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.
Today’s lecture
Complex Reactions
A +2B  C
A + 3C  D
 Liquid Phase PFR
 Liquid Phase CSTR
 Gas Phase PFR
 Gas Phase Membrane Reactor
Sweep Gas Concentration Essentially Zero
Sweep Gas Concentration Increases with Distance
 Semi Batch Reactor
2
Reactor Mole Balance Summary
Reactor Type
Gas Phase
Batch
dN A
 rA V
dt
Semibatch
dN A
 rA V
dt
dN B
 rB V  FB0
dt
3
Liquid Phase
dC A
 rA
dt
0 C A
dC A
 rA 
dt
V
0 C B0  C B 
dC B
 rB 
dt
V
Reactor Mole Balance Summary
Reactor Type
4
Gas Phase
Liquid Phase
CA0  CA 
CSTR
FA 0  FA
V
 rA
V  0
PFR
dFA
 rA
dV
dC A
0
 rA
dV
PBR
dFA
 rA
dW
dC A
0
 rA
dW
 rA
Note: The reaction rates in the above mole balances are net rates.
Reactor Mole Balance Summary
The new things for multiple reactions are:
Rates:
1. Rate Law for every reaction
2. Relative Rates for every reaction
3. Net Rates of Reaction
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Gas Phase
Multiple Reactions
6
Note: We could use the gas phase mole balance for liquids and
then just express the concentration as:
Flow: CA=FA/v0
Batch: CA=NA/V0
Note: The reaction rates in the above mole balances are net rates.
The new things for multiple reactions are:
1. Rate Law for every reaction
2. Relative Rates for every reaction
3. Net Rates of Reaction
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Net Rate of Reaction for species A
For N reactions, the net rate of formation of species A is:
N
rA   riA
i 1
For a given reaction i:
(i) aiA+biB ciC+diD:
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riC riD
riA
riB



 a i  bi ci d i
Batch
Flow
NB
CB 
V
NT P0 T0
V  V0
NT0 P T
FB
CB 

FT P0 T0
  0
FT 0 P T
N B N T 0 P T0
CB 
N T V0 P0 T
FB FT 0 P T0
CB 
FT 0 P0 T
N B P T0
N T P0 T
FB P T0
C B  CT 0
FT P0 T
C B  CT 0
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Last Lecture
Example A: Liquid Phase PFR
The complex liquid phase reactions follow elementary rate
laws
2

r

k
C
C
A  2B  C (1)
1A
1A A B
NOTE: The specific reaction rate k1A is defined with respect to
species A.
3C  2A  D (2)
 r2C  k 2CC3CC2A
NOTE: The specific reaction rate k2C is defined with respect to
species C.
10
Complex Reactions
Example B: Liquid Phase CSTR
Same reactions, rate laws, and rate constants as example A
A  2B  C (1)
 r1A  k1ACAC2B
NOTE: The specific reaction rate k1A is defined with respect to
species A.
3C  2A  D (2)
 r2C  k 2CC3CC2A
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
reation 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 ofV
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Example B: Liquid Phase CSTR
CSTR (1) A + 2B →C
(2) 2A + 3C → D
r1A  k1A C A C 2B
r2C  k 2C C C
2
A
1) Mole balance:
A
3
C
V0 C A 0  V0 C A  rA V  0 1
B
V0 C B0  V0 C B  rB V  0 2
C
0  V0 CC  rC V  0
D
0  V0 C D  rD V  0
3
4
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2) Rates:
r1A   k1A C A C 2B
r2 C   k 2 C C 2A C3C
rA  r1A  r2 A
rC  r1C  r2 C
5
6
7 
9
r1A r1B r1C


1  2 1
11
r1B  2r1A
r2 A
2
 r2 C
3
SC / D 
13
rB  r2 B
rD  r2 D
8
10
r2 A r2 C r2 D


2 3
1
12
r1C   r1A
r2 D
1
  r2 C
3
14
FC
0CC
15

FD  0.0001 0CD  0.0001
3) Parameters:
k1A , k 2C , CA0 , CB0 , V, V014
Complex Reactions
Example C: Gas Phase PFR, No ΔP
Same reactions, rate laws, and rate constants as example A
A  2B  C (1)
 r1A  k1ACAC2B
NOTE: The specific reaction rate k1A is defined with respect to
species A.
3C  2A  D (2)
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 r2C  k 2CC3CC2A
NOTE: The specific reaction rate k2C is defined with respect to
species C.
Example C: Gas Phase PFR, No ΔP
1) Mole balance:
A
B
dFC
dFA
 rA 1 C
 rC  R C
dV
dV
dFB
dFD
 rB 2  D
 rD 4
dV
dV
3
2) Rates: Same as CSTR (5)-(14)
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Example C: Gas Phase PFR, No ΔP
3) Stoich:
C A  CT 0
FA
y 15
FT
CC  CT 0
FC
y 17
FT
C B  CT 0
FB
y 16
FT
C D  CT 0
FD
y 18
FT
FT  FA  FB  FC  FD 19
4) Selectivity:
 FC 
FC
S
 if V  0.00001 then  else 0 20
FD
 FD 
y  1 21
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Complex Reactions
Example D: Membrane Reactor with ΔP
Same reactions, rate laws, and rate constants as example A
A  2B  C (1)
 r1A  k1ACAC2B
NOTE: The specific reaction rate k1A is defined with respect to
species A.
3C  2A  D (2)
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 r2C  k 2CC3CC2A
NOTE: The specific reaction rate k2C is defined with respect to
species C.
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 in
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. 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:
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Fi MWi  A C1
G 2 m 2 A C2  Fi MWi  A C2




G1 m1 A C1  Fi 0 MWi  A C1  Fi 0 MWi  A C2
Example D: Membrane Reactor with ΔP
Because the smallest molecule is the one diffusing out and has
the lowest molecular weight, we will neglect the changes in the
mass flow rate down the reactor and will take as first
0 m
)
approximation. (m
1) Mole Balance:
dFC
dFA
A
 rA 1 C
 rC  R C 3
dV
dV
dFB
dFD
B
 rB 2  D
 rD 4
dV
dV
We also need to account for the molar rate of desired product
C leaving in the sweep gas FCsg
dFCsg
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dV
 RC
Example D: Membrane Reactor with ΔP
2) Rates: Same (5)-(14)
3) Stoich: Same (15)-(20)
dy
 FT

;
dW
2y FT 0
dy
 FT

dV
2 y FT 0
21
R C  k C CC  CCSweep 
4) Sweep Gas Balance:
FCsg
V
dFCsg
dV
 FCsg
V  V
 R C V  0
 RC
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Example D: Membrane Reactor with ΔP
Case 1 Large sweep gas velocity
CC  CCsg
CCsg  0
RC  kCC CC
Case 2 Moderate to small sweep gas velocity

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C Csg 
FCsg
sg
 F0sg  FCsg 

sg  sg 0 
 F

0
sg


Vary υsg to see changes in profiles
Complex Reactions
Example E: Liquid Semibatch
Same reactions, rate laws, and rate constants as example A
A  2B  C (1)
 r1A  k1ACAC
2
B
NOTE: The specific reaction rate k1A is defined with respect to
species A.
3C  2A  D (2)
 r2C  k 2CC3CC2A
NOTE: The specific reaction rate k2C is defined with respect to
species C.
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Example E: Liquid 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 Semibatch
FA0
(1) A + 2B →C
(2) 2A + 3C → D
1) Mole balance:
dN A
dt
dN B
dt
dN C
dt
dN D
dt
B
 rA V  FA 0
NA0  0
 rB V
NB0  CB0V0  2.000
 rC V
N C0  0
 rD V
N D0  0
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Example E: Liquid Semibatch
2) Rates: Same (5)-(14)
Net Rates, Rate Laws and relative rates – are the same as Liquid
and Gas Phase PFR and Liquid Phase CSTR
V  V0  v 0 t 15
NA
16
CA 
V
NC
18
CC 
V
NB
CB 
V
ND
CD 
V
17
19
3) Selectivity:
 NC 
 else (0) 20
SC / D  if ( t  0.0001) then
ND 

4) Parameters:
3
0  10dm3 min V0  100dm
FA0  3 mol min
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End of Lecture 13
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