Transcript Lecture 4
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
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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
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Review Lecture 2
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Review Lecture 2
Two steps to get
Step 1: Rate Law
rA gC
i
Step 2: Stoichiometry
Ci hX
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
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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
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T 0 k 0
A 1013
Review Lecture 3
Mole Balance
Rate Laws
These topics build upon one another
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Stoichiometry
Review Lecture 3
rA f X
Step 1: Rate Law
Step 2: Stoichiometry
Step 3: Combine
to get
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rA gCi
Ci hX
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.
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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:
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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 00 Ci 0
yi 0 and d c b 1
Where: i
a a a
N A0 C A00 C A0 y A0
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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
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N B N A 0
b
b
CB
B X CA 0B 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
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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
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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
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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
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C Be
0C
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
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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
2200.2 8
2
1 Xe
X eb 0.703
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Flow System Stochiometric Table
Species Symbol Reactor Feed Change
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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
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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:
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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 0B X
0
a
a
NB
etc.
We will consider CA and CB for gas phase reactions in the
next lecture
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Heat Effects
Isothermal Design
Stoichiometry
Rate Laws
Mole Balance
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End of Lecture 4
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