Chapter 5: Extent of Reaction

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Godfrey McLaughlin
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1 Chapter 5: Extent of Reaction Accumulation = input - output + generation - consumption Steady state system: 0 = input - output + generation - consumption output = input + generation - consumption Input (1) Output (2) Reactor n 1,N2 n 1,H2 n 2,NH3 N 2 + 3H 2 2NH 3 n N2,out = n N2,in Consumption n NH3,out = n NH3,in + Generation n A,out = n A,in + υ A ξ Extent of reaction: ξ = n A,out n A,in υ A 27

2 Chapter 5: Species Mole Balances, SMB Extent of reaction (Batch process): ξ = n A,final n A,ini υ A Extent of reaction (Continuous process): ξ = n A,out n A,in υ A f =Conversion factor of the limited reactant n A,final = n A,ini + υ A ξ n A,out = n A,in + υ A ξ Species Material Balances f = f DC = moles(mass) of limited reactant reacted moles(mass) of limted reactant fed 100% ξ = f n in,limeted reactant υ LR ξ = n A,out n A,in υ A 28

Cl + HCl You are asked to determine the product composition if the conversion of the limiting reactant is 67%, and the feed composition

3 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 CH 4 + Cl 2 CH 3 Cl + 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% CH 4, 50%Cl2,and 10% N 2. Assumptions: The reactor is Open, Steady state process Solution: Species moles % feed CH 4 40% Cl 2 50% N 2 10% n 1,Cl2 n 1,CH4 n 1,N2 1 2 Reactor n2, Cl2 n 2,CH4 n2, N2 f LR 67% 29

4 Chapter 5: Species Mole Balances 30 CH 4 + Cl 2 CH 3 Cl + HCl 1 2 n 1,Cl2 n 1,CH4 n 1,N2 Reactor n2, Cl2 n 2,CH4 n2, N2 n2, HCl n2, CH3 Cl

A proposed process to remove H 2 S 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

5 Chapter 5: Species Mole Balances 31 Example 10.2: A Reaction in Which the Fraction Conversion is to Be Calculated: H 2 S is toxic in very small quantities and is quite corrosive to process equipment. A proposed process to remove H 2 S 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 reactant, and the feed rates of the H2S and SO2 streams. Assumptions: The reactor is Open, Steady state process. Solution: the limiting

6 32 Chapter 5: Species Mole Balances Species CH4 80% H2S 20% m 4,S n 3,SO2 /n 3,H2S 3 n 3,H2O /n 3,H2S 10 moles % feed 5000 kg 2H2S(g) + SO2(g) 3S(s) + 2H2O(g) n 1 y 1,H2S =0.20 y 1,CH4 =0.80 n 1,H2S n 1,CH4 n 2,SO2 2 Product 1 3 Reactor 4 m 4,S =5000 kg n4,s= mol n 3,H2S n 3,CH4 n 3,H2O n 3,SO2 Input Output Moles, Kmol MW Mass, Kg n1 H2S n1 CH n2 SO Sum input Kmol Kg n3 H2S n3 CH n3 H2O n3 SO n4 S Sum out Kmol Kg

+ 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

7 Chapter 5: Process Involving Multiple Reactions 33 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:

8 34 Chapter 5: Process Involving Multiple Reactions y 2,O2 21% y 2,N2 79% Conv., f CH3OH 90% Yiled CH2O 75% 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) n 1,CH3OH n2(air) y 2,O2 =0.21 y 2,N2 =0.79 n 2,O2 n 2,N2 2 Product 1 3 Formaldehyde Reactor n 3,CH3OH n 3,CH2O n 3,H2O n 3,CO n 3,O2 n 3,N2

Chapter 5: Process Involving Multiple Reactions 35 Example 5.9: Analysis of a Bioreactor A bioreactor is a vessel in which biological conversion is carried out.

12% solution of glucose(c 6 H 12 O 6 ) in water. After fermentation, 120 kg of CO 2 are produced and 90 kg of unreacted glucose(c 6 H 12 O 6 ) remains in the solution.

9 Chapter 5: Process Involving Multiple Reactions 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: C 6 H 12 O 6 2C 2 H 5 OH + 2CO 2 Reaction 2: C 6 H 12 O 6 2C 2 H 3 CO 2 H + 2H 2 O In a batch process, a tank is charged with 4000 kg of a 12% solution of glucose(c 6 H 12 O 6 ) in water. After fermentation, 120 kg of CO 2 are produced and 90 kg of unreacted glucose(c 6 H 12 O 6 ) remains in the solution. What are the weight (mass) percent of ethanol(c 2 H 5 OH) and propenoic acid(c 2 H 3 CO 2 H) in the solution at the end of the fermentation process? Assume that none of the glucose(c 6 H 12 O 6 ) is assimilated(digested) into the bacteria. Solution:

10 Chapter 5: Process Involving Multiple Reactions 36 C6H12O6 2C2H5OH + 2CO2 (1) C6H12O6 2C2H3CO2H + 2H2O (2) m ini,soln =4000 kg Soln m final,co2 =120 kg CO2 x ini,c6h12o6 =0.120 Unreacted C 6 H 12 O 6 x ini,h2o =0.88 m final,c6h12o6 =90 kg C 6 H 12 O 6 H2O CO 2 C2H 5 OH C 2 H 3 CO 2 H MW(g/mol) m ini,soln =4000 kg n ini,h2o n ini,c6h12o6 n final n final,h2o n final,c6h12o6 n final,c2h3co2h n final,co2 n final,c2h5oh

11 Chapter 5: Element Material Balances, EMB 37 Species Moles Balances: n A,out = n A,in + υ A ξ Element Material Balances, EMB: Number of atoms enter the reaction EQUAL number of atoms leave the reaction Input (atoms)= Output (atoms) CO 2 + H 2 O H 2 CO 3 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.

12 Chapter 5: Element Material Balances, EMB 38 Example 5: Carbon dioxide is absorbed in water in the process shown below. The reaction is CO 2 + H 2 O H 2 CO 3 Apply the element balance to find the unknowns in the flow chart? 1 n 1,H 2 O n 2,CO 2 2 Absorber 3 n 3 =100 mol y 3,H 2 CO 3 =0.05 y 3,H 2 O = 0.95 n 3,H 2 CO 3 n 3,H 2 O

13 Chapter 5: Element Material Balances 39 Example 5.11: Hydrocracking Researchers in the field oil industry study the hydrocracking of pure components, such as octane (C 8 H 18 ) to understand the behavior of cracking reactions. In one such experiment for the hydrocracking of octane (C 8 H 18 ), the cracked products had the following composition in mole percent: 19.5% C3H8, 59.4% C 4 H 10, and 21.1% C 5 H 12. You are asked to determine the molar ratio of hydrogen consumed to octane reacted for this process. Solution:

Chapter 5: Element Material Balances 40 Species C3H8 19.5% C4H10 59.4% Product Moles percentage C5H12 21.

14 Chapter 5: Element Material Balances 40 Species C3H8 19.5% C4H % Product Moles percentage C5H % 1 n 1,C8H18 n 2,H2 2 Lab Reactor 3 n 3 n 3,C3H8 n 3,C4H10 n 3,C5H12 y 3,C3H8 =0.195 y 3,C4H10 =0.594 y 3,C5H12 =0.211

15 Chapter 5: Material Balances Involving Combustion CH 4 + Air CO 2 (g)+ 2H 2 O (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 producing CO 2, SO 2, and H 2 O. the hydrocarbon fuel 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

16 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 = required air 100 excess O = required O = excess O2 required O2 100 O2 excess = O2,in (enter the process) O2 required O2 in(enter the process) O2 required % Excess air = 100 O2 required 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

17 Chapter 5: Material Balances Involving Combustion 43 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 C 3 H 8 is burned with 400 kg of air to produce 44 kg of CO 2 and 12 kg of CO. What was the percent excess air? Solution: C 3 H 8 + 5O 2 3CO 2 + 4H 2 O (g)

18 Chapter 5: Material Balances Involving Combustion 44 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:

19 Chapter 5: Material Balances Involving Combustion 45 Air H2O CO 2 CH4 MW(g/mol) m 1,CH4 =16.0 kg n 1,CH4 Lab Reactor 3 n 3 n 3,CO2 n 3,N2 n 3,O2 n 3,H2O 2 m 2,air =300 kg n 2,air n 2,O2 n 2,N2 y 2,O2 =0.21 y 2,N2 =0.79

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%.

20 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 n 1,C2H6 =100 mol 1 3 Combustion Unit 2 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 46

21 Chapter 5: Material Balances Involving Combustion 47 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 n 1,C2H6 =100 mol 1 3 Combustion Unit 2 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

22 Chapter 5: Material Balances Involving Combustion 48 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 1,C n 1,H 1 3 Combustion Unit 2 n 2,air y 2,O2 =0.21 y 2,N2 =0.79 n 3 =100 mol (dry gas ) n 3,CO2 n 3,CO n 3,O2 n 3,N2 n 4,H2O Output composition on a dry basis Elements Input Moles % CO 1.5% CO2 6.0% O2 8.2% N2 84.3%

23 Chapter 5: Material Balances Involving Combustion 49 Output composition on a dry basis n 1,C n 1,H 1 3 Combustion Unit 2 n 2,air y 2,O2 =0.21 y 2,N2 =0.79 n 3 =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

24 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 kg H 2 O/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 % C 83.05% H 4.45 % O 3.36% N 1.08% S 0.70% Ash 7.36% Output composition on a dry basis Elements Input Moles % CO 2 +SO % CO 0 O 2 4% N % 50

25 Chapter 5: Material Balances Involving Combustion 51 m 1 =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 m tot 1 3 Lab Reactor 2 m 2,air =300 kg n 2,air y 2,O2 =0.21 y 2,N2 =0.79 n 3 n 3,CO2 n 3,N2 n 3,O2 n 3,H2O

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Chapter 10 Material Balances for Processes Involving Reaction 10.1 Species Material Balances Processes Involving a Single Reaction

Chapter 10 Material Balances for Processes Involving Reaction 10.1 Species Material Balances 10.1.1 Processes Involving a Single Reaction The material balance for a species must be augmented to include