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

ChE 551 Lecture 18
Non-linear Collisions
1
Background: Collision Theory
Key equation
rABC  ZABCPreaction
Method
 Use molecular dynamics to simulate the
collisions
 Integrate using statistical mechanics
2
Summary Of Collision Theory
For Linear Collisions



One can approximate 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 saddle point 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.
3
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.
It is necessary 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.

Want banked curves, that funnel molecules to
saddle point.
4
Summary Continued

Leads to very complex behavior, but not
big rate changes.



Latter four effects only cause perhaps a factor
of 10 or 100 in rate.
There are always some trajectories which
make it over the barrier, even though the
molecules have barely enough energy to cross
the barrier.
If 1% of the trajectories make it those
trajectories will have an important effect on
the rate.
5
Today: Non-linear Case
Non-linear case:
Reaction probability varies with the impact
parameter
v
v
A
v
bA

BC
B
C
Figure 8.27 The typical trajectory for the collision of an A atom with a BC molecule.
6
Variation Of Reaction Probability With
Impact Parameter
Reaction
Probability
1.0
Leads to finite
cross section.
0.5
0.0
0
1
2
Impact Parameter Å
3
Figure 8.26 The variation in PABC with changing impact
parameter.
7
Why Does Reaction Probability
Vary With Impact Parameter?


Molecules still have enough energy to
go over the barrier
Energy is not coupled to the bond that
breaks.


Reactants miss.
Angular momentum, Coriolis force
carries molecules apart.
8
Angular Momentum Creates Extra
Barrier To Non-Linear Collisions


Component of velocity carries reactants apart.
Must overcome that velocity component for
reaction to occur.
v
v
A
v
bA

BC
B
C
Figure 8.27 The typical trajectory for the collision of an A atom with a BC molecule.
9
Derivation Of The Angular Momentum
Barrier To Reaction A+BProducts
Classical equations of motion of A and B.
mA
d 2R A
dt 2

 FA   A V(R A , R B )
Pages of Algebra
2 

E AB b 2
1
 dR AB 
E   AB 
 V(R AB ) 
  2
 dt 
2
 R

AB
Kinetic
energy
Extra energy barrier
due to Coriolis force
Potential
energy
10
Result: Atom Moves In Effective
Potential:
AB
R AB
dt 2
Veff

 Feff
R AB
(8.144)
Veff (R AB ) 
E AB b 2
R 2AB
 V(R AB )
(8.56)
11
Angular Momentum Barrier To
Reaction
The angular momentum
barrier prevents reactions
from occurring when
molecules approach with
large impact parameters.
As a result, no reaction
occurs unless the
reactants get close to
each other.
VEff
b =6
0
b =0
1
2
3
4
5
R
6
7
8
9
10
AB
12
Angular Momentum Barrier Results In Drop Off
Of Reaction Probability At Large b.
Reaction
Probability
1.0
0.5
0.0
0
1
2
Impact Parameter Å
3
Figure 8.26 The variation in PABC with changing impact
parameter.
13
Next: Quantifying The Effects: Preaction
Varies with:
 vABC, the velocity that the A molecule approaches
the BC molecule;
 EBC, the internal state (i.e., vibrational, rotational
energy) of the BC molecule before collision occurs;
 The “impact parameter” bABC,which is a measure
of how closely A collides with BC;
 “The angle of approach” where  is a measure of
the angle of the collision
 The initial position RBC and velocity vBC of B
relative to C when collision occurs.
14
Next Equation
 A  BC  v A  BC , E BC     Preaction  v A  BC , E BC , b A  BC ,  R BC , v BC  
D1R BC , v BC b A  BC ,db A  BCddR BCdv BC
(8.57)
Structureless molecules
r
A
BC vABC ,EBC   2 PABC ( bABC ,vABC ,EBC )bABCdbABC
(8.58)
15
Example 8.E Calculating The Cross
Section
A program called ReactMD is available
from Dr. Masel’s website. Assume you
used the program to calculate the
reaction probability as a function of
impact parameter, and the data in
Table 8.E.1 were obtained. Calculate
the cross section for the reaction.
16
Table 8.E.1
bABC,Å
0
.2
.4
.6
.8
1.0
P(b)
.84
.83
.85
.78
.80
.75
bABC,Å
1.2
1.4
1.6
1.8
2.0
2.2
P(b)
.80
.83
.72
.21
.10
0
17
Solution: The Cross Section Is
Given By:

  2  P(b)bdb
0
We can integrate using the trapezoid rule.
18
Spreadsheet To Do The
Calculations:
03
04
05
06
07
08
09
10
11
12
13
14
15
16
17
18
A
b
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
2.2
B
P(b)
0.84
0.83
0.85
0.78
0.8
0.75
0.8
0.83
0.72
0.21
0.1
0
C
integral=
=
=0.5*(C4+C15)+SUM(C5:C14)
=2*PI()*C17
=A4*B4
=A5*B5
=A6*B6
=A7*B7
=A8*B8
=A9*B9
=A10*B10
=A11*B11
=A12*B12
=A13*B13
=A14*B14
=A15*B15
19
Numerical Results
A
03
04
05
06
07
08
07
10
11
12
13
14
15
16
17
18
b
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
2.2
B
C
P(b)
0.84
0.83
0.85
0.78
0.8
0.75
0.8
0.83
0.72
0.21
0.1
0
0
0.166
0.34
0.468
0.64
0.75
0.96
1.162
1.152
0.378
0.2
0
integral=
=
6.216
39.05628
20
Example 8.D: Calculating The Rate
Constant Using Equation 8.60
Figure 8.17 shows some data for the cross section
for the reaction H+H2H2+H as a function of ET,
the translational energy of H approaching H2.
Assume that the cross section follows:
r
A
 BC  0 for E T  0.35eV
  

2
r
 A BC  12 Å 1  exp0.35eV - E T  / 5.08eV 
exp E T / 2eV
for E T  0.35eV
(8.D.2)
where ET is the translational energy. Calculate the rate
constant for the reaction at 300K.
21
Solution:
Step 1: Derive Equation
According to equation 8.60, if there is no
EBC dependence

k
ABC
v
0
 D(v
r
ABC
AB
ABC
)dv
ABC
(8.D.3)
22
Derive Equation Continued
According to results in Example 6.D:
D v A  BC  
v

A  BC

2
  v A  BC 

  ABC
2
exp 
 v A BC  
 2 kBB T

2
  ABC
2
exp 
v A  BC  



2
(8.D.3)
Where µABC is the reduced mass of ABC, kB is
Boltzman’s constant and T is the temperature.
23
Derive Equation Continued
Looking up the integral in the CRC yields:
4   ABC 
D v ABC  


2

T

kBBBT 
 2
3/ 2
v ABC 
2
  ABC
2
exp 
v ABC  
 2 kBB T

(8.D.5)
Combining equations (8.D.3) and (8.D.5)
1
2
E


v


and substituting T 2 ABC ABC yields:
8
kB
T 
r
B
k A BC 

kBB T d E T / 
kBB T
o  E T / kBB T A BC exp- E T / 
 AB
(8.D.7)
24
Final Equation
Note
8
kBB T
= v A BC
 AB
(8.D.6)
Therefore,
r
kB TdE T / kBB T
k ABC = v ABC 0 E T / k
 BB T A
BC exp   / 
(8.D.8)
25
Define Average Cross-Section IABC
r
IABC = 0 ET / 
kBB TA
kBB TdET / 
kBB T
BC exp   / 
(8.D.8)
Equation 8.C.8 becomes:
k ABC  vABCIABC
(8.D.9)
26
Step 2: Simplify Equation
Let’s define a new variable W by:
W   E T  0.35eV / 
kB

B T
(8.D.10)
Substituting equation (8.C.6) into equation
(8.C.5) yields:
k A BC  v A BC e

0.35eV / 
kBB T



r

kBB T  A BC exp- WdW
-0.35eV/ 

kBB T  0.35eV / 
kBB T W
(8.D.11)
27
Step 3: Substitute In Numbers
For future reference, it is useful to note:
kBT = 0.6 kcal/mole = 0.026 eV/molecule
(8.D.12)
1
1
1
1.5



 ABC 2AMU / mole 1AMU / mole AMU
(8.D.13)
28
Step 4: Calculate The Velocity
According to equation (7.A.4):


 T 
v A  BC  2.4  10 Å / sec 

 300K 

13
1/ 2
 1AMU 


  
1/ 2

1/ 2
1/ 2
300K
1AMU




13
 2.4  10 Å / sec 



 300K 
 1AMU  15
 
(8.D.15)
 2.94  1013 Å / sec
29
Step 5: Further Algebra
Substituting equation 8.C.10 into 8.C.8 and
adding the appropriate conversion
factors yields:
k ABC  2.94  10 Å / sec exp 0.35eV / kBB T
13
r

W
+
0.35eV
/

T

k
-0.35eV/ kBB T 
B  ABC exp- WdW
B
(8.D.16)
30
Step 6: Further Simplify Equation
Note:  = 0 for ET < 0.35eV.
Therefore,
2.94  1013 Å
k A  BC 
exp- 0.35eV / kB
B T 
molecule - sec
r

W
+
0.35eV
/

T

k


A  BC exp- WdW
-o.
B
B
-0.35eV/k T
B
(8.D.15)
31
More Simplification
Let’s define IABC and F(W) by:
r

I ABC  o. W + 0.35eV / kB
B T ABC exp- WdW
(8.D.17)
r
F(W)  W + 0.35eV / kB
T


B
A  BC
(8.D.18)
32
Combining Equations (8.D.15) And (8.D.16)
Yields:

I A  BC  o. FW
exp- WdW
(8.D.19)
33
One Can Conveniently Integrate Equation 8.D.17
Using The Laguere Integration Formula

o. FWexp- WdW =  Bi FWi 
i
(8.D.20)
Where the Bi and Wis are given in the
spreadsheet on the next page.
34
Spreadsheet For The
Calculations
A
02 kbT=
03
04 W
05 0.22285
B
=D2*0.00198/23.05 T=
Et
=A5*kbT+0.35
06 1.118893 =A6*kbT+0.35
07 2.99273
=A7*kbT+0.35
08 5.77514
=A8*kbT+0.35
09 9.83747
=A9*kbT+0.35
10 15.98287 =A10*kbT+0.35
C
s
=12*(1-EXP((0.35$B5)/5.08))*EXP(-$B5/2)
=12*(1-EXP((0.35$B6)/5.08))*EXP(-$B6/2)
=12*(1-EXP((0.35$B7)/5.08))*EXP(-$B7/2)
=12*(1-EXP((0.35$B8)/5.08))*EXP(-$B8/2)
=12*(1-EXP((0.35$B9)/5.08))*EXP(-$B9/2)
=12*(1-EXP((0.35$B10)/5.08))*EXP($B10/2)
D
E
F
300
F(w)
=$B5*$C5/kbt
I=
Bi
0.458964
=SUM(F5:F10)
term in sum
=D5*E5
=$B6*$C6/kbt
0.417
=D6*E6
=$B7*$C7/kbt
0.113373
=D7*E7
=$B8*$C8/kbt
0.0103991
=D8*E8
=$B9*$C9/kbt
0.000261017 =D9*E9
=$B10*$C10/kbt 8.98547e-7
=D10*E10
35
Results:
A
02
03
04
05
06
07
08
09
10
kbT=
W
0.22285
1.118893
2.99273
5.77514
9.83747
15.98287
B
0.02577
Et
0.355743
0.378834
0.427123
0.498826
0.603512
0.76188
C
T=
s
0.011349
0.056199
0.146036
0.26998
0.43199
0.6385
D
E
F(w)
0.156665
0.826153
2.420454
5.225937
10.11683
18.87694
I=
Bi
0.458964
0.417
0.113373
0.010399
2.61E-04
8.99E-07
F
300
0.747826
term in sum
0.071904
0.34506
0.274414
0.054345
0.002641
1.7E-5
36
Solution
I ABC  0.748Å
13 Å
k A  BC  2.94  10
sec
2

(8.D.21)

 0.748Å 2 exp 0.35eV / kB
T
B 
(8.D.24)
k A  BC  2.2  1013
Å 
3
molecule  sec
exp 0.35eV / k B T
B
(8.D.25)
37
Notes



Notice that the activation barrier is about 0.35
eV (i.e., the minimum energy to get reaction)
even though the reaction probability is small
below 0.5eV.
It is not exactly 0.35eV though.
In the problem set, we ask the reader to
calculate the rate constant at other
temperatures. If you make an Arrhenius plot of
the data, you find that the activation barrier is
not exactly 0.35eV, but close to 0.35eV, even
though the reaction probability is negligible at E
= 0.35eV.
38
Summary



Can use MD to calculate reaction
probability as a function of b.
Can integrate to get reaction rate.
Results differ from TST due to dynamic
corrections (turning and angular
momentum barriers to reaction).
39
Question

What did you learn new in this lecture?
40