developing and using stio tables notes

8/18/2019 Developing and Using Stio Tables Notes

1/27

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


2/27

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


3/27


•

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


4/27

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


5/27


•

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


6/27


•

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


7/27


•





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


8/27


•





[ ]

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


9/27

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.


10/27

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


11/27

Example





Calculate ε. δ=ε Aoy

0

1212

1a

b

a

c

a

d

=−−+=

−−+=δ

0=ε


24 + → + ν


12/27

Example

Evaluate –r A in terms of conversion of chlorine,

the specific reaction rate, and the initial chlorine

concentration.


24 + → + ν






13/27

Example

0.2 dm6/(s⋅

mol2)→

3rd Order (assume elementary)

B2AA CkCr =− Need stoichiometry for C A & CB


24 + → + ν






14/27

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


24 + → + ν






15/27

Example

Substitute stoichiometry into rate law

( )[ ] ( )( )[ ]( )3A

3Ao

AAo

2

AAo

B2AA

X1C2

kX12CX1Ck

CkCr

−=

−−=

=−


24 + → + ν






16/27

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

==−=


24 + → + ν






17/27

Example

What is the rate of reaction at X = 60%?

( )

( ) ( )

sdmmol1026.1

6.01dmmol027.02

molsdm2.0

X1C2

kr

37

33326

3A

3AoA

⋅×=

−⋅=

−=−

−


24 + → + ν






18/27

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


24 + → + ν






19/27

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 ⋅=×=


24 + → + ν






20/27

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 +→+


21/27


HCl4CClCHCl4 442 +→+

1.0950

95

P

Py

atm0.94kPa95P

atm39.9kPa950P

K348C75T

V

e,CCl

CCl,V

4

4

===

====

=°=


22/27


• 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


23/27


• When condensation begins,

• Therefore, condensation begins at 50% conversion

5.0X9.0

X5FF5

'FF

c

cAoAo

TT

=

−=

=


24/27


• 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=

===

===


25/27


• Expressing stoichiometric table column “remaining (PD


26/27


• 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


27/27


• 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

developing and using stio tables notes

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