developing and using stio tables notes
TRANSCRIPT
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8/18/2019 Developing and Using Stio Tables Notes
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Stoichiometric Table Batch System
• Stoichiometry set up of equations with A as basis
dDcCbBaA +↔+
A N Ao -N Ao X N A = N Ao(1 – X)
B NBo = N Ao B -(b/a)N Ao X NB = N Ao[ B –(b/a)X]
C NCo = N Ao C +(c/a)N Ao X NC = N Ao[ C +(c/a)X]
D NDo = N Ao D +(d/a)N Ao X ND = N Ao[ D +(d/a)X]
I NI = N Ao I NI = N Ao I
specie initial change remaining
NTo = N Ao i NT = NTo +δN AoX
i = Nio/N Ao
δ = (d/a) + (c/a) – (b/a) - 1
i
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Stoichiometric Tables
•
Express table in terms of concentrations – Concentration (batch):
– Constant Volume Batch
V
NC ii =
0VV =
( )( )
−Θ=
−Θ==
−=−
==
Xa
bCX
a
b
V
N
V
NC
X1CV
X1N
V
NC
B0AB
0
0ABB
0A
0
0AAA
+Θ=
+Θ==
+Θ=
+Θ==
Xa
dCX
a
d
V
N
V
NC
Xa
cCX
a
c
V
N
V
NC
D0AD
0
0ADD
C0AC
0
0ADC
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Stoichiometric Tables
•
Elementary rate law substitution:
( )( )
−Θ=
−Θ==
−=−
==
Xa
bCX
a
b
V
N
V
NC
X1CV
X1N
V
NC
B0AB
0
0ABB
0A
0
0AAA
+Θ=
+Θ==
+Θ=
+Θ==
Xa
dCX
a
d
V
N
V
NC
Xa
cCX
a
c
V
N
V
NC
D0AD
0
0ADD
C0AC
0
0ADC
dDcCbBaA +↔+
−=−
C
d
D
c
Cb
B
a
AAA K
CCCCkr
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Stoichiometric Table Flow System
• Stoichiometry set up of equations with A as basis
dDcCbBaA +↔+
A F Ao -F Ao X F A = F Ao(1 – X)
B FBo = F Ao B -(b/a)F Ao X FB = F Ao[ B –(b/a)X]
C FCo = F Ao C +(c/a)F Ao X FC = F Ao[ C +(c/a)X]
D FDo = F Ao D +(d/a)F Ao X FD = F Ao[ D +(d/a)X]
I FI = F Ao I FI = F Ao I
specie initial change remaining
FTo = F Ao i FT = FTo +δF AoX
i = Fio/Fao= Cioνo/Cao νo= yio/yao
δ = (d/a) + (c/a) – (b/a) - 1
i
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Stoichiometric Tables
•
Express table in terms of concentrations – concentration (liquid flow):
– constant volumetric flowrate
ν= iiF
C
( )( )
−Θ=
−Θ
ν=
ν=
−=ν
−=
ν=
Xa
bCX
a
bFFC
X1CX1FF
C
B0AB
0
0ABB
0A
0
0AAA
0ν=ν
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Stoichiometric Tables
•
Express table in terms of concentrations
– concentration for variable
volumetric gas flow:
– From compressibility factor EOS
δ+ν=ν
P
P
T
TX
F
F1 0
00T
0A0
ν= iiF
C
ν=ν
P
P
T
T
F
F 0
00T
T0
( )
δ+ν=ν
P
P
T
TXy1 0
0
0A0
( )
ε+ν=ν
P
P
T
TX1 0
0
0
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Stoichiometric Tables
•
Express table in terms of concentrations
– concentration for variable
volumetric gas flow:
– From compressibility factor EOS
ε+
−=
0
00AA
P
P
T
T
X1
X1CC
ε+
−
ν=
0
0
0
0AA
P
P
T
T
X1
X1FC( )
ε+ν=ν
P
P
T
TX1 0
0
0
ν= AAF
C
ν= iiF
C
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Stoichiometric Tables
•
Express table in terms of concentrations
– concentration for variable
volumetric gas flow:
– From compressibility factor EOS
[ ]
ε+
−Θ=
0
0B0AB
P
P
T
T
X1
XabCC( )
ε+ν=ν
P
P
T
TX1 0
0
0
ν= AAF
C
ν= iiF
C
ε+
−=
0
00AA
P
P
T
T
X1
X1CC
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8/18/2019 Developing and Using Stio Tables Notes
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Example
Set up a stoichiometric table, assuming the feed
is in stoichiometric proportions and comprised
only of reactants. Use chlorine as the basis forcalculation.
)g(HCl2)l,g(ClCH)g(Cl2)g(CH 22h
24 + → + ν
• The following elementary gas-phase reaction takes place in a constant-
pressure isothermal vessel (1 atm, 25°C):
– k = 0.2 dm6/(s⋅mol2) at 25°C (estimated)
– CH2Cl2: Pvap = 400 mm Hg (53 kPa) at 25°C.
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Example
Component Symbol Initial Change Out
Cl2
A 1 -X 1-X
CH4 B 0.5 -0.5X 0.5(1-X)
CH2Cl2 C 0 +0.5X 0.5X
HCl D 0 +X X
Total T 1.5 1.5
)g(HCl2)l,g(ClCH)g(Cl2)g(CH 22h24 + → + ν
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8/18/2019 Developing and Using Stio Tables Notes
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Example
• The following elementary gas-phase reaction takes place in a constant-
pressure isothermal vessel (1 atm, 25°C):
– k = 0.2 dm6/(s⋅mol2) at 25°C (estimated)
– CH2Cl2: Pvap = 400 mm Hg (53 kPa) at 25°C.
Calculate ε. δ=ε Aoy
0
1212
1a
b
a
c
a
d
=−−+=
−−+=δ
0=ε
)g(HCl2)l,g(ClCH)g(Cl2)g(CH 22h
24 + → + ν
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Example
Evaluate –r A in terms of conversion of chlorine,
the specific reaction rate, and the initial chlorine
concentration.
)g(HCl2)l,g(ClCH)g(Cl2)g(CH 22h
24 + → + ν
• The following elementary gas-phase reaction takes place in a constant-
pressure isothermal vessel (1 atm, 25°C):
– k = 0.2 dm6/(s⋅mol2) at 25°C (estimated)
– CH2Cl2: Pvap = 400 mm Hg (53 kPa) at 25°C.
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Example
0.2 dm6/(s⋅
mol2)→
3rd Order (assume elementary)
B2AA CkCr =− Need stoichiometry for C A & CB
)g(HCl2)l,g(ClCH)g(Cl2)g(CH 22h
24 + → + ν
• The following elementary gas-phase reaction takes place in a constant-
pressure isothermal vessel (1 atm, 25°C):
– k = 0.2 dm6/(s⋅mol2) at 25°C (estimated)
– CH2Cl2: Pvap = 400 mm Hg (53 kPa) at 25°C.
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Example
( )( ) ( )( )58.1'X
400'X5.07605.1
PnPn 2211
=⋅=
=
Therefore, CH2Cl2 does not condense in this system
( )AAoA X1CC −=
( )AAoB X1C21C −=
System is gas phase unti l Pc = 400 mm Hg. Need to find
conversion at which CH2Cl2 condenses
)g(HCl2)l,g(ClCH)g(Cl2)g(CH 22h
24 + → + ν
• The following elementary gas-phase reaction takes place in a constant-
pressure isothermal vessel (1 atm, 25°C):
– k = 0.2 dm6/(s⋅mol2) at 25°C (estimated)
– CH2Cl2: Pvap = 400 mm Hg (53 kPa) at 25°C.
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8/18/2019 Developing and Using Stio Tables Notes
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Example
Substitute stoichiometry into rate law
( )[ ] ( )( )[ ]( )3A
3Ao
AAo
2
AAo
B2AA
X1C2
kX12CX1Ck
CkCr
−=
−−=
=−
)g(HCl2)l,g(ClCH)g(Cl2)g(CH 22h
24 + → + ν
• The following elementary gas-phase reaction takes place in a constant-
pressure isothermal vessel (1 atm, 25°C):
– k = 0.2 dm6/(s⋅mol2) at 25°C (estimated)
– CH2Cl2: Pvap = 400 mm Hg (53 kPa) at 25°C.
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Example
What is the concentration of chlorine at X = 60% ?
( )AAoA X1CC −=
( )( )3
AoAo
dmmol027.0
K2.298KmolatmL082.0atm1
5.11
RT
PyC
=
⋅⋅=
=
( )3
3
A
dmmol011.06.01dmmol027.0C
==−=
)g(HCl2)l,g(ClCH)g(Cl2)g(CH 22h
24 + → + ν
• The following elementary gas-phase reaction takes place in a constant-
pressure isothermal vessel (1 atm, 25°C):
– k = 0.2 dm6/(s⋅mol2) at 25°C (estimated)
– CH2Cl2: Pvap = 400 mm Hg (53 kPa) at 25°C.
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8/18/2019 Developing and Using Stio Tables Notes
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Example
What is the rate of reaction at X = 60%?
( )
( ) ( )
sdmmol1026.1
6.01dmmol027.02
molsdm2.0
X1C2
kr
37
33326
3A
3AoA
⋅×=
−⋅=
−=−
−
)g(HCl2)l,g(ClCH)g(Cl2)g(CH 22h
24 + → + ν
• The following elementary gas-phase reaction takes place in a constant-
pressure isothermal vessel (1 atm, 25°C):
– k = 0.2 dm6/(s⋅mol2) at 25°C (estimated)
– CH2Cl2: Pvap = 400 mm Hg (53 kPa) at 25°C.
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8/18/2019 Developing and Using Stio Tables Notes
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Example
What is the activation energy if the frequency
factor is 2 x 1012 dm6/s⋅mol2 ?
RTEAek −=
( ) ( )( )K2.298KmolJ314.8E261226 emolsdm102molsdm2.0 ⋅−⋅×=⋅molJ74212E =
Arrhenius Equation
)g(HCl2)l,g(ClCH)g(Cl2)g(CH 22h
24 + → + ν
• The following elementary gas-phase reaction takes place in a constant-
pressure isothermal vessel (1 atm, 25°C):
– k = 0.2 dm6/(s⋅mol2) at 25°C (estimated)
– CH2Cl2: Pvap = 400 mm Hg (53 kPa) at 25°C.
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8/18/2019 Developing and Using Stio Tables Notes
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Example
What is the specific reaction rate at 100°C ?
410
2.298
1
2.373
1
R
Eexp
k
k
25
100 =
−−=
Arrhenius
Equation
2625100 molsdm95.81410kk ⋅=×=
)g(HCl2)l,g(ClCH)g(Cl2)g(CH 22h
24 + → + ν
• The following elementary gas-phase reaction takes place in a constant-
pressure isothermal vessel (1 atm, 25°C):
– k = 0.2 dm6/(s⋅mol2) at 25°C (estimated)
– CH2Cl2: Pvap = 400 mm Hg (53 kPa) at 25°C.
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8/18/2019 Developing and Using Stio Tables Notes
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Example (with phase change)
• The gas-phase reaction between chlorine and methane to
form carbon tetrachloride and hydrochloric acid is to be
carried out at 75°C and at 950 kPa in a continuous-flow
reactor.
• The vapor pressure of carbon tetrachloride at 75°C is 95 kPa.
Volumetric flow rate is 0.4 dm3/s
– Set up a stoichiometric table for this reaction with phase change. Calculate
the conversion of methane at which condensation begins.
– Plot the concentrations and molar flow rates of each species as well as the
total molar flow rate as a function of conversion for a stoichiometric feed .
HCl4CClCHCl4 442 +→+
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Example (with phase change)
HCl4CClCHCl4 442 +→+
1.0950
95
P
Py
atm0.94kPa95P
atm39.9kPa950P
K348C75T
V
e,CCl
CCl,V
4
4
===
====
=°=
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Example (with phase change)
• Stoichiometry set up of equations with CH4 as basis
A - CH4
F Ao
-F Ao
X F A
= F Ao
(1 – X)
B - Cl2 4 F Ao -4F Ao X FB = 4F Ao[1 –X]
C - HCl4 0 +4F Ao X FC = 4F AoX
D(g) - CCl4 (g) 0 +F Ao X FD = F AoX
specie initial change
remaining
(PD
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Example (with phase change)
• When condensation begins,
• Therefore, condensation begins at 50% conversion
5.0X9.0
X5FF5
'FF
c
cAoAo
TT
=
−=
=
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Example (with phase change)
• Before Condensation
• Total concentration is constant δ=ε Aoy 0=δ 0=ε
0PT =∆=∆
( ) oi
o
iii
oAoAo
ToAo
o
oToT
F
X1
FFC
s
gmol02631.0
s
L4.0
L
gmol0658.0CF
L
gmol0658.0
5
329.0C
5
1C
L
gmol329.0
RT
PCC
ν=
ε+ν=
ν=
=×=ν=
===
===
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Example (with phase change)
• Expressing stoichiometric table column “remaining (PD
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Example (with phase change)
• After Condensation
( )( )
−ν=
−ν=
ν=ν
5.4
X5
9.0F5
X5F
F
'F
o
Ao
Aoo
To
To
−ν
=ν
=5.4
X5FFC
o
iii
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Example (with phase change)
• Expressing stoichiometric table column “remaining (PD=PV) after
condensation” in terms of
A - CH4
B - Cl2
C - HCl4
D(g) - CCl4 (g)
specie
F Ao(1 – X)
4F Ao[1 –X]
4F AoX
0.1 FT
F Ao(5-X)/0.9
Fi Ci
C Ao(1 – X) (4.5)/(5-X)
4C Ao(1 – X)(4.5)/(5-X)
4C AoX(4.5)/(5-X)
C Ao(5-X)(4.5)/(5-X) = 0.5C Ao
C Ao(5-X)(4.5)/(5-X) = 0.5C Aototal
−
ν=
5.4
X5FC
o
ii