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```MATERIAL BALANCES WITH CHEMICAL REACTIONS
Introduction:
Chemical reactions play a vital role in manufacturing process. For design of chemical
process equipment, the operating conditions such as pressure, temperature,
composition and flow of the streams should be known. The material balance and energy
balance calculations come to the rescue of the designer and allows him to calculate the
various flow rates and temperature of the streams. Assuming that the kinetic data of the
reaction is available, the overall material balance of the steady state condition will be
discussed here.
Material balances:
The general mathematical statement can be written as
Total mass entering the unit = Total mass of products leaving the unit
It should be noted that in chemical reactions, the total mass of the input remains
constant, but the total moles may or may not remain constant.
Example: Consider the shift reaction
CO +H2O
CO2 +H2
In this, it can be observed that two moles of reactants react with each other and
produce also two moles, thus the number of moles of the reactants entering the reaction
equals the number of the products leaving the reaction. The ammonia synthesis
reaction can be written as
N2 +3H2
2NH3
It can be observed that four moles of reactant only produce two moles of ammonia. The
total mass of the reactant entering and the product leaving the reaction are equal. For
the reaction of first type
1 mole CO
≡ 1 mole H2O
≡ 1 mol H2
≡ 1 mole CO2
Similarly for second type reaction
1 mole of N2
≡ 3 mole of H2
≡ 2 mole of NH3
The above equalities decide the stoichiometric requirements of the components. In the
processes involving chemical reaction, the total mass of various compounds entering a
reaction is equal to the total mass of various components leaving the reaction, but
entering moles of components need not be equal to the moles of components leaving.
While doing material balance calculations in such case, it is very convenient to use
basis of calculations in molar units. Generally calculations should be based on limiting
reactant and quantity of new products formed should be calculated with the help
of chemical reactions and amount of limiting reactants reacted.
For any reactant material the balance of material can be written as
Material entering= material reacted +material un reacted
For products we can write
Material leaving = material produced by the reaction.
If the material is produced by more than one reaction, then the material leaving is sum
of the materials produced by all he reactions.
Definitions:
Stoichiometry:
It is a theory of the proportions in which chemical species combine with one another.
Stoichiometric equations:
Stoichiometric equation of a chemical reaction is a statement indicating relative moles of
reactant and products that take part in the reaction. Any balanced reaction equation is a
stoichiometric equation.
For example consider the stoichiometric equation
CO + 2H2
CH3OH
According to the equation one molecule (mol or kmol) of CO reacts with two molecules
(mol or kmol) of hydrogen to produce one molecule (mol or kmol) of methanol.
Stoichiometric co-efficient:
It is the number that precedes the formula of each component involved in a chemical
reaction.
Thus in the above example, the Stoichiometric co-efficient of CO is one. The
stoichiometric co-efficient of H2 is 2 and stoichiometric co-efficient of methanol is one.
The stoichiometric requirements of components are given as
1kmol of CO ≡ 1kmol of CH3OH ≡ 2kmol of H2
Stoichiometric ratio:
Two reactants A and B are said to be present in Stoichiometric proportions if the ratio of
moles of A present to the moles of B present is equal to the Stoichiometric ratio
obtained from the balanced equation.
Consider a reaction
CO + 2H2
CH3OH
For the reactants in the above equation to be present in Stoichiometric proportion there
must be 2 moles of H2 for every mole of CO (then nH2/ n CO = 2/1 = 2) present in the
feed to the reactor.
When the reactants are fed to a reactor in Stoichiometric proportion and the reaction
goes to completion all of the reactants will be consumed. In case of above cited
reaction, for example, if 200 kmol of H2 and 100kmol of CO are initially present, the H2
and CO would disappear at the same instant. It follows that if we start with 100ml of CO
and less than 200 mol of H2 (i.e., of H2 is present in less than its Stoichiometric
proportion) H2 disappears before the CO. On the other hand, if there are more than 200
moles of H2 initially present, the CO disappears first.
Limiting reactant / component:
Limiting reactant is the reactant which is present in such proportions that its
consumption will limit the extent to which the reaction can proceed. This reactant will not
be present in excess of that required to combine with any of the other reacting
materials.
It is the reactant which is present in such proportion that it’s compete consumption by
the reaction will limit the extent to which the reaction can proceed.
It is the reactant that would disappear first if a reaction goes to completion.
Excess Reactant:
Excess reactant is the reactant which is taken more than the requirement to combine
with other reactant as per the Stoichiometry.
It is the reactant which will be present in the product even if the reaction goes to
completion. It is the one which is in excess of theoretical or stoichiometric requirement.
E.g.: Consider the reaction
C2H4 + ½O2
C2H4O
In industrial practice of producing ethylene oxide by oxidation of ethylene, oxygen/ air
fed to the reactor is always in excess of theoretically required, thus ethylene is a limiting
reactant and oxygen/air is an excess reactant.
When all reactants are present in Stoichiometric proportion then none of the reactants
involved is limiting.
Percentage excess:
Percentage excess reactant is defined as % excess quantity taken based on theoretical
requirement.
It is the amount in excess of Stoichiometric (theoretical) requirements expressed as the
percentage of Stoichiometric / theoretical requirement.
Consider a reaction
A+B
C
Where B is the excess reactant, then
Percent excess of B = mole of B supplied or fed – moles of B theoretically required x100
Moles of B theoretically required
Moles of B theoretically required are the moles of B that would correspond to the
stoichiometric proportions.
The above formula can be rearranged to calculate quantity of excess reactant actually
supplied.
Moles of B actually fed = stoichiometric requirement x (1 + fraction of excess of B)
The quantity of excess reactant must be found based on quantity of limiting reactant fed
to the reactor.
Consider for example, the reaction
SO2 + 1/2O2
SO3
And suppose that 100 moles SO2 /hr and 75 moles O2 /h are fed to the reactor. SO2 is
clearly the limiting reactant and to be in stoichiometric proportions, moles of O2 would
have to be 50 kmol / hr. The percent excess of O2 is therefore
= [(75 – 50) / 50] x 100 = 50
CONVERSION:
Conversion or fractional conversion or degree of completion is defined with respect to
limiting reactant and it gives idea regarding degree of completion of reaction. The
unreacted quantities of raw materials are easily obtained knowing the charged
quantities with the help of conversion then in turn it gives idea in case of unit process
whether recycling is to be done or not for process to be economically feasible.
Consider a chemical reaction
A+B
C
Where A = a limiting reactant
B = the excess reactant.
Then the conversion or fractional conversion of A is the ratio of amount of A reacted to
the amount of A charged or fed to a reactor. The percentage conversion of A is the
amount of A reacted expressed as the percentage of amount of A charged or fed to a
reactor.
The amount of A can be expressed in moles or weight of the amount of A is expressed
in moles then
% conversion of A =
Moles of A reacted
x 100
Moles of A charged or fed
In case of recycle operations, the term namely per pass conversion is commonly used.
The per pass conversion is defined as the quantity of limiting reactant reacted /
consumed expressed as a percentage of limiting reactant in the mixed feed.
If 100 moles of a reactant are fed and 75 moles of the reactant reacts, the fractional
conversion is 0.75 (the % conversion is 75%) and the fraction unreacted is 0.25. If 50
moles of a reactant are fed and the percentage conversion is 80% then (50) x0.8 = 40
moles have reacted and (50) (0.20) = 10 moles remain reacted.
YIELD and SELECTIVITY:
In most of the chemical processes, though the objective is to produce a desired product
by reacting raw materials, the raw material may also undergo a series of parallel
reactions (side reaction) resulting into production of undesired material which has
reverse effect on economics of the process. In such cases, in industrial practice, the
steps are taken to depress the side reaction by use of selective catalyst or by other
means like maintaining proper concentration of reactants and by varying the
temperature of reaction. In addition the inhibitors are also used.
The terms yield and selectivity are used in case of multiple reactions to give information
regarding the degree to which a desired reaction predominates over side reaction or
reactions involved.
Consider the multiple reactions, namely a series parallel reaction, a series reaction and
a parallel reaction
Series parallel reaction
Series reaction
A+B
C
C+B
D
A
Parallel reaction
C
D
A
C
A
D
Where C is a desired product, D is an undesired product, A is a limiting reactant.
Then yield of C is given as
Yield of C = moles of A reacted to produce C
x 100
Total moles of A reacted
Yield of desired product is the quantity of limiting reactant reacted to produce the
desired product. It is expressed as a percentage of quantity of limiting reactant reacted
totally to produce the desired product .
Consider parallel reaction
A
C
and A
D
Where C is a desired component, D is an undesired component
In such cases, the selectivity is given as
Selectivity of C relative to D =
moles of C (desired product) formed
Moles of D (undesired product) formed
UNIT PROCESSES:
A Unit Process is a step in manufacturing in which chemical reaction takes place like
oxidation of toluene to benzaldehyde, hydrogenation of oils to vanaspathi.
Following are some of the examples of unit processes
Oxidation, hydrogenation, nitration, hydrolysis, esterification, Neutralization, alkylation,
amination etc.
OXIDATION:
Oxidation is defined in many ways and one simple definition of oxidation is the
interaction between oxygen molecules and all the different substances they may contact
CH3CHO
+
Acetaldehyde
C6H6
1/2O2
CH3COOH
oxygen
Acetic acid
1/2O2
C4H2O3
+
Benzene
+
2CO2
+
2H2O
maleic anhydride
4HCl
+ O2
2Cl2
+
2 H2O
HYDROGENATION:
This unit process specifically refers to the chemical reaction of a substance with
molecular hydrogen in the presence of a catalyst.
CO2
+ 3H2
CH3OH
R CH = CH R' + H2
+
H2O
RCH2 – CH2 R'
NITRATION:
Those reactions where one or more nitro group (- NO2)n are introduced into the reacting
molecule.
CH3
C6H5 CH3 +
HNO3
C6H4
+
H2O
NO2
Toluene
nitric acid
mono nitro toluene
HYDROLYSIS:
It is applied to the reaction where in water effect a double decomposition with another
compound, hydrogen going to one component hydronyl to other.
C5H10Cl2
Mono chloro pentane
+
H2O
HCl
+
hydro chloride
C5H11OH
n – pentanol
KCN
H2O
+
HCN
Pot.cyanide
+
KOH
hydrogen cyanide
pot.hydroxide
NEUTRALISATION:
The reaction taking place between an acid and alkali resulting in the formation of salt
and water is called neutralization.
CH3COOH
+
Acetic acid
NaOH
sodium hydroxide
CH3COONa
+
H2O
sodium acetate
ALKYLATION:
It may be defined as introduction of alkyl radical by substitution or addition into an
organic compound.
CH3Br
+
Methyl bromide
2Na
sodium
+
C2H5Br
CH3C2H5
ethyl bromide
propane
+
2NaBr
sodium bromide
AMINATION:
The amination reaction is one which results in the formation of amines. This may be by
reduction or by ammonolysis
Amination by reduction:
It refers to those reactions which involve the synthesis of amines by reductive
methods.
Amines may be defined as derivatives of ammonia where one or more of the hydrogen
is replaced by alkyl, aryl, hydronyl, aralkyl or heterocyclic groups.
NO2
NH2
Zn
Acid
Nitro benzene
aniline
Amination by ammonolysis:
It is the process of forming amines by the action of ammonia. It also includes the
use of the primary and secondary amines as aminating agents.
H2
RCHO + NH3
R CN (H2O + H2)
RCH2 NH2
SULFONATION:
It may be defined as any chemical processes by which the sulfonic acid group (SO2OH) or the corresponding salt or sulfonyl halide group (- SO2Cl) is introduced into
an organic compound.
R CH2COOH
NH2COOH
+ SO3
+
R CH (SO2OH) COOH
H2S2O7
NH2SO3H
+
CO2
EMULSIFICATION:
All chemical reactions resulting in the formation of esters are all under this category.
CH3COOH
+
Acetic acid
CH3CH2OH
n-propanol
CH3COOC2H5
+
H2O
ethyl acetate
water
HALOGENATION (Chlorination):
It can be defined as the process where by one or more halogen atom is introduced into
an organic compound.
C6H6
+
Benzene
C6H5Cl
Cl2
chlorine
+
Cl2
Mono chloro benzene
chlorine
CH2 = CH2
Br2
Ethylene
+
bromide
C6H5Cl
+
HCl
mono chloro benzene
C6H4Cl2
+
dichlorobenzene
CH2Br - CH2Br
ethylene bromide
HCl
Oxidation reactions:
Note:
Air is used as source of oxygen
Air is a mixture of oxygen and nitrogen
Air contains 79% nitrogen and 21% oxygen by mole
Solution Procedure:
Assume a suitable basis
Write the possible reactions
Calculate the oxygen required assuming complete oxidation
Calculate oxygen supplied from the excess
Calculate nitrogen entering along with oxygen
Identify the products of oxidation for the problem
Calculate the quantities of each of the products
Calculate the other required quantities as per problem requirement
1. Ethylene oxide is produced by oxidation of ethylene. 100 kmol of ethylene are fed to
a reactor and the product is found to contain 80 kmol ethylene oxide and 10 kmol CO2.
Calculate
a) the % conversion of ethylene and b) the percent yield to ethylene oxide.
Basis: 100 kmol ethylene fed to the reactor.
Reactions:
C2H4+1/2 O2C2H4O
(1)
C2H4+3O22CO2+2H2O (2)
80 kmol ethylene oxide is produced and this is possible only by reaction 1. As per
Stoichiometry 1kmol ethylene oxide will be produced per kmol of ehylene. Therefore
ethylene reacted for reaction 1 is 80 kmoles.
10 kmol of CO2 is produced and this possible through reaction 2. As per Stoichiometry 2
moles CO2 will be produced per mol ethylene. Therefore kmol ethylene reacted by
reaction 2 is 10/2=5 kmol.
Total ethylene reacted towards equation1and 2 = 80 + 5 = 85 kmol.
Total kmol of ethylene taken = 100 kmol
Therefore % conversion = (ethylene reacted / ethylene taken) x100
= (85/100) x100= 85%
% yield of ethylene oxide= (moles ethylene reacted to ethylene oxide /Total moles of
ethylene reacted) x100
= (80/85)100=94.12%
2. In the production of chlorine gas by oxidation of hydrochloric acid gas, air is used
30% in excess of that theoretically required. Based on 4 kmol HCl, Calculate a) the
weight ratio of air to hydrochloric aid gas in feed. b) if the oxidation is 80% complete,
find the composition of the product stream on mole basis.
Basis: 4 kmol HCl
4HCl+ O2 2Cl2 + 2 H2O
O2 required = 1 kmol
O2 supplied= 1x1.30 = 1.3 kmol because 30% excess air is used
N2 supplied along with O2= 1.3 x 79/21=4.89
Air supplied= O2 supplied + N2 supplied= 1.3+4.89=6.19 kmol
Weight of air supplied= 6.19x29=179.51 kg
HCl taken= 4 kmol= 4x36.5= 146 kg
Weight ratio of air to hydrochloric acid in feed = 179.51/ 146=1.23
If oxidation is 80% complete composition of the gases in mole%
Gases leaving:
Cl2, H2O, HCl,O2, N2
HCl reacted= 4x0.8=3.2 kmol
HCl leaving unreacted= 4 - 3.2=0.8 kmol
O2 leaving = O2 supplied – O2 consumed= 1.3 – 3.2x1/4
=1.3-0.8= 0.5kmol
N2 leaving = N2 entering= 4.89 kmol
Cl2 leaving = Cl2 produced= 3.2x2/4= 1.6 kmol
H2O leaving = H2O produced= 3.2x2/4= 1.6 kmol
Composition of gases leaving:
Components
Kmoles
Mole%
Cl2
1.60
17.04
H2O
1.60
17.04
HCl
0.80
8.52
O2
0.50
5.32
N2
4.89
52.08
Total
9.39
100.00
3. Calculate the composition of gases obtained by the burning of pure FeS2 with 60%
excess air. The oxidation proceeds according to the reaction
4FeS2 +11O2 2Fe2O3 + 8 SO2 .
Assume complete conversion of the reaction
Basis: 4 kmol of FeS2
O2 required= 11 kmol
O2 supplied =11x1.6= 17.6 kmol
N2 supplied= 17.6x79/21=66.21 kmol
Gases leaving: SO2,O2,N2
SO2 leaving = SO2 produced= 8 kmol
O2 leaving= O2 supplied- O2 utilized= 17.6- 11 =6.6 moles (O2 required= O2 utilized
because of complete conversion of reaction)
N2 leaving = N2 supplied=66.21kmol
Composition of gases leaving:
Component
moles
Mole%
SO
8.00
9.90
6.60
8.17
N
66.21
81.93
Total
80.81
100.00
O
2
2
2
4. Iron pyrites FeS2 is burnt with air 100 % in excess of that required to oxidize all iron
to Fe2O3 and all sulphur to sulphur dioxide. Calculate the composition of the exit gases
in mole% and weight %, if 80% of sulphur is oxidized to sulphur dioxide and the rest to
sulphur trioxide. All iron is oxidized to Fe2O3.
Basis: 4 moles of FeS2
4FeS2 +11O2 2Fe2O3 + 8 SO2
SO2 +1/2O2 SO3
O2 required= 11 moles
O2 supplied =11x2= 22 moles
N2 supplied= 22x79/21=82.76 moles
Gases leaving: SO2, SO3,O2,N2
SO2 leaving= SO2 produced- SO2 converted to SO3
= 8 – 8 x(1-0.8) = 8 -1.6 = 6.4 moles
SO3 leaving = SO3 produced= 1.6 moles
O2 leaving = O2 supplied – O2 consumed
= 22 - 11- (1.6 x1/2 ) = 22-11- 0.8 =10.2
N2 leaving = N2 supplied= 82.76 moles
Compositions of gases leaving
Component
Moles
Mole%
Mol.wt
weight
Weight%
SO
6.40
6.34
64
409.46
12.87
1.60
1.58
80
128.00
4.02
10.20
10.10
32
326.40
10.26
N
82.76
81.97
28
2317.28
72.84
Total
100.96
99.99
3181.14
99.99
SO
O
2
3
2
2
5. Orthoxylene on oxidation gives Phthalic anhydride as per the reaction
C8H10 +3O2 C8H4O3+3H2O.
20% excess air is used. Conversion is 50% and yield of phthalic anhydride is
90%.Calculate the requirement of orthoxylene and air for 100 kmole of phthalic anhydrid
Basis: 1 mol orthoxylene
C8H10 +3O2 C8H4O3+3H2O
C8H10+21/2 O2 8CO2+5H2O
O2 required= 3 kmol
O2 supplied=3x1.2= 3.6 kmol
N2 supplied= 3.6x79/21=13.54 kmol
From Stoichiometry kmol phthalic anhydride produced assuming 100% yield= 0.5 kmol
Since yield is 90%, phthalic anhydride produced=0.5x0.9=0.45 kmol
To produce 0.45 kmol phthalic anhydride , kmol of orthoxyene used = 1 kmol
Therefore to produce 100 kmol phthalic anhydride, orthoxylene required=
(1/0.45)x100=222.22 kmol
Weight of orthoxylene required= 222.22x106=23555.55 kg.
6. In the oxidation of SO2 to SO3, the conversion is 75% by using 70% excess air.
Calculate a) composition of gases leaving the reactor in mole basis b) kg mole air fed
per kg mole SO2.
Basis: 1mol SO2
SO2+1/2O2 SO3
Mol O2 required= 0.5 mol
O2 supplied= 0.5x1.7= 0.85
N2 supplied=0.85x79/21=3.20
Air supplied= O2 supplied +N2 supplied= 0.85+3.2=4.05
Gases leaving the reactor
SO2,SO3,O2,N2
Moles SO2 reacted=1x0.75= 0.75 kmol
Moes SO2 leaving = SO2 unreacted= 1x(1-0.75)=0.25
Moles SO3 leaving = mol SO3 produced= 0.75x1/1=0.75
Mol O2 leaving= mol O2 supplied-O2 consumed = 0.85-0.75x1/2= 0.85-0.375=0.475
Mol N2 leaving= mol N2 supplied= 3.20
Composition of gases leaving
Component
Mol
Mol%
SO2
0.250
5.35
SO3
0.750
16.04
O2
0.475
10.16
N2
3.200
68.45
Total
4.675
100.00
7. In production of sulphur trioxide (SO3) 100 k mol of SO2 and 200 k mol of O2 are fed
to the reactor. The product stream is found to contain 80 k mol of SO3. Find the percent
conversion of SO2.
BASIS: 100 k mol of SO2 entering the reactor
100 k mol of SO2
200 k mol O2
product
REACTOR
SO3, SO2 ,O2
Reaction
SO2 + 1/2O2
1 k mol of SO2 ≡
SO3
1 k mol of SO3
1 k mol SO3 requires 1 k mol of SO2 to be reacted by above reaction
1 k mol of SO3 ≡
80 k mol of SO3 =
1 k mol of SO2
?
SO2 reacted = 80 x (1/1) = 80 k mol
% conversion of SO2 = k mol of SO2 reacted * 100 = 80 * 100 = 80%
K mol of SO2 fed
100
8.10 kg of PbS and 3 kgs of O2 react to yield 6 kgs of Pb and 1 kg of PbO2 according to
the reaction
PbS + O2
Pb+ SO2
PbS+2O2
PbO2+SO2
Calculate a. the amount of PbS that does not react b. %excess O2 based on the amount
of PbS that actually reacts c. amounts of SO2 formed d. % conversion of PbS to Pb.
Basis: 10 kg PbS and 3 kg O2 taken
PbS + O2
Pb+ SO2
( 1)
PbS+2O2
PbO2+SO2 (2)
The atomic weight values are Pb-207.2, S-32, O-16. Therefore weights of PbS = 239.2,
PbO2= 239.2 SO2=64, O2=32
, 2O2=64
Pb leaving= 6 kg = Pb produced according to reaction 1.
PbS reacted for reaction 1 = (239.2/ 207.2)x6= 6.93 kg
PbS reacted for reaction 2= (239.2/239.2)x1= 1 kg
Total PbS reacted= 6.93+1=7.93 kg
Amount of PbS that does not react= 10-7.93= 2.07 kg
O2 required= (32/239.2)6.93 + (64/239.2)1=0.927+0.268=1.195
O2 taken =3 kg
% excess O2= [(O2 taken –O2 required)/ O2 required] x100
= [(3-1.195) / 1.195] x 100 = 151.05%
SO2 formed by reaction 1= (64/207.2) x 6.93 = 2.14
SO2 formed by reaction 2 = (64/207.2) x 1 = 0.31
Total SO2 formed= 2.14+0.31=2.45 kg
Conversion of PbS to Pb = (PbS reacted to Pb / PbS taken) x100 = (6.93/10) x100
= 69.3%.
9. Vapor phase oxidation of ethyl alcohol using Cu catalyst at 550°C results into formation
of acetaldehyde
C2H5OH + 1/2O2
CH3CHO + H2O
The conversion is 30% and excess air used is 10%. Calculate the composition of
product stream if 100 kg of C2H5OH per hour is fed to the reactor.
BASIS: 1 hr operation
100 kg/hr of C2H5OH is fed which is equal to = 100/46 = 2.1739 k mole
O2 required for 2.1739 k mole of C2H5OH = 2.1739x1/2 =1.087
O2 supplied = 1.087 x 1.1 = 1.1957 k mole
N2 accompanying = 1.1957 x (79/21) = 4.4981 k mole
Quantity of aldehyde formed = 2.1739 x 0.3 = 0.6522 (the conversion is 30%)
Alcohol (C2H5OH) remaining unconverted = supplied – consumed
= 2.1739 – (2.1739 x 0.3) = 1.5217 k mole
Water formed during the reaction = 0.6522 k mole
Oxygen leaving= oxygen supplied- oxygen utilized
=1.1957-2.1739x0.3x0.5=1.1957-0.3261=0.8696 kmol
Nitrogen leaving= 4.4981 kmol
Component
Moles
Mole%
Alcohol
1.5217
18.57
Formaldehyde
0.6522
7.96
Water
0.6522
7.96
Oxygen
0.8696
10.61
Nitrogen
4.4981
54.90
8.1938
100.00
10. Ethylene and oxygen are fed to a tubular reactor filled with silver catalyst at 250°C.
The product ethylene oxide is recovered from gaseous effluent by absorption in water in
presence of acid at 60°C. The reaction taking place is as follows
C2H4 + 1/2O2
C2H4O
C2H4O + H2O
- (1)
CH2OH
- (2)
CH2OH
Conversion for reaction (1) is 90% which for reaction (2) it is 65%. If 100 kg of ethylene
glycol (CH2OH - CH2OH) is formed, Calculate the quantity of ethylene supplied to the
tubular reactor.
BASIS: 100 kg of CH2OH - CH2OH formed
Ethylene glycol formed= 100/62 = 1.6129 k mol
As per reaction (2) the conversion is 65%
Quantity of C2H4O required for 1.6129 k mole of (CH2OH) 2 will be
= 1.6129 x1/1x (100/65)= 2.4814 k mol
As per reaction (1) conversion is 90%
Quantity of C2H4 required for the formation of 2.4814 k mole of C2H4O will be
= 2.4814 x (100/90) = 2.757 kmol
Mass of C2H4 required = 2.757 x 28 = 77.196 kg
Mass of C2H4 required for 100 kg of ethylene glycol formed is 77.196 kg.
11. Ethylene oxide is produced by oxidation of ethylene .100 k mole of ethylene and 100
k mole of oxygen is charged to the reactor. The conversion of ethylene is 85% and
percent yield of C2H4O is 94.12.Calculate the composition of product leaving the
reactor.
BASIS: 100 k mol each of C2H4 and O2 charged
100 k mole of C2H4
100 k mole of O2
REACTOR
Product
C2H4, C2H4O, CO2, H2O
C2H4 + 1/2O2
C2H4O
C2H4 + 3O2
2CO2 + 2H2O
Percent conversion = 85
Moles of C2H4 totally reacted = 0.85 x 100 = 85 k mol
Percent yield of C2H4O = 94.12
Moles of C2H4 reacted to C2H4O by reaction (1)
= 0.9412 x85 = 80 kmol
Moles of C2H4 reacted by reaction (2) = 85 – 80 = 5 k mole
1 kmol C2H4
≡
2 kmol of CO2
Moles of CO2 produced = (2/1) x 5 = 10 k moles
From reaction 1, 1 kmol of C2H4 ≡ 0.5 k mol O2
O2 reacted by reaction 1 = (0.5/1) x 80 = 40 kmol
From reaction 2, 1 k mol C2H4 ≡ 3 kmol of O2
O2 reacted by reaction 2 = (3/1) x 5 = 15 k moles
O2 totally reacted = 40 + 15 = 55 k moles
O2 fed = O2 reacted + O2 unreacted
O2 unreacted = 100 – 55 = 45 kmoles
1 kmol of C2H4 ≡ 2 kmol of H2O
Moles of H2O produced = (2/1) x 5 = 10 kmoles
C2H4 fed = C2H4 reacted + C2H4 unreacted
C2H4 unreacted = 100 – 85 = 15 kmoles
Composition of product stream:
Component
Quantity, kmol
C2H4
15
9.38
C2H4O
85
50.00
O2
45
28.12
CO2
10
6.25
H2O
10
6.25
Total
160
Mole%
100.00
12. Oxidation of ethylene to produce ethylene oxide is given by reaction
C2H4 + 1/2 O2
C2H4O
If air is used 20%b in excess of that theoretically required. Calculate the quantity of air
supplied based on 100 k mol of ethylene fed to the reactor.
BASIS: 100 kmol of ethylene fed to the reactor.
Reaction:
C2H4 + 1/2 O2
C2H4O
From reaction we have
1 kmol of C2H4 ≡
0.5 kmol of O2
Stoichiometric or theoretical requirement of oxygen for
1 kmol of C2H4 ≡
0.5 kmol of O2
Theoretical requirement of oxygen for
100 kmol of C2H4 = (0.5/1) x 100 = 50 k moles
Oxygen supplied 20% in excess of theoretical requirements,
Oxygen in supplied air = theoretical requirement of oxygen
= 50 x 1.2= 60 kmol
Moles of air supplied = (100/21)x 60 = 285.71 kmoles
Amount of air supplied = 285.71 x 29 = 8285.59 kg
13.In manufacture of acetic acid by oxidation of acetaldehyde, acetaldehyde and
oxygen were fed to reactor and the product leaving the reactor contains 14.81%
acetaldehyde, 59.26% acetic acid and rest oxygen (mole basis). Find the % conversion
of acetaldehyde.
BASIS: 100 k mol of products leaving
Acetaldehyde
Acetic acid- 59.26%
REACTOR
Oxygen
acetaldehyde-14.81%
Oxygen-25.93
CH3CHO + 1/2O2
Reaction:
CH3COOH
Acetic acid formed= 59.26 kmoles
Acetaldehyde reacted= 59.26x1/1= 59.26 kmol
Acetaldehyde unreacted= 14.81 kmol
Total acetaldehyde entering= reacted +unreacted= 59.26+14.81= 74.07 kmol
% conversion of acetaldehyde= (59.26/ 74.07) x100= 80%
14. In production of sulphur trioxide, 100 k mol of SO2 and 100 k mol of O2 are fed to the
reactor. If the % conversion of SO2 is 80, calculate the composition of the product
stream on mole basis
BASIS: 100 k mol of SO2 and 100 k mole of O2 charged to the reactor
100 k mol of SO2
100 Kmol of O2
REACTOR
REACTOR
Product SO3, SO2, O2
Reaction
SO2 + 1/2O2
SO3
Percent conversion = 80%
Moles of SO2 reacted = 100x0.8= 80 kmol
SO2 unreacted = 100 – 80 = 20 k mol
From reaction
1 k mole of SO2 ≡ 1 k mole of SO3
SO3 produced = (1/1) x 80 = 80 k mol
From reaction
1 k mol of SO2 ≡ ½ k mol of O2
O2 reacted = (0.5 / 1) 80 = 40 k molx
O2 unreacted = 100 – 40 = 60 k mol
Composition of product stream
Component
k mol
mole%
SO2
20
12.5
SO3
80
50.0
O2
60
37.5
Total
160
100.0
15. In manufacture of SO3,the feed to the reactor consists of 50 k moles of SO2 and 150
k moles of air. Calculate % excess of air used.
BASIS: 50 k moles of SO2 fed to the reactor.
Reaction:
SO2 + 1/2 O2
SO3
Air used = 150 k mole
Oxygen in supplied air = 150 x 0.21 = 31.5 kmol
From reaction 1 kmol of SO2 ≡
0.5 kmol of O2
Theoretical (stoichiometric) requirement of O2 for 50 kmol SO2
= (0.5/1) x 50 = 25 kmol
% excess O2 used = oxygen supplied – oxygen theoretically required
Oxygen theoretically required
= [(31.5 – 25) / 25] x100 = 26
16.In the manufacture of benzaldehyde a mixture containing dry air and toluene is fed to
the reactor (converter) at a temperature of 448°K (175°C) and 100 kpa of pressure. The
mixture passes through the catalyst bed in the converter. The product gas stream
leaves the converter at 468°K (195°C). The reaction takes place as follows
C6H5CH3 (g) + O2 (g)
C6H5CHO (g) + H2O (g)
The dry air is supplied at 100% excess so as to maintain the high yield of
benzaldehyde. The side reaction taking place is
C6H5CH3 (g) + 9O2 (g)
7CO2 (g) + 4H2O (g)
At the above mentioned conditions the overall conversion is 13% based on toluene.
Approximately 0.5 of the toluene charged burns to CO2 and H2O. calculate the
composition of gas stream leaving the converter.
BASIS: 100 kmol of toluene charged
C6H5CH3 (g) + O2 (g)
C6H5CHO (g) + H2O (g)
C6H5CH3 (g) + 9O2 (g)
7CO2 (g) + 4H2O (g)
Overall conversion = 13%
Toluene totally reacted = 0.13 x 100 =13 kmol
Toluene unreacted = 100 – 13 = 87 kmol
Toluene
Air
C6H5CHO, C6H5CH3, CO2, H2,N2
Toluene reacted by reaction 2 = (0.5/100) x 100 = 0.5 kmol
Toluene reacted by reaction 1 = 13 – 0.5 = 12.5 kmol
From reaction
1 kmol of toluene ≡ 1 kmol of benzaldehyde
≡ 1 kmol of water
Benzaldehyde produced = (1/1) x 12.5 = 12.5 kmol
Water produced by reaction 1 = (1/1) x 12.5 = 12.5 kmol
Carbon dioxide produced by reaction 2 = (7/1) x 0.5 = 3.5 kmol
Water produced by reaction 2 = (4/1) x 0.5 = 2 kmol
Total water produced = 12.5 + 2 = 14.5 kmol
Theoretical oxygen required = (1/1)x 100 = 100 kmol
Oxygen in supplied air = 100 x2 = 200 kmol (100% excess )
Nitrogen in supplied air = 200 x 79 /21 = 752.4 kmol
Oxygen reacted
= oxygen reacted by reaction 1 + oxygen reacted by reaction 2
= (1/1) x 12.5 + (9/1)x 0.5 = 17 kmol
Oxygen unreacted = 200 – 17 = 193 kmol
Component
kmol
mole%
C6H5CH3
87.0
8.18
C6H5CHO
72.5
1.18
N2
752.4
70.79
O2
193.0
18.16
CO2
3.5
0.33
H2O
14.5
1.36
1062.9
100.00
Hydrogenation
17.A mixture of pure carbon dioxide and hydrogen is passed over a nickel catalyst. The
temperature of the catalyst bed is 588K and the reactor pressure is 2.02 mPag. The gas
mixture leaving the reactor is analyzed to contain 57.1% CO2, 41.1% H2, 1.68% CH4
and 0.12% CO by volume on dry basis. The reactions taking place in the reactor are
CO2+4H2 CH4+2H2O
CO2+H2CO+H2O
Find a) the conversion of CO2 per pass b) the yield of CH4 in terms of CO2 reacted and
c) the composition of feed.
Basis: 100 kmol of dry gases leaving the reactor
CO+4H2 CH4+2H2O (1)
CO2+H2CO+H2O (2)
Methane leaving = 1.68 kmol
Therefore CO2 reacted by equation 1= 1/1x1.68= 1.68 kmol
Co leaving =1.2 kmol
Therefore CO2 reacted by equation 2= 1/1x0.12= 0.12 kmol
Total CO2 reacted = 1.68+0.12= 1.8 kmol
CO2 supplied= CO2 reacted + CO2 un reacted = 1.8 + 57.1= 58.9 kmol
% conversion of CO2 per pass= (1.8/58.9)x100= 3.056 %
% yield of CH4 = (kmol CO2 reaced to CH4/ kmol CO2 totally reacted) x100
= (1.68/1.8)x100=93.33
Hydrogen reacted by equation 1= 4/1x1.68= 6.72 kmol
H2 reacted by reaction 2= 1/1x0.12= 0.12
Total hydrogen reacted= 6.72 + 0.12 = 6.84
Hydrogen supplied= hydrogen reacted + hydrogen un reacted = 6.84+41.1= 47.94 kmol
Analysis of feed
Component
Mol
Mol%
H2
47.94
44.87
CO2
58.90
55.13
Total
106.84
100.00
18.Gaseous benzene C6H6 reacts with hydrogen in presence of Ni catalyst as per the
reaction
C6H6 +
Benzene
3H2
C6H12
cyclohexane
30% excess hydrogen is used above that required by the above reaction. Conversion is
50% and yield is 90%. Calculate the requirement of benzene and hydrogen gas for 100
moles of cyclohexane.
Basis: 1 mol benzene
H2 required= 3 mol
Hydrogen supplied= 3x1.3= 3.9 mol
Cyclohexane produced for a 50% conversion and 100 % yield= (1/1) x 1 x 0.5 = 0.5 mol
Since yield is 90 %, cyclohexane produced = 0.5 x 0.9 = 0.45 mol
0.45 mol cyclohexane is produce by 1 mol hexane
Therefore mole hexane required t produce 100 mole cyclohexane= (1/0.45) x100
= 222.22 mol
Hydrogen supplied to produce 100 mol cyclohexane = (3.9/1)222.22 = 866.658 mol
19.An organic ester of formula C19H36O2 is to be hydrogenated at a rate of 100 kg / hr to
yield C19H38O2. The hydrogen required for the plant, which run continuously, is available
as 50 lit cylinders in which the gas is contained at 70 bar and 300 K. How many cylinder
should the company order per week?
Basis: 100 kg / hr of C19H36O2
C19H36O2 + H2 C19 H38O2
Mol weight of C19H36O2 = 228 + 36 + 32 = 296
Mol weight of C19H38O2 = 228 + 38 + 32 = 298
Mol C19H36O2 charged = 100/ 296 kmol
Hydrogen required per hour= (1/1)100/296= 100/296 kmol
Volume of hydrogen required= (100/296) 22.414 x 1000 x 300 /273 x1.013 / 70
= 120.42 lit/hr
Volume hydrogen required per week= 120.42 x 24 x 7= 20230.56 lit
Volume of 1 cylinder= 50 lit
Therefore number of cylinders required= 20230.56 / 50= 404.61
20. Methanol is produced by the reaction of carbon monoxide with hydrogen.
CO+2H2CH3OH
(1)
CO+3H2CH4+H2O
.(2)
The side reaction is
The conversion per pass is 12.5%.Of this 87.5% is reached via equation (1) and 12.5%
reach via equation (2). The feed mixture entering the reactor contains 32 mol% CO and
68 mol% hydrogen. The stream leaving the reactor passes through a condenser and a
separator. The carbon monoxide and hydrogen leaving this unit are recycled. The
methane leaves as a gas and the liquid mixture of methanol and water passes to a
distillation column for the concentration of methanol.Compute a) the analysis of hot
gases leaving the reactor by mol% and weight % b)the methanol content weight% of
the liquid leaving the condenser.
Basis: 100 mol feed to the reactor
Conversion per pass= 12.5%
Gase leaving: CO,
Therefore CO consumed = 32 x 0.125 = 4 kmol
CO consumed by reaction (1) is = 4 x 0.875 = 3.5 kmol
CO consumed by reaction (2) is = 4 x 0.125 = 0.5 kmol
Total CO reacted = 3.5 + 0.5 = 4 k mol
CO leaving= CO unreacted = CO entering – CO reacted= 32- 4 = 28 kmol
H2 consumed in reaction (1) = (2/1) 3.5 = 7 kmol
H2 consumed in reaction (2) = (3/1)0.5 = 1.5 kmol
Total H2 consumed = 7+1.5 = 8.5 kmol
H2 leaving = H2 unreacted = H2 supplied – H2 consumed = 68 - 8.5 = 59.5 kmol
Methanol leaving the reactor= methanol formed = (1/1) x 3.5 = 3.5 kmol
Water leaving the reactor = Water formed = (1/1) x 0.5 = 0.5 kmol
Methane leaving= methane formed = (1/1) x 0.5 = 0.5 kmol
Analysis of gases leaving:
Component
mol
Mol%
Mol
weight
Weight%
weight
CO
28
30.43
28
784
75.97
H2
59.5
64.68
2
119
11.53
CH3OH
3.5
3.81
32
112
10.85
H2O
0.5
0.54
18
9
0.87
CH4
0.5
0.54
16
8
0.78
Total
92
100.00
1032
100.00
Methanol content of the liquid =
x100 = 92.56% by weight
21. The carbon monoxide is reacted with hydrogen to produce methanol. Calculate from
the reaction, the stoichiometric ratio of H2 to CO. k mol of CH3OH produced per k mol
CO reacted. The weight ratios of CO to H2 if both are fed to the reactor in stoichiometric
proportions. The quantity of CO required to produce 1000 kg of CH3OH.
reaction
CO + 2H2
CH3OH
Stoichiometric co-efficient of CO = a =1
Stoichiometric co-efficient of H2 = b = 2
Stoichiometric ratio of H2 to CO = b/a = 2/1 = 2
From the reaction 1 k mol of CO ≡
1 k mol of CH3OH
CO reacted to produce 1 k mol of CH3OH = 1/1 * 1 = 1 k mol.
CO = feed to the reactor = 100 kg
CO = feed to the reactor = 1 * 28 = 28 kg
H2 fed to the reactor in stoichiometric proportion with CO = 2 k mol.
H2 fed to the reactor = 2 * 2 = 4
Weight ratio of CO to H2 fed = 28/4 = 70
For reaction 28 kg CO ≡
32 kg CH3OH
Amount of CO required to produce 1000 kg CH3OH = 28/32 * 100 = 875 kg.
22. Ammonia is produced by following reaction
N2 +
H2
2NH3
Calculate the molal flow rate of hydrogen corresponding to nitrogen feed rate of 25
k mol/hr if they are at the stochiometric proportions. The kg of ammonia produced per
hour if percent conversion is 25 and nitrogen feed rate is 25 k mol/hr.
BASIS: 25 kmol / hr of N2 fed to the reactor.
Reaction:
N2 +
H2
2NH3
Stoichiometric proportion of N2 to H2 from the reaction is 1:3
1 k mol of N2
≡
3 k mol of H2
Molal flow rate of hydrogen in stoichiometric proportion to nitrogen flow rate
= 3/1x25 = 75 k mol/hr
No of moles of N2 reacted = 25x0.25 = 6.25 k mol/hr
From reaction we have 1 k mol of N2 ≡ 2 k mol of NH3
2 k mol of NH3 is produced per 1 k mol of N2 reacted
NH3 produced = (2/1) x 6.25 = 12.50 k mol/hr
Amount of NH3 produced = 12.5x molecular weight of NH3 = 12.5x 17 = 212.5 kg/hr.
23.Gaseous benzene (C6H6) reacts with hydrogen in presence of Ni catalyst as per the
reaction
C6H6 (g) + 3H2 (g)
C6H12 (g)
30% excess hydrogen is used above that required by the above reaction. Conversion is
50% and yield is 90%. Calculate the requirement of benzene and hydrogen gas for 100
kmol of cyclo hexane.
BASIS: 100 kmol of cyclo hexane
Reaction:
C6H6 (g) + 3H2 (g)
C6H12 (g)
Cyclo hexane produced = 100 kmol
1 kmol of cyclo hexane needs 1 kmol of benzene
Benzene reacted to cyclo hexane = 1/ 1x100 =100kmol
% yield of cyclo hexane = kmol of benzene reacted to produce cyclo hexane x100
kmol of benzene totally reacted
90
=
100 x 100
Kmol of benzene totally reacted
Kmoles of benzene totally reacted = 100 x 100 = 111.11 kmol
90
% conversion of benzene = 50
% conversion of benzene = kmol of benzene totally reacted x 100
Kmol of benzene totally fed
50 =
111.11
x 100
Kmol of benzene charged
Kmol of benzene charged = 100 x 111.11 = 222.22 kmol
50
From reaction
1 kmol of benzene ≡ 3 kmol of hydrogen
Theoretical requirement of hydrogen gas = 3/1x 222.22 = 666.66 kmol
Requirement of benzene = 222.22 kmol
Requirement of hydrogen = 666.66 kmol
Nitration
24.Benzene reacts with nitric acid to produce nitrobenzene and water
C6H6+HNO3 C6H5NO2 +H2O
Nitrobenzene undergo further nitration to form dinitrobenzene
C6H5NO2 +HNO3C6H4 (NO2)2+H2O
The percent conversion of benzene is 90 and acid used is 65% excess over theoretical
required by reaction 1. If the mole ratio of benzene to dinitrobenzene in product stream
is 17:1, calculate the quantities of benzene and nitric acid required for production of
2000 kg/h of nitrobenzene.
Basis: 1kmol / h benzene fed to the reactor
C6H6+HNO3 C6H5NO2 +H2O
C6H5NO2 +HNO3C6H4 (NO2)2+H2O
Benzene converted= 0.9 kmol because conversion is 90%
Kmol nitrobenzene produced = 0.9 x 17/18= 0.85
Kmol dinitrobenzene produced= 0.9 x 1/18= 0.05
(Ratio of nitrobenzene to dinitrobenzene in product is 17:1)
Nitric acid required for the reaction= 1kmol
Nitric acid fed = 1 x 1.65 = 1.65 kmol
Weight of Nitric acid fed = 1.65 x 63 = 103.95 kg/h
Weight of benzene fed to the reactor = 1x78= 78 kg/h
Nitrobenzene produce /h = 0.85 x 123 = 104.55 kg/h
Benzene required for the production of 2000 kg/h of nitrobenzene
= (78/104.55) x 2000 = 1492.11kg/h
Nitric acid required for the production of 2000 kg/h of nitrobenzene
= (103.95/104.55) x 2000 = 1988.52 kg/h
25. Benzene reacts with nitric acid to produce nitrobenzene and water. Some of the
nitrobenzene may further undergo nitration to form dinitrobenzene. If the product
streams contain 10 mol nitrobenzene,2 mol dinitrobenzene and 4mol benzene, calculate
the percent conversion of benzene, yield of nitrobenzene and selectivity to
nitrobenzene.
Basis: 10 mol nitrobenzene, 2 mol dinitrobenzene and 4 mol benzene in the
product stream
C6H6+HNO3 C6H5NO2 +H2O (1)
C6H5NO2 +HNO3C6H4 (NO2)2+H2O (2)
Benzene reacted to give nitrobenzene
= (1/1) x10= 10 mol
Benzene reacted to give dinitrobenzene
= (1/1) x 2 = 2 mol
Total benzene reacted= 10+2 =12 mol
Benzene unreacted = 4 mol
Benzene charged = benzene reacted + benzene un reacted = 12+4=16 mol
Conversion benzene = (benzene reacted/ benzene charged) x100 = (12/16) x100=75%
Yield of nitrobenzene= (benzene reacted to nitrobenzene/ total benzene reacted) x 100
= (10/12) x 100= 83.33%
Selectivity to nitrobenzene = (mol nitrobenzene formed /mol dinitrobenzene formed)
=5/2
26. Picric acid is obtained by nitrating phenol according to the reaction
C6H5OH+ 3HNO3 C6H2OH (NO2)3+3H2O.
In a particular operation, 100 kg of phenol was reacted with 250 kg of HNO3 to produce
200 kg of picric acid. Which is the limiting reactant? What is the conversion of the
reaction? Find the composition of the product stream by weight% and mol %.
Basis: 100 Phenol
C6H5OH+ 3HNO3 C6H2OH (NO2)3+3H2O
94
189
229
54
Mol weight of phenol= 94 , mol weight of nitric acid=63 ,molecular weight of picric
acid=229, molecular weight of water=18.
Weight of HNO3 required to react with 100 kg phenol= (189 /94)100= 201.06 kg
HNO3 used = 250 kg. Since HNO3 used is more than the required, it is an excess
reactant. Therefore phenol is the limiting reactant.
200 kg of picric acid is produced.
Therefore weight of phenol reacted to give 200 kg picric acid as per stoichometry
= (94/229) x 200 = 82.1 kg.
Conversion of the reaction= (limiting reactant reacted / limiting reactant taken) x100
= (82.1/100)100= 82.1 %
Components leaving the reactor: Picric acid, phenol, nitric acid, water
Picric acid leaving = picric acid produced = 200 kg
Water leaving = water produced = (54/229)x200 = 47.16 kg
Phenol leaving = phenol unreacted = phenol taken –phenol reacted= 100- 82.1= 17.9 kg
Nitric acid leaving = nitric acid unreacted = nitric acid taken - nitric acid reacted
= 250 - (189/94) x 82.1= 250-165.07= 84.93 kg
Composition of the product stream:
Component
Weight
Weight %
Mol weight
mol
Mol%
Picric acid
200.00
57.14
229
0.873
17.70
Phenol
17.90
5.11
94
0.190
3.85
Water
47.16
13.47
18
2.620
53.13
Nitric acid
84.93
24.27
63
1.248
25.31
Total
349.99
99.99
4.931
99.99
27. Toluene reacts with nitric acid to form ortho nitro toluene and water. After carrying
out the nitration operation which can be represented by the equation
C6H5CH3 +HNO3 C6H4CH3NO2+H2O
The product mixture was analyzed and found to contain 75% ortho nitro toluene 10%
water, 15% nitric acid by weight. Comment about the conversion of the reaction. What is
the % excess of the reactant used? What is the quantity of toluene used for the
reaction?
Basis: 100 kg of products
The product does not contain toluene and therefore the conversion is 100% and also
toluene is the limiting reactant.
Amount of ortho nitro toluene produced= 75 kg
HNO3 reacted to produce 75 kg o nitro toluene= (63/137) x 75 = 34.49 kg
HNO3 unreacted= 15 kg
HNO3 fed= HNO3 reacted+HNO3 unreacted= 34.49+15= 49.49 kg
% excess HNO3= (15/ 34.49) x100= 43.49%
Toluene fed for the reaction= (92/ 137) x75= 50.36 kg.
28.Benzoic acid is nitrated to obtain nitro benzoic acid according to the reaction
C6H5COOH+HNO3 C6H4 (COOH) NO2 + H2O.
122
63
167
18
Calculate the weight of 90% HNO3 required for reacting with 250 kg benzoic acid. If
50% excess acid is used and if the conversion is 85% find the composition of the
products leaving the reactor.
Basis: 250 kg benzoic acid
90% HNO3 required to react with 250 kg benzoic acid = (63/122) x250 x 100/90 =143.44
Acid used = 143.44 x 1.5 = 215.16 kg
100% acid = 215.16 x 0.9 = 193.64 kg
Water entering with acid = 215.16 x 0.1= 21.52kg
Conversion is 85%
Benzoic acid reacted = 250 x 0.85 = 212.5 kg
Benzoic acid unreacted = 250 x 0.15 = 37.5 kg
HNO3 reacted
= (63/122) x 212.5 = 109.73 kg
HNO3 unreacted = 193.64-109.73 = 83.91 kg
Water leaving = 21.52 kg
Nitrobenzoic acid produced= (267/122) x 212.5= 465.06 kg
Water produced= (18/122) x 212.5 = 31.35 kg
Water leaving= water produced +water in acid = 31.35+21.52 = 52.87 kg
Composition of the products leaving the reactor:
Component
Weight
Weight%
Benzoic acid
37.50
5.87
Nitric acid
83.91
13.12
Nitro benzoic acid
465.06
72.74
Water
52.87
8.27
Total
639.34
100.00
29. Chlorobenzene is nitrated using a mixture of nitric acid and sulphuric acid. During
the pilot plant studies, a charge of 100 kg chlorobenzene 106.5 kg 65.5% (by weight)
nitric acid and 108.0 kg 93.6% (by weight) sulphuric acid. After two hours of operation,
the final mixture was analysed.It was found that the final product contained 2%
unreacted chlorobenzene. Also the product distribution was found to be 66% p – nitro
chloro benzene and 34% O – nitro chloro benzene. Calculate a) The analysis of charge
b)The % conversion of chloro benzene and c)The composition of product mixture.
BASIS: 100 kg Chloro benzene
The charge consists of chloro benzene and mixed acid
HNO3 charged = 106.5 x 0.655 = 69.758 kg
H2SO4 charged = 108.8x 0.936 = 101.088 kg
Water in the charge = 106.5 x 0.345 + 108.0 x 0.064 = 43.655 kg
The analysis of the reactant can be tabulated as shown
Component
Chloro benzene
mol weight
Qty Charged
Weight.%
112.5
100.00
31.80
Nitric acid
63.0
69.76
22.18
Sulphuric acid
98.0
101.09
32.14
Water
18.0
43.66
13.88
314.51
100 .00
The reaction taking place are
Cl
Cl
NO2
+
H2O
+ HNO3
Chloro benzene nitric acid
O – nitro chloro benzene
water
Cl
Cl
+
H2O
+ HNO3
NO2
Chloro benzene nitric acid
P – nitro chloro benzene
water
As given in the problem, the yield of P – NCB is 66%. Since the total charge (weight)
remains constant.
Unreacted C.B in the product = 314.5 x 0.02 = 6.29 kg.
Amount of C.B that has reacted = 100 – 6.29 = 93.71 kg.
Conversion of C.B = 93.71 x 100 = 93.71%
100
Sulphuric acid remains unreacted
From reaction I it is clear that 1 kmol of C.B
≡ 1 kg mol HNO3
≡1 kg mol NCB
≡1 kg mol H2O
112.5 kg of CB ≡ 63 kg of HNO3
Thus 63 kg of HNO3 will be consumed for converting 112.5 kg C.B into NCB
Total HNO3 consumed = (63/112.5) x 93.71
=52.478 kg
Unreacted HNO3 = 69.758-52.478 =17.28 kg
Mol wt of NCB=157.5
Total NCB produced = (157.5/112.5) x 93.71=131.194 kg
p-NCB = 0.66x131.194=86.588 KG
o-NCB = 0.34x131.194 =44.606 kg
Water produced = (15/112.5)x 93.71 =14.994 kg
Total water produced =43.665+14.994=58.649 kg
Final product analysis:
Component
Weight
Weight%
CB
6.27
2.00
P NCB
86.59
27.33
O NCB
44.61
14.18
HNO3
17.28
5.49
H2SO4
101.09
32.15
H2O
58.65
18.65
TOTAL
314.48
100.00
30. Acetylene is produced according to the following reaction:
CaC2+2H2OC2H2+ Ca(OH)2
If 100 lit of gas is burnt per hour at 298 K (25°C) and 98.68 kpa pressure, calculate the
amount of CaC2 in kg which must be used in the acetylene lamp, at the above
temperature and pressure, to get 15 hours service of the lamp. CaC2 reacts to give
acetylene gas to burn in the lam.
Basis: 15 h of operation.
Gas burning rate of acetylene= 100 lit/ hr
Acetylene gas burned= 100x15= 1500 lit
C2H2 generated=C2H2 burned=1500 lit
31.Ethyl alcohol is produced industrially by fermentation of molasses. The molasses
sample contains 45% by weight fermentable sugar in the form of sucrose. The reactions
taking lace in the fermenter are.
Sucrose
(Monosaccharide)
(d-Glucose) (d-Fructose)
Alcohol
Calculate the theoretical production of rectified spirit of density (0.785 kg/l) in liters per
ton of molasses.
Basis: 1000 kg of molasses.
Sucrose
Sucrose in molasses = 0.45 X 1000 = 450 kg
From the reactions, we have
(d-Glucose)
(d-Fructose)
32.100 kmol of ethyl acetate is charged and hydrolyzed to produce ethyl alcohol and
acetic acid. The product stream is fond to contain 45 kmol ethyl alcohol. What is the
conversion? What will be the amount of acetic acid produced?
Basis: 100 kmol ethyl acetate
CH3COOC2H5 + H2O C2H5OH +CH3COOH
88
18
46
60
Kmol ethyl alcohol produced= 45 kmol
As per Stoichiometry 1kmol ethyl acetate gives 1 kmol ethyl alcohol
Therefore kmol ethyl acetate reacted = (1/1) x 45 = 45 kmol
Conversion of ethyl acetate= (ethyl acetate reacted / ethyl acetate taken) x100
= (45/100) x 100=45%
Acetic acid produced / 45 kmol ethyl acetate = (1/1) x 45 = 45 kmol
33. Acetyl chloride is hydrolyzed for the production of acetic acid and hydrochloric acid.
The product stream is found to contain 38.46 mol% acetic acid, 38.46% hydrochloric
acid and the remaining water. Find the ratio of acetyl chloride to water used for reaction.
Basis: 100 kmol products
CH3COCl + H2O CH3COOH + HCl
kmol acetic acid produced = 38.46 kmol
As epr Stoichiometry 1 kmol acetic acid will be produced from 1 kmol acetyl chloride
Therefore actyl chloride reacted = (1/1) x 38.46 = 38.46 kmol
Water reacted= (1/1) x 38.46 = 38.46 kmol
Water present in product = 100- 38.46-38.46 = 23.08 kmol = water unreacted
Water fed = water reacted + water unreacted = 38.46+23.08 = 61.54 kmol
Ratio of acetyl chloride to water in feed = 38.46 / 61.54 = 0.625
Neutralization
34.100 kg hydrochloric acid is neutralized by 150 kg of sodium hydroxide. What is the
quantity of sodium chloride that will be produced? Find the composition of neutralized
mixture by eight and mol.
Basis: 100 kg HCl and 150 kg NaOH
HCl+ NaOH NaCl + H2O
36.5
40
58.5
18
NaOH required to react with 100 kg HCl= (40/36.5) x 100 = 109.59 kg
NaOH taken= 150 kg
Since NaOH is taken more than the required, it is excess reactant
As per Stoichiometry 58.5 kg sodium chloride is produced for 36.5 kg HCl
Therefore sodium chloride produced= (58.5/36.5) x 100 =160.27 kg
Components leaving
NaCl,H2O,NaOH
Water leaving=Water produced = (18/36.5) x 100 = 49.59 kg
NaOH leaving = NaOH entering- NaOH reacted = 150-109.59 = 40.41 kg
Sodium chloride leaving = sodium chloride produce = 160.27 kg
Composition of neutralized mixture
Component
weight
Weight %
Mol wt
mol
Mol%
NaCl
160.27
64.11
58.5
2.74
42.22
NaOH
40.41
16.16
40
1.01
15.56
H2O
49.32
19.73
18
2.74
42.22
Total
250
100.00
6.49
100.00
35. Ammonium hydroxide is used for neutralizing acetic acid according to the reaction.
CH3COOH +NH4OH CH3COONH4 +H2O
If 120 kg of acetic acid is to be neutralized what will be the quantity of ammonium
hydroxide required? If 25% excess NH4OH is used and if the reaction is 90% complete,
what will be the composition of the product streams?
Basis: 120 kg acetic acid fed
CH3COOH + NH4OH CH3COONH4 +H2O
60
35
77
18
As per Stoichiometry 35 kg NH4OH s required for 60 kg acetic acid
Therefore NH4OH required to react with 120 kg acetic acid = (35/60) x 120 =70 kg
NH4OH supplied = 70 x 1.25 = 87.5 kg
Reaction is 90% complete
Acetic acid reacted = 120 x 0.9 = 108 kg
Acetic acid unreacted = acetic acid taken- acetic acid reacted = 120-108 = 12 kg
Ammonium hydroxide reacted = (35/60) x 108 = 63 kg
Ammonium hydroxide unreacted = ammonium hydroxide supplied- reacted
= 87.5-63 = 24.5 kg
Ammonium acetate produced = (77/60) x 108 = 32.4 kg
Composition of products
Component
Weight
Weight %
Ammonium acetate
138.6
66.80
Water
32.4
15.61
Ammonium hydroxide
24.5
11.81
Acetic Acid
12.0
5.78
Total
207.5
100.00
Esterifcation
36.Ethyl acetate ester is produced by the reaction between acetic acid and ethanol
using catalyst. The reaction goes to completion. The acetic acid used contained 10%
water. Calculate the quantity of ethyl acetate produced for 200 kg ethanol and complete
analysis of products.
Basis: 200 kg ethanol
CH3COOH+ C2H5OH CH3COOC2H5 +H2O
60
46
88
18
As per Stoichiometry 88 kg ethyl acetate produced for 46 kg ethanol
Therefore ethyl acetate produced for 200 kg ethanol= (88/46) x 200 = 382.61 kg
46 kg ethanol requires 60 kg acetic acid
Therefore 200 kg ethanol requires = (60/46) x 200 = 260.87 kg
Acetic acid used is 90% pure
90% acetic acid used = 260.87x100/90 = 289.86 kg
Water in acetic acid = 189.86 - 260.87 = 28.99 kg
Water produced= (18/46) x 200 = 78.26 kg
Water leaving= water produced+ water with acid = 78.26 + 28.99 = 107.25 kg
Complete analysis of products
Component
Weight
Weight%
Ethyl acetate
382.61
78.11
Water
107.25
21.89
Total
489.86
100.00
37. In the manufacture of ethyl benzoate, benzoic acid is reacted
with ethanol in
presence of catalyst. The feed mixture is taken in a batch reactor along with catalyst
and at the end of reaction the products were analyzed and found to contain 35% ethyl
benzoate, 35% water, 10% ether and 20% benzoic acid. What is the % conversion of
the reaction? Which is the excess reactant and what is the % excess?
Basis: 100 kmol reaction products
C6H5 COOH+ C2H5OH C6H5COOC2H5+ H2O
Benzoic acid
ethanol
ethyl benzoate
Ethyl benzoate in products = 35 kmol
As per Stoichiometry 1 kmol ethyl benzoate requires I kmol benzoic acid and 1 kmol
ethanol
Therefore kmol benzoic acid reacted = (1/1) x 35 = 35 kmol
Kmol ethanol reacted= (1/1) x 35 = 35 kmol
Kmol benzoic acid fed = benzoic acid reacted + benzoic acid unreacted
= 35+20= 55 kmol
Kmol ethanol fed = kmol ethanol reacted + kmol ethanol unreacted = 35+10= 45 kmol
Since benzoic acid fed is more than the required to react with ethanol, benzoic acid is
excess reactant and ethanol is the limiting reactant
Conversion= (ethanol reacted/ ethanol fed) x100 = (35/45) x 100 = 77.78%
% excess benzoic acid
= [(benzoic acid fed- benzoic acid required) / benzoic acid required] x 100
= [(55-45)/45] x 100 = (10/45) x 100 = 22.22%
Neutralization
38.100 kg hydrochloric acid is neutralized by 150 kg of sodium hydroxide. What is the
quantity of sodium chloride that will be produced? Find the composition of neutralized
mixture by eight and mol.
Basis: 100 kg HCl and 150 kg NaOH
HCl+ NaOH NaCl + H2O
36.5
40
58.5
18
NaOH required to react with 100 kg HCl= (40/36.5) x 100 = 109.59 kg
NaOH taken= 150 kg
Since NaOH is taken more than the required, it is excess reactant
As per Stoichiometry 58.5 kg sodium chloride is produced for 36.5 kg HCl
Therefore sodium chloride produced= (58.5/36.5) x 100 =160.27 kg
Components leaving
NaCl,H2O,NaOH
Water leaving=Water produced = (18/36.5) x 100 = 49.59 kg
NaOH leaving = NaOH entering- NaOH reacted = 150-109.59 = 40.41 kg
Sodium chloride leaving = sodium chloride produce = 160.27 kg
Composition of neutralized mixture
Component
weight
Weight %
Mol wt
mol
Mol%
NaCl
160.27
64.11
58.5
2.74
42.22
NaOH
40.41
16.16
40
1.01
15.56
H2O
49.32
19.73
18
2.74
42.22
Total
250
100.00
6.49
100.00
39. Ammonium hydroxide is used for neutralizing acetic acid according to the reaction.
CH3COOH +NH4OH CH3COONH4 +H2O
If 120 kg of acetic acid is to be neutralized what will be the quantity of ammonium
hydroxide required? If 25% excess NH4OH is used and if the reaction is 90% complete,
what will be the composition of the product streams?
Basis: 120 kg acetic acid fed
CH3COOH + NH4OH CH3COONH4 +H2O
60
35
77
18
As per Stoichiometry 35 kg NH4OH s required for 60 kg acetic acid
Therefore NH4OH required to react with 120 kg acetic acid = (35/60) x 120 =70 kg
NH4OH supplied = 70 x 1.25 = 87.5 kg
Reaction is 90% complete
Acetic acid reacted = 120 x 0.9 = 108 kg
Acetic acid unreacted = acetic acid taken- acetic acid reacted = 120-108 = 12 kg
Ammonium hydroxide reacted = (35/60) x 108 = 63 kg
Ammonium hydroxide unreacted = ammonium hydroxide supplied- reacted
= 87.5-63 = 24.5 kg
Ammonium acetate produced = (77/60) x 108 = 32.4 kg
Composition of products
Component
Weight
Weight %
Ammonium acetate
138.6
66.80
Water
32.4
15.61
Ammonium hydroxide
24.5
11.81
Acetic Acid
12.0
5.78
Total
207.5
100.00
Esterifcation
40.Ethyl acetate ester is produced by the reaction between acetic acid and ethanol
using catalyst. The reaction goes to completion. The acetic acid used contained 10%
water. Calculate the quantity of ethyl acetate produced for 200 kg ethanol and complete
analysis of products.
Basis: 200 kg ethanol
CH3COOH+ C2H5OH CH3COOC2H5 +H2O
60
46
88
18
As per Stoichiometry 88 kg ethyl acetate produced for 46 kg ethanol
Therefore ethyl acetate produced for 200 kg ethanol= (88/46) x 200 = 382.61 kg
46 kg ethanol requires 60 kg acetic acid
Therefore 200 kg ethanol requires = (60/46) x 200 = 260.87 kg
Acetic acid used is 90% pure
90% acetic acid used = 260.87x100/90 = 289.86 kg
Water in acetic acid = 189.86 - 260.87 = 28.99 kg
Water produced= (18/46) x 200 = 78.26 kg
Water leaving= water produced+ water with acid = 78.26 + 28.99 = 107.25 kg
Complete analysis of products
Component
Weight
Weight%
Ethyl acetate
382.61
78.11
Water
107.25
21.89
Total
489.86
100.00
41. In the manufacture of ethyl benzoate, benzoic acid is reacted
with ethanol in
presence of catalyst. The feed mixture is taken in a batch reactor along with catalyst
and at the end of reaction the products were analyzed and found to contain 35% ethyl
benzoate, 35% water, 10% ether and 20% benzoic acid. What is the % conversion of
the reaction? Which is the excess reactant and what is the % excess?
Basis: 100 kmol reaction products
C6H5 COOH+ C2H5OHC6H5COOC2H5+ H2O
Benzoic acid
ethanol ethyl benzoate
Ethyl benzoate in products = 35 kmol
As per Stoichiometry 1 kmol ethyl benzoate requires I kmol benzoic acid and 1 kmol
ethanol
Therefore kmol benzoic acid reacted = (1/1) x 35 = 35 kmol
Kmol ethanol reacted= (1/1) x 35 = 35 kmol
Kmol benzoic acid fed = benzoic acid reacted + benzoic acid unreacted
= 35+20= 55 kmol
Kmol ethanol fed = kmol ethanol reacted + kmol ethanol unreacted = 35+10= 45 kmol
Since benzoic acid fed is more than the required to react with ethanol, benzoic acid is
excess reactant and ethanol is the limiting reactant
Conversion= (ethanol reacted/ ethanol fed) x100 = (35/45) x 100 = 77.78%
% excess benzoic acid
= [(benzoic acid fed- benzoic acid required) / benzoic acid required] x 100
= [(55-45)/45] x 100 = (10/45) x 100 = 22.22%
Alkylation
42. Ethyl benzene is manufactured by the action of benzene with ethyl chloride using
anhydrous aluminium chloride as catalyst.The reaction is
C6H6
Benzene
+
ClC2H5
C6H5C2H5 +
ethyl chloride
HCl
ethyl benzene
Ethyl chloride is used 15% in excess of theoretically required. The reaction is 95%
complete. During alkylation the liberated HCl is scrubbed with water in order to obtain
20%HCl solution by weight. Calculate the raw materials required for 3000 kg of ethyl
benzene produced and the amount of 20% HCl produced.
Basis: 1 kmol benzene
C6H6 + C2H5Cl C6H5C2H5 + HCl
78
64.5
106
36.5
As per stoichometry I kmol benzene requires 1 kmol ethyl chloride
Ethyl chloride required= (1/1) x 1 = 1 kmol
Ethyl chloride fed= 1x1.15 = 1.15 kmol
Weight of ethyl chloride fed= 1.15 x 64.5= 74.175 kg
Reaction is 95% complete
Therefore ethyl benzene produced = (1/1) x 1 x 0.95 = 0.95 kmol
Weight of ethyl benzene produced = 0.95 x 106 = 100.7 kg
HCl produced = (1/1) x 1 x 0.95 = 0.95 kmol
Weight of HCl produced = 0.95 x 36.5 = 34.675 kg
Benzene unreacted = 1-0.95 = 0.05 kmol = 0.05 x 78 = 3.9 kg
To produce 20% solution of HCl, water required = (80/20) x 34.675 = 138.7 kg
Total weight of HCl produced = 138.7+ 34.675 = 173.375 kg
Benzene required for 3000 kg ethyl benzene = (78/100.7) x 3000 = 2323.73 kg
Ethyl chloride fed for 3000 kg ethyl benzene = (74.175/100.7) x 3000 = 2209.78 kg
43. One method of producing toluene is by alkylation of benzene by chloromethane
using aluminium chloride catalyst. In a particular operation 200 kg of benzene and 150
kg of chloro methane are used to produce 200 kg of toluene. The HCl produced is
absorbed using 200 kg water to obtain HCl solution. Calculate a) the % excess
chloromethane used b)the composition of raw materials entering c)the degree of
completion of reaction and d) concentration of HCl solution produced by completely
absorbing the HCl gas produced.
Basis: 200 kg benzene and 150 kg chloromethane fed
C6H6+ CH3Cl C6H5CH3 +HCl
78
50.5
92
36.5
Chloromethane required to react with 200 kg benzene = (50.5/78) x 200 = 129.49 kg
% excess chloromethane used = [(150-129.49) / 129.49] x 100 = 15.84%
Composition of raw material entering
Component
Weight
Weight %
Benzene
200
57.14
Chloromethane
150
42.86
Total
350
100.00
Toluene produced = 200 kg
Benzene reacted to produce 200 kg toluene = (78/92) x 200 = 169.57 kg
Conversion of benzene= (benzene reacted/benzene fed) x 100
= (169.57/200) x 100 = 84.785%
HCl produced = (36.5/92) x 200 = 79.35 kg
Water used for absorption= 200 g
Total HCl solution produced = 200 + 79.35 = 279.35 kg
Concentration of HCl solution = (79.35/279.35) x 100 = 28.41%
AMINATION
44. The reaction between acetone and ammonia to form amino propane is carried out in
a batch reactor. The final products stream is analyzed and found to contain 35 mol%
amino propane, 35 mol% water 20 mol% ammonia and the remaining acetone. The
feed to the reactor contains only acetone and ammonia. Calculate the fractional
conversion of the limiting reactant and the % by which the other reactant is in excess.
Basis: 100 kmol of products formed
CH3COCH3+ NH3 CH3-CH-NH2-CH3+H2O
Acetone
Amino propane
The products contain 20 mol% NH3 and 10% acetone
As per Stoichiometry the requirement of acetone and ammonia are same, but the
product stream contains 20% unreacted NH3 and 10% unreacted acetone
Therefore NH3 is the excess reactant and acetone is limiting reactant
Mole amino propane in product = 35 kmol
Therefore acetone reacted = (1/1) x 35 = 35 kmol
Acetone fed= acetone reacted+ acetone unreacted = 35 +10 = 45 kmol
% conversion= (35/45) x 100 = 77.78%
NH3 reacted= (1/1) x 35 = 35 kmol
NH3 fed = NH3 reacted + NH3 unreacted = 35 + 20 = 55 kmol
Ammonia required to react with 45 kmol acetone = (1/1) x 45 = 45 kmol
% excess NH3 fed = [(55-45)/45] x 100 = (10/45) x 100 = 22.22%
45. Calculate the composition of products obtained by amination of benzyl chloride with
60% excess NH3.Assume the reaction proceeds according to the reaction
C6H5COCl+NH3 C6H5CONH2+HCl
Benzyl chloride
Benzmide
The conversion is 90%.
Basis: 1 kmol benzyl chloride
C6H5COCl+NH3 C6H5CONH2+HCl
NH3 required= (1/1) x 1 = 1 kmol
NH3 fed= 1x1.6= 1.6 kmol
Conversion is 90%
Kmol benzyl chloride leaving = benzyl chloride unreacted = 1 x (1-0.9) = 0.1 kmol
Ammonia leaving = ammonia unreacted = ammonia fed- ammonia reacted
=1.6- (1/1) x 0.9 = 0.7 kmol
Benzamide leaving = benzamide produced = (1/1) x 0.9 = 0.9 kmol
HCl leaving = HCl produced = (1/1) x 0.9 = 0.9 kmol
Composition of products
Component
Mol
mol %
Benzamide
0.9
34.62
Hydrochloric acid
0.9
34.62
Benzyl chloride
0.1
3.85
Ammonia
0.7
26.92
Total
2.6
100.01
```