Chap04-Stoichiometry-Chemical reaction engineering-20241010.pdf
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Oct 10, 2024
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About This Presentation
ㅇㅇ
Slide Content
1
2
3
Review-What we have learned
In Chapter 1, we have learned about General Mole Balance
Reactor Comment Differential Algebraic Integral Graph
BatchNo spatial variations
CSTR
No spatial variations,
Steady-state
- -
PFR Steady-state
PBR Steady-state
N
A
t
F
A
V
�=න
�??????1
�
??????0��
�
−�
��
��
�
��
=�
��
�=
??????
�0−??????
�
−�
�
�??????
�
��
=�
��
�??????
�
��
=�′
��
�
1=න
????????????1
????????????0�??????
�
−�
��
�
1=න
??????
??????1
????????????0�??????
�
−�′
��
F
A
W
Review-What we have learned
In Chapter 1, we have learned about General Mole Balance
Reactor Comment Differential Algebraic Integral Graph
BatchNo spatial variations
CSTR
No spatial variations,
Steady-state
- -
PFR Steady-state
PBR Steady-state
N
A
t
F
A
V
�=න
�??????1
�
??????0��
�
−�
��
��
�
��
=�
��
�=
??????
�0−??????
�
−�
�
�??????
�
��
=�
��
�??????
�
��
=�′
��
�
1=න
????????????1
????????????0�??????
�
−�
��
�
1=න
??????
??????1
????????????0�??????
�
−�′
��
F
A
W
4
Review-What we have learned
In Chapter 2, we have learned about Conversion(X)
•Mole balance which is in terms of X
•How to calculate reactor volume
•Comparison reactor size of CSTR and PFR
•Series of reactors.
Canyou explain
what “??????” means?
Review-What we have learned
In Chapter 2, we have learned about Conversion(X)
•Mole balance which is in terms of X
•How to calculate reactor volume
•Comparison reactor size of CSTR and PFR
•Series of reactors.
Canyou explain
what “??????” means?
5
Review-What we have learned
In Chapter 3, we have learned about Rate Laws
•Relationship of reaction rate for each species,
•Writing rate law in terms of concentration,
•Using Arrhenius equation to find the rate constant as a function of temp.
•Activation energy
−�
�=�
�(�)��
��
��
��
�…
Review-What we have learned
In Chapter 3, we have learned about Rate Laws
•Relationship of reaction rate for each species,
•Writing rate law in terms of concentration,
•Using Arrhenius equation to find the rate constant as a function of temp.
•Activation energy
−�
�=�
�(�)��
��
��
��
�…
6
Review-What we have learned
�=
�
�0−�
�
�
�
�=
??????
�0−??????
�
??????
�
For batch systemsFor flow systems
In this chapter, we learned that
•How to define Conversion(X)
Reactor Differential Algebraic Integral Concentration Conversion
Batch -
CSTR - - - -
PFR -
PBR -
N
A
t
F
A
V
�=�
�0න
0
??????
��
−�
��
�
�0
��
��
=−�
��
??????
�0
��
��
=−�
�� �=න
??????
??????�
??????
���??????
�0��
−�
�
F
A
W
X
t
X
W
X
V
??????
�0
��
��
=−�′
�
�=න
??????
??????�
??????���??????
�0��
−�′
�
�=
??????
�0∙(�
���−�
��)
−�
����
�
�=
��������������������������??????��??????
������������ℎ�1���������
•The reactor design equation in terms of the Conversion
Review-What we have learned
�=
�
�0−�
�
�
�
�=
??????
�0−??????
�
??????
�
For batch systemsFor flow systems
In this chapter, we learned that
•How to define Conversion(X)
Reactor Differential Algebraic Integral Concentration Conversion
Batch -
CSTR - - - -
PFR -
PBR -
N
A
t
F
A
V
�=�
�0න
0
??????
��
−�
��
�
�0
��
��
=−�
��
??????
�0
��
��
=−�
�� �=න
??????
??????�
??????
���??????
�0��
−�
�
F
A
W
X
t
X
W
X
V
??????
�0
��
��
=−�′
�
�=න
??????
??????�
??????���??????
�0��
−�′
�
�=
??????
�0∙(�
���−�
��)
−�
����
�
�=
��������������������������??????��??????
������������ℎ�1���������
•The reactor design equation in terms of the Conversion
7
Lecture goal
8
We will learn about
•Relationship between concentrationand conversion.
•Definition of concentration.
•Use stoichiometry tables to find the concentration as a function of conversion.
•Finally, we will be able to write the rate of reaction as a function of conversion,
•And can calculate the equilibrium conversion for both batch and flow reactor.
Rate law → Designequation(It was NOT a function of conversion, X)
Why? It was written in k, C
j
∴We will learn how to express in Stoichiometry coefficient.
8
We will learn about
•Relationship between concentrationand conversion.
•Definition of concentration.
•Use stoichiometry tables to find the concentration as a function of conversion.
•Finally, we will be able to write the rate of reaction as a function of conversion,
•And can calculate the equilibrium conversion for both batch and flow reactor.
Rate law → Designequation(It was NOT a function of conversion, X)
Why? It was written in k, C
j
∴We will learn how to express in Stoichiometry coefficient.
9
10
4.1. Batch systems
aA+ bB→cC+ dD
−??????�
�
=
−??????�
�
=
??????�
�
=
??????�
�
A+
b
a
B
c
a
C+
d
a
D
Why stoichiometry?
N
A0
N
B0
N
C0
N
D0
N
I0
N
A
N
B
N
C
N
D
N
I
���=0 ���=�
Baissofcalculation
Inert
(i.e. water, solvent)
�
�=�
�0
−�
�0
�=�
�0
(1−�)
4.1. Batch systems
aA+ bB→cC+ dD
−??????�
�
=
−??????�
�
=
??????�
�
=
??????�
�
A+
b
a
B
c
a
C+
d
a
D
Why stoichiometry?
N
A0
N
B0
N
C0
N
D0
N
I0
N
A
N
B
N
C
N
D
N
I
���=0 ���=�
Baissofcalculation
Inert
(i.e. water, solvent)
�
�=�
�0
−�
�0
�=�
�0
(1−�)
11
4.1. Batch systems
Stoichiometry table
�
�=�
�
0
−�
�
0
�=�
�
0
(1−�)
A+
b
a
B
c
a
C+
d
a
D
δ=
Cℎ����??????�������������������
��������������
∴�
�=�
�
0
=�(�
�
0
�)
4.1. Batch systems
Stoichiometry table
�
�=�
�
0
−�
�
0
�=�
�
0
(1−�)
A+
b
a
B
c
a
C+
d
a
D
δ=
Cℎ����??????�������������������
��������������
∴�
�=�
�
0
=�(�
�
0
�)
12
To determine the reaction rate as a function of
conversion(X),
We need to know the concentrations of
the reacting species as a function of
conversion(X).
4.1. Batch systems
−??????
�☞??????
??????(??????)
All I Want For Chap. 4 is �
??????=??????
??????(??????)
To determine the reaction rate as a function of
conversion(X),
We need to know the concentrations of
the reacting species as a function of
conversion(X).
4.1. Batch systems
−??????
�☞??????
??????(??????)
All I Want For Chap. 4 is �
??????=??????
??????(??????)
4.1.1. Batch concentration for generic reaction
13
What is concentration?�
�=
??????
�
??????
�
�=
�
�
�
=
�
�
0
(1−�)
�
�
�=
�
�
�
=
�
�
0
−(Τ��)�
�
0
�
�
�
�=
�
�
�
=
�
�0
+(��)�
�0
�
�
�
�=
�
�
�
=
�
�
0
+(��)�
�
0
�
�
From Stoichiometric table
13
What is concentration?�
�=
??????
�
??????
�
�=
�
�
�
=
�
�
0
(1−�)
�
�
�=
�
�
�
=
�
�
0
−(Τ��)�
�
0
�
�
�
�=
�
�
�
=
�
�0
+(��)�
�0
�
�
�
�=
�
�
�
=
�
�
0
+(��)�
�
0
�
�
From Stoichiometric table
14
Almost batch reactors are rigid vessel
4.1.1. Batch concentration for generic reaction
�=�
0
☞
�
�=
�
�
�
0
=
�
�
0
(1−�)
�
0
�
�=�
�
0
(1−�)
Θ
�=
�
�
0
�
�
0
=
�
�
0
�
�
0
=
??????
�
0
??????
�
0
Θ
�=
�����������??????��"??????"??????�??????�??????���??????
�����������??????���??????�??????�??????���??????
Mole fraction
�
�=
�
�
�
=
�
�
0
−(Τ��)�
�
0
�
�
=
�
�
0
Τ�
�
0
�
�
0
−(Τ��)�
�
=
�
�
0
Θ
�−(Τ��)�
�
�
�=�
�
0
Θ
�−
�
�
�Θ
�=
�
�0
�
�0
�
�=�
�
0
Θ
�+
�
�
�
Θ
�=
�
�0
�
�0
Θ
�=
�
�0
�
�0
�
�=�
�0
Θ
�+
�
�
�
☞
If we know the rate law,
nowwe can now obtain –r
A= f(X)
따란~!
Almost batch reactors are rigid vessel
4.1.1. Batch concentration for generic reaction
�=�
0
☞
�
�=
�
�
�
0
=
�
�
0
(1−�)
�
0
�
�=�
�
0
(1−�)
Θ
�=
�
�
0
�
�
0
=
�
�
0
�
�
0
=
??????
�
0
??????
�
0
Θ
�=
�����������??????��"??????"??????�??????�??????���??????
�����������??????���??????�??????�??????���??????
Mole fraction
�
�=
�
�
�
=
�
�
0
−(Τ��)�
�
0
�
�
=
�
�
0
Τ�
�
0
�
�
0
−(Τ��)�
�
=
�
�
0
Θ
�−(Τ��)�
�
�
�=�
�
0
Θ
�−
�
�
�Θ
�=
�
�0
�
�0
�
�=�
�
0
Θ
�+
�
�
�
Θ
�=
�
�0
�
�0
Θ
�=
�
�0
�
�0
�
�=�
�0
Θ
�+
�
�
�
☞
If we know the rate law,
nowwe can now obtain –r
A= f(X)
따란~!
15
4.1.1. Batch concentration for generic reaction
For liquid-phase reactions taking place in solution, the solvent usually dominates
the situation.
As a result, changes in the density of the solute do NOT affect the overall density of
the solution significantly.
Now we have concentration as
afunction of conversion!
For example, most liquid-phase organic reactions do not change density during the
course of the reactionand represent still another case for which the constant-
volumesimplifications apply.
Therefore it is essentially a constant-volume reaction process, V = V
0and v = v
0.
−�
�=��
��
�
aA+ bB→cC+ dD
−�
�=��
�
2
1−�Θ
�−
�
�
�
�
�=�
�
0
Θ
�−
�
�
�
�
�=�
�
0
(1−�)
=�(�)
4.1.1. Batch concentration for generic reaction
For liquid-phase reactions taking place in solution, the solvent usually dominates
the situation.
As a result, changes in the density of the solute do NOT affect the overall density of
the solution significantly.
Now we have concentration as
afunction of conversion!
For example, most liquid-phase organic reactions do not change density during the
course of the reactionand represent still another case for which the constant-
volumesimplifications apply.
Therefore it is essentially a constant-volume reaction process, V = V
0and v = v
0.
−�
�=��
��
�
aA+ bB→cC+ dD
−�
�=��
�
2
1−�Θ
�−
�
�
�
�
�=�
�
0
Θ
�−
�
�
�
�
�=�
�
0
(1−�)
=�(�)
16
4.1.1. Batch concentration for generic reaction
Example 4-1 Saponification(비누화반응 )
4.1.1. Batch concentration for generic reaction
Example 4-1 Saponification(비누화반응 )
17
4.1.1. Batch concentration for generic reaction
Example 4-1
3���??????
(??????�)+�
17??????
35���
3�
3??????
5⟶3�
17??????
35�����+�
3??????
5(�??????)
3
Saponification(비누화반응 )
Expressing Cjhj(X) for a Liquid-Phase Batch Reaction
caustic soda glyceryl stearate sodiumstrearate(soap) glycerol
���??????
(??????�)+
1
3
�
17??????
35���
3�
3??????
5⟶�
17??????
35�����+
1
3
�
3??????
5(�??????)
3
�+
1
3
�
⟶ �+
1
3
�
÷
1
3
�
�=
�
�
�
=
�
�0
(1−�)
�
=�
�0
(1−�)
Liquid reaction
(constant density)
Θ
�=
�
�0
�
�0
Θ
�=
�
�0
�
�0
Θ
�=
�
�0
�
�0
4.1.1. Batch concentration for generic reaction
Example 4-1
3���??????
(??????�)+�
17??????
35���
3�
3??????
5⟶3�
17??????
35�����+�
3??????
5(�??????)
3
Saponification(비누화반응 )
Expressing Cjhj(X) for a Liquid-Phase Batch Reaction
caustic soda glyceryl stearate sodiumstrearate(soap) glycerol
���??????
(??????�)+
1
3
�
17??????
35���
3�
3??????
5⟶�
17??????
35�����+
1
3
�
3??????
5(�??????)
3
�+
1
3
�
⟶ �+
1
3
�
÷
1
3
�
�=
�
�
�
=
�
�0
(1−�)
�
=�
�0
(1−�)
Liquid reaction
(constant density)
Θ
�=
�
�0
�
�0
Θ
�=
�
�0
�
�0
Θ
�=
�
�0
�
�0
18
4.1.1. Batch concentration for generic reaction
Example 4-2 What is the Limiting reactant?
If the initial mixture consists of sodium hydroxide at a concentration of 10 mol/dm
3
and glyceryl stearate at a
concentration of 2 mol/dm
3
, what are the concentrations of glyceryl stearate(B), and of glycerine(D), when the
conversion of sodium hydroxide is (a) 20% and (b) 90%?
sol(a) For 20% conversion of NaOH
�
�=�
�0
�
3
=10
0.2
3
=0.67���/�(=0.67 �����
3
)
�
�=�
�0
Θ
�−
�
3
=10
2
3
−
0.2
3
=10(0.133)=1.33���/�(=1.33 �����
3
)
(b) For 90% conversion of NaOH
�
�=�
�0
�
3
=10
0.9
3
=3���/�(=3 �����
3
)
�
�=�
�0
Θ
�−
�
3
=10
2
3
−
0.9
3
=100.2−0.3=−??????�??????�/??????(=−??????Τ�??????���
??????
)
���??????
(??????�)+
1
3
�
17??????
35���
3�
3??????
5⟶�
17??????
35�����+
1
3
�
3??????
5(�??????)
3
�
+
1
3
�⟶ �+
1
3
�
4.1.1. Batch concentration for generic reaction
Example 4-2 What is the Limiting reactant?
If the initial mixture consists of sodium hydroxide at a concentration of 10 mol/dm
3
and glyceryl stearate at a
concentration of 2 mol/dm
3
, what are the concentrations of glyceryl stearate(B), and of glycerine(D), when the
conversion of sodium hydroxide is (a) 20% and (b) 90%?
sol(a) For 20% conversion of NaOH
�
�=�
�0
�
3
=10
0.2
3
=0.67���/�(=0.67 �����
3
)
�
�=�
�0
Θ
�−
�
3
=10
2
3
−
0.2
3
=10(0.133)=1.33���/�(=1.33 �����
3
)
(b) For 90% conversion of NaOH
�
�=�
�0
�
3
=10
0.9
3
=3���/�(=3 �����
3
)
�
�=�
�0
Θ
�−
�
3
=10
2
3
−
0.9
3
=100.2−0.3=−??????�??????�/??????(=−??????Τ�??????���
??????
)
���??????
(??????�)+
1
3
�
17??????
35���
3�
3??????
5⟶�
17??????
35�����+
1
3
�
3??????
5(�??????)
3
�
+
1
3
�⟶ �+
1
3
�
19
20
4.2. Flow systems
A+
b
a
B
c
a
C+
d
a
D
??????
A
??????
B
??????
C
??????
D
??????
I
Baissofcalculation
Inert
(i.e. water, solvent)
??????
A0
??????
B0
??????
C0
??????
D0
??????
I0
aA+ bB→cC+ dD
Θ
�=
??????
�0
??????
�0
=
�
�0
�
0
�
�0
�
0
=
�
�0
�
�0
=
??????
�0
??????
�0
where,
�=
�
�
+
�
�
−
�
�
−1
δ=
Cℎ����??????�������������������
��������������
4.2. Flow systems
A+
b
a
B
c
a
C+
d
a
D
??????
A
??????
B
??????
C
??????
D
??????
I
Baissofcalculation
Inert
(i.e. water, solvent)
??????
A0
??????
B0
??????
C0
??????
D0
??????
I0
aA+ bB→cC+ dD
Θ
�=
??????
�0
??????
�0
=
�
�0
�
0
�
�0
�
0
=
�
�0
�
�0
=
??????
�0
??????
�0
where,
�=
�
�
+
�
�
−
�
�
−1
δ=
Cℎ����??????�������������������
��������������
21
4.2.1. Equations for concentrations in Flow systems
�
�=
??????
�
�
=
������??????��
�??????�����??????��
=
�����
�??????���
??????
A
??????
B
??????
C
??????
D
??????
I
??????
A0
??????
B0
??????
C0
??????
D0
??????
I0
�
�=
??????
�
�
=
??????
�
0
�
(1−�)
�
�=
??????
�
�
=
??????
�0
−(Τ��)??????
�0
(1−�)
�
�
�=
??????
�
�
=
??????
�0
+(��)??????
�0
(1−�)
�
�
�=
??????
�
�
=
??????
�
0
+(��)??????
�
0
(1−�)
�
??????.�.�??????������,��
3
���, ��
3
�??????�
4.2.1. Equations for concentrations in Flow systems
�
�=
??????
�
�
=
������??????��
�??????�����??????��
=
�����
�??????���
??????
A
??????
B
??????
C
??????
D
??????
I
??????
A0
??????
B0
??????
C0
??????
D0
??????
I0
�
�=
??????
�
�
=
??????
�
0
�
(1−�)
�
�=
??????
�
�
=
??????
�0
−(Τ��)??????
�0
(1−�)
�
�
�=
??????
�
�
=
??????
�0
+(��)??????
�0
(1−�)
�
�
�=
??????
�
�
=
??????
�
0
+(��)??????
�
0
(1−�)
�
??????.�.�??????������,��
3
���, ��
3
�??????�
22
4.2.2. Liquid-phase concentrations
For liquids, the fluid volume change with reaction is negligiblewhen no phase changes are taking
place.
☞�=�
0
�
�=
??????
�
0
�
0
1−�=�
�0
(1−�)
�
�=�
�0
Θ
�−
�
�
�
�
�=�
�0
Θ
�+
�
�
�
�
�=�
�0
Θ
�+
�
�
�
☞Now, we can find –r
A= f(X) for liquid-phase reactions using
any one of the rate lawsin Chapter 3.
What about gas-phasereactions?
(volumetric flow rate is often changed during the reaction)
4.2.2. Liquid-phase concentrations
For liquids, the fluid volume change with reaction is negligiblewhen no phase changes are taking
place.
☞�=�
0
�
�=
??????
�
0
�
0
1−�=�
�0
(1−�)
�
�=�
�0
Θ
�−
�
�
�
�
�=�
�0
Θ
�+
�
�
�
�
�=�
�0
Θ
�+
�
�
�
☞Now, we can find –r
A= f(X) for liquid-phase reactions using
any one of the rate lawsin Chapter 3.
What about gas-phasereactions?
(volumetric flow rate is often changed during the reaction)
23
4.2.3. Gas-phase concentrations
�
2+3??????
2⇄2�??????
3
There are other systems, though, in which either V or does VARY, and these will now
be considered.
We considered primarily systems in which the reaction volume or volumetric flow rate
did NOT varyas the reaction progressed.
(Most batch and liquid-phase and some gas-phase systems fall into this category.)
1���+3���⟶2���(4���⟶2���)
The molar flow rate will be changing as the reaction progresses.
☞The volumetric flow rate will also change
(equal numbers of moles occupy equal volumes in the gas phase at the same T, P)
4.2.3. Gas-phase concentrations
�
2+3??????
2⇄2�??????
3
There are other systems, though, in which either V or does VARY, and these will now
be considered.
We considered primarily systems in which the reaction volume or volumetric flow rate
did NOT varyas the reaction progressed.
(Most batch and liquid-phase and some gas-phase systems fall into this category.)
1���+3���⟶2���(4���⟶2���)
The molar flow rate will be changing as the reaction progresses.
☞The volumetric flow rate will also change
(equal numbers of moles occupy equal volumes in the gas phase at the same T, P)
24
4.2.3. Gas-phase concentrations
Flow Reactors with Variable Volumetric Flow Rate.
�
�=
??????
�
�
=
�
���
total
�
�=P/ZRT(the gas law)
Compressibility factor
�
�
0
=
??????
�0
�
0
=
�
0
�
0��
0
�=�
0
??????
�
??????
�0
�
0
�
�
�
0�≅�
0
(4-15) (4-16)
The ratio of (4-14) to (4-15)
And rearrangement,
Z=
�
�∙��??????�
�
�∙���??????�
=
��������������������
�������������??????�������
Z > 1: High pressure, attraction < repulsion
Z < 1: High pressure, attraction > repulsion
at the entrance
�
�
�
�0
=
??????
�
??????
�0
�
0
�
=
�
0
�
�
�
0
�
0
�
(4-14)
4.2.3. Gas-phase concentrations
Flow Reactors with Variable Volumetric Flow Rate.
�
�=
??????
�
�
=
�
���
total
�
�=P/ZRT(the gas law)
Compressibility factor
�
�
0
=
??????
�0
�
0
=
�
0
�
0��
0
�=�
0
??????
�
??????
�0
�
0
�
�
�
0�≅�
0
(4-15) (4-16)
The ratio of (4-14) to (4-15)
And rearrangement,
Z=
�
�∙��??????�
�
�∙���??????�
=
��������������������
�������������??????�������
Z > 1: High pressure, attraction < repulsion
Z < 1: High pressure, attraction > repulsion
at the entrance
�
�
�
�0
=
??????
�
??????
�0
�
0
�
=
�
0
�
�
�
0
�
0
�
(4-14)
25
4.2.3. Gas-phase concentrations
We can now express the concentration of species j for a flow system in terms of
its flow rate, F
j, T, P.
�
�=
??????
�
�
=
??????
�
�
0
??????
�
??????
�0
�
0
�
�
�
0
=
??????
�
0
�
0
??????
�
??????
�0
�
�
0
�
0
�
�=�
0
??????
�
??????
�0
�
0
�
�
�
0
(4-16)
�
�=�
�
0
??????
�
??????
�0
�
�
0
�
0
�
(4-17)
This concentration equation will be used for membrane
reactors (Chap. 6) and for multiple reactions (Chap. 8).
4.2.3. Gas-phase concentrations
We can now express the concentration of species j for a flow system in terms of
its flow rate, F
j, T, P.
�
�=
??????
�
�
=
??????
�
�
0
??????
�
??????
�0
�
0
�
�
�
0
=
??????
�
0
�
0
??????
�
??????
�0
�
�
0
�
0
�
�=�
0
??????
�
??????
�0
�
0
�
�
�
0
(4-16)
�
�=�
�
0
??????
�
??????
�0
�
�
0
�
0
�
(4-17)
This concentration equation will be used for membrane
reactors (Chap. 6) and for multiple reactions (Chap. 8).
26
4.2.3. Gas-phase concentrations
The total molar flow rate
= the sum of the molar flow rates of each of the species
??????
�=??????
�+??????
�+??????
�+??????
�+??????
�⋯=
�=1
�
??????
�
�
�=�
�
0
??????
�
??????
�0
�
�
0
�
0
�
(4-17)
We can also write equation (4-17) in terms of the mole fraction of species j, y
j, and the pressure ratio, p,
with respect to the initial or entering conditions, i.e.,sub“0”
??????
�=
??????
�
??????
�
�=
�
�
0
�
�=�
�
0
??????
��
�
0
�
※The molar flow rates, Fj, are found by solving the mole
balance equations.
mole fractionpressure ratio
Need to consider the inert flowrate
4.2.3. Gas-phase concentrations
The total molar flow rate
= the sum of the molar flow rates of each of the species
??????
�=??????
�+??????
�+??????
�+??????
�+??????
�⋯=
�=1
�
??????
�
�
�=�
�
0
??????
�
??????
�0
�
�
0
�
0
�
(4-17)
We can also write equation (4-17) in terms of the mole fraction of species j, y
j, and the pressure ratio, p,
with respect to the initial or entering conditions, i.e.,sub“0”
??????
�=
??????
�
??????
�
�=
�
�
0
�
�=�
�
0
??????
��
�
0
�
※The molar flow rates, Fj, are found by solving the mole
balance equations.
mole fractionpressure ratio
Need to consider the inert flowrate
27
4.2.3. Gas-phase concentrations
Let’s express the concentration in terms of conversion
for gas flow systems.
??????
�=??????
�
0
+�??????
�
0
�
??????
�
??????
�
0
=1+�
??????
�0
�
??????
�
0
=1+�??????
�0
�=1+��
÷??????
�
0
�=
�
�
+
�
�
−
�
�
−1
??????
�0
??????
�0
☞�����??????���ℎ??????�������������
(완전전환율)
�=�??????
�
0
���??????����??????�????????????��????????????��??????��
??????
�
0
:���������??????����??????�??????��??????��
(���ℎ���������)
�=
�ℎ����??????������������������
��������������
�=
??????
�
??????
−??????
�0
??????
�
0
X=1, ??????
�=??????
�
??????
�=�
0(1+��)
�
0
�
�
�
0
�=�
0
??????
�
??????
�
0
�
0
�
�
�
0
(4-16)
(4-23)
☞Function of X
4.2.3. Gas-phase concentrations
Let’s express the concentration in terms of conversion
for gas flow systems.
??????
�=??????
�
0
+�??????
�
0
�
??????
�
??????
�
0
=1+�
??????
�0
�
??????
�
0
=1+�??????
�0
�=1+��
÷??????
�
0
�=
�
�
+
�
�
−
�
�
−1
??????
�0
??????
�0
☞�����??????���ℎ??????�������������
(완전전환율)
�=�??????
�
0
���??????����??????�????????????��????????????��??????��
??????
�
0
:���������??????����??????�??????��??????��
(���ℎ���������)
�=
�ℎ����??????������������������
��������������
�=
??????
�
??????
−??????
�0
??????
�
0
X=1, ??????
�=??????
�
??????
�=�
0(1+��)
�
0
�
�
�
0
�=�
0
??????
�
??????
�
0
�
0
�
�
�
0
(4-16)
(4-23)
☞Function of X
28
4.2.3. Gas-phase concentrations
The concentration of species j in a flow system is�
�=
??????
�
�
??????
�=??????
�0
+�
�??????
�0
�=??????
�0
(Θ
�+�
��)
A+
b
a
B
c
a
C+
d
a
D
The molar flow rate of species j is
�
�=−1,�
�=−Τ��,�
�=��,�
�=Τ��,Θ
�=Τ??????
�
0
??????
�
0
�=�
0(1+��)
�
�
0
�
0
�
�
�=
??????
�0
(Θ
�+�
��)
�
0(1+��)
�
�
0
�
0
�
�
�=
�
�
0
(Θ
�+�
��)
(1+��)
�
�
0
�
0
�
??????
�=
??????
�
??????
�
�
�0
=??????
�0
�
�0
�=�??????
�
0
4.2.3. Gas-phase concentrations
The concentration of species j in a flow system is�
�=
??????
�
�
??????
�=??????
�0
+�
�??????
�0
�=??????
�0
(Θ
�+�
��)
A+
b
a
B
c
a
C+
d
a
D
The molar flow rate of species j is
�
�=−1,�
�=−Τ��,�
�=��,�
�=Τ��,Θ
�=Τ??????
�
0
??????
�
0
�=�
0(1+��)
�
�
0
�
0
�
�
�=
??????
�0
(Θ
�+�
��)
�
0(1+��)
�
�
0
�
0
�
�
�=
�
�
0
(Θ
�+�
��)
(1+��)
�
�
0
�
0
�
??????
�=
??????
�
??????
�
�
�0
=??????
�0
�
�0
�=�??????
�
0
29
4.2.3. Gas-phase concentrations
4.2.3. Gas-phase concentrations
30
4.2.3. Gas-phase concentrations
aA+ bB→cC+ dD
Liquid phase Gas phase
Flow Batch Batch Flow
�
�=
??????
�
�
�
�=
�
�
�
�
�=
�
�
�
�
�=
??????
�
�
�=�
0 �=�
0
�
�=�
�0
Θ
�−
�
�
�
�=�
0
�
�
�
�0
�
0
�
�
�
0
�=�
0
??????
�
??????
�0
�
0
�
�
�
0
�
�=
�
�
�
�
�
�0
V
0
�
�
0
�
0
�
�
�=�
�0
�
�
�
�
�
�
0
�
0
�
�
�=
??????
�
??????
�
??????
�0
�
0
�
�
0
�
0
�
�
�=�
�0
??????
�
??????
�
�
�
0
�
0
�
NO phase change
NO semipermeable membrane
�
�=
�
�0
(Θ
�+�
��)
(1+��)
�
�
0
�
0
�
�=�
0(1+��)
�
0
�
�
�
0
NO phase change
isothermalNeglect P drop
where,�=�??????
�0
�=
�
�
+
�
�
−
�
�
−1
C
T0
=
P
0
RT
0
,C
A0
=??????
�0
C
T0
Figure 4-3 Expressing concentration as a function of conversion.
4.2.3. Gas-phase concentrations
aA+ bB→cC+ dD
Liquid phase Gas phase
Flow Batch Batch Flow
�
�=
??????
�
�
�
�=
�
�
�
�
�=
�
�
�
�
�=
??????
�
�
�=�
0 �=�
0
�
�=�
�0
Θ
�−
�
�
�
�=�
0
�
�
�
�0
�
0
�
�
�
0
�=�
0
??????
�
??????
�0
�
0
�
�
�
0
�
�=
�
�
�
�
�
�0
V
0
�
�
0
�
0
�
�
�=�
�0
�
�
�
�
�
�
0
�
0
�
�
�=
??????
�
??????
�
??????
�0
�
0
�
�
0
�
0
�
�
�=�
�0
??????
�
??????
�
�
�
0
�
0
�
NO phase change
NO semipermeable membrane
�
�=
�
�0
(Θ
�+�
��)
(1+��)
�
�
0
�
0
�
�=�
0(1+��)
�
0
�
�
�
0
NO phase change
isothermalNeglect P drop
where,�=�??????
�0
�=
�
�
+
�
�
−
�
�
−1
C
T0
=
P
0
RT
0
,C
A0
=??????
�0
C
T0
Figure 4-3 Expressing concentration as a function of conversion.
31
Determining �
??????=??????
??????(??????)for a Gas-phase reaction
2��
2+O
2→2�O
3
Example 4-3
Sulfur dioxide
(a) First, set up a stoichiometric table using only the symbols (i.e., Θ
�, ??????
�).
(b) Next, prepare a second table evaluating the species concentrations as a function of conversionfor the case
when the total pressure is 1485 kPa (14.7 atm) and the temperature is constant at 227℃.
(c) Evaluate the parameters and make a plotof each of the concentrations SO
2, SO
3, N
2as a function of
conversion.
A mixture of 28% ��
2and 72% air is charged to a flow reactor in which ��
2is oxidized.
4.2.3. Gas-phase concentrations
Determining �
??????=??????
??????(??????)for a Gas-phase reaction
2��
2+O
2→2�O
3
Example 4-3
Sulfur dioxide
(a) First, set up a stoichiometric table using only the symbols (i.e., Θ
�, ??????
�).
(b) Next, prepare a second table evaluating the species concentrations as a function of conversionfor the case
when the total pressure is 1485 kPa (14.7 atm) and the temperature is constant at 227℃.
(c) Evaluate the parameters and make a plotof each of the concentrations SO
2, SO
3, N
2as a function of
conversion.
A mixture of 28% ��
2and 72% air is charged to a flow reactor in which ��
2is oxidized.
4.2.3. Gas-phase concentrations
32
4.2.3. Gas-phase concentrations
2��
2+O
2→2�O
3
(a) First, set up a stoichiometric table using only the symbols (i.e., Θ
�, ??????
�).
��
2+
1
2
O
2→�O
3
�+
1
2
�→�
4.2.3. Gas-phase concentrations
2��
2+O
2→2�O
3
(a) First, set up a stoichiometric table using only the symbols (i.e., Θ
�, ??????
�).
��
2+
1
2
O
2→�O
3
�+
1
2
�→�
33
4.2.3. Gas-phase concentrations
(b) Next, prepare a second table evaluating the species concentrations as a function of conversionfor the
case when the total pressure is 1485 kPa (14.7 atm) and the temperature is constant at 227℃.
�
�=
??????
�
0
�
0
1−�=�
�
0
(1−�) �=�
0(1+��)
�
0
�
�
�
0
=�
0(1+��)
Neglecting pressure dropIsothermal
�
�=
�
�
0
(Θ
B+
�
�
X)
(1+��)
�
�
0
�
0
�
=
�
�
0
(Θ
B+
1
2
X)
(1+��)
�+
1
2
�→�
For species A (SO
2)
For species B (O
2)
4.2.3. Gas-phase concentrations
(b) Next, prepare a second table evaluating the species concentrations as a function of conversionfor the
case when the total pressure is 1485 kPa (14.7 atm) and the temperature is constant at 227℃.
�
�=
??????
�
0
�
0
1−�=�
�
0
(1−�) �=�
0(1+��)
�
0
�
�
�
0
=�
0(1+��)
Neglecting pressure dropIsothermal
�
�=
�
�
0
(Θ
B+
�
�
X)
(1+��)
�
�
0
�
0
�
=
�
�
0
(Θ
B+
1
2
X)
(1+��)
�+
1
2
�→�
For species A (SO
2)
For species B (O
2)
34
4.2.3. Gas-phase concentrations
(c) Evaluate the parameters and make a plotof each of the concentrations SO
2, SO
3, N
2as a function of
conversion.
�
�0=??????
�0�
�0=??????
�0
�
0
��
0
=0.28
1485���
8.314���∙��
3
/(���∙�)×500�
=0.1�����
3
The inlet concentration of A = (the inlet mole fraction of A) x (the total inlet molar concentration)
The total concentration ☞the ideal gas law. (yA0 = 0.28, T0 = 500 K, and P0 = 1485 kPa.)
The total concentration at constant temperature and pressure
�
�=
??????
�
�
=
??????
�0+??????
�0��??????
�0
�
0(1+��)
=
??????
�0(1+��)
�
0(1+��)
=
??????
�0
�
0
=�
�0
�
�
�0=
�
0
��
0
=
1485���
8.314���∙��
3
/(���∙�)×500�
=0.357 �����
3
�=??????
�0�=0.281−1−
1
2
=−0.14
�+
1
2
�→���
2+
1
2
O
2→�O
3
4.2.3. Gas-phase concentrations
(c) Evaluate the parameters and make a plotof each of the concentrations SO
2, SO
3, N
2as a function of
conversion.
�
�0=??????
�0�
�0=??????
�0
�
0
��
0
=0.28
1485���
8.314���∙��
3
/(���∙�)×500�
=0.1�����
3
The inlet concentration of A = (the inlet mole fraction of A) x (the total inlet molar concentration)
The total concentration ☞the ideal gas law. (yA0 = 0.28, T0 = 500 K, and P0 = 1485 kPa.)
The total concentration at constant temperature and pressure
�
�=
??????
�
�
=
??????
�0+??????
�0��??????
�0
�
0(1+��)
=
??????
�0(1+��)
�
0(1+��)
=
??????
�0
�
0
=�
�0
�
�
�0=
�
0
��
0
=
1485���
8.314���∙��
3
/(���∙�)×500�
=0.357 �����
3
�=??????
�0�=0.281−1−
1
2
=−0.14
�+
1
2
�→���
2+
1
2
O
2→�O
3
35
4.2.3. Gas-phase concentrations
Initially, 72% of the total number of moles is air containing 21% and 79% N2, along with 28% SO
2.
??????
�0=(0.28)(??????
�0) ??????
�0=(0.72)(0.21)(??????
�0)
��
2+
1
2
O
2→�O
3
Θ
B=
??????
�0
??????
�0
=
(0.72)(0.21)
0.28
=0.54 Θ
I=
??????
�0
??????
�0
=
(0.72)(0.79)
0.28
=2.03
C
A=C
A
0
1−X
1+εX
=0.1
1−X
1−0.14X
Τmoldm
3
SO
2
O
2
SO
3
N
2
C
B=
�
�
0
(Θ
B+
1
2
X)
1+��
=0.1
0.54−0.5X
1−0.14X
Τmoldm
3
C
C=
�
�
0
�
1+��
=
0.1�
1−0.14X
Τmoldm
3
C
I=
�
�
0
Θ
I
1+��
=
0.1(2.03)
1−0.14X
Τmoldm
3
4.2.3. Gas-phase concentrations
Initially, 72% of the total number of moles is air containing 21% and 79% N2, along with 28% SO
2.
??????
�0=(0.28)(??????
�0) ??????
�0=(0.72)(0.21)(??????
�0)
��
2+
1
2
O
2→�O
3
Θ
B=
??????
�0
??????
�0
=
(0.72)(0.21)
0.28
=0.54 Θ
I=
??????
�0
??????
�0
=
(0.72)(0.79)
0.28
=2.03
C
A=C
A
0
1−X
1+εX
=0.1
1−X
1−0.14X
Τmoldm
3
SO
2
O
2
SO
3
N
2
C
B=
�
�
0
(Θ
B+
1
2
X)
1+��
=0.1
0.54−0.5X
1−0.14X
Τmoldm
3
C
C=
�
�
0
�
1+��
=
0.1�
1−0.14X
Τmoldm
3
C
I=
�
�
0
Θ
I
1+��
=
0.1(2.03)
1−0.14X
Τmoldm
3
36
4.2.3. Gas-phase concentrations
2��
2+O
2→2�O
3
A mixture of 28% ��
2and 72% air is charged to a flow reactor in which ��
2is oxidized.
4.2.3. Gas-phase concentrations
2��
2+O
2→2�O
3
A mixture of 28% ��
2and 72% air is charged to a flow reactor in which ��
2is oxidized.
37
4.2.3. Gas-phase concentrations
�
�=�
���=�
�0
??????
�
??????
�0
�
�
0
�
0
�
��=�
�0��
0
??????
�
??????
�0
�=�
�0
??????
�
??????
�0
�
�
�0??????
�
p (pressure ratio)
☞For mole balance is written
in terms of molar flow rates
☞For mole balance is written
in terms of conversion
�
�=�
���=�
�0
Θ
�+�
��
(1+��)
�
�
0
�
0
�
��=�
�0��
0
Θ
�+�
��
(1+��)
�=�
�0
Θ
�+�
��
(1+��)
�
�
�0
�
�=�
�0
??????
�
??????
�0
�
�
�=�
�0
Θ
�+�
��
(1+��)
�
Rate laws for catalyst reactions
��=���
4.2.3. Gas-phase concentrations
�
�=�
���=�
�0
??????
�
??????
�0
�
�
0
�
0
�
��=�
�0��
0
??????
�
??????
�0
�=�
�0
??????
�
??????
�0
�
�
�0??????
�
p (pressure ratio)
☞For mole balance is written
in terms of molar flow rates
☞For mole balance is written
in terms of conversion
�
�=�
���=�
�0
Θ
�+�
��
(1+��)
�
�
0
�
0
�
��=�
�0��
0
Θ
�+�
��
(1+��)
�=�
�0
Θ
�+�
��
(1+��)
�
�
�0
�
�=�
�0
??????
�
??????
�0
�
�
�=�
�0
Θ
�+�
��
(1+��)
�
Rate laws for catalyst reactions
��=���
38
4.2.3. Gas-phase concentrations
Example 4-4
Expressing the rate law for SO
2oxidation in terms of partial pressures and conversions
2��
2+O
2→2�O
3
catalyst
solidgas
−�′
��2
=
��
��2
�
�2
−
�
��3
�
�
1+�
�2
�
�2
+�
��2
�
��2
2
�����
2oxidized/(h)(g−���)
Isothermal reaction at 400 ℃. �=9.7 Τ�����
2 ���
Τ32
Τℎ�−���, �
�2
=38.5���
−1
, �
��2
=42.5���
−1
, �
�=930���
−1
�+
1
2
�→�
☞
The rate law for this SO2 oxidation was found experimentally.
The total pressure and the feed composition (e.g., 28% SO2) are the same as in Example 4-3.
The entering partial pressure of SO2 is 4.1 atm. (No pressure drop.)
Write the rate law as a function of conversion.
(E4-4.1)
4.2.3. Gas-phase concentrations
Example 4-4
Expressing the rate law for SO
2oxidation in terms of partial pressures and conversions
2��
2+O
2→2�O
3
catalyst
solidgas
−�′
��2
=
��
��2
�
�2
−
�
��3
�
�
1+�
�2
�
�2
+�
��2
�
��2
2
�����
2oxidized/(h)(g−���)
Isothermal reaction at 400 ℃. �=9.7 Τ�����
2 ���
Τ32
Τℎ�−���, �
�2
=38.5���
−1
, �
��2
=42.5���
−1
, �
�=930���
−1
�+
1
2
�→�
☞
The rate law for this SO2 oxidation was found experimentally.
The total pressure and the feed composition (e.g., 28% SO2) are the same as in Example 4-3.
The entering partial pressure of SO2 is 4.1 atm. (No pressure drop.)
Write the rate law as a function of conversion.
(E4-4.1)
39
4.2.3. Gas-phase concentrations
�
��2
=�
��2
��=
??????
��2
�
��=
??????
��2,01−���
�
0(1+��)
�
�
0
�
0
�
=
??????
��2,0
�
0
��
0(1−�)
�
�
0
(1+��)
=
�
��2,0(1−�)
�
�
0
(1+��)
SO
2
Solution (No Pressure Drop, Isothermal Operation)
Step 1) Remind the relationship between partial pressure and concentration
Step 2) Express partial pressure as a function of conversion (X)
No pressure drop
�=�
0
�=1
�
��2,0=??????
��2,0�
0=0.281485���=415���(4.1���)
From example 4-3
SO
3
�
��3
=�
��3
��=
�
��2,0��
0�
1+��
=
�
��2,0�
1+��
=
�
��2,01−��
(1+��)
=
�
��2,01−�
(1+��)
4.2.3. Gas-phase concentrations
�
��2
=�
��2
��=
??????
��2
�
��=
??????
��2,01−���
�
0(1+��)
�
�
0
�
0
�
=
??????
��2,0
�
0
��
0(1−�)
�
�
0
(1+��)
=
�
��2,0(1−�)
�
�
0
(1+��)
SO
2
Solution (No Pressure Drop, Isothermal Operation)
Step 1) Remind the relationship between partial pressure and concentration
Step 2) Express partial pressure as a function of conversion (X)
No pressure drop
�=�
0
�=1
�
��2,0=??????
��2,0�
0=0.281485���=415���(4.1���)
From example 4-3
SO
3
�
��3
=�
��3
��=
�
��2,0��
0�
1+��
=
�
��2,0�
1+��
=
�
��2,01−��
(1+��)
=
�
��2,01−�
(1+��)
40
4.2.3. Gas-phase concentrations
O
2
�
�3
=�
�2
��=
�
��2,0Θ
�−
1
2
���
0
1+��
=
�
��2,0Θ
�−
1
2
�
1+��
=
1
2
∙
�
��2,01.08−�
1+��
From example 4-3
Θ
�=0.54
From example 4-3
�=−0.14
−�′
��2
=
��
��2
�
�2
−
�
��3
�
�
1+�
�2
�
�2
+�
��2
�
��2
2
=�
�
��2
Τ321−�
1−0.14�
(1.08−�)
2(1−0.14�)
−
�
��2,0�
1−0.14�
1
930���
Τ−12
1+38.5∙
1
2
∙
�
��2,01.08−�
1+��
+42.5∙
�
��2,0�
(1−0.14�)
2
Substitute for the partial pressure in the rate-law equation (E4-4.1)
�
�2
=38.5���
−1
�
��2
=42.5���
−1
,
�=9.7 �����
2 ���
Τ32
Τℎ�−���, �
��2,0=4.1���, �
��2
Τ32
=8.3���
−�′
��2
=9.7
���
ℎ∙�−���∙���
Τ32
8.3���
Τ321−�
1−0.14�
(1.08−�)
2(1−0.14�)
−
0.0044���
Τ32
�
1−0.14�
1+38.5∙
1
2
∙
4.11.08−�
1−0.14�
+42.5∙
174�
(1−0.14�)
2
4.2.3. Gas-phase concentrations
O
2
�
�3
=�
�2
��=
�
��2,0Θ
�−
1
2
���
0
1+��
=
�
��2,0Θ
�−
1
2
�
1+��
=
1
2
∙
�
��2,01.08−�
1+��
From example 4-3
Θ
�=0.54
From example 4-3
�=−0.14
−�′
��2
=
��
��2
�
�2
−
�
��3
�
�
1+�
�2
�
�2
+�
��2
�
��2
2
=�
�
��2
Τ321−�
1−0.14�
(1.08−�)
2(1−0.14�)
−
�
��2,0�
1−0.14�
1
930���
Τ−12
1+38.5∙
1
2
∙
�
��2,01.08−�
1+��
+42.5∙
�
��2,0�
(1−0.14�)
2
Substitute for the partial pressure in the rate-law equation (E4-4.1)
�
�2
=38.5���
−1
�
��2
=42.5���
−1
,
�=9.7 �����
2 ���
Τ32
Τℎ�−���, �
��2,0=4.1���, �
��2
Τ32
=8.3���
−�′
��2
=9.7
���
ℎ∙�−���∙���
Τ32
8.3���
Τ321−�
1−0.14�
(1.08−�)
2(1−0.14�)
−
0.0044���
Τ32
�
1−0.14�
1+38.5∙
1
2
∙
4.11.08−�
1−0.14�
+42.5∙
174�
(1−0.14�)
2
41
4.2.3. Gas-phase concentrations
4.2.3. Gas-phase concentrations
42
4.3. Reversible reactions & Equilibrium conversion
43
The only difference is the maximum conversion, X
e
N
2O
4⇄2NO
2
Example 4-5
nitrogen tetroxide
(사산화이질소 )
nitrogen dioxide
(이산화질소 )
This reaction is to be carried out at constant temperature.
The feed consists of pure N
2O
4at 340 Kand 202.6 kPa (2 atm).
The concentration equilibrium constant, K
C, at 340 K is 0.1 mol /dm
3
The rate constant is 0.5min
–1
.
(a) Set up a stoichiometric table and then calculate the equilibrium conversion of in a constant-volume batch
reactor.
(b) Calculate the equilibrium conversion of in a flow reactor.
(c) Assuming the reaction is elementary, express the rate of reaction solely as a function of conversionfor a
flow system and for a batch system.
(d) Determine the CSTR volume necessary to achieve 80% of the equilibrium conversion.
43
The only difference is the maximum conversion, X
e
N
2O
4⇄2NO
2
Example 4-5
nitrogen tetroxide
(사산화이질소 )
nitrogen dioxide
(이산화질소 )
This reaction is to be carried out at constant temperature.
The feed consists of pure N
2O
4at 340 Kand 202.6 kPa (2 atm).
The concentration equilibrium constant, K
C, at 340 K is 0.1 mol /dm
3
The rate constant is 0.5min
–1
.
(a) Set up a stoichiometric table and then calculate the equilibrium conversion of in a constant-volume batch
reactor.
(b) Calculate the equilibrium conversion of in a flow reactor.
(c) Assuming the reaction is elementary, express the rate of reaction solely as a function of conversionfor a
flow system and for a batch system.
(d) Determine the CSTR volume necessary to achieve 80% of the equilibrium conversion.
44
4.3. Reversible reactions & Equilibrium conversion
Example 4-5
(a) Set up a stoichiometric table and then calculate the equilibrium conversion of in a constant-volume
batch reactor.
N
2O
4⇄2NO
2 A⇄2B
K
C=
C
Be
2
C
Ae
From (3-10)
forbatchsystem,C
i=ΤN
iV
C
A=
N
A
V
=
N
A
V
0
=
N
A0
1−X
V
0
=C
A0
1−X
C
B=
N
B
V
=
N
B
V
0
=
N
A0
X
V
0
=2C
A0
X
C
A0
=
y
A0
P
0
RT
0
=
12atm
0.082 Τatm∙dm
3
mol∙K340K
=0.07174 Τmoldm
3
atequilibrium,X=X
e
X
eb=
K
c1−X
eb
4�
�0
4.3. Reversible reactions & Equilibrium conversion
Example 4-5
(a) Set up a stoichiometric table and then calculate the equilibrium conversion of in a constant-volume
batch reactor.
N
2O
4⇄2NO
2 A⇄2B
K
C=
C
Be
2
C
Ae
From (3-10)
forbatchsystem,C
i=ΤN
iV
C
A=
N
A
V
=
N
A
V
0
=
N
A0
1−X
V
0
=C
A0
1−X
C
B=
N
B
V
=
N
B
V
0
=
N
A0
X
V
0
=2C
A0
X
C
A0
=
y
A0
P
0
RT
0
=
12atm
0.082 Τatm∙dm
3
mol∙K340K
=0.07174 Τmoldm
3
atequilibrium,X=X
e
X
eb=
K
c1−X
eb
4�
�0
45
4.3. Reversible reactions & Equilibrium conversion
Input prompt for Chat GPT Solving the problem
Solution
X
eb=
K
c1−X
eb
4�
�0
4.3. Reversible reactions & Equilibrium conversion
Input prompt for Chat GPT Solving the problem
Solution
X
eb=
K
c1−X
eb
4�
�0
46
4.3. Reversible reactions & Equilibrium conversion
Example 4-5
(b) Calculate the equilibrium conversion of in a flowreactor.
JustreplaceN
itoF
i
�
�=
??????
�
�
=
F
�0
1−�
v
=
F
�0
1−�
�
01+��
=
C
�0
1−�
1+��
For constant temperature and pressure, the volumetric flow
rate is �=�
01+��
�
B=
??????
B
�
=
2F
�0
X
�
01+��
=
2C
�0
X
1+��
K
C=
C
Be
2
C
Ae
=
ΤC
�0
1−X
e1+�X
e
2
Τ2C
�0
X
e1+�X
e
=
4C
�0
X
e
2
1−X
e1+�X
e
X
e=
K
C1−X
e1+�X
e
4C
�0
For a flow system with pure feed, �=�??????
�0
=2−11=1
4.3. Reversible reactions & Equilibrium conversion
Example 4-5
(b) Calculate the equilibrium conversion of in a flowreactor.
JustreplaceN
itoF
i
�
�=
??????
�
�
=
F
�0
1−�
v
=
F
�0
1−�
�
01+��
=
C
�0
1−�
1+��
For constant temperature and pressure, the volumetric flow
rate is �=�
01+��
�
B=
??????
B
�
=
2F
�0
X
�
01+��
=
2C
�0
X
1+��
K
C=
C
Be
2
C
Ae
=
ΤC
�0
1−X
e1+�X
e
2
Τ2C
�0
X
e1+�X
e
=
4C
�0
X
e
2
1−X
e1+�X
e
X
e=
K
C1−X
e1+�X
e
4C
�0
For a flow system with pure feed, �=�??????
�0
=2−11=1
47
Input prompt for Co-polit
Solving the problem
Solution
4.3. Reversible reactions & Equilibrium conversion
X
e=
K
C1−X
e1+�X
e
4C
�
0
Input prompt for Co-polit
Solving the problem
Solution
4.3. Reversible reactions & Equilibrium conversion
X
e=
K
C1−X
e1+�X
e
4C
�
0
48
4.3. Reversible reactions & Equilibrium conversion
Example 4-5
(c) Assuming the reaction is elementary, express the rate of reaction solely as a function of
conversionfor a flow system and for a batch system.
−�
�=�
��
�−
�
�
2
�
�
Assuming that the reaction follows an elementary rate law, then
−�
�=�
��
�−
�
�
2
�
�
=�
��
�0
(1−�)−
4�
�0
2
�
2
�
�
1. For a constant volume (V=V0) batch system,
Here, CA=NA/V0 and CB=NB/V0. Substituting Equations (E4-5.2) and (E4-5.3) into the rate law, we
obtain the rate of disappearance of A as a function of conversion
4.3. Reversible reactions & Equilibrium conversion
Example 4-5
(c) Assuming the reaction is elementary, express the rate of reaction solely as a function of
conversionfor a flow system and for a batch system.
−�
�=�
��
�−
�
�
2
�
�
Assuming that the reaction follows an elementary rate law, then
−�
�=�
��
�−
�
�
2
�
�
=�
��
�0
(1−�)−
4�
�0
2
�
2
�
�
1. For a constant volume (V=V0) batch system,
Here, CA=NA/V0 and CB=NB/V0. Substituting Equations (E4-5.2) and (E4-5.3) into the rate law, we
obtain the rate of disappearance of A as a function of conversion
49
4.3. Reversible reactions & Equilibrium conversion
Example 4-5
−r
A=0.036
1−X
1+X
−
2.88X
2
1+X
2
mol
dm
3
∙min
−�
�=�
�
�
�0
(1−�)
1+��
−
4�
�0
2
�
2
�
�1+��
2
2. For a flow system,
Here, CA=FA/vand CB=FB/v. with v=v
0(1 +εX). Consequently, we can substitute Equations (E4-5.5) and
(E4-5.6) into Equation (E4-5.9) to obtain
As expected, the dependence of reaction rate on conversion for a constant volume batch system [i.e.,
Equation (E4-5.10)] is different than that for a flow system [Equation (E4-5.11)] for gas-phase reactions.
If we substitute the values for C
A0, K
C, ε, and k
A= 0.5 min
–1
in Equation(E4-5.11), we obtain –r
Asolely
as a function of X for the flow system.
−r
A=
0.5
�??????�
0.072
���1−X
���1+X
−
40.072 �����
32
X
2
0.01�����
3
1+X
2
4.3. Reversible reactions & Equilibrium conversion
Example 4-5
−r
A=0.036
1−X
1+X
−
2.88X
2
1+X
2
mol
dm
3
∙min
−�
�=�
�
�
�0
(1−�)
1+��
−
4�
�0
2
�
2
�
�1+��
2
2. For a flow system,
Here, CA=FA/vand CB=FB/v. with v=v
0(1 +εX). Consequently, we can substitute Equations (E4-5.5) and
(E4-5.6) into Equation (E4-5.9) to obtain
As expected, the dependence of reaction rate on conversion for a constant volume batch system [i.e.,
Equation (E4-5.10)] is different than that for a flow system [Equation (E4-5.11)] for gas-phase reactions.
If we substitute the values for C
A0, K
C, ε, and k
A= 0.5 min
–1
in Equation(E4-5.11), we obtain –r
Asolely
as a function of X for the flow system.
−r
A=
0.5
�??????�
0.072
���1−X
���1+X
−
40.072 �����
32
X
2
0.01�����
3
1+X
2
50
4.3. Reversible reactions & Equilibrium conversion
Example 4-5
(d) Determine the CSTR volume necessary to achieve 80% of the equilibrium conversion.
Let’s calculate the CSTR reactor volume necessary to achieve 80% of the equilibrium conversion of 51%
(i.e., X= 0.8Xe=(0.8)(0.51) = 0.4) for a molar feed rate of A of 3 mol/min.
−r
A=0.036
1−0.4
1+0.4
−
2.88(0.4)
2
1+(0.4)
2
=0.0070
mol
dm
3
∙min
�=
??????
�0
�
−�
�,??????
=
??????
�0
(0.4)
−�
�,0.4
=
0.3 ����??????�0.4
0.0070 �����
3
∙�??????�
=171��
3
(0.171�
3
)
4.3. Reversible reactions & Equilibrium conversion
Example 4-5
(d) Determine the CSTR volume necessary to achieve 80% of the equilibrium conversion.
Let’s calculate the CSTR reactor volume necessary to achieve 80% of the equilibrium conversion of 51%
(i.e., X= 0.8Xe=(0.8)(0.51) = 0.4) for a molar feed rate of A of 3 mol/min.
−r
A=0.036
1−0.4
1+0.4
−
2.88(0.4)
2
1+(0.4)
2
=0.0070
mol
dm
3
∙min
�=
??????
�0
�
−�
�,??????
=
??????
�0
(0.4)
−�
�,0.4
=
0.3 ����??????�0.4
0.0070 �����
3
∙�??????�
=171��
3
(0.171�
3
)
51
4.3. Reversible reactions & Equilibrium conversion
4.3. Reversible reactions & Equilibrium conversion
52
4.3. Reversible reactions & Equilibrium conversion
4.3. Reversible reactions & Equilibrium conversion
53
54
Summary
Having completed this chapter, you should be able to write the rate law solely in
terms of conversion and the reaction-rate parameters, (e.g., k, K
C)for both liquid-
phase and gas-phase reactions.
Once expressing r
A= f(X) is accomplished, you can proceed to use the techniques in
Chapter 2 to calculate reactor sizes and conversionfor single CSTRs, PFRs, and
PBRs, as well as those connected in series.
After studying this chapter you should also be able to calculate the equilibrium
conversion for both constant-volume batch reactors and for constant-pressure flow
reactors.
�
�=
�
�
0
(Θ
�+�
��)
(1+��)
�
�
0
�
0
�
−�
�=��
�
2
1−�Θ
�−
�
�
�For batch system For flow system
Summary
Having completed this chapter, you should be able to write the rate law solely in
terms of conversion and the reaction-rate parameters, (e.g., k, K
C)for both liquid-
phase and gas-phase reactions.
Once expressing r
A= f(X) is accomplished, you can proceed to use the techniques in
Chapter 2 to calculate reactor sizes and conversionfor single CSTRs, PFRs, and
PBRs, as well as those connected in series.
After studying this chapter you should also be able to calculate the equilibrium
conversion for both constant-volume batch reactors and for constant-pressure flow
reactors.
�
�=
�
�
0
(Θ
�+�
��)
(1+��)
�
�
0
�
0
�
−�
�=��
�
2
1−�Θ
�−
�
�
�For batch system For flow system
55
Summary
Q) Write the rate law for the elementary liquid phase reaction
3A + 2B → 4C
solely in terms of conversion.
The feed to the batch reactor is equal molar A and B with C
A0= 2 mol/dm
3
and k
A=0.01/s.
1. What is the rate law?
�)−�
�=�
��
��
�
Τ23
�)−�
�=�
��
��
� �)−�
�=�
��
�
3
�
�
2
2. What is the concentration of A?
�)�
�=�
�0
(1−�) �)�
�=�
�0
(2−�) �)�
�=�
�0
(1−3�)
3. What is the concentration of B?
�)�
�=�
�
0
(1−
2
3
�) b)�
�=�
�
0
(
2
3
−
2
3
�) �)�
�=�
�0
(1−2�)
Summary
Q) Write the rate law for the elementary liquid phase reaction
3A + 2B → 4C
solely in terms of conversion.
The feed to the batch reactor is equal molar A and B with C
A0= 2 mol/dm
3
and k
A=0.01/s.
1. What is the rate law?
�)−�
�=�
��
��
�
Τ23
�)−�
�=�
��
��
� �)−�
�=�
��
�
3
�
�
2
2. What is the concentration of A?
�)�
�=�
�0
(1−�) �)�
�=�
�0
(2−�) �)�
�=�
�0
(1−3�)
3. What is the concentration of B?
�)�
�=�
�
0
(1−
2
3
�) b)�
�=�
�
0
(
2
3
−
2
3
�) �)�
�=�
�0
(1−2�)
56
Summary
1. Rate law: �)−�
�=�
��
�
3
�
�
2
Solution)
2. What is the concentration of A?
�)�
�=�
�
0
(1−�)�
�=
�
�
�
=
�
�
�
0
=
�
�
0
�
0
1−�=�
�
0
(1−�)
3. What is the concentration of B?
�
�=
�
�
�
=
�
�
�
0
=
�
�
0
�
0
Θ
�−
�
�
�=�
�
0
(Θ
�−
�
�
�) Θ
�=
�
�0
�
�0
=
1
1
=1
�+
2
3
�→
4
3
�
�
�
=
Τ23
1
=
2
3
�
�=�
�0
(1−
2
3
�)
−�
�=�
��
�
3
�
�
2
3A + 2B → 4C
Summary
1. Rate law: �)−�
�=�
��
�
3
�
�
2
Solution)
2. What is the concentration of A?
�)�
�=�
�
0
(1−�)�
�=
�
�
�
=
�
�
�
0
=
�
�
0
�
0
1−�=�
�
0
(1−�)
3. What is the concentration of B?
�
�=
�
�
�
=
�
�
�
0
=
�
�
0
�
0
Θ
�−
�
�
�=�
�
0
(Θ
�−
�
�
�) Θ
�=
�
�0
�
�0
=
1
1
=1
�+
2
3
�→
4
3
�
�
�
=
Τ23
1
=
2
3
�
�=�
�0
(1−
2
3
�)
−�
�=�
��
�
3
�
�
2
3A + 2B → 4C
57
58
What’s next?
•Combine reactions and reactors as we bring all the material in the previous four
chapters together (to arrive at a logical structure for the design of various types of
reactors).
•Size BR, CSTRs, PFRs, PBRs, and microreactors for isothermal operation
given the rate law and feed conditions.
•Account for the effects of pressure drop on conversion in packed bed tubular
reactors.
What’s next?
•Combine reactions and reactors as we bring all the material in the previous four
chapters together (to arrive at a logical structure for the design of various types of
reactors).
•Size BR, CSTRs, PFRs, PBRs, and microreactors for isothermal operation
given the rate law and feed conditions.
•Account for the effects of pressure drop on conversion in packed bed tubular
reactors.
59
Homework
•What?
✓Summarize our class contentsfrom 1
st
week to 4
th
week
•How?
✓Assume that you are presenting to GET a JOB using this material.
✓You have to make the interviewer understand in five minutes what you learned.
✓The first page have to be ONE PAGE summary.
✓Rest of pages are free format.
(No limitation, BUT don’t forget, you only have FIVE minutes)
•How to get a GOOD score?
✓WELL ARRANGED (Essential contents have to be included-SIMPLE and EASY).
✓Assumethat you explain your report to your parents or a boy/girl friend.
(complex, boring, too much information are not good.)
•Submission: Tuesday, October 15 2024, 18:00 ([email protected])
✓(Caution) Late submission: score x 50%
Format: A4 (portrait), *.hwp, *.docx
e.g. Name-12206789-CRE summary-1.hwp
Refer to this video ☞https://youtu.be/DOoWPOJR5WM?si=JLgNycfoQu2h4oDt
Homework
•What?
✓Summarize our class contentsfrom 1
st
week to 4
th
week
•How?
✓Assume that you are presenting to GET a JOB using this material.
✓You have to make the interviewer understand in five minutes what you learned.
✓The first page have to be ONE PAGE summary.
✓Rest of pages are free format.
(No limitation, BUT don’t forget, you only have FIVE minutes)
•How to get a GOOD score?
✓WELL ARRANGED (Essential contents have to be included-SIMPLE and EASY).
✓Assumethat you explain your report to your parents or a boy/girl friend.
(complex, boring, too much information are not good.)
•Submission: Tuesday, October 15 2024, 18:00 ([email protected])
✓(Caution) Late submission: score x 50%
Format: A4 (portrait), *.hwp, *.docx
e.g. Name-12206789-CRE summary-1.hwp
Refer to this video ☞https://youtu.be/DOoWPOJR5WM?si=JLgNycfoQu2h4oDt
60https://blog.naver.com/chorong-dream
Ho-Joong KimPh.D.
Don’t think,
JUST DO!!!
If you do NOTHING,
NOTHING will happen.
Only one step is enough!
Ho-Joong KimPh.D.
Don’t think,
JUST DO!!!
If you do NOTHING,
NOTHING will happen.
Only one step is enough!
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