Episode 62 : MATERIAL BALANCE FOR REACTING SYSTEM Many chemical reactions are irreversible and occur in one direction only, namely forward Reversible reactions occur in both directions i.e. forward and backward SAJJAD KHUDHUR ABBAS Ceo , Founder & Head of SHacademy Chemical Engineering , Al-Muthanna University, Iraq Oil & Gas Safety and Health Professional – OSHACADEMY Trainer of Trainers (TOT) - Canadian Center of Human Development

1. SAJJAD KHUDHUR ABBAS
Ceo , Founder & Head of SHacademy
Chemical Engineering , Al-Muthanna University, Iraq
Oil & Gas Safety and Health Professional – OSHACADEMY
Trainer of Trainers (TOT) - Canadian Center of Human
Development
Episode 62 : MATERIAL
BALANCE FOR REACTING
SYSTEM

2. EQUILIBRIUM REACTIONS
• Many chemical reactions are irreversible and occur in one
direction only, namely forward
Reversible reactions occur in both directions i.e. forward and
backward
Example: hydrolysis reaction of ethylene (E) to ethanol (A)
• C2H4 + H2O  C2H5OH
• There are actually 2 opposing reactions
•
•
•
C2H4 + H2O  C2H5OH
C2H4 + H2O  C2H5OH
r1 = k1xExW
r2 = k2xA
When the reaction rate of the forward direction is equal to
the backwards reaction r1 = r2 , chemical equilibrium is
achieved
• Equilibrium constant
Ke

k1xE xW
r2 k2 xA
r1
 1  K
E W

k1 xA
e
k x x

2

3. 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 Ck  0
k1
S
S1 S
S1 S
k
k1
x x x ...x
1 2
1 2
Nik Nok
xik
Ni
= Nik + k r
xok
No
Equilibrium
reactor
Nok

Nik k r
o
S
k
 ok
ok
N
N
x 

4. 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)i2i1
m
o
s
k1
 
k
k1
e
N  r
N
x K 
S
k
k


21
21
S
m  k
k1
K  N  r
k1
kike
k

5. 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 that
temperature, 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-1
NiW = 2.0 mole h-1
xoM
xoW
xoC
xoH
No
Water shift
reactor
Equilibrium
conversion Xe

6. 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 m
Then
S
m  k  1111  0
k1
K  N  r
k1
N  1rN  1r
 1rNiW  1rNiM 1 r2  r
r
 1 0
2
.
s
iC iH
kike
k

7. EXAMPLE
• Solve forr
• Then r = 0.6667 mol j-1
Total flow rate out No
No  Nik k r NiM  1r  NiW  1r  1r  1r
•
Equilibrium composition
• CO
• H2O
• CO2
• H2
• Limiting reactant
• CO is limiting reactant
• Equilibrium CO
conversion
 2  3r  r2
r2
 N  N  1 2  3 mole h1
k1
iM iW
S
 NiM
 NiW
 NiC
M r No
W r No
C r No
 1 10.6667 3  0.111
 2  10.6667 3  0.444
 0  10.6667 3  0.222
 0  10.6667 3  0.222
xoM
xoW
xoC
xoH
 NiH H r No
1 N
M  1
1iMN 2
W  1
 2iW
 r  10.667  0.667
N 1
X 
iM
M
M

8. 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

9. MATERIAL BALANCE WITH MULTIPLE
REACTIONS
•
•
•
• 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!!
C2H6
C2H6
C2H4
 C2H4 + H2
2CH4
r1
r2
r3
+ H2 
+ C2H6  C3H6 + 2CH4

10. MATERIAL BALANCE WITH MULTIPLE
REACTIONS
• Three additional unknowns r1 r2 r3
• Ethane conversion
where 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 reacted
ethane
• Ethylene selectivity:
Hydrogen selectivity:
• Selectivity of CH4 :
• Selectivity of C3H6:
 r r1  r2  r3
1 3
Ethylene selectivity  SL 
r r  r
E
L

 rE r1  r2  r3
r1  r2
Hydrogen Selectivity  S 
rH
H 
rM

2r2  r3 
E 1 2 3
Methane selectivity  S 
 r r  r  r
M
 rE r1  r2  r3
3
Propylene selectivity  S  
rrP
P
XE  r1  r2  r3  NiE

11. 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
Nik  Nok  kr rr
r1
• Conversion of limiting reactant
R
r1
• Selectivity of reaction product h with respect to the limiting
 kr rr
rk
Nikj  Nokj  kr rr
j1 j r1
L M R
X  N  N N    pr r ipp ip op ip  r N
  
R
 hr rr X p Nip
r1 Nop
Nip
 Nih
Noh
Sh 
R
 prrr rqmaks
r1
rqmaks
p
Ypq
r


12. 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
Sh
• Product selectivity
X N p ip
R
 hr rr
r1 Nop
Nip
 Nih
Noh
 
rqmaks
R
 prrr
r1
rqmaks
p
Ypq
r


13. 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 inert
components (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
xi E = 0.85
xiI = 0.15
NoL
NoH
NoM
NoI
Ethane
dehydrogenation
reactor

14. EXAMPLE
• 5 components and 5 independent mass balance equations
• 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 E2 2
r1
E
2
  
r1  r2  42.585
2
 r N   r  r N x Lr r iE L1 1 L2 2 i iE
r1
L S 
0.471  1r  0r 0.851001 2
r1  40.035
r2  2.55
0.501  1r  1r 0.851001 2

15. EXAMPLE
• Component balance
• C2H6
• C2H4
• H2
N  85  42.585  42.415 mol h-1
oE
 NoL L1r1 L2r2
NiL
N  40.035 mol h-1
oL
H1r1 H 2r2
NiH  NoH
N  37.485 mol h-1
oH
NiE  NoE E1r1 E2r2
0.85100 NoE  140.035 12.55
0  NoL  140.035 02.55
0  NoH  140.035 12.55

16. EXAMPLE
• Component balance
• CH4
•
• Check with total mass balance
Balanced!
NiM  NoM M 1r1 M 2r2
N  5.10mol h-1
M
2550 15MI  2550 15MI
1000.85301000.15MI
 42.41530 40.03528 37.4852 5.11615MI
 Ni xiI MI  NoE ME  NoLML  NoH M H  NoM MM  Ni xiI M I
Ni xiE M E
0  N  040.035 22.55oM

17. 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 = 0.90
Eight unknown :molar flow rate out CH4, NoM molar flow rate out
CO, NoX molar flow rate out CO2, NoC molar 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
xiM = 0.078
xiO = 0.194
xiN = 0.728
NoC /NoX
NoN = NiN
NoO
= 8
NoW
Natural gas burner

18. 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 M1 1 M 2 2 i iMXM  Mr rr NiM
r1
2
0.9  1r  1r 0.0781001 2
r1  r2  7.02
 NoC C1r1 C2r2
NiC
 r2
NoC
0  NoC  0r1  1r2

19. EXAMPLE
• CO balance
•
But Then
• In matrix form
Hence
• CO2 balance
• CO balance
• CH4 balance
 NoX X1r1 X 2r2
NiX
 r1
NoX
 8NoX
NoC r2  8r1
r  7.02 9  0.78 mole h1
1
r  80.78 6.24 mole h
1
2
0  NoX  1r1  0r2
   

1 1r1  7.02
 
01r2 8
-1
 6.24mole hNoC
N  0.78 mole h-1
 NoM M1r1 M 2r2
oX
NiM
7.8  NoM  10.79 16.24
N  0.78 mole h-1
oM

20. EXAMPLE
• O2
•
• H2O
•
•
N2
• Check with total balance:
• Balanced!
19.4  NoO  1.5r1   2r2
N  5.75 mole h-1
 NoW W1r1 W 2r2
oO
0  NoW  2r1  2r2
N  14.04 mol h-1
oW
N  N  72.8 mol h-1
oN iN
765.984  765.984
 NoW MW 


  
 NoOMO  NoN M N
NoC MC  NoLMLL  NoM MM

Ni xiM MM  Ni xiOMO 

 Ni xiN M N
1000.078161000.19432 6.244 0.7828 0.781814.0418
1000.72828
   5.75321000.72828 
  
NiO  NoO O1r1 O2r2
NiW

21. 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
xiB = 1.0
No
xoB
xoC = 0.25
xoT
Isomerizer

22. EXAMPLE
• 3 components and 3 independent mass balance equations
• 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 balance NiC
r N   r  r  r  NiBB3 3B2 2iB B1 1
r1
 Br rBX   
3
r1  r3  65
 NoC C1r1 C2r2  0
 r1  r2
NoC
0.65  1r  0r  1r 100
1 2 3
0  NoC  1r1  1r2  0

23. 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 B2r2 B3r3
NiB
 100  r1  r3
NoB
NiT  NoT T 2r2 T 3r3
 r2  r3
0  NoT  1r2  1r3
NoT
 100  r  r  r  r  r  r  100 mole h
1
1 3 1 2 2 3N  NoB  NoC  NoTo
100  NoB  1r1  0r2  1r3
o

NoC
oC
N
x
100
r1  r2  25
0.25 
r1  r2

24. EXAMPLE
• Then
• Composition of product
• Check
10056 35(56)  25(56)  40(56)
•
Balanced!
N  r  r  
N
r  r  r  r  65  25  40 mole h
1
2 3 1 3 1 2oT
 x N  0.25100 25 mole h1
oC ooC
 N  N  100  25  40  35 mole h1NoB  No oToC
 35 100  0 35xoB
xoT  40 100  0.40
100 100
 NoB M B  NoC MC  NoT MT
NiB M B

25. 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-1
xiE = 0.1
xiO
xiN
No
xoE
xoL
x
x
oO
oN
xoW
xoC
Catalyrtic
reactor
Conversion 25 %
Yield of EtO 80%
• 2C2H4 + O2  2C2H4O
• C2H4 + 3O2  2CO2 + 2H2O

26. EXAMPLE
• 5 components and 5 independent mass balance equations
• 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 stream
Ethylene conversion
• Or
• In order to use information on the product yield, maximum rate
of reaction of ethylene is needed with other rates of reaction set
to zero e.g. rate of CO2
N'iC  N'oC C 2r'2
0  0  r'2
r'2  0
  r  r N x E1 1 E2 2 i iEX E  Errr NiE
r1
2
2r1  r2  2.5
0.25   2r  1r 0.11001 2

27. EXAMPLE
• Ethylene balance at this condition
•
Then
• Yield of ethylene
oxide
• Hence
• Ethylene balance
 NoE E1r'1E2r'2
NiE
0.1100  7.5  2r'10
1
r'1  1.25 mole h
1
 2r'1  2.5 mole hrEmaks
0.8  2r  0r  2.51 2
r1 1.0 mole h
rEmaks
R
r1
YL E  Lr rr
1
r  0.5 mole h1
2
NiE  NoE E1r1 E2r2
N  7.5 mole h1
oE
0.1100  NoE   21.0 10.5

28. EXAMPLE
• Ethylene oxide
• Oxygen
• Water
• CO2
0  NoL  21 00.5
N  2.0 mole h1
oL
0.210.9100  NoO  11  30.5
N  16.4 mole h1
 NoW W1r1 W 2r2
oO
0  NoW  01.0 20.5
N  1.0 mole h1
oW
 NoL L1r1 L2r2
NiL
 NoO O1r1 O2r2
NiO
NiW
0  NoC  01 20.5
C1r1 C 2r2
NiC  NoC
N  1.0 mole h1
oC

29. EXAMPLE
• N2
• Total flow rate product stream
• Composition of product streams
 0.790.9100 71.1 mole h1NiN  NoN
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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