1. CHAPTER 3 material balance part iI
Sem 1, 2015/2016
ERT 214 Material and Energy Balance / Imbangan Bahan dan Tenaga

2. Content Stoichiometry
Limiting and Excess Reactant, Fractional Conversion and Extent of Reaction
Chemical Equilibrium
Multiple Reaction, Yield and Selectivity
Balance on Reactive System

3. Stoichiometry
Stoichiometry – theory of proportions in which chemical species combine with one another.
Stoichiometric equation of chemical reaction – statement of the relative number of molecules or moles of reactants and products that participate in the reaction.
2 SO2 + O2 ---> 2 SO3
Stoichiometric ratio
ratio of species stoichiometry coefficients in the balanced reaction equation
can be used as a conversion factor to calculate the amount of particular reactant (or product) that was consumed (produced). Eg:
2 mol SO3 generated 2 mol SO2 consumed
2 mol SO2 consumed or 1 mol O2 consumed or etc

4. C4H8 + 6 O2 --------> 4 CO2 + 4 H2O
Example:
C4H8 + 6 O > 4 CO2 + 4 H2O
Is the stochiometric equation balance?
Yes
What is stochiometric coefficient for CO2 4
What is stochiometric ratio of H2O to O2 including it unit
4 mol H2O generated/ 6 mol O2 consumed
How many lb-moles of O2 reacted to form 400lb-moles CO2
400lb-moles CO moles O2 reacted = 600 lb-moles O2 reacted
4 moles CO2 produced
100 mol/min C4H8 fed into reactor and 50% is reacted. At what rate water is formed?
100 mol /min fed 4 mol water generated = 200 mol/min water generated
mol C4H8 reacted

5. Limiting Reactant & Excess Reactant
Limiting reactant : The reactant that would run out if a reaction proceeded to completion, and the other reactants are termed excess reactants.
A reactant is limiting if it is present in less than its stoichiometric proportion relative to every other reactant.
If all reactants are present in stoichiometric proportion, then no reactant is limiting.
Is the amount by which the A in the feed exceeds the amount needed to react completely if the reaction goes to completion.
(nA) feed = is the number of moles of an excess reactant A, present in the feed to a reactor.
(nA) stoich = stoichiometric requirement of A or the amount needed to react completely with the limiting reactant

6. Example C2H2 + 2H2 ------> C2H6
Inlet condition: 20 kmol/h C2H2 and 50 kmol/h H2
What is limiting reactant and fractional excess?
(H2:C2H2) feed = 2.5 : 1 (50:20)
(H2:C2H2) stoich = 2 : 1
H2 is excess reactant and C2H2 is limiting reactant
Fractional excess of H2 = (50-40)/40 = 0.25

7. Fractional Conversion
Fractional Conversion (f)
The fraction unreacted ; 1- f
Example:
If 100 moles of a reactant are fed and 90 moles react, the fractional conversion is 0.90 (the percentage conversion is 90%) and the fraction unreacted is 0.10.
if20 mol/min of a reactant is fed and the percentage conversion is 80%, then (20)(0.80) = 16 mol/min has reacted and (20)( ) = 4 mol/min remains unreacted.

8. Extent of Reaction Extent of Reaction, ξ 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

9. N2 + 3H2 ------------> 2NH3
Example
N2 + 3H > 2NH3
Reactor inlet: 100 mol N2/s; 300 mol H2/s; 1 mol Argon/s (inert gas)
If fractional conversion of H2 0.6, calculate extent of reaction and the outlet composition.
Unreacted H2 or H2 outlet= (1-0.6) 300 = 120 mol H2/s
Solve for extent of reaction : 60 mol/s
120 = 300 – 3 ξ
3 ξ =
3 ξ = 180
ξ = 60

10. Test Yourself C2H4 is limiting because [ ]f < [ ]stoi
The oxidation of ethylene to produce ethylene oxide proceeds according to the equation
2 C2H4 + O >2 C2H4O
The feed to a reactors contains 100kmol C2H4 and 100kmol O2.
a) which is limiting reactant?
[C2H4]f = 100
[O2]f = 100
[C2H4 / O2 ]f = 100 / 100 = 1
[C2H4 / O2 ]stoi = 2 / 1 = 2
[O2 / C2H4 ]f = 100 / 100 = 1
[O2 / C2H4 ]stoi = 1/ 2 = 0.5
C2H4 is limiting because [ ]f < [ ]stoi
O2 is excess because [ ]f > [ ]stoi

11. The feed to a reactors contains 100kmol C2H4 and 100kmol O2.
2 C2H4 + O >2 C2H4O
The feed to a reactors contains 100kmol C2H4 and 100kmol O2.
b) Percentage of excess of the other reactant?
[O2 ]stoi = 100 kmol C2H4 1 mol O2 react = 50 kmol O2 react
2 mol C2H4 react
% excess = [O2 ]f - [O2]stoi = x 100% = 100% (because an AIR)
[O2]stoi

12. The feed to a reactors contains 100kmol C2H4 and 100kmol O2.
2 C2H4 + O >2 C2H4O
The feed to a reactors contains 100kmol C2H4 and 100kmol O2.
c) If the reaction proceeds to completion, how much of the excess reactant will be left; how much C2H40 will be formed; and what is the extent of reaction?
2 mol C2H4 react with 1 mol O2
If 100kmol C2H (½) x 100 kmol = 50 kmol ‘02 react
100kmol O2 feed - 50 kmol ‘02 react = 50 kmol ‘02 left
2 mol C2H4 react produced 2 mol C2H4O
If 100kmol C2H (2/2) x 100 kmol = 100 kmol ‘C2H4O formed
Extent of reaction (y masih tinggal)
[O2 ] = 100 kmol O2 feed – 50 kmol O2 left 1
= 50 kmol O2
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

13. The feed to a reactors contains 100kmol C2H4 and 100kmol O2.
2 C2H4 + O >2 C2H4O
The feed to a reactors contains 100kmol C2H4 and 100kmol O2.
d) if fractional conversion for limiting reactant is 50%, what is outlet composition and extent of reaction?
so, = X/100 = 50kmol C2H4 react
2 C2H4 + O >2 C2H4O
50kmol C2H4 react so, 2mol C2H4 react with 1 mol O2
50kmol C2H4 react with (1//2)x 50 = 25 kmol O2 react
O2 produced (not react) = 100 – 25 = 75 kmol O2
50kmol C2H4 react so, 2mol C2H4 produced 2 mol C2H4
50kmol C2H4 produced (2/2)x 50 = 50 kmol C2H4 produced
Extent of reaction
[O2 ] = 100 kmol O2 feed – 75 kmol O2 left = 25 kmol O2 1

14. The feed to a reactors contains 100kmol C2H4 and 100kmol O2.
2 C2H4 + O >2 C2H4O
The feed to a reactors contains 100kmol C2H4 and 100kmol O2.
e) if reaction proceed to a point where 60kmol O2 left, what is fractional conversion for C2H4? Fractional conversion of O2 and extent of reaction?
-100 kmol O2 feed ….but 60kmol O2 left….so O2 react = =40kmol O2 react
Based on stoi,
1 mol O2 react with 2 mol C2H4
if 40 kmol O x 2 = 80 kmol C2H4 react
fC2H4= 80/100 =0.8
fO2= 40/100= 0.4
Extent of reaction
[O2 ] = 100 kmol O2 feed – 60kmol O2 left = 40 kmol O2 1

15. Chemical Equilibrium
For a given set reactive species and reaction condition, two fundamental question might be ask:
1. What will be the final (equilibrium) composition of the reaction mixture?
– chemical engineering thermodynamics deals with this
2. How long will the system take to reach a specified state short of equilibrium?
– chemical kinetics deals with this
Irreversible reaction
reaction proceeds only in a single direction (from reactants to products)
the concentration of the limiting reactant eventually approaches zero.
Reversible reaction
reactants form products for forward reaction and products undergo the reverse reactions to reform the reactants.
Equilibrium point is a rate of forward reaction and reverse reaction are equal

16. Multiple Reaction
Some of the chemical reaction has a side reaction which is formed undesired product- multiple reaction occurred.
Effects of this side reaction might be:
Economic loss
Less of desired product is obtained for a given quantity of raw materials
Greater quantity of raw materials must be fed to the reactor to obtain a specified product yield.

17. Multiple Reaction

18. Selectivity selectivity = moles of desired product
moles of undesired product

19. Yield 3 definition of yield with different working definition Yield =
Moles of desired product formed
Moles that would have been formed if there were no side reaction and the limiting reactant had reacted completely
Yield =
Moles of desired product formed
Moles of reactant fed
Yield =
Moles of desired product formed
Moles of reactant consumed

20. Extent of Reaction for Multiple Reaction
Concept of extent of reaction can also be applied for multiple reaction
only now each independent reaction has its own extent.

21. Example: Yield and Selectivity in a Dehydrogenation Reactor
The reactions
take place in a continuous reactor at steady state. The feed contains 85.0 mole% ethane (C2H6)
and the balance inerts (I). The fractional conversion of ethane is 0.501, and the fractional yield of
ethylene is Calculate the molar composition of the product gas and the selectivity of ethylene to methane production.
SOLUTION:
Basis: 100 mol Feed
fC2H6= 0.501
Yield, Y C2H4 = 0.471

22. the outlet component amounts in terms of extents of reaction are as follows:
nl (mol C2H6) = 85.0 mol C2H6 – ξ1- ξ2
n2(mol C2H4) = ξ1
n3(mol H2) = ξ1- ξ2
n4(mol CH4 ) = 2ξ2
n5(mol I) = 15.0 mol I(Inert) ξ1 ξ2

23. Ethane Conversion If the fractional conversion of ethane is 0
Ethane Conversion If the fractional conversion of ethane is 0.501, the fraction unconverted (and hence leaving the reactor)must be ( ). Ethylene Yield
Given : fC2H6= 0.501
Given : Yield, Y C2H4 = 0.471

24. Product:

25. balance on reactive process
Sem 1, 2015/2016
ERT 214 Material and Energy Balance / Imbangan Bahan dan Tenaga

26. Balance of Reactive Processes
Balance on reactive process can be solved based on three method:
Atomic Species Balance
Extent of Reaction
Molecular Species Balance
Each approach leads to the same results, but anyone of them may be more convenient for a given calculation so it is a good idea to become comfortable with all three.

27. 1. Atomic Species Balance
No. of unknowns variables
- No. of independent atomic species balance
- No. of molecular balance on indep. nonreactive species
- No. of other equation relating the variable
= No. of degree of freedom

28. 2. Extent of Reaction No. of unknowns variables
+ No. of independent chemical reaction
- No. of independent reactive species
- No. of independent nonreactive species
- No. of other equation relating the variable
= No. of degree of freedom

29. 3. Molecular Species Balance
No. of unknowns variables
+ No. of independent chemical reaction
- No. of independent molecular species balance
- No. of other equation relating the variable
= No. of degree of freedom

30. Degree-of-Freedom Analysis
To carry out degree-of-freedom analyses of reactive systems you must first understand the concepts of :
independent equations,
independent species, and
independent chemical reactions. ?

31. 1. Independent Equation
Algebraic equation are independent if we cannot obtain any one of them by adding and subtracting multiples of any of the others
Eg:
x + 2y = 4 [1]
3x + 6y = 12 [2]
Only one independent equation because [2]= 3 x [1]
2x – z= 2 [2]
4y + z= 6 [3]
Although 3 equation, but only two independent equation exist because [3]=2x[1] –[2]

32. 2. Independent Species
a) If two MOLECULAR species are in the SAME RATIO to each other wherever they appear in a process, balance on those species will not be independent (i.e. only one independent equation is get)
Similarly
b) If two ATOMIC species are in the SAME RATIO to each other wherever they appear in a process, balance on those species will not be independent (i.e. only one independent equation is get)

33. a) Independent Molecular Species
If two MOLECULAR species are in the SAME RATIO to each other wherever they appear in a process, balance on those species will not be independent (i.e. only one independent equation is get)

34. Let make a molecular balance on both species to prove it
Example:
Since N2 and O2 have a same ratio wherever they appear on the flowchart (3.76 mol N2/ mol O2), only ONE independent balance can obtain.
Let make a molecular balance on both species to prove it
Balance on O2:
n1=n3 [1]
Balance on N2:
3.76 n1=3.76n3
n1=n3 [2]
Eq. [1] and [2] are SAME. Only ONE INDEPENDENT EQUATION OBTAIN although two species involved.

35. b) Independent Atomic Species
If two ATOMIC species are in the SAME RATIO to each other wherever they appear in a process, balance on those species will not be independent (i.e. only one independent equation is get)

36. Atomic N and O are always in same proportion to each other in the process (3.76:1), similar for atom C and Cl which always same ratio too (1:4).
Although FOUR atomic species exist, only TWO independent equation can obtain for this cases.
Prove:
Balance on atomic O: 2n1=2n3
n1=n3 [1]
Balance on atomic N: 2(3.76)n1=2(3.76)n3
n1=n3 [2]
Balance on atomic C: n2=n4+n5 [3]
Balance on atomic Cl: 4n2=4n4 +4n5
n2=n4+n5 [4]
Eq. [1]=[2] and [3]=[4], only TWO independent equation (Eq n1=n3 and n2=n4+n5 )obtained

37. 3. Independent Reaction
Used when we using either molecular species balance or extent of reaction method to analyze a balance on reactive process
Chemical reaction are independent if the stoichiometric equation of any one of them cannot be obtained by adding and subtracting multiples of the stoichiometric equations of the others
Only TWO independent eq. can obtained although three equation exist sice [3]=[1] + 2[2].

38. Example: Atomic Species Balance
Dehydrogenation of ethane in a steady-state continuous reactor. The reaction is
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
All atomic balance is INPUT=OUTPUT
Degree-of-freedom analysis
40 kmol H2/min
n1 kmol C2H6/min
n2 kmol C2H4/min
Reactor
100 kmol C2H6/min

40. Example: Molecular Species Balance
Dehydrogenation of ethane in a steady-state continuous reactor. The reaction is
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
40 kmol H2/min
n1 kmol C2H6/min
n2 kmol C2H4/min
Reactor
100 kmol C2H6/min
Degree-of-freedom analysis
2 unknowns variables (n1, n2)
+1 independent chemical reaction
- 3 independent molecular species balance (C2H6, C2H4, H2)
- 0 other equation relating the variable
=============================
0 No. of degree of freedom

41. Reactor 100 kmol C2H6/min 40 kmol H2/min n1 kmol C2H6/min
Dehydrogenation of ethane in a steady-state continuous reactor. The reaction is
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
40 kmol H2/min
n1 kmol C2H6/min
n2 kmol C2H4/min
Reactor
100 kmol C2H6/min

42. Example: Extent of Reaction
Dehydrogenation of ethane in a steady-state continuous reactor. The reaction is
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
40 kmol H2/min
n1 kmol C2H6/min
n2 kmol C2H4/min
Reactor
100 kmol C2H6/min
Degree-of-freedom analysis
2 unknowns variables (n1,n2)
+ 1 independent chemical reaction
- 3 independent reactive species (C2H6, C2H4, H2)
- 0 independent nonreactive species
- 0 other equation relating the variable
=============================
0 No. of degree of freedom

43. Write extent of reaction for each species C2H6 :n1 = 100-ξ C2H4 :n2= ξ
Dehydrogenation of ethane in a steady-state continuous reactor. The reaction is
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
40 kmol H2/min
n1 kmol C2H6/min
n2 kmol C2H4/min
Reactor
100 kmol C2H6/min
Write extent of reaction for each species
C2H6 :n1 = 100-ξ
C2H4 :n2= ξ
H2 :40= ξ
Solve for n1 and n2 (ξ =40)
n1= 60 kmol C2H6/min; n2= 40 kmol C2H4/min

44. Product Separation & Recycle
Overall Conversion
Reactant input to Process – reactant output from Process
Reactant input to Process
Single Pass Conversion
Reactant input to Reactor – reactant output from Reactor
Reactant input to Reactor

45. Example: Product Separation & Recycle
For example, consider the following labeled flowchart for a simple chemical process based on the reaction A B:
75 mol B/min
100 mol A/min
75 mol A/min
Reactor
Product Separation Unit
25 mol A/min

46. Product Separation Unit
Purging
To prevent any inert or insoluble substance build up and accumulate in the system
Purge stream and recycle stream before and after the purge have a same composition.
Product
Fresh Feed
Reactor
Product Separation Unit
Recycle
Purge

47. THANK YOU