chapter 5 balances on reactive process

8/16/2019 Chapter 5 Balances on Reactive Process

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Material Balance Page 4-1© copyright PCS-FKKKSA

Balances on Reactive Process(Part A)

Azeman Mustafa, PhDKamarul ‘Asri Ibrahim, PhD


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Chemical Reaction Stoichiometry

Stoichiometry

The theory of proportions in which chemical species combine

with one another.

Stoichiometric equation

input output

2SO

2

+ 1O

2

2 SO

3

Stoichiometric coefficients, (v i)

SO2

O2SO3


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Write the stoichiometric reaction of gaseous methane with pure

oxygen to produce carbon dioxide and water vapour

Stoichiometric Ratio :

Ratio of stoichiometric coefficients can be used as a conversion

factor.

It can be used to calculate the amount or reactant (or product)

that was consumed (or produced) given another quantity of

another reactant or product that participated in reaction.


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2SO

2

+ O

2

-----> 2SO

3

(A) (B) (C)

Stoichiometry equation :

Stoichiometry ratio ; 2 mol (or kg-mole, Ib-mole ) SO3 produced

1 mol (or kg-mole, Ib-mole ) O2 reacted

2 mol (or kg-mole, Ib-mole ) SO2 reacted2 mol (or kg-mole, Ib-mole ) SO3 produced

Two reactants, A and B are in Stoichiometry proportion when:

mole A present = Stoichiometry proportion ratio obtained from the

mole B present the balanced equation


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For the production of 1600 kg/hour of SO

3

, calculate the rate of oxygen

needed:

Recall stoichiometric equation : 2SO2

+ O

2

2SO

3

1600 kg SO3 produced 1 kmol SO3 1 kmol O2 reactedhour 80 kg SO3 2 kmol SO3 produced

= 10 kmol O2/hour


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Working Session I

Consider the reaction

C4H8 + O2 CO2 + H2O

1. Write the stoichiometric equation of the above reaction?2. What is the stoichiometric coefficient of CO2?

3. What is the stoichiometric ratio of H2O to O2? (Include units)4. Does the total moles of the reactants equal that of the products?5. Does the total mass of the reactants equal that of the products?6. How many lb-moles of O2 react to form 400 lb-moles of CO2 (Use a

dimensional equation.)7. One hundred g-moles of C4H8 are fed into a reactor, and 50% reacts.

At what rate is water formed?


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Limiting and Excess Reactants

Limiting reactant

Reactant that its present is less than its stoichiometricproportion relative to every other reactant

Reactant that would be first fully consumed / reacted

Excess reactants

Reactant that if its is more than its stoichiometric proportionrelative to every other reactant

Reactant that would have some unconsumed / unreacted afterthe reaction is complete

Fractional Excess = n nstoic

Excess= 100 x n nstoic

n

stoic

n

stoic


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Limiting and Excess Reactants – Example

.Limiting reactant will disappear first for a complete reaction

2SO

2

+ O

2

2 SO

3

n

i = 200 mol 100 mol 0 moln

f

= 0 mol 0 mol 200 mol

n

i

= 180 mol 100 mol 0 mol

n

f = 0 10 mol 180 mol

Limiting reactant = SO2

excess reactant = O2

Fractional excess of O

2

= 0.1 Excess of O

2

= 10

.

From the stoichiometric equation- What can you say about the total moles of input and output?- What can you say about the total mass of input and output?


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2SO

2

+ O

2

-----> 2SO

3

n

initial

= 200 mol 100 mol

n

reacted

= ? mol ? mol

n

final

= ? mol ? mol 150 mol

Fractional Conversion, f = mol reacted

moles feed

Fractional Conversion

Percentage Conversion =mol reacted x 100%

moles feed

2

and O

2

for the following

reaction.

Example : Calculate the percentage conversion of SO


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Extent of Reaction

Suppose we start with 100 mol of H2 , 50 mol of Br 2 and 30 mol of HBr.

30 mol of H2 reacts with Br 2 to form HBr.

a) Which reactant is limiting?

b) What is the percentage excess of other reactant?

c) If 30 mol of H2 reacts with Br 2 to form HBr, calculate the molar

compositions of the product?

H2 + Br 2 2HBr

100 mol H250 mol Br 230 mol HBr

? mol H2? mol Br 2? mol HBr


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Extent of Reaction

If 30 mol of H2 reacts with Br 2 to form HBr, calculate the molar compositions of the

product?

H2 + Br 2 2HBr

n

initial

= 100 mol 50 mol 30 mol

n

reacted


n

final



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Extent of Reaction

2nn

nn

nn

initial HBr final HBr

initial Br final Br

initial H final H

22

22

ξ is the extent of reaction

(30 mol of H

2

reacted)

reactionof extent:ξ

(inert) 0β

)(reactants νβ

(products) νβwhere

ξβnn

i

ii

ii

ii0i

=

ecall stoichiometric coefficient (

v

i

) :-

v

i

= negative for reactant

v

i

= positive for products

Example

1H2 + 1Br 2 2HBr v H2 = -1 v Br2 = -1 v HBr = +2


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Working Session II

2C

2

H

4

+ O

2

-----> 2C

2

H

4

O

ethylene oxygen ethylene oxide

The oxidation of ethylene to produce ethylene oxide proceeds according to the

equation:

The feed to the reactor contains 100 kmol C2H4 and 100 kmol O2.

(1) Which reactant is limiting?

(2) What is the percentage excess of the excess reactant?

(3) If the reaction proceeds to completion, how much of the excess reactant will be left;

how much C2H4O will be formed; and what is the extent of reaction?

(4) If the reaction proceeds to a point where the fractional conversion of the limiting

reactant is 50%, how much of each reactant and product is present at the end, and

what is the extent of reaction?(5) If the reaction proceeds to a point where 60 mol O2 are left, what is the fractional

conversion of C2H4? The fractional conversion of O2? The extent of reaction?


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Balances on Reactive Systems

Example 4.6-1

C

3

H

6

+ NH

3

+ 3/2 O

2

----> C

3

H

3

N + 3H

2

O

Acetonitrile is produced by the reaction of propylene, ammonia and

oxygen.

The feed contains 10 mole% propylene, 12% ammonia and 78% air. A

fractional conversion of 30 % of the limiting reactant is achieved. Determine

which reactant is limiting, the percentage by which each of the reactants is

in excess, and the molar flow rates of all product gas constituents for a

30% conversion of the limiting reactants, taking 100 mol of feed as basis.


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Balances on Reactive SystemsExample 4.6-1

100 mol

0.100 mol C3H6/mol

0.120 mol NH3/mol0.780 mol air/mol

(0.21 mol O2/mol0.79 mol N2/mol)

NC3H6 mol C3H6NNH3 mol NH3NO2 mol O2NN2 mol N2NC3H3N mol C3H3NN

H2Omol H

2O

Reactor

The feed to the reactor contains:

(C3H6)feed = 10.0 mol (NH3/C3H6)feed = 1.2, NH3 is in excess

(NH3)feed = 12.0 mol (O2/C3H6)feed = 1.64, O2 is in excess

(O2)feed = 78.0 mol air 0.210 mol O2mol air

= 16.4 mol


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(% excess) NH3 = (NH3)feed - (NH3)stoich x 100%

(NH3)stoich

= (12 - 10)/10 x 100 = 20% excess NH3

(% excess) O2 = (O2)feed - (O2)stoich(O2)stoich

x 100

= (16.4 - 15.0)/15.0 x 100 = 9.3% excess O2

If the fractional conversion of C3H6 is 30%, then

(C3H6)out = 0.700(C3H6)feed = 7.0 mol C3H6

Extent of reaction, ξ = 3.0

NC3H6 = 12.0 - ξ = 9.0 mol NH3; NO2 = 16.4 - 1.5 ξ = 11.9 mol O2NC3H3N = ξ = 3.00 mol C3H3 ; NN2 = (N2)o = 61.6 mol N2NH2O = 3ξ = 9.0 mol H2O

Balances on Reactive SystemsExample 4.6-1


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Working Session III

In the Deacon process for the manufacture of chlorine (Cl2),

hydrochloride acid (HCl) and oxygen (O2) react to form Cl2and water (H2O). Sufficient air (21 mole % O2, 79% N2) is fed

to provide 35% excess oxygen and the fractional conversion

of HCl is 85%. Calculate the mole fractions of the product

stream components using the extent of reaction.


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Solution to Working Session III

100 mol HCl

nair mol Air

0.21 mol O2/mol

0.79 mol N2/mol

(35% excess O2)

np mol

n1 mol Cl2n2 mol H2On3 mol HCl

n4 mol O2n5 mol N2

f = (100 - n3)/100 = 0.85

2HCl + 0.5O

2

Cl

2

+ H

2

O

ii) Basis : 100 mol HCl (WHY???)

i) Draw a process flow chart

y1 mol Cl2/mol

y2 mol H2O/moy3 mol HCl/mo

y4 mol O2/mol

y5mol N2/mol


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IV.

Total moles of air required with 35 excess O

2

V. Moles of HCl reacted ( conversion = 85 )


lmo85d mol HCl fe

eacted mol HCl r 85 .0ed xmol HCl f 100 =

⎟

⎠

⎞

⎜

⎝

⎛

mol 2 .160

mol O21 .0

mol air x

mol O

mol O35 .1 x

mol HCl2

mol O5 .0 x HCl mol 100n

22

22air

=

=


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II. Extent of Reaction

mol 5 .42

2100n .bal HCl

vnn

3

i o ,i i

=

mol 120n79 .0n .bal N mol 5 .125 .0n21 .0n .bal O

mol 5 .420n .bal H

mol 5 .420n .bal Cl

air 52

air 42

2

12

O

2

=

=

=

=

5312 .0n

n y ,0522 .0

n

n y

0626 .0n

n y ,177 .0

n

n y ,177 .0

n

n y

mol 5 .239n

p

55

p

44

p

33

p

22

p

11

p

=

=

=


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Chemical Equilibrium

Chemical Reaction

Engineering

Final equilibrium

composition

How long the system take

to reach a specified

state of equilibrium

Chemical EquilibriumThermodynamics

Chemical ReactionKinetics


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Chemical Equilibrium

Reversible reaction

• The rates of forward and reverse reactions are identical when the equilibrium is reached.

• The compositions of product and reactant do no change when the reaction mixture is in

chemical equilibrium.

CO(g) + H2O(g) CO2(g) + H2(g)

yco 2 y H 2

yco y H 2O

= K (T )

Irreversible reaction:

• The reaction proceeds only in a single direction (reactants products)

• The reaction ceases and hence equilibrium composition is attained when the limiting

reactant is fully consumed.

2C2H4 + O2 ===> 2C2H4O


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Equilibrium Composition – Example 4.6-2

If the water-gas shift reaction

CO(g) + H2O(g) CO2(g) + H2(g)

Proceeds to equilibrium at a temperature T(K), the mole fractions of the fourreactive species satisfy the relation

yco2 y H 2

yco y H 2O= K (T )

where K (T) is the reaction equilibrium constant. At T = 1105 K, K = 1.0

Suppose the feed to a reactor contains 1 mol of CO, 2 mol of H2O and no CO

2or

H2, and the reaction mixture comes to equilibrium at 1105 K. Calculate the

equilibrium composition and the fractional conversion of the limiting reactant.


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Strategy:

1. Express all mole fraction in terms of a single variable ξe (extent of reaction)

2. Substitute ξe in the equilibrium relation and solve for ξe.

3. Use ξe to calculate mole fractions and any other desired quantity.

1.00 mol CO

2.00 mol H2O

COH2O

CO2H2

T = 1105 K

yco 2 y H 2

yco y H 2O = K (T )= 1.0


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1. Express all moles and mole fractions in terms ξenCO = 1.00 - ξenH2O = 2.00 - ξenCO2 = ξe

nH2 = ξe _____________ ntotal = 3.00

=====>

yCO = (1.00 - ξe)/3.00yH2O = (2.00 - ξe)/3.00

yCO2 = ξe /3.00yH2 = ξe /3.00

====> (1)

2. Substitute mole fractions from (1) in the equilibrium reaction

ξe2 = 1.00;(1.00 - ξe)(2.00 - ξe)

ξe = 0.667

3. yCO = 0.111; yH2O = 0.444; yCO2 = 0.222; yH2 = 0.222

(a) limiting reactant CO;

(b) At equilibrium, nCO = 1.00 - 0.667 = 0.333

Fractional conversion = f co = (1.00 - 0.333) mol reacted = 0.667at equilibrium 1.00 mol fed


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Multiple Reaction, Yield, Selectivity

Multiple reaction : one or more reaction

Side Reaction : undesired reaction

Example :- Production of ethylene (dehydrogenation of ethane)Main reaction

C2H6 C2H4 + H2Side Reactions

C2H6 + H2 2CH4C2H4 + C2H6 C3H6 + CH4

Design Objective Maximize desired products (C2H4)

Minimize undesired products (CH4, C3H6)


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Yield

(moles of desired product formed)

(moles of product formed, assuming no side

reactions and the limiting reactant is completely

reacted)

Selectivity

(moles of desired product formed)(moles of undesired product formed)


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Calculation of molar flow rates for multiple reactions

j i 0v

j i

j i

vnn

ij

ij

ij

j

j ij 0i i

reactioninappearnotdoesif

reactioninreactantais if

reactioninproductais if

=

∑

M lti l R ti Yi ld S l ti it


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Multiple Reaction, Yield, SelectivityTest Your Self – pg. 125

Consider the following pair of reactions

A 2B (desired)

A C (undesired)

Suppose 100 mol of A is fed to a batch reactor and the final productcontains 10 mol of A, 160 mol of B and 10 mol of C. Calculate

a) The fractional conversion of A. (f=0.9) b) The percentage yield of B. ( Y = 80%)

c) The selectivity of B relative to C. (S = 16 mol B/mol C) d) The extents of the first and second reactions. (80, 10)


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Working Session IV

Methane (CH4) and oxygen (O2) react in the presence of a catalyst toform formaldehyde (HCHO). In a parallel reaction methane is oxidizedto carbon dioxide (CO2) and water (H2O) :

CH4 + O2 HCHO + H2O

CH4 + 2O2 CO2 + 2H2O

Suppose 100 mol/s of equimolar amount of methane and oxygen is fedto a continuous reactor. The fractional conversion of methane is 0.9and the fractional yield of formaldehyde is 0.855. Calculate the molar

composition of the reactor output stream and the selectivity offormaldehyde production relative to carbon dioxide production.


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Solution to Working Session IV

100 mol /s

0.5 mol CH4/mol

0.5 mol O2/mol

n1 mol/s CH4n2 mol/s HCHO

n3 mol/s H2O

n4 mol/s CO2

n5 mol/s O2

Overall process :- CH4 + 1.5O2 0.5HCHO + 0.5CO2 + 1.5H2O

Fractional conversion of methane = 0.9Fractional yield of formaldehyde = 0.855

i) Draw a process flow chart


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II. Mass balance analysis using the extents of reactions

)5 ..( ξ 2ξ 50bal. nO

)4 .. ( ξ 0bal. nCO

)3 .. ( ξ 2ξ 0nO bal. H

)2 .. ( ξ 0 n HCHO bal.

)1 .. ( ξ ξ 50bal. nCH

ξ vnn

j i 52

j 42

j i 32

i 2

j i 14

j

j ij i,oi

∑


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25

24

23

2

41

j i i

j i 4

mol/s O75 .2 )25 .2( 275 .4250 , n5From eq.

mol/s CO25 .2 , n4From eq.

Omol/s H 25 .47 )25 .2( 275 .42 , n3From eq.

HOmol/s HC 75 .42 , n2From eq.

mol/s CH 5 )50( 9 .0 –50 , n1From eq.

mol 25 .2 ξ mol and75 .42 ξ .....50

ξ 855 .0Thus,

mol 50elyed complet t is react taniting reaclimif theCHO formed Moles of H

855 .0 HO yieldGiven : HC

45 )50( 9 .0ξ ξ ,9 .0nconversioGiven : CH

=

=

=

=

=

=

=

=

=


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24

2

2

25

2423

241

mol CO

mol HCHO 0 .19

n

n

n productioto COrelative production y of HCHO Selectivit

/mol mol/s O0275 .0 y /mol mol/s CO0225 .0 yO/mol mol/s H 4725 .0 y

HO/mol mol/s HC 4275 .0 y /mol mol/s CH 05 .0 y

ream?)output st hat of theequal to t streamthe inputowrate ofe molar fl m, must thtive syste(In a reac? ) f the feed as that o(WHY same

mol/s100t streamof producr flowrateTotal mola

stream product theof position Molar com

=>

=

=

=

=

W ki S i V


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Working Session V

Methane (CH4) and oxygen (O2) react in the presence of a catalyst toform formaldehyde (HCHO). In a parallel reaction methane is oxidizedto carbon dioxide (CO2) and water (H2O) :

CH4 + O2 HCHO + H2OCH4 + 2O2 CO2 + 2H2O

Suppose 50 mol/s of methane and 240 mol/s of air (21 mole % O2, 79%N2) are fed to a continuous reactor. The fractional conversion ofmethane is 0.9 and the fractional yield of formaldehyde is 0.855.Calculate the molar composition of the reactor output stream and the

selectivity of formaldehyde production relative to carbon dioxideproduction.

chapter 5 balances on reactive process

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