Chemical Reaction Engineering
Asynchronous Video Series
Chapter 3, Part 1:
Rate Laws
H. Scott Fogler, Ph.D.
Why a Rate Law?
•
If we have
FA 0
 rA
X
•
Then we can size a number of CSTR and PFR reaction systems
Summary
Why a Rate Law?
•
If we have
FA 0
 rA
X
•
Then we can size a number of CSTR and PFR reaction systems
2 CSTRs
CSTR
PFR
FA 0
 rA
FA 0
 rA
X
FA 0
 rA
X
X
Why a Rate Law?
•
If we have
FA 0
 rA
X
•
Then we can size a number of CSTR and PFR reaction systems
FA 0
 rA
FA 0
 rA
X
•
2 CSTRs
CSTR
PFR
FA 0
 rA
X
To find -rA= f(X)
1) Need the rate law, -rA=f(CA, CB)
2) Need the reaction stoichiometry, CA=CA0(1-X)
X
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.
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)
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)
•
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)
Rate Law Basics
•
To find the activation energy, plot k as a function of temperature:
Where:
Reaction Order
•
You can tell the overall reaction order by the units of k
CA
-rA
(mol/dm3) (mol/dm3*s)
Reaction Order
Rate Law
k
zero
-rA = k
(mol /dm3*s)
Reaction Order
•
You can tell the overall reaction order by the units of k
CA
-rA
(mol/dm3) (mol/dm3*s)
Reaction Order
Rate Law
k
zero
-rA = k
(mol /dm3*s)
1st
-rA = kC A
s -1
Reaction Order
•
You can tell the overall reaction order by the units of k
CA
-rA
(mol/dm3) (mol/dm3*s)
Reaction Order
Rate Law
k
zero
-rA = k
(mol /dm3*s)
1st
-rA = kC A
s -1
2nd
-rA = kC A2
(dm3/mol*s)
Examples of Rate Laws
• First Order Reactions
(1) Homogeneous irreversible elementary gas phase reaction
C2 H 6  C2 H 4  H 2
 rA  kCC2H6
with
82kcal 1
1
 

1 mol  1000 T 
k  0.072s e
Examples of Rate Laws
• First Order Reactions
(1) Homogeneous irreversible elementary gas phase reaction
C2 H 6  C2 H 4  H 2
with
 rA  kCC2H6
82kcal 1
1
 

1 mol  1000 T 
k  0.072s e
(2) Homogeneous reversible elementary reaction
n  C4 H10  i  C4 H10

 rn  k C nC4  CiC4 K C
with


 T  360 
k  31.1exp 7906

 360T 

and

 T  333 
K C  3.03exp  830.3

 333T 

Examples of Rate Laws
• First Order Reactions
(1) Homogeneous irreversible elementary gas phase reaction
C2 H 6  C2 H 4  H 2
82kcal 1
1
 

1 mol  1000 T 
k  0.072s e
with
 rA  kCC2H6
(2) Homogeneous reversible elementary reaction
n  C4 H10  i  C4 H10

 rn  k C nC4  CiC4 K C
with


 T  360 
k  31.1exp 7906

 360T 

and

 T  333 
K C  3.03exp  830.3

 333T 

• Second Order Reactions
(1) Homogeneous irreversible non-elementary reaction
with
m3
k  0.0017
kmol.min
and
E  11273
cal
At 188˚C
mol
 rA  kCONCBC NH3
This is first order in ONCB, first order in ammonia and overall second order.
Examples of Rate Laws
• Second Order Reactions
(2) Homogeneous irreversible elementary reaction
CNBr  CH3 NH2  CH3Br  NCNH2
 rA  kCCNBr CCH 3 NH2
with
2.2dm 3
k
s.mol
Examples of Rate Laws
• Second Order Reactions
(2) Homogeneous irreversible elementary reaction
CNBr  CH3 NH2  CH3Br  NCNH2
 rA  kCCNBr CCH 3 NH2
with
2.2dm 3
k
s.mol
This reaction is first order in CNBr, first order in CH3NH2 and overall second order.
(3) Heterogeneous catalytic reaction: The following reaction takes place over a solid
catalyst: