Chapter 5: Extent of Reaction

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27

Chapter 5: Extent of Reaction

𝜉 = 𝑛

_{𝐴,𝑜𝑢𝑡}

− 𝑛

_{𝐴,𝑖𝑛}

𝜐

_𝐴

Extent of reaction:

𝑛

= 𝑛

_{𝐴,𝑖𝑛}

+ 𝜐

_𝐴

𝜉 n

_N2,out

= n

_N2,in

− Consumption

𝐴𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛 = 𝑖𝑛𝑝𝑢𝑡 - 𝑜𝑢𝑡𝑝𝑢𝑡 + 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 - 𝑐𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛

0 = 𝑖𝑛𝑝𝑢𝑡 - 𝑜𝑢𝑡𝑝𝑢𝑡 + 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 - 𝑐𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛

Steady state system:

𝑛_1,𝑁2 𝑛_1,𝐻2

Reactor

Input (1) Output (2)

𝑛_2,𝑁𝐻3

N₂ + 3H₂  2NH₃ 𝑜𝑢𝑡𝑝𝑢𝑡 = 𝑖𝑛𝑝𝑢𝑡 + 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 - 𝑐𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛

n

_NH3,out

= n

_NH3,in

+ Generation

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28

Chapter 5: Species Mole Balances, SMB

𝜉 = 𝑛

− 𝑛

_{𝐴,𝑖𝑛}

𝜐

_𝐴

𝜉 = 𝑛

_{𝐴,𝑓𝑖𝑛𝑎𝑙}

− 𝑛

_{𝐴,𝑖𝑛𝑖}

𝜐

_𝐴

Extent of reaction (Batch process):

Extent of reaction (Continuous process):

𝜉 = −𝑓 ∙ 𝑛

𝑖𝑛,𝑙𝑖𝑚𝑒𝑡𝑒𝑑 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡

𝜐

_𝐿𝑅

f =Conversion factor of the limited reactant

𝑓 = 𝑓

_𝐷𝐶

= 𝑚𝑜𝑙𝑒𝑠(𝑚𝑎𝑠𝑠) 𝑜𝑓 𝑙𝑖𝑚𝑖𝑡𝑒𝑑 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑟𝑒𝑎𝑐𝑡𝑒𝑑

𝑚𝑜𝑙𝑒𝑠(𝑚𝑎𝑠𝑠) 𝑜𝑓 𝑙𝑖𝑚𝑡𝑒𝑑 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑓𝑒𝑑 × 100%

𝑛

= 𝑛

_{𝐴,𝑖𝑛}

+ 𝜐

_𝐴

𝜉 𝑛

_{𝐴,𝑓𝑖𝑛𝑎𝑙}

= 𝑛

_{𝐴,𝑖𝑛𝑖}

+ 𝜐

_𝐴

𝜉

Species Material Balances

𝜉 = 𝑛

− 𝑛

_{𝐴,𝑖𝑛}

𝜐

_𝐴

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29

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

:

𝑛1,Cl₂ 𝑛1,CH₄ 𝑛1,N₂

1 2

Reactor

Species moles % feed

CH₄ 40%

Cl₂ 50%

N₂ 10%

f_LR 67%

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30

Chapter 5: Species Mole Balances

CH₄ + Cl₂  CH₃Cl + HCl

1 2

𝑛2,Cl₂ 𝑛2,CH₄ 𝑛2,N₂ 𝑛2,HCl 𝑛2,CH₃Cl

Reactor

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31

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:

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32

Chapter 5: Species Mole Balances

Species moles % feed

CH4 80%

H2S 20%

𝑚4,S 5000 kg

𝑛3,SO2/𝑛3,H2S 3 𝑛3,H2O/𝑛3,H2S 10

𝑛1𝑦1,H2S=0.20 𝑦1,CH⁴=0.80 𝑛1,H2S

𝑛1,CH⁴

1 3

𝑛3,H2S 𝑛3,CH⁴ 𝑛3,H2O 𝑛3,SO2

Reactor

2

4

𝑛2,SO2

n4,s= 156.25 mol 𝑚4,S=5000 kg

Product

2H2S(g) + SO2(g)  3S(s) + 2H2O(g)

Moles, Kmol MW Mass, Kg

Input

n1 H2S 114.6 34 3896.4

n1 CH4 458.48 16 7335.68

n2 SO2 83.3 64 5331.2

Sum input 656.38 Kmol 16563.3 Kg

Output

n3 H2S 10.42 34 354.28

n3 CH4 458.48 16 7335.68

n3 H2O 104.2 18 1875.6

n3 SO2 31.2 64 1996.8

n4 S 156.25 32 5000

Sum out 760.55 Kmol 16562.4 Kg

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33

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

:

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34

Chapter 5: Process Involving Multiple Reactions

𝑦2,O2 21%

𝑦2,N2 79%

𝐶𝑜𝑛𝑣. , 𝑓CH3OH 90%

𝑌𝑖𝑙𝑒𝑑CH2O 75%

𝑛1,CH3OH

1 3

𝑛3,CH3OH 𝑛3,CH2O 𝑛3,H2O 𝑛3,CO 𝑛3,O2 𝑛3,N2

Formaldehyde

Reactor

𝑛2(Air)

𝑦2,O2=0.21 𝑦2,N2 =0.79 𝑛2,O2

𝑛2,N2

Product 2

• Assume that CH3OH requires twice the stoichiometric amount of Air are fedto the reactor.

CH3OH + ½O2  CH2O + H2O (1) CH2O + ½ O2  CO + H2O (2)

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35 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

:

Chapter 5: Process Involving Multiple Reactions

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36

Chapter 5: Process Involving Multiple Reactions

C6H12O6  2C2H5OH + 2CO2 (1) C6H12O6  2C2H3CO2H + 2H2O (2)

𝑚_{𝑖𝑛𝑖,𝑠𝑜𝑙𝑛}=4000 kg 𝑛_{𝑖𝑛𝑖,𝐻2𝑂}

𝑛𝑖𝑛𝑖,𝐶6𝐻12𝑂6

𝑛_{𝑓𝑖𝑛𝑎𝑙} 𝑛_{𝑓𝑖𝑛𝑎𝑙,𝐻2𝑂} 𝑛𝑓𝑖𝑛𝑎𝑙,𝐶6𝐻12𝑂6

𝑛𝑓𝑖𝑛𝑎𝑙,𝐶2𝐻3𝐶𝑂2𝐻

𝑛_{𝑓𝑖𝑛𝑎𝑙,𝐶𝑂2} 𝑛𝑓𝑖𝑛𝑎𝑙,𝐶2𝐻5𝑂𝐻

𝑚_{𝑖𝑛𝑖,𝑠𝑜𝑙𝑛}=4000kg Soln 𝑥𝑖𝑛𝑖,𝐶6𝐻12𝑂6 =0.120 𝑥_{𝑖𝑛𝑖,𝐻2𝑂} =0.88

𝑚_{𝑓𝑖𝑛𝑎𝑙,𝐶𝑂2} =120 kg CO2 Unreacted C₆H₁₂O₆ 𝑚𝑓𝑖𝑛𝑎𝑙,𝐶6𝐻12𝑂6 =90 kg

C₆H₁₂O₆ H2O CO₂C2H₅OH C₂H₃CO₂H MW(g/mol) 180 18 44.0 46 72

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37

Chapter 5: Element Material Balances, EMB

Input (atoms)= Output (atoms)

CO₂ + H₂O  H₂CO₃

𝑛

= 𝑛

_{𝐴,𝑖𝑛}

+ 𝜐

_𝐴

𝜉 Species Moles Balances:

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.

Element Material Balances, EMB: Number of atoms enter the

reaction EQUAL number of atoms leave the reaction

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38

Chapter 5: Element Material Balances, EMB

𝑛_1,H₂O

𝑛₃=100 mol

𝑦_3,H₂CO₃ =0.05 𝑦_3,H₂O = 0.95 𝑛_3,H₂CO₃

𝑛_3,H₂O

Absorber

𝑛_2,CO₂

1

2

3

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?

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39

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

:

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40

Chapter 5: Element Material Balances

𝑛_1,C8H18

𝑛₃

𝑛_3,C3H8 𝑛_3,C4H10 𝑛_3,C5H12

𝑦_3,C3H8=0.195 𝑦_3,C4H10 =0.594 𝑦_3,C5H12 =0.211

Lab Reactor

𝑛_2,H2

1

2

3

Species ^Product Moles percentage

C3H8 19.5%

C4H10 59.4%

C5H12 21.1%

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41

Chapter 5: Material Balances Involving Combustion

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).

CH₄+ Air  CO₂ (g)+ 2H₂O (g) + Energy

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42

Chapter 5: Material Balances Involving Combustion

% Excess air = excess air

required air ∙ 100

= excess O2 ^0.21

required O2 ^0.21 ∙ 100 = excess O2

required O2 ∙ 100

% Excess air =O2 in(enter the process)−O2 required

O2 required ∙ 100

O2 excess = O2,in (enter the process) – O2 required

Excess air (or excess oxygen): excess air (or oxygen) is the amount of air (or oxygen) in excess of that required for complete combustion.

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.

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43

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: C₃H₈+ 5O₂  3CO₂ + 4H₂O (g)

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44

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:

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45

Chapter 5: Material Balances Involving Combustion

𝑚_1,CH4=16.0 kg 𝑛_1,CH4

𝑚_{2,𝑎𝑖𝑟}=300 kg 𝑛_{2,𝑎𝑖𝑟}

𝑛_2,O2 𝑛_2,N2

𝑦_2,O2=0.21 𝑦_2,N2 =0.79

Lab Reactor

𝑛₃

𝑛_3,CO2 𝑛_3,N2 𝑛_3,O2 𝑛_3,H2O

1

3

Air H2O CO₂CH4

2

MW(g/mol) 29 18 44.0 16

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46

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?

50% 𝑒𝑥𝑐𝑒𝑠𝑠 𝑎𝑖𝑟 𝑛_{2,𝑎𝑖𝑟}

𝑦_2,O2=0.21 𝑦_2,N2 =0.79

Combustion Unit

𝑛_3,𝐶2𝐻6 𝑛_3,CO2 𝑛_3,CO 𝑛_3,N2 𝑛_3,O2 𝑛_3,H2O

1 3

2

𝑛_1,𝐶2𝐻6=100 mol

C2H6+ ⁷ ₂O2  2CO2 + 3H2O C2H6+ ⁵ ₂O2  2CO + 3H2O

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47

Chapter 5: Material Balances Involving Combustion

50% 𝑒𝑥𝑐𝑒𝑠𝑠 𝑎𝑖𝑟 𝑛_{2,𝑎𝑖𝑟}

𝑦_2,O2=0.21 𝑦_2,N2 =0.79

𝑛_3,𝐶2𝐻6 𝑛_3,CO2 𝑛_3,CO 𝑛_3,N2 𝑛_3,O2 𝑛_3,H2O

1 3

2

𝑛_1,𝐶2𝐻6=100 mol

C2H6+ ⁷ ₂O2  2CO2 + 3H2O C2H6+ ⁵ ₂O2  2CO + 3H2O

excess air = 50%

f_C2H6 =90%

25% C2H6 reacts to form CO 75 % C2H6 reacts to form CO2

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48

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?

𝑛_{2,𝑎𝑖𝑟}

𝑦_2,O2=0.21 𝑦_2,N2 =0.79

𝑛₃=100 mol (dry gas ) 𝑛_3,CO2

𝑛_3,CO 𝑛_3,O2 𝑛_3,N2 𝑛_4,H2O

1 3

2

𝑛_1,𝐶 𝑛_1,𝐻

Elements Input

Moles %

CO 1.5%

CO2 6.0%

O2 8.2%

N2 84.3%

Output composition on a dry basis

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49

Chapter 5: Material Balances Involving Combustion

𝑛_{2,𝑎𝑖𝑟}

𝑦_2,O2=0.21 𝑦_2,N2 =0.79

𝑛₃=100 mol (dry gas ) 𝑛_3,CO2 = 1.5 mol

𝑛_3,CO = 6.0 mole 𝑛_3,O2 = 8.20 mol 𝑛_3,N2 =84.3 mol 𝑛_4,H2O

1 3

2

𝑛_1,𝐶 𝑛_1,𝐻 Output composition on a dry basis

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50

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?

Elements Input Moles %

C 83.05%

H 4.45 %

O 3.36%

N 1.08%

S 0.70%

Ash 7.36%

Input composition on a dry basis

Elements Input

Moles

% CO₂+SO₂ 15.4%

CO 0

O₂ 4%

N₂ 80.6%

Output composition on a dry basis

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51

Chapter 5: Material Balances Involving Combustion

𝑚_𝑡𝑜𝑡

𝑚_{2,𝑎𝑖𝑟}=300 kg 𝑛_{2,𝑎𝑖𝑟}

𝑦_2,O2=0.21 𝑦_2,N2 =0.79

Lab Reactor

𝑛₃

𝑛_3,CO2 𝑛_3,N2 𝑛_3,O2 𝑛_3,H2O

1 3

2

𝑚₁=100 kg 𝑚_1,𝐶=83.05 kg 𝑚_1,𝐻=4.45 kg 𝑚_1,O=3.36 kg 𝑚_1,N= 1.08 kg 𝑚_1,S=0.70 kg 𝑚_1,Ash =7.36 kg

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52