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 1/18/2011
 Block 1
 Size CSTRs and PFRs given –rA=f(X)
 Block 2
 Reaction Orders
 Arrhenius Equation
 Block 3
 Stoichiometric Table
 Definitions of Concentration
 Calculate the Equilibrium Conversion, Xe
2
Review Lecture 2
Reactor
Differential
Algebraic
Integral
X
X
Batch
N A0
V
CSTR
PFR
dX
t  N A0 
 rAV
0
dX
  r AV
dt
dX
FA0
 rA
dV
t
FA0 X
 rA
X
V  FA0 
0
dX
 rA
X
PBR
3
dX
FA0
  rA
dW
X
W  FA 0 
0
dX
 rA
W
Review Lecture 2
FA 0
rA
X


Review Lecture 2
Area = Volume of PFR
FA 0
rA
V
X1


X1
0
FA 0 

d X
rA 
Review Lecture 2
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
7
Review Lecture 2
Two steps to get
Step 1: Rate Law
rA  gC
i
Step 2: Stoichiometry
Ci   hX 

Step 3: Combine to get

rA  f X 
rA  f X 
Review Lecture 3
Power Law Model
α order in A
 
 rA  kCACB
β order in B
OverallRectionOrder  α  β
2A  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
 rA  k AC2ACB
9
2nd order in A, 1st order in B, overall third order
Review Lecture 3
k

Ae
E = Activation energy (cal/mol)
 E RT
T  k  A
k
R = Gas constant (cal/mol*K)
T
T = Temperature (K)
A = Frequency factor (same units as rate constant k)
 order)
(units of A, and k, depend on overall reaction
10
T 0 k 0
A  1013
Review Lecture 3
Mole Balance
Rate Laws
These topics build upon one another
11
Stoichiometry
Review Lecture 3
rA  f X 
Step 1: Rate Law

Step 2: Stoichiometry
Step 3: Combine
 to get


12
rA  gCi 
Ci   hX 
rA  f X 
We shall set up Stoichiometry Tables using species A as our basis
of calculation in the following reaction. We will use the
stochiometric 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.
13
NA  NA 0  NA 0 X
For every mole of A that react, b/a moles of B react. Therefore
moles of B remaining:
N B 0 b 
b
N B  N B 0  N A 0  N A 0 
 X 
a
N A 0 a 
Let ΘB = NB0/NA0
Then:
14

b 
N B  N A 0 B  X 

a 

c
c 
NC  NC 0  N A 0 X  N A 0 C  X 

a
a 
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
NT=NT0+δNA0X
N i 0 Ci 00 Ci 0
yi 0 and   d  c  b  1



Where: i 
a a a
N A0 C A00 C A0 y A0
15
δ = change in total number of mol per mol A reacted
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
N A N A 0 1  X 
CA 

 CA 0 1  X 
V
V0


16

N B N A 0 
b 
b 
CB 

B  X  CA 0B  X 

V
V0 
a 
a 
etc.
Suppose rA  kA CA2 CB
Batch:
V  V0
rA  k A CA 0
2

b 
1  X  B  X 

a 
Equimolar feed:
B  1
Stoichiometric feed:
b
B 
a

17
2
if rA  kA CA2 CB then
rA  CA 0

3

b 
1  X  B  X  ConstantVolume Batch

a 
2
and we have r  f X 
A

18
1
 rA
X
Calculating the equilibrium conversion
for gas phase reaction,Xe
Consider the following elementary reaction with KC=20 dm3/mol
and CA0=0.2 mol/dm3.
Xe’ for both a batch reactor and a flow reactor.
2A  B
 2 CB 
 rA  k A CA 

K
C

19
Calculate Xe
CA0  0.2 mol dm
3
KC  20 dm3 mol
dX rAV

dt
NA 0
Step 2: rate law,
 rA  k AC2A  k BCB
Step 1:

20
 2 CB 
 rA  k A CA 

KC 

kA
KC 
kB
Calculate Xe
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
CBe
Ke  2
CAe
21
C Be
0C 
KC
2
Ae
NAe
CAe 
 CA 0 1 X e 
V
Xe
CBe  CA 0
2
Calculating the equilibrium conversion
for gas phase reaction
Solution:
At equilibrium
Stoichiometry
Constant volume
Batch
Species
A
B
22
 2 C Be 
 rA  0  k A C Ae 

K
C 

C Be
KC  2
C Ae
A  B/ 2
V  V0
Initial
NA0
0
NT0=NA0
Change
-NA0X
+NA0X/2
Remaining
NA=NA0(1-X)
NB=NA0X/2
NT=NA0-NA0X/2
Xe
CA0
Xe
2
Ke 

2
2
CA 0 1  X e  2CA 0 1  X e 
2K e C A 0
Xe

 2200.2  8
2
1  Xe 
X eb  0.703
23
Flow System Stochiometric Table
Species Symbol Reactor Feed Change
24
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
CA 0 0
CA 0
yA 0
Flow System Stochiometric Table
Species Symbol Reactor Feed Change
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=A0ΘI
----------
FT0
FA 0
and
C A 0 0
FI=FA0ΘI
FT=FT0+δFA0X
Where: i  Fi 0  Ci 0 0  Ci 0  yi 0
25
Reactor Effluent
CA 0
yA0
d c b
    1
a a a
FA
Concentration – Flow System C A 

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
FT0
Where:
26
i 
Fi 0
C 
C
y
 i0 0  i0  i0
FA 0 CA 0 0 CA 0 y A 0
Concentration – Flow System C A 
FT=FT0+δFA0X
and
FA


d c b
  1
a a a
Concentration Flow System: C A 
Liquid Phase Flow System:
CA 
FA


FA 0 1  X 
0
FA

  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.
We will consider CA and CB for gas phase reactions in the
next lecture

27
Heat Effects
Isothermal Design
Stoichiometry
Rate Laws
Mole Balance
28
End of Lecture 4
29