1. 제3장
Rate Laws
and Stoichiometry
Chemical Reaction Engineering 1
반응공학 1

2. An African-American Scholar and Inventor
Booker T. Washington
“Success is measured not so much by the position one has reached in life, as by the obstacles one has overcome while trying to succeed”
An African-American Scholar and Inventor

3. Objectives After completing Chapter 3, reader will be able to:
 Write a rate law and define reaction order and activation energy.
 Set up a stoichiometric table for both batch and flow systems and express concentration as a function or conversion.
 Calculate the equilibrium conversion for both gas and liquid phase reactions.
 Write the combined mole balance and rate law in measures other than conversion.
 Set up a stoichiometric table for reactions with phase change.

4. Basic Definitions
Homogeneous reaction is one that involves only one phase
Heterogeneous reaction involves more than one phase. (reaction usually occurs at or very near the interface between the phase)
Irreversible reaction is one that proceeds in only one direction and continues in that direction until the reactants are exhausted.
Reversible reaction can proceed in either direction, depending on the concentration of reactants and products relative to the corresponding equilibrium concentration.

5. Example of Elementary Reaction
Basic Definitions
Molecularity of a reaction is the number of atoms, ions or molecules involved in a reaction step.
The term, unimolecular, bimolecular, and termolecular refer to reactions involving, respectively, one, two, or three atoms (or molecules) interacting in any one reaction step.
Example of Elementary Reaction

6. Relative Rates of Reactions
If the rate law depends on more than one species, we MUST relate the concentrations of different species to each other. A stoichiometric table presents the stoichiometric relationships between reacting molecules for a single reaction.
aA + bB cC + dD
(2-1)
In formulating our stoichiometric table, we shall take species A as our basis of calculation (i.e., limiting reactant) and then divide through by the stoichiometric coefficient of A. In order to put everything on a basis of “per mole of A.”
(2-2)
The relationship can be expressed directly from the stoichiometry of the reaction.
(3-1)

7. Relative Rates of Reactions
Example:
2NO + O NO2
rNO
rO2
rNO2 = = -1 2 -2
If rNO2= 4 mol/m3-sec
rNO= -4 mol/m3-sec
rO2= -2 mol/m3-sec

8. Reaction Order and Rate Law
Let take A as the basis of calculation
a species A is one of the reactants that is disappearing as a result of the reaction. The limiting reactant is usually chosen as our basis for calculation.
The rate of disappearance of A, -rA, depends on temperature and composition and it can be written as the product of the reaction constant k and
Rate raw (Kinetic expression) : the algebraic equation that relates –rA to the species concentration

9. Reaction Order and Rate Law
The dependence of the reaction rate –rA on the concentration of the species is almost without exception determined by experimental observation.
The order of a reaction refers to the powers to which the concentrations are raised in the kinetic rate law.
(3-3)
a order with respect to reactant A
b order with respect to reactant B
n (=a+b) : the overall order of the reaction

10. Unit of Specific Reaction Rate
The unit of the specific reaction rate, kA, vary with the order of the reaction.
(Concentration)1-n k=
A  products
Time
(3-4)
(3-5)
(3-6)
(3-7)

11. Elementary and Nonelementary Reaction
Kinetic rate raw
Elementary reaction
Non-elementary reaction
2NO + O2  2NO2
CO + Cl2  COCl2
2nd order w.r.t. nitric oxide
1st order w.r.t. oxygen
overall is 3rd order reaction
1st order w.r.t. carbon monoxide
3/2 order w.r.t. chorine
overall is 5/2 order reaction
In general, first- and second-order reactions are more commonly observed.

12. Example of Rate Law

13. Example of Rate Law

14. Apparent Reaction Order
Sometimes reactions have complex rate expressions that cannot be separated into solely temperature-dependent and concentration-dependent portions.
2N2O N O2 Pt
Kinetic rate raw
Limiting conditions: depending on oxygen concentration
(1)
“apparent” 1st-order w.r.t. N2O
1st order overall
“apparent” reaction-order
-1 w.r.t. O2 , +1 w.r.t. N2O
(2)

15. Reversible Reactions aA + bB cC + dD
All rate raws for for reversible reactions must reduce to the thermodynamic relationship relating the reacting species concentrations at equilibrium. At equilibrium, the rate of reaction is identically zero for all species (i.e., -rA=0). For the general reaction
aA + bB cC + dD
The concentrations at equilibrium are related by the thermodynamic relationship

16. 2C6H C12H10 +H2
2B D +H2
The rate of disappearance of benzene The rate of formation of benzene
The net rate of formation of benzene

17. The rate law for the rate of disappearance of benzene
Exothermic
reaction KC KC
Endothermic
reaction T T

18. Thermodynamically Consistent with Equilibrium Constant
2B D +H2
We need to check
whether the above rate law is thermodynamically consistent at equilibrium.
At equilibrium, -rB=0,
Rearranging, we obtain
identical

19. Thermodynamically Consistent with Equilibrium Constant
2B D +H2
The rate of formation of diphenyl is
(3-15)
Using the relationship
We can obtain the relationship between the various specific reaction rates,
kB, kD:
(3-16)

20. kA: Reaction Rate Constant
The reaction rate constant k is not truly a constant, but is merely independent of the concentrations of the species involved in the reaction.
The quantity k is also referred to as the specific reaction rate (constant). It is almost always strongly dependent on temperature.
In gas-phase reactions, it depends on the catalyst and total pressure.
In liquid systems, it depends on the total pressure, ionic strength and solvent.
In this text, it will be assumed that kA depends only on temperature.

21. Arrhenius equation Activation energy, J/mol or cal/mol
Specific reaction rate (constant)
Absolute
Temperature, K
Arrhenius frequency factor or
pre-exponential factor
Gas constant
8.314 J/mol · K
1.987 cal/mol · K
8.314 kPa · dm3/mol · K
mathematical number
e= …

22. Arrhenius Equation A
T  0, kA  0
T  , kA  A kA
T (K)

23. SVANTE AUGUST ARRHENIUS
Svante August Arrhenius was born in Vik, Sweden in At age 25 he turned in his PhD thesis at the University of Uppsala, Sweden. His PhD examining committee did not think very highly of his thesis and rated it 4th class. His oral thesis defense did not fair much better as they rated it as only 3rd class. Arrhenius left Sweden for five years to work with Oswald, Boltzmann and van't Hoff. In 1889 his interpretation of temperature-dependent equation by van't Hoff led to the universal accepted Arrhenius equation for kinetic rate laws in chemistry. He received the Nobel Prize in From 1905 until his death in 1927 he was director of Physical Chemistry at the Nobel Institute

24. Activation energy Activation energy E :
a minimum energy that must be possessed by reacting molecules before the reaction will occur.
The fraction of the collisions between molecules that together have this minimum energy E (the kinetic theory of gases)
Activation energy E is determined experimentally by carrying out the reaction at several different temperature.

25. Activation energy
A+BC A-B-C AB+C

26. Example 3-1: Determination of Activation energy
Decomposition of benzene diazonium chloride to give chlorobenzene and N2 Cl
N=N
Calculate the activation energy using following information for this first-order

27. Finding the activation energy
Plot (ln k) vs (1/T)
1/T (K-1)
0.01
0.001
0.0001
0.0030
0.0031
0.0032
k (sec-1)
Slope = -E/R

28. Finding the activation energy
Plot (ln k) vs (1/T)
[Solution 1]
Select two data points,
(1/T1, k1) and (1/T2, k2),
and then calculate.
Therefore, E  116 kJ/mole
[Solution 2]
Find a best fit from data point
ln k=14,017/T
Slope=E/R=14,017
E=116.5 kJ/mole 1 1 2 2
( )
( ) k2 E 1 1 ln -
= - k1 R T2 T1
임의의 온도에서
속도상수를 구할 수 있다.

29. Finding Activation Energy
Plot (ln k) vs (1/T)
0.01
0.001
0.0001
0.0030
0.0031
0.0032
1/T (K-1)
k (sec-1)
0.005
0.0005
ln k=14,017/T

30. Activation Energy k (sec-1) 1/T (K-1)
The larger the activation energy, the more temperature-sensitive is the rate of reaction.
Typical values of 1st order gas-phase reaction
Frequency factor ~ 1013 s-1
Activation energy ~ 300 kJ/mol
0.01
k (sec-1)
0.001
Low E
High E
0.0001
0.0030
0.0031
0.0032
1/T (K-1)

31. Specific Reaction Rate
E를 알고 나면 실험하지 않은 어느 온도에서의 k값을 구할 수 있다.
Taking the ratio

32. Present Status of Our Approach
Design Equations
Differential
form
Algebraic
form
Integral
form
Batch
CSTR
PFR
PBR

33. Stoichiometry aA + bB cC + dD (2-1) (3-1) (2-2)
The relationship can be expressed directly from the stoichiometry of the reaction.
(3-1)

34. Batch System NA = NA0 - NA0X = NA0(1-X)
NB0
NC0
ND0
NI0
At time t=0, we will open the reactor and place a number of moles of species A, B, C, D, and I (NA0, NB0, NC0, ND0, and NI0, respectively)
Species A is our basis of calculation.
t = t NA NB NC ND Ni
NA0 is the number of moles of A initially present in the reactor.
NA0X moles of A are consumed as a result of the chemical reaction.
NA0-NA0X moles of A leave in the system.
The number of moles of A remaining in the reactor after conversion X
NA = NA0 - NA0X = NA0(1-X)

35. The number of moles of B remaining in the system, NB
To determine the number of moles of species B remaining at time t (after NA0X moles of A have reacted). For every mole of A that reacts, b/a moles of B must reacted;
The number of moles of B remaining in the system, NB
moles of A reacted
moles of B initially
moles of B disappeared

36. This stoichiometric table presents the following information.
To determine the number of moles of each species remaining after NA0X moles of A have reacted, we form the stoichiometric table (Table 3-3).
This stoichiometric table presents the following information.
Column 1: the particular species
Column 2: the number of moles of each species initially present
Column 3: the change the number of moles brought about by reaction
Column 4: the number of moles remaining in the system at time t.

37. Stoichiometric Table for a Batch System

38. Total Number of Moles per Mole of A Reacted
The total number of moles in the system, NT
Change in the total number of moles
mole of A reacted

39. Equation for Batch Concentration
CA =f(X)
NA =f(X)

40. Concentration of Each Species

41. Volume as a Function of Conversion
We need V(X) to obtain CB=f(X)
- For liquids, volume change with reaction is negligible
when no phase changes are taking place.
For gas-phase reactions, the volumetric flow rate most often
changes during the course of the reaction due to a change
in the total number of moles or in temp. or pressure.

42. Constant Volume Batch Reactor

43. Example 3-2: Liquid-Phase Reaction
Soap consists of the sodium and potassium salts of various fatty acids as oleic, stearic, palmitic, lauric, and myristic acids. The saponification for the formation of soap from aqueous caustic soda and glycol stearate is as follow. Letting X represent the conversion of sodium hydroxide (the mole of sodium hydroxide reacted per mole of sodium hydroxide initially present), set up a stoichiometric table expressing the concentration of each species in terms of its initial concentration and the conversion of X.
3NaOH + (C17H35COO)3C3H5  3C17H35COONa+C3H5(OH)3

44. Example 3-2: Liquid-Phase Reaction

45. Flow Systems CA =f(X) FA =f(X) Entering Leaving FA0 FA FB0 FB FC0 FC
FD0
FI0 FA FB FC FD FI
Definition of
concentration
for flow system
CA =f(X)
FA =f(X)

46. Flow Systems If v=vo CA=CAo(1-X) CB=CAo(B -(b/a)X) Batch System

47. Stoichiometric Table for a Flow System

48. Volume Change with Reaction
Gas-phase reactions that do not have an equal number
of product and reactant moles.
N2 + 3H NH3
-The combustion chamber of the internal-combustion engine
-The expanding gases within the breech and barrel

49. Batch Reactor with Variable Volume
Individual concentrations can be determined by expressing the volume V for batch system (or volumetric flow rate v for a flow system) as a function of conversion using the following equation of state.
(3-30)
T = temperature, K
P = total pressure, atm
Z = compressibility factor
R = gas constant = dm3·atm/gmol · K
This equation is valid at any point in the system at any time. At time t=0,
(3-31)
Dividing (3-30) by (3-31) and rearranging yields
(3-32)

50. The total number of moles in the system, NT
(3-33)
We divide through by NT0
(3-34)
where yA0 is mole fraction of A initially present. If all the species in the generalized reaction are in the gas phase, then
(3-23)

51. Eq (3-34) is further simplified by letting
(3-35)
(3-36)
Eq (3-32) now becomes
(3-37)

52. In gas-phase systems that we shall be studying, the temperature and pressure are such that the compressibility factor will not change significantly during the course of the reaction; hence Z0~Z. For a batch system the volume of the gas at any time t is
(3-38)
Eq (3-38) applies only to a variable-volume batch reactor. If the reactor is a rigid steel container of constant volume, then of course V=V0. For a constant-volume container, V=V0, and Eq. (3-38) can be used to calculate the pressure inside the reactor as a function of temperature and conversion.

53. Flow Reactor with Variable Volumetric Flow Rate
To derive the concentration of the species in terms of conversion for a variable-volume flow system, we shall use the relationships for the total concentration. The total concentration at any point in the reactor is
(3-39)
At the entrance to the reactor
(3-40)
Taking Eq (3-40)/Eq(3-39) and assuming Z~Z0,
(3-41)

54. We can express the concentration equation of species j for a flow system in terms of conversion:
(3-42)

55. From Table 3-4, the total molar flow rate can be written in terms of X
(3-43)
(3-44)
(3-45)

56. Fj=Fjo+vj(FAoX)=FAo(j+vjX)
Cj = v
The molar flow rate of species j is
Fj=Fjo+vj(FAoX)=FAo(j+vjX)
where vi is the stoichiometric coefficient, which is negative for reactants and positive for products.
For example, for the reaction
vA = -1, vB = -b/a, vA = c/a, and vA = d/a.

57. vo((1+X)(Po/P)(T/To))
Substituting for v using Equation (3-42) and for Fj, we have
FAo(j+vjX) Fj
Cj = = v
vo((1+X)(Po/P)(T/To))
Recalling yA0=FA0/FT0 and CA0=yA0CT0, then
(3-46)

58. Concentration as a Function of Conversion
variable-volume gas flow system
Table 3-5

59. Expressing concentration as a function of conversion
Liquid Phase
Gas Phase
Flow
Batch
Batch
Flow
No Phase Change
Constant
Volume
No Phase Change or No Semipermeable Membranes
Isothermal
Isothermal + No Pressure Drop
No Pressure Drop

60. Example 3-4: Manipulation of the Equation Cj=hj(X)
Show under what conditions and manipulation the expression for CB for a gas
flow system reduces to the following equation in Table 3-5. CB FB
For a flow system,
CB = v
From Table 3-4,
Combining two equations yields

61. Using Equation (3-45) gives
By combing above two equations,
Recalling FAo/vo=CAo, CB

62. Example 3-5: Determination of Cj=hj(X) for a Gas-Phase Reaction

68. Homework
P3-7B
P3-15B
P3-16B
Due Date: Next Week