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
dX
dt
  r AV
0
V 
CSTR
PFR
t  N A0 
FA 0
dX
 r AV
FA 0 X
 rA
X
dX
  rA
dV

t
V  FA0 
0
dX
 rA
X
3
PBR

FA 0
dX
dW
X
  rA
W  FA0 
0
dX
 r A
W
Review Lecture 2
Levenspiel Plots
FA 0
 rA
X

4

Review Lecture 2
PFR
5
Review Lecture 2
Reactors in Series
Xi 
moles of A reacted up to point i
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 C i 

Step 2: Stoichiometry
Step 3: Combine to get
8
C i   h  X 
 rA  f  X

Review Lecture 3
Building Block 2: Rate Laws
Power Law Model:


 rA  kC A C B
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
2
 rA  k A C A C B
9
2nd order in A, 1st order in B, overall third order
Review Lecture 3
Arrhenius Equation
k  Ae
T  k A
 E RT
k
T 0 k 0
E = Activation energy (cal/mol)
13
A  10
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
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 C i 
Step 2: Stoichiometry
Step 3: Combine to get
12
C i   h  X 
 rA  f  X

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).
A 
b
B
a
A is the limiting reactant.
13
c
a
C
d
a
D
Stoichiometry
N A  N A0  N A0X
For every mole of A that reacts, b/a moles of B react. Therefore
moles of B remaining:
NB  NB0 
b
a
NA0
N

b
B0
 N A 0 
 X 
a 
N A 0
Let ΘB = NB0/NA0
Then:
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b


N B  N A0   B  X 
a



c
c 
N C  N C 0  N A 0 X  N A 0  C  X 

a
a 
Batch System - Stoichiometry Table
Species
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
Where:
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i 
N i0
N A0

C i 0 0
C A 0 0
NT=NT0+δNA0X

C i0
C A0

yi0
y A0
and 

d
a
δ = change in total number of mol per mol A reacted

c
a

b
a
1
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  V 0
CA 
CB 

16

etc.
NA
V
NB
V

N A 0 1  X 
V0
N A 0 


 B
V 0 
 C A 0 1  X 


b 

X   C A 0  B 
X 

a 
a 
b
Stoichiometry Constant Volume Batch
2
Suppose  rA  k A C A C B
Batch: V  V 0
2

 rA  k A C A 0 1  X
Equimolar feed:

2
b


B  X 
a


B 1
Stoichiometric feed:  B 
17
b
a
Stoichiometry Constant Volume Batch
If  rA  k A C A2 C B , then
 rA  C A 0 1  X
3

2
b


  B  X  Constant Volume Batch
a


and we have  rA  f  X 
1

18
 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

CB 
2
 rA  k A  C A 

KC 

19
Batch Reactor - Example
C A 0  0 . 2 mol dm
Calculate Xe
K C  20 dm
Step 1:
dX
dt

NA0
 rA  k A C A  k B C B
2

CB 
2
 rA  k A  C A 

K
C 

KC 
20
mol
 rA V
Step 2: rate law:

3
3
kA
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
0C
@ equilibrium: -rA=0
Ke 
21
C Be
C
2
Ae
2
Ae
C Ae 

C Be
KC
N Ae
V
C Be  C A 0
 C A 0 1  X e 
Xe
2
Batch Reactor - Example
Solution:
At equilibrium

C Be 
2
 rA  0  k A  C Ae 

K
C 

Stoichiometry:
Constant Volume:
KC 
A  B/2
V  V0
Batch
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Species
Initial
Change
Remaining
A
B
NA0
0
NT0=NA0
-NA0X
+NA0X/2
NA=NA0(1-X)
NB=NA0X/2
NT=NA0-NA0X/2
C Be
2
C Ae
Batch Reactor - Example
C A0
Xe
Xe
2
Ke 

2
2
C A 0 1  X e  2 C A 0 1  X e 
2K eC A 0 
Xe
1  X e 
X eb  0 .703
23
2
 2  20 0 . 2   8
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:
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i 
Fi0
FA 0

C i 0 0
C A 0 0

C i0
CA0

y i0
yA0
Flow System – Stoichiometry Table
Species
Symbol
Reactor Feed
Change
Reactor Effluent
C
C
FC0=FA0ΘC
+c/aFA0X
D
D
FD0=FA0ΘD
+d/aFA0X FD=FA0(ΘD+d/aX)
Inert
I
FI0=A0ΘI
FI=FA0ΘI
----------
FT=FT0+δFA0X
FT0
Where:  i

Fi 0
FA 0

C i0  0
C A00

C i0
CA0

y i0
and
yA0
Concentration – Flow System
25
FC=FA0(ΘC+c/aX)

d
a
CA 
FA


c
a

b
a
1
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
FA 0

C i0  0
C A00

C i0
CA0

y i0
and
yA0
CA 
Concentration – Flow System
26
FA


d
a

c
a

b
a
1
Stoichiometry
FA
Concentration Flow System:
CA 
Liquid Phase Flow System:
  0
FA
CA 
CB


F A 0 1  X 
0

 C A 0 1  X  Flow Liquid Phase

N A 0 
b 
b 


 B  X   C A 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
28
End of Lecture 4
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