Transcript che 551 lectures - Classnotes For Professor Masel's Classes

ChE 551 Lecture 17
Reactions As Collisions
1
According To Collision Theory
rA  BC  ZABC Preaction
(Equation 7.10)
2
Today Advanced Collision Theory
Method
 Simulate the collisions
 Integrate using statistical mechanics
rABC   rABC (vABC , EBC , , bABC , R, vBC ) 
DvABCEBC , , bABC , R, vBC dv ABCdE BCd dbABCdR BCdvBC
(Equation 8.20)
3
Define Our Variables
d(area)
A
bA

BC
B

C
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
4
Yields Very Imposing Equations
dr
ABC
=C C V
A
BC
ABC
P
d(area)
reaction
rABC  CACB  Preaction(vABC , EBC , , bABC , R, vBC ) 
DvABCEBC , , bABC , R, vBC dv ABCdE BCd dbABCdR BCdvBC
k 2   Preaction(vABC , EBC , , bABC , R, vBC ) 
DvABCEBC , , bABC , R, vBC dv ABCdE BCd dbABCdR BCdvBC
If we can calculate reaction probabilities we can
calculate rates.
5
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
6
Picture Of The Collisions
A
A
B
B
C
C
A
A
B
B
C
C
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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)
8
Saddle
Point
37
34
31
28
25
22
19
16
13
10
7
4
1
20.0
18.0
16.0
14.0
12.0
10.0
8.0
6.0
4.0
2.0
0.0
PE Surface
S31
S28
S25
S22
S19
S16
S13
S10
S7
S4
S1
9
MD For Motion
Idea
Numerically integrate Newton’s equations of
motion for motion of all atoms.
Newton’s Equation
Forces from PE surface


F =m a

F =E
10
Computer Program
React MD available from course web site.
11
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
C
Non Reactive
Trajectory
B
A
Time
C
Reactive
Trajectory
B
A
Time
Figure 8.10 A series of trajectories during the reaction A+BCAB+C
with Ma=Mc=1, Mb=19 and various initial reactant configurations
13
Motion On PE Surface
Non Reactive
Trajectory
RAB
Reactive
Trajectory
RBC
Figure 8.11 A series of typical trajectories for
motion over a potential energy contour.
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Example For Today
Consider exchange reactions
reactants
A  BC  AB  C
0.5 Å
(8.45)
(8.46)
r
AB
Transition State
13.86 Kcal/mole
F   CH 3Cl  FCH 3  Cl 
24 kcal/mole
15 kcal/mole
(8.47)
HO   CH 3CH 2 CH 3  HOCH 3  CH 2 CH 3
(8.48)
9 kcal/mole
r
products
D  +CH 3CH 3  DH  CH 2 CH 3
BC
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!
E= 13.6
E= 13.7
E= 13.9
E= 13.8
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.
16
Not Every Collision With Enough Energy
Makes It Through
Collisions with 14.5 kcal/mole
Trans E=14.5 kcal/mol
Vib E= 0.01 kcal/mol
Trans E=12. kcal/mol
Vib E= 2.5 kcal/mol
Trans E=9.7 kcal/mol
Vib E= 4.8 kcal/mol
Trans E=7.25 kcal/mol
Vib E= 7.25 kcal/mol
Trans E=4.8 kcal/mol
Vib E= 9.7 kcal/mol
Trans E=2.4 kcal/mol
Vib E= 12.1 kcal/mol
Need to turn at the right time.
17
Generally Excess Energy Is
Needed
.
4
1
.
2
0
.
8
0
.
6
0
.
4
0
.
2
Opening For
Higher Energy
Opening For
Low Energy
Cross-Section, Å
2
1
1
0
Figure 8.17 A blow up of the
tope of the barrier.
0
1
02
03
04
05
06
Collision Energy, kcal/mole
07
08
0
Figure 8.18 The cross section for reaction
H+H2H2+H. Adapted from Tsukiyama et. al.
[1988] and Levine and Bernstein[1987].
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(a)
C
B
repulsion
A
Position
Position
Get Higher Rate If C Is Moving Away When
A Hits
(b)
C
B
A
Time
(c)
C
B
A
Position
Position
Time
B
A
Time
(e)
B
A
Time
Position
Position
Time
C
(d)
C
(f)
C
B
A
Time
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
(a)
(c)
(e)
(b)
(d)
(f)
rAB
rAB
rBC
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
E= 13.6
E= 13.7
E= 13.9
E= 13.8
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:
(a)
(c)
(e)
(b)
(d)
(f)
rAB
rAB
rBC
rBC
rBC
Figure 8.20 A replot of the results in Figure 8.19 on a potential
energy surface.
22
Also Need Correct Distribution Between
Translation And Vibration:
Trans E=14.5 kcal/mol
Vib E= 0.01 kcal/mol
Trans E=12. kcal/mol
Vib E= 2.5 kcal/mol
Trans E=9.7 kcal/mol
Vib E= 4.8 kcal/mol
Trans E=7.25 kcal/mol
Vib E= 7.25 kcal/mol
Trans E=4.8 kcal/mol
Vib E= 9.7 kcal/mol
Trans E=2.4 kcal/mol
Vib 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
23
Polanyi Rules:
Can use vibration/translation to probe
structure of transition state:
Early
Middle
Late
R BC
R BC
R AB
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:
1
Boltzman Distribution
Probability
0.8
Experiment
0.6
n=3
0.4
n=4
n=2
0.2
n=1
0
n=5
n=0
0
10
20
30
40
Energy, kcal/mole
Figure 8.25 The distribution of vibrational energy produced
during the reaction F+H2HF+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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