ChE 553 Lecture 17 Prediction of Mechanisms 1. Objectives Develop methods to predict mechanisms...
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Transcript of ChE 553 Lecture 17 Prediction of Mechanisms 1. Objectives Develop methods to predict mechanisms...
ChE 553 Lecture 17 Prediction of Mechanisms
1
Objectives
• Develop methods to predict mechanisms
• Apply the ideas for a simple reaction
2
You Already Learned About the Mechanisms of Reactions in Organic Chemistry
Organic view of mechanisms – things to memorize
Masel view of mechanisms – things to calculate
3
Key: Activation Barriers Control Mechanisms
• Reaction goes by the pathway that has the lowest activation barrier between reactants and products– Catalytic cycles used to lower barriers
4
The Idea of Computing a Mechanism
5
1) write down all possible reactions2) Use rules to make sure no important
reactions are missing.3) Use rules to eliminate excess reactions.
General Rules for Mechanisms
All commercially important mechanisms are basically the same !!
• Step 1 Create reactive species
• Step 2 Catalytic cycle to pump out
product
• Step 3 Reactive species lost:
6
Example: H2 + Br2 2HBr
7
X Br 2Br X
Br H HBr H
H Br HBr Br
X 2Br Br X
H HBr H Br
21
22
23
42
52
Br
H
HBr
Br 2HBr
H 2
Br2
Figure 5.1 A cycle for HBr formation via reaction (5.3).
Initiation-Propagation Mechanisms
• Initiation step: create reactive species
• Transfer step: convert initial radical into a more reactive species
• Propagation step: go around cycle to
produce product
• Termination step: destroy radicals 8
Consider: H2 + Br2 2HBr
9
X Br 2Br X
Br H HBr H
H Br HBr Br
X 2Br Br X
H HBr H Br
21
22
23
42
52
Br
H
HBr
Br 2HBr
H 2
Br2
Figure 5.1 A cycle for HBr formation via reaction (5.3).
Discussion Problem: The reaction CH3CH3CH2CH2 + H2
Goes By the Following Mechanism
10
Label each step as being a a) initiation b) propagation c) termination d) transfer.
CH3CH3 +X1 2CH3+X
CH3 + CH3CH3 2 CH4 + CH2CH3
CH2CH3 + X 3 CH2CH2 + H +X
H+ CH3CH3 4 H2 + CH2CH3
2 CH2CH3 +X 5 CH3 CH2CH2CH3 + X
Examples of Initiation Propagation Mechanisms
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Reaction Example Mechanism
Combustion e.g., CH4 + O2 CO2 +2H2O + other products
O2 2O O + CH4
CH3 + OH OH + CH4
H2O + CH3 CH3 + O2
CH3 + O + O CH3 + OH CH2 + H2O + other products
OH walls CH3
walls Free Radical Polymerization
e.g. ethylene polyethylene with a free radical catalyst, R2
R2 2R
R+C2H4 R(C2H4)
RC2H4+C2H4
R(C2H4)2
R(C2H4)n+C2H4
R(C2H4)n+1
R(C2H4)m+R(C2H4)n R(C2H4)m+nR
Ozone Depletion O2+h1 2O
O+O2+X O3 O3+h2
O2+O Cl+O3
O2+ClO ClO+O O2+Cl
Hydrocarbon Pyrolysis X+CH3COH CH3+COH+X CH3+CH3OH CH3CO+CH4
CH3CO+CH3OH CH4+CH3CO COH+X CO+H+X
H+CH3COH CH4+COH H+CH3COH CH3+CO+H2
2CH3+X C2H6+X H+CH3+X CH4+X
H+CH3CO+X CH3COH+X
General Approach to Finding a Mechanism
• Guess or predict all of the species that are likely to form during the reaction.
• Write down all of the possible reaction of those species (only include 7 generic types of reactions).
• Use various rules to pare down the list to manageable of steps.
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Rules for Initiation Propagation Reactions
• There must be at least one initiation reaction• The propagation reactions must occur in a
cycle where radicals react with the reactants to form new radicals and then the new radicals react to from the original radicals again
• All of the steps in the catalytic cycle must have low barriers
• There should be at least on termination reaction where two radicals combine to yield stable species
13
Example: The Reaction CH3CH3H2C=CH2+H2 Obeys the Following Mechanism:
XCH2XCHCH 31
33 3242
333 CHCHCHCHCHCH XHCHCHXCHCH 223
32 3224
33 CHCHHCHCHH 335
3 CHCHXCH2 (+ other reactions)
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Verify that it follows the rules
Step 1: Make a Diagram of the Reaction Similar to That in Figure 5.3
XCH2XCHCH 31
33 3242
333 CHCHCHCHCHCH XHCHCHXCHCH 223
32 3224
33 CHCHHCHCHH 335
3 CHCHXCH2 (+ other reactions)
15
H
H 2 X
2 H C=CH 2
3 CH CH 2
X+ 3 CH CH 3
CH 4
3 CH CH 3
CH 3
3 CH CH 3
+X
Catalytic Cycl
e
Chain Transf
er
initiation
Step 2: Identify the Initiation Step, the Transfer Step, the Propagation Steps, the Termination Steps
XCH2XCHCH 31
33 3242
333 CHCHCHCHCHCH XHCHCHXCHCH 223
32 3224
33 CHCHHCHCHH 335
3 CHCHXCH2 (+ other reactions)
16
H
H 2 X
2 H C=CH 2
3 CH CH 2
X+ 3 CH CH 3
CH 4
3 CH CH 3
CH 3
3 CH CH 3
+X
Catalytic Cycl
e
Chain Transf
er
initiation
Step 2: Continued
b) Reaction 1 – initiation
Reaction 2 – chain transfer
Reaction 3 – propagation (-hydrogen elimination)
Step 4 – propagation (hydrogen transfer)
Reaction 5 - termination
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The Mechanisms
Does the mechanisms follow the rules?• There is an initiation step (step 1)• There is a catalytic cycle (steps 3 and 4)• There is a termination step (step 5)• Still need to verify that the activation
barriers are low enough
18
Empirical Rules for Activation Barriers
Practical reactions Ea<0.15 T initiation
reactions Set minimum T
Ea<0.05 T catalytic For Reaction
cycle
Ea<0.07 T Transfer reactions and side reactions 19
KMole
Kcal
KMole
Kcal
KMole
Kcal
Methods to Estimate Ea
Polanyi relationship
20
rP0aa HEE
Blowers Masel Equation
when
when
when
1E4
H0a
r
1E4
H1
0a
r
1E4
H0a
r
(10.63)
0a0
0a0
0PEw
EWw2V
(10.65)
0
H
2r
20
2P
2r0Pr0
a
r
)H()w(4)V(
)Hw2V)(H5.0w(E
Ea= Activation Energy
Ea0= Intrinsic Activation Barrier
P= Transfer coefficent
Hr= Heat of Reaction Memorize this equation
Intrinsic Barriers and Transfer. Coefficients for Different Types of Neutral Species
21
Reaction Example Actual EAO
kcal/mole EA
O to assume when predicting mechanisms kcal/mole
Actual P P to assume
when predicting mechanisms
Simple bond scission
AB+X A+B+X X=a collision partner
0-1 1 1.0 1.0
Recombination A+B+X AB+X X=a collision partner
0-1 1 1.0 1.0
Exothermic atom transfer reaction
R x + R1 R + x-R1
x = an atom
8-16 12 0.2 to 0.6 0.3
Endothermic atom transfer reaction
R- x + R R + x-R1 x=an atom
8-16 12 0.4 to 0.8 0.7
Ligand transfer reaction to hydrogen
H+R-R1 HR + R1 40-50 45 0.4 to 0.6 0.5
Other ligand transfer reactions
x + R-R1 xR+ R1
x=an atom 50 or more 50 0.3 to 0.7 0.5
Next: Estimate the Activation Barriers
Consider
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XCH2XCHCH 333 First estimate rH
Next estimate EA using Table 5.4. This is a simple bond scission reaction. From Table 5.4 mole/kcal6.90H1E rA
From NIST Web book (http://webbook.nist.gov) 0.20)CHCH(H 33f
Therefore mole/kcal8.34)CH(H 3f
mole/kcal6.89)0.20()8.34(2Hr
Next: CH3•+CH3CH3CH4+•CH2CH3
From the NIST web book mole/kcal0.20)CHCH(H 33f
mole/kcal8.34)CH(H 3f
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This is an atom transfer reaction. From Table 5.4
mole/kcal7.8)3.4(3.0mole/kcal10EA
mole/kcal9.17)CH(H 4f mole/kcal4.28)CHCH(H 32f
Therefore mole/kcal3.4)0.20(8.344.289.17Hr
CH2CH3+XCH3CH2+H+X
From the NIST web book mole/kcal4.28)CHCH(H 32f mole/kcal5.12)CHCH(H 22f
mole/kcal1.52)H(Hf mole/kcal2.364.285.121.52H
24
From Table 5.4 mole/kcal3.40)2.36(7.015EA
H•CH3CH3H2+•CH2CH3
From the NIST web book mole/kcal1.52)H(Hf
mole/kcal0.20)CHCH(H 33f mole/kcal4.28)CHCH(H 32f
mole/kcal0)H(H 2f Therefore
mole/kcal7.3)0.20(1.524.280Hr 25
This is a hydrogen transfer reaction. From Table 5.4
mole/kcal9.5)7.3(3.010EA
•CH3+ •CH3+X CH3 CH3+X
mole/kcal6.89Hr (reverse reaction 1) This is recombination reaction. From Table 5.4
mole/kcal1EA
26
Next: Calculate Temperature to Meet Constants
Kmolkcal
15.0EA for initiation
27
Kmole
kcal07.0E
rA T for all propagation
moleK
kcal05.0EA T for initiation
Solution
For initiation
28
For propagation
K80605.0/3.40T
Therefore any temperature above 806K will satisfy all constraints.
mole/kcal6.90EA
K60415.0/6.90T
Example 5.B Consider the Following Alternate Mechanism for Ethylene Production from
Ethane
29
31
33 CH2XCHCH
3242
333 CHCHCHCHCHCH
XHCHCHXCHCH 223
32
346
33 CHCHCHCHH
XCHCHXCH2 335
3
a) Does this mechanism follow all of the rule at 810K?
b) Is this mechanism more or less likely than the mechanism in example 5.A?
Solution
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a) This does follow the rules! 1)There is an initiation step (step 1) 2) There is a catalytic cycle (steps 2,3,6) 3) There is a termination step (step 5)
Check all steps obey constraint in equation 5.36 Steps 1,2,3,5 do (see example 5.A) Check step 6
H•CH3 CH3CH4+ •CH3
31
From the Nist webbook mole/kcal1.52)H(Hf
mole/kcal0.20)CHCH(H 33f
mole/kcal9.17)CH(H 4f 8.34)CH(H 3f
mole/kcal2.15)0.20(1.52)9.17(8.34Hr This is a ligand transfer reaction to hydrogen. From Table 5.4
mole/kcal4.37)2.15(5.00.45EA
This Reaction is a Reaction in the Catalytic Cycle
32
mole
kcal5.40)k810)(05.0(T)
mole
kcal05.0(EA
Therefore all constraints are satisfied
Which Mechanism is Better
33
3224
33 CHCHHCHCHH
346
33 CHCHCHCHH
Ea=8.9Ea=37.4
Summary
Today derived a series of rules for reactions
• Must be an initiation reaction
• Must have a catalytic cycle
• Should have termination
• Barriers low enough
Next time: Use rules to predict mechanisms.34