che 452 lecture 24 reactions as collisions 1. according to collision theory 2 (equation 7.10)
TRANSCRIPT
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ChE 452 Lecture 24 Reactions As Collisions
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According To Collision Theory
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(Equation 7.10)
r Z PA BC ABC reaction
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Today Advanced Collision Theory
Method Simulate the collisions Integrate using statistical mechanics
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BCBCBCABCBCABCBCABCBCA
BCBCABCBCABCABCA
dvddb ddEdvv,,b,,EvD
)v,,b,,E,(vrr
RR
R
(Equation 8.20)
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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)
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Yields Very Imposing Equations
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d(area)PVCC=rdreactionBCABCABCA
If we can calculate reaction probabilities we can calculate rates.
BCBCBCABCBCABCBCABCBCA
BCBCABCBCAreaction2
dvddb ddEdvv,,b,,EvD
)v,,b,,E,(vPk
RR
R
BCBCBCABCBCABCBCABCBCA
BCBCABCBCAreactionBABCA
dvddb ddEdvv,,b,,EvD
)v,,b,,E,(vPCCr
RR
R
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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
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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
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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
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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
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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
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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
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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.
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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
Trans E=9.7 kcal/molVib E= 4.8 kcal/mol
Trans E=7.25 kcal/molVib E= 7.25 kcal/mol
Trans E=4.8 kcal/molVib E= 9.7 kcal/mol
Trans E=2.4 kcal/molVib E= 12.1 kcal/mol
Collisions with 14.5 kcal/mole
Need to turn at the right time.
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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].
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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.
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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.
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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.
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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.
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Also Need Correct Distribution Between Translation And Vibration:
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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
Trans E=9.7 kcal/molVib E= 4.8 kcal/mol
Trans E=7.25 kcal/molVib E= 7.25 kcal/mol
Trans E=4.8 kcal/molVib E= 9.7 kcal/mol
Trans E=2.4 kcal/molVib E= 12.1 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
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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.
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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].
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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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