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Mass Balance on Reactive Processes

System

• Chapter 2.2

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CONTENT

• Stoichiometry• Limiting and Excess Reactant,

Fractional Conversion and Extent of Reaction

• Chemical Equilibrium• Multiple Reaction, Yield and Selectivity• Balance on Reactive System

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I. Stoichiometry• Stoichiometry – theory of proportions in which chemical

species combine with one another.• Stoichiometric equation of chemical reaction – statement

of the relative number of molecules or moles of reactants and products that participate in the reaction.

2 SO2 + O2 ---> 2 SO3• Stoichiometric ratio

– ratio of species stoichiometry coefficients in the balanced reaction equation

– can be used as a conversion factor to calculate the amount of particular reactant (or product) that was consumed (produced).

2 mol SO3 generated 2 mol SO2 consumed

2 mol SO2 consumed 1 mol O2 consumed33

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Test YourselfC4H8 + 6 O2 --------> 4 CO2 + 4 H2O

1. Is the stochiometric equation balance? Yes 2. What is stochiometric coefficient for CO2 ; 43. What is stochiometric ratio of H2O to O2 including it

unit–4 mol H2O generated/ 6 mol O2 consumed

1. How many lb-moles of O2 reacted to form 400lb-moles CO2 ; 600 lb-moles O2 reacted

1. 100 mol/min C4H8 fed into reactor and 50% is reacted. At what rate water is formed?

–200 mol/min water generated44

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II. LIMITING REACTANT & EXCESS REACTANT• The reactant that would run out if a reaction

proceeded to completion is called the limiting reactant, and the other reactants are termed excess reactants.

• A reactant is limiting if it is present in less than its stoichiometric proportion relative to every other reactant.

• If all reactants are present in stoichiometric proportion, then no reactant is limiting.

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%

stoich n

stoich n -

feed n

Excess Percentage

stoich n

stoich n -

feed n

Excess Fractional

111×=

=

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Example

C2H2 + 2H2 ------> C2H6Inlet condition: 20 kmol/h C2H2 and 50 kmol/h H2What is limiting reactant and fractional excess?

(H2:C2H2) o = 2.5 : 1(H2:C2H2) stoich = 2 : 1 H2 is excess reactant and C2H2 is limiting reactantFractional excess of H2 = (50-40)/40 = 0.25

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II. FRACTIONAL CONVERSION

• Fractional Conversion (f)

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%111fed mole

reacted moles f ,Conversion Percentage

fed mole

reacted moles f ,Conversion Fractional

×=

=

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II. EXTENT OF REACTION

• Extent of Reaction, ξ

ξ = extent of reactionni = moles of species i present in the

system after the reaction occurrednio = moles of species i in the system when

the reaction starts.vi = stoichiometry coefficient for species i

in the particular chemical reaction equation88

ξ

ξ

iioi

iioi

vnn

vnn

+=

+=or

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ExampleN2 + 3H2 ------------> 2NH3

Reactor inlet: 100 mol N2/s; 300 mol H2/s; 1 mol Ar/sIf fractional conversion of H2 0.6, calculate extent of reaction and the outlet composition.

Unreacted H2 or H2 outlet= (1-0.6) 300 = 120 mol H2/sSolve for extent of reaction : 60 mol/s

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ξ

ξξ

1

1

111

1111

1

1

1

==

−=−=

NH

Ar

N

H

n

n

n

n

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Test Yourself Page 1192 C2H4 + O2 ------->2 C2H4OThe feed to a reactors contains 100kmol C2H4 and 100 kmol O2.a) which is limiting reactant? C2H4b) Percentage of excess?

{ (100-50)/50} x 100% = 100%c) O2 out? C2H4 formed? Extent

of reaction?50kmol 100kmol C2H4 50kmol

d) if fractional conversion for limiting reactant is 50%, what is outlet composition and extent of reaction?50kmol C2H4; extent of reaction = 25 kmol;75 kmol O2 ;50 kmol C2H4O

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Test Yourself Page 1192 C2H4 + O2 ------->2 C2H4OThe feed to a reactors contains 100kmol C2H4 and 100kmol O2.

e) if reaction proceed to a point where 60kmol O2 left, what is fractional conversion for C2H4? Fractional conversion of O2 and extent of reaction?

fC2H4=0.8 fO2=0.4 extent of rxn=40 kmol

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Class DiscussionExample 4.6-1

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III. CHEMICAL EQUILIBRIUM• For a given set reactive species and

reaction condition, two fundamental question might be ask:

1. What will be the final (equilibrium) composition of the reaction mixture? – chemical engineering thermodynamics

2. How long will the system take to reach a specified state short of equilibrium? – chemical kinetics

• Irreversible reaction–. reaction proceeds only in a single direction

(from reactants to products)–. the concentration of the limiting reactant

eventually approaches zero.

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Chemical Equilibrium• Reversible reaction

– reactants form products for forward reaction and products undergo the reverse reactions to reform the reactants.

– Equilibrium point is a rate of forward reaction and reverse reaction are equal

• However the discussion to get the chemical equilibrium point is not covered in this text- learn in chemical engineering thermodynamic

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Class Discussion Example 4.6-2CO + H2O <----> CO2 + H2

nco = 1-ξnH2O = 2- ξn CO2 = ξnH2 = ξntotal = 3

K=yCO2 yH2 / y CO y H2O=1yY CO2 = ξ/3yH2 = ξ/3y CO = (1- ξ)/3y H2O = (2- ξ)/3 1515

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MULTIPLES REACTION, YIELD & SELECTIVITY

• Some of the chemical reaction has a side reaction which is formed undesired product- multiple reaction occurred.

• Effects of this side reaction might be:1. Economic loss2. Less of desired product is obtained for a

given quantity of raw materials3. Greater quantity of raw materials must be fed

to the reactor to obtain a specified product yield.

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selectivity =moles of desired product

moles of undesired product

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YIELD

• 3 definition of yield with different working definition

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Yield =Moles of desired product formed

Moles that would have been formed if there were no side reaction and the limiting

reactant had reacted completely

Yield =Moles of desired product formed

Moles of reactant fed

Yield =Moles of desired product formed

Moles of reactant consumed

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EXTENT OF REACTION FOR MULTIPLE REACTION

• Concept of extent of reaction can also be applied for multiple reaction

• only now each independent reaction has its own extent.

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ijjjiioi vnn ∑+= ξ

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Class DiscussionExample 4.6-3

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BALANCE OF REACTIVE PROCESSES

• Balance on reactive process can be solved based on three method:

1. Atomic Species Balance

1. Extent of Reaction

1. Molecular Species Balance

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Atomic Species Balance

– No. of unknowns variables– No. of independent atomic species

balance– No. of molecular balance on indep.

nonreactive species– No. of other equation relating the

variable =============================

No. of degree of freedom =============================

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Extent of Reaction

No. of unknowns variables+ No. of independent chemical reaction- No. of independent reactive species- No. of independent nonreactive species- No. of other equation relating the variable

=============================No. of degree of freedom

=============================

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Molecular Species Balance

No. of unknowns variables+ No. of independent chemical reaction- No. of independent molecular species balance- No. of other equation relating the variable

=============================No. of degree of freedom=============================

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Application of Method

C2H6 -------> C2H4 + H2

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40 kmol H2/min

n1 kmol C2H6/min

n2 kmol C2H4/min

Reactor

100 kmol C2H6/min

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Method 1: Atomic Species Balance

• All atomic balance is INPUT=OUTPUT• Degree-of-freedom analysis

2 unknowns variables (n1, n2)

- 2 independent atomic species balance (C, H)

- 0 molecular balance on indep. nonreactive species

- 0 other equation relating the variable

=============================

0 No. of degree of freedom

=============================

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Method 1: Atomic Species Balance

• Balance on atomic C (input= output)

200=2n1 + 2n2100=n1 + n2 [1]

• Balance on atomic H (input = output)100(6)=40(2) + 6n1+4n2520 = 6n1 + 4n2 [2]

Solve simultaneous equation, n1= 60 kmol C2H6/min; n2= 40

kmol C2H4/min

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100 kmol C2H6 2 knol C

=

n1 kmol C2H6 2 kmol C

+ n2(2)1 kmol C2H6 1 kmol C2H6

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Method 2: Extent of Reaction

• Degree-of-freedom analysis

2 unknowns variables (n1,n2)

+ 1 independent chemical reaction

- 3 independent reactive species (C2H6, C2H4, H2)

- 0 independent nonreactive species

- 0 other equation relating the variable

=============================

0 No. of degree of freedom

=============================2727

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Method 2: Extent of Reaction

• Write extent of reaction for each species

C2H6 : n1 = 100-ξC2H4 : n2= ξH2 : 40= ξ

Solve for n1 and n2 (ξ =40)

n1= 60 kmol C2H6/min; n2= 40 kmol C2H4/min

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Method 3: Molecular Species Balance

• Degree-of-freedom analysis

2 unknowns variables (n1, n2)

+1 independent chemical reaction

- 3 independent molecular species balance (C2H6, C2H4, H2)

- 0 other equation relating the variable

=============================

0 No. of degree of freedom

=============================

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Method 3: Molecular Species Balance

H2 balance (Gen=Output):H2 Gen= 40 kmol H2/min

C2H6 Balance (input=output + cons.):100 kmol C2H6/min = n1 kmol C2H6/min

+ 40 kmol H2 gen X (1 kmol C2H4 gen/1

kmol H2 gen)n1= 60 kmol C2H6/min

C2H4 balance (Gen.=Ouput):40 kmol H2 gen x (1 kmol C2H4 gen./ 1

kmol H2 gen) = n2n2= 40 kmol C2H4/min

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CLASS DISCUSSION

EXAMPLE 4.7-1

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Product Separation & Recycle

Overall ConversionReactant input to Process – reactant output

from ProcessReactant input to Process

Single Pass ConversionReactant input to Reactor – reactant output

from ReactorReactant input to Reactor

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75 mol B/min100 mol A/min75 mol A/minReactor Product

Separation Unit

25 mol A/min75 mol B/min

25 mol A/min

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Purging

• To prevent any inert or insoluble substance build up and accumulate in the system

• Purge stream and recycle stream before and after the purge have a same composition.

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ProductFresh FeedReactor Product

Separation Unit

Recycle Purge

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CLASS DISCUSSION

EXAMPLE 4.7-2 & 4.7-3

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Balance on Reactive Processes System: COMBUSTION REACTIONS

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COMBUSTION REACTIONS• Combustion – Rapid reaction of a fuel with

oxygen. This reaction releases tremendous quantities of energy that can be manipulated to boil water to produce steam.

• Combustion releases products such as CO, CO2 and SO2 and as chemical engineers, we are tasked to monitor and analyze the production of these noxious gases.

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COMBUSTION REACTIONS• Combustion fuels could be coal (carbon,

some hydrogen, and sulfur and various noncombustible materials), fuel oil (mostly high molecular weight hydrocarbons, some sulfur), gaseous fuel (such as natural gas, which is primarily methane) or it could be liquefied petroleum gas, which is usually propane or butane.

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COMBUSTION CHEMISTRYFrom Fuel Perspective

• Fuel contains carbonaceous material that will form either CO2 or CO, Hydrogen forming H2O and Sulfur forming SO2.

From O2 Source Perspective• For economic reason, AIR is the source of

oxygen in most combustion reactions. Dry air has the following average molar composition:

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N2 78.03%O2 20.99%Ar 0.94%CO2 0.03%H2, He, Ne, Kr, Xe 0.01%

Average MolecularWeight = 29.0

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• However, in most combustion calculations, it is acceptable to simplify this composition to 79% N2, 21% O2

• For combustion reaction, generally we have TWO type of expressions to express the mole composition of a gas, that is Composition On A Wet Basis and Composition On A Dry Basis.

• Composition On A Wet Basis is commonly used to denote mole fractions of a gas that contains water.

• Composition On A Dry Basis can also be used to denote mole fractions of the same gas that contains water but by excluding the presence of water in the calculation.

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• Example: A gas contains 33.3 mole% CO2, 33.3%

N2 and 33.3% H2O on wet basis is deemed to have a composition of 50.0 mol% CO2 and 50.0 mol% H2O on dry basis.

• It is important to have a grasp of knowledge on the correct technique to calculate these two type of compositions or to convert from dry basis to wet basis or vice versa.

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• The MAIN REASON is whenever flow rate of a product gas leaving the stack (stack gas or flue gas refers to product gas that leaves a combustion furnace) is measured, the measurement is for the total flow rate that also involved the product H2O, while on the other hand, common techniques for analyzing stack gases provide compositions on a dry basis.

• The procedure to convert Dry Basis to Wet Basis or vice versa follows exactly the same procedure outlined earlier on for converting Mass Compositions to Mole Compositions or vice versa in previous chapter.

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EXAMPLE 4.8-1:

A stack gas contains 60.0 mole% N2, 15.0% CO2, 10.0% O2 and the balance O2. Calculate the molar composition of the gas on a dry basis.

SOLUTION• Basis: 100 mol Wet Gas 60.0 mol N2

15.0 mol CO210.0 mol O285.0 mol dry gas (DG)

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60.0/85.0 = 0.706 mol N2/mol DG15.0/85.0 = 0.176 mol CO2/mol DG10.0/85.0 = 0.118 mol O2/mol DG

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Theoretical and Excess Air• Theoretical Oxygen: The moles (batch

process) or molar flow rate (continuous process) of O2 needed for complete combustion of all the fuel fed to the reactor, assuming that all the carbon in the fuel is oxidized to CO2 and all the hydrogen is oxidized to H2O.

• Theoretical Air: The quantity of air that contains the theoretical oxygen.

• Excess Air: The amount by which the air fed to the reactor exceeds the theoretical air.

• Percent Excess Air:4343

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Material Balances on Combustion Process

• The procedure for writing and solving material balances for a combustion reactor is the same as that for any other reactive system. However, be extra cautious with these:

• When you draw and label flow chart, be sure that the outlet stream (the stack gas) includes:

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Unreacted fuel unless you are told otherwise

Unreacted O2

H2O and CO2, also CO if the problem statement says so

N2 if the fuel burned With AIR and Not PURE O2

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• To calculate the O2 feed rate from a specified percent oxygen excess, first is to calculate the theoretical O2 from the fuel feed rate and the reaction stoichiometry for COMPLETE COMBUSTION, then calculate the oxygen feed rate by multiplying the theoretical O2 by (1+fractional excess of O2).

• Atomic balances are usually most convenient for use in the calculation.

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EXAMPLE 4.8-3

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