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7/21/2019 Lecture 5 - Stoichiometry for Constant Volume Systems (1)
http://slidepdf.com/reader/full/lecture-5-stoichiometry-for-constant-volume-systems-1 1/12
Lecture 5Stoichiometry, Liquid Systems
(constant volume systems)
Fall 2015
CHE 314
Chemical Reaction Engineering
1 Slides adapted from Fogler’s PowerPoint slides
7/21/2019 Lecture 5 - Stoichiometry for Constant Volume Systems (1)
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2
• Recap algorithm to get reaction rate as a
function of conversion
• Stoichiometric table for constant volume batch
reactors
• Stoichiometric table for flow reactors with
constant volumetric flow rate.
Outline:
7/21/2019 Lecture 5 - Stoichiometry for Constant Volume Systems (1)
http://slidepdf.com/reader/full/lecture-5-stoichiometry-for-constant-volume-systems-1 3/12
Algorithm
( )i A C gr =−Step 1: Rate Law
( ) ( ) X hC i =Step 2: Stoichiometry
( ) X f r A =−Step 3: Combine to get
How to find ( ) X f r A =−
7/21/2019 Lecture 5 - Stoichiometry for Constant Volume Systems (1)
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Stoichiometry
We shall set up Stoichiometric tables using
species A as our basis of calculation (i.e. assume A
is the limiting reactant).
We will use the stoichiometric tables to express theconcentration as a function of conversion.
We will combine Ci = f(X) with the appropriate rate
law to obtain -r A = f(X).
Da
d C
a
cB
a
bA +→+
7/21/2019 Lecture 5 - Stoichiometry for Constant Volume Systems (1)
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Batch System – Stoichiometric Table 1)
t = 0
NA0
NB0
NC0
ND0
NI0
t = t NA
NB
NC
ND
NI
Batch reactor
5
Da
d C
a
cB
a
bA +→+
7/21/2019 Lecture 5 - Stoichiometry for Constant Volume Systems (1)
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Batch System – Stoichiometric Table 2)
6
Species Symbol Initial Change Remaining
B B N B0=N A0ΘB -b/aN A0X N B=N A0( ΘB-b/aX)
A A N A0 -N A0X N A=N A0(1-X)
Inert I N I0=N A0ΘI ---------- N I =N A0ΘI
N T0 N T =N T0+δN A0X
Where:0
0
0
0
00
00
0
0
A
i
A
i
A
i
A
ii
y
y
C
C
V C
V C
N
N ====Θ 1−−+=
a
b
a
c
a
d δ and
C C N C0=N A0ΘC +c/aN A0X N C =N A0( ΘC +c/aX)
D D N D0=N A0ΘD +d/aN A0X N D=N A0( ΘD+d/aX)
δ = change in total number of moles per mol A reacted
7/21/2019 Lecture 5 - Stoichiometry for Constant Volume Systems (1)
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Batch System – Stoichiometric Table 3)
• If the reaction occurs in the liquid phase
• If a gas phase reaction occurs in a rigid (e.g. steel)
batch reactor
V =V 0
Then
C A= N
A
V
= N
A 0 1− X ( )V 0
= C A 0 1− X ( )
C B= N
B
V = N
A 0
V 0
Θ B− b
a X
= C
A 0Θ
B− b
a X
etc.
Constant Volume System
7/21/2019 Lecture 5 - Stoichiometry for Constant Volume Systems (1)
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Batch System – rate as a function of X
Suppose −r A = k AC A2C B
Batch (constant volume): 0
V V =
( )
−Θ−=− X
ab X C k r B A A A
230 1
−r A = f X ( )we have
Ar −1
X
7/21/2019 Lecture 5 - Stoichiometry for Constant Volume Systems (1)
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Practice Problem 1 - Batch System
Consider the following elementary reaction with
KC=20 dm
3
/mol and C A0=0.2 mol/dm
3
.
The reaction takes place in a constant volume
batch reactor.
Calculate the equilibrium conversion for gasphase reaction, Xe .
BA2 ⇔
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Flow System – Stoichiometric Table 1)
entering
FA0
FB0
FC0
FD0
FI0
Leaving
FA
FB FC
FD
FI
Continuous-flow
reactor
Da
d C
a
cB
a
bA +→+
7/21/2019 Lecture 5 - Stoichiometry for Constant Volume Systems (1)
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Flow System – Stoichiometric Table 2)
11
Species Symbol Initial Change Remaining
B B F B0=F A0ΘB -b/aF A0X F B=F A0( ΘB-b/aX)
A A F A0 -F A0X F A=F A0(1-X)
Inert I F I0=F A0ΘI ---------- F I =F A0ΘI
F T0 F T =F T0+δF A0X
Where:0
0
0
0
00
00
0
0
A
i
A
i
A
i
A
ii
y
y
C
C
vC
vC
F
F ====Θ 1−−+=
a
b
a
c
a
d δ and
C C F C0=F A0ΘC +c/aF A0X F C =F A0( ΘC +c/aX)
D D F D0=F A0ΘD +d/aF A0X N D=F A0( ΘD+d/aX)
δ = change in total number of moles per mol A reacted
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Practice Problem 2 - Flow System liquid)
Consider the following elementary reaction with
KC= 10 dm3/mol and C A0 = CB0 = 2 mol/dm3.
The reaction takes place in a flow reactor.
Calculate the equilibrium conversion and
concentrations for the liquid phase reaction
below .
C B A ⇔+