Chapter 5: Extent of Reaction
𝐴𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛 = 𝑖𝑛𝑝𝑢𝑡 - 𝑜𝑢𝑡𝑝𝑢𝑡 + 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 - 𝑐𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛
Steady state system:
0 = 𝑖𝑛𝑝𝑢𝑡 - 𝑜𝑢𝑡𝑝𝑢𝑡 + 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 - 𝑐𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛
𝑜𝑢𝑡𝑝𝑢𝑡 = 𝑖𝑛𝑝𝑢𝑡 + 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 - 𝑐𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛
Input (1)
𝑛1,𝑁2
𝑛1,𝐻2
Reactor
Output (2)
𝑛2,𝑁𝐻3
nN2,out = nN2,in − Consumption
nNH3,out = nNH3,in + Generation
Extent of reaction:
𝑛𝐴,𝑜𝑢𝑡 − 𝑛𝐴,𝑖𝑛
𝜉=
𝜐𝐴
N2 + 3H2  2NH3
𝑛𝐴,𝑜𝑢𝑡 = 𝑛𝐴,𝑖𝑛 + 𝜐𝐴 𝜉
27
Chapter 5: Species Mole Balances, SMB
Extent of reaction (Batch process):
𝑛𝐴,𝑓𝑖𝑛𝑎𝑙 − 𝑛𝐴,𝑖𝑛𝑖
𝜉=
𝜐𝐴
Extent of reaction (Continuous process):
𝑛𝐴,𝑜𝑢𝑡 − 𝑛𝐴,𝑖𝑛
𝜉=
𝜐𝐴
𝑛𝐴,𝑓𝑖𝑛𝑎𝑙 = 𝑛𝐴,𝑖𝑛𝑖 + 𝜐𝐴 𝜉
𝑛𝐴,𝑜𝑢𝑡 = 𝑛𝐴,𝑖𝑛 + 𝜐𝐴 𝜉
Species Material Balances
f =Conversion factor of the limited reactant
𝑓 = 𝑓𝐷𝐶
𝑚𝑜𝑙𝑒𝑠(𝑚𝑎𝑠𝑠) 𝑜𝑓 𝑙𝑖𝑚𝑖𝑡𝑒𝑑 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑟𝑒𝑎𝑐𝑡𝑒𝑑
=
× 100%
𝑚𝑜𝑙𝑒𝑠(𝑚𝑎𝑠𝑠) 𝑜𝑓 𝑙𝑖𝑚𝑡𝑒𝑑 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑓𝑒𝑑
−𝑓 ∙ 𝑛𝑖𝑛,𝑙𝑖𝑚𝑒𝑡𝑒𝑑 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡
𝜉=
𝜐𝐿𝑅
𝑛𝐴,𝑜𝑢𝑡 − 𝑛𝐴,𝑖𝑛
𝜉=
𝜐𝐴
28
Chapter 5: Species Mole Balances
Example 5.7: Reaction in Which the Fraction Conversion is Specified:
The chlorination of methane occurs by the following reaction
CH4 + Cl2  CH3Cl + HCl
You are asked to determine the product composition if the conversion of
the limiting reactant is 67%, and the feed composition in mole % is given as:
40% CH4, 50%Cl2,and 10% N2.
Assumptions: The reactor is Open, Steady state process
Solution:
Species
moles %
feed
CH4
40%
Cl2
50%
N2
10%
fLR
67%
1
𝑛1,Cl
2
𝑛1,CH
4
𝑛1,N
2
Reactor
2
𝑛2,Cl
2
𝑛2,CH
4
𝑛2,N
2
29
Chapter 5: Species Mole Balances
1
CH4 + Cl2  CH3Cl + HCl
𝑛1,Cl
2
𝑛1,CH
4
𝑛1,N
2
Reactor
2
𝑛2,Cl
2
𝑛2,CH
4
𝑛2,N
2
𝑛2,HCl
𝑛2,CH Cl
3
30
Chapter 5: Species Mole Balances
Example 10.2: A Reaction in Which the Fraction Conversion is to Be
Calculated:
H2S is toxic in very small quantities and is quite corrosive to process
equipment.
A proposed process to remove H2S is by reaction with SO2:
2H2S(g) + SO2(g)  3S(s) + 2H2O(g)
In a test of the process, a gas stream containing 20% H2S and 80% CH4
was combined with a stream of pure SO2.
The process produced 5000 kg of S(s), and in the product gas the ratio of
SO2 to H2S was equal to 3, and the ratio of H2O to H2S was 10.
You are asked to determine the fractional conversion of the limiting
reactant, and the feed rates of the H2S and SO2 streams.
Assumptions: The reactor is Open, Steady state process.
Solution:
31
Chapter 5: Species Mole Balances
𝑛2,SO2
moles % feed
Species
CH4
80%
H2S
20%
𝑚4,S
5000 kg
𝑛3,SO2 /𝑛3,H2S
3
𝑛3,H2O /𝑛3,H2S
10
1
𝑛1
𝑦1,H2S =0.20
𝑦1,CH4 =0.80
𝑛1,H2S
𝑛1,CH4
2H2S(g) + SO2(g)  3S(s) + 2H2O(g)
Input
Output
n1 H2S
n1 CH4
n2 SO2
Sum input
n3 H2S
n3 CH4
n3 H2O
n3 SO2
n4 S
Sum out
Moles, Kmol
114.6
458.48
83.3
MW
34
16
64
656.38 Kmol
10.42
458.48
104.2
31.2
156.25
760.55 Kmol
2
Reactor
4
𝑚4,S =5000 kg
Product
3
𝑛3,H2S
𝑛3,CH4
𝑛3,H2O
𝑛3,SO2
n4,s= 156.25 mol
Mass, Kg
3896.4
7335.68
5331.2
16563.3 Kg
34
16
18
64
32
354.28
7335.68
1875.6
1996.8
5000
16562.4 Kg
32
Chapter 5: Process Involving Multiple Reactions
Example 5.8: Material Balances Involving Two Ongoing Reactions
Formaldehyde (CH2O) is produced industrially by the catalytic oxidation of
methanol (CH3OH) according to the following reaction:
CH3OH + ½O2  CH2O + H2O
(1)
Unfortunately, under the conditions used to produce formaldehyde an
undesired reaction occurs, that is:
CH2O + ½ O2  CO + H2O
(2)
Assume that methanol and twice the stoichiometric amount of air needed
for complete conversion of the CH3OH to the desired products (CH2O and
H2O) are fed to the reactor.
Also assume that 90% conversion of the methanol results, and that a 75%
yield of formaldehyde occurs based on the theoretical production of CH2O
by Reaction 1.
Determine the composition of the product gas leaving the reactor.
Solution:
33
Chapter 5: Process Involving Multiple Reactions
𝑦2,O2
21%
𝑦2,N2
79%
𝐶𝑜𝑛𝑣. , 𝑓CH3OH
90%
𝑌𝑖𝑙𝑒𝑑CH2O
75%
•
𝑛2 (Air)
𝑦2,O2 =0.21
𝑦2,N2 =0.79
𝑛2,O2
2
𝑛2,N2
Assume that CH3OH
requires
twice
the
stoichiometric amount of
Air are fed to the reactor.
CH3OH + ½O2  CH2O + H2O (1)
CH2O + ½ O2  CO + H2O (2)
1
𝑛1,CH3OH
Formaldehyde
Reactor
Product
3
𝑛3,CH3OH
𝑛3,CH2O
𝑛3,H2O
𝑛3,CO
𝑛3,O2
𝑛3,N2
34
Chapter 5: Process Involving Multiple Reactions
Example 5.9: Analysis of a Bioreactor
A bioreactor is a vessel in which biological conversion is carried out. The
following overall reactions occurs:
Reaction 1: C6H12O6  2C2H5OH
+ 2CO2
Reaction 2: C6H12O6  2C2H3CO2H + 2H2O
In a batch process, a tank is charged with 4000 kg of a 12% solution of
glucose(C6H12O6) in water. After fermentation, 120 kg of CO2 are produced
and 90 kg of unreacted glucose(C6H12O6) remains in the solution. What are
the weight (mass) percent of ethanol(C2H5OH) and propenoic
acid(C2H3CO2H) in the solution at the end of the fermentation process?
Assume that none of the glucose(C6H12O6) is assimilated(digested) into the
bacteria.
Solution:
35
Chapter 5: Process Involving Multiple Reactions
C6H12O6  2C2H5OH
C6H12O6  2C2H3CO2H
𝑚𝑖𝑛𝑖,𝑠𝑜𝑙𝑛 =4000 kg Soln
𝑥𝑖𝑛𝑖,𝐶6𝐻12𝑂6 =0.120
𝑥𝑖𝑛𝑖,𝐻2𝑂
=0.88
C6H12O6 H2O
MW(g/mol)
180
18
+ 2CO2
(1)
+ 2H2O (2)
𝑚𝑓𝑖𝑛𝑎𝑙,𝐶𝑂2 =120 kg CO2
Unreacted C6H12O6
𝑚𝑓𝑖𝑛𝑎𝑙,𝐶6𝐻12𝑂6 =90 kg
CO2 C2H5OH
44.0 46
C2H3CO2H
72
𝑚𝑖𝑛𝑖,𝑠𝑜𝑙𝑛 =4000 kg
𝑛𝑖𝑛𝑖,𝐻2𝑂
𝑛𝑖𝑛𝑖,𝐶6𝐻12𝑂6
𝑛𝑓𝑖𝑛𝑎𝑙
𝑛𝑓𝑖𝑛𝑎𝑙,𝐻2𝑂
𝑛𝑓𝑖𝑛𝑎𝑙,𝐶6𝐻12𝑂6
𝑛𝑓𝑖𝑛𝑎𝑙,𝐶2𝐻3𝐶𝑂2𝐻
𝑛𝑓𝑖𝑛𝑎𝑙,𝐶𝑂2
𝑛𝑓𝑖𝑛𝑎𝑙,𝐶2𝐻5𝑂𝐻
36
Chapter 5: Element Material Balances, EMB
Species Moles Balances:
𝑛𝐴,𝑜𝑢𝑡 = 𝑛𝐴,𝑖𝑛 + 𝜐𝐴 𝜉
Element Material Balances, EMB: Number of atoms enter the
reaction EQUAL number of atoms leave the reaction
Input (atoms)= Output (atoms)
CO2 + H2O  H2CO3
For most problems it is easier to apply mole balances, but for some
problems, such as problems with complex or unknown reaction
equations, element balances are preferred.
37
Chapter 5: Element Material Balances, EMB
Example 5: Carbon dioxide is absorbed in water in the process shown
below. The reaction is
CO2 + H2O  H2CO3
Apply the element balance to find the unknowns in the flow chart?
𝑛2,CO
2
2
1
𝑛1,H O
2
Absorber
3
𝑛3 =100 mol
𝑦3,H CO =0.05
2
3
𝑦3,H O = 0.95
2
𝑛3,H CO
2
3
𝑛3,H O
2
38
Chapter 5: Element Material Balances
Example 5.11: Hydrocracking
Researchers in the field oil industry study the hydrocracking of pure components,
such as octane (C8H18) to understand the behavior of cracking reactions.
In one such experiment for the hydrocracking of octane (C8H18) , the cracked
products had the following composition in mole percent: 19.5% C3H8, 59.4% C4H10,
and 21.1% C5H12.
You are asked to determine the molar ratio of hydrogen consumed to octane reacted
for this process.
Solution:
39
Chapter 5: Element Material Balances
Species
Product
Moles
percentage
C3H8
19.5%
C4H10
59.4%
C5H12
21.1%
𝑛2,H2
1
𝑛1,C8H18
2
Lab
Reactor
3
𝑛3
𝑛3,C3H8
𝑛3,C4H10
𝑛3,C5H12
𝑦3,C3H8 =0.195
𝑦3,C4H10 =0.594
𝑦3,C5H12 =0.211
40
Chapter 5: Material Balances Involving Combustion
CH4+ Air  CO2 (g)+ 2H2O (g) + Energy
Wet basis: all the gases resulting from a combustion process including the
water vapor, known as Flue or stack gas.
Dry basis: all the gases resulting from a combustion process not9including
the water vapor.
Complete combustion: the complete reaction of the hydrocarbon fuel
producing CO2, SO2, and H2O.
Partial combustion: the combustion of the fuel producing at least some
CO.
Theoretical air (or theoretical oxygen): the minimum amount of air (or
oxygen) required to be brought into the process for complete combustion.
Sometimes this quantity is called the required air (or oxygen).
41
Chapter 5: Material Balances Involving Combustion
Excess air (or excess oxygen): excess air (or oxygen) is the amount of air
(or oxygen) in excess of that required for complete combustion.
excess air
% Excess air =
∙ 100
required air
excess O2 0.21
excess O2
=
∙ 100 =
∙ 100
required O2 0.21
required O2
O2 excess
= O2,in (enter the process) – O2 required
O2 in(enter the process)−O2 required
∙ 100
O2 required
% Excess air =
The calculated amount of excess air does not depend on how much material is
actually burned but what is possible to be burned. Even if only partial
combustion takes place.
42
Chapter 5: Material Balances Involving Combustion
Example 5.12: Excess Air
Fuels other than gasoline are being eyed for motor vehicles because they
generate lower levels of pollutants than does gasoline. Compressed propane
is one such proposed fuel.
Suppose that in a test 20 kg of C3H8 is burned with 400 kg of air to
produce 44 kg of CO2 and 12 kg of CO.
What was the percent excess air?
Solution:
C H + 5O  3CO + 4H O (g)
3
8
2
2
2
43
Chapter 5: Material Balances Involving Combustion
Example 5.13: A Fuel Cell to Generate Electricity From Methane
Fuel cell is an open system into which fuel and air are fed, and the outcome
are electricity and waste products. Figure bellow is a sketch of a fuel cell in
which a continuous flow of methane (CH4) and air (O2 plus N2) produce
electricity plus CO2 and H2O.
Special membranes and catalysts are needed to promote the reaction of
CH4.
Based on the data given in Flow chart, you are asked to calculate the
composition of the products in stream 3.
Solution:
44
Chapter 5: Material Balances Involving Combustion
Air
MW(g/mol) 29
H2O
18
CO2 CH4
44.0 16
1
𝑚1,CH4 =16.0 kg
𝑛1,CH4
Lab
Reactor
3
𝑛3
𝑛3,CO2
𝑛3,N2
𝑛3,O2
𝑛3,H2O
2
𝑚2,𝑎𝑖𝑟 =300 kg
𝑛2,𝑎𝑖𝑟
𝑛2,O2
𝑛2,N2
𝑦2,O2 =0.21
𝑦2,N2 =0.79
45
Chapter 5: Material Balances Involving Combustion
Example 6: Combustion of Ethane
Ethane is burned with 50% excess air. The percentage conversion of the
ethane is 90%.
The ethane burned: 25% reacts to form CO and 75 % reacts to form CO2?
Calculate the molar composition of the stack gas on a dry and wet basis and
the mole ratio of water to dry stack gas?
C2H6+ 7 2O2  2CO2 + 3H2O
C2H6+ 5 2O2  2CO + 3H2O
1
Combustion
Unit
𝑛1,𝐶2𝐻6 =100 mol
2
50% 𝑒𝑥𝑐𝑒𝑠𝑠 𝑎𝑖𝑟
𝑛2,𝑎𝑖𝑟
𝑦2,O2 =0.21
𝑦2,N2 =0.79
3
𝑛3,𝐶2𝐻6
𝑛3,CO2
𝑛3,CO
𝑛3,N2
𝑛3,O2
𝑛3,H2O
46
Chapter 5: Material Balances Involving Combustion
C2H6+ 7 2O2  2CO2 + 3H2O
C2H6+ 5 2O2  2CO + 3H2O
excess air = 50%
fC2H6 =90%
25% C2H6 reacts to form CO
75 % C2H6 reacts to form CO2
1
Combustion
Unit
𝑛1,𝐶2𝐻6 =100 mol
2
50% 𝑒𝑥𝑐𝑒𝑠𝑠 𝑎𝑖𝑟
𝑛2,𝑎𝑖𝑟
𝑦2,O2 =0.21
𝑦2,N2 =0.79
3
𝑛3,𝐶2𝐻6
𝑛3,CO2
𝑛3,CO
𝑛3,N2
𝑛3,O2
𝑛3,H2O
47
Chapter 5: Material Balances Involving Combustion
Example 7: Combustion of a Hydrocarbon Fuel of Unknown Composition
A hydrocarbon gas is burned with air. The dry-basis product gas
composition is 1.5 mole% CO, 6.0% C02,8.2% 02, and 84.3% N2. There is no
atomic oxygen in the fuel.
Calculate the ratio of hydrogen to carbon in the fuel gas and speculate on
what the fuel might be?
Calculate the percent excess air fed to the reactor?
1
𝑛1,𝐶
𝑛1,𝐻
Combustion
Unit
2
𝑛2,𝑎𝑖𝑟
𝑦2,O2 =0.21
𝑦2,N2 =0.79
3
𝑛3 =100 mol (dry gas )
𝑛3,CO2
𝑛3,CO
𝑛3,O2
𝑛3,N2
𝑛4,H2O
Output composition
on a dry basis
Elements
Input
Moles %
CO
1.5%
CO2
6.0%
O2
8.2%
N2
84.3%
48
Chapter 5: Material Balances Involving Combustion
Output composition on a dry basis
𝑛1,𝐶
𝑛1,𝐻
1
Combustion
Unit
2
𝑛2,𝑎𝑖𝑟
𝑦2,O2 =0.21
𝑦2,N2 =0.79
3
𝑛3 =100 mol (dry gas )
𝑛3,CO2 = 1.5 mol
𝑛3,CO = 6.0 mole
𝑛3,O2 = 8.20 mol
𝑛3,N2 =84.3 mol
𝑛4,H2O
49
Chapter 5: Material Balances Involving Combustion
Example 5.14: Combustion of Coal
A local utility burns, the moisture in the input fuel was 3.90%.
The air on the average contained 0.0048 kg H2O/kg dry air.
The refuse showed 14.0% unburned coal, with the remainder being ash.
You are asked to check the consistency of the data before they are stored in
a database. is the consistency satisfactory?
What was the average percent excess air used?
Input composition on a dry basis
Elements Input
Moles %
Output composition
on a dry basis
C
83.05%
H
4.45 %
O
3.36%
CO2+SO2
15.4%
N
1.08%
CO
0
S
0.70%
O2
4%
Ash
7.36%
N2
80.6%
Elements
Input
Moles
%
50
Chapter 5: Material Balances Involving Combustion
𝑚1 =100 kg
𝑚1,𝐶 =83.05 kg
𝑚1,𝐻 =4.45 kg
𝑚1,O =3.36 kg
𝑚1,N = 1.08 kg
𝑚1,S =0.70 kg
𝑚1,Ash =7.36 kg
1
𝑚𝑡𝑜𝑡
Lab
Reactor
2
3
𝑛3
𝑛3,CO2
𝑛3,N2
𝑛3,O2
𝑛3,H2O
𝑚2,𝑎𝑖𝑟 =300 kg
𝑛2,𝑎𝑖𝑟
𝑦2,O2 =0.21
𝑦2,N2 =0.79
51
52