1. Rate Law
Chapter 3

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 Basics 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
k is the specific reaction rate (constant)

5. 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:

6. Non-Elementary Rate Laws
If the rate law for the non-elementary reaction
is may 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).

7. Rate Law Basics 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

8. 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)

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

10. For the reaction
the reaction coordinate is

11. 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.

12. We see that 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.

13. 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?

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

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

16. 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

17. 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.

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

19. 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

20. 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

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

22. 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
NT=NT0+δNA0X

23. 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

24. Example

25. 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

26. Flow System Stoichiometric Table

27. 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

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

29. 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:

30. 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

33. 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.

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

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

36. (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.

41. (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

45. Tutorial 3

46. Kinetics of Homogeneous Reactions: reaction rate, stoichiometry, reaction order
A reaction has the stoichiometric equation A + B 2R.
What is the order of reaction?
2. Given the reaction: NO2 + 1/2 O2 N2O5
What is the relation between the rates of formation and disappearance of the three reaction components?
3. A reaction with stoichiometric equation: 1/2A + B  R + 1/2S has the following rate expression : rA = kCA0.5CB
What is the rate expression for this reaction
if the stoichiometric equation is written as A + 2B  2R + S?

47. 4. For the complex reaction with stoichiometry
A + 3B  2R + S
and with second-order rate expression:
-rA = k1CACB
are the reaction rates related as follows: rA = rB = rR?
If the rates are not so related, then how are they related? Account for the sign, + or -.
5. A certain reaction has a rate given by:
- rA =0.005C2A mol/cm3.min
If the concentration is to be expressed in mol/l and time in hours, what would be the value and units of the rate constant?

49. Section A: Choose the correct answer from the following questions.
1. Which equation is used in arriving at the design equation for a batch reactor?
A. Gj = Vrj
B. dNj/dt = 0
C. Fjo = Fj = 0
D. E=mc2

50. 2. What assumptions are made when modeling an ideal tubular reactor?
A. Steady sate and no radial variations
B. Plug flow and liquid systems
C. Gas flow and steady state
D. That the reactor will photograph well
A. Steady sate and no radial variations

51. 3. What does the mole balance for a CSTR become if rj = - kCj?
A. Cj = (Fjo-Fj)/V
B. too complicated…
C. V = (Fjo – Fj)/(kCj)
D. V = (Fjo-Fj)/k

52. 4. Which reactor is modeled by the equation
V = (Fjo-Fj)/(kCj) if rj = -Cjk?
CSTR
B. The red one with polka dots
C. Batch reactor
D. Tubular reactor

53. 5. What happens during a decomposition reaction?
A. Species’ molecular configuration changes only
B. A molecule rots
C. Two molecules combine to give one molecule
D. A molecule breaks down into smaller molecules

54. 6. What assumption is made when studying an IDEAL CSTR?
A. Innocent until proven guilty
B. Adiabatic operation
C. Perfect mixing
D. Constant volume

55. 7. What are the dimensions of k in the equation
–ra = k Ca?
A. moles2/volume2/time
B. time/mole/volume
C. three
D. 1/time

56. 8. What type of mathematical equation is used to express the rate law?
A. Irreversible equation
B. Differential equation
C. Unsolvable equation
D. Algebraic equation

57. P3-10(a) Provide the correct solutions to the following Write the rate law for the below elementary reactions

58. P3-10(a)

59. P3-10(b)

60. Problem 3a
Setup a stoichiometric table for the following reactions and apply in reaction rate law assuming elementary rates
P0 = 6 atm
T0 = 1100K
Constant volume batch reactor

61. Species
Symbols
Initial
Change
Remaining
C2H6 A
NA0
-NA0X
NA=NA0(1–X)
C2H4 B
NB0=NA0ΘB
NA0X
NB=NA0(ΘB+X) H2 C
NC0=NA0ΘC
NC=NA0(ΘC+X)
NT0
NT=NT0+δNA0X

63. Problem 3b
Setup a stoichiometric table for the following reactions and apply in reaction rate law assuming elementary rates

64. Species
Symbols
Initial
Change
Remaining
C2H4 A
FA0
-FA0X
FA=FA0(1–X) O2 B
FB0=FA0ΘB
- (b/a)FA0X
FB=FA0(ΘB-X)
C2H4O C
FC0=FA0ΘC=0
+FA0X
FC=FA0X
FT0
FT=FT0+δFA0X

66. Problem 4
The trimerization 3A(g) A3(g,l) is carried out isothermally and without pressure drop in a PFR at 298K and 2atm. As the concentration of A3 increases down the reactor, A3 begins to condense. The vapor pressure of A3 at 298K is 0.5atm. If an equal molar mixture of A and inert, I, is fed to the reactor, at what conversion of A will A3 begin to condense?

67. Stoichiometric Table species Entering change leaving A FA0 -FA0X FA=
FI=FA0 A3
FA0X/3
FA3=
(yA3)(FT)
FT=
FA(2-2X/3)

69. P3-13
The formation of nitoanalyne is formed from the reaction of ONCB and ammonia.
The liquid phase reaction is 1st order in both reactants with k= m3/kmol.min at 188C and activation energy = 11.2 kcal/mol
The concentrations of nitoanalyne and ONCB are 1.8 and 6.6 kmol/m3 respectively.

70. P3-13a
Write the rate law of the disappearance of ONCB in terms of concentration?

71. P3-13b Setup a stoiciometric table for flow system Species Entering
Change
Leaving A
FA0
-FA0X
FA0(1-X) B
FB0=ΘBFA0
=(6.6/1.8)FA0
-2FA0X
FB=
FA0(ΘB-2X) C
FA0X
FC=FA0X D
FD=FA0X

72. P3-13(c)

73. P3-13(d)

74. P3-13(e)

75. P3-13(g)

78. Problem Q4. P4-5 (CDROM)
The liquid phase reaction A + B C follows an elementary rate law and is carried out isothermally in a flow system. The concentrations of the A and B feed streams are 2M before mixing. The volumetric flow rate of each stream is 5dm3/min and the entering temperature is 300K. The streams are mixed immediately before entering. Two reactors are available, a 200dm3 CSTR (that can be heated to 77°C or cooled to 0°C, and the other a PFR operated at 300K.
Note that k = 0.07 dm3/mol.min at 300K and E = 20kcal/mol.
(a) What reactor and what conditions would you choose. Hint – first calculate the value of the constant k at 77°C (350K).
(b) How long would it take to achieve 90% conversion in a 200dm3 batch reactor with CAO = CBO = 1M after mixing at a temperature of 77°C.

79. Given information CA0=CB0=2mol/dm3 vA=vB=5dm3/min k1=0.07dm3/mol.min
EA=20kcal
T1=300K
Rate law: kCACB,
CA=CA0(1-X)
Cj=FJ0/Vj

83. Problem 3.2