MATERIAL BALANCE 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.
1 mole CO= 1 mole H2O
1 mole H2
1 mole CO2
Limiting reactant / component:
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
Excess Reactant:
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 requiredx100
Moles of B theoretically required
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.
ANS. The percent excess of O2 is therefore
= [(75 – 50) / 50] x 100 = 50
CONVERSION:
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
YIELD and SELECTIVITY:
Consider the multiple reactions, namely a series parallel reaction, a series reaction and
a parallel reactionSeries parallel reaction
A + B--> C
C + B--> D
Series reaction A --> C -->D
Parallel reaction 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
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
EXAMPLE:
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.
ANS: Basis: 100 kmol ethylene fed to the reactor.
Reactions:
C2H4+1/2 O2 --> C2H4O........ ( 1 )
C2H4+3O2 --> 2CO2+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%
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.
1 mole CO= 1 mole H2O
1 mole H2
1 mole CO2
Limiting reactant / component:
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
Excess Reactant:
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 requiredx100
Moles of B theoretically required
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.
ANS. The percent excess of O2 is therefore
= [(75 – 50) / 50] x 100 = 50
CONVERSION:
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
YIELD and SELECTIVITY:
Consider the multiple reactions, namely a series parallel reaction, a series reaction and
a parallel reactionSeries parallel reaction
A + B--> C
C + B--> D
Series reaction A --> C -->D
Parallel reaction 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
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
EXAMPLE:
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.
ANS: Basis: 100 kmol ethylene fed to the reactor.
Reactions:
C2H4+1/2 O2 --> C2H4O........ ( 1 )
C2H4+3O2 --> 2CO2+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%
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