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Lecture 6
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 6 – Tuesday 1/29/2013




Block 1:
Block 2:
Block 3:
Block 4:
Mole Balances
Rate Laws
Stoichiometry
Combine
Review of Blocks 1, 2 and 3
Examples : Undergraduate Reactor Experiments
CSTR
PFR
BR
Gas Phase Reaction with Change in
the Total Number of Moles
2
Review Lecture 2
Building Block 1: Mole Balances
in terms of conversion, X
Reactor
Differential
Algebraic
Integral
X
X
Batch
N A0
t
FA 0 X
V
rA
CSTR
PFR
dX
t  N A0 
 rAV
0
dX
  r AV
dt
dX
FA 0
 rA
dV 
X
V  FA0 
0
dX
 rA
X
PBR
3

dX
FA 0
 rA
dW
X
W  FA 0 
0
dX
 rA
W
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
OverallRectionOrder  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
4
2nd order in A, 1st order in B, overall third order
Review Lecture 4
Building Block 3:
Stoichiometry
5
Review Lecture 5
Building Block 4: Combine
6
Review Lecture 5
Building Block 4:
Combine
7
Today’s lecture
 Example for Liquid Phase Undergraduate Laboratory
Experiment
(CH2CO)2O + H2O  2CH3COOH
A
+ B

2C
8
Entering
Volumetric flow rate
Acetic Anhydride
Water
Elementary with k’
v0 = 0.0033 dm3/s
7.8% (1M)
92.2% (51.2M)
1.95x10-4 dm3/(mol.s)
Case I
Case II
V = 1dm3
V = 0.311 dm3
CSTR
PFR
Today’s lecture
 Example for Gas Phase : PFR and Batch Calculation
2NOCl  2NO + Cl2
2A
 2B + C
Pure NOCl fed with CNOCl,0 = 0.2 mol/dm3 follows an
elementary rate law with k = 0.29 dm3/(mol.s)
Case I
Case II
9
PFR with
v0 = 10 dm3/s
Find space time,  with X = 0.9
Find reactor volume, V for X = 0.9
Batch constant volume
Find the time, t, necessary to achieve 90%
conversion. Compare  and t.
Part 1: Mole Balances in terms of
Conversion
Algorithm for Isothermal Reactor Design
1. Mole Balances and Design Equation
2. Rate Laws
3. Stoichiometry
4. Combine
5. Evaluate
A. Graphically (Chapter 2 plots)
B. Numerical (Quadrature Formulas Chapter 2 and
appendices)
C. Analytical (Integral Tables in Appendix)
D. Software Packages (Appendix- Polymath)
10
CSTR Laboratory Experiment
Example: CH3CO2 + H20  2CH3OOH
CA0  1M
CB0  51.2 M
X?
V  1 dm 3
 0  3.3 10 3
dm 3
s
 2C
A+B
1) MoleBalance:
11
FA0 X
CSTR: V 
 rA
CSTR Laboratory Experiment
 rA  k AC ACB
2) Rate Law:
3) Stoichiometry:
12
A
FA0
-FA0X
FA=FA0(1-X)
B
FA0ΘB
-FA0X
FB=FA0(ΘB-X)
C
0
2FA0X
FC=2FA0X
CSTR Laboratory Experiment
CA 
CB 
FA


FA0 1  X 
0
FA0  B  X 
0
 C A0 1  X 
 C A0  B  X 
51.2
B 
 51 .2
1
CB  CA0 51.2  X   CA0 51.2  CB0
13
CSTR Laboratory Experiment
 rA  k ' CB 0 C A0 1  X   kCA0 1  X 

k
0 kCA0 X
kX
V
kX
V




C A0 1  X 
0 1  X 
0 1  X 
k
X 
1  k
3.03
X
 0.75
4.03
14
V
PFR Laboratory Experiment
A + B  2C
dm3
0.00324
s
0.311 dm 3
1) Mole Balance:

2) Rate Law:
3) Stoichiometry:
15
dX  rA

dV FA0
 rA  kCACB
CA  CA0 1  X 
CB  CB 0
X ?
PFR Laboratory Experiment
4) Combine:
 rA  k ' CB0CA0 1  X   kCA0 1  X 
dX kCA0 1  X 

dV
C A00
dX
k
 dV  kd
1  X  0
1
ln
 k
1 X
X  1  e  k
0.311dm 3
 
 96.0 sec
3
0 0.00324 dm sec
V
16
X  0.61
k  0.01 s 1
Gas Flow PFR Example
2 NOCl  2 NO + Cl2
2A  2B + C
dm3
dm3
k  0.29
0  10
m ol s
s
P  P0
X  0.9
1) Mole Balance:


dX  rA

dV FA0
2) Rate Law:
 rA  kCA2
T  T0
17
C A0
mol
 0 .2
L
V ?
Gas Flow PFR Example
3) Stoichiometry:
(Gas Flow)
  0 1  X 
C A0 1  X 
CA 
1  X 
A  B + ½C
kC 1  X 
 rA 
1  X 2
2
A0
4) Combine:
2
kCA2 0 1  X 
dX

dV C A0 0 1  X 2
2
1  X 
0 1  X 2
X

18`1
2
V
dX  
0
kCA0
0
Da


kCA0V
dV 
 kCA0
0
Gas Flow PFR Example
2


1


X
2
kCA0  2 1    ln1  X    X 
1 X
1 1
  y A0  1  
2 2
kCA0  17.02
17.02

 294sec
kCA0
V  V0  2940L
19
Constant Volume Batch Example
Gas Phase 2A  2B + C
t=?
1) Mole Balance: dX   rAV0   rA   rA
dt
N A0
N A0 V0 C A0
2) Rate Law:
3) Stoichiometry:
(Gas Flow)
20
 rA  kCA2
V  V0
N A0 1  X 
CA 
 C A0 1  X 
V0
 rA  kC 1  X 
2
A0
2
Constant Volume Batch Example
4) Combine:
dX kC 1  X 
2

 kCA0 1  X 
dt
C A0
2
A0
2
dX
2
 kC A0 1  X 
dt
dX
 kCA0 dt
2
1  X 
1
 kC A0t
1 X
21
t  155 sec
Heat Effects
Isothermal Design
Stoichiometry
Rate Laws
Mole Balance
22
End of Lecture 6
23

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