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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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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 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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Chemical Reaction Stoichiometry
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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Chemical Reaction Stoichiometry
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
= 30 mol 30 mol 60 mol
n
final
= 70 mol 20 mol 90 mol
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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 )
Solution to Working Session III
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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Solution to Working Session III
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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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 2 y H 2
yco y H 2O = K (T )= 1.0
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Equilibrium Composition – Example 4.6-2
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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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)
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Multiple Reaction, Yield, Selectivity
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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Solution to Working Session IV
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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Solution to Working Session IV
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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Solution to Working Session IV
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.