Lecture 4
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 4 – Tuesday 1/22/2013
 Block 1
 Mole Balances
 Size CSTRs and PFRs given –rA=f(X)
 Block 2
 Rate Laws
 Reaction Orders
 Arrhenius Equation
 Block 3




2
Stoichiometry
Stoichiometric Table
Definitions of Concentration
Calculate the Equilibrium Conversion, Xe
Review Lecture 2
Reactor Mole Balances Summary
in terms of conversion, X
Reactor
Differential
Algebraic
Integral
X
X
Batch
N A0
t
FA 0 X
V
rA
CSTR
PFR
dX
 rAV
0
t  N A0 
dX
  r AV
dt
dX
FA 0
 rA
dV 
X
dX
 rA
0
V  FA0 
X
3
PBR

dX
FA 0
 rA
dW
X
dX
 rA
0
W  FA0 
W
Review Lecture 2
Levenspiel Plots
FA 0
rA
X

4

Review Lecture 2
PFR
5
Review Lecture 2
Reactors in Series
moles of A reacted up to point i
Xi 
moles of A fed to first reactor
Only valid if there are no side streams
6
Review Lecture 2
Reactors in Series
7
Review Lecture 2
Two steps to get rA  f X 
Step 1: Rate Law  rA  g Ci 

Step 2: Stoichiometry
Ci   h X 
Step 3: Combine to get  rA  f  X 
8
Review Lecture 3
Building Block 2: Rate Laws
Power Law Model:


 rA  kCA CB
2 A  B  3C
α order in A
β order in B
Overall Rection Order  α  β
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 AC A2CB
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2nd order in A, 1st order in B, overall third order
Review Lecture 3
Arrhenius Equation
k  Ae
T  k  A
 E RT
k
E = Activation energy (cal/mol)
R = Gas constant (cal/mol*K)
T
T = Temperature (K)
A = Frequency factor (same units as rate constant k)
(units of A, and k, depend on overall reaction
 order)
10
T 0 k 0
A  1013
Review Lecture 3
Reaction Engineering
Mole Balance
Rate Laws
Stoichiometry
These topics build upon one another
11
Review Lecture 3
Algorithm
How to find
 rA  f  X 
Step 1: Rate Law  rA  g Ci 
Step 2: Stoichiometry
Ci   h X 
Step 3: Combine to get  rA  f  X 
12
Building Block 3: Stoichiometry
We shall set up Stoichiometry Tables using species
A as our basis of calculation in the following
reaction. We will use the stoichiometric 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
A is the limiting reactant.
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Stoichiometry
NA  NA0  NA 0 X
For every mole of A that reacts, b/a moles of B react. Therefore
moles of B remaining:
æ N B0 b ö
b
N B = N B0 - N A0 X = N A0 ç
- X÷
a
è N A0 a ø
Let ΘB = NB0/NA0
Then:
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b 

N B  N A0   B  X 
a 


c
c 
NC  NC 0  N A 0 X  N A 0 C  X 

a
a 
Batch System - Stoichiometry Table
Species
Symbol
Initial
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
Where:
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Change
NT=NT0+δNA0X
Ni0
Ci 00
Ci 0
yi 0
d c b
i 



    1
N A0 C A00 C A0 y A0 and
a a a
δ = change in total number of mol per mol A reacted
Stoichiometry 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
NA
N A 0 1  X 
CA 

 CA 0 1  X 
V
V0

NB
N A 0 
b 
b 
CB 

B  X  CA 0B  X 


V
V0 
a 
a 

16
etc.
Stoichiometry Constant Volume Batch
2
Suppose rA  k A CA CB
Batch: V  V0
b 
2
2

 rA  k AC A0 1  X    B  X 
a 

Equimolar feed:
B  1
b
Stoichiometric feed:  B 
a
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Stoichiometry Constant Volume Batch
If  rA  k AC A2CB , then
 rA  C A0
3
b 

1  X    B  X  Constant Volume Batch
a 

2
and we have rA  f X 

18
1
 rA
X
Batch Reactor - Example
Calculate the equilibrium conversion for gas phase
reaction, Xe .
Consider the following elementary reaction with
KC=20 dm3/mol and CA0=0.2 mol/dm3.
Find Xe for both a batch reactor and a flow reactor.
2A  B
 2 CB 
 rA  k A CA 

K
C

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Batch Reactor - Example
Calculate Xe
C A0  0.2 mol dm3
K C  20 dm 3 mol
dX
r
V
A
Step 1:

dt
NA 0
Step 2: rate law:  rA  k A C2A  k BCB

20
 2 CB 
 rA  k A CA 

K
C

kA
KC 
kB
Batch Reactor - Example
Symbol
Initial
Change
Remaining
A
NA0
-NA0X
NA0(1-X)
B
0
½ NA0X
NA0 X/2
Totals: NT0=NA0
NT=NA0 -NA0 X/2
@ equilibrium: -rA=0
C Be
Ke  2
C Ae
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C Be
0C 
KC
2
Ae
N Ae
CAe 
 CA 0 1  X e 
V
Xe
CBe  CA 0
2
Batch Reactor - Example
Solution:
At equilibrium
 2 CBe 
 rA  0  k A CAe 

K
C 

Stoichiometry:
Constant Volume:
C Be
KC  2
C Ae
A  B/ 2
V  V0
Batch
Species
A
Initial
NA0
Change
-NA0X
Remaining
NA=NA0(1-X)
B
0
+NA0X/2
NB=NA0X/2
NT0=NA0
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NT=NA0-NA0X/2
Batch Reactor - Example
Xe
CA0
Xe
2
Ke 

2
2
C A 0 1  X e  2C A 0 1  X e 
2K e C A 0
Xe

 2200.2  8
2
1  Xe 
Xeb  0.703
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Flow System – Stoichiometry Table
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  Ci0 0  Ci0  y i0
FA 0
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CA 0 0
CA 0
yA 0
Flow System – Stoichiometry Table
Species
Symbol
Reactor Feed
Change
C
C
FC0=FA0ΘC
+c/aFA0X
D
D
FD0=FA0ΘD
+d/aFA0X FD=FA0(ΘD+d/aX)
Inert
I
FI0=A0ΘI
FT=FT0+δFA0X
Where: i  Fi 0  Ci 0 0  Ci 0  yi 0
FA 0
C A 0 0
FC=FA0(ΘC+c/aX)
FI=FA0ΘI
----------
FT0
CA0
yA0
Concentration – Flow System
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Reactor Effluent
d
a
c
a
b
a
and      1
CA 
FA

Flow System – Stoichiometry Table
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
CA 
Concentration – Flow System
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and
FA


d c b
  1
a a a
Stoichiometry
Concentration Flow System:
FA
CA 

Liquid Phase Flow System:
  0
FA
CA 


FA 0 1  X 
0
 CA 0 1  X  Flow Liquid Phase

N A 0 
b 
b 
CB 

B  X  CA 0B  X 


0 
a 
a 
NB
etc.
27
We will consider CA and CB for gas phase
reactions in the next lecture
Heat Effects
Isothermal Design
Stoichiometry
Rate Laws
Mole Balance
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End of Lecture 4
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