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Lecture 13
Chemical Reaction Engineering (CRE) is the
field that studies the rates and mechanisms of
chemical reactions and the design of the reactors in
which they take place.
Lecture 13 – Tuesday 2/26/2013
 Complex Reactions:
A +2B  C
A + 3C  D
 Example A: Liquid Phase PFR
 Example B: Liquid Phase CSTR
 Example C: Gas Phase PFR
 Example D: Gas Phase Membrane Reactors
Sweep Gas Concentration Essentially Zero
Sweep Gas Concentration Increases with Distance
 Example E: Semibatch Reactor
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Gas Phase
Multiple Reactions
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New things for multiple reactions are:
1. Number Every Reaction
2. Mole Balance on every species
3. Rate Laws
(a) Net Rates of Reaction for every species
N
rA 
r
iA
i 1
(b) Rate Laws for every reaction
r1 A   k 1 A C A C B
2
r2 C   k 2 C C A C C
2
4
3
(c) Relative Rates of Reaction for every reaction
For a given reaction i: (i) aiA+biB ciC+diD:
riA
riB
riC
riD



 ai
 bi
ci
di
Reactor Mole Balance Summary
Reactor Type
Gas Phase
Batch
dN
A
dt
dN
Semibatch
dt
dN
dt
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B
A
Liquid Phase
dC
 r AV
dt
dC A
 r AV
 rB V  F B 0
dt
dC B
dt
A
 rA
 rA 
 rB 
 0C A
V
 0 C B 0  C B 
V
Reactor Mole Balance Summary
Reactor Type
Gas Phase
CSTR
V 
PFR
dF A
dV
PBR
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dF A
dW
FA0  FA
 rA
 rA
 rA
Liquid Phase
C A 0  C A 
V  0
0
dC
0
dC
 rA
A
 rA
A
 r A
dV
dW
Note: The reaction rates in the above mole balances are net rates.
Batch
CB 
V  V0
CB 
V
N T P0 T 0
NT0 P T
N B N T 0 P T0
N T V 0 P0 T
C B  CT 0
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NB
N B P T0
N T P0 T
Flow
CB 
  0
CB 
FB

FT P0 T 0
FT 0 P T
F B FT 0 P T 0
FT  0 P0 T
C B  CT 0
F B P T0
FT P0 T
Stoichiometry
Concentration of Gas:
 F A   T0 
 y 
C A  C T 0 
 T
F
 T   
FT  F A  F B  FC  F D
Note: We could use the gas phase mole balances for
liquids and then just express the concentration as:
Flow:
Batch:
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CA 
CA 
FA
0
NA
V0
Example A: Liquid Phase PFR
The complex liquid phase reactions follow
elementary rate laws:
(1) A  2 B  C
 r1 A  k 1 A C A C
2
B
NOTE: The specific reaction rate k1A is defined with
respect to species A.
( 2 ) 3C  2 A  D
 r2 C  k 2 C C C C A
3
2
NOTE: The specific reaction rate k2C is defined with
respect to species C.
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Example A: Liquid Phase PFR
Complex Reactions
(1)
A  2B  C
(2) A  3C  D
1) Mole Balance on each and every species

(1)
dF A
dV
(3)
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dF C
dV
 rA
 rC
(2)
dF B
dV
(4)
dF D
dV
 rB
 rD
Example A: Liquid Phase PFR
2) Rate Laws:
Net Rates
Rate Laws
( 5 ) rA  r1 A  r2 A
( 7 ) rB  r1 B  r2 B
( 6 ) rC  r1C  r2 C
( 8 ) rD  0  r2 D
( 9 ) r1 A   k 1 A C A C B
2
(10 ) r2 C   k 2 C C A C C
2
Relative Rates
Reaction 1
r1 A
1

r1 B
2
(11) r1 B  2 r1 A
11
(12) r1 C   r1 A

3
r1 C
1
Example A: Liquid Phase PFR
Relative Rates
Reaction 2
r2 A
2

r2 C
3
(13 ) r2 A 

r2 D
1
2
3
(14 ) r2 D  
r2 C
r2 C
3
rA   k 1 A C A C 
2
B
2
3
2
rB   2 k 1 A C A C B
2
rC  k 1 A C A C B  k 2 C C A C C
2
12
rD 
k 2C
3
2
3
C AC C
3
k 2C C AC C
3
Example A: Liquid Phase PFR
3) Stoichiometry
Liquid
(15 ) C A  F A  0
(16 ) C B  F B  0
(17 ) C C  FC  0
(18 ) C D  F D  0
~
(19 ) S C
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D
 FC 
 else 0
 if V  0 . 00001  then 

F
 D 
Example A: Liquid Phase PFR
Others
FT  Liquid
– Not Needed
(19 )   Liquid
( 20 ) C T 0  Liquid
4) Parameters
( 21 ) k 1 A  10
( 22 ) k 2 C  20
( 23 )   Liquid
( 24 ) C T 0  Liquid
( 25 ) V f  2500
( 26 ) F A 0  200
( 28 ) F B 0  200
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( 26 )  0  100
– Not Needed
– Not Needed
Example B: Liquid Phase CSTR
Same reactions, rate laws, and rate constants as
Example A
(1) A  2 B  C
 r1 A  k 1 A C A C B
2
NOTE: The specific reaction rate k1A is defined with
respect to species A.
( 2 ) 3C  2 A  D
 r2 C  k 2 C C C C A
3
2
NOTE: The specific reaction rate k2C is defined with
respect to species C.
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Example B: Liquid Phase CSTR
The complex liquid phase reactions take place in
a 2,500 dm3 CSTR. The feed is equal molar in A
and B with FA0=200 mol/min, the volumetric flow
rate is 100 dm3/min and the reaction volume is 50
dm3.
Find the concentrations of A, B, C and D existing
in the reactor along with the existing selectivity.
Plot FA, FB, FC, FD and SC/D as a function of V
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Example B: Liquid Phase CSTR
(1) A + 2B →C
(2) 2A + 3C → D
r1 A   k 1 A C A C B
2
r2 C   k 2 C C A C C
2
3
1) Mole Balance
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(1)
A
 0 C A 0   0 C A  r AV  0
(2)
B
 0 C B 0   0 C B  rB V  0
(3)
C
0   0 C C  rC V  0
(4)
D
0   0 C D  rD V  0
Example B: Liquid Phase CSTR
2) Rate Laws: (5)-(14) same as PFR
3) Stoichiometry: (15)-(18)
same as Liquid Phase PFR
(19 ) S C / D 
FC
F D  0 . 0001

 0C C
 0 C D  0 . 0001
4) Parameters:
k1 A , k 2 C , C A 0 , C B 0 , V ,  0
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Example B: Liquid Phase CSTR
In terms of molar flow rates
(1) A + 2B →C
(2) 2A + 3C → D
r1 A   k 1 A C A C B
2
r2 C   k 2 C C A C C
2
1) Mole Balance (1–4)
(1) f  F A   F A 0  F A  r AV (=0)
( 2 ) f  F B   F B 0  F B  rBV
(=0)
( 3 ) f  FC   0  FC  rC V (=0)
( 4 ) f  F D   0  F D  rD V
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(=0)
3
2) Rates (5–14) 3) Stoichiometry: (15–19)
Same as
Example A
(15 )
C A  FA  0
(16 )
C B  FB  0
(17 )
C C  FC  0
(18 )
C D  FD  0
(19 )
SC
D

FC
F D  0 . 00001
Example B: Liquid Phase CSTR
In terms of concentration
(1) A + 2B →C
(2) 2A + 3C → D
r1 A   k 1 A C A C B
2
r2 C   k 2 C C A C C
2
1) Mole Balance (1–4)
(1) f C A    0 C A 0   0 C A  rAV
2) Rates (5–14)
(=0)
( 2 ) f C B    0 C B 0   0 C B  rBV
(=0)
( 3 ) f C C   0   0 C C  rC V
(=0)
( 4 ) f C D   0   0 C D  rDV
(=0)
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3
Same as
Example A
3) Stoichiometry: (15–19)
(15 ) S C
D

FC
F D  0 . 00001
Example C: Gas Phase PFR, No ΔP
Same reactions, rate laws, and rate constants as
Example A:
(1) A  2 B  C
 r1 A  k 1 A C A C B
2
NOTE: The specific reaction rate k1A is defined with respect to
species A.
( 2 ) 3C  2 A  D
 r2 C  k 2 C C C C A
3
2
NOTE: The specific reaction rate k2C is defined with respect to
species C.
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Example C: Gas Phase PFR, No ΔP
1) Mole Balance
(1)
dF A
dV
(2)
dF B
dV
 rA
 rB
(3)
dF C
dV
(4)
dF D
dV
 rC
 rD
2) Rate Laws: (5)-(14) same as CSTR
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Example C: Gas Phase PFR, No ΔP
3) Stoichiometry:
Gas: Isothermal T = T0
(15 ) C A  C T 0
(17 ) C C  C T 0
FA
FT
FC
y
y
FT
(16 ) C B  C T 0
(18 ) C D  C T 0
(19 ) FT  F A  F B  FC  F D
Packed Bed with Pressure Drop
dy
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dW

  FT   T 
 FT



2 y  FT 0   T 0 
2 y FT 0
FB
FT
FD
FT
y
y
Example C: Gas Phase PFR, No ΔP
4) Selectivity
S 
FC
FD
y 1
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 FC 
 else 0 
 if V  0 . 00001  then 

 FD 
 21 
 20 
Example D: Membrane Reactor with ΔP
Same reactions, rate laws, and rate constants as
Example A:
(1) A  2 B  C
 r1 A  k 1 A C A C B
2
NOTE: The specific reaction rate k1A is defined with respect to
species A.
( 2 ) 3C  2 A  D
 r2 C  k 2 C C C C A
3
2
NOTE: The specific reaction rate k2C is defined with respect to
species C.
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Example D: Membrane Reactor with ΔP
Because the smallest molecule, and the one with the
lowest molecular weight, is the one diffusing out, we will
neglect the changes in the mass flow rate down the
reactor and will take as first approximation: m 0  m
1) Mole Balances
A
dF A
dV
B
dF B
dV
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 rA
 rB
1
2 
C
dF C
dV
D
dF D
dV
 rC  R C
 rD
3 
4 
We also need to account for the molar rate of desired
product C leaving in the sweep gas FCsg dF Csg
 RC
dV
Example D: Membrane Reactor with ΔP
We need to reconsider our pressure drop equation.
When mass diffuses out of a membrane reactor there
will be a decrease in the superficial mass flow rate, G.
To account for this decrease when calculating our
pressure drop parameter, we will take the ratio of the
superficial mass velocity at any point in the reactor to
the superficial mass velocity at the entrance to the
reactor.
  Fi  MW i 
  0
 0

G0
  Fi 0  MW i 
G
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Example D: Membrane Reactor with ΔP
The superficial mass flow rates can be obtained by
multiplying the species molar flow rates, Fi, by their
respective molecular weights, Mwi, and then
summing over all species:
G
G0
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
m AC 1
m 0 AC 1
F   MW  A


 F   MW  A
i
i0
i
C1
i
C1
F  MW 


 F  MW 
i
i0
i
i
Example D: Membrane Reactor with ΔP
2) Rate Laws: (5)-(14) same as Examples A, B, and C.
3) Stoichiometry: (15)-(20) same as Examples A and B
(T=T0)
dy

dW

FT
dy
2 y FT 0
dV
R C  k C C C  C CSweep

4) Sweep Gas Balance:
FCsg
V
dF Csg
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dV
 FCsg
 RC
V  V
 RC  V  0

 FT
2 y FT 0
 21 
Example E: Liquid Phase Semibatch
Same reactions, rate laws, and rate constants as
Example A:
(1) A  2 B  C
 r1 A  k 1 A C A C B
2
NOTE: The specific reaction rate k1A is defined with respect to
species A.
( 2 ) 3C  2 A  D
 r2 C  k 2 C C C C A
3
2
NOTE: The specific reaction rate k2C is defined with respect to
species C.
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Example E: Liquid Phase Semibatch
The complex liquid phase reactions take place
in a semibatch reactor where A is fed to B with
FA0= 3 mol/min. The volumetric flow rate is 10
dm3/min and the initial reactor volume is 1,000
dm3.
The maximum volume is 2,000 dm3 and
CA0=0.3 mol/dm3 and CB0=0.2 mol/dm3. Plot CA,
CB, CC, CD and SS/D as a function of time.
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Example E: Liquid Phase Semibatch
(1) A + 2B →C
(2) 2A + 3C → D FA0
1) Mole Balances:
dN
A
dt
dN B
dt
dN C
dt
dN D
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dt
B
 rAV  F A 0
N A0  0
 rB V
N B 0  C B 0V 0  2 . 000
 rC V
NC0  0
 rD V
N D0  0
Example E: Liquid Phase Semibatch
2) Rate Laws: (5)-(14)
Net Rate, Rate Laws and relative rate – are the same
as Liquid and Gas Phase PFR and Liquid Phase CSTR
V  V 0  v 0 t 15 
CA 
NA
CC 
NC
V
V
16 
18 
CB 
NB
CD 
ND
V
V
17 
19 
3) Selectivity and Parameters:
SC /D
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 NC
 if ( t  0 . 0001 ) then 
 ND
 0  10 dm
3
min
V 0  100 dm
3

 else ( 0 )


 20 
FA 0  3 mol min
End of Lecture 13
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