Solving Material Balances Problems Involving Reactive Processes

CHE 31. INTRODUCTION TO CHEMICAL ENGINEERING CALCULATIONS

Lecture 10

Solving 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

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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Degrees of Freedom of Analysis for Reactive Processes Atomic Species Balance

+ No. identified/labeled unknowns – No. independent atomic species – No. of independent nonreactive molecular species – No. other equations relating unknown variables ----------------------------------------------------------------------------= No. degrees of freedom


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Degrees of Freedom of Analysis for Reactive Processes Extent of Reaction

+ No. identified/labeled unknowns + No. independent chemical reactions – No. of independent reactive molecular species – No. of independent nonreactive molecular species – No. other equations relating unknown variables ----------------------------------------------------------------------------= No. degrees of freedom


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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.


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Consider the following reactions:

A =======> 2B B =======> C A =======> 2C Are these chemical reactions independent?


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


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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)


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

In the Deacon process for the manufacture of chlorine, HCl and 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 balances b. atomic species balances c. extent of reaction


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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


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To get mole fractions of components in the product stream:

yi = ni/nt For the identified components:

yHCl = n2/nt yO2 = n3/nt yN2 = n4/nt yCl2 = n5/nt yH2O = n6/nt where nt = n2 + n3 + n4 + n5 + n6


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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


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Method I: Molecular Species Balance

35% excess O2:  0.5 molO 2  (O 2 )T  100mol HCl   25molO 2   2 mol HCl  (O 2 ) A  25mol O 2 1.35   33.75molO 2  1molair  n1  33.75mol O 2   160.7 molair   0.21mol O 2  160.7 mol air molair Required air   1.607 100 mol HCl mol HCl Prof. Manolito E Bambase Jr. Department of Chemical Engineering. University of the Philippines Los Baños

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Example 10-1. Production of Chlorine (Deacon Process) HCl Balance: Input – Consumed – Output = 0 (100 mol) – 0.85(100 mol) – n2 = 0 n2 = 15 mol HCl O2 Balance:

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

N2 Balance:

Output = Input n4 = 160.7 mol air (0.79 mol N2/1 mol air) n4 = 127 mol N2


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Example 10-1. Production of Chlorine (Deacon Process) Cl2 Balance:

Generated – Output = 0 85 mol HCl react (1/2) – n5 = 0 n5 = 42.5 mol Cl2

H2O Balance:

Generated – Output = 0 85 mol HCl react (1/2) – n6 = 0 n6 = 42.5 mol H2O


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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


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DEGREES OF FREEDOM ANALYSIS: Atomic Balance


+6

independent atomic specie(s)

–3

independent nonreactive molecular specie(s)

–1


–2

Degrees of freedom

0


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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


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DEGREES OF FREEDOM ANALYSIS: Extent of Reaction


+6

independent chemical reaction(s)

+1

independent reactive molecular species

–4

independent nonreactive molecular species

–1


–2

Degrees of freedom

0


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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)


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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?


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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

Degrees of freedom

0


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Determine the limiting reactant:

Stoichiometric Ratio :

 C2 H 4     1.0  HBr S

Actual Ratio :

 x 165mol / s   x  C2 H 4        HBr   A  (1  x)(165mol / s)  1  x

Solve x and n2 using any 2 of the 3 atomic species balances: C-Balance H-Balance Br-Balance


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C-Balance:

165

mol  x molC2 H 4   2mol C   n 2  0.310  2   n 2  0.517  2     s  mol   1mol C2 H 4 

330x  1.654n 2 Br-Balance:

mol  1  x  mol HBr   1mol Br  165  n 2  0.1731  n 2  0.517 1     s  mol  1mol HBr  165(1  x)  0.69n 2


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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:

0.545  C2 H 4   1.0     HBr A 1  0.545 Therefore, HBr is limiting.

% excess C 2 H 4 

actual  stoichiometric  100 actual


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Actual feed for C2H4:

(165 mol/s)(0.545) = 89.93 mol/s Theoretical requirement for C2H4 based on stoichiometry:

mol  1  0.545 mol HBr  1mol C 2 H 4  mol 165  75.08   1mol HBr  s  mol s   % excess C2 H 4 

89.93  75.08  100  19.8% 75.08


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Fractional conversion of HBr:

X HBr

amount reacted input  output   amount fed input

165 1  0.545  108.77  0.173 X HBr   0.749 1651  0.545  The  can be determined based on C2H4, HBr, C2H5Br: C2H4: HBr: C2H5Br: Solving for :

0.310(108.77) = (165)(0.545) –  0.173(108.77) = (165)(1-0.545) –  0.517(108.77) = 0 – 

 = 56.2 mol/s


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