Lecture 10 Solving Material Balances Problems Involving Reactive Processes

Lecture 10

Solving Material Balances 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.

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

Dehydrogenation of Ethane

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

Dehydrogenation of Ethane

Total Balance: Input = Output Molecular Species Balance:

C₂H₆: Input – Consumed = Output C₂H₄: Generated = Output

H₂: Generated = Output

Atomic (Elemental) Species Balance:

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

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

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

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.

Independent Chemical Reactions, Molecular and Atomic Species

Consider the following reactions:

A =======> 2B

B =======> C

A =======> 2C

Independent Chemical Reactions, Molecular and Atomic Species

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

Independent Chemical Reactions, Molecular and Atomic Species

Molecular Species Analysis

Total: 3 (O₂, N₂, CCl₄) Independent: 2 (O₂ or N₂, CCl₄)

Atomic Species Analysis

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

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

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

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

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

To get mole fractions of components in the product stream:

y

= n

/n

For the identified components:

y

= n

/n

y

= n

/n

y

= n

/n

y

= n

/n

y

= n

/n

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

DEGREES OF FREEDOM ANALYSIS: Molecular Balance

Unit:

Reactor

unknowns (

n

,n

,n

,n

,n

,n

)

+6

independent chemical reaction

+1

independent molecular species

–

5

other equations:

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

Method I: Molecular Species Balance

35% excess O

:





0.5 mol O

(O )

100 mol HCl

25 mol O

2 mol HCl

(O )

25mol O 1.35

33.75mol O

1mol air

n

33.75mol O

160.7 mol air

0.21mol O

160.7 mol air

mol air

Required air

1.607







_

_

















_

_











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) n = 127 mol N

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

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

Calculation for mole fractions:

Component i ni (moles) y

HCl

15.0

(15.0/239.5) = 0.063

O

12.5

(12.5/239.5) = 0.052

N

127.0

(127.0/239.5) = 0.530

Cl

42.5

(42.5/239.5) = 0.177

H

O

42.5

(42.5/239.5) = 0.177

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

DEGREES OF FREEDOM ANALYSIS: Atomic Balance

Unit:

Reactor

unknowns (

n

,n

,n

,n

,n

,n

)

+6

independent atomic specie(s)

–

3

independent nonreactive molecular specie(s)

–

1

other equations:

35% excess O

& fractional HCl conversion

–2

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

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

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

DEGREES OF FREEDOM ANALYSIS: Extent of Reaction

Unit:

Reactor

unknowns (

n

,n

,n

,n

,n

,n

)

+6

independent chemical reaction(s)

+1

independent reactive molecular species

–

4

independent nonreactive molecular species

–

1

other equations:

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

From % excess O₂ ======> n₁ From fractional conversion ======> n₂ Extent of Reaction:

HCl: n₂ = 100 – (2) Cl₂: n₅ = 0 + (1) H₂O: n₆ = 0 + (1)

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.

C

H

+ HBr

=====>

C

H

Br

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?

Example 10-2. Production of Ethyl Bromide

DEGREES OF FREEDOM OF ANALYSIS: Atomic Species

Unit:

Reactor

unknowns (

x and n

)

+2

independent atomic specie(s)

–

2

independent nonreactive molecular specie(s)

0

other equations

0

Example 10-2. Production of Ethyl Bromide

Determine the limiting reactant:





C H

Stoichiometric Ratio :

1.0

HBr

x 165mol / s

C H

x

Actual Ratio :

HBr

(1 x)(165mol / s)

1 x































_

_









_

_

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

C-Balance H-Balance

Example 10-2. Production of Ethyl Bromide C-Balance:



 



 

x mol C H

mol

2 mol C

165

n

0.310 2

n

0.517 2

s

mol

1mol C H

330x

1.654n















_

_



_

_









 



 

1 x mol HBr

mol

1mol Br

165

n

0.173 1

n

0.517 1

s

mol

1mol HBr

165(1 x)

0.69n

































Example 10-2. Production of Ethyl Bromide

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

1.0

HBr

1 0.545



















actual stoichiometric

% excess C H







100

Example 10-2. Production of Ethyl Bromide

Actual feed for C₂H₄:

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

Theoretical requirement for C₂H₄ based on stoichiometry:



1 0.545 mol HBr



1mol C H

mol

mol

165

75.08

s

mol

1mol HBr

s























 



89.93 75.08

% excess C H

100 19.8%

75.08









Example 10-2. Production of Ethyl Bromide Fractional conversion of HBr:





 











amount reacted

input

output

X

amount fed

input

165 1 0.545

108.77 0.173

X

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 – 