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

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

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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

qq

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:

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

qq

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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