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Instructor: Dr. Azeman Mustafa
SKF 1113 - Material Balances
Material Balance Page 4-1© copyright PCS-FKKKSA
Balances on Reactive Process(Part A)
Azeman Mustafa, PhDKamarul ‘Asri Ibrahim, PhD
Material Balance Page 4-2© copyright PCS-FKKKSA
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
Instructor: Dr. Azeman Mustafa
SKF 1113 - Material Balances
Material Balance Page 4-3© copyright PCS-FKKKSA
Chemical Reaction Stoichiometry
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.
Material Balance Page 4-4© copyright PCS-FKKKSA
Chemical Reaction Stoichiometry
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
Instructor: Dr. Azeman Mustafa
SKF 1113 - Material Balances
Material Balance Page 4-5© copyright PCS-FKKKSA
Chemical Reaction Stoichiometry
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
Material Balance Page 4-6© copyright PCS-FKKKSA
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?
Instructor: Dr. Azeman Mustafa
SKF 1113 - Material Balances
Material Balance Page 4-7© copyright PCS-FKKKSA
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
Material Balance Page 4-8© copyright PCS-FKKKSA
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?
Instructor: Dr. Azeman Mustafa
SKF 1113 - Material Balances
Material Balance Page 4-9© copyright PCS-FKKKSA
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.
Material Balance Page 4-10© copyright PCS-FKKKSA
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
Instructor: Dr. Azeman Mustafa
SKF 1113 - Material Balances
Material Balance Page 4-11© copyright PCS-FKKKSA
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
Material Balance Page 4-12© copyright PCS-FKKKSA
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
=
−=+=
+=
Instructor: Dr. Azeman Mustafa
SKF 1113 - Material Balances
Material Balance Page 4-13© copyright PCS-FKKKSA
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?
Material Balance Page 4-14© copyright PCS-FKKKSA
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.
Instructor: Dr. Azeman Mustafa
SKF 1113 - Material Balances
Material Balance Page 4-15© copyright PCS-FKKKSA
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
Material Balance Page 4-16© copyright PCS-FKKKSA
(% 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
Instructor: Dr. Azeman Mustafa
SKF 1113 - Material Balances
Material Balance Page 4-17© copyright PCS-FKKKSA
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.
Material Balance Page 4-18© copyright PCS-FKKKSA
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
Instructor: Dr. Azeman Mustafa
SKF 1113 - Material Balances
Material Balance Page 4-19© copyright PCS-FKKKSA
IV. Total moles of air required with 35% excess O2
V. Moles of HCl reacted (% conversion = 85%)
Solution to Working Session III
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
=
=
Material Balance Page 4-20© copyright PCS-FKKKSA
Solution to Working Session III
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
====
======
=
Instructor: Dr. Azeman Mustafa
SKF 1113 - Material Balances
Material Balance Page 4-21© copyright PCS-FKKKSA
Chemical Equilibrium
Chemical Reaction Engineering
Final equilibriumcomposition
How long the system taketo reach a specified state of equilibrium
Chemical EquilibriumThermodynamics
Chemical Reaction Kinetics
Material Balance Page 4-22© copyright PCS-FKKKSA
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 )
Instructor: Dr. Azeman Mustafa
SKF 1113 - Material Balances
Material Balance Page 4-23© copyright PCS-FKKKSA
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 )
Material Balance Page 4-24© copyright PCS-FKKKSA
Equilibrium Composition – Example 4.6-2
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
Instructor: Dr. Azeman Mustafa
SKF 1113 - Material Balances
Material Balance Page 4-25© copyright PCS-FKKKSA
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
Equilibrium Composition – Example 4.6-2
Material Balance Page 4-26© copyright PCS-FKKKSA
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)
Instructor: Dr. Azeman Mustafa
SKF 1113 - Material Balances
Material Balance Page 4-27© copyright PCS-FKKKSA
Multiple Reaction, Yield, Selectivity
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)
Material Balance Page 4-28© copyright PCS-FKKKSA
Multiple Reaction, Yield, Selectivity
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
Instructor: Dr. Azeman Mustafa
SKF 1113 - Material Balances
Material Balance Page 4-29© copyright PCS-FKKKSA
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)
Material Balance Page 4-30© copyright PCS-FKKKSA
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.
Instructor: Dr. Azeman Mustafa
SKF 1113 - Material Balances
Material Balance Page 4-31© copyright PCS-FKKKSA
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
Material Balance Page 4-32© copyright PCS-FKKKSA
Solution to Working Session IV
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
−−=
+=
++=
+=
−−=
+= ∑
Instructor: Dr. Azeman Mustafa
SKF 1113 - Material Balances
Material Balance Page 4-33© copyright PCS-FKKKSA
Solution to Working Session IV
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
=−−=
=
=+==
==
===
==
==−∴=
Material Balance Page 4-34© copyright PCS-FKKKSA
Solution to Working Session IV
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
==>
=
==
==
=
Instructor: Dr. Azeman Mustafa
SKF 1113 - Material Balances
Material Balance Page 4-35© copyright PCS-FKKKSA
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.