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.

2




Block 1: Mole Balances
Block 2: Rate Laws
Block 3: Stoichiometry
Block 4: 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

3
Review Lecture 2
in terms of conversion, X
Reactor
Batch
Differential
N
A 0 dX dt
  r
A
V
Algebraic t

Integral
N
A 0

0
X dX
 r
A
V
X
CSTR V 
F
A 0
 r
A
X
PFR
F
A 0 dX dV
  r

A
F
A 0 dX dW
V

F
A 0
X

0 dX
 r
A
W

F
A 0
X

0 dX
 r
A

X t
W


4
Review Lecture 3
Power Law Model:
 r

A kC

A
C
B
 α order in A
β order in B
Overall Rection Order
 α  β
2 A

B

3 C
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
 r
A
 k
A
C
2
A
C
B
2nd order in A, 1st order in B, overall third order

5
Review Lecture 4

6
Review Lecture 5

Review Lecture 5
Building Block 4:
Combine
7

8
 Example for Liquid Phase Undergraduate Laboratory
Experiment
(CH
2
CO)
2
O + H
2
O  2CH
A + B 
3
COOH
2C
Entering
Volumetric flow rate
Acetic Anhydride
Water
Elementary with k’ v
0
= 0.0033 dm
7.8% (1M)
3 /s
92.2% (51.2M)
1.95x10
-4 dm 3 /(mol.s)
Case I
Case II
CSTR
PFR
V = 1dm 3
V = 0.311 dm 3

9
 Example for Gas Phase : PFR and Batch Calculation
2NOCl  2NO + Cl
2
2A  2B + C
Pure NOCl fed with C
NOCl,0
= 0.2 mol/dm elementary rate law with k = 0.29 dm 3 /(mol.s)
3 follows an
Case I PFR with v
0
= 10 dm3/s
Find space time, 𝜏 with X = 0.9
Find reactor volume, V for X = 0.9
Case II Batch constant volume
Find the time, t, necessary to achieve 90% conversion. Compare 𝜏 and t.

10
Part 1: Mole Balances in terms of
Conversion
Algorithm for Isothermal Reactor Design
1.
Mole Balances and Design Equation
2.
3.
Rate Laws
Stoichiometry
4.
5.
Combine
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)

11
CSTR Laboratory Experiment
Example: CH
3
CO
2
+ H
2
0  2CH
3
OOH
C
A 0

1 M
C
B 0

51 .
2 M
X

?
V  1 dm 3

0
 3.3
 10  3 dm 3 s
A + B  2C
CSTR : V

F

A 0 r
A
X

12
CSTR Laboratory Experiment
2) Rate Law:  r

A k
A
C
A
C
B
3) Stoichiometry:
A
B
F
A0
F
A0
Θ
B
C 0
-F
A0
X
-F
A0
X
2F
A0
X
F
A
=F
A0
(1-X)
F
B
=F
A0
(Θ
B
-X)
F
C
=2F
A0
X

13
CSTR Laboratory Experiment
C
A

F
A


F
A 0

1

X


0

C
A 0

1

X

C
B

F
A 0



B
0

X


C
A 0


B

X


B

51 .
2
1

51 .
2
C
B

C
A 0

51 .
2

X


C
 
A 0
51 .
2

C
B 0

14
CSTR Laboratory Experiment
 r
A

'
 0
C
A 0

1

X

 k kC
A 0

1

X

V

C

0
A 0 kC
A 0
1

X

X


V

0


1 kX

X

  
V

0


1 kX

X

X

1
 k
  k
X

3 .
03
4 .
03

0 .
75

15
PFR Laboratory Experiment
A + B  2C
0 .
00324 dm
3 s
0.311 dm 3
1) Mole Balance:
 dX

 dV F r
A
A 0
X

?
2) Rate Law:  r

A kC
A
C
B
3) Stoichiometry: C
A

C
A 0

1

X

C

B
C
B 0

16
PFR Laboratory Experiment
4) Combine:
 r
A
 k ' C
B 0
C
A 0

1

X

 kC
A 0

1

X
 dX dV
 kC
A 0
C

1

X

A 0

0

1 dX

X

 k

0 dV ln
1

1
X
X


1
 e
 k
 k

 kd

 
V

0

0 .
311 dm
0 .
00324 dm
3
3 sec

96 .
0 sec k

0 .
01 s

1
X

0 .
61

17
Gas Flow PFR Example

0
2 NOCl  2 NO + Cl
2
2A  2B + C

10 dm
3 s
T

T
0
P k

0 .
29 dm
3 mol
 s
 P
0
X  0.9
C
A 0

0 .
2 mol
L
V

?
1) Mole Balance:


2) Rate Law: dX

 dV F
A r
A
0
 r

A kC
A
2

18`1
Gas Flow PFR Example
3) Stoichiometry:
(Gas Flow)
4) Combine:

C
A
 
0

1
 
X


C
A 0
1


1


X
X
 

A  B + ½C
 r
A
 kC
A
2
1
0


1


X
X
 
2

2 dX dV

C
2 kC
A 0
A 0

0


1
1


X

X

2

2
 
X
0


1
1



X
X

2

2 dX
 
V
0 kC
A 0

0 dV
 kC
A

0
0
V

  kC
A 0


Gas Flow PFR Example
19 kC
A 0
 
2
 
1
   
1

X

 
2
X


1
  
2
1

X
X
  y
A 0
 
1
2

1
2 kC
A 0
 
17 .
02
 
17 .
02 kC
A 0

294 sec
V

V
0
 
2940 L

20
Constant Volume
Example
Gas Phase 2A  2B + C t=?
1) Mole Balance:
2) Rate Law: dX dt

 r
A
V
0
N
A 0


N
A 0 r
A
V
0

 r
C
A 0
A
 r

A kC
A
2
3) Stoichiometry:
(Gas Flow)
V

V
0
C
A
 r
A

N
A 0

1

X


C
A 0

1

X


V
0 kC
2
A 0

1

X

2

Constant Volume
Example
21
4) Combine: dX dt

2 kC
A 0

1

X

2
C
A 0
 kC
A 0

1

X

2 dX
 kC
A 0

1

X

2 dt

1 dX

X

2
 kC
A 0 dt
1

1
X
 kC
A 0 t t

155 sec

22
Heat Effects
Isothermal Design
Stoichiometry
Rate Laws
Mole Balance

23