L9b-1
Review: Analysis of Rate Data
Goal: determine reaction order, a, and specific reaction rate constant, k
• Data collection is done in the lab so we can simplify BMB, stoichiometry,
and fluid dynamic considerations
• Want ideal conditions → well-mixed (data is easiest to interpret)
• Constant-volume batch reactor for homogeneous reactions: make
concentration vs time measurements during unsteady-state operation
• Differential reactor for solid-fluid reactions: monitor product concentration
for different feed conditions during steady state operation







Method of Excess
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.
L9b-2
Review: Method of Half-lives
Half-life of a reaction (t1/2): time it takes
for the concentration of the reactant to
drop to half of its initial value
ln (t1/2)
Slope = 1- a
A  products  dCA  kC a
A
a
dt
rA  kCA
 1
1
1 
t
 a 1 

a

1
k a  1  CA
CA0

1
CA  CA0 at t = t1 2
2
2a 1  1  1 
t1 2 


k a  1  CA0a 1 
ln CA0
Plot ln(t1/2) vs ln CA0. Get a straight
line with a slope of 1-α
 
ln t1 2
2a 1  1
 ln
 1  a  lnCA0
k a  1
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L9b-3
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  kCA0a
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L9b-4
Review: Differential Catalyst Bed
Conversion of reactants
& change in reactant
concentration in the bed
is extremely small
L
r’A: rate of reaction per unit mass of catalyst
flow in - flow out + rate of gen = rate of accum.
FA0  FAe  rA W  0
FA0  FAe 0 CA0   CAe
rA 

W
W
FAe
FA0
CA0
Cp
Fp
W
When constant flow rate, 0 = :
0  CA0  CAe  0Cp
rA 

W
W
Product
concentration
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.
L9b-5
Review: Multiple Rxns & Selectivity
1) Parallel / competing rxns
k1
B
k1
A
Desired product
2) Series rxns
B
A
k2
C
3) Complex rxns
instantaneous rate selectivity, SD/U
rate of formation of D rD
SD U 

rate of formation of U rU
A+B
k1
k2
C
C+D A+C
k2
E
instantaneous yield, YD
(at any point or time in reactor)
r
rate of formation of D
YD 
 D
rate of consumption of A rA
overall rate selectivity, S
DU
ND
Final moles of desired product
SD U 

NU Final moles of undesired product
F
Exit molar flow rate of desired product
SD U  D 
FU Exit molar flow rate of undesired product
overall yield, Y
D
FD
flow YD 
FA0  FA
at
exit
ND
batch Y 
D
NA0  NA
at
tfinal
Maximize selectivity / yield to maximize production of desired product
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Review: Maximizing SD/U for
Parallel Rxns
SD U 
ED EU 
AD
e RT
AU

L9b-6

What reactor conditions and
a1a2
1 2
CA
CB
configuration maximize selectivity?
Specific rate of desired reaction kD increases:
a) If ED > EU
b) If ED < EU
more rapidly with increasing T
less rapidly with increasing T
Use lower temperature(not so low
Use higher temperature
that the reaction rate is tiny)
To favor production of the desired product
Now evaluate concentration:
a) a1  a2  a1  a2  0
b) a1  a2  a1  a2  0
→ Use large CA
→ Use small CA
c) 1  2  1  2  0
d) 1  2  1  2  0
→ Use large CB
→ Use small CB
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Concentration Requirements &
Reactor
Selection
k
D
D
A+B
kU
How do concentration requirements play
into reactor selection?
CA00
CB00
U
PFR
PFR (or PBR): concentration is
high at the inlet & progressively
drops to the outlet concentration
CA(t)
CB(t)
L9b-7
Batch:
concentration is
high at t=0 &
progressively drops
with increasing time
CB0
CA
CA0
CB0
CSTR:
concentration is
always at its
lowest value
(that at outlet)
Semi-batch: concentration
of one reactant (A as
shown) is high at t=0 &
progressively drops with
increasing time, whereas
concentration of B can be
kept low at all times
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
kD
A+B
kU
L9b-8
D
High CA favors undesired
High
C
favors
desired
A
a1  a2
a1  a2
product formation
product formation
(keep CA low)
U
1  2
High CB
favors
desired
product
formation
1  2
Batch reactor
When CA & CB are low (end time
or position), all rxns will be slow
PFR/PBR
High P for gas-phase rxn, do not
add inert gas (dilutes reactants)
PFR/PBR
Side streams feed low CA
CA
Semi-batch
reactor slowly feed
←High CB
A to large amt of B
CA
CA
CA
CSTRs in
series
CA00
CB00
CA0
CB0
CSTR
PFR/PBR w/ side streams feeding
High CB
low CB
CB
favors
Semi-batch
←High CA
undesired reactor, slowly
PFR/PBR
PFR/PBR
product
w/ high
feed B to large amount of A
recycle
CB
CB
formation
CB
• Dilute feed with inerts that are
(keep CB CSTRs in
series
easily separated from product
low)
B consumed
before&leaving
CSTR
• Low
P if gas
phaseUrbana-Champaign.
Slides courtesy of Prof
M L Kraft, Chemical
Biomolecular
Engr
University
of Illinois,
n Dept,
L9b-9
Different Types of Selectivity
instantaneous rate selectivity, SD/U
SD U 
rate of formation of D rD

rate of formation of U rU
overall rate selectivity, SD U
F
Exit molar flow rate of desired product
SD U  D 
FU Exit molar flow rate of undesired product
N
Final moles of desired product
SD U  D 
NU Final moles of undesired product
instantaneous yield, YD
(at any point or time in reactor)
YD 
r
rate of formation of D
 D
rate of consumption of A rA
overall yield, YD
FD
flow YD  F  F
A0
A
Evaluated
at outlet
batch YD 
ND
NA0  NA
Evaluated
at tfinal
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L9b-10
Series (Consecutive) Reactions
A
k1
k2
U
D
(desired) (undesired)
Spacetime t for a flow reactor
Time is the key factor here!!!
Real time t for a batch reactor
To maximize the production of D, use:
CSTRs in series
Batch
or
PFR/PBR
or
n
and carefully select the time (batch) or spacetime (flow)
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L9b-11
Concentrations in Series Reactions
A
k1
B
k2
C
-rA = k1CA
rB,net = k1CA – k2CB
How does CA depend on t?
dFA
dC A
 k1C A  0
 k1C A  C A  C A0e k1t
dV
dV
How does CB depend on t?
dFB
dCB
 k1C A  k 2CB  0
 k1 C A0e k1t  k 2CB
dV
dV






Substitute
dCB
dCB
 k1 CA0ek1t  k 2CB 
 k 2CB  k1 CA0ek1t
dt
dt
Use integrating
factor (reviewed
on Compass)


d CBek 2t
dt
 k C
1
V
0
t

 ek1t  ek 2t
 k 2 k1t  CB  k1CA0 
e
 k 2  k1
A0




CC  CA0  CA  CB
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L9b-12
Reactions in Series: Cj & Yield
CA  CA0ek1t
B
C
A
 ek1t  ek 2t
CB  k1CA0 
 k 2  k1




CC  CA0  CA  CB
topt
The reactor V (for a given 0) and t that maximizes CB occurs when dCB/dt=0


dCB  k1CA0 
k1t
k 2t


k
e

k
e
0
1
2

dt
 k 2  k1 
t opt 
k
1
ln 1
k1  k 2 k 2
V
 t so Vopt  0t opt
0
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
What reactor/reactors scheme and conditions would you use to maximize L9b-13
the selectivity parameters for the following parallel reaction?
A+C
A+C
kD
kU1
D desired
U1 undesired
Need to maximize SD/U1
SD U1 
rD

rU1
Plug in
800
S

numbers: D U1 10 e
E/R
300
2000
rU1  10e T CA CC
rD  800e T CA 0.5 CC
E
k T  AeRT
 ED EU
1
aD aU 
D  U
AD

T
1
1
e
 CA
 CC
AU1



 2000300 
T

 
CA0.51 CC11
1700
 SD U1  80e T CA 0.5
To maximize the production of the desired product, the temperature should be
a) As high as possible (without decomposing the reactant or product)
b) Neither very high or very low
ED > EU, so use higher T
c) As low as possible (but not so low the rate = 0)
d) Doesn’t matter, T doesn’t affect the selectivity
e) Not enough info to answer the question
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
What reactor/reactors scheme and conditions would you use to maximize L9b-14
the selectivity parameters for the following parallel reaction?
A+C
A+C
kD
kU1
D desired
U1 undesired
rD 
Need to maximize SD/U1
2000
800e T CA 0.5 CC
rU1
k T 
E
AeRT

 ED EU
1
AD
T
e
r
SD U1  D 
rU1 AU
1
Plug in
800
S

numbers: D U1 10 e
a)
b)
c)
d)
e)
 2000300 
T

300
 10e T C
A CC
 C aD aU1  C D  U1
 A
 C


CA0.51 CC11
1700
 SD U1  80e T CA 0.5
To maximize the production of the desired product, CA should be
As high as possible
αD < αU1, so high CA favors undesired
Neither very high or very low
product formation (keep CA low)
As low as possible
Doesn’t matter, CA doesn’t affect the selectivity
Not enough info to answer the question
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L9b-15
What reactor/reactors scheme and conditions would you use to maximize
the selectivity parameters for the following parallel reaction?
A+C
A+C
kD
kU1
D desired
U1 undesired
2000
rD  800e T CA 0.5CC
Need to maximize SD/U1

 ED EU
1
AD
T
e
r
SD U1  D 
rU1 AU
1
Plug in
800
S

numbers: D U1 10 e
 2000300 
T

rU1
300
 10e T C
A CC
 C aD aU1  C D  U1
 A
 C


CA0.51 CC11
1700
 SD U1  80e T CA 0.5
• Since ED>EU1, kD increases faster than kU1 as the temperature increases
• Operate at a high temperature to maximize CD with respect to CU1
• aD<aU1, keep CA low to maximize CD with respect to CU1
• rD and rU1 are 1st order in CC, so changing CC does not influence selectivity
• HOWEVER, high CC will increase the reaction rate and offset the slow
reaction rate that is caused by low CA (that’s a good thing)
What reactor should we use?
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L9b-16
What reactor/reactors scheme and conditions would you use to maximize
the selectivity parameters for the following parallel reaction?
kD
300
2000
A+C
D desired
rU1  10e T CA CC
rD  800e T CA 0.5CC
kU1
A+C
U1 undesired
Need to maximize SD/U1
1700
SD U1  80e T CA 0.5
• ED>EU1, operate at a high temperature to maximize CD with respect to CU1
• aD<aU1, keep CA low to maximize CD with respect to CU1
• rD and rU1 are 1st order in CC, so changing CC does not influence selectivity
• HOWEVER, high CC will increase the reaction rate and offset the slow
reaction rate that is caused by low CA (that’s a good thing)
What reactor should we use?
C
PFR
A
PFR/PBR w/ side streams feeding low CA
CA
Semi-batch reactor
slowly feed A to large
amount of C
←High CC
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
How does the selection of reactor/reactors scheme and conditions
change if D can react with C and form another undesired product?
A+C
kD
D desired
2000
rD  800e T CA 0.5CC
A+C
rU1
kU1
U1 undesired
300
 10e T C
Need to maximize SD/U1 and SD/U2
SD U2 
rD

rU2
A CC
D+C
kU2
L9b-17
U2 undesired
8000
rU2  106 e T CCCD
• ED>EU1, operate at a high T
1700
• aD<aU1, keep CA low

0.5
SD U1  80e T CA
• High CC increases rxn rate &
offsets slow rxn from low CA
2000
6000
800e T C A 0.5CC
4
T C 0.5C 1

S

8

10
e
D U2
A
D
8000
106 e T CCCD
• Since ED<EU21, kD increases slower than kU2 as T increases ⇨ operate at low
T to maximize CD Conflicts with maximizing SD/U1!
• aD>aU2, keep CA high to maximize CD
Conflicts with maximizing SD/U1!
• rD, rU1 & rU2 are all 1st order in CC, so changing CC does not influence
selectivity, but high CC will offset the rate decrease due to low CA
• Low CD reduces the production of U2 Conflicts with producing the product D!!!
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
A+C
kD
D desired
1700
SD U1  80e T CA 0.5
A+C
kU1
U1 undesired
Maximize SD/U1 & SD/U2
D+C
kU2
L9b-18
U2 undesired
6000
SD U2  8  104 e T CA 0.5CD1
• ED>EU1, operate at a high T
• ED<EU2, operate at low T
• aD<aU1, keep CA low
• aD>aU2, keep CA high
• Want to maximize CD
• Low CD reduces production of U2
• High CC increases rxn rate & offsets slow rate caused by low CA
Consider relative magnitude of SD/U1 and DD/U2 as a function of position in PFR
PFR w/ side streams feeding low CA
C
PFR, high T
PFR 2, low T
A
High T, CC is initially high, CA is low
→ high SD/U1
Initially CD=0 → rU2=0. Both gradually
increase down reactor
Initially high SD/U2 (because CD is low),
but SD/U2 gradually decreases down
reactor
• At some distance down the reactor,
significant amounts of D have formed
• SD/U2 becomes significant with respect
to SD/U1
• At this point, want low T, high CA &
low CC
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L9b-19
If a CSTR were used with CA = 1 mol/L and CD= 1 mol/L, at what
temperature should the reactor be operated?
A+C
kD
D desired
A+C
2000
rD  800e T CA 0.5CC
rU1
kU1
300
 10e T C
Need to maximize SD/(U1+U2)
rD
SD U U  
1 2
rU1  rU2
S
D
U1U2

2000
800
e T
10
 300 

10  T 
e
10
1
U1 undesired
8000
rU2  106 e T CCCD
A CC
0.5
8000
106

e T
U2 undesired
2000
800e T CA 0.5CC

 300 
8000


10e  T CA CC  106 e T CCCD
1
10
D+C
kU2
S
1
D
U1U2

CA=1
CD=1
2000
80e T
 300 


T


e
8000
 105 e T
Plot SD/(U1+U2) vs temperature to find the temperature that maximizes SD/(U1+U2)
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L9b-20
If a CSTR were used with CA = 1 mol/L and CD= 1 mol/L, at what
temperature should the reactor be operated?
A+C
kD
D desired
A+C
2000
rD  800e T CA 0.5CC
rU1
kU1
U1 undesired
300
 10e T C
Need to maximize SD/(U1+U2)
4
3.5
S
3
SD/(U1+U2)
2.5
D
U1U2

D+C
kU2
U2 undesired
8000
rU2  106 e T CCCD
A CC
600K
2000
80e T
 300 


T


e
8000
 105 e T
2
1.5
1
0.5
0
0
200
400
600
Temperature (K)
800
1000
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Calculate the yield of forming B in a CSTR and PFR when the conversion L9b-21
of A is 90% and CA0 = 4 mol/L. The following reactions occur in the reactor:
mol
kB
k
r

k

2
A
B B
A C C rC  kCCA kC  1 min1
B
L  min
What is the expression for the yield of B for a CSTR?
CB0
CB
FB
Y


Y

YB 
B
B
(overall yield)
CA00  CA0
CA0  CA
FA0  FA
We know CA0 and CA when XA=0.9. How do we get CB?
In - Out + Gen. = Accum.
C 
dNB
FB0  FB  rB V 
 FB  rB V  0  rB  B 0
V
dt
0
0
mol
CB
mol
CB
2
t  CB
 rB  2

 rB 
L

min
L  min t
t
Use the mole balance on A to find t (at 90% conversion)
In - Out + Gen. = Accum.
dNA
 CA00  CA0  rA V
FA0  FA  rA V 
dt 0
CA0  CA  rA
V
0
 CA0  CA  rAt 
CA0  CA
t
rA
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Calculate the yield of forming B in a CSTR and PFR when the conversion L9b-22
of A is 90% and CA0 = 4 mol/L. The following reactions occur in the reactor:
mol
kB
k
r

k

2
A
B B
A C C rC  kCCA kC  1 min1
B
L  min
YB 
CB
CA0  CA
mol
2
t  CB
L  min
CA0  CA
t
rA
What is –rA?
mol
1


r

2

CA
 rA  kB  kCCA
A
L  min min
CA0  CA
t 
t 
rA
CA0  CA
mol
1
2

CA
L  min min
rA  rB  rC
Plug -rA back into
expression for t
CA0 = 4 mol/L, and at
XA=0.9, CA= 0.4 mol/L
mol
mol
 0.4
L
L
t 
 t  1.5 min Residence time for XA = 0.9
mol
1 
mol 
2

 0.4

L  min min 
L 
mol
2
1.5min 
rBt
YB 
 YB  L  min
 YB  0.83
CA 0  CA
mol
mol
4
 0.4
L
L
4
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Calculate the yield of forming B in a CSTR and PFR when the conversion L9b-23
of A is 90% and CA0 = 4 mol/L. The following reactions occur in the reactor:
kC
mol
kB
1
r

k
C
A
C
r

k

2
k

1
min
A
B B
C
C
A
B
C
L  min
mol
1

CA
 rA  kB  kCCA  rA  2
L  min min
What is the expression for the yield of B for a PFR?
CB0
CB
FB
(overall
yield)
Y


Y

YB 
B
B
CA00  CA0
CA0  CA
FA0  FA
Use the mass balance to get CB
dCB
mol
dCB0
dCB


2

 rB

 rB
dt
L  min
dV
dt
CB
mol t
mol
mol
  dCB  2
t
t  0   CB  2
 dt  CB  CB0  2

L

min
L  min
L  min
CB0
0
0
dFB
 rB
dV
Use the mole balance on A to find t (at 90% conversion)
dCA0

 rA
dV
dCA
1
dCA
 mol




2

C

 rA

A
dt
dt
 L  min min

CA
dCA
1 t
dCA
 mol
 1
 

 dt

 2
 CA 
dt
CA0   2mol L  C A  min 0
 L
 min
dFA
 rA
dV
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Calculate the yield of forming B in a CSTR and PFR when the conversion L9b-24
of A is 90% and CA0 = 4 mol/L. The following reactions occur in the reactor:
mol
kB
kC
r

k

2
A
B B
A
C rC  kCCA kC  1 min1
B
L  min
mol
1


r

2

CA
 rA  kB  kCCA
A
L  min min
CB
mol
YB 
C

2
t
Use mole balance on A to find t (at XA = 0.9)
B
CA0  CA
L  min
 mol


2

C
CA

A0 
dCA
1 t
L

  1 t 0
 

d
t

ln



mol
 min 0
 mol
 min
CA0   2

C

2

C

A

A
 L

 L

mol 
 mol
2

4


L
L

  1 t
CA0 = 4 mol/L  ln
 0.92 min  t
mo
l
mol

 min
CA = 0.4 mol/L
2

0.
4


L 
 L
Yield was better
mol
2
0.92min
in the CSTR, but
CB
rBt
L

min
YB 

 YB 
 YB  0.51 the residence
mol
mol
CA0  CA CA0  CA
4
 0.4
time was longer
L
L
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.