# Lec6_non ```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 1/25/2011




Block 1:
Block 2:
Block 3:
Block 4:
Review of Blocks 1, 2 and 3
CSTR
PFR
BR
Gas Phase Reaction with Change in
the Total Number of Moles
2
Previous Lectures
Reactor
Differential
Algebraic
Integral
X
Batch
N A0
dX
dt
  r AV
0
V 
CSTR
PFR
t  N A0 
FA0
dX
dV
dX
 r AV
t
FA0 X
 rA
X
  rA
X
V  FA0 
0
dX
 rA
X
PBR
3
FA0
dX
dW
X
  rA
W  FA0 
0
dX
 r A
W
Previous Lectures
Power Law Model


 rA  kC A C B
α order in A
β order in B
Overall Rection
4
Order  α  β
2 A  B  3C
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
2nd order in A, 1st order in B, overall third order
Previous Lectures
5
Previous Lectures
6
Previous Lectures
7
Today’s lecture
 Example for Liquid Phase Undergraduate Laboratory Experiment
(CH2CO)2O + H2O
A
+ B
Entering
Volumetric flow rate
Acetic Anhydride
Water
Elementary with k’
Case I
Case II
8
CSTR
PFR
 2CH3COOH

2C
v0 = 0.0033 dm3/s
7.8% (1M)
92.2% (51.2M)
1.95x10-4 dm3/(mol.s)
V = 1dm3
V = 0.311 dm3
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
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.

9
Lecture 6
Algorithm for Isothermal Reactor Design
1. Mole Balance and Design Equation
2. Rate Law
3. Stoichiometry
4. Combine
5. Evaluate
A. Graphically (Chapter 2 plots)
10
B. Numerical (Quadrature Formulas Chapter 2 and appendices)
C. Analytical (Integral Tables in Appendix)
D. Software Packages (Appendix- Polymath)
Example: CH3CO2 + H20  2CH3OOH
C A 0  1M
C B 0  51 . 2 M
X ?
V  1 dm
3
 0  3.3 10
3
dm
3
s
 2C
A+B
1) Mole 
Balance:
11
CSTR:
V 
FA0 X
 rA
2) Rate Law:
 rA  k A C A C B
3) Stoichiometry:
12
A
FA0
-FA0X
FA=FA0(1-X)
B
FA0ΘB
-FA0X
FB=FA0(ΘB-X)
C
0
2FA0X
FC=2FA0X
CA 
CB 
B 
FA


F A 0 1  X
0
F A 0  B  X
0
51 . 2


 C A 0 1  X
 C A 0  B  X


 51 . 2
1
C B  C A 0 51 . 2  X   C A 0 51 . 2   C B 0
13
 rA  k ' C B 0 C A 0 1  X   kC A 0 1  X

k
V 
X 
 0 kC A 0 X
C A 0 1  X
k
1  k
X 
3 . 03
4 . 03
14

 0 . 75

V
0

kX
1 
X


  
V
0

kX
1 
X

A + B  2C
0 . 00324
dm
3
0.311 dm
s
1) Mole Balance:
dX

2) Rate Law:
3) Stoichiometry:
15

dV
X
3
 rA

FA0
 r A  kC A C B
C A  C A 0 1  X
CB  CB0

 rA  k ' C B 0 C A 0 1  X   kC A 0 1  X
4) Combine:
dX
kC A 0 1  X

dX
ln
 
16

C A 0 0
dV
1 
X


1
1 X
k
0
dV  kd 
 k
X  1 e
 k
V
0 . 311 dm
0


0 . 00324 dm
X  0 . 61
3
3
96 . 0 sec
sec
k  0 . 01 s
1
2 NOCl  2 NO + Cl2
2A  2B + C
 0  10
dm
3
k  0 . 29
s
T  T0
2) Rate Law:
17
3
C A 0  0 .2
mol  s
P  P0
1) Mole Balance:

dm
X  0.9
dX
dV


 rA
FA0
 rA  kC A
2
mol
L
   0 1   X
3) Stoich: Gas

C A 0 1  X
CA 

1   X 
A  B + &frac12;C
4) Combine:
 rA 
1   X 
 1  X 2
0

18`1
V
dX 
1 
X

1   X 2

dV
2
kC
2
kC A 0 1  X
2
2
C A 0 0 1   X 
2
dX
X
2
A0

0
kC A 0
0
Da

kC A 0V
dV 
 kC A 0
0
kC A 0  2  1    ln 1  X    X 
2
1 1
  y A 0   1   
2 2
kC A 0  17 . 02
 
17 . 02
 294 sec
kC A 0
V  V 0  2940 L
19
1   2 X
1 X
Gas Phase 2A  2B + C
1) Mole Balance:
dX

dt
2) Rate Law:
3) Stoich:
 rAV 0
N A0

N A0 V0

 rA
C A0
 rA  kC A
2
Gas, but:
CA 
V  V0
N A 0 1  X

V0
 r A  kC A 0 1  X
2
20
 rA
 C A 0 1  X
2

4) Combine:
dX
X

 kC A 0 1  X
2

kC
dt
1 
2
C A0
dX
dt
dX
1 
X
1
1 X
21
2
A0

2
 kC A 0 dt
 kC A 0 t
t  155 sec
 kC A 0 1  X

2
Heat Effects
Isothermal Design
Stoichiometry
Rate Laws
Mole Balance
22
End of Lecture 6
23
```