Lecture 4
Kjemisk reaksjonsteknikk
Chemical Reaction Engineering
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Review of previous lectures
Stoichiometry
Stoichiometric Table
Definitions of Concentration
Calculate the Equilibrium Conversion, Xe
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Reactor Mole Balances in terms of conversion
Reactor
Differential
Algebraic
Integral
X
X
Batch
N A0
V
CSTR
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PFR
dX
t  N A0 
 rA V
0
dX
  r AV
dt
FA 0
dX
 rA
dV
t
FA 0 X
 rA
X
V  FA 0 
0
dX
 rA
X
PBR
FA 0
dX
  rA
dW
X
W  FA 0 
0
dX
 rA
W
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Last Lecture
Relative Rates of Reaction
aA  bB  cC  dD
b
c
d
A B C D
a
a
a
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rC rD
rA
rB



a b c
d
3
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Last Lecture
Rate Laws - Power Law Model

A

B
 rA  kC C
α order in A
β order in B
OverallRect ionOrder  α  β
2A  B  3C
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A reactor follows an elementary rate law if the reaction orders just
happens to agree with the stoichiometric coefficients for the reaction
as written.
e.g. If the above reaction follows an elementary rate law
 rA  k AC2ACB
2nd order in A, 1st order in B, overall third order
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Last Lecture
Arrhenius Equation
k is the specific reaction rate (constant) and is given by the Arrhenius
Equation.
Where:
k  Ae
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 E RT
T  k  A
k
T 0 k 0
A  1013
T
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Reaction Engineering
Mole Balance
Rate Laws
Stoichiometry
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These topics build upon one another
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How to find
rA  f X 
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Step 1: Rate Law
rA  gCi 
Step
2: Stoichiometry
Ci   hX 
Step 3: Combine
 to get
rA  f X 


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We shall set up Stoichiometry Tables using species A as our basis of
calculation in the following reaction. We will use the stochiometric tables to
express the concentration as a function of conversion. We will combine Ci = f(X)
with the appropriate rate law to obtain -rA = f(X).
b
c
d
A  B C  D
a
a
a
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A is the limiting Reactant.
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NA  NA 0  NA 0 X
For every mole of A that react, b/a moles of B react. Therefore moles
of B remaining:
 N B0 b 
b
N B  N B 0  N A0 X  N A0 
 X 
a
 N A0 a 
Let ΘB = NB0/NA0
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Then:

b 
N B  N A 0 B  X 

a 

c
c 
NC  NC 0  N A 0 X  N A 0 C  X 

a
a 
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Batch System Stoichiometry Table
Species
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Symbol
Initial
Change
Remaining
A
A
NA0
-NA0X
NA=NA0(1-X)
B
B
NB0=NA0ΘB
-b/aNA0X
NB=NA0(ΘB-b/aX)
C
C
NC0=NA0ΘC
+c/aNA0X
NC=NA0(ΘC+c/aX)
D
D
ND0=NA0ΘD
+d/aNA0X
ND=NA0(ΘD+d/aX)
Inert
I
NI0=NA0ΘI
----------
NI=NA0ΘI
FT0
N
C 
C
y
Where: i  i 0  i 0 0  i 0  i 0 and
N A 0 CA 0 0 CA 0 y A 0
NT=NT0+δNA0X
d c b
    1
a a a
δ = change in total number of mol per mol A reacted
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Constant Volume Batch
Note:
If the reaction occurs in the liquid phase
or
if a gas phase reaction occurs in a rigid (e.g. steel) batch reactor
Then
V  V0
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N A N A 0 1  X 
CA 

 CA 0 1  X 
V
V0



N B N A 0 
b 
b 
CB 

B  X  CA 0B  X 

V
V0 
a 
a 
etc.
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Stoichiometry
Suppose
rA  kA C CB
Batch:
V  V0
2
A
rA  k A CA 0
2

b 
1  X  B  X 

a 
2
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Equimolar feed:
B  1
Stoichiometric feed:
b
B 
a

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Flow System Stochiometric Table
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Species
Symbol
Reactor Feed
Change
Reactor Effluent
A
A
FA0
-FA0X
FA=FA0(1-X)
B
B
FB0=FA0ΘB
-b/aFA0X
FB=FA0(ΘB-b/aX)
Where:
i 
Fi0
C 
C
y
 i0 0  i0  i0
FA 0 CA 0 0 CA 0 y A 0
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Flow System Stochiometric Table
Species
Symbol
Reactor Feed
Change
C
C
FC0=FA0ΘC
+c/aFA0X
FC=FA0(ΘC+c/aX)
D
D
FD0=FA0ΘD
+d/aFA0X
FD=FA0(ΘD+d/aX)
Inert
I
FI0=A0ΘI
----------
FI=FA0ΘI
FT=FT0+δFA0X
FT0
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Reactor Effluent
Fi 0
C 
C
y
 i0 0  i0  i0
FA 0 C A 0 0 C A 0 y A 0
Where:
i 
and
d c b
    1
a a a
Concentration – Flow System
FA
CA 

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Flow System Stochiometric Table
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Species
Symbol
Reactor Feed
Change
Reactor Effluent
A
A
FA0
-FA0X
FA=FA0(1-X)
B
B
FB0=FA0ΘB
-b/aFA0X
FB=FA0(ΘB-b/aX)
C
C
FC0=FA0ΘC
+c/aFA0X
FC=FA0(ΘC+c/aX)
D
D
FD0=FA0ΘD
+d/aFA0X
FD=FA0(ΘD+d/aX)
Inert
I
FI0=FA0ΘI
----------
FI=FA0ΘI
FT=FT0+δFA0X
FT0
Where:
i 
Fi 0
C 
C
y
 i0 0  i0  i0
FA 0 C A 0 0 C A 0 y A 0
Concentration – Flow System
CA 
and

d c b
  1
a a a
FA

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Concentration Flow System:
Liquid Phase Flow System:
FA
CA 

  0
Liquid Systems
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FA FA 0 1  X 
CA 

 CA 0 1  X  Flow Liquid Phase

0
CB 
NB NA0 
b 
b 




X

C


X
 B

A0 
B
V
V0 
a 
a 

etc.
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Liquid Systems
 rA  kCA CB
If the rate of reaction were
 rA  CA 0
then we would have
This gives us
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2
b 

1  X  B  X 
a 

 rA  f X
FA 0
rA
X
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For Gas Phase Flow Systems
Combining the compressibility factor equation of state with Z = Z0
Stoichiometry:
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P
CT 
ZRT
P0
CT 0 
Z 0 R0T0
FT  CT V
FT 0  CT 0V0
We obtain:

FT P0 T
  0
FT 0 P T0
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 P  T0  FT 0  F j  P  T0 
   

  
C j  Fj  
 FT   P0  T   0  FT  P0  T 

 0 
 FT 0 
Fj
CT 0  FT 0  0
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
For Gas Phase Flow Systems
FB P T0 
CB  CT 0    
FT P0 T 
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For Gas Phase Flow Systems
The total molar flow rate is:
Substituting FT gives:
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FT  FT0  FA0X
 FT 0  FA 0X  T P0

  0 
FT 0

 T0 P

 T P0
FA 0
  0 1 
X 
FT 0

 T0 P
T P0
  0 1  y A 0X 
T0 P
T P0
  0 1  X 
T0 P
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For Gas Phase Flow Systems
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For Gas Phase Flow Systems
Concentration Flow System:
Gas Phase Flow System:
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CA 
FA
CA 

T P0
  0 1  X 
T0 P
FA 0 1  X 
C 1  X  T0 P
FA

 A0
T P0
1  X  T P0

0 1  X 
T0 P
b 
b 


FA 0   B  X  C A 0   B  X 
FB
a 
a  T0 P


CB 


P
T
1  X 
  1  X 
T P0
0
0
T0 P
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For Gas Phase Flow Systems
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b 
b 


FA0   j  X  C A0   j  X 
Fj
a 
a  T0 P


Cj 


1  X  T P0
  1  X  T P0
0
T0 P
Cj=f(Fj, T, P) =f(x, T,P)
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For Gas Phase Flow Systems
If –rA=kCACB


b 

 1  X    B  a X   P T  2 


2
0

 
 rA  k A C A 0 
 1  X  1  X   P0 T  


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This gives us
FA0/-rA
X
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Example: Calculating the equilibrium conversion for gas
phase reaction in a flow reactor, Xef
Consider the following elementary reaction with KC=20 dm3/mol and
CA0=0.2 mol/dm3.
Calculate Equilibrium Conversion or both a batch reactor (Xeb) and a
flow reactor (Xef).
2A  B
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 2 CB 
 rA  k A CA 

KC 

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Batch Reactor Example
Calculating the equilibrium conversion for gas phase
reaction,Xe
Consider the following elementary reaction with KC=20 m3/mol and
CA0=0.2 mol/m3.
Xe’ for both a batch reactor and a flow reactor.
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2A  B
 2 CB 
 rA  k A CA 

K
C

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Example
Calculate Xe
Batch Reactor
CA0  0.2 mol L
KC  20 L mol
dX rAV

dt
NA 0
Step 2: rate law,
 rA  k AC2A  k BCB
Step 1:
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
 2 CB 
 rA  k A CA 

KC 

kA
KC 
kB
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Example
Calculate Xe
Batch Reactor
Symbol
Change
Remaining
A
NA0
-NA0X
NA0(1-X)
B
0
½ NA0X
NA0 X/2
Totals:
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Initial
NT0=NA0
NT=NA0 -NA0 X/2
@ equilibrium: -rA=0
CBe
Ke  2
CAe
0C
CAe
CBe
2
Ae
C Be

KC
NAe

 CA 0 1  X e 
V
Xe
 CA 0
2
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Batch Reactor
Example
Calculating the equilibrium conversion for gas phase
reaction
Solution:
At equilibrium
Stoichiometry

C Be 
2
 rA  0  k A C Ae 

K
C 

C Be
KC  2
C Ae
A  B/ 2
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V  V0
Constant volume
Batch
Species
Initial
Change
Remaining
A
NA0
-NA0X
NA=NA0(1-X)
B
0
+NA0X/2
NB=NA0X/2
NT0=NA0
NT=NA0-NA0X/2
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BR Example Xeb
Xe
CA0
Xe
2
Ke 

2
2
C A 0 1  X e  2C A 0 1  X e 
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2K e C A 0
Xe

 2200.2  8
2
1  Xe 
X eb  0.703
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Gas Phase Example Xef
2A  B
X eb  0.703
X ef  ?


Solution:
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
Rate law:
 2 CB 
 rA  k A C A 

KC 

1
A B
2
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Gas Flow Example Xef
Species
Fed
Change
Remaining
A
FA0
-FA0X
FA=FA0(1-X)
B
0
+FA0X/2
FB=FA0X/2
FT0=FA0
FT=FA0-FA0X/2
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Gas Flow Example Xef
A
FA0
-FA0X
FA=FA0(1-X)
B
0
FA0X/2
FB=FA0X/2
Stoichiometry: Gas isothermal T=T0, isobaric P=P0
V  V0 1 X 
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FA 0 1  X  CA 0 1  X 
CA 

V0 1 X 
1 X 
FA 0 X 2
CA 0 1  X 
CB 

V0 1 X 
21 X 
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Gas Flow Example Xef
 C 1  X   2

C
X
A0
 
 rA  k A  A 0

21  X K C 
 1  X  
Pure A  yA0=1, CA0=yA0P0/RT0, CA0=P0/RT0
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1 
1
  y A 0  1 1 
2 
2
@ eq: -rA=0
2K C C A 0
X e 1  X e 

1  X e 2
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2K C C A 0
 dm 3 
mol 


 2 20
8
 0.2
3 

dm 
 mol 
1 
1
  y A 0  1 1 
2 
2
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
Gas Flow Example Xef
36
8

X e  0.5X e 2
1  2X e  X e 2

8.5X e 2 17X e  8  0
Flow: X ef  0.757
Recall
Batch: X eb  0.70
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