ChE 452 Lecture 20 Collision Theory 1. So Far This Course Has Shown 2.
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Transcript of ChE 452 Lecture 20 Collision Theory 1. So Far This Course Has Shown 2.
ChE 452 Lecture 20 Collision Theory
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So Far This Course Has Shown
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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:
Theory Of Reaction Rates Has Two Parts
Theory of PreexponentialsCollision Theory, Transition State Theory, RRKM, Molecular Dynamics
Theory of Activation BarriersPolanyi Relationship, Marcus Equation, Blowers-Masel, Quantum Methods
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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
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Plan For Today
Describe Arrhenius’ Model (1889) Describe Trautz and Lewis model
(1918) Show limitations
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Arrhenius Model For AB
Cold unreactive molecules Hot reactive molecules
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0 1 2 3 4 5
Velocity, cm/sec x 1000
Num
ber
Of M
olec
ules
Hot Reactive Molecules
Cold Unreactive Molecules
Figure 7.1 The Boltzmann distribution of molecular velocities.
Divides molecules into two populations
Next Derive Equation For Rate
Equilibrium:
Rate equ
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B
ΔG†-
k T1 ok =k e
(7.4)
B
ΔG†-
k T† uA AC =C e
(7.2)
Derivation Continued
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††
B B
ΔHΔS-
k k T1 ok = k e e
(7.6)
†
B
(ΔS )
k
0 ok =k e
a
B
E-k T
1 0k =k e
(7.7)
(7.8)
††† STHG
(7.4)
(7.5)
B
ΔG†-k T
1 ok =k e
Result of Arrhenius’ Model
Rate constantvaries exponentially
with T-1
No expression for Ko
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(7.8)
†
B
(ΔS )
k
0 ok =k e
Collision Theory
Assume Ko
equals the collision rate
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(7.2)
B
ΔG†-k T† u
A AC =C e
Collision Theory
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A
A
B B
C
C
AA
B B
CC
Figure 7.2 A collision between an A molecule and BC molecules.
†
B
ΔG-
k TA BC ABCr =Z e
(7.12)
(7.10)
(7.11)
†
B
ΔG-
k TreactionP =e
A BC ABC reactionr =Z P
Next: Consider Billiard Ball Collisions
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BC
A
A
BC
A
BC
A
A
A
Collisions occur whenever molecules get close
Figure 7.3 Some typical billiard ball collisions
Next: Calculate How Many Collisions Occur
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A
A
BC
bcoll
XY
LABC
AA
A
A
AA
Consider the volume swept out by a BC molecule in time to
LABC = vABC tc(7.13)
Next: Calculate How Many Collisions Occur
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# collisions
of the given
BC molecule
Volume of
cylinderCA
(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
cylinderb ) Lcoll
2ABC
(
(7.15)
Pages Of Algebra Yields Trautz & Lewis’ Approximation
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k = v 0 A BC A BCc
(7.26)
Equation (7.26) is the key result for simple collision theory.
Derivation
Additional Assumption
Calculate the molecular velocity ignoring that molecules are hot.
Where:
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v =8 T
A BCB
ABC
1 2/
(7.27)
1 1
m
1
m mABC A B C
(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).
k
Simplified Equation
In lecture 14 we showed
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v 2.52 10Å
sec
T
300K
1AMUABC
131/2
ABC
1/2
(7.29)
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:
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According to collision theory:
ABCcoll2
0 vdk (7.A.2)
Step 1: Calculate VABC
According to equation (7.26):
with
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v 2.4 10 Å / secT
300K
1AMUABC
131/2
ABC
1/2
BCA
ABC
M
1
M
11
(7.A.3)
(7.A.4)
Step 1 Continued
For reaction (7.A.1)
(7.A.5)
Substituting the numbers shows that 500K:
(7.A.6)
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A-BC1
1
1AMU
1
30AMU
0.968AMU
1/213 13
ABC
500K 1AMUv 2.52 10 3.31 10 Å/sec
300K 0.968AMU
Step 2: Estimate dcoll
Trautz’s approximation
Were dA and dB are the Van der Waals radii of A and B
Therefore
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2
ddd BA
coll
Å53d Å,51d6H2C2H ..
Å522
3.5Å Å51dcoll .
. (7.A.7)
Solution Continued
Substituting (7.A.5) and (7.A.6) into equation (7.A.2) yields:
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(7.A.8)
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13 140
2.5Å Å Åk =π 3.31 10 =6.49 10
molecule second molecule second
Discussion Problem
Use collision theory to calculate the rate constant for the reaction
F + H2 H + HF
Assume a collision diameter of 2.3Å
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Solution: Step 1 Calculate
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AMU19
1
AMU2
1
m
1
m
11
FH2
= 1.81 AMU
Step 2: Calculate v
V xAMU T
K
V xK
K
V x
2 52 101
300
2 52 101
181
300
300
187 10
131 2 1 2
131 2 1 2
13
.sec
..
.
/ /
/ /
Å
Å
sec
Å
sec
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v
v
v
Solution
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2o collk =v π(d )
ko=(41012Å/sec) ((3Å)2) = 1.1 × 1014 Å3/sec
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?
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Table 7.2 a selection of the preexponentials reported by Wesley [1980]
Reaction Preexponential
Å3/molecule Sec
Reaction Preexponential
Å3/molecule SecH+C2H6
C2H5+H2 1.6 1014 O+C2H6 OH+C2H5 2.5 1013
H+CH H2+C 1.1 1012 O+C3H8 (CH3)2CH+OH 1.4 1010
H+CH4 H2+CH3 1 1014 O2+H OH+O 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
Comparisons Between Collision Theory And Experiments
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Calculated Preexponential
assuming bcoll=van Der Waals radius
Calculated Preexponential assuming bcoll=covalent radius
Experimental
Å3/molec sec Å3/molec sec Preexponential
6.2 1014 2.0 1014 1.6 1014
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
Table 7.3 Preexponentials calculated from equation (7.30) for a number of reactions compared to experimental data.
Reaction
25262 HHCHCH
H CH H C2
O C H OH C H2 6 2 5
OH OH H O+O2
H O OH O2
Cases Where Collision Theory Fails
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CH CH CH O: CH C HCH + OH3 2 3 3 3
(7.30)
ko=1.41010 Å3/molecule-sec
2O 2O O2 2 (7.31)
ko=5.81015 Å3/molecule-sec
Why Does Collision Theory Fail For Reaction 7.30
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Reaction 7.30 requires a special collision geometry:
(7.33)
(7.34)
•
3 2 3 3 3
3 2 3 2 2 3
CH CH CH +O: CH C HCH +•OH (7.32a)
CH CH CH +O: CH CH CH +•OH (7.32b)
B
S
kConfigurations = e
†
B
ΔS
k configurations which lead to reactions e =
average number of configurations of the reactants
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
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Class Question
What did you learn new today?
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