1. Process Classification
Fundamentals of Material Balances
Process Classification

2. Balances on Reactive Systems
Material balance no longer takes the form
INPUT = OUTPUT
Must account for the disappearance of reactants and appearance of products through stoichiometry.

3. Stoichiometric Equations
The stoichiometric equation of a chemical reaction is a statement of the relative amounts of reactants and products that participate in the reaction.
2 SO2 + O2 → 2 SO3
A stoichiometric equation is valid only if the number of atoms of each atomic species is balanced.
2 S → 2 S
4 O + 2 O → 6 O

4. Stoichiometric Equations
The stoichiometric equation of a chemical reaction is a statement of the relative amounts of reactants and products that participate in the reaction.
2 SO2 + O2 → 2 SO3
A stoichiometric ratio of two molecular species participating in a reaction is the ratio of their stoichiometric coefficients:
2 mol SO3 generated / 1 mol O2 consumed
2 mol SO3 generated / 2 mol SO2 consumed

5. Stoichiometric Equations
C4H8 + 6 O2 → 4 CO2 + 4 H2O
Is this stoichiometric equation balanced?
What is the stoichiometric coefficient of CO2?
What is the stoichiometric ratio of H2O to O2?
How many lb-mol O2 react to form 400 lb-mol CO2?
100 lbmol/min C4H8 is fed and 50% reacts. At what rate is water formed?

6. Limiting and Excess Reactants
Two reactants are said to be in stoichiometric proportion if the ratio (moles A present/moles B present) equals the stoichiometric ratio from the balanced reaction equation.
2 SO2 + O2 → 2 SO3
the feed ratio that would represent stoichiometric proportion is nSO2/nO2 = 2:1
If reactants are fed in stoichometric proportion, and the reaction proceeds to completion, all reactants are consumed.

7. Limiting and Excess Reactants
Stoichiometric Proportion – Reactants are present in a ratio equivalent to the ratio of the stoichiometric coefficients.
A + 2B → 2C

8. Limiting and Excess Reactants
Limiting reactant – A reactant is limiting if it is present in less than stoichiometric proportion relative to every other reactant.
A + 2B → 2C
Excess reactant – All other reactants besides the limiting reactant.

9. Limiting and Excess Reactants
fractional excess (fXS) – ratio of the excess to the stoichiometric proportion.
A + 2B → 2C

10. Limiting and Excess Reactants
fractional conversion (f) – ratio of the amount of a reactant reacted, to the amount fed.
A + 2B → 2C

11. Limiting and Excess Reactants
fractional conversion (f) – ratio of the amount of a reactant reacted, to the amount fed.
A + 2B → 2C

12. Limiting and Excess Reactants
fractional conversion (f) – ratio of the amount of a reactant reacted, to the amount fed.
A + 2B → 2C

13. Limiting and Excess Reactants
fractional conversion (f) – ratio of the amount of a reactant reacted, to the amount fed.
A + 2B → 2C

14. Limiting and Excess Reactants
fractional conversion (f) – ratio of the amount of a reactant reacted, to the amount fed.
A + 2B → 2C

15. Extent of Reaction A + 2B → 2C
extent of reaction (ξ) – an extensive quantity describing the progress of a chemical reaction ν – stoichiometric coefficients: νA = -1, νB = -2, νC = 2
A + 2B → 2C

16. Extent of Reaction A + 2B → 2C
extent of reaction (ξ) – an extensive quantity describing the progress of a chemical reaction ν – stoichiometric coefficients: νA = -1, νB = -2, νC = 2
A + 2B → 2C

17. Extent of Reaction A + 2B → 2C
extent of reaction (ξ) – an extensive quantity describing the progress of a chemical reaction ν – stoichiometric coefficients: νA = -1, νB = -2, νC = 2
A + 2B → 2C

18. Extent of Reaction A + 2B → 2C
extent of reaction (ξ) – an extensive quantity describing the progress of a chemical reaction ν – stoichiometric coefficients: νA = -1, νB = -2, νC = 2
A + 2B → 2C

19. Extent of Reaction A + 2B → 2C
extent of reaction (ξ) – an extensive quantity describing the progress of a chemical reaction ν – stoichiometric coefficients: νA = -1, νB = -2, νC = 2
A + 2B → 2C

20. ethylene Assume an equimolar reactant feed of 100 kmol:
What is the limiting reactant?
A reactant is limiting if it is present in less than stoichiometric proportion relative to every other reactant.
ethylene

21. Assume an equimolar reactant feed of 100 kmol:
What is the percentage excess of each reactant?

22. Assume an equimolar reactant feed of 100 kmol:
If the reaction proceeds to completion:
(a) how much of the excess reactant will be left?
(b) How much C2H4O will be formed?
(c) What is the extent of reaction?

23. Assume an equimolar reactant feed of 100 kmol:
If the reaction proceeds to a point where the fractional conversion of the limiting reactant is 50%, how much of each reactant and product is present at the end? What is ξ?

24. Assume an equimolar reactant feed of 100 kmol:
If the reaction proceeds to a point where 60 mol of O2 is left, what is the fractional conversion of C2H4? What is ξ?

29. Reaction Stoichiometry
Acrylonitrile produced by reaction of ammonia, propylene, and O2 at 30% conversion of limiting reactant:
determine limiting reactant
limiting

30. Acrylonitrile produced by reaction of ammonia, propylene, and O2 at 30% conversion of limiting reactant:
determine fractional excesses
limiting

31. Acrylonitrile produced by reaction of ammonia, propylene, and O2 at 30% conversion of limiting reactant:
limiting
use fractional conversion to determine amount of propylene that leaves the reactor

32. Acrylonitrile produced by reaction of ammonia, propylene, and O2 at 30% conversion of limiting reactant:
limiting
determine extent of reaction by applying mole balance to propylene

33. Acrylonitrile produced by reaction of ammonia, propylene, and O2 at 30% conversion of limiting reactant:
apply mole balance to all remaining species

34. Fuel for motor vehicles other than gasoline are being eyed because they generate lower levels of pollutants than does gasoline. Compressed propane has been suggested as a source of economic power for vehicles. Suppose that in a test 22 kg of C3H8 is burned with 400 kg of air to produce 44 kg of CO2 and 12 kg of CO.
a. What was the percent excess air?
b. The components in the product stream.

35. C3H8 + 5O2 3CO2 + 4H2O Component Amount (kg) Molecular weight Input
Output
C3H8 22 44
0.50 ?
Air
400 29
13.80 O2
0.21*13.80=2.90 N2
0.79*13.80=10.9
CO2 - 1 CO 12 28
0.43
H2O. 18
Total
29.87
Percent excess air = (O2 in – O2 used)/(O2 theory)
= (2.90 – 5*0.5)/(5*0.5)=16%

36. C3H8 + 5O2 3CO2 + 4H2O C3H8 + 7/2O2 3CO + 4H2O [1] [2] Component Input
(kg-mol)
Consumed [1]
Consumed [2]
Generated [1]
Generated [2]
Total Output
C3H8
0.50
0.33
0.43/3=0.14 -
0.03 O2
0.21*13.80=2.90
5*0.33= 1.65
3.5*0.14= 0.49
0.76 N2
0.79*13.80=10.9
10.9
CO2 1 CO
0.43
H2O.
4*0.33= 1.32
4*0.14= 0.56
1.88
Total
29.87 15

37. Example C2H6 + 7/2O2 2CO2 + 3H2O [1] C2H6 + 5/2O2 2CO + 3H2O [2]
A gas stream containing 80 mol% C2H6 and 20 mol% O2 is burned in an engine with 200% excess air. If 80% of the ethane goes to CO2, 10% goes to CO and the remaining unburned, determine the amount of the excess air per 100 moles of the feed gas.
C2H /2O CO H2O [1]
C2H /2O CO + 3H2O [2]

38. C2H6 + 7/2O2 2CO2 + 3H2O Basis=100 mole feed gas [1]
Component
Input Gas Stream
Output
(kg-mol)
C2H6 80 O2
*Gair= 799.7
Air
Gair= 4.76*780 = 3,712.8 N2
0.79*Gair= 2,933.1
CO2 CO
H2O.
Total
Percent excess air (based on air):
(O2 in – O2 used)/(O2 theory)=2
O2 theory (Eq 1) = 3.5*80=280
O2 used (air) =O2 theory (Eq 1)-20=260 moles
(O2 in (air)-260)/(260)=2
O2in (air)=260+(2*260)=780moles
N2 in = 3.76*780=2,932.8
Air stream=2, =4.76*780=3,712.8

39. C2H6 + 7/2O2 2CO2 + 3H2O [1] C2H6 + 5/2O2 2CO + 3H2O [2] Component
Input
(mol)
Consumed [1]
Consumed [2]
Generated [1]
Generated [2]
Total Output
C2H6 80 64 8 - O2
799.7
224 20
204 N2
2,933.1
2933.1
CO2
128 CO 16
H2O.
192 24
216
Total
3812.8
3,505.1

40. Chemical Equilibrium Given Determine a set of reactive species, and
reaction conditions
Determine
the final (equilibrium) composition of the reaction mixture
how long the system takes to reach a specified state short of equilibrium

41. Chemical Equilibrium Irreversible reaction
reaction proceeds only in a single direction A → B
concentration of the limiting reactant eventually approaches zero (time duration can vary widely)
Equilibrium composition of an irreversible reaction is that which corresponds to complete conversion.

42. Chemical Equilibrium Reversible reaction
reaction proceeds in both directions A ↔ B
net rate (forward – backward) eventually approaches zero (again, time can vary widely)
Equilibrium composition of a reversible reaction is that which corresponds to the equilibrium conversion.

43. Equilibrium Composition
An equilibrium reaction proceeds to an extent at temperature T based on the equilibrium constant, K(T).
where yi is the mole fraction of species i
Water-gas shift reaction:
Assume 1 mole CO and 2 mole H2O
K(1105 K) = 1.00

44. Equilibrium Composition
Water-gas shift reaction:
Assume 1 mole CO and 2 mole H2O
K(1105 K) = 1.00

45. Equilibrium Composition
Water-gas shift reaction:
Assume 1 mole CO and 2 mole H2O
K(1105 K) = 1.00

46. Equilibrium Composition
Water-gas shift reaction:
Assume 1 mole CO and 2 mole H2O
K(1105 K) = 1.00

47. Water-gas shift reaction:
limiting reactant is CO
Water-gas shift reaction:
Assume 1 mole CO and 2 mole H2O
K(1105 K) = 1.00
at equilibrium,
fractional conversion at equilibrium

48. Multiple Reactions
for j reactions of i species,
mole balance becomes

49. Multiple Reactions
for j reactions of i species,
mole balance becomes

50. Multiple Reactions 100 moles A fed to a batch reactor
product composition: 10 mol A, 160 B, 10 C
What is
fA?
YB?
SB/C?
ξ1, ξ2

51. Multiple Reactions 100 moles A fed to a batch reactor
product composition: 10 mol A, 160 B, 10 C
What is
fA?
YB?
SB/C?
ξ1, ξ2

52. 100 moles A fed to a batch reactor
product composition: 10 mol A, 160 B, 10 C
What is
fA?
YB?
SB/C?
ξ1, ξ2

53. What is fA? YB? SB/C? ξ1, ξ2 100 moles A fed to a batch reactor
product composition: 10 mol A, 160 B, 10 C
What is
fA?
YB?
SB/C?
ξ1, ξ2

54. Balances on Reactive Processes
Continuous, steady-state dehydrogenation of ethane
Total mass balance still has INPUT = OUTPUT form
Molecular balances contain consumption/generation
Atomic balances (H and C) also have simple form

55. Balances on Reactive Processes
Continuous, steady-state dehydrogenation of ethane
First consider molecular balances:
Molecular H2 balance: generation = output
generation H2 = 40 kmol H2/min
C2H6 balance: input = output + consumption
100 kmol C2H6/min = n1 + (C2H6 consumed)
C2H4 balance: generation = output
(C2H4 generated) = n2

56. Balances on Reactive Processes
Continuous, steady-state dehydrogenation of ethane
Atomic C balance: input = output
Atomic H balance: input = output

57. Independent Equations
To understand the number of independent species balances in a reacting system requires an understanding of independent algebraic equations.
Algebraic equations are independent if you cannot obtain any of them by adding/subtracting multiples of the others.
3×[1] = [2]
2×[3] – [4] = [5]

58. Independent Equations
To understand the number of independent species balances in a reacting system requires an understanding of independent algebraic equations.
Algebraic equations are independent if you cannot obtain any of them by adding/subtracting multiples of the others.

59. Independent Species
If two molecular or atomic species are in the same ratio to each other where ever they appear in a process and this ratio is incorporated in the flowchart labeling, balances on those species will not be independent equations.

60. Independent Chemical Reactions
When using molecular species balances or extents of reaction to analyze a reactive system, the degree of freedom analysis must account for the number of independent chemical reactions among the species entering and leaving the system.

61. Independent Chemical Reactions
Chemical reactions 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.
2×[2] + [1] = [3]

62. Solving Reactive Systems
There are 3 possible methods for solving balances around a reactive system:
Molecular species balances require more complex calculations than the other methods and should be used only for simple (single reaction) systems.
Atomic species balances generally lead to the most straightforward solution procedure, especially when more than one reaction is involved
Extents of reaction are convenient for chemical equilibrium problems.

63. Molecular Species Balances
To use molecular species balances to analyze a reactive system, the balances must contain generation and/or consumption terms.
The degree-of-freedom analysis is as follows:
# unknown labeled variables
+ # independent chemical reactions
- # independent molecular species balances
- # other equations relating unknown variables
# of degrees of freedom

64. Molecular Species Balances
2 unknown labeled variables
+ 1 independent chemical reactions
- 3 independent molecular species balances
- 0 other equations relating unknown variables
0 degrees of freedom
at steady state

65. Molecular Species Balances
at steady state
H2 Balance: generation = output
C2H6 Balance: input = output + consumption
C2H4 Balance: generation = output

66. Atomic Species Balance
All atomic species balances take the form
INPUT = OUTPUT
Degree-of-freedom analysis, ndf =
# unknown labeled variables
- # independent atomic species balances
- # molecular balances on independent nonreactive species
- # other equations relating unknown variables

67. Atomic Species Balance
All atomic species balances take the form
INPUT = OUTPUT
Degree-of-freedom analysis, ndf = 0 =
2 unknown labeled variables
- 2 independent atomic species balances
- 0 molecular balances on independent nonreactive species
- 0 other equations relating unknown variables

68. Atomic Species Balance
C Balance: input = output
H Balance: input = output

69. Atomic Species Balance
Solve simultaneously

70. Extent of Reaction
The 3rd method by which to determine molar flows in a reactive system is using expressions for each species flow rate in terms of extents of reaction (ξ).
Degree-of-freedom analysis for such an approach:
ndf = # of unknown labeled variables
+ # independent reactions
- # independent nonreactive species
- # other relationships or specifications

71. Incomplete Combustion of CH4
Methane is burned with air in a continuous steady-state reactor to yield a mixture of carbon monoxide, carbon dioxide, and water.
The feed to the reactor contains 7.80 mol% CH4, 19.4 mol% O2, 72.8 mol% N2. Methane undergoes 90.0% conversion, and the effluent gas contains 8 mol CO2 per mole CO.

72. + 2 independent reactions
fCH4 = 0.9
The feed to the reactor contains 7.80 mol% CH4, 19.4 mol% O2, 72.8 mol% N2. Methane undergoes 90.0% conversion, and the effluent gas contains 8 mol CO2 per mole CO.
ndf = 5 unknowns
+ 2 independent reactions
- 5 expressions for ξ (CH4, O2, CO, CO2, H2O)
- 1 nonreactive species balance (N2)
- 1 specified methane conversion =0

73. fCH4 = 0.9
N2 balance: nonreactive species, INPUT = OUTPUT CH4 conversion specification:

74. Extent of reaction mole balances:

75. Product Separation and Recycle
Two definitions of reactant conversion are used in the analysis of chemical reactors with product separation and recycle of unconsumed reactants.

77. Catalytic Propane Dehydrogenation
95% overall
conversion

78. Catalytic Propane Dehydrogenation
95% overall
conversion
Overall
Process
ndf = 3 unknowns (n6, n7, n8)
– 2 independent atomic balances (C and H)
– 1 relation (overall conversion)
= 0
consider n6, n7, n8 known for further DOF analyses

79. Catalytic Propane Dehydrogenation
95% overall
conversion
ndf = 2
Mixing
point
ndf = 4 unknowns (n9, n10, n1, n2)
– 2 balances (C3H8 and C3H6)
= 2

80. Catalytic Propane Dehydrogenation
95% overall
conversion
ndf = 3
ndf = 2
reactor
ndf = 5 unknowns (n3, n4, n5, n1, n2)
– 2 balances (C and H)
= 3

81. Catalytic Propane Dehydrogenation
95% overall
conversion
ndf = 3
ndf = 0
ndf = 2
separator
ndf = 5 unknowns (n3, n4, n5, n9, n10)
– 3 balances (C3H8, C3H6, and H2)
– 2 relations (reactant and product recovery fractions)
= 0

82. Catalytic Propane Dehydrogenation
95% overall
conversion
overall
conversion
relationship

83. Catalytic Propane Dehydrogenation
95% overall
conversion
overall
C atomic balance

84. Catalytic Propane Dehydrogenation
95% overall
conversion
overall
H atomic balance

85. Catalytic Propane Dehydrogenation
95% overall
conversion
separator
given relations

86. Catalytic Propane Dehydrogenation
95% overall
conversion
separator
propane balance

87. Catalytic Propane Dehydrogenation
95% overall
conversion
mixer
propane balance

88. Catalytic Propane Dehydrogenation
95% overall
conversion
mixer
propylene balance

89. Catalytic Propane Dehydrogenation
95% overall
conversion
reactor
C atomic balance

90. Catalytic Propane Dehydrogenation
95% overall
conversion
reactor
H atomic balance

91. Catalytic Propane Dehydrogenation
95% overall
conversion
single-pass
conversion

92. Catalytic Propane Dehydrogenation
95% overall
conversion
Recycle ratio

93. Purging
Necessary with recycle to prevent accumulation of a species that is both present in the fresh feed and is recycled rather than separated with the product.
mixed fresh feed
and recycle is a
convenient
basis selection

94. Methanol Synthesis
ndf = 7 unknowns (n0, x0C, np, x5C, x5H, n3, n4) + 1 rxn
- 5 independent species balances = 3

95. Methanol Synthesis ndf = 4 unknowns (n1, n2, n3, n4) + 1 rxn
– 4 independent species balances
– 1 single pass conversion = 0

96. ndf = 3 unknowns (n5, x5C, X5H)
– 3 independent species balances
= 0

97. ndf = 3 unknowns (n0, x0C, nr)
– 3 independent species balances
= 0

98. – 1 independent species balance = 0
investigate
mole balances and
their solution in
the text
ndf = 1 unknowns (np)
– 1 independent species balance
= 0

99. Combustion Reactions
Combustion - rapid reaction of a fuel with oxygen.
Valuable class of reactions due to the tremendous amount of heat liberated, subsequently used to produce steam used to drive turbines which generates most of the world’s electrical power.
Common fuels used in power plants:
coal
fuel oil (high MW hydrocarbons)
gaseous fuel (natural gas)
liquified petroleum gas (propane and/or butane)

100. Combustion Chemistry When a fuel is burned
C forms CO2 (complete) or CO (partial combustion)
H forms H2O
S forms SO2
N forms NO2 (above 1800°C)
Air is used as the source of oxygen. DRY air analysis:
78.03 mol% N2
20.99 mol% O2
0.94 mol% Ar
0.03 mol% CO2
0.01 mol% H2, He, Ne, Kr, Xe
usually safe to assume:
79 mol% N2
21 mol% O2

101. Combustion Chemistry
Stack (flue) gas – product gas that leaves a furnace.
Composition analysis:
wet basis – water is included in mole fractions
dry basis – does not include water in mole fractions
Stack gas contains (mol) on a wet basis:
60.0% N2, 15.0% CO2, 10.0% O2, 15.0% H2O
Dry basis analysis:
60/( ) = mol N2/mol
15/( ) = mol CO2/mol
10/( ) = mol O2/mol

102. Combustion Chemistry assume 100 mole dry gas basis
Stack gas contains (mol) on a dry basis:
65% N2, 14% CO2, 10% O2, 11% CO
assume 100 mole dry gas basis
7.53 mole H2O
65 mole N2
14 mole CO2
10 mole O2
11 mole CO
total = mole

103. Theoretical and Excess Air
The less expensive reactant is commonly fed in excess of stoichiometric ratio relative to the more valuable reactant, thereby increasing conversion of the more expensive reactant at the expense of increased use of excess reactant.
In a combustion reaction, the less expensive reactant is oxygen, obtained from the air. Conseqently, air is fed in excess to the fuel.
Theoretical oxygen is the exact amount of O2 needed to completely combust the fuel to CO2 and H2O.
Theoretical air is that amount of air that contains the amount of theoretical oxygen.

104. Theoretical and Excess Air
Excess air is the amount by which the air fed to the reactor exceeds the theoretical air.

105. Theoretical and Excess Air
nC4H10 = 100 mol/hr; nair = 5000 mol/hr

106. Combustion Reactors
Procedure for writing/solving material balances for a combustion reactor
When you draw and label the flowchart, be sure the outlet stream (the stack gas) includes
unreacted fuel (unless the fuel is completely consumed)
unreacted oxygen
water and carbon dioxide (and CO if combustion is incomplete)
nitrogen (if air is used as the oxygen source)
Calculate the O2 feed rate from the specified percent excess oxygen or air
If multiple reactions, use atomic balances

107. Combustion of Ethane degree-of-freedom analysis ndf = 7 unknowns
fC2H6 = 0.9
degree-of-freedom
analysis
ndf = unknowns
- 3 atomic balances
- 1 nitrogen balance
- 1 excess air specification
- 1 ethane conversion specification
- 1 CO/CO2 ratio specification
= 0
25% of the ethane burned forms CO

108. excess air specification
fC2H6 = 0.9
25% of the ethane burned forms CO
Ethane conversion specification

109. CO/CO2 ratio specification
fC2H6 = 0.9
25% of the ethane burned forms CO
Nitrogen balance
Atomic C balance

110. fC2H6 = 0.9
atomic H balance
atomic O balance

111. fC2H6 = 0.9
25% of the ethane burned forms CO
atomic O balance

112. stack gas composition (dry basis)
fC2H6 = 0.9
25% of the ethane burned forms CO
stack gas composition (dry basis)