Episode 62 : MATERIAL BALANCE FOR REACTING SYSTEM

SAJJAD KHUDHUR ABBASCeo , Founder & Head of SHacademyChemical Engineering , Al-Muthanna University, IraqOil & Gas Safety and Health Professional – OSHACADEMYTrainer of Trainers (TOT) - Canadian Center of Human Development

Episode 62 : MATERIAL BALANCE FOR REACTING

SYSTEM

EQUILIBRIUM REACTIONS• Many chemical reactions are irreversible and

occur in one direction only, namely forwardReversible reactions occur in both directions i.e. forward and backwardExample: hydrolysis reaction of ethylene (E) to ethanol (A)

• C2H4 + H2O C2H5OH

• There are actually 2 opposing reactions•••

C2H4 + H2O

C2H5OH C2H4+ H2O

C2H5OH

r1 = k 1x Ex W

r2 = k 2x AWhen the reaction rate of the forward direction is equal to the backwards reaction r1 = r2 , chemical equilibrium isachieved

• Equilibrium constantKe

k1 xE xW

r2 k2 xA

r1

1 KE W

k1 xAe k x x

2

MATERIAL BALANCE FOR EQUILIBRIUM REACTIONS• In general for equilibrium

reaction:

•xk = equilibrium mole fraction of component k

• Material balance of equilibrium reactor

Equilibrium constant

S

Ke x

• Mole fraction of component k out of reactor

k C k 0

k 1

S

S 1

SS 1 S

k

k 1

x x x ...x1

21 2

NikNokx ikNi

= Nik + k

rxokNo

Equilibriumreacto

r

Nok N ik k r

oS

k ok

ok NNx

MATERIAL BALANCE FOR EQUILIBRIUM REACTIONS• Substitute mole fraction into the equilibrium

equation

• and•Non linear equation of r

• If the number of moles of reactants is equal to the number of moles of products in the stoichiometry, then

s

N r

... N r S 1

N

r

N mo

SiS( S 1)i ( S 1)

i 2

i1

m o

s

k 1

kk

1

e

N r

Nx

K

S

k

k

21

21

S

m

kk 1

K

N

r k

1

kikek

EXAMPLE

• Example 3.11 Let the following water shift reaction

•CO(g)

+ H2O(g) CO2(g) +H2(g)

occurs until equilibrium is achieved at T

= 1105 K.At thattemperature, Ke for the reaction is 1.0. If the flow rates

in of CO(M) is 1.0 mole h-1 and water (W) is 2.0 mole h-1 and both CO2(C) and H2 (H) are not present in the incoming stream, calculate the equilibrium composition in the reactor and the equilibrium conversion of the limiting reactant.NiM = 1.0 mole h-1NiW = 2.0 mole h-1

xo

Mxo

WxoCxo

HNo

Water shiftreacto

rEquilibriu

m conversion Xe

EXAMPLE• Four unknown :

– mole fraction of CO xoM out of the reactor

– mole fraction of CO2 xoC out of the reactor

– mole fraction of H2 xoH out of the reactor

– mole fraction of water, xoW out of the reactor

• 4 components and 4 independent mass balance equations

• Degree of freedom= 4 – 4 = 0••

Calculate mThen

S

m k

1 1 1 1 0

k 1

K N r

k 1

N 1r N

1r 1r N iW 1r

N iM 1 r 2

r

r 1 0

2

.s

iC iHkike

k

EXAMPLE

• Solve forr•

Then r = 0.6667 mol j-1Total flow rate out No

No N ik k r N iM 1r N iW 1r

1r 1r•Equilibrium composition• CO

• H2O

• CO2

• H2

• Limiting reactant• CO is limiting

reactant• Equilibrium

CO conversion

2 3r r 2r 2

N N 1 2 3 mole h 1

k 1

iM iW

S

NiM

NiW

N iC

M r No

W r No

C r No

1 10.6667 3 0.111

2 10.6667 3 0.444

0 10.6667 3 0.222

0 10.6667 3 0.222

xoM

xoW

xoC

xoH

N iH H r No 1

N M 1

1

iMN 2 W 1

2

iW

r 10.667

0.667N 1

X iM

MM

MATERIAL BALANCE WITH MULTIPLE REACTIONS

• Industrial chemical processes often involve more than 1 reaction

• The reaction producing the desired product competes with side reaction producing undesired products

• Example production of ethylene by dehydrogenation of ethane

• C2H6 C2H4 + H2 r1

• Hydrogen reacts with ethane to produce methane• C2H6 + H2 2CH4 r2

• Ethylene reacts with ethane to produce propylene and methane

C2H4 + C2H6 C3H6 + 2CH4

r3• Only the first reaction produces the desired product

• Reactions 2 & 3 produce undesired products


•••• C2H6 (E) is used in all reactions. Rate of reaction of C2H6 :

rE r1 r2 r3

C2H4 (L) is produced by reaction 1 and used in reaction 3. Rate of reaction of C2H4 :

rL r1 r3

• H2 (H) is produced by reaction 1 and used in reaction 2. Rate of reaction of H2 :

rH r1 r2

• CH4 (M) is produced by reactions 2 and 3. Rate of reaction of CH4

rM 2r2 2r3

• Not all reactions are used to produce ethylene!!

C2H

6

C2H

6

C2H

4

C2H4 + H2

2CH4

r1

r2

r3

+ H2

+ C2H6 C3H6 +2CH4


• Three additional unknowns r1 r2

r3

• Ethane conversionwhere NiEt = flow rate in of ethane

• We need 2 more equations to get degree of freedom zero: Information of production of any 2 products per mole of reactedethane

• Ethylene selectivity:

Hydrogen selectivity:

• Selectivity of CH4 :

• Selectivity of C3H6:

r r1 r2 r3

1 3Ethylene selectivity SL

r r r

E

L

rEr1 r2 r3

r1 r2Hydrogen Selectivity S

rHH

rM 2r2 r3

E 1 2 3

Methane selectivity S r r r

r

M

rEr1 r2 r3

3Propylene selectivity S

rrPP

X E r1 r2 r3 N iE

STOICHIOMETRIC BALANCE OF MULTIPLE REACTIONS

reactant

• Total rate of reaction of component k in R

R

reactions

r 1

• Component mole balance of multiple reactions

• Yield

• If there is one inlet and one outlet streams:

R

N ik N ok kr rrr 1

• Conversion of limiting reactantR

r 1

• Selectivity of reaction product h with respect to the limiting

kr

rr

rk

N ikj N okj kr rrj 1 j r 1

L M R

X N N N

pr r ipp ip op ip r N

R

hr rr X p N ipr 1

N opN ip

N ih

N oh

S h

R

pr rr

rqmaksr 1

rqmaks

pYpqr

STOICHIOMETRIC BALANCE OF MULTIPLE REACTIONS

• Solution strategy for material balance of single reaction by calculating the rate of reaction r, adds an unknown r and degrees of freedom = 1

• Conversion Xp imposes relationship constraint between r and molar flow rates in and out of the limiting reactants that reduce the degree of freedom = 0

• The same strategy can be extended to the multi-reaction system

• If there are S components involved in R reactions, component balances produce S independent equation with additional R unknown: rr with r = 1, 2, 3, ..., R..

• For systems of R reactions, at least R - 1 additional equations

Or product yield

are required to solve it

S h

• Product selectivity

X N

pip

R

hr

rr

r 1

N opN ip

N ih

N oh

rqmaks

R

pr

rr r 1

rqmaks

pYpqr

EXAMPLE

• Example 3.12 Ethylene (L) can be produced by dehydrogenation of ethane. Two important reactions involved are

C2H6

C2H4 + H2

C2H6 + H2

2CH4

••• An ethane feed contains 85% ethane (E) and the rest

are inertcomponents (I). Ethane conversion, XE is 0.501 and selectivity of ethylene, SL is 0.471 mole of ethylene produced per mol of ethane feed. Calculate the molar composition of the gas products NoE

Conversion of C2H6 = 0.501

• Seven unknowns: molar flow rate out C2H6, NoE molar flow rate out C2H4, NoL molar flow rate out H2, NoH molar flow rate out CH4, NoM molar flow rate out I, NoI & 2 reaction rates, r1 and r2

Ni = 100 mole h-1

x i E = 0.85x iI = 0.15

NoLNo

HNo

MNoI

Ethanedehydrogenation reactor

EXAMPLE


• 2 reactions, ethane conversion & selectivity of ethylene• Degree of freedom= 7 – 7 = 0 , the system has a unique

solution•••••

Basis of calculation 100 mole h-1 ethane flow

•••

Then

Ethane conversion X

Ethylene selectivity

r N

r r N x

i iE Er r iE E1 1 E 2

2r 1

E

2

r1 r2 42.585

2 r N r r N x

Lr r iE L1 1 L 2 2 i iE

r 1

L S0.471 1r 0r 0.85100

1 2

r1 40.035

r2 2.55

0.501 1r 1r 0.85100

1 2

EXAMPLE

• Component balance

•C2H6

• C2H4

• H2

N 85 42.585 42.415 mol h -1

oE

NoL L1r1 L

2r2

N iL

N 40.035 mol h -

1oL

H 1r1 H

2r2

N iH NoH

N 37.485 mol h -

1oH

N iE NoE E1r1 E 2r2

0.85100 NoE 140.035 12.55

0 NoL 140.035 02.55

0 NoH 140.035 12.55

EXAMPLE

• Component balance

• CH4

•

• Check with total mass balance

Balanced!

N iM N oM M 1r1 M 2 r2

N 5.10 mol h -1

M

2550 15M I 2550 15M I

1000.8530 1000.15M I

42.41530 40.03528 37.4852 5.116 15M I

N i xiI M I NoE M E NoL M L NoH M H NoM M M N i xiI M IN i xiE M E

0 N 040.035 22.55

oM

EXAMPLE

• Example 3.13 Methane (M) is burned in a continuous burner to produce a mixture of CO (X), CO2 (C) & water (W).•

•• The feed contains 7.8% mole methane, 19.4% mole

oxygen and

CH4 + (3/2)O2 CO +2H2O

CO2 + 2H2OCH4 +2O2

•

the remainder nitrogen. Methane conversion is 90%. Ratio CO2 product /CO product = 8

NoM

Conversion CH4 = 0.90Eight unknown :molar flow rate out CH4, NoM molar flow rate out CO, NoX molar flow rate out CO2, NoCmolar flow rate out O2, NoO molar flow rate out air, NoW molar flow rate out nitrogen, NoN and 2 rates of reaction, r1 and r2

Ni = 100 mole h-1

x iM = 0.078x iO = 0.194x iN = 0.728

NoC /NoX

NoN =

NiNNoO

= 8

No

W

Natural gas burner

EXAMPLE• 6 components and 6 independent mass balance

equations• 2 reactions & ethane conversion & product ratio

CO2/CO• Degree of freedom= 8 – 8 = 0••••

Basis 100 mole h-1 ethane stream.Methane conversion

Or• CO2

balance•

r r N x

M 1 1 M 2 2iiM

X M Mr rr N iMr 1

2

0.9 1r 1r 0.078100

1 2

r1 r2 7.02 NoC C1r1 C 2r2

N iC

r2

N oC

0 NoC 0r1

1r2

EXAMPLE

• CO balance

•But

Then• In matrix

formHence

• CO2 balance

• CO balance

• CH4 balance

NoX X 1r1 X

2r2

N iX

r1

NoX

8NoX

NoC r2 8r1

r 7.02 9 0.78 moleh 1

1

r 80.78 6.24 moleh

12

0 NoX 1r1

0r2

1 1 r1 7.02

0 1 r2 8

-1 6.24 mole h

N oC

N 0.78 mole h -1

NoM M 1r1 M 2r2

oX

N iM

7.8 NoM 10.79 16.24N 0.78 mole h -

1oM

EXAMPLE

• O2

•

• H2O•

•N2• Check with total balance:

• Balanced!

19.4 NoO 1.5r1 2r2N 5.75 mole h -1

NoW W 1r1 W 2r2

oO

0 NoW 2r1

2r2N 14.04 mol h -1

oW

N N 72.8 mol h -1

oN iN

765.984 765.984

NoW MW

No O M O NoN M N

N oC M C N oLMLL N oM M M

N i xiM M M N i xiO M O N i xiN M N

1000.078161000.19432 6.244 0.7828

0.781814.04181000.72828

5.75321000.72828

N iO NoO O1r1 O 2r2

N iW

EXAMPLE

• • •

• 60% of a stream of pure 1-butena is converted in a reactor into

• Example 3.14 1-butene (B) is converted into 2 other isomers: cis-2-butene (C) & trans-butene (T) on alumina catalyst:

1-butene cis-2-butene

cis-2- butene trans-2-butene

1-butene

trans-2- butenae

a product containing 25% mole cis-2-butene. Calculate the concentration of other components

• Conversion 60 %

Six unknowns : total molar flow rate out, No mole fraction out 1- butene, xoB mole fraction out trans-2-butene, NoT and 3 rates of reaction r1 r2 and r3

1-butene

Ni = 100 mole h-1

x iB = 1.0

No

xo

BxoC = 0.25xoT

Isomerizer

EXAMPLE


• 3 reactions, 1 conversion & concentration of cis-2-butene

• Degree of freedom = 6 – 5 = 1• Specify basis so that Degree of freedom becomes

zero•••

Basis 100 mole h-1 1-butene stream.Conversion 1-butene or

•

• Cis-2-butene balanceN iC

r N

r r r N

iBB3 3B 2 2

iB B1 1r

1

Br r

BX

3

r1 r3 65

NoC C1r1 C 2r2 0

r1 r2

NoC

0.65 1r 0r 1r 100

123

0 NoC 1r1 1r2

0

EXAMPLE

• 1-butenae

• Then• Trans-2-

butene

• Then• Total mole going out of reactor by combining all relationship

• Mole fraction out cis-2-butene 0.25

•

• Hence

NoB B1r1 B 2 r2 B 3r3

N iB

100 r1 r3

NoB

N iT NoT T 2 r2 T 3r3

r2 r3

0 NoT 1r2 1r3

NoT

100 r r r r r r 100 mole h

11 3 1 2 2 3N N oB N oC

N oT

o

100 NoB 1r1 0r2

1r3

o

NoC

oC Nx

100r1 r2 25

0.25 r1

r2

EXAMPLE

• Then

• Composition of product

• Check

10056 35(56) 25(56) 40(56)

• Balanced!

N r r N

r r r r 65 25 40 mole h

12 3 1 3 1 2oT

x N 0.25100 25 mole h 1

oC ooC

N N 100 25 40 35 mole h 1

N oB N o oToC

35 100 0 35

xoB

xoT 40 100 0.40

100 100

NoB M B NoC M C NoT M TN iB M B

EXAMPLE• Example 3.15 Ethylene oxide (L) is produced

through partial oxidation of ethylene (E) in air

• At ethylene conversion of 25%, feed contaminating 10% ethylene produce an ethylene oxide yield of 80% from ethylene. Calculate the composition of the product

• Eight unknowns: total molar flow rate out, No mole fraction out ethylene, xoE mole fraction out ethylene oxide, xoL mole fraction out oxygen, xoO mole fraction out nitrogen, xoN mole fraction out water, xoW & 2 rate of reaction r1 & r2

Ni = 100 mol h-1x iE

= 0.1x

iOx iN

Noxo

ExoLx

x

oOoN

xoWxoC

Catalyrticreactor

Conversion 25 % Yield of EtO 80%

• 2C2H4 + O2 2C2H4O• C2H4 + 3O2 2CO2 +

2H2O

EXAMPLE


• 3 reactions, ethylene conversion, sum of mole fractions and ethylene oxide composition of output stream

• Degree of freedom = 5 – 5 = 0••

Basis 100 mole h-1 feed streamEthylene conversion

• Or• In order to use information on the product yield, maximum rate of reaction of ethylene is needed with other rates of reactionset to zero e.g. rate of CO2

N 'iC N 'oC C 2r'20 0 r'2r'2 0

r r N x

E1 1 E 2 2iiE

X E Er rr N iEr 1

2

2r1 r2 2.5

0.25 2r 1r 0.1100

1 2

EXAMPLE• Ethylene balance at this

condition

•Then

• Yield of ethylene oxide

• Hence• Ethylene

balance

NoE E1r'1 E

2r'2

N iE

0.1100 7.5 2r'1

0 1r'1 1.25 mole h1 2r'1 2.5 mole

h

rEmaks

0.8 2r 0r

2.51

2

r1 1.0 mole h

rEmaks

R

r 1

YL E Lr

rr

1

r 0.5 mole h 12

N iE NoE E1r1 E

2r2

N 7.5 mole h 1

oE

0.1100 NoE 21.0 10.5

EXAMPLE

• Ethylene oxide

• Oxygen

• Water

• CO2

0 NoL 21 00.5N 2.0 mole h 1

oL

0.210.9100 NoO 11 30.5N 16.4 mole h 1

NoW W 1r1 W 2r2

oO

0 NoW 01.0 20.5N 1.0 mole h 1

oW

NoL L1r1 L 2

r2

N iL

NoO O1r1 O 2 r2N iO

N iW

0 NoC 01 20.5

C1r1 C 2 r2N iC NoC

N 1.0 mole h 1

oC

EXAMPLE

• N2

• Total flow rate product stream

• Composition of product streams

0.790.9100 71.1 mole h 1

N iN N oN

N 7.5 2.0 16.4 1.0 1.0 71.1 99 mole h 1

o

7.5 99 0.0758

16.4 99 0.1657

1.0 99 0.0101

1.0 99 0.0101

71.1 99 0.7182

xoL

xoE

xoO

xoW

xoC

xoN

2.0 99 0.0202

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