ACRE2a Non Elementary Reaction Kinetics Rev
ACRE2a Non Elementary Reaction Kinetics Rev
ACRE2a Non Elementary Reaction Kinetics Rev
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L10-2
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L10-4
rA k A CaA CB
b c d
k A CCCD
consumed generated
At equilibrium, the reaction rate is zero, rA=0
rA 0 k ACaACB
b c d
k ACCCD
k ACaACB
b c d
k ACCCD
kA CcCCD
d
Thermodynamic equilibrium relationship
a b KC
k A CA CB KC: concentration equilibrium constant (capital K)
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L10-5
Now derive a rate equation for the postulated mechanism and check if
it describes the experimentally observed rate equation
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L10-8
k k C k C k 1
rNO k1CNOCO2 1 2 NO 2 NO Add fraction
k 1 k 2CNO k 1 k 2CNO
2k 2CNO Multiply 2k1k 2CNO2CO2
rNO k1CNOCO2
k
1 2 NO
k C
rNO
k 1 k 2CNO
2k1k 2 k 1 CNO2CO2 Conventional to reduce
rNO the additive constant in
1 k 2 k 1 CNO the denominator to 1
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L10-12
13
Active Intermediates and PSSH
14
Example
The rate law for the reaction
A B C
is found from experiment to be
2
kC
rA A
1 k C A
15
Example
For reactions with active intermediates, the reaction
coordinate now has trough in it and the active
intermediate, A*, sits in this trough
16
Active Intermediates/Free Radicals
and PSSH
Hall of Fame Reaction
-rNO2
17
T
Why does the rate law decrease with increasing
temperature?
Mechanism:
NO O2
k1
NO3* (1)
NO3*
k2
NO O2 (2)
NO3* NO
k3
2 NO2 (3)
18
Define k with respect to NO3*
Assume that all reactions are elementary
reactions, such that:
rNO = 2 é -r3NO* ù
2 ë 3 û
19
The net reaction rate for NO3* is the sum of the
individual reaction rates for NO3*:
rNO * 0
3
k1 [ NO ][O2 ]
éë NO ùû =
*
k2 + k3 [ NO ]
3
21
Pseudo Steady State Hypothesis (PSSH)
𝑘1 𝑁𝑂 𝑂2
[𝑁𝑂3∗ ] =
𝑘2 + 𝑘3 𝑁𝑂
rNO2 2r3 NO* 2 NO3* NO
3
k1 k 3 NO 2 O2
𝑟𝑁𝑂2 =2
k 2 + k 3 NO
22
Pseudo Steady State Hypothesis (PSSH)
k2 >> k3 [ NO]
E2 E1 E3
rNO2 2
k1k3
NO O2 2
2 A1 A3
e RT
NO O2
2
k2 A2
E2 E1 E3
24
To find a mechanism that is consistent with the experimental
observations, we use the following steps.
TABLE 9-1 STEPS TO DEDUCE A RATE LAW
1. Propose an active intermediate(s).
2. Propose a mechanism, utilizing the rate law obtained from experimental
data, if possible.
3. Model each reaction in the mechanism sequence as an elementary
reaction.
4. After writing rate laws for the rate of formation of desired product, write
the rate laws for each of the active intermediate
5. Write the net rate of reaction for the active intermediate & use the PSSH.
6. Eliminate the concentrations of the intermediate species in the rate laws
by solving the simultaneous equations developed in Steps 4 and 5.
7. If the derived rate law does not agree with experimental observation,
assume a new mechanism and/or intermediates and go to Step 3.
A strong background in organic and inorganic chemistry is helpful in
predicting the activated intermediate for the reaction under consideration.
25
Step 1. Propose an active intermediate. We will choose as an
active intermediate an azomethane molecule that has been excited
through molecular collisions, to form AZO*,
Step 2. Propose a mechanism.
In reaction 1, two AZO molecules collide and the kinetic energy of one
AZO molecule is transferred to internal rotational and vibrational
energies of the other AZO molecule, and it becomes activated and
highly reactive (i.e., AZO*).
In reaction 2, the activated molecule (AZO*) is deactivated through
collision with another AZO by transferring its internal energy to
increase the kinetic energy of the molecules with which AZO* collides.
In reaction 3, this highly activated AZO* molecule, which is wildly
vibrating, spontaneously decomposes into ethane and nitrogen.
26
Step 3. Write rate laws.
Because each of the reaction steps is elementary, the
corresponding rate laws for the active intermediate AZO* in
reactions (1), (2), and (3) are
These rate laws [Equations (9-3) through (9-5)] are pretty much
useless in the design of any reaction system because the
concentration of the active intermediate AZO* is not readily
measurable.
Consequently, we will use the Pseudo-Steady-State-Hypothesis
(PSSH) to obtain a rate law in terms of measurable concentrations.
27
Step 4. Write rate of formation of product.
28
Step 6. Eliminate the concentration of the active intermediate species
in the rate laws by solving the simultaneous equations developed in Steps 4
and 5. Substituting Equation (9-8) into Equation (9-6)
Chain Reactions
• A chain reaction consists of the following sequence:
– Initiation
• formation of an active intermediate (radicals)
– Propagation or chain transfer
• interaction of an active intermediate with the
reactant or product to produce another active
intermediate (a radical species)
– Termination
• deactivation of the active intermediate
Common in radical polymerizations and cracking of ethane
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L10-31
C2H5 •
k
3 C H H • r3,C2H4 k3CC2H5
2 4
H • C2H6
k
4 C H • H r4,C2H6 k 4CHCC2H6
2 5 2
k 2
Termination: 2C2H5
5 C H
4 10 r5,C2H5 k5 CC2H5
(a) Use the PSSH to derive a rate law for the rate of formation of ethylene
(b) Compare the PSSH solution in Part (a) to that obtained by solving the complete set
of ODE mole balance
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
The rate of formation of ethylene rC2H4 k 3 CC2H5
L10-33
Goal: replace CC2H5•
The net rates of reaction of reactive intermediates C2H5•, CH3•, H• are:
C2H5• was formed in rxns 2 & 4, and consumed in r1,C2H6 k1CC2H6
rxns 3 & 5:
rC H r2,C H r3,C H r4,C H r5,C H r k C C
2 5 2 6 2 4 2 6 2 5 2,C2H6 2 CH3 C2H6
CH3• was formed in rxn 1 and consumed in rxn 2:
-r3,C2H4 -k 3 C C 2H5
rCH3 2r1,C2H6 r2,C2H6
•
r4,C2H6 k 4CHCC2H6
H• was formed in rxn 3 and consumed in rxn 4:
2
rH r3,C2H4 r4,C2H6 r5,C2H5 k5 CC2H5
Plug rate eqs into eq above, assume rate = 0 (PSSH) & solve for reactive species
2
rC H k 2CCH CC H k3CC H k 4CHCC H k5 CC H 0
2 5 3 2 6 2 5 2 6 2 5
Need to replace CCH3• and CH•
rCH3 0 2k1CC2H6 k 2CCH3 CC2H6 2k1CC2H6 k 2CCH3 CC2H6
k1CC2H6 k3CC
2k 2H5
CCH3 CCH 1 Do the same for CH• CH
k 2CC2H6 3 k2 k 4CC
2H6
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
The rate of formation of ethylene rC2H4 k 3 CC2H5
L10-34
Goal: replace CC2H5•
The net rates of reaction of active intermediates C2H5•, CH3•, H• are (PSSH):
C2H5• was formed in rxns 2 & 4, and consumed in r1,C2H6 k1CC2H6
rxns 3 & 5:
rC H r2,C H r3,C H r4,C H r5,C H r k C C
2 5 2 6 2 4 2 6 2 5 2,C2H6 2 CH3 C2H6
CH3• was formed in rxn 1 and consumed in rxn 2:
-r3,C2H4 -k 3 C C 2H5
rCH3 2r1,C2H6 r2,C2H6
•
r4,C2H6 k 4CHCC2H6
H• was formed in rxn 3 and consumed in rxn 4:
2
rH r3,C2H4 r4,C2H6 r5,C2H5 k5 CC2H5
Plug rate eqs into eq above, assume rate=0 (PSSH) & solve for reactive species
2
rC H k 2CCH CC H k3CC H k 4CHCC H k5 CC H 0
2 5 3 2 6 2 5 2 6 2 5
Plug in expressions for CCH3• and CH• into k3CC
2k 2H5
C2H5• rate eq for C2H5• & solve for C2H5• CCH3 • 1 CH
k2 k 4CC
assuming rate = 0 (PSSH) 2H6
2k1 k3CC H 2
rC k2 CC H k3CC H k 4 2 5 C
C2H6 k 5 CC2H5 0
2H5 k2 2 6 2 5 k 4CC H
2 6
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
The rate of formation of ethylene rC2H4 k 3 CC2H5
L10-35
Goal: replace CC2H5•
The net rates of reaction of active intermediates C2H5•, CH3•, H• are (PSSH):
C2H5• was formed in rxns 2 & 4, and consumed in r1,C2H6 k1CC2H6
rxns 3 & 5:
rC H r2,C H r3,C H r4,C H r5,C H r k C C
2 5 2 6 2 4 2 6 2 5 2,C2H6 2 CH3 C2H6
CH3• was formed in rxn 1 and consumed in rxn 2:
-r3,C2H4 -k 3 C C 2H5
rCH3 2r1,C2H6 r2,C2H6
•
r4,C2H6 k 4CHCC2H6
H• was formed in rxn 3 and consumed in rxn 4:
2
rH r3,C2H4 r4,C2H6 r5,C2H5 k5 CC2H5
Plug rate eqs into eq above, assume rate=0 (PSSH) & solve for reactive species
k3CC H
2 2k1
rC H k 2CCH CC H k3CC H k 4CHCC H k5 CC H 0 CCH3 • CH 2 5
2 5 3 2 6 2 5 2 6 2 5 k2 k 4CC H
2 6
2k1 k3CC H 2
k2 CC H k3CC H k 4 2 5 C
C2H6 k 5 CC2H5 0
k2 2 6 2 5 k 4CC H
2 6
2 2
2k1CC H k3CC H k3CC H k5 CC H 0 2k1CC2H6 k5 CC2H5
2 6 2 5 2 5 2 5
2k1CC 2k1CC H
2H6 CC 2
2H5 2 6 C
C2H5
k5 k5
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
The rate of formation of ethylene rC2H4 k 3 CC2H5
L10-36
Goal: replace CC2H5•
The net rates of reaction of active intermediates C2H5•, CH3•, H• are (PSSH):
C2H5• was formed in rxns 2 & 4, and consumed in r1,C2H6 k1CC2H6
rxns 3 & 5:
rC H r2,C H r3,C H r4,C H r5,C H r k C C
2 5 2 6 2 4 2 6 2 5 2,C2H6 2 CH3 C2H6
CH3• was formed in rxn 1 and consumed in rxn 2:
-r3,C2H4 -k 3 C C 2H5
rCH3 2r1,C2H6 r2,C2H6
•
r4,C2H6 k 4CHCC2H6
H• was formed in rxn 3 and consumed in rxn 4:
2
rH r3,C2H4 r4,C2H6 r5,C2H5 k5 CC2H5
Other methods can also be used - solve the complete set of ODE mole balances
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L10-39
C2H6
k1
2CH3 •
H • C2H6
k4
C2H5 • H2
CH3 • C2H6
k2
CH4 C2H5 •
2C2H5 •
k5
C4H10
C2H5 •
k3
C2H4 H •
1. Mole balances: 2. Rate laws for each species:
(Batch)
dC1
= r1 r1 = -k1C1 - k 2C1C2 - k 4C1C6
C2H6 dt
dC2
CH3• dt = r2 r2 = 2k1C1 - k 2C2C1 All these O.D.Es can be
dC3 r3 k 2C1C2 solved simultaneously
CH4 dt = r3
dC4
C2H5• = r4 r4 = k 2C1C2 - k3C4 + k 4C1C6 - k5C42
dt
dC5
C2H4 dt = r5 r5 = k3C4 The comparisons of the results
dC6
r6 = k3C4 - k 4C1C6 obtained from the two methods are
H• = r6
dt shown on Fig.E7.2 of textbook 3rd Ed.
dC7 The two results are identical,
H2 = r7 r7 = k 4C1C6
dt indicating the validity of the PSSH
C4H10 dC8 = r8 1
r8 = k 5C42 under these conditions
dt 2
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L10-40
Fig.E7.2 of textbook 3rd Ed- Comparison of
Concentration-time Trajectories
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L10-41
Fig.E7.2 of textbook 3rd Ed- Comparison of
Temperature-time Trajectories for Ethylene
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L10-42
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L10-43
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Postulating Mechanism for O3 L10-44
Decomposition (Step 1)
1. If CB appears in the denominator of the rate law, then one elementary rxn
step is probably: reactive intermediate
B A * Collision products
2. If the denominator contains a constant term, then one rxn step is probably:
A * Decomposition products
3. If the numerator contains a species concentration, then one rxn step is
probably:
Cspecies other species? A * other products?
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Postulating Mechanism for O3 L10-46
Now derive a rate equation for the postulated mechanism and check if
it describes the experimentally observed rate equation
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L10-48
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L10-49
rO3 k1CO3 CM k 2CO3 k 1CMCO2 CO
1) Write rO•
2) Rearrange to get CO• in terms of measurable species
3) plug eq for CO• back into –rO3
rO k1CO3 CM k 1CMCO2 CO k 2CO3 CO
CO• is very small, and O• is so reactive that it is consumed as fast as it is
formed, so apply pseudo-steady state hypothesis: rO 0
rO 0 k1CO3 CM CO k 1CMCO2 k 2CO3
Put concentration of reactive intermediate O• in terms of other species
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L10-50
Concentration of Reactive O•
Postulated mechanism: 1.) O M k1 Reactive intermediate,
3 M O2 O must replace CO• in the
k 1
k2 rate equation
2.) O3 O
2O2
kCO3 2CM Observed rate equation
rO3
CO2 CM k 'CO3 (nonelementary)
rO3 k1CO3 CM k 2CO3 k 1CMCO2 CO
rO 0 k1CO3 CM CO k 1CMCO2 k 2CO3 Solve for CO• in terms
of other species
k 1CMCO2 k 2CO3 CO k1CO3 CM
k1CO3 CM
CO Plug CO• into –rO3
k 1CMCO2 k 2CO3
k1CO3 CM
rO3 k1CO3 CM k 2CO3 k 1CMCO2
k 1CMCO2 k 2CO3
Now we will rearrange and simplify to see if it matches the experimental data
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L10-51
rO3
k1CO3 CM k 1CMCO2 k 2CO3
k1k 2CO3 2CM k 1k1CM2CO2 CO3
k 1CMCO2 k 2CO3 k 1CMCO2 k 2CO3
k1k 1CM2CO2 CO3 k1k 2CO3 2CM k1k 2CO3 2CM k 1k1CM2CO2 CO3
rO3
k 1CMCO2 k 2CO3
2k1k 2CO3 2CM Conventional to 2k1k 2 k 1 CO3 2CM
rO3 remove constant rO3
k 1CMCO2 k 2CO3 CMCO2 k 2 k 1 CO3
from 1st term
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L10-52
2k1k 2 k2
k k'
k 1 k 1
These are the same → postulated rate law explains the experimental data
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L10-53
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.