ChE 452 Lecture 24 Reactions As Collisions 1. According To Collision Theory 2 (Equation 7.10)

ChE 452 Lecture 24 Reactions As Collisions

1

According To Collision Theory

2

(Equation 7.10)

r Z PA BC ABC reaction

Today Advanced Collision Theory

Method Simulate the collisions Integrate using statistical mechanics

3

BCBCBCABCBCABCBCABCBCA

BCBCABCBCABCABCA

dvddb ddEdvv,,b,,EvD

)v,,b,,E,(vrr

RR

R

(Equation 8.20)

Define Our Variables

Figure 8.2 A typical trajectory for the collision of an A atom with a BC molecule as calculated by the methods in section 8.3.2

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b

A

B

C

A BC

d(area)

Yields Very Imposing Equations

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d(area)PVCC=rdreactionBCABCABCA

If we can calculate reaction probabilities we can calculate rates.


BCBCABCBCAreaction2


)v,,b,,E,(vPk

RR

R


BCBCABCBCAreactionBABCA


)v,,b,,E,(vPCCr

RR

R

Molecular Dynamics As A Way Of Calculating Reaction Probabilities

Idea:Treat atoms as billiard balls moving in the

force field created by all of the other atoms.

Assume molecules start moving toward each other.

Solve Newton’s equation of motion (F=ma) to calculate the motion of the atoms during the collision.

See if reaction occurs

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Picture Of The Collisions

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A

A

B B

C

C

AA

B B

CC

Next Set Up The Background To Do The Simulation

What do we need to know to do the simulations?

We need to know intermolecular forces/PE surface already have from lecture 19

We need to have a way to integrate equation of motion (already have one from chapter 4)

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PE Surface

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1 4 7 10 13 16 19 22 25 28 31 34 37

S1

S4

S7

S10

S13

S16

S19

S22

S25

S28

S31

0.0

2.0

4.0

6.0

8.0

10.0

12.0

14.0

16.0

18.0

20.0

SaddlePoint

MD For Motion

IdeaNumerically integrate Newton’s

equations of motion for motion of all atoms.

Newton’s EquationForces from PE surface

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F =m a

F = E

Computer Program

React MD available from course web site.

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Ways To Visualize Motion

As a trajectory in space where all the atoms move.

As a trajectory on the potential energy surface in Figure 8.9 where the bond lengths evolve.

As a plot of the motion of the atoms versus time.

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Position Of Atoms During Collisions

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Time

B

A

C

Pos

ition

Time

B

A

C

Pos

ition

Non Reactive

Trajectory

Reactive Trajectory

Figure 8.10 A series of trajectories during the reaction A+BCAB+C with Ma=Mc=1, Mb=19 and various initial

reactant configurations

Motion On PE Surface

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RBC

RAB

Figure 8.11 A series of typical trajectories for motion over a potential energy contour.

Figure 8.11 A series of typical trajectories for motion over a potential energy contour.

Non Reactive Trajectory

Reactive Trajector

y

Example For Today

Consider exchange reactions

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A BC AB C (8.45)

D +CH CH DH CH CH3 3 2 3 (8.46)

F CH Cl FCH Cl3 3 (8.47)

HO CH CH CH HOCH CH CH3 2 3 3 2 3

(8.48)

rAB

reactants

rBC

prod

ucts

Transition State13.86 Kcal/mole

0.5

24 kcal/mole

Å

15 kcal/mole

9 kcal/mole

Figure 8.12 An idealized potential energy surface for the reaction A + B AB + C

If the System Has Enough Energy To Make It Over The Saddle Point, Reaction Can

Occur!

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E= 13.6 E= 13.7 E= 13.8

E= 13.9 E= 14.0

Figure 8.16 A series of trajectories calculated by fixing the total energy of the reactants and then optimizing all of the other parameters. The barrier is 13.88 kcal/mole for this example.

Not Every Collision With Enough Energy Makes It Through

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Trans E=14.5 kcal/molVib E= 0.01 kcal/mol

Trans E=12. kcal/molVib E= 2.5 kcal/mol





Collisions with 14.5 kcal/mole

Need to turn at the right time.

Generally Excess Energy Is Needed

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Opening ForLow Energy

Opening ForHigher Energy

Figure 8.17 A blow up of the tope of the barrier.

0 10 20 30 40 50 60 70 800

0.2

0.4

0.6

0.8

1

1.2

1.4

Cro

ss-S

ectio

n, Å

Collision Energy, kcal/mole

2

Figure 8.18 The cross section for reaction H+H2H2+H. Adapted from Tsukiyama et. al.

[1988] and Levine and Bernstein[1987].

Get Higher Rate If C Is Moving Away When A Hits

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Time

B

A

C

Pos

ition

Time

B

A

C

Time

B

A

C

Pos

ition

Time

B

A

C

Time

B

A

C

Pos

ition

Time

B

A

C

(a) (b)

(d)(c)

(e) (f)

repulsion

Pos

ition

Pos

ition

Pos

ition

Figure 8.19 A series of cases calculated by fixing free energy at 18 kcal/mole, fixing the vibrational energy at 6 kcal/mole and varying whether A hits when C is vibrating in toward B or out away from B.

Corresponds To Turning At The Right Time On PE Surface

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(d)

(c)

(f)

(e)

(b)

(a)

rAB

rBC

rAB

rBC

rBC

Figure 8.20 A replot of the results in Figure 8.19 on a potential energy surface.

What Do We Need To Get Reaction – Linear Case?

Need enough energy to get over the barrier

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E= 13.6 E= 13.7 E= 13.8

E= 13.9 E= 14.0

Figure 8.16 A series of trajectories calculated by fixing the total energy of the reactants and then optimizing all of the other parameters. The barrier is 13.88 kcal/mole for this example.

Need Coordinated Motion Of The Atoms:

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(d)

(c)

(f)

(e)

(b)

(a)

rAB

rBC

rAB

rBC

rBC

Figure 8.20 A replot of the results in Figure 8.19 on a potential energy surface.

Also Need Correct Distribution Between Translation And Vibration:

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Trans E=12. kcal/molVib E= 2.5 kcal/mol





Figure 8.21 A series of cases where the molecules have enough energy to get over the barrier, but they do not make it, unless partitioning the energy between translation and rotation is correct.

Again turning analogy important

Polanyi Rules:

Can use vibration/translation to probe structure of transition state:

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Early Middle Late

R AB

R BC R BC R BC

Figure 8.24 Potential energy surfaces with early, middle and late transition states.

Leads To Excess Energy In Product:

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0 10 20 30 400

0.2

0.4

0.6

0.8

1

Energy, kcal/mole

Pro

ba

bili

ty

Boltzman Distribution

Experiment

n=0

n=1

n=2

n=3

n=4

n=5

Figure 8.25 The distribution of vibrational energy produced during the reaction F+H2HF+H. Results of Polanyi and Woodall [1972].

Summary Of Linear Collisions

At this point, it is useful to summarize what we have learned about linear A + BC collisions.

First, we found that to a reasonable approximation one

can treat the collision of two molecules as a collision between two classical particles following Newton’s equations of motion.

The reactants have to have enough total energy to get over the transition state (or Col) in the potential energy surface.

It is not good enough for the molecules to just have enough energy. Rather, the energy needs to be correctly distributed between vibration and transition.

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Summary Continued

Coordinated motions of the atoms are needed. In particular, it helps to have C moving away from B when A collides with BC.

We also find that we need to localize energy and momentum into the B-C bond for reaction to happen.

The detailed shape of the potential energy surface has a large influence on the rate.

These effects are a factor of ~10~100 in rate – ignored in TST

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Conclusion

Can use molecular dynamics to calculate rate of reaction

Solve Newton's equations of motion Integrate to get rate Reaction probabilities are subtle:

vary with many factors. Transition state theory ignores

dynamics.

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Question

What did you learn new in this lecture

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ChE 452 Lecture 24 Reactions As Collisions 1. According To Collision Theory 2 (Equation 7.10)

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