Solving Material Balances Problems Involving Reactive Processes

CHE 31. INTRODUCTION TO CHEMICAL ENGINEERING CALCULATIONS

Lecture 10Solving Material Balances Problems Involving Reactive Processes

Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños

LECTURE 10. Solving Material Balance Problems Involving Reactive Processes

Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE

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Material Balances on Reactive Processes

Material balances on processes involving chemical reactions may be solved by applying:

1. Molecular Species Balance – a material balance equation is applied to each chemical compound appearing in the process.

2. Atomic Species Balance – the balance is applied to each element appearing in the process.

3. Extent of Reaction – expressions for each reactive species is written involving the extent of reaction.



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Molecular and Elemental Balances

For steady-state reactive processes,

Input + Generation = Output + Consumption

The generation and consumption terms in the molecular balance equation is usually obtained from chemical stoichiometry.

But for an atomic balance, for all cases

Input = Output



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Dehydrogenation of Ethane

Consider the dehydrogenation of ethane in a steady-state continuous reactor,



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Dehydrogenation of Ethane

Total Balance: Input = Output

Molecular Species Balance:

C2H6: Input – Consumed = Output

C2H4: Generated = Output

H2: Generated = Output

Atomic (Elemental) Species Balance:

C-Balance: Input = Output

H-Balance: Input = Output



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Degrees of Freedom of Analysis for Reactive Processes

Molecular Species Balance

+ No. identified/labeled unknowns

+ No. independent chemical reactions

– No. of independent molecular species

– No. other equations relating unknown variables-------------------------------------------------------------------------

= No. degrees of freedom



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Atomic Species Balance


– No. independent atomic species

– No. of independent nonreactive molecular species

– No. other equations relating unknown variables-----------------------------------------------------------------------------




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Extent of Reaction


+ No. independent chemical reactions

– No. of independent reactive molecular species

– No. of independent nonreactive molecular species

– No. other equations relating unknown variables-----------------------------------------------------------------------------




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Independent Chemical Reactions, Molecular and Atomic Species

Chemical reaction: A chemical reaction is independent if it cannot be obtained algebraically from other chemical reactions involved in the same process.

Molecular Species: If two molecular species are in the same ratio to each other wherever they appear in a process, then these molecular species are not independent.

Atomic Species: If two atomic species occur in the same ration wherever they appear in a process, balances on those species will not be independent equations.


Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños SLIDE10


Consider the following reactions:

A =======> 2B

B =======> C

A =======> 2C

Are these chemical reactions independent?




Consider a continuous process in which a stream of liquid carbon tetrachloride (CCl4) is vaporized into a stream of air.




Molecular Species Analysis

Total: 3 (O2, N2, CCl4)

Independent: 2 (O2 or N2, CCl4)

Atomic Species Analysis

Total: 4 (O, N, C, Cl)

Independent 2 (O or N, Cl or C)



Example 10-1. Production of Chlorine (Deacon Process)

In the Deacon process for the manufacture of chlorine, HCland O2 react to form Cl2 and H2O.

Sufficient air (21 mole% O2, 79% N2) is fed to provide 35% excess oxygen and the fractional conversion of HCl is 85%.

Determine the amount of air required per mole of HCl fed into the process.Calculate the mole fractions of the product stream components using:

a. molecular species balancesb. atomic species balancesc. extent of reaction




Identify the components of the product stream:

HCl since not all will be converted (based on fractional conversion)

O2 since it is supplied in excess

N2 it goes with the O2 in air but not consumed during the reaction

Cl2 produced during the process

H2O produced during the process




To get mole fractions of components in the product stream:

yi = ni/nt

For the identified components:

yHCl = n2/ntyO2 = n3/ntyN2 = n4/ntyCl2 = n5/ntyH2O = n6/nt

where nt = n2 + n3 + n4 + n5 + n6




DEGREES OF FREEDOM ANALYSIS: Molecular Balance

Unit: Reactor

unknowns (n1,n2,n3,n4,n5,n6) +6

independent chemical reaction +1

independent molecular species –5

other equations:35% excess O2 & fractional HCl conversion –2

Degrees of freedom 0




Method I: Molecular Species Balance

35% excess O2:

22 T 2

2 A 2 2

1 22

0.5 molO(O ) 100molHCl 25molO2molHCl

(O ) 25molO 1.35 33.75molO

1molairn 33.75molO 160.7molair0.21molO

160.7molair molairRequiredair 1.607100molHCl molHCl




HCl Balance: Input – Consumed – Output = 0(100 mol) – 0.85(100 mol) – n2 = 0n2 = 15 mol HCl

O2 Balance: Input – Consumed – Output = 0(33.75 mol) – 85 mol HCl react (0.5/2) – n3 = 0n3 = 12.5 mol O2

N2 Balance: Output = Inputn4 = 160.7 mol air (0.79 mol N2/1 mol air)n4 = 127 mol N2




Cl2 Balance: Generated – Output = 085 mol HCl react (1/2) – n5 = 0n5 = 42.5 mol Cl2

H2O Balance: Generated – Output = 085 mol HCl react (1/2) – n6 = 0n6 = 42.5 mol H2O




Calculation for mole fractions:

Component i ni (moles) y

HCl 15.0 (15.0/239.5) = 0.063

O2 12.5 (12.5/239.5) = 0.052

N2 127.0 (127.0/239.5) = 0.530

Cl2 42.5 (42.5/239.5) = 0.177

H2O 42.5 (42.5/239.5) = 0.177

Total 239.5 1.000




DEGREES OF FREEDOM ANALYSIS: Atomic Balance

Unit: Reactor


independent atomic specie(s) –3

independent nonreactive molecular specie(s) –1

other equations:35% excess O2 & fractional HCl conversion –2





From % excess O2 ======> n1

From fractional conversion ======> n2

Atomic Species Balance:

H-Balance: 100(1) = n2 + 2n6

O-Balance: n1(0.21)(2) = 2n3 + n6

Cl-Balance: 100(1) = n2 + 2n5

N-Balance: n1(0.79)(2) = 2n4




DEGREES OF FREEDOM ANALYSIS: Extent of Reaction

Unit: Reactor


independent chemical reaction(s) +1

independent reactive molecular species –4

independent nonreactive molecular species –1other equations:35% excess O2 & fractional HCl conversion –2





From % excess O2 ======> n1

From fractional conversion ======> n2

Extent of Reaction:

HCl: n2 = 100 – (2)

Cl2: n5 = 0 + (1)

H2O: n6 = 0 + (1)

N2: n4 = 0.79n1 ± (0)

O2: n3 = 0.21n1 – (0.5)



Example 10-2. Production of Ethyl Bromide

The reaction between ethylene and hydrogen bromide to form ethyl bromide is carried out in a continuous reactor.

C2H4 + HBr =====> C2H5Br

The product stream is analyzed and found to contain 51.7 mole% C2H5Br and 17.3% HBr. The feed to the reactor contains only ethylene and hydrogen bromide.

Calculate the fractional conversion of the limiting reactant and the percentage by which the other reactant is in excess. If the molar flow rate of the feed stream is 165 mol/s, what is the extent of reaction?




DEGREES OF FREEDOM OF ANALYSIS: Atomic Species

Unit: Reactor

unknowns (x and n2) +2

independent atomic specie(s) –2

independent nonreactive molecular specie(s) 0

other equations 0





Determine the limiting reactant:

2 4

S

2 4

A

C HStoichiometricRatio : 1.0HBr

x 165mol / sC H xActualRatio :HBr (1 x)(165mol / s) 1 x

Solve x and n2 using any 2 of the 3 atomic species balances:

C-BalanceH-BalanceBr-Balance




C-Balance:

2 42 2

2 4

2

x molC Hmol 2molC165 n 0.310 2 n 0.517 2s mol 1molC H

330x 1.654n

Br-Balance:

2 2

2

1 x molHBrmol 1molBr165 n 0.173 1 n 0.517 1s mol 1molHBr

165(1 x) 0.69n




Solving simulateneously,

x = 0.545 mol C2H4/mol ; n2 = 108.77 mol/s

Solving for the actual ratio of C2H4 and HBr in the feed:

2 4

A

C H 0.545 1.0HBr 1 0.545

Therefore, HBr is limiting.

2 4actual stoichiometric% excessC H 100

actual




Actual feed for C2H4:

(165 mol/s)(0.545) = 89.93 mol/s

Theoretical requirement for C2H4 based on stoichiometry:

2 41 0.545 mol HBr 1molC Hmol mol165 75.08s mol 1molHBr s

2 489.93 75.08% excess C H 100 19.8%

75.08




Fractional conversion of HBr:

HBr

HBr

amount reacted input outputXamount fed input

165 1 0.545 108.77 0.173X 0.749

165 1 0.545

The can be determined based on C2H4, HBr, C2H5Br:

C2H4: 0.310(108.77) = (165)(0.545) – HBr: 0.173(108.77) = (165)(1-0.545) – C2H5Br: 0.517(108.77) = 0 –

Solving for : = 56.2 mol/s

Solving Material Balances Problems Involving Reactive Processes

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