
Assume concentration of hot reactive molecules in equilibrium with reactants.
Assume reaction occurs whenever hot molecules collide in right configuration to react.
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Can Derive Tolman’s Equation
5
Where kABC is the rate constant for reaction (9.1), is Boltzmann’s constant; T is the absolute temperature; hP is Plank’s constant; qA is the microcanonical partition function per unit volume of the reactant A; qBC 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; E+ is the average energy of the molecules which react and, q+ is the average partition function of the molecules which react, divided by the partition function for the translation of A toward BC.
Tolman’s Equation is almost exact.
+
+BA BC B
P A BC
k T qk = exp -E /k T
h q q
A Comparison of Tolman’s Equation and Transition State Theory
Equation
Definitions
Tolman’s
E+=Average energy of the hot molecules before they react.q+=partition function for the hot molecules before they react. The partition function includes all of the normal modes except the mode bringing the species together.
TST E‡ = Saddle Point energy.
= partition function for molecules at the
saddle point in the potential energy
surface. The partition function includes all
of the normal modes of the AB-C complex
except the normal mode carrying the
species over the barrier.
6
+ +B
A BCP A BC B
k T q Ek = exp -
h q q k T
‡ ‡B T
A BCP A BC B
k T q Ek = exp -
h q q k T
‡Tq
Approximate Derivation Of TST:
Let’s go back to the statistical mechanics definition of the partition functions.
7
q
P gE E
T
VH ot
NH ot
NN
BN
exp
(9.40)
q
P gE E
T
VT S T
NT S T
NN
BN
e x p
‡
(9.41)
Need to assume Phot = PTST (Distribution of hot molecules poised to react equals the distribution of states at the transition state).
Assumptions
8
rBC
0.5
24 kcal/mole
Å
15 kcal/mole
rBC
prod
ucts
0.5
24 kcal/mole
Å
15 kcal/mole
9 kcal/mole
reactants
rABr
Ignores: Re-crossing trajectories Changes in q’s Dynamics
Figure 9.7 A recrossing trajectory.
Let Me Examine The Assumption That q’s Are Equal
9
Rx
Ry
Figure 9.3 The minimum energy pathway for motion over the barrier as determined by the trajectory calculations in Chapter 8.
(9.38)
Assume! Vibration q = translation q
q q q qHotX Y everything else _
Additional Assumptions In Conventional State Theory
All the in equation (9.40) are 0 for states with energies below the barrier and unity for states with energy above the barrier so that .
Motion along the Rx direction is pure translation.
Motion perpendicular to the Rx direction is pure vibration.
The are determined only by the properties of the transition state; and independent of shape at barrier.
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PNHot
P PNHot
NTST
PNHot
HNC HCN(9.46)
Example HNCHCN
11
H
/ \
C N
(9.47)
q q q q q q qHNC t3
r2
sym asym BendA BendB
(9.48)
q = q q q q qT‡
t‡ 3
r‡ 2
sym‡
asym‡
BendB‡
(9.50)
TST assume that the bending mode is pure translation, the other modes are pure vibration.
Cancellation of error.
Successes Of Transition State Theory
Transition state theory generally gives preexponentials of the correct order of magnitude.
Transition state theory is able to relate barriers to the saddle point energy in the potential energy surface;
Transition state theory is able to consider isotope effects;
Transition state theory is able to make useful prediction in parallel reactions like reactions (7.27) and (7.29).
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Key Prediction
Experimentally, if one lowers the energy of the transition state, one usually lowers the activation energy for the reaction.
13
Example Of Isotope Effects
Table 9.5 The relative rates of a series of reactions at 300K.
Reaction Experimental Relative Rate
CTST Relative Rate
D+H2 DH+H 6 12.3
H+H2 H2+H 4 6.7
D+D2 D2+D 2 1.7
H+D2 HD+D 1 1
14
Isotope Effects Continued
15
Table 9.6 The various contributions to the relative rate of the reactions in Table 9.5at 300K.
Reaction Symmetric Stretching Frequency,
cm-1
Bending Frequen
cy cm-1
Zero Point
Energy kJ/Mole
Relative Rate 300K
D + H2 DH + H
1732 924 4.79 12
H + H2 H2
+ H 2012 965 6.96 6.7
D + D2 D2 + D
1423 683 7.60 1.7
H + D2 HD + D
1730 737 10.16 1
Limitations Of TST
0.05
0.1
0.2
0.5
1
2
5
0.01 0.02 0.05 0.1 0.2 0.5 1 2 5 10
Exp
erim
ent
Transition State Theory
16
Correction to collision theory.
TST ignores Tunneling
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Barrier
B A
Tail of WavefunctionExtends through barrier
Center of Wavefunctionbefore barrier
Wavefunction
Figure 9.9 A diagram showing the extent of the wavefunction for a molecule. In A the molecule is by itself. In B the molecule is near a barrier. Notice that the wavefunction has a finite size (i.e. there is some uncertainty in the position of the molecule.) As a result, when a molecule approaches a barrier, there a component of the molecule on the other side of the barrier.
Tunneling Data For H+H2H2+H
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Rat
e
2 2.2 2.4 2.6 2.8 3 3.2 3.4
1000
10000
1000/T
300 K350 K400K450 K
Figure 9.13 An Arrhenius plot for the reaction H+H2
H2+H.
Modern Versions Of Transition State Theory
Variational transition state theory. Tunneling corrections.
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Variational Transition State Theory
Key idea: TST is saddle point in the free energy plot, not the energy plot
20
Distance Over the Potential Energy Surface
En
erg
y
Zero Point Of Reactants
Zero Point Energy
Zero Point OfTransition state
Figure 9.4 The activation barriers for the reactions in Table 9.6.
Example: X+H+CH3HCH3+X
1000
3000
2000
2.0 2.5 3.0 3.5 4.01.5
C-H Distance, Angstroms
Ene
rgy,
kca
l/mol
e
-1
Vibrational Frequency
1.0 2.0 3.0 4.0-20
-10
0
10
20
The Potential
Zero point energy
Barrier
C-H Distance, Angstroms
Vib
ratio
nal F
requ
enci
es, c
m
Sum
Figure 9.17 The vibrational frequency, zero point energy potential energy and the sum of poential and zero point energy for reaction (9.51) as a function of the C-H bond length. After Hase [1997].
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Little improvement over TST except for methane.
Tunneling Corrections:
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Barrier
B A
Tail of WavefunctionExtends through barrier
Center of Wavefunctionbefore barrier
Wavefunction
Figure 9.9 A diagram showing the extent of the wavefunction for a molecule. In A the molecule is by itself. In B the molecule is near a barrier. Notice that the wavefunction has a finite size (i.e. there is some uncertainty in the position of the molecule.) As a result, when a molecule approaches a barrier, there a component of the molecule on the other side of the barrier.
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Barrier
HydrogenWavefunction
Figure 9.8 Tunneling through a barrier.
Tunneling through a barrier.
Derive: An Equation For Tunneling
24
xA0-de
Pot
enti
al E
nerg
y
0
VB
Figure 9.11 A plot of the square-well barrier.
(9.73)
Solve Shroedinger equation for Motion near a barrier.
2 2
B i i2i A
h d+V ψ =Eψ
2m dX
ħ
Solution:
25
P(-d ) =2
pd
2m (V Ee
i
i ie
i B i2
KK
exp
)2
(9.81)
PotentialBarrier
TotalWavefunction
IncidentWavefunction
ScatteredWavefunction
X A
Figure 9.12 The real part of the incident (i)
scattered (r) and total wavefunction (T) for the
square-well barrier.
ħ2
Eckart Barriers Better Assumption
26
V1
DistanceE
nerg
y
V0
Figure 9.14 The Eckart potential.
Significant improvement over TST.
‡ ‡B T
A BCP A BC B
k T q Ek = exp -
h q q k TTk
2
P i B‡
B
h ν k T1k(T)=1+ 1+
24 k T E
(9.63)
(9.89)
Example 9.A Tunneling Corrections Using The Eckart Barrier
In problem 7.C we calculated the preexponential for the reaction:
F+H2 HF+H
(9.A.1) How much will the pre-exponential
change at 300 K if we consider tunneling?
27
Solution
28
From Equation 9.69
k TT
h
q
q qexp - E TA BC
B
p
‡
A BC
‡B
/
(9.A.2)
We already evaluated the term in brackets in Equation (9.A.1) in Example 7.C. Substituting that result:
k T Å molecule-secA BC3
2 1014 / (9.A.3)
Bk TBk T
Next Evaluate К(T)
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According to Equation (9.101)
Where according to Table 7.C.1 =310 cm-1.
Th
Tp i
B
B‡
11
241
2
T
E
(9.A.4)
Bk T
Solution Continued
Equation (7.C.17) says:
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h
T4.784 10 cm
300K
Tp i
B
-3
(9.A.5)
Substituting i = 310 cm-1 at T=300K into Equation (9.A.5) yields
h
Tcm cm
K
300Kp i
B
-1
310 4784 10
3001483. . .
(9.A.6)
(9.A.5)
Substituting Into Equation (9.A.4) Yields
Therefore the rate will only go up by 10% at 300K.
31
T
1
1
24148 1
00198 300
561102.
. /
. /.
kcal mole K K
kcal mole
(9.A.7)
At 100K
Substituting into Equation (9.A.9)
32
hcm cm
300K
100Kp i -1
310 4784 10 4453. .
T
1
1
24445 1
00198 100
561852.
.
..
(9.A.8)
(9.A.9)
Note
The tunneling correction in this example is smaller than normal, because the barrier has such a small curvature. A typical number would be 1000 cm-1. Still it illustrates the point that tunneling becomes more important as the temperature drops.
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If We Replace The Hydrogen With A Deuterium:
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F+DH FD+H
(9.A.10)
the curvature drops to 215 cm-1. In that case
h
Tcm cm
300K
100Kp i
B
-1
215 4784 10 3083. .
(9.A.11)
Th
T
T
E
p i
11
241 1 41
2
B
B‡ .
(9.A.12)
so the tunneling correction for deuterium is half that of hydrogen.
Bk T
Semiclassical Approximation For Tunneling
35
P Sm V E
dSi B iSS
o
exp 2
2
(9.96)
Marcus-Coltrin Path
MinimumEnergy path
Figure 9.15 Some of the paths people assume to calculate tunneling rates.
ħ2
Exact Calculations
36
rBC
0.5
24 kcal/mole
Å
15 kcal/mole
rBC
pro
du
cts
0.5
24 kcal/mole
Å
15 kcal/mole
9 kcal/mole
reactants
rAB
r
Figure 9.18 Some trajectories from the reactants to products.
Conclusions:
37
Conclusions: Key assumptions in TST1) no recrossing trajectories2) qhot=qtst
3) no tunneling OK to factor of 20 for bimolecular reactionsHard to do better
Question
What did you learn new in this lecture?
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