CHE 351:
Reactor Design
Chapter 9a. Reaction Mechanisms
Review: Analysis Methods
 Differential method
 Integral method
 Half-lives method
• Initial rate method
• Differential reactor
• More complex kinetics
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Review: Method of Initial Rates
• When the reaction is reversible, the method of initial rates
can be used to determine the reaction order and the specific
rate constant
• Very little product is initially present, so rate of reverse
reaction is negligible
– A series of experiments is carried out at different initial
concentrations
– Initial rate of reaction is determined for each run
– Initial rate can be found by differentiating the data and
extrapolating to zero time
– By various plotting or numerical analysis techniques relating -rA0
to CA0, we can obtain the appropriate rate law:
rA0  kC A0
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Evaluating the mole balance on a constant V batch reactor at t = 0:
 dCHCl 


  (rHCl )0  kCHCl,0
dt 0

Plot of ln (-rHCl,0) vs ln
CHCl,0 will give reaction
order  & k
 dC 
ln   HCl   lnk   lnCHCl,0
dt 0

CHCl, 0
Initial reaction rate –rHCl,0
(N)
(mol/cm2 s) x 107
1
4
2
0.1
0.5
1.2
2.0
1.36
0.36
0.74
ln (-rHCl,0)
Slope = 
ln (CHCl)
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Review: Analysis Methods
 Differential method
 Integral method
 Half-lives method
 Initial rate method
• Differential reactor
• More Complex Kinetics
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Review: Differential Reactors
• The criterion for a reactor being differential is that the conversion
of the reactants in the bed is extremely small, as is the change in
reactant concentration through the bed
• Reactant concentration through the reactor is essentially
constant (i.e. the reactor is considered to be gradient-less)
• Can treat the mole balance like a CSTR
• Rate of reaction determined for a specified number of predetermined initial or entering reactant concentrations
• Determine rate of reaction as a function concentration or partial
pressure
• Operate isothermally
CA0
CA
CAe
CA0 ~ CA~ CAe
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Review: Differential Catalyst Bed
The rate of reaction per unit mass of catalyst, r’A
flow rate in - flow rate out + rate of generation = rate of accumulation
FA0  FAe  rA W  0
FA0  FAe 0 C A0   C Ae
rA 

W
W
When constant flow rate, 0 = :
0  C A0  C Ae  0 Cp Product
concentration
rA 

W
W
The reaction rate is determined by measuring product concentration, Cp
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Review: More Complex Kinetics
• Carry out batch experiments
• Use optimization software to compute kinetic parameters by least squares
(covered in process control)
# data pts

i1
 rA,measurement i  rA,calc,i 
2
Sum of squares difference between the measured values and calculated values
• Investigate errors by calculating standard deviations of parameters and
looking at magnitudes of ra,meas,i –rA, calc,i to look for outliers (will learn in
process design, this is FYI for this class
• If parameters are sufficiently accurate, then stop. If not, keep repeating
the procedure
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Review: Non-Linear Least-Square Analysis
Now we could use MATLAB to find the values of α and k that would
minimize the sum of squares of differences between the measured (CAm)
and calculated (CAc) concentrations.
That is, for N data points,
Similarly, one can calculate the time at a specified concentration, tc
and compare it with the measured time, tm, at that same concentration.
That is, we find the values of k and α that minimize:
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Review: Non-Linear Least-Square Analysis
Example
Ch7_example_1.m:
clc
clear
t=[0;50;100;150;200;250;300];%Experimental time(min)value
Ca=[0.05;0.038;0.0306;0.0256;0.0222;0.0195;0.0174];% Experimental Ca(mol/dm^3) value
yo=[3;0.1];%Initial guess vector for alpha and k
yfinal= nlinfit(Ca,t,@ODEfun,yo);
function t = ODEfun(yo,Ca)
a=yfinal(1);
a=yo(1);
k=yfinal(2);
k=yo(2);
disp('The final value of alpha and k(dm^3/mol.min) are')
t=(.05^(1-a)-Ca.^(1-a))/(k*(1-a));% Model
disp(a)
The final value of alpha and k(dm^3/mol.min) are
end
disp(k)
2.0447
0.1467
Ch9. Nonelementary Reaction Kinetics
In practice, knowledge of the reaction mechanism helps use to design better
catalyst, trouble shoot, aid in troubleshooting poor reactor performance
• Rate law is typically determined from experimental data
• Goal: Use the experimental rate law to postulate a reaction mechanism
• Elementary: the reaction orders and stoichiometric coefficients are identical
• Nonelementary reaction kinetics: no direct correspondence between
reaction order and stoichiometry
– Nonelementary kinetics
– Pseudo-steady-state hypothesis (PSSH)
– Chain reactions: cracking ethane or polymerizations
– Enzymatic reactions
– Bioreactors
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Determining Mechanism to Describe
Nonelementary Reaction Kinetics
Overall Reaction: 2NO  O2  2NO2
• If the reaction were elementary, the reaction kinetics would follow:
rNO  k1CO CNO 2
2
•
•
•
• Instead experiments show that the kinetics are:
k iCNO 2CO
2
Nonelementary
rNO 
1  k iiCNO
• Nonelementary kinetics are the result of multiple elementary reaction
steps and reactive intermediates (an intermediate that is so reactive it is
consumed as fast as it is formed)
How do we determine the mechanism?
Postulate a reaction mechanism that is a series of elementary reactions
Derive a rate equation for the postulated mechanism
Is the rate equation for the postulated mechanism consistent with the
experimental results?
For example…
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Postulating a Reaction Mechanism
1. If CB appears in the denominator of the rate law, then one elementary rxn
step is probably:
B  A *  Collision products A* is a reactive intermediate
2. If the denominator contains a constant that is not multiplied by a
concentration, then one rxn step is probably:
A *  Decomposition products
3. If the numerator contains a species concentration, then one step is probably:
Cspecies   other species?   A *   other products? 
Apply:
rNO 
k iCNO 2CO
2
1  k iiCNO
Experimentally observed rate equation for
 overall reaction  : 2NO  O 2  2NO 2
CNO in denominator:
NO collides with reactive intermediate, NO3
NO3  NO  2NO2
CNO & CO2 in numerator: NO and O2 produce NO3 in one reaction step
NO  O2  NO3
Constant in denominator: NO3 produces NO and O2 (reverse of previous?)
NO3  NO  O2
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
A Reaction Mechanism for Observed Kinetics?
Apply:
rNO 
k iCNO 2CO
2
1  k iiCNO
Experimentally observed rate equation for
 overall reaction  : 2NO  O 2  2NO 2
CNO in denominator:
NO3  NO  2NO2
CNO & CO2 in numerator:
NO  O2  NO3
Constant in denominator:
NO3  NO  O2
Postulated mechanism:
NO + O2
k1
k-1
Reactive
intermediate
NO3
k
2  2NO
NO3  NO 
2
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.
Postulated Mechanism for Nonelementary
Reaction Kinetics
Nonelementary kinetics, result of
2
rNO 
multiple elementary rxns & active
1  k iiCNO
intermediates
k1
Reactive


Postulated mechanism: NO  O 2 
NO

3 intermediate
k 1
k iCNO 2CO
k
2  2NO
NO3  NO 
2
Write –rNO for the postulated reaction mechanism
rNO   rxns that form NO  - rxns that consume NO 
rNO  k 1CNO  k1CNOCO  k 2CNO CNO
3
2
3
Consumption of NO: rNO  k1CNOCO2  k 1CNO3  k 2CNO3 CNO
(change signs)
• -rNO is in terms of CNO3, which is not measurable species because it is a
reactive intermediate (so reactive it is consumed as fast as it is formed)
• Need to get CNO3 in terms of measurable species and plug into -rNO
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Pseudo-Steady State Hypothesis
k1
Postulated mechanism 1.) NO  O 

 NO 3

2
k 1
k2
2.) NO3  NO 
 2NO 2
Reactive intermediate,
must replace CNO3 in
the rate equation
rNO  k1CNO CO  k 1CNO  k 2 CNO CNO Factor out to simplify
2
3
3
 rNO  k1CNO CO   k 2CNO  k 1  CNO
2
3
1) Write rNO3
2) Rearrange to get CNO3 in terms of measurable species
3) plug eq for CNO3 back into -rNO
rNO  k1CNO CO  k 1CNO  k 2 CNO CNO
3
2
3
3
CNO3 is very small, and NO3 is assumed to be so reactive that it is consumed
rNO  0
as fast as it is formed, so
3
Pseudo-Steady State Hypothesis: Net formation of reactive intermediate ≈0
rNO  0  k1CNO CO  k 1CNO  k 2 CNO CNO
3
2
3
3
Solve for concentration of reactive intermediate NO3 in terms of other species
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Concentration of Reactive Intermediate
k1
Reactive


Postulated mechanism 1.) NO  O 2 
 NO 3 intermediate
k 1
k
2  2NO
2.) NO3  NO 
2
rNO 
k iCNO 2CO
Observed rate equation
(nonelementary)
2
1  k iiCNO
rNO  k1CNO CO   k 2CNO  k 1  CNO
2
rNO  0  k1CNO CO  k 1CNO  k 2 CNO CNO
3
2
3
3
3
Solve for CNO3 in
terms of other species
 k 1CNO  k 2 CNO CNO  k1CNO CO  CNO  k 1  k 2CNO   k1CNO CO
3
3
2
3
2
 CNO 
3
k1CNO CO
2
k 1  k 2CNO
 rNO  k1CNO CO   k 2 CNO  k 1 
2
Plug CNO3 into -rNO
k1CNO CO
2
k 1  k 2 CNO
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.
Rearranging the Postulated Rate Eq.
k1
Reactive
Postulated mechanism 1.) NO  O 

NO

2 
3 intermediate
k 1
k
2  2NO
2.) NO3  NO 
2
Plug in CNO3  rNO  k1CNO CO2   k 2CNO  k 1 
k1CNO CO
2
k 1  k 2 CNO

k 2CNO  k 1  Common
Factor out  r

k
C
C
1

 denominator
NO
1 NO O2 
k1CNO CO
k

k
C



1
2
NO
2


 k 1  k 2CNO k 2CNO  k 1 
 rNO  k1CNOCO 

 Add fraction
2
 k 1  k 2CNO k 1  k 2CNO 


2k1k 2CNO 2CO
 2k 2CNO  Multiply
2
 rNO  k1CNOCO 
 rNO 

2
k 1  k 2CNO
 k 1  k 2CNO 


 2k1k 2 k 1  CNO 2CO2
 rNO 
1   k 2 k 1  CNO
Conventional to reduce
the additive constant in
the numerator to 1
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Comparison Between Postulated and
Experimental Rate Equation
k1
Reactive
Postulated mechanism 1.) NO  O 

NO

2 
3 intermediate
k 1
k
2  2NO
2.) NO3  NO 
2
Compare rate eq for postulated mechanism to the experimental rate eq
 2k1k 2 k 1  CNO 2CO2
 rNO 
1   k 2 k 1  CNO
Rate equation for
postulated mechanism
rNO 
k iCNO 2CO
2
1  k iiCNO
Experimentally observed
rate equation
Clicker Q: Are these rate equations the same?
a) Yes
2k1k 2
k2
 ki
 k ii
b) No
k 1
k 1
Yes, 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.
Ozone (O3 ) Example Problem
Overall Reaction for Ozone Decomposition: 2O3  3O2
When ozone decomposes in the presence of an inert gas, M, the
following kinetics are observed:
rO 
3
kCO 2 CM
3
CO CM  k 'CO
2
3
Postulate a reaction mechanism that is consistent with this rate law
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Postulating Mechanism for O3 Decomposition
(Step 1)
1. If CB appears in the denominator of the rate law, then one elementary rxn
reactive intermediate
step is probably:
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? 
Apply: r
O3 
kCO 2 CM
3
CO CM  k 'CO
2
3
Experimentally observed rate equation for
 overall reaction  : 2O3  3O2
CO2, CM & CO3 in denominator, so O2, M, and O3 must each collide a with
reactive intermediate. What is the reactive intermediate?
Since one oxygen atom is lost from the ozone molecule (O3), oxygen
radicals (O•) are likely the reactive intermediate:
O2  O  O3 (Possible Rxn 1) O3  O  2O2 (PR2)
M  O  M  ? (PR3)
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Postulating Mechanism for O3 Decomposition
(Step 1 continued)
1. If CB appears in the denominator of the rate law, then one elementary rxn
reactive intermediate
step is probably:
B  A *  Collision products
Apply: r
O3 
kCO 2 CM
3
CO CM  k 'CO
2
3
Experimentally observed rate equation for
 overall reaction  : 2O3  3O2
CO2, CM & CO3 in denominator; they must collide with reactive intermediate O•
O2  O  O3 (possible rxn 1) O3  O  2O2 (PR2)
M  O  M  ? (PR3)
Wait, M is inert, so it cannot react with O• to create a new chemical species
• An inert molecule can provide kinetic energy in another reaction
Which one?
→ M must participate in one of the other reactions
CM is multiplied by CO2 in the denominator, so combine PR3 with PR1
M  O2  O  O3  M (PR1b)
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Postulating Mechanism for O3 Decomposition
(Steps 2 & 3)
1. If CB appears in the denominator of the rate law, then one elementary rxn
reactive intermediate
step is probably:
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? 
Apply: r
O3 
kCO 2 CM
3
CO CM  k 'CO
2
3
Experimentally observed rate equation for
 overall reaction  : 2O3  3O2
M  O2  O  O3  M (PR1b)
O3  O  2O2 (PR2)
Denominator doesn’t contain a constant, so step 2 isn’t in our mechanism
Both CO3 and CM appear in the numerator, so one reaction step may be:
O3  M  M  O2  O  (PR4)
Note that PR4 is the reverse of PR1
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Postulated Mechanism for Ozone
Decomposition
Apply: r
O3 
kCO 2 CM
Experimentally observed rate equation for
 overall reaction  : 2O3  3O2
3
CO CM  k 'CO
2
CM and CO2 in denominator:
CO3 in denominator:
CO3 and CM in numerator:
3
M  O2  O  O3  M
O3  O  2O2
Reactive
O3  M  M  O2  O  intermediate
Postulated mechanism: O3 + M
k1
k-1
M + O2 + O●
k
2  2O
O3  O  
2
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.
Postulated Mechanism for Nonelementary
Reaction Kinetics
kCO 2 CM
Nonelementary kinetics, result of
3
rO 
multiple elementary rxns & active
3
CO CM  k 'CO
2
3
intermediates
k1
Reactive
M + O2 + O●
Postulated mechanism: O3 + M
intermediate
k-1
k
2  2O
O3  O  
2
Write –rO3 for the postulated reaction mechanism
rO   rxns that consume O3  - rxns that form O 3 
3
rO  k1CO CM  k 1CMCO CO  k 2 CO CO
3
3
2
3
• -rO3 is in terms of CO•, which is not measurable species because it is a
reactive intermediate (so reactive it is consumed as fast as it is formed)
• Need to get CO• in terms of measurable species and plug into –rO3
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Pseudo-Steady State Hypothesis
Postulated mechanism: 1.) O3 + M
k1
Reactive intermediate,
must replace CO• in the
rate equation
M + O2 + O●
k-1
k2
2.) O3  O  
 2O 2
rO  k1CO CM  k 1CMCO CO  k 2CO CO
3
3
2
3
Factor out to simplify


 rO  k1CO CM  k 2 CO  k 1CMCO CO
3
3
3
2
1) Write rO•
2) Rearrange to get CO• in terms of measurable species
3) plug eq for CO• back into –rO3
rO  k1CO CM  k 1CMCO CO  k 2CO CO
3
2
3
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  k1CO CM  CO k 1CMCO  k 2 CO
3
2
3

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.
Concentration of Reactive O•
k1
Postulated mechanism: 1.) O3 + M
Reactive intermediate,
must replace CO• in the
rate equation
M + O2 + O●
k-1
k2
2.) O3  O  
 2O 2
rO 
3
kCO 2 CM
Observed rate equation
(nonelementary)
3
CO CM  k 'CO
2

3

rO  k1CO CM  k 2CO  k 1CMCO CO
3
3
3
2

rO  0  k1CO CM  CO k 1CMCO  k 2 CO
3
2
3

for C in terms
 Solve
of other species
O•

 k 1CMCO  k 2 CO CO  k1CO CM
2
3
3
 CO  
k1CO CM
3
k 1CMCO  k 2 CO
2

 rO  k1CO CM  k 2 CO  k 1CMCO
3
3
3
Plug CO• into –rO3
3
k1CO CM
2
k C C
3
 1 M O 2  k 2 CO 3
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.
Rearranging the Postulated Rate Eq.
k1
Postulated mechanism: 1.) O3 + M
M + O2 + O●
k-1
k2
2.) O3  O  
 2O 2
Reactive intermediate,
must replace CO• in the
rate equation


k1CO CM
3

Plug in CO•  rO3  k1CO3 CM  k 2CO3  k 1CMCO2 
 k 1CMCO  k 2CO 

2
3 

Multiply out
 rO  k1CO CM 
3
3
brackets
 rO 
3


k1k 2 CO 2 CM  k 1k1CM2CO CO
3
k 1CMCO  k 2 CO
2
3
k1CO CM k 1CMCO  k 2 CO
3
2
k 1CMCO  k 2CO
2
2
3
3
Get common
denominator
  k1k 2CO 2CM  k 1k1CM2CO CO
3
2
k 1CMCO  k 2 CO
3
2
3
3
k1k 1CM2 CO CO  k1k 2 CO 2 CM  k1k 2 CO 2 CM  k 1k1CM2CO CO
2
3
3
3
2
3
 rO 
3
k 1CMCO  k 2 CO
2
2k1k 2 CO 2 CM
3
Conventional to
2k1k 2 k 1  CO 2 CM

3
 rO 


r

remove
constant
3
O
k 1CMCO  k 2 CO
3
CMCO   k 2 k 1  CO
2
3 from 1st term
2
3
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
3
Comparison Between Postulated and
Experimental Rate Equation
Postulated mechanism: 1.) O3 + M
k1
k-1
k2
Reactive intermediate,
cannot appear in the
rate equation
M + O2 + O●
2.) O3  O  
 2O 2
Compare rate eq for postulated mechanism to the experimental rate eq
 2k1k 2 k 1  CO3 2CM
rO 
3
CMCO   k 2 k 1  CO
2
3
Rate equation for
postulated mechanism
2k1k 2
k
k 1
rO 
3
kCO 2 CM
3
CO CM  k 'CO
2
3
Experimentally observed
rate equation
k2
 k'
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.