Balance On Reactive Systems LECTURE 5

1/6/2021
LECTURE 5
Balances on Reactive Systems
Balances on Reactive Systems, Limiting & Excess Reactants

Extent of Reaction and the Calculation of Extent of Reaction
Multiple Reactions, Yield & Selectivity2
Extent of Reaction for Multiple Reactions with Calculation
Balances on Molecular & Atomic Species
Degree of Freedom Analysis on Reactive Balances
 BALANCES ON REACTIVE SYSTEMS
 Stoichiometry
- Theory of the proportions in which chemical species combine with one another
- The stoichiometric equation of a chemical reaction is a statement of the relative

number of molecules or moles of reactants and products that participate in the
reaction
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1/6/2021
𝑆𝑂2 + 𝑂2 → 𝑆𝑂3
Equation isn't valid. Why?

1
𝑆𝑂2 + 2 𝑂2 → 𝑆𝑂3
(1 𝑆 → 1 𝑆) (3 O → 3 𝑂 )
and,
2𝑆𝑂2 + 𝑂2 → 2 𝑆𝑂3
(2 𝑆 → 2 𝑆) (6 𝑂 → 6𝑂 )
For the reaction,

2𝑆𝑂2 + 𝑂2 → 2𝑆𝑂3
you can write the stoichiometric ratios
2 𝑚𝑜𝑙 𝑆𝑂3 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 , 2 𝑙𝑏 − 𝑚𝑜𝑙𝑒𝑠 𝑆𝑂2 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑

2 𝑚𝑜𝑙 𝑆𝑂3 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 , 2 𝑙𝑏 − 𝑚𝑜𝑙𝑒𝑠 𝑆𝑂2 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑
and so on.

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If you know, for example, that 1600 kg/h of 𝑆𝑂3 is to be produced, you can calculate the
amount of oxygen required as
2𝑆𝑂2 + 𝑂2 → 2𝑆𝑂3
1600 kg 𝑆𝑂3 generated 1 kmol 𝑆𝑂3 1 kmol 𝑂2 consumed = 10 kmol 𝑂2/h

h 80 kg 𝑆𝑂3 2 kmol 𝑆𝑂3 generated
⇒ 10 kmol 𝑂2 32 kg 𝑂2 = 320 kg 𝑂2/h

h 1 kmol 𝑂2
TEST YOURSELF
𝐶4𝐻8 + 6𝑂2 → 4𝐶𝑂2 + 4𝐻2𝑂
1. Is the stoichiometric equation balanced?

2. What is the stoichiometric coefficient of 𝐶𝑂2?
3. What is the stoichiometric ratio of 𝐻2𝑂 to 𝑂2? (include units)
4. How many lb-moles of 𝑂2 react to form 400 lb-moles of 𝐶𝑂2? (use a
dimensional equation)
5. One hundred mol/min of 𝐶4𝐻8 is fed into a reactor, and 50% reacts. At what
rate is the water formed?

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 Limiting and Excess Reactants
- The reactant that would run out if a reaction proceeded to completion is called the
limiting reactant, and the other reactants are termed as excess reactants
- A reactant is limiting if it is present in less than its stoichiometric proportion relative to

every other reactant
- If no reactant is present in stoichiometric proportion, then no reactant is limiting
𝒏𝒇𝒆𝒆𝒅 − 𝒏𝒔𝒕𝒐𝒊𝒄𝒉.
𝑭𝒓𝒂𝒄𝒕𝒊𝒐𝒏𝒂𝒍 𝑬𝒙𝒄𝒆𝒔𝒔 =
𝒏𝒔𝒕𝒐𝒊𝒄𝒉.
𝒏𝒇𝒆𝒆𝒅 − 𝒏𝒔𝒕𝒐𝒊𝒄𝒉.
% 𝑬𝒙𝒄𝒆𝒔𝒔 = 𝒙𝟏𝟎𝟎
𝒏𝒔𝒕𝒐𝒊𝒄𝒉.

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Example,
𝐶2𝐻2 + 2𝐻2 → 𝐶2𝐻6
- Suppose that 20 kmol/h of acetylene and 50 kmol of hydrogen are fed to a reactor
(𝐶2 𝐻2 : 𝐻2 )𝑜 = 1 : 2.5
(𝐶2 𝐻2 : 𝐻2 )𝑠𝑡𝑜𝑖𝑐ℎ. = 1:2
𝐻2 is excess reactant and 𝐶2𝐻2 limiting reactant
- Since it would take 40 kmol 𝐻2/h to react completely with acetylene fed to the reactor
𝟓𝟎 − 𝟒𝟎 𝒌𝒎𝒐𝒍/𝒉
𝑭𝒓𝒂𝒄𝒕𝒊𝒐𝒏𝒂𝒍 𝒆𝒙𝒄𝒆𝒔𝒔 𝑯𝟐 = = 𝟎. 𝟐𝟓
𝟒𝟎 𝒌𝒎𝒐𝒍/𝒉
 Fractional Conversion
- It’s the ratio of reactants,
𝒎𝒐𝒍𝒆𝒔 𝒓𝒆𝒂𝒄𝒕𝒆𝒅
𝒇=
𝒎𝒐𝒍𝒆𝒔 𝒇𝒆𝒅
- The fraction unreacted accordingly would be 1 − 𝑓
- If 100 moles of a reactant is fed and 90 moles react, the fractional conversion is 0.90 (the
percentage conversion is 90%) and the fraction unreacted is 0.10
- Similarly, if 20 mol/min of a reactant is fed and the percentage conversion is 80%, then
𝟐𝟎 𝟎. 𝟖𝟎 = 𝟏𝟔 𝒎𝒐𝒍/𝒎𝒊𝒏 has reacted
𝟐𝟎 𝟏 − 𝟎. 𝟖𝟎 = 𝟒 𝒎𝒐𝒍/𝒎𝒊𝒏 remains unreacted

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 Extent of reaction
𝐶2𝐻2 + 2𝐻2 → 𝐶2𝐻6
- Suppose, 20 kmol of acetylene (𝐶2𝐻2)

50 kmol of hydrogen
50 kmol of ethane (𝐶2𝐻6)
are charged in to a batch reactor.
After sometime, 30 kmol of hydrogen has reacted. How much of each species will be
present in the reactor?
- We started with 20 kmol of 𝐻2

Out of which 30 kmol has reacted
We’re left with 20 kmol of 𝑯𝟐
- Also, if 20 kmol of 𝐻2 has reacted

15 kmol of 𝐶2𝐻2 also reacts
leaving 5 kmol of 𝑪𝟐𝑯𝟐
- Finally, 30 kmol of 𝐻2 reacts to form

15 kmol of 𝐶2 𝐻6
added to 50 kmol of 𝐶2 𝐻6
gives 65 kmol of 𝑪𝟐𝑯𝟔

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Generally, if ξ (kmol of 𝐻2) reacts, we may follow the same reasoning and write,
𝑛𝐻2 = (𝑛𝐻2 )𝑜 − ξ
𝑛𝐶2 𝐻2 = (𝑛𝐶2 𝐻2 )𝑜 −1/2 ξ
𝑛𝐶2 𝐻6 = (𝑛𝐶2 𝐻6 )𝑜 + ξ
Once you know how much hydrogen or acetylene reacts or how much ethane is
formed, you can determine ξ which has same units as n, is called extent of
reaction.
Let 𝑣𝑖 be the stoichiometric coefficient, then

𝑣𝐶2𝐻2 = −1 𝑣𝐻2 = −2 𝑣𝐶2𝐻6 = +1
Then if 𝑛 𝑖 𝑜 is the number of moles (batch) or molar flow rate (continuous) of species i
in the feed to a batch or continuous steady state process,
𝒏𝒊 = 𝒏𝒊𝒐 + 𝒗𝒊ξ or 𝒏𝒐 = 𝒏𝒊𝒐 + 𝒗𝒊ξ
where,
ξ = Extent of reaction
𝒏𝒊 = moles of species i present in the system after the reaction occurred
𝒏𝒊𝒐 = moles of species i in the system when the reaction starts
𝒗𝒊 = stoichiometry coefficient for species i in the particular chemical reaction equation

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Example, 𝑁2 + 3𝐻2 → 2𝑁𝐻3
Rector Inlet: 100 mol 𝑁2/s; 300 mol 𝐻2/s; 1 mol Ar/s.
If fractional conversion of 𝐻2 is 0.6, calculate extent of reaction and the outlet
composition.
Solution:
𝑚𝑜𝑙
𝑛𝑁2 = 100 −𝜉
𝑠
𝑚𝑜𝑙
𝑛𝐻2 = 300 − 3𝜉
𝑠
𝑛𝑁𝐻3 = 2𝜉
𝑛 𝐴 = 1 𝑚𝑜𝑙/𝑠
- Hydrogen consumed = 300 (0.6) = 180 mol 𝐻2/s
- Hydrogen unreacted or at the outlet = 300 – 180 = 120 mol 𝐻2/s
𝑛𝐻2 = 120mol 𝐻2/s
Now solve for extent of reaction

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Example 4.6-1
Acrylonitrile is produced in the reaction of propylene, ammonia and oxygen

3
𝐶3 𝐻6 + 𝑁𝐻3 + 𝑂2 → 𝐶3 𝐻3 𝑁 + 3𝐻2 𝑂
2
The feed contains 10 mole% propylene, 12 mole% ammonia and 78 mole% air. A
fractional conversion of 30% of the limiting reactant is achieved. Taking 100 mol of
feed as a basis, determine which reactant is limiting, the percentage by which each of
the other reactants is in excess, and the molar amounts of all gas constituents for a
30% conversion of the limiting reactant.
SOLUTION
Basis: 100 mol of feed
100 mol
0.100 mol 𝑛𝐶3𝐻6(mol 𝐶3 𝐻6 )
𝐶3𝐻6/𝑚𝑜𝑙 𝑛𝑁𝐻3 (mol 𝑁𝐻3)
0.120 mol 𝑁𝐻3/𝑚𝑜𝑙 𝑛𝑂2(mol air)
Inlet: 0.780 mol air/mol 𝑛𝐶3𝐻3𝑁 (mol 𝐶3𝐻3𝑁)
𝐶3𝐻6 = 10 mol 𝑛𝐻2𝑂 (mol 𝐻2𝑂)
𝑁𝐻3 = 12 mol
𝐴𝑖𝑟 = 78 mol
(16.4 mol 𝑂2, 61.6 mol 𝑁2)

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 For limiting reactant
𝑛𝑁𝐻3 12 𝑚𝑜𝑙
(𝑛 )𝑜 = 10 𝑚𝑜𝑙 = 1.2
𝐶3𝐻6
𝑛𝑁𝐻3 1𝑚𝑜𝑙
(𝑛 )𝑠𝑡. = 1𝑚 𝑜 = 1 (1.2 > 1)
𝐶3𝐻6
𝑛𝑂2 𝑙 𝑚𝑜𝑙
16.4
( )𝑜 = =1.64
𝑛𝐶3𝐻6 10 𝑚𝑜𝑙
𝑛𝑂2 1.5 𝑚𝑜𝑙
(𝑛 )𝑠𝑡. = 1 𝑚𝑜𝑙 = 1.5 (1.6 4 > 1.5)
𝐶3𝐻6
Propylene is the limiting reactant
 % Excess Reactant
𝑛𝑁𝐻3 𝑠𝑡. = 10 mol 𝐶3𝐻6 1 mol 𝑁𝐻3 = 10 mol 𝑁𝐻3

1 mol 𝐶3𝐻6
𝑛𝑂2 𝑠𝑡.= 10 mol 𝐶3𝐻6 1.5 mol 𝑂2 = 15 mol 𝑂2
1 mol 𝐶3𝐻6
𝑛 𝑁 𝐻 3 𝑂 − 𝑛 𝑁 𝐻 3 𝑠𝑡. 12−10
% 𝑒𝑥𝑐𝑒𝑠𝑠𝑁𝐻3 = x 100 = 𝑥100 = 20%
𝑛𝑁𝐻3 10
𝑠𝑡
𝑛𝑂2 𝑂− 𝑛 𝑂 2 .𝑠𝑡. 16.4−15
% 𝑒𝑥𝑐𝑒𝑠𝑠𝑂2 = x 100 = 𝑥100 = 9.3%
𝑛𝑂2 15
𝑠𝑡
.

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 Fractional Conversion
𝑛𝐶3 𝐻6 𝑓𝑒𝑒𝑑
= 10 mol
𝑛𝐶3𝐻6 𝑜𝑢𝑡 = 10 mol x 70 % or 10 mol (100% – 30%) = 7 mol
 Extent of reaction
𝑛𝐶3𝐻6 𝑜𝑢𝑡 = 𝑛𝐶3𝐻6 𝑜𝑢𝑡 − 𝜉
7 𝑚𝑜𝑙 = 10 𝑚𝑜𝑙 − 𝜉 ⇒ 𝜉 = 3 𝑚𝑜𝑙
𝑛𝑁𝐻3 𝑜𝑢𝑡 = 𝑛𝑁𝐻3 𝑓𝑒𝑒𝑑

−𝜉
𝑛𝑁𝐻3 = 12 𝑚𝑜𝑙 − 𝜉  (1)
𝑜𝑢𝑡
𝑛𝑂2 𝑜𝑢𝑡 = 𝑛𝑂2 𝑓𝑒𝑒𝑑 − 1.5 𝜉
𝑛𝑂2 𝑜𝑢𝑡 = 16.4 𝑚𝑜𝑙 − 1.5 𝜉  (2)
 Multiple reactions, Yield and Selectivity
- Some of the chemical reaction has a side reaction which forms undesired products –
multiple reaction occurs
- Effects of this side reaction might be:
1. Economic loss
2. Less of desired product is obtained for a given quantity of raw materials
3. Greater quantity of raw materials must be fed to the reactor to obtain a specified
product yield
𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒅𝒆𝒔𝒊𝒓𝒆𝒅 𝒑𝒓𝒐𝒅𝒖𝒄𝒕

𝑺𝒆𝒍𝒆𝒄𝒕𝒊𝒗𝒊𝒕𝒚 =
𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒖𝒏𝒅𝒆𝒔𝒊𝒓𝒆𝒅 𝒑𝒓𝒐𝒅𝒖𝒄𝒕 𝒇𝒐𝒓𝒎𝒆𝒅

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 Yield
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑓𝑜𝑟𝑚𝑒𝑑

𝑌𝑖𝑒𝑙𝑑 = 𝑚𝑜𝑙𝑒𝑠 𝑡ℎ𝑎𝑡 𝑤𝑜𝑢𝑙𝑑 ℎ𝑎𝑣𝑒 𝑏𝑒𝑒𝑛 𝑓𝑜𝑟𝑚𝑒𝑑 𝑖𝑓 𝑡ℎ𝑒𝑟𝑒 𝑤𝑒𝑟𝑒
𝑛𝑜 𝑠𝑖𝑑𝑒 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛𝑠 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑙𝑖𝑚𝑖𝑡𝑖𝑛𝑔 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 ℎ𝑎𝑑
𝑟𝑒𝑎𝑐𝑡𝑒𝑑 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒𝑙𝑦
- High values of the yield and selectivity signify that the undesired side reactions have
been successfully suppressed relative to the desired reaction

𝑌𝑖𝑒𝑙𝑑 =
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑓𝑒𝑑

𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑
 Extent of reaction for multiple reactions
- Consider the pair of reactions in which in which ethylene is oxidized either to ethylene
oxide (desired) or to carbon dioxide (undesired)
1
𝐶2 𝐻4 + 𝑂2 → 𝐶2 𝐻4 𝑂
2
𝐶2𝐻4 + 3𝑂2 → 2𝐶𝑂2 + 2𝐻2𝑂
The moles (molar flow rate) of each of the five species involved in these rxns can be
expressed in terms of the feed values and extents of rxns:
(𝑛𝐶2 𝐻4 )𝑜 𝑢 𝑡 = (𝑛𝐶2 𝐻4 )𝑜 −𝜉1 − 𝜉2
(𝑛𝑜 2 )𝑜 𝑢 𝑡 = (𝑛𝑂2 )𝑜 −0.5𝜉1 − 3𝜉2
(𝑛 𝐶 2 𝐻 4 𝑂 ) 𝑜𝑢𝑡 = (𝑛𝐶2𝐻4𝑂)𝑜+𝜉1
(𝑛𝐶𝑂2 )𝑜 𝑢 𝑡 = (𝑛𝐶𝑂2 )𝑜 +2𝜉2 (
𝑛 𝐻 2 𝑂 ) 𝑜𝑢𝑡 = (𝑛𝐻2𝑂)𝑜+2𝜉2
If values of any two outlet amounts are given, the values of 𝜉1, 𝜉2 may be determined from
corresponding two equations

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Example 4.6-3
The reactions 𝐶2𝐻6 → 𝐶2𝐻4 + 𝐻2
𝐶2𝐻6 + 𝐻2 → 2𝐶𝐻4
take place in a continuous reactor at ss. The feed contains 85 mole% ethane and the
balance inert. The fractional conversion of ethane is 0.501, and the fractional yield of
ethylene is 0.471. Calculate the molar composition of the product gas and the selectivity
of ethylene to methane production.
Solution
Basis : 100 mol of feed
100 mol
0.850 mol 𝒏𝟏(mol 𝑪𝟐𝑯𝟔)
𝑪𝟐𝑯𝟔/mol 𝒏𝟐 (mol
0.150 mol 𝑪𝟐𝑯𝟒)
I/mol 𝒏𝟑 (mol 𝑯𝟐)
𝒏𝟒 (mol 𝑪𝑯𝟒)
(mol )
The outlet component amounts in terms of extents of reactions are as follows:
(𝑛𝐶2 𝐻6 )𝑜 𝑢 𝑡 = (𝑛𝐶2 𝐻6 )𝑜 − 𝜉1 − 𝜉
2 (𝑛𝐶2 𝐻4 )𝑜 𝑢 𝑡 = (𝑛𝐶2 𝐻4 )𝑜 + 𝜉1
(𝑛𝐻2 )𝑜 𝑢 𝑡 = (𝑛𝐻2 )𝑜 + 𝜉1 − 𝜉2
(𝑛𝐶𝐻4 )𝑜 𝑢 𝑡 = (𝑛𝐶𝐻4 )𝑜 + 2𝜉2
(𝑛𝐼)𝑜𝑢𝑡 = 15 𝑚𝑜𝑙 𝐼
(𝑛𝐶2 𝐻6 )𝑜 𝑢 𝑡 = 85 𝑚𝑜𝑙 − 𝜉1 − 𝜉2
(𝑛𝐶2 𝐻4 )𝑜 𝑢 𝑡 = 𝜉1 (
𝑛𝐻2 )𝑜 𝑢 𝑡 = 𝜉1 − 𝜉2
(𝑛𝐶𝐻4 )𝑜 𝑢 𝑡 = 2𝜉2

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Ethane Conversion
(𝑛 𝐶2 𝐻6 )𝑜 𝑢 𝑡 = (1 − 0.501) 𝑚𝑜𝑙 𝐶2 𝐻6 𝑢𝑛𝑟𝑒𝑎𝑐𝑡𝑒𝑑 85 𝑚𝑜𝑙 𝐶2 𝐻6 𝑓𝑒𝑑

𝑚𝑜𝑙 𝐶2𝐻6 𝑓𝑒𝑑
⇒ (𝑛 𝐶2 𝐻6 )𝑜 𝑢 𝑡 = 42.4 𝑚𝑜𝑙 𝐶2 𝐻6
Ethylene Yield
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑒𝑡ℎ𝑦𝑙𝑒𝑛𝑒 𝑓𝑜𝑟𝑚𝑒𝑑

𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑒𝑡ℎ𝑦𝑙𝑒𝑛𝑒 𝑟𝑒𝑎𝑐𝑡𝑒𝑑
(𝑛𝐶2 𝐻4 )𝑜 𝑢 𝑡 = 0.471 85 𝑚𝑜𝑙𝑒𝑠 = 40 𝑚𝑜𝑙𝑒𝑠

From extent of reaction for 𝐶2𝐻4 we get,
𝜉1 = 40 𝑚𝑜𝑙
Similarly substituting this value in other equation and solve for 𝜉2 and then solve for
molar compositions of gas at the outlet
 BALANCES ON REACTIVE PROCESSES

 Balances on Molecular and Atomic Species
The reaction for the dehydrogenation of ethane in a ss continuous reactor,

𝐶2𝐻6 → 𝐶2𝐻4 + 𝐻2
One hundred kmol/min of ethane is fed to the reactor. The molar flow rate of 𝐻2 in the
product stream is 40 kmol/min.
100 kmol 𝑪𝟐𝑯𝟔/ 40 kmol 𝑯𝟐/ min

min 𝒏 𝟏 (𝒌𝒎𝒐𝒍𝑪𝟐𝑯𝟔/min)
𝒏 𝟐 (𝒌𝒎𝒐𝒍𝑪𝟐𝑯𝟒/min)

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Input + Generation = Output + Consumption
Molecular 𝑯𝟐 Balance generation = output

𝑚𝑜𝑙 𝐻2𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝐻2
𝐺𝑒𝑛𝐻2 = 40 𝑚𝑜𝑙
𝑚𝑖𝑛 𝑚𝑖𝑛
𝑪𝟐𝑯𝟔 Balance input = output + consumption

100 𝑚𝑜𝑙 𝐶2𝐻6 𝑚𝑜𝑙 𝐶2 𝐻 𝑚𝑜𝑙 𝐶2𝐻6
=𝑛 1 6 + 𝐶𝑜𝑛𝑠 𝐶2𝐻6
𝑚𝑖𝑛 𝑚𝑖𝑛 𝑚𝑖𝑛
𝑪𝟐𝑯𝟒 Balance generation = output

𝑚𝑜𝑙 𝐶2𝐻4 𝑚𝑜𝑙 𝐶2𝐻4
𝐺𝑒𝑛𝐶2𝐻4 = 𝑛2
Atomic C Balance input = output
100 𝑚𝑜𝑙 𝐶2𝐻6 𝑚𝑜𝑙𝐶2 𝐻6 2 𝑚𝑜𝑙 𝐶 𝑚𝑜𝑙𝐶2𝐻4 2 𝑚𝑜𝑙 𝐶

=𝑛1 + 𝑛2
𝑚𝑖𝑛 𝑚𝑖𝑛 1 𝑚𝑜𝑙 𝐶2𝐻6 𝑚𝑖𝑛 1 𝑚𝑜𝑙 𝐶2𝐻4
⇒ 100 𝑚𝑜𝑙 𝐶/𝑚𝑖𝑛 = 𝑛1 + 𝑛2

Atomic H Balance input = output
100 𝑚𝑜𝑙 𝐶2𝐻6 6 𝑚𝑜𝑙 𝐻 40 𝑚𝑜𝑙 𝐻2 2 𝑚𝑜𝑙 𝐻

= +
𝑚𝑖𝑛 1 𝑚𝑜𝑙 𝐶2𝐻6 𝑚𝑖𝑛 1 𝑚𝑜𝑙 𝐻2
𝑚𝑜𝑙𝐶2 𝐻6 6 𝑚𝑜𝑙 𝐻 𝑚𝑜𝑙𝐶2𝐻4 4 𝑚𝑜𝑙 𝐶

𝑛1 + 𝑛2
𝑚𝑖𝑛 1 𝑚𝑜𝑙 𝐶2𝐻6 𝑚𝑖𝑛 1 𝑚𝑜𝑙 𝐶2𝐻4
⇒ 600 𝑚𝑜𝑙 𝐻/𝑚𝑖𝑛 = 80 𝑚𝑜𝑙 𝐻/𝑚𝑖𝑛 +6 𝑛1 + 4 𝑛2

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 Degrees of Freedom Analysis on Reactive Balances
 Molecular Species Balances
No. of unknown labeled variables

+ No. of independent chemical rxns
- No. of independent molecular species balances
- No. of other equations relating unknown variables
= No. degrees of freedom
Example 4.7-1
Methane is burned with air in a continuous ss combustion reactor to yield a mixture of

𝐶𝑂, 𝐶𝑂2 and water. The reactions taking place are:
3
𝐶𝐻4 + 𝑂2 → 𝐶𝑂 + 2𝐻2 𝑂
2
𝐶𝐻4 + 2𝑂2 → 𝐶𝑂2 + 2𝐻2𝑂
The feed to the reactor contains 7.8 mole% 𝐶𝐻4, 19.4% 𝑂2, and 72.8% 𝑁2. The percentage
conversion of methane is 90% and the gas leaving the reactor contains 8 mol 𝐶𝑂2/mol CO.
Carry out a dof analysis on the process. Then calculate the molar composition of the
product stream using molecular species balances, atomic species balances and extents of
reaction.

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Solution:
𝒏𝑪𝑯𝟒 (𝒎𝒐𝒍 𝑪𝑯𝟒)

𝒏𝑪𝑶 (𝒎𝒐𝒍 𝑪𝑶)
8 𝒏𝑪𝑶𝟐 (𝒎𝒐𝒍 𝑪𝑶𝟐)
100 mol
0.078 mol
𝒏𝑯𝟐𝑶 𝒎𝒐𝒍 𝑯𝟐𝑶
𝑪𝑯𝟒/mol
𝒏𝑶𝟐 𝒎𝒐𝒍 𝑶𝟐
0.194 mol 𝑶 𝟐/mol
𝒏𝑵𝟐 𝒎𝒐𝒍 𝑵𝟐
0.728 mol 𝑵𝟐/mol
Degree-of-Freedom Analysis
The analysis can be based on any of the three solution methods:
• Molecular species balances
5
• Atomic species balances
• Extents of reaction

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LECTURE 5
Balances on Reactive Systems
(Part 2)
 Multiple reactions, Yield and Selectivity
- Some of the chemical reaction has a side reaction which forms undesired
products – multiple reaction occurs.
- Effects of this side reaction might be:
1. Economic loss
2. Less of desired product is obtained for a given quantity of raw materials
3. Greater quantity of raw materials must be fed to the reactor to obtain a
specified product yield
𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒅𝒆𝒔𝒊𝒓𝒆𝒅 𝒑𝒓𝒐𝒅𝒖𝒄𝒕

𝑺𝒆𝒍𝒆𝒄𝒕𝒊𝒗𝒊𝒕𝒚 =
𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒖𝒏𝒅𝒆𝒔𝒊𝒓𝒆𝒅 𝒑𝒓𝒐𝒅𝒖𝒄𝒕 𝒇𝒐𝒓𝒎𝒆𝒅

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 Yield

𝑌𝑖𝑒𝑙𝑑 = 𝑚𝑜𝑙𝑒𝑠 𝑡ℎ𝑎𝑡 𝑤𝑜𝑢𝑙𝑑 ℎ𝑎𝑣𝑒 𝑏𝑒𝑒𝑛 𝑓𝑜𝑟𝑚𝑒𝑑 𝑖𝑓 𝑡ℎ𝑒𝑟𝑒 𝑤𝑒𝑟𝑒
𝑛𝑜 𝑠𝑖𝑑𝑒 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛𝑠 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑙𝑖𝑚𝑖𝑡𝑖𝑛𝑔 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 ℎ𝑎𝑑
𝑟𝑒𝑎𝑐𝑡𝑒𝑑 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒𝑙𝑦
- High values of the yield and selectivity signify that the undesired side reactions
have been successfully suppressed relative to the desired reaction.

𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑓𝑒𝑑

𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑
 Extent of reaction for multiple reactions
-Consider the pair of reactions in which in which ethylene is oxidized either to ethylene
oxide (desired) or to carbon dioxide (undesired),
1
𝐶2𝐻4 + 𝑂2 → 𝐶2𝐻4𝑂
2
𝐶2𝐻4 + 3𝑂2 → 2𝐶𝑂2 + 2𝐻2𝑂
The moles (molar flow rate) of each of the five species involved in these rxns can be
expressed in terms of the feed values and extents of rxns:
(𝑛𝐶2 𝐻4 )𝑜𝑢 𝑡 = (𝑛𝐶2 𝐻4 )𝑜 −𝜉1 − 𝜉2
(𝑛𝑜 2 )𝑜𝑢 𝑡 = (𝑛𝑂2 )𝑜 −0.5𝜉1 − 3𝜉2
(𝑛𝐶2𝐻4𝑂)𝑜𝑢𝑡= (𝑛𝐶2𝐻4𝑂)𝑜+𝜉1
(𝑛𝐶𝑂2 )𝑜 𝑢 𝑡 = (𝑛𝐶𝑂2 )𝑜 +2𝜉2 (
𝑛𝐻2𝑂)𝑜𝑢𝑡= (𝑛𝐻2𝑂)𝑜+2𝜉2
If values of any two outlet amounts are given, the values of 𝜉1, 𝜉2 may be determined
from corresponding two equations.

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Example 4.6-3
The reactions 𝐶2𝐻6 → 𝐶2𝐻4 + 𝐻2
𝐶2𝐻6 + 𝐻2 → 2𝐶𝐻4
take place in a continuous reactor at ss. The feed contains 85 mole% ethane and
the balance inert. The fractional conversion of ethane is 0.501, and the fractional
yield of ethylene is 0.471. Calculate the molar composition of the product gas and
the selectivity of ethylene to methane production.
Solution
Basis : 100 mol of feed
100 mol
0.850 mol 𝑪𝟐𝑯𝟔/mol 𝒏𝟏(mol 𝑪𝟐 𝑯𝟔)
0.150 mol I/mol 𝒏𝟐 (mol 𝑪𝟐𝑯𝟒)
𝒏𝟑 (mol 𝑯𝟐)
𝒏𝟒 (mol 𝑪𝑯𝟒)
𝒏𝟓 (mol 𝑰)
The outlet component amounts in terms of extents of reactions are as follows:
(𝑛𝐶2 𝐻6 )𝑜 𝑢 𝑡 = (𝑛𝐶2 𝐻6 )𝑜 − 𝜉1 − 𝜉
2 (𝑛𝐶2 𝐻4 )𝑜𝑢 𝑡 = (𝑛𝐶2 𝐻4 )𝑜 + 𝜉1
(𝑛𝐻2 )𝑜𝑢 𝑡 = (𝑛𝐻2 )𝑜 + 𝜉1 − 𝜉2 (
𝑛𝐶𝐻4)𝑜𝑢𝑡 = (𝑛𝐶𝐻4)𝑜 + 2𝜉2
(𝑛𝐼)𝑜𝑢𝑡 = 15 𝑚𝑜𝑙 𝐼
(𝑛𝐶2 𝐻6 )𝑜 𝑢 𝑡 = 85 𝑚𝑜𝑙 − 𝜉1 − 𝜉2
(𝑛𝐶2 𝐻4 )𝑜𝑢 𝑡 = 𝜉1
(𝑛𝐻2 )𝑜 𝑢 𝑡 = 𝜉1 − 𝜉2
(𝑛𝐶𝐻4 )𝑜𝑢 𝑡 = 2𝜉2

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Ethane Conversion
(𝑛 𝐶2𝐻6 ) 𝑜𝑢𝑡 = (1 − 0.501)(𝑚𝑜𝑙 𝑜𝑓 𝐶2𝐻6) 𝑢𝑛𝑟𝑒𝑎𝑐𝑡𝑒𝑑 85 𝑚𝑜𝑙 𝑜𝑓 𝐶2𝐻6 𝑓𝑒𝑑

𝑚𝑜𝑙 𝑜𝑓 𝐶2𝐻6 𝑓𝑒𝑑
⇒ (𝑛 𝐶2 𝐻6 )𝑜 𝑢 𝑡 = 42.4 𝑚𝑜𝑙 𝐶2 𝐻6
Ethylene Yield
𝑀𝑎𝑥 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑒𝑡ℎ𝑦𝑙𝑒𝑛𝑒 𝑓𝑜𝑟𝑚𝑒𝑑 = 85 𝑚𝑜𝑙 𝑜𝑓 𝐶2𝐻6 𝑓𝑒𝑑 1 𝑚𝑜𝑙 𝐶2𝐻4 = 85 𝑚𝑜𝑙
1 𝑚𝑜𝑙 𝐶2𝐻6
(𝑛𝐶2 𝐻4 )𝑜𝑢 𝑡 = 0.471 85 𝑚𝑜𝑙𝑒𝑠 = 40 𝑚𝑜𝑙𝑒𝑠

From extent of reaction for 𝐶2𝐻4 we get,
𝜉1 = 40 𝑚𝑜𝑙
Similarly substituting this value in other equation and solve for 𝜉2 and then solve for
molar compositions of gas at the outlet
(𝑛𝐶2 𝐻6 )𝑜𝑢 𝑡 = 85 𝑚𝑜𝑙 − 𝜉1 − 𝜉2
(𝑛𝐶2 𝐻4 )𝑜𝑢 𝑡 = 𝜉1
𝜉1 = 40 𝑚𝑜𝑙
(𝑛𝐻2)𝑜𝑢𝑡 = 𝜉1 − 𝜉2
(𝑛𝐶𝐻4 )𝑜𝑢 𝑡 = 2𝜉2

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40 𝑚𝑜𝑙 𝑜𝑓 𝐶2𝐻4
𝑆𝑒𝑙𝑒𝑐𝑡𝑖𝑣𝑖𝑡𝑦 =
5.2 𝑚𝑜𝑙 𝑜𝑓 𝐶𝐻4
𝒎𝒐𝒍 𝒐𝒇 𝑪𝟐𝑯𝟒
𝑺𝒆𝒍𝒆𝒄𝒕𝒊𝒗𝒊𝒕𝒚 = 𝟕. 𝟕
𝒎𝒐𝒍 𝒐𝒇 𝑪𝑯𝟒
 BALANCES ON REACTIVE PROCESSES

 Balances on Molecular and Atomic Species
The reaction for the dehydrogenation of ethane in a ss continuous reactor,

𝐶2𝐻6 → 𝐶2𝐻4 + 𝐻2
One hundred kmol/min of ethane is fed to the reactor. The molar flow rate of 𝐻2 in
the product stream is 40 kmol/min.
100 kmol 𝑪𝟐𝑯𝟔/ min 40 kmol 𝑯𝟐/ min

𝒏 𝟏 (𝒌𝒎𝒐𝒍𝑪𝟐𝑯𝟔/min)
𝒏 𝟐 (𝒌𝒎𝒐𝒍𝑪𝟐𝑯𝟒/min)

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Input + Generation = Output + Consumption

𝐶2𝐻6 → 𝐶2𝐻4 + 𝐻2
Molecular 𝑯𝟐 Balance generation = output
𝑚𝑜𝑙 𝐻2𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝐻2
𝐺𝑒𝑛𝐻2 = 40 𝑚𝑜𝑙
𝑪𝟐𝑯𝟔 Balance input = output + consumption

100 𝑚𝑜𝑙 𝐶2𝐻6 𝑚𝑜𝑙 𝐶2 𝐻6 𝑚𝑜𝑙 𝐶2𝐻6
=𝑛 1 + 𝐶𝑜𝑛𝑠 𝐶2𝐻6
𝑚𝑖𝑛 𝑚𝑖𝑛 𝑚𝑖𝑛
𝑪𝟐𝑯𝟒 Balance generation = output

𝑚𝑜𝑙 𝐶2𝐻4 𝑚𝑜𝑙 𝐶2𝐻4
𝐺𝑒𝑛𝐶2𝐻4 = 𝑛2
𝐶2𝐻6 → 𝐶2𝐻4 + 𝐻2
• Atomic C Balance: input = output
100 𝑚𝑜𝑙 𝐶2𝐻6 2 𝑚𝑜𝑙 𝐶 = 𝑛1 𝑚𝑜𝑙 𝐶2𝐻6 2 𝑚𝑜𝑙 𝐶 + 𝑛 2 𝑚𝑜𝑙 𝐶2𝐻4 2 𝑚𝑜𝑙 𝐶
𝑚𝑖𝑛 1 𝑚𝑜𝑙 𝐶2𝐻6 𝑚𝑖𝑛 1 𝑚𝑜𝑙 𝐶2𝐻6 𝑚𝑖𝑛 1 𝑚𝑜𝑙 𝐶2𝐻4
⇒ 100 𝑚𝑜𝑙 𝐶/ min = 𝑛 1 + 𝑛

2
input = output
• Atomic H Balance:
= 40 𝑚𝑜𝑙 𝐻2 2 𝑚𝑜𝑙 𝐻 + 𝑛1 𝑚𝑜𝑙 𝐶2 𝐻 6 𝑚𝑜𝑙 𝐻

100 𝑚𝑜𝑙 𝐶2𝐻6 6 𝑚𝑜𝑙 𝐻
𝑚𝑖𝑛 1 𝑚𝑜𝑙 𝐶2𝐻6 𝑚𝑖𝑛 6
1 𝑚𝑜𝑙 𝐻2 𝑚𝑖𝑛 1 𝑚𝑜𝑙 𝐶2𝐻6
+ 𝑛 2 𝑚𝑜𝑙 𝐶2𝐻4 4 𝑚𝑜𝑙 𝐻
𝑚𝑖𝑛 1 𝑚𝑜𝑙 𝐶2𝐻4
⇒ 600 𝑚𝑜𝑙 𝐻/ min = 80 𝑚𝑜𝑙 𝐻/ min + 6𝑛 1 + 4𝑛 2

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 Degrees of Freedom Analysis on Reactive Balances
 Molecular Species Balances

+ No. of independent chemical rxns
- No. of independent molecular species balances
 Atomic Species Balances

- No. of independent atomic species balance
- No. of molecular species balances on independent non-reactive species

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 Extent of reaction

+ No. of independent reaction (one extent of rxn for each)
- No. of independent reactive species
- No. independent non-reactive
Example 4.7-1
Methane is burned with air in a continuous ss combustion reactor to yield a

mixture of 𝐶𝑂, 𝐶𝑂2 and water. The reactions taking place are:
3
𝐶𝐻4 + 𝑂2 → 𝐶𝑂 + 2𝐻2 𝑂
2
𝐶𝐻4 + 2𝑂2 → 𝐶𝑂2 + 2𝐻2𝑂
The feed to the reactor contains 7.8 mole% 𝐶𝐻4, 19.4% 𝑂2, and 72.8% 𝑁2. The
percentage conversion of methane is 90% and the gas leaving the reactor
contains 8 mol 𝐶𝑂2/mol CO. Carry out a dof analysis on the process. Then
calculate the molar composition of the product stream using molecular species
balances, atomic species balances and extents of reaction.

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Solution:

𝒏𝑪𝑯𝟒 (𝒎𝒐𝒍 𝑪𝑯𝟒)
𝒏𝑪𝑶 (𝒎𝒐𝒍 𝑪𝑶)
8 𝒏𝑪𝑶 (𝒎𝒐𝒍 𝑪𝑶𝟐)
100 mol
0.078 mol
𝒏𝑯𝟐𝑶 𝒎𝒐𝒍 𝑯𝟐𝑶
𝑪𝑯𝟒/mol
𝒏𝑶𝟐 𝒎𝒐𝒍 𝑶𝟐
0.194 mol 𝑶𝟐/mol
0.728 mol 𝑵𝟐/mol 𝒏𝑵𝟐 𝒎𝒐𝒍 𝑵𝟐

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What you will Learn in This Video Course
• Learn the common processes in Chemical Engineering industry and process
variables and how to calculate them.
• Will Understand about Pressure (Fluid Pressure and Hydrostatic head) and
types of pressure and way to measure Fluid Pressure Measurement by
Monometer.
• Get the knowledge about Ideal Gas and the Laws of Gases.
• Able to understand Law of Conservation of mass and know how to apply it
on Different Systems.
• Will Understand about Pressure and Pressure Scales and the Conversion
Factor and the Measurement of Pressure.
• Learn what system energy is and perform energy balance calculation.
• Know how to Use Differential & Integral Balances for Material Balance.
• Learn how to perform material balance and energy balance on Reactive
Processes.
• Know how to use Flow Chart Streams and Flow Chart Scaling for the
Calculation of Material Balance.
• Extent of Reaction and the Calculation of Extent of Reaction for Multiple
Reactions.
• Calculation of Balances on Molecular & Atomic Species
• Understand about the Energy Balance on Open & Closed Systems
• On completion of the course, students will be able to apply mass balance
equation for different systems and can solve numerical based on mass
transfer.
• On completion of the course, students will be able to apply thermal energy
balance equation for different systems and can solve numerical on Energy
Transfer
• On completion of the course, students will be able to calculate fluid pressure.

Balance On Reactive Systems LECTURE 5

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1/6/2021

Balances on Reactive Systems, Limiting & Excess Reactants

 BALANCES ON REACTIVE SYSTEMS

- The stoichiometric equation of a chemical reaction is a statement of the relative

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Equation isn't valid. Why?

For the reaction,

you can write the stoichiometric ratios

2 𝑚𝑜𝑙 𝑆𝑂3 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 , 2 𝑙𝑏 − 𝑚𝑜𝑙𝑒𝑠 𝑆𝑂2 𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑

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1600 kg 𝑆𝑂3 generated 1 kmol 𝑆𝑂3 1 kmol 𝑂2 consumed = 10 kmol 𝑂2/h

⇒ 10 kmol 𝑂2 32 kg 𝑂2 = 320 kg 𝑂2/h

𝐶4𝐻8 + 6𝑂2 → 4𝐶𝑂2 + 4𝐻2𝑂

1. Is the stoichiometric equation balanced?

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 Limiting and Excess Reactants

- A reactant is limiting if it is present in less than its stoichiometric proportion relative to

- If no reactant is present in stoichiometric proportion, then no reactant is limiting

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𝐻2 is excess reactant and 𝐶2𝐻2 limiting reactant

- It’s the ratio of reactants,

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𝐶2𝐻2 + 2𝐻2 → 𝐶2𝐻6

- Suppose, 20 kmol of acetylene (𝐶2𝐻2)

- We started with 20 kmol of 𝐻2

- Also, if 20 kmol of 𝐻2 has reacted

- Finally, 30 kmol of 𝐻2 reacts to form

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Let 𝑣𝑖 be the stoichiometric coefficient, then

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Example, 𝑁2 + 3𝐻2 → 2𝑁𝐻3

- Hydrogen consumed = 300 (0.6) = 180 mol 𝐻2/s

- Hydrogen unreacted or at the outlet = 300 – 180 = 120 mol 𝐻2/s

𝑛𝐻2 = 120mol 𝐻2/s

Now solve for extent of reaction

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Acrylonitrile is produced in the reaction of propylene, ammonia and oxygen

Basis: 100 mol of feed

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 For limiting reactant

Propylene is the limiting reactant

𝑛𝑁𝐻3 𝑠𝑡. = 10 mol 𝐶3𝐻6 1 mol 𝑁𝐻3 = 10 mol 𝑁𝐻3

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𝑛𝑁𝐻3 𝑜𝑢𝑡 = 𝑛𝑁𝐻3 𝑓𝑒𝑒𝑑

 Multiple reactions, Yield and Selectivity

𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒅𝒆𝒔𝒊𝒓𝒆𝒅 𝒑𝒓𝒐𝒅𝒖𝒄𝒕

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𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑓𝑜𝑟𝑚𝑒𝑑

𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑓𝑜𝑟𝑚𝑒𝑑

𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑓𝑜𝑟𝑚𝑒𝑑

 Extent of reaction for multiple reactions

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The outlet component amounts in terms of extents of reactions are as follows:

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(𝑛 𝐶2 𝐻6 )𝑜 𝑢 𝑡 = (1 − 0.501) 𝑚𝑜𝑙 𝐶2 𝐻6 𝑢𝑛𝑟𝑒𝑎𝑐𝑡𝑒𝑑 85 𝑚𝑜𝑙 𝐶2 𝐻6 𝑓𝑒𝑑

𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑒𝑡ℎ𝑦𝑙𝑒𝑛𝑒 𝑓𝑜𝑟𝑚𝑒𝑑

(𝑛𝐶2 𝐻4 )𝑜 𝑢 𝑡 = 0.471 85 𝑚𝑜𝑙𝑒𝑠 = 40 𝑚𝑜𝑙𝑒𝑠

 BALANCES ON REACTIVE PROCESSES