ChE 452 Lecture 20
Collision Theory
1
So Far This Course Has Shown
1) How to get rate equations from
experiments
2) How to get rate equations from
mechanisms
3) How to predict the mechanism
Next part of the course: Theory of rate
constants:
2
Theory Of Reaction Rates Has Two Parts
Theory of Preexponentials
Collision Theory, Transition State Theory,
RRKM, Molecular Dynamics
Theory of Activation Barriers
Polanyi Relationship, Marcus Equation,
Blowers-Masel, Quantum Methods
3
Models For Preexponentials




Collision theory (old collision theory) – simple
model for preexponential - ~1013/sec,
~1013Å3/sec, ~1013A6/sec
Transition state theory – slightly better model
for preexponential – bimolecular (small
correction to collision theory).
RRKM – better model for preexponential –
unimolecular-explains rate constraints at
1018/sec
Molecular Dynamics & Tunneling – accurate
method, but time consuming
4
Plan For Today



Describe Arrhenius’ Model (1889)
Describe Trautz and Lewis model (1918)
Show limitations
5
Divides molecules
into two
populations
Number Of Molecules
Arrhenius Model For AB
Hot Reactive
Molecules
Cold Unreactive
Molecules
0
1
2
3
4
5
Velocity, cm/sec x 1000
Figure 7.1
The Boltzmann
distribution of molecular velocities.


Cold unreactive molecules
Hot reactive molecules
6
Next Derive Equation For Rate
Equilibrium:
†
A
u
A
C =C e
Rate equ
 ΔG† 

k
T

B

(7.2)
k1 =k oe
 ΔG† 

 kBT 
(7.4)
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Derivation Continued
k1 =k oe
 ΔG† 

k
T

B

k 1 =k 0 e
(7.7)
(7.4)
G  H  TS
†
 Ea 

k
T
 B 
†
†
(7.5)
ΔS

k
k1 =  k oe B


†
(7.6)
 - 
 e kBT


ΔH
†
(Δ S )
†
k 0 =k oe
k
B
(7.8)
8
Result of Arrhenius’ Model


Rate constant
varies exponentially
with T-1
No expression for Ko
†
(Δ S )
k 0 =k oe
k
B
(7.8)
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Collision Theory

Assume Ko
equals the
collision rate
†
A
u
A
C =C e
 ΔG† 

k
T

B

(7.2)
10
Collision Theory
A
A
rA  B C = Z A B C Preaction
B
(7.10)
-
Preactio n = e
C
ΔG
B
C
A
†
A
kBT
B
B
C
(7.11)
-
rA  B C = Z A B C e
(7.12)
ΔG
†
k BT
C
Figure 7.2 A collision between an A
molecule and BC molecules.
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Next: Consider Billiard Ball Collisions
AA
BC
AA
Collisions occur
whenever molecules
get close
BC
AA
BC
Figure 7.3 Some typical billiard ball collisions
12
Next: Calculate How Many Collisions
Occur
Consider the volume swept out by a BC molecule
in time to
LABC = vABC tc
(7.13)
A
A
A
A
L
AB
C
b
co
ll
X
Y
A
A
BC
A
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Next: Calculate How Many Collisions
Occur
 # collisions 

  Volume of 
 of the given    cylinder  C A


 BC molecule
(7.14)
where CA is the concentration of the A
molecules in the reacting mixture, measured
in molecules/cm3.
The volume of the cylinder is
 Volume of 
2

  ( b coll ) L ABC
 cylinder 
(7.15)
14
Pages Of Algebra Yields Trautz & Lewis’
Approximation
k 0 = vA  BC cA  BC
Derivation
(7.26)
Equation (7.26) is the key result for simple collision
theory.
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Additional Assumption
Calculate the molecular velocity ignoring that
molecules are hot.
vA  BC
 8 k B T 
=

  ABC 
1/ 2
(7.27)
Where:
1
 ABC

1
1

mA mB  mC
(7.28)
and mA, mB and mC are the masses of A, B and C in
atomic mass units (1 AMU = 1.66  10-24g).
22
Simplified Equation
In lecture 14 we showed
vABC
Å  T 
 2.52  10


sec  300K
13
1/2
 1AMU


  ABC 
1/2
(7.29)
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Example 7.A A Collision Theory Calculation
Use collision theory to calculate the
preexponential for the reaction:
H+CH3CH3H2+CH2CH3
(7.A.1)
at 500K.
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Solution:
According to collision theory:
k 0  d
2
coll
v ABC
(7.A.2)
25
Step 1: Calculate VABC
According to equation (7.26):
vABC
 T 
 2.4  10 Å / sec

 300K
13
(7.A.3)
with

ABC

1/2
 1AMU


  ABC 
1/2
1
1
M

A
1
M
(7.A.4)
BC
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Step 1 Continued
For reaction (7.A.1)
 A -BC 
1
1
1

1A M U
 0.968A M U
30A M U
(7.A.5)
Substituting the numbers shows that 500K:
13  500K 
v A B C  2.52  10 

300K


1/2
(7.A.6)
 1A M U 
13

3.31

10
Å /sec


 0.968A M U 
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Step 2: Estimate dcoll
Trautz’s approximation
dA  dB
d coll 
2
Were dA and dB are the Van der Waals radii of A
and B
d H 2  1 . 5 Å, d C 2 H 6  3 . 5 Å
Therefore
d coll 
1 . 5 Å  3.5Å
2
 2.5Å
(7.A.7)
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Solution Continued
Substituting (7.A.5) and (7.A.6) into
equation (7.A.2) yields:
 2.5Å 
2

Å

Å
13
14
k 0 =π
 3.31  10
 =6.49  10
m olecule 
second 
m olecule  second
3
(7.A.8)
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Discussion Problem
Use collision theory to calculate the rate
constant for the reaction
F + H2  H + HF
Assume a collision diameter of 2.3Å
30
Solution: Step 1 Calculate 
1 1
1
1
1




 m H 2 m F 2AMU 19 AMU
= 1.81 AMU
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Step 2: Calculate v
13 Å  1AMU 
v
V  2.52 x10


sec 

13 Å  1 
v
V  2.52 x10


sec  181
. 
13 Å
v
V  187
. x10
sec
1/ 2

 T 


 300K 
1/ 2
1/ 2
 300K 


 300K 
1/ 2
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Solution
k o = v π(d co ll )
2
ko=(41012Å/sec)  ((3Å)2) = 1.1 × 1014 Å3/sec
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Key Predictions Of Collision Theory




Preexponentials always between 1013 and
1014/sec for small molecules
No special configurations effects
Lighter species (i.e. H atoms tend to react
faster).
Larger molecules have larger cross
sections than smaller molecules
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Preexponentials Usually The Same Order As
Collision Theory?
Table 7.2 a selection of the preexponentials reported by Wesley [1980]
Reaction
H+C2H6C2H5+H2
Preexponential
Å3/molecule Sec
1.6  1014
Reaction
Preexponential
Å3/molecule Sec
2.5  1013
O+C2H6OH+C2H5
H+CHH2+C
1.1  1012
O+C3H8(CH3)2CH+OH
H+CH4H2+CH3
1  1014
O2+HOH+O
1.4  1010
1.5  1014
O+H2OH+H
1.8  1013
OH+OHH2O+O
1  1013
O+OHO2+H
2.3  1013
OH+CH4H2O+CH3
5  1013
O+CH4CH3+OH
2.1  1013
OH+H2COH2O+HCO
5  1013
O+CH3H+CH3O
5  1013
OH+CH3H+CH3O
1  1013
O+HCOH+CO2
5  1012
OH+CH3H2O+CH2
1  1013
35
Comparisons Between Collision Theory And
Experiments
Table 7.3 Preexponentials calculated from equation (7.30) for a number of reactions
compared to experimental data.
Reaction
H  C2H6 C2H5  H2
H  CH  H 2  C
O  C 2 H 6  OH  C 2 H 5
OH  OH  H 2 O+ O
H  O 2  OH  O
Calculated
Calculated Preexponential
Preexponential
assuming bcoll=covalent radius
assuming bcoll=van
Der Waals radius
Experimental
Å3/molec sec
6.2 1014
Å3/molec sec
2.0 1014
Preexponential
4 1014
2.0 1014
1.1 1012
1.9 1014
7.6 1013
2.5 1013
1.25 1014
5.8 1013
1 1013
4.0 1014
2 1014
1.5 1014
1.6 1014
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Cases Where Collision Theory Fails

CH 3CH 2 CH 3  O:  CH 3 C HCH 3 + OH
(7.30)
ko=1.41010 Å3/molecule-sec
2O2  2O   O2
(7.31)
ko=5.81015 Å3/molecule-sec
37
Why Does Collision Theory Fail For Reaction
7.30
Reaction 7.30 requires a special collision geometry:
•
C H 3 C H 2 C H 3 +O :  C H 3 C H C H 3 +•O H
(7.32a)
C H 3 C H 2 C H 3 +O :   C H 2 C H 2 C H 3 +•O H
(7.32b)
S
C onfigurations = e
kB
(7.33)
ΔS
e
†
kB
=
configurations w hich lead to reactions
average num ber of configurations of the reactants
(7.34)
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Summary
Collision theory: reaction occurs whenever
reactants collide.
Gives correct order of magnitude or slightly
high pre-exponential
Some spectacular failures
TST theory after exam
39
Class Question

What did you learn new today?
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