1. Chemical Reaction Engineering
Chapter 3, Part 1:
Rate Laws
Dr. Khashayar Nasrifar

2. Rate Law: Objectives Part I
Write the relationship between the relative rates of reaction.
Write a rate law and define reaction order and activation energy.
Part II
Set up a stoichiometric table for both batch and flow systems and express concentration as a function or conversion.
Write -rA solely as a function of conversion given the rate law and then entering concentration.
Calculate the equilibrium conversion for both gas and liquid phase reactions.

3. Why a Rate Law? X
If we have
Then we can size a number of CSTR and PFR reaction systems
To find -rA= f(X)
Need the rate law, -rA=f(CA, CB)
Need the reaction stoichiometry, CA=CA0(1-X)

4. Rate Law: Definitions
Homogeneous reaction: any of a class of chemical reactions that occur in a single phase (gaseous, liquid, or solid). The most important of homogeneous reactions are the reactions between gases (e.g., the combination of common household gas and oxygen to produce a flame) and the reactions between liquids or substances dissolved in liquids (e.g., the reactions between aqueous solutions of acids and bases).
Heterogeneous reaction: are those reactions involve more than one phase

5. Elementary Reaction: a reaction that cannot be broken down into smaller steps. A reaction for which no reaction intermediates have been detected or need to be postulated in order to describe the chemical reaction on a molecular scale. An elementary reaction is assumed to occur in a single step and to pass through a single transition state.
Non-elementary Reaction

6. Rate Law: Definitions
Reversible reaction: a chemical reaction that can proceed in either direction
AB
Irreversible reaction:
AB

7. Rate Law: Definitions
Molecularity: The number of molecules consumed in a chemical reaction
Unimolecular: When a single molecules is consumed
92U238  90Th He4
Bimolecular: When two molecules are consumed.
Br• + C2H6  HBr + C2H5•
Termolecular: when three or more reactants and consumed in single step of the reaction.
2NO +O2  2NO2

8. Rate Law: Relative rates of reaction
Consider the general equation
The relative rate of reaction of the various species involved in the reaction can be obtained from the ratio of the stoichiometric coefficients.
For every mole of A consumed c/a moles of C appears.
Rate of formation of C= (c/a) rate of disappearance of A
The relationship can be expressed directly from the stoichiometry of the reaction

9. Rate Law: Reaction Order
A rate law describes the behavior of a reaction.
The rate of a reaction is a function of temperature (through the rate constant) and concentration.
Power Law Model
= n

10. Reaction Order
You can tell the overall reaction order by the units of k C -r
Reaction Order
Rate Law k A A
(mol/dm 3 )
(mol/dm 3
*s)
zero -r
= k
(mol
/dm3*s) A
1st s -1 -r
= kC A A
2nd -r
= kC 2
(dm3
/mol*s) A A
k is the specific reaction rate (constant)

11. Rate Law Basics k is given by the Arrhenius Equation: Where:
E = activation energy (cal/mol)
R = gas constant (cal/mol*K)
T = temperature (K)
A = frequency factor (units of A, and k, depend on overall reaction order)

12. Rate Law Basics
To find the activation energy, plot k as a function of temperature:
Where:

13. One can also write the specific reaction rate as:
Taking the ratio:
     

14. Activation Energy
The activation energy can be thought of as a barrier to the reaction.
One way to view the barrier to a reaction is through the reaction coordinates. These coordinates denote the energy of the system as a function of progress along the reaction path.
For the reaction
the reaction coordinate is

15. For the reaction to occur, the reactants must overcome an energy barrier EB. The energy barrier is related to the activation energy, E. The barrier height is a result of
(1) the molecules needing energy to distort or stretch their bonds in order to break them and to thus form new bonds,
(2) the reaction molecules come close together they must overcome both steric and electron repulsion forces, and
(3) the quantum effects that can in some cases produce a barrier.

16. Activation Energy
The activation energy is a measure of how temperature sensitive the reaction is. Reactions with large activations energies are very temperature sensitive.

17. Example Consider the following elementary reactions
1)   Which reaction has the higher activation energy?
2)   Which reactions have the same activation energy?
3)   Which reaction is virtually temperature insensitive?
4)   Which reaction will dominate (i.e. take place the fastest) at high temperatures?

18. Elementary rate law
A reaction follows an elementary rate law if and only if the stoichiometric coefficients are the same as the individual reaction order of each species.
For the reaction in the previous example,
The rate law would be:

19. Non-Elementary Rate Laws
If the rate law for the non-elementary reaction
is found to be
then the reaction is said to be 2nd order in A, 1st order in B,
and order overall is 2+1 =3 (3rd order).
Most reactions are non-elementary, especially catalytic reactions. This means the kinetic equation is empirical in nature. A generic kinetic equation has the form of Langmuir-Hinshelwood kinetics.

20. KC = kforward / kreverse
Reversible Reactions
aA + bB   cC + dD
Reversible reactions consist of 2 reactions (forward and backward or reverse reactions).
The overall reaction is the net reaction of 2 reactions.
The rate at equilibrium is Zero
KC = kforward / kreverse

21. Examples of Rate Laws First Order Reactions
(1) Homogeneous irreversible elementary gas phase reaction
(2) Homogeneous reversible elementary reaction

22. Examples of Rate Laws Second Order Reactions
(1) Homogeneous irreversible non-elementary reaction
This is first order in ONCB, first order in ammonia and overall second order.
At 188˚C

23. Examples of Rate Laws Second Order Reactions
(2) Homogeneous irreversible elementary reaction
This reaction is first order in CNBr, first order in CH3NH2 and overall second order.

24. Examples of Rate Laws
(3) Heterogeneous catalytic reaction: The following reaction takes place over a solid catalyst:

25. Part II
Stoichiometry

26. Stoichiometry
If the rate law depends on more than one species we must relate the concentrations of different species to each other
This relationship is most easily established with the aid of a stoichiometric table
The stoichiometric table presents the stoichiometric relationship between reacting molecules for a single reaction

27. Reaction Stoichiometry
The relative rate of reaction of the various species involved in the reaction can be obtained from the ratio of the stoichiometric coefficients.
For every mole of A consumed c/a moles of C appears.
Rate of formation of C= (c/a) × rate of disappearance of A
The relationship can be expressed directly from the stoichiometry of the reaction

28. Batch Stoichiometric Table
NA0
NB0
NC0
ND0
NI0
t=t NA NB NC ND NI

29. Batch Stoichiometric Table
Species
Symbols
Initial
Change
Remaining A
NA0
-NA0X
NA=NA0(1–X) B
NB0=NA0ΘB
-(b/a)NA0X
NB=NA0(ΘB –(b/a)X) C
NC0=NA0ΘC
(c/a)NA0X
NC=NA0(ΘC+(c/a)X) D
ND0=NA0ΘD
(d/a)NA0X
ND=NA0(ΘD+(d/a)X)
Inert I
NI=NA0ΘI
NI = NA0ΘI
NT0
δNA0X
NT=NT0+δNA0X

30. Concentration: Batch Systems
Constant Volume Batch:
Note:  if the reaction occurs in the liquid phase or if a gas phase reaction occurs in a rigid (e.g., steel) batch reactor Then

31. Example

32. Species Symbols Initial Change Remaining NaOH A NA0 -NA0X NA=NA0(1–X)
C17H35COO)3-C3H5 B
NB0=NA0ΘB
-(1/3)NA0X
NB=NA0(ΘB –(1/3)X)
C17H35COONa C
NC0=NA0ΘC
NA0X
NC=NA0(ΘC+X)
C3H5(OH)3 D
ND0=NA0ΘD
(1/3)NA0X
ND=NA0(ΘD+(1/3)X)
Water (Inert) I
NI=NA0ΘI
NI = NA0ΘI
NT0
NT=NT0+δNA0X

33. Flow System Stoichiometric Table

34. Flow System Stoichiometric Table
Species
Symbols
Initial
Change
Remaining A
FA0
-FA0X
FA=FA0(1–X) B
FB0=FA0ΘB
-(b/a)FA0X
FB=FA0(ΘB –(b/a)X) C
FC0=FA0ΘC
(c/a)FA0X
FC=FA0(ΘC+(c/a)X) D
FD0=FA0ΘD
(d/a)FA0X
FD=FA0(ΘD+(d/a)X)
Inert I
FI=FA0ΘI
FI = FA0ΘI
FT0
FT=FT0+δFA0X

35. Concentration: Flow System
Liquid Phase Flow System:
This gives us -rA = f(X)
If the rate of reaction were:
then we would have:

36. Concentration: Gas Flow System
For gaseous reactants and assuming ideal gases, therefore from ideal gas law we may get
From Stoichiometric Table
Substitute in v
Where:

37. Concentration: Gas Flow System
Combining the compressibility factor equation of state with Z = Z0
with
and
For example if the gas phase reaction has the rate law
then k
with

40. Example Production of Nitric acid
Nitric acid is made commercially from nitric oxide. Nitric oxide is produced by the gas-phase oxidation of ammonia.
4NH3 + 5O2  4NO + 6H2O
The feed consists of 15 mol% ammonia in air at 8.2 atm and 227°C.

41. What is the total entering concentration? Assume ideal gas behavior.

42. (b) What is the entering concentration of ammonia?

43. (c) Set up a stoichiometric table with ammonia as your basis of calculation.
Express Ci for all species as functions of conversion for a constant-volume batch reactor.
Express PT as a function of X.
Express Pi and Ci for all species as functions of conversion for a flow reactor.

47. (d) Write the combined mole balance and rate law solely in terms of the molar flow rates and rate law parameters for C1 and C2 above. Assume the reaction is first order in both reactants

50. 2

51. Measures Other Than Conversion
Uses: A. Membrane reactors
B. Multiple reaction
Liquids: Use concentrations, i.e., CA
For the elementary liquid phase reaction
carried out in a CSTR, where V, vo,CAo,k, and Kc are given and the feed is pure A, the combined mole balance, rate laws, and stoichiometry are:

52. Measures Other Than Conversion
For Reactant A:
For CSTR

53. Measures Other Than Conversion
For Reactant B:
There are two equations, two unknowns, CA and CB

54. Measures Other Than Conversion
Gases: Use molar flow rates, i.e., FT
A < === > B

55. Measures Other Than Conversion
If the reaction, A < === > B
carried out in the gas phase in a PFR, where V, vo,CAo,k, and Kc are given and the feed is pure A, the combined mole balance, rate laws, and stoichiometry yield, for isothermal operation (T=To) and no pressure drop (DP=0) are:

56. Summary
At the start of the chapter we saw we needed -rA=f(X). This result is achieved in two steps.
Rate Laws
-rA=k f(Ci)
1st order A  B or st order
2nd order A+B ==> C
Rate laws are found by experiment
Stoichiometry
Liquid:
Gas:
A < === > B
-rA=kCA
-rA=kACACB