Notes 5 Reactive System 1

Instructor: Dr. Azeman Mustafa

SKF 1113 - Material Balances

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Balances on Reactive Process(Part A)

Azeman Mustafa, PhDKamarul ‘Asri Ibrahim, PhD


Chemical Reaction Stoichiometry

StoichiometryThe theory of proportions in which chemical species combine with one another.

Stoichiometric equation

input output2SO2 + 1O2 2 SO3

Stoichiometric coefficients, (vi)

SO2

O2

SO3





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 conversionfactor.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.



Stoichiometry equation : 2SO2 + O2 -----> 2SO3(A) (B) (C)

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 reacted

2 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





For the production of 1600 kg/hour of SO3, calculate the rate of oxygen needed:

Recall stoichiometric equation : 2SO2 + O2 2SO3

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

= 10 kmol O2/hour


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 ReactantsLimiting reactant

Reactant that its present is less than its stoichiometric proportion relative to every other reactantReactant that would be first fully consumed / reacted

Excess reactants

Reactant that if its is more than its stoichiometric proportion relative to every other reactantReactant that would have some unconsumed / unreacted after the reaction is complete

Fractional Excess = n - nstoic % Excess = 100 x n - nstoicnstoic nstoic


Limiting and Excess Reactants – Example.Limiting reactant will disappear first for a complete reaction

.Limiting reactant = SO2 excess reactant = O2

Fractional excess of O2 = 0.1 % Excess of O2 = 10%

2SO2 + O2 2 SO3

ni = 200 mol 100 mol 0 molnf = 0 mol 0 mol 200 mol

ni = 180 mol 100 mol 0 molnf = 0 10 mol 180 mol

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?




2SO2 + O2 -----> 2SO3

ninitial = 200 mol 100 mol nreacted = ? mol ? mol nfinal = ? mol ? mol 150 mol

Fractional Conversion, f = mol reactedmoles feed

Fractional Conversion

Percentage Conversion = mol reacted x 100%moles feed

Example : Calculate the percentage conversion of SO2 and O2 for the following reaction.


Extent of Reaction

Suppose we start with 100 mol of H2 , 50 mol of Br2 and 30 mol of HBr. 30 mol of H2 reacts with Br2 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 Br2 to form HBr, calculate the molar

compositions of the product?

H2 + Br2 2HBr

100 mol H250 mol Br230 mol HBr

? mol H2? mol Br2? mol HBr




Extent of Reaction

If 30 mol of H2 reacts with Br2 to form HBr, calculate the molar compositions of the product?

H2 + Br2 2HBrninitial = 100 mol 50 mol 30 molnreacted = 30 mol 30 mol 60 molnfinal = 70 mol 20 mol 90 mol


Extent of Reaction( ) ( )( ) ( )( ) ( ) ξ

ξ

ξ

2nn

nn

nn

initialHBrfinalHBr

initialBrfinalBr

initialHfinalH

22

22

+=

−=

−=ξ is the extent of reaction

(30 mol of H2 reacted)

Recall stoichiometric coefficient (vi) :-

vi = negative for reactant vi = positive for products

Example

1H2 + 1Br2 2HBrvH2 = -1 vBr2 = -1 vHBr = +2 reaction ofextent :ξ

(inert) 0β)(reactants νβ

(products) νβwhere

ξβnn

i

ii

ii

ii0i

=

−=+=

+=




Working Session II

2C2H4 + O2 -----> 2C2H4Oethylene 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?


Balances on Reactive Systems Example 4.6-1

C3H6 + NH3 + 3/2 O2 ----> C3H3N + 3H2O

Acetonitrile is produced by the reaction of propylene, ammonia and oxygen.

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




100 mol

0.100 mol C3H6/mol0.120 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 C3H3NNH2O mol H2O

The feed to the reactor contains:

(C3H6)feed = 10.0 mol(NH3)feed = 12.0 mol(O2)feed = 78.0 mol air 0.210 mol O2

mol air= 16.4 mol

(NH3/C3H6)feed = 1.2, NH3 is in excess(O2/C3H6)feed = 1.64, O2 is in excess

Balances on Reactive Systems Example 4.6-1

Reactor


(% 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




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.


Solution to Working Session III

100 mol HCl

nair mol Air

0.21 mol O2/mol0.79 mol N2/mol

(35% excess O2)

np mol

n1 mol Cl2n2 mol H2On3 mol HCln4 mol O2n5 mol N2

f = (100 - n3)/100 = 0.85

2HCl + 0.5O2 Cl2 + H2O

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

i) Draw a process flow chart

y1 mol Cl2/mol y2 mol H2O/moly3 mol HCl/moly4 mol O2/moly5mol N2/mol




IV. Total moles of air required with 35% excess O2

V. Moles of HCl reacted (% conversion = 85%)


l mo85dmol HCl feeacted mol HCl r85.0ed x mol HCl f100 =

mol2.160

mol O21.0

mol airx mol O

mol O35.1x mol HCl 2

mol O5.0xHClmol100n22

22air

=

=



II. Extent of Reaction

mol5.422100n.balHCl

vnn

3

io,ii

=

−=

+=

ξξ

ξ

( )( ) mol120n79.0n.balN

mol5.125.0n21.0n.balOmol5.420n.balH

mol5.420n.balCl

air52

air42

2

12

O

2

==

=−==+=

=+=

ξξ

ξ

5312.0nn

y,0522.0nny

0626.0nn

y,177.0nny,177.0

nny

mol5.239 n

p

55

p

44

p

33

p

22

p

11

p

====

======

=




Chemical Equilibrium

Chemical Reaction Engineering

Final equilibriumcomposition

How long the system taketo reach a specified state of equilibrium

Chemical EquilibriumThermodynamics

Chemical Reaction Kinetics


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.

Chemical Equilibrium

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.

CO(g) + H2O(g) <===> CO2(g) + H2(g)

2C2H4 + O2 ===> 2C2H4O

yco 2yH2

yco yH2O= K (T )




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 four reactive species satisfy the relation

where K (T) is the reaction equilibrium constant. At T = 1105 K, K = 1.0Suppose the feed to a reactor contains 1 mol of CO, 2 mol of H2O and no CO2 or H2, and the reaction mixture comes to equilibrium at 1105 K. Calculate the equilibrium composition and the fractional conversion of the limiting reactant.

Equilibrium Composition – Example 4.6-2

yco 2yH2

yco yH2O= K (T )



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 2yH2

yco yH2O= K (T ) = 1.0




1. Express all moles and mole fractions in terms ξenCO = 1.00 - ξenH2O = 2.00 - ξenCO2 = ξenH2 = ξe_____________ntotal = 3.00

=====>yCO = (1.00 - ξe)/3.00yH2O = (2.00 - ξe)/3.00yCO2 = ξe /3.00yH2 = ξe /3.00

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

====> (1)

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= 0.667

at equilibrium 1.00 mol fed



Multiple Reaction, Yield, Selectivity

Multiple reaction : one or more reaction

Side Reaction : undesired reaction Example :- Production of ethylene (dehydrogenation of ethane)

Main reactionC2H6 C2H4 + H2

Side ReactionsC2H6 + H2 2CH4C2H4 + C2H6 C3H6 + CH4

Design ObjectiveMaximize desired products (C2H4)Minimize undesired products (CH4, C3H6)





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)



Calculation of molar flow rates for multiple reactions

ji0v

ji

ji

vnn

ij

ij

ij

j

jij0ii

reaction inappear not does if

reaction inreactant a is if

reaction inproduct a is if

=

−

+

+= ∑

ν

ν

ξ




Multiple Reaction, Yield, Selectivity Test 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 product contains 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)


Working Session IV

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

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

Suppose 100 mol/s of equimolar amount of methane and oxygen is fed to a continuous reactor. The fractional conversion of methane 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 dioxide production.




Solution to Working Session IV

100 mol /s

0.5 mol CH4/mol0.5 mol O2/mol

n1 mol/s CH4n2 mol/s HCHOn3 mol/s H2On4 mol/s CO2n5 mol/s O2

Overall process :- CH4 + 1.5O2 0.5HCHO + 0.5CO2 + 1.5H2OFractional conversion of methane = 0.9Fractional yield of formaldehyde = 0.855

i) Draw a process flow chart



II. Mass balance analysis using the extents of reactions

)5 ..(ξ2ξ50 bal. nO

)4 .. ( ξ0 bal. nCO

)3 .. (ξ2ξ0nO bal. H)2.. ( ξ0 nHCHO bal.

)1 .. (ξξ50 bal. nCH

ξvnn

ji52

j42

ji32

i2

ji14

j

jiji,oi

−−=

+=

++=

+=

−−=

+= ∑





25

24

23

2

4 1

jii

ji4

mol/s O75.2)25.2(275.4250 , n5From eq. mol/s CO25.2 , n4From eq.

O mol/s H25.47 ) 25.2(2 75.42 , n3From eq. HO mol/s HC75.42 , n2From eq.

mol/s CH5 ) 50(9.0 – 50 , n1From eq.

mol25.2 ξmol and 75.42 ξ .....50ξ

855.0Thus,

mol50 ely ed complett is reacttaniting reaclim if the CHO formedMoles of H 855.0 HO yield Given : HC

45)50(9.0ξ ξ , 9.0n conversioGiven : CH

=−−=

=

=+==

==

===

==

==−∴=



24

2

2

25

2423

24 1

mol COmol HCHO 0.19

nn

n productioto CO relative productiony of HCHO Selectivit

/mol mol/s O0275.0 y/mol mol/s CO0225.0 yO/mol mol/s H4725.0 yHO/mol mol/s HC4275.0 y/mol mol/s CH05.0 y

ream?) output sthat of theequal to tstreamthe input owrate of e molar flm, must thtive syste(In a reac? )f the feed as that o(WHY same

mol/s 100 t stream of producr flowrateTotal mola

streamproducttheofpositionMolar com

==>

=

==

==

=




Working Session V

Methane (CH4) and oxygen (O2) react in the presence of a catalyst to form formaldehyde (HCHO). In a parallel reaction methane is oxidized to 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 of methane 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 dioxide production.

Notes 5 Reactive System 1

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