Transcript Animated PowerPoint

Lecture 14
Chemical Reaction Engineering (CRE) is the
field that studies the rates and mechanisms of
chemical reactions and the design of the reactors in
which they take place.
Lecture 14 – Thursday 2/28/2013
 Pseudo Steady State Hypothesis (PSSH)
 Net rate of reaction of active intermediates is zero
 Hall of Fame Reaction: 2NO +O2 2NO2
 Introduction to Enzyme Kinetics
 Begin non-Isothermal reactor design
2
Active Intermediates and PSSH
An active intermediate is a molecule that is in a
highly energetic and reactive state It is short
lived as it disappears virtually as fast as it is
formed. That is, the net rate of reaction of an
active intermediate, A*, is zero.
The assumption that the net rate of reaction is
zero is called the Pseudo Steady State
Hypothesis (PSSH)
3
Active Intermediates and PSSH
4
Example
The rate law for the reaction
A  BC
is found from experiment to be
2
A
kC
 rA 
1  k C A
How did this rate law come about? Suggest a
mechanism consistent with the rate law.
5
Example
For reactions with active intermediates, the reaction
coordinated now has trough in it and the active
intermediate, A*, sits in this trough
6
Example - Solution
7
1
A  A  A * A
2
A * A  A  A
r2 A*  k2C A*C A
3
A *  B  C
r3 A*  k3C A*
k1
k2
k3
r1 A*  k1C
2
A
Rate Laws:
k3 is defined w.r.t. A*
Reaction (1)
r1A*  k1C
Reaction (2)
r2 A*  k2C AC A*
(2)
Reaction (3)
r3 A*  k3CA*
(3)
2
A
(1)
But C*A cannot be measured since it is so small
Relative Rates:
8
r1A  r1A* , r3B  r3 A*
Net Rates: Rate of Formation of Product
rB  r3B  r3 A*  k3CA*
Pseudo Steady State Hypothesis
r*A = 0
rA*   rA*  r1A*  r2 A*  r3 A*
(5)
 k1CA2  k2CACA*  k3CA*  0
(6)
Solving for C A*
9
(4)
2
A
k1C
C A* 
k3  k 2C A
Substituting for C A* in Equation (4) the rate of
formation of B is
k1k3C A2
(8)
rB 
k3  k 2 C A

Relative rates overall
A  BC
rA rB

1 1
k1k3C A2
rA  rB  
k3  k 2 C A
(9)
10
For high concentrations of A, we can neglect k 3 in
the denominator with respect to k2C A
k 2C A  k3
and the rate law becomes
k1k3
rA  
C A  kCA
k2
(10)
(apparent first order)
11
For low concentrations of A, we can neglect k 2C A in
the denominator with respect to k3.
k3  k 2C A
and the rate law becomes
k3k1 2
rA  
C A  k1C A2
k3
(apparent second order)
(11)
Dividing by k3 and letting k’=k2/k3 and k=k1 we have
the rate law we were asked to derive
kCA2
 rA  
1  k C A
(12)
Active Intermediates
Why do so many reactions follow elementary rate laws?
What about  rA  kCA
kC A
rA 
1  k ' CI
(1)
( I  Inert )
k1
A  I 
A* I
(2) A *  I  A  I
k2
13
k3
(3) A 
BC
k1k3C A
- rA 
k 2  k 2C I
Active Intermediates/Free Radicals
and PSSH
Hall of Fame Reaction
The reaction:
2NO +O2 2NO2
has an elementary rate law
2
rNO2  kCNO
CO2
However… Look what happens to the rate as the
temperature increases:
-rNO2
14
T
Why does the rate law decrease with increasing
temperature?
Mechanism:
k1
NO  O2 
NO3*
(1)
k2
NO3* 
NO  O2
(2)
k3
NO3*  NO 
2 NO2
(3)
Write rate of formation of product rNO2
Note: k3 is defined w.r.t. NO3*
15
Define k with respect to NO3*
Assume that all reactions are elementary
reactions, such that:
r1NO3 = k1CNOCO2 = k1 [ NO][O2 ]
r2 NO* = -k2CNO* = k2 éë NO ùû
3
3
r3NO* = -k3CNO* CNO = k3 éë NO3* ùû [ NO]
*
3
3
3
rNO = 2 é -r3NO* ù
ë
2
3 û
16
The net reaction rate for NO3* is the sum of the
individual reaction rates for NO3*:
r1NO* = r1 Þ r1NO* = k1[ NO] [O2 ]
3
3
-r2NO* = r2 Þ r2NO * = -k2 [ NO
3
*
3
3
]
r3 NO*  r3  r3 NO*  k3  NO3*   NO 
3
3
rNO* = r1NO* + r2NO* + r3NO*
3
3
3
3
rNO*  k1  NO  O2   k2  NO3*   k3  NO3*   NO 
3
17
Pseudo Steady State Hypothesis (PSSH)
The PSSH assumes that the net rate of species A*
(in this case NO3*) is zero.
rNO*  0
3
0  k1  NO  O2   k2  NO3*   k3  NO3*   NO 
0 = k1 [ NO] [O2 ] - éë NO3* ùû ( k2 + [ NO])
k1 [ NO ][O2 ]
éë NO ùû =
k2 + k3 [ NO ]
*
3
18
Pseudo Steady State Hypothesis (PSSH)
[𝑁𝑂3∗ ]
𝑘1 𝑁𝑂 𝑂2
=
𝑘2 + 𝑘3 𝑁𝑂


rNO2  2r3 NO*  2 NO3* NO 
3
𝑟𝑁𝑂2
19
k1 k 3 NO 2 O2
=2
k 2 + k 3 NO
Pseudo Steady State Hypothesis (PSSH)
k2 >> k3 [ NO]
rNO2
k1k3
A1 A3
2
NO  O2   2
2
e
k2
A2
E2   E1  E3 
RT
NO  O2 
2
E2  E1  E3 
This result shows why the
rate decreases as
temperature increases.
20
-rNO2
T
End Lecture 14
21