L9b Selectivity example problems.pptx

L9b-1
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Review: Analysis of Rate Data
• 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
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)
 Method of Excess
 Differential method
 Integral method
 Half-lives method
 Initial rate method
 Differential reactor
 More complex kinetics

L9b-2
A
A
dC
kC
dt
a
 
A  products
  1 1
A A0
1 1 1
t
k 1 C C
a a
a  
 
 
 
 
  
A A0 1 2
1
C C at t = t
2

 
1
1 2 1
A0
2 1 1
t
k 1 C
a
a
a


 

  
 
  
   
 
1
1 2 A0
2 1
ln t ln 1 lnC
k 1
a
a
a


  

ln (t1/2)
ln CA0
Slope = 1- a
A A
r kC a
 
Plot ln(t1/2) vs ln CA0. Get a straight
line with a slope of 1-α
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

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:
A0 A0
r kC a
 

L9b-4
Conversion of reactants
& change in reactant
concentration in the bed
is extremely small
Review: Differential Catalyst Bed
FA0
Cp
CA0
FAe
Fp
W
L
r’A: rate of reaction per unit mass of catalyst
flow in - flow out + rate of gen = rate of accum.
A0 Ae A
F F r W 0

   
A0 Ae 0 A0 Ae
A
F F C C
r
W W
 
 

  
 
When constant flow rate, 0 = :
  0 p
0 A0 Ae
A
C
C C
r
W W

 

  
 
Product
concentration
The reaction rate is determined by measuring
product concentration, Cp

L9b-5
Review: Multiple Rxns & Selectivity
A B C
k1 k2
A
C
B
k2
k1
2) Series rxns
3) Complex rxns
1) Parallel / competing rxns
Desired product
A+B C+D A+C E
k1 k2
D
D U
U
F molar flow rate of desired product
S
F molar flow rate of undesir
Exit
E ed pr
x oduct
it
 
D
D U
U
N moles of desired product
S
N moles of undes
Fi
ir
nal
Fi ed pr u
nal od ct
 
instantaneous rate selectivity, SD/U
D
D U
U
r
rate of formation of D
S
rate of formation of U r
 
overall rate selectivity,
D U
S
D
D
A
r
Y
rate of consumption of A r
 

instantaneous yield, YD
(at any point or time in reactor)
overall yield,
D
Y
D
D
A0 A
F
Y
F F


flow
D
D
A0 A
N
Y
N N


batch
at
exit
at
tfinal
Maximize selectivity / yield to maximize production of desired product

L9b-6
Review: Maximizing SD/U for
Parallel Rxns
 
 
E E
D U
D RT 1 2 1 2
D U A B
U
A
S e C C
A
a a  
 
 

What reactor conditions and
configuration maximize selectivity?
a) If ED > EU
Specific rate of desired reaction kD increases:
Use higher temperature
b) If ED < EU
less rapidly with increasing T
Use lower temperature(not so low
that the reaction rate is tiny)
more rapidly with increasing T
To favor production of the desired product
1 2 1 2
a) 0
a a a a
   
Now evaluate concentration:
→ Use large CA
1 2 1 2
b) 0
a a a a
   
→ Use small CA
1 2 1 2
c) 0
   
   
→ Use large CB
1 2 1 2
d) 0
   
   
→ Use small CB

L9b-7
Concentration Requirements &
Reactor Selection
CA00
CB00
CA0
CB0
How do concentration requirements play
into reactor selection?
CSTR:
concentration is
always at its
lowest value
(that at outlet)
PFR
PFR (or PBR): concentration is
high at the inlet & progressively
drops to the outlet concentration
CA(t)
CB(t)
D
kD
A+B
U
kU
Batch:
concentration is
high at t=0 &
progressively drops
with increasing time
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
CB0
CA

L9b-8
D
kD
A+B
U
kU
1  2
High CB
favors
desired
product
formation
1  2
High CB
favors
undesired
product
formation
(keep CB
low)
a1  a2
High CA favors desired
product formation
a1  a2
High CA favors undesired
product formation
(keep CA low)
PFR/PBR
Batch reactor
When CA & CB are low (end time
or position), all rxns will be slow
High P for gas-phase rxn, do not
add inert gas (dilutes reactants)
PFR/PBR w/ side streams feeding
low CB CB
←High CA
Semi-batch
reactor, slowly
feed B to large amount of A
CB
CB CB
CSTRs in
series
B consumed before leaving CSTRn
CA00
CB00
CA0
CB0
CSTR
PFR/PBR
PFR/PBR
w/ high
recycle
• Dilute feed with inerts that are
easily separated from product
• Low P if gas phase
PFR/PBR
Side streams feed low CA
←High CB
CA
Semi-batch
reactor slowly feed
A to large amt of B
CA
CA CA
CSTRs in
series

L9b-9
Different Types of Selectivity
instantaneous rate selectivity, SD/U
D
D U
U
r
S
rate of formation of U r
 
D
D U
U
F molar flow rate of desired product
S
F molar flow rate of undesir
Exit
E ed pr
x oduct
it
 
overall rate selectivity, D U
S
D
D U
U
N moles of desired product
S
N moles of undes
Fi
ir
nal
Fi ed pr u
nal od ct
 
D
D
A
r
Y
rate of consumption of A r
 

instantaneous yield, YD
(at any point or time in reactor)
overall yield, D
Y
D
D
A0 A
F
Y
F F


flow
D
D
A0 A
N
Y
N N


batch
Evaluated
at outlet
Evaluated
at tfinal

L9b-10
Series (Consecutive) Reactions
(desired) (undesired)
A D U
k1 k2
Time is the key factor here!!!
Spacetime t for a flow reactor Real time t for a batch reactor
To maximize the production of D, use:
Batch
or PFR/PBR or
n
CSTRs in series
and carefully select the time (batch) or spacetime (flow)

L9b-11
Concentrations in Series Reactions
A B C
k1 k2
-rA = k1CA
rB,net = k1CA – k2CB
k1
A A
1 A
A
0
0
A 1
A
dF dC
dV dV
C C e
k C k C 
       t

How does CA depend on t?
How does CB depend on t?
1 A
B
2 B
k C
dF
C
d
k
V
   
k1
A0
1
B
2 B
0 C e
k
dC
d
k C
V

  
t

 
k
B 1
1 A0 2 B
dC
k C e k C
d
t
t

  
Substitute
 
k
B 1
2 B 1 A0
dC
k C k C e
d
t
t

  
Use integrating
factor (reviewed
on Compass)
   
k2
B k k
2 1
1 A0
d C e
k C e
d
t
t
t

 
k k
1 2
B 1 A0
2 1
e e
C k C
k k
t t
 
 

   
 

 
0
V
t


C A0 A B
C C C C
  

L9b-12
The reactor V (for a given 0) and t that maximizes CB occurs when dCB/dt=0
 
k k
1 A0
B 1 2
1 2
2 1
k C
dC
k e k e 0
d k k
t t
t
 
 
   
 

 
1
opt
1 2 2
k
1
ln
k k k
t 

0
V
t

 so opt 0 opt
V  t

A
B
C
k1
A A0
C C e t


k k
1 2
B 1 A0
2 1
e e
C k C
k k
t t
 
 

  
 

 
C A0 A B
C C C C
  
topt
Reactions in Series: Cj & Yield

L9b-13
 
E E
D U1
D U D U
D T 1 1
A C
U1
A
e C C
A
a a  
 
 
 
  
 
What reactor/reactors scheme and conditions would you use to maximize
the selectivity parameters for the following parallel reaction?


2000
T 0.5
D A C
r C
00e C
8


300
T
U A C
1
10
r C C
e
D
D U1
U1
r
S
r

Need to maximize SD/U1
 
 
2000 300
0.5 1 1 1
T
D U A C
1
800
S e C C
10
 
 

Plug in
numbers:
1700
0.5
T
D U A
1
S 80e C


 
A+C D desired
kD
A+C U1 undesired
kU1
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
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
 


E
RT
k T Ae
E/R
ED > EU, so use higher T

L9b-14
 
E E
D U1
D U D U
D T 1 1
A C
U1
A
e C C
A
a a  
 
 
 
  
 


2000
T 0.5
D A C
r C
00e C
8


300
T
U A C
1
10
r C C
e
D
D U1
U1
r
S
r

 
 
2000 300
0.5 1 1 1
T
D U A C
1
800
S e C C
10
 
 

Plug in
numbers:
1700
0.5
T
D U A
1
S 80e C


 
A+C D desired
kD
A+C U1 undesired
kU1
To maximize the production of the desired product, CA should be
a) As high as possible
b) Neither very high or very low
c) As low as possible
d) Doesn’t matter, CA doesn’t affect the selectivity
e) Not enough info to answer the question
 


E
RT
k T Ae
αD < αU1, so high CA favors undesired
product formation (keep CA low)

L9b-15
 
E E
D U1
D U D U
D T 1 1
A C
U1
A
e C C
A
a a  
 
 
 
  
 
2000
0.5
T
D A C
r 800e C C


300
T
U A C
1
r 10e C C


D
D U1
U1
r
S
r

 
 
2000 300
0.5 1 1 1
T
D U A C
1
800
S e C C
10
 
 

Plug in
numbers:
1700
0.5
T
D U A
1
S 80e C


 
• Since ED>EU1, kD increases faster than kU1 as the temperature increases
• 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
A+C D desired
kD
A+C U1 undesired
kU1
• Operate at a high temperature to maximize CD with respect to CU1
• 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?

L9b-16
2000
0.5
T
D A C
r 800e C C


300
T
U A C
1
r 10e C C


1700
0.5
T
D U A
1
S 80e C



• 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
A+C D desired
kD
A+C U1 undesired
kU1
• ED>EU1, operate at a high temperature to maximize CD with respect to CU1
• 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?
←High CC
Semi-batch reactor
slowly feed A to large
amount of C
CA
PFR/PBR w/ side streams feeding low CA
A
C PFR

L9b-17
How does the selection of reactor/reactors scheme and conditions
change if D can react with C and form another undesired product?
2000
0.5
T
D A C
r 800e C C


300
T
U A C
1
r 10e C C


Need to maximize SD/U1 and SD/U2
1700
0.5
T
D U A
1
S 80e C



A+C D desired
kD
A+C U1 undesired
kU1
D+C U2 undesired
kU2
8000
6 T
U C D
2
r 10 e C C


D
D U2
U2
r
S
r

2000
0.5
T
A C
8000
6 T
C D
800e C C
10 e C C



6000
4 0.5 1
T
D U A D
2
S 8 10 e C C
 
  
• aD<aU1, keep CA low
• ED>EU1, operate at a high T
• High CC increases rxn rate &
offsets slow rxn from low CA
• Since ED<EU21, kD increases slower than kU2 as T increases ⇨ operate at low
T to maximize CD
• aD>aU2, keep CA high to maximize CD
• 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 maximizing SD/U1!
Conflicts with maximizing SD/U1!
Conflicts with producing the product D!!!

L9b-18
A
C PFR, high T
Maximize SD/U1 & SD/U2
1700
0.5
T
D U A
1
S 80e C



A+C D desired
kD
A+C U1 undesired
kU1
D+C U2 undesired
kU2
6000
4 0.5 1
T
D U A D
2
S 8 10 e C C
 
 
• aD<aU1, keep CA low
• ED>EU1, operate at a high T
• Want to maximize CD
• ED<EU2, operate at low T
• aD>aU2, keep CA high
• 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
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
PFR 2, low T

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?
2000
0.5
T
D A C
r 800e C C


300
T
U A C
1
r 10e C C


Need to maximize SD/(U1+U2)
A+C D desired
kD
A+C U1 undesired
kU1
D+C U2 undesired
kU2
8000
6 T
U C D
2
r 10 e C C


 
D
D U U
1 2
U U
1 2
r
S
r r
 

2000
0.5
T
A C
300 8000
6
T T
A C C D
800e C C
10e C C 10 e C C

  
 
 


 
   
2000
0.5
T
D 300 8000
6
U U
1 2 T T
800
e 1
10
S
10 10
e 1 e 1
10 10

  
 
  


CA=1
CD=1
2000
T
D 300 8000
U U 5
1 2 T T
80e
S
e 10 e

  
 
  
 

Plot SD/(U1+U2) vs temperature to find the temperature that maximizes SD/(U1+U2)

L9b-20
0
0.5
1
1.5
2
2.5
3
3.5
4
0 200 400 600 800 1000
Temperature (K)
If a CSTR were used with CA = 1 mol/L and CD= 1 mol/L, at what
temperature should the reactor be operated?
2000
0.5
T
D A C
r 800e C C


300
T
U A C
1
r 10e C C


Need to maximize SD/(U1+U2)
A+C D desired
kD
A+C U1 undesired
kU1
D+C U2 undesired
kU2
8000
6 T
U C D
2
r 10 e C C


2000
T
D 300 8000
U U 5
1 2 T T
80e
S
e 10 e

  
 
  


S
D/(U1+U2)
600K

L9b-21
Calculate the yield of forming B in a CSTR and PFR when the conversion
of A is 90% and CA0 = 4 mol/L. The following reactions occur in the reactor:
A B
kB
B B
mol
r k 2
L min
 

A C
kC
C C A
r k C
 1
C
k 1 min

What is the expression for the yield of B for a CSTR?
B
B
A0 A
F
Y
F F

 (overall yield)
B 0 B
B B
A0 0 A 0 A0 A
C C
Y Y
C C C C

 
  
 
We know CA0 and CA when XA=0.9. How do we get CB?
In - Out + Gen. = Accum.
B
B0 B B
dN
F F r V
dt
  
0 0
   
B
B r V
F 0   B 0
B
C
V
r

  B
B
C
r
t
B
B
C
mol
r 2
L min t
  

Use the mole balance on A to find t (at 90% conversion)
In - Out + Gen. = Accum.
   A
A
A0 A
d
V
F
N
r
F
dt 0
   
A0 A A
0 0 V
C C r
 
      
0
A0 A A A0 A A
C C r C C
V
r

t A0 A
A
C C
r
t

 

B
mol
2 C
L min
t
 


L9b-22
A B
kB
B B
mol
r k 2
L min
 

A C
kC
C C A
r k C
 1
C
k 1 min

B
A
B
A0
C
Y
C
C


A0 A
A
C C
r
t



What is –rA?
CA0 = 4 mol/L, and at
XA=0.9, CA= 0.4 mol/L
A B C
r r r
  
A B C A
r k k C
    A A
mol 1
r 2 C
L min min
   

A
A
A
A0 A0
A C
C C
mol 1
r
2 C
L min m
C
in
t t
 
  



Plug -rA back into
expression for t
mol 1 mol
2 0.4
L min mi
mol
4
L
mol
0.
L
n
4
L
t


 
  
  
1.5 min
t
  Residence time for XA = 0.9
B
mol
2
L min
C
t 

 
B
1.5min
mol
4
mol
2
L
L
mi
mo
L
n
4
Y
l
0.
 


B
Y 0.83
 
B
A
B
A
0
r
C
C
Y
t



L9b-23
A B
kB
B B
mol
r k 2
L min
 

A C
kC
C C A
r k C
 1
C
k 1 min

What is the expression for the yield of B for a PFR?
B
B
A0 A
F
Y
F F


(overall yield) B 0 B
B B
A0 0 A 0 A0 A
C C
Y Y
C C C C

 
  
 
Use the mass balance to get CB
B
B
dF
r
dV
 B 0
B
dC
r
dV

  B
B
dC
r
dt
  B
dC mol
2
d L min
t
 

CB
B
C 0
B0
mol
dC 2 d
L min
t
t
 
 

 
   

B B0
mol
C C 2 0
L min
t
Use the mole balance on A to find t (at 90% conversion)
A
A
dF
r
dV
 A 0
A
dC
r
dV

  A
A
dC
r
dt
 
A B C A
r k k C
    A A
mol 1
r 2 C
L min min
   

 
   
 

 
A
A
dC mol 1
2 C
d L min min
t
 
 
 
 
CA
A
A
C 0
A0
dC 1
d
2mol L C min
t
t
 
   
 
 
A
A
dC mol 1
2 C
d L min
t
B
mol
mi
C 2
L n
t

 
0

L9b-24
A B
kB
B B
mol
r k 2
L min
 

A C
kC
C C A
r k C
 1
C
k 1 min

B
B
A0 A
C
Y
C C

 B
mol
in
C 2
L m
t

 Use mole balance on A to find t (at XA = 0.9)
A B C A
r k k C
    A A
mol 1
r 2 C
L min min
   

CA0 = 4 mol/L
CA = 0.4 mol/L
 
 
 
 
 
 
CA
A
C 0
A0 A
dC 1
d
mol min
2 C
L
t
t  
A0
A
mol
2
1
L
ln 0
m
C
ol min
2
L
C
t
 
 
 
 
  
 
 
 
 
mol
2
1
L
ln
mo mol
0.
mol
l mi
L
n
2
L L
4
4
t
 

 
 
 
 

 
 
0.92 min t
 
B B
A0 A0
A
B
A
B
B
mol
C C
0.4
L
0
Y Y Y 0
.92m
m
i
ol
n
mol
C C
4
.5
L
2
r L min
C
1
t 
     
 

Yield was better
in the CSTR, but
the residence
time was longer

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