ChE 452 Lecture 26

Transition State Theory Revisited

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Last Time We Discussed Advanced Collision Theory:

Method  Use molecular dynamics to simulate the collisions  Integrate using statistical mechanics r A  BC    r A  BC (v A  BC , E BC A  BC E BC A  BC , R BC A  , v BC BC , R  dv , v A BC  ) BC  dE BC (8.20) A  BC d R BC dv BC 2

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?

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Next: Review Arrhenius’ Model vs Transition State Theory

Arrhenius’ model:  1.

Consider two populations of molecules: 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

k A  BC =   k T B h P   q + q q A BC  B  Where k A  BC is the rate constant for reaction (9.1), is Boltzmann’s constant; T is the absolute temperature; h P is Plank’s constant; q A is the microcanonical partition function per unit volume of the reactant A; q BC is the microcanonical partition function per unit volume of the reactant A; q BC 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.

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A Comparison of Tolman’s Equation and Transition State Theory

Equation Tolman’s k A  BC =   h P     q + BC     E + Definitions   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 k A  BC =   k T B h P     q ‡ T q q A BC     E ‡ k T B   E ‡ = Saddle Point energy.

q ‡ = 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.

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Approximate Derivation Of TST:

Let’s go back to the statistical mechanics definition of the partition functions.

q H o t   N P N H o t g N ex p     E N   E     V (9.40) q T S T   N P N T S T g N e x p   V E  E N  ‡  (9.41) Need to assume P hot = P TST (Distribution of hot molecules poised to react equals the distribution of states at the transition state).

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r r AB reactants

Assumptions

Ignores:  Re-crossing trajectories  Changes in q’s  Dynamics 9 kcal/mole r r BC Figure 9.7 A recrossing trajectory.

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Let Me Examine The Assumption That q’s Are Equal

R y R x Figure 9.3 The minimum energy pathway for motion over the barrier as determined by the trajectory calculations in Chapter 8.

q

Hot 

q q q

X (9.38) Assume! Vibration q = translation q 9

Additional Assumptions In Conventional State Theory

All the P Hot N in equation (9.40) are 0 for states with energies below the barrier and unity for states with energy above the barrier so that P N Hot  TST P N .

Motion along the R x direction is pure translation.

Motion perpendicular to the R x direction is pure vibration.

The P Hot N are determined only by the properties of the transition state; and independent of shape at barrier.

HNC  HCN (9.46) 10

Example HNC



HCN

H q HNC  3 q q 2 r q ‡ T / \ sym C  N (9.47) q asym   q BendA   q BendB  (9.48) q ‡ sym q ‡ asym q ‡ BendB (9.50) TST assume that the bending mode is pure translation, the other modes are pure vibration.

Cancellation of error.

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Successes Of Transition State Theory

    Transition magnitude.

state theory generally gives preexponentials of the correct order of 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.

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Example Of Isotope Effects Table 9.5

The relative rates of a series of reactions at 300K.

Reaction D+H H+H D+D H+D 2 2 2 2     DH+H H D 2 2 +H +D HD+D Experimental Relative Rate 6 4 2 1 CTST Relative Rate 12.3 6.7 1.7 1 14

Isotope Effects Continued Table 9.6

The various contributions to the relative rate of the reactions in Table 9.5at 300K.

Reaction 1732 924  Zero Point Energy kJ/Mole 4.79 D + H 2  DH + H H + H 2  H 2 + H D + D 2  D 2 + D H + D 2  HD + D Symmetric Stretching Frequency, cm -1 2012 1423 1730 Bending Frequen cy cm -1 965 683 737 6.96 7.60 10.16 Relative Rate 300K 12 6.7 1.7 1 15

Limitations Of TST

5 2 1 0.5

0.2

0.1

0.05

0.01 0.02 0.05 0.1 0.2

0.5 1 Transition State Theory 2 5 10 Correction to collision theory.

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Tail of Wavefunction Extends through barrier

TST ignores Tunneling

B Center of Wavefunction before barrier A Wavefunction Barrier

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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Tunneling Data For H+H 2



H 2 +H 10000

450 K 400K 350 K 300 K

1000 2 2.2

2.4

2.6

2.8

1000/T

3 3.2

Figure 9.13

An Arrhenius plot for the reaction H+H 2  H 2 +H.

3.4

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 

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 Zero Point Of Transition state  Zero Point Energy Zero Point Of Reactants Distance Over the Potential Energy Surface

Figure 9.4

The activation barriers for the reactions in Table 9.6.

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Example: X+H+CH 3



HCH 3 +X

Vibrational Frequency

20

3000

10

Zero point energy Barrier 2000

0

1000

-10

Sum The Potential

-20

1.5

2.0

2.5

3.0

3.5

C-H Distance, Angstroms 4.0

1.0

2.0

3.0

C-H Distance, Angstroms

4.0

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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Tunneling Corrections:

Tail of Wavefunction Extends through barrier B Center of Wavefunction before barrier A Wavefunction Barrier

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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Tunneling through a barrier.

Hydrogen Wavefunction Barrier

Figure 9.8

Tunneling through a barrier.

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Derive: An Equation For Tunneling

Solve Shroedinger equation for Motion near a barrier.

V B 0 -d e x A 0

Figure 9.11

A plot of the square-well barrier.

  ħ h 2 d 2 2m dX i 2 A +V B   ψ =Eψ i i (9.73) 24

Solution:

Scattered Wavefunction Incident Wavefunction  K i 2 K i  p i  exp    2 d e 2m (V B ħ 2  E i )  (9.81) Potential Barrier Total Wavefunction

X A Figure 9.12

The real part of the incident (  i ) scattered (  r ) and total wavefunction (  T ) for the square-well barrier.

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Eckart Barriers Better Assumption

k A  BC =   k T B h P  

k T

   q ‡ T q q A BC     exp E ‡ k T B   (9.63)  V 0 k(T)=1+ 1   h ν P B i (9.89) 1+ k T B ‡ E Distance

Figure 9.14

The Eckart potential.

Significant improvement over TST.

V 1 26

Example 9.A Tunneling Corrections Using The Eckart Barrier

In problem 7.C we calculated the preexponential for the reaction: F+H 2  HF+H (9.A.1) How much will the pre-exponential change at 300 K if we consider tunneling?

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Solution

From Equation 9.69 k A  BC        h p  q ‡ (

9.A.2)

‡ /      We already evaluated the term in brackets in Equation (9.A.1) in Example 7.C. Substituting that result: k       14 Å 3 / molecule-sec 

(9.A.3)

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Next Evaluate К(T)

According to Equation (9.101)



1 24

  

h

 k T

T

   2  

1

  k T

T E

‡   (9.A.4)

Where according to Table 7.C.1 =310 cm

-1

.

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Solution Continued

Equation (7.C.17) says: h p   k T B T i    4 .7 8 4  1 0 -3 c m    3 0 0 K T  

(9.A.5)

Substituting

 i

= 310 cm

-1

at T=300K into Equation (9.A.5) yields

h    310 cm -1  4784 10  3 cm    300 K 300K   

.

(9.A.6)

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(9.A.5)

Substituting Into Equation (9.A.4) Yields

   1 24   2  148 1   .

00198

(9.A.7)

  300 K   Therefore the rate will only go up by 10% at 300K.

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At 100K

h

   

310 cm

-1  (9.A.8)  3

cm

  

300K 100K

   Substituting into Equation (9.A.9)   

1 24

  2  

1



.

00198 100

   (9.A.9) 32

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:

F+DH  FD+H

(9.A.10)

the curvature drops to 215 cm -1 . In that case h   T   215 cm -1   3 cm    300K 100K   

(9.A.11)

 1 24  h p  i  k T T  2   1   k T T E ‡   

(9.A.12)

so the tunneling correction for deuterium is half that of hydrogen. 34

Semiclassical Approximation For Tunneling

 exp   2  S S o 2 (9.96)  B  ħ   E i  dS  Marcus-Coltrin Path Minimum Energy path

Figure 9.15

Some of the paths people assume to calculate tunneling rates.

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Exact Calculations

reactants r r AB 9 kcal/mole r r BC

Figure 9.18

products.

Some trajectories from the reactants to 36

Conclusions:

Conclusions: Key assumptions in TST 1) no recrossing trajectories 2) q

hot

=q

tst

3) no tunneling OK to factor of 20 for bimolecular reactions Hard to do better

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Question

 What did you learn new in this lecture?

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