Nonelementary Reaction Kinetics - Dicky Dermawan

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Transcript Nonelementary Reaction Kinetics - Dicky Dermawan

ITK-329 Kinetika & Katalisis
Chapter 4
Nonelementary Reaction Kinetics
Dicky Dermawan
www.dickydermawan.net78.net
[email protected]
65
Historical Perspective
Dobereiner (1829), Wilhelmy (1850) supposed that reaction
rates would be simply related to the stoichiometry of the
reaction
1878: Van’t Hoff showed that the rate equation had little
correlation to stoichiometry.
Reaction
Rate Equation
4 PH3  P4 + 6 H2
rPH3  k 3  [PH3 ]
2 AsH3  As2 + 3 H2
rAsH3  k 4  [ AsH3 ]
2 PH3 + 4 O2  P2O5 + 3 H2O
rPH3  k 5  [PH3 ]  [O 2 ]1/ 2

rS  k 6  [suksrose ]  [H ]
H
rAc  k 7  [CH3COOR]  [H ]
H
C12H22O11 + H2O 
 C6H12O6 + C5H9O5CH2OH
CH3COOR + H2O 
 CH3COOH + ROH

H
CH3COOH + ROH 
 CH3COOR + H2O
ClCH2COOH + H2O  HOCH2COOH + HCl
rAc  k 8  [CH3 COOH]  [ROH]  [H ]
rC 2H3 ClO 2  k 9  [C2H3ClO2 ]
2 FeCl3 + SnCl2  FeCl2 + SnCl4
r
KClO3 + 6 FeO  KCl + 3 Fe2O3
r
Fe
3
Fe 3 
 k10  [Fe3  ]2  [Sn2  ]

 k11  [Fe2  ]  [ClO3 ]
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Historical Perspective
Van’t Hoff: the kinetics of a reaction related to molecularity, i.e. the
number of molecules participating in some critical step in the
reaction
Unimolecular reaction:
Cyclopropane  Propylene
Bimolecular reaction:
*OH + C2H6  H2O + C2H5*
Termolecular reaction:
CH3* + CH3* + N2  C2H6 + N2
: all first-order reactions are unimolecular
: all second-order reactions are bimolecular
: all third-order reactions are termolecular
Critical step: what about?
4 PH3  P4 + 6 H2
rPH3  k 3  [PH3 ]
67
Historical Findings
When a reaction involves the formations and subsequent reactions of
intermediate species, it is not uncommon to find a non-integer order or
other type of kinetic expression:
CH3CHO  CH4 + CO
At +/- 500oC: -rCH3CHO = k.CCH3CHO3/2
H2 + I2  2 HI
 rHI 
k1  k 3  CI2  CH2
k 2  k 3  CH2
(CH3)2N2  C2H6 + N2
At low pressures below 50 mmHg:
-rN2 ~ CAZO2
At high pressures greater than 1 atm:
-rN2 ~ CAZO
An elementary reaction is defined as a chemical reaction going from reactants to products
without going through any stable intermediates.
In this context, a species is said to be stable if it has lifetime longer than ~10-11 sec
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Reactive Intermediates
Reactive Intermediates are by definition reactive. The undergo many reactions
David Chapman (1913), Muriel Chapman & Max Bodenstein (1907):
H2 + Cl2  2 HCl
Cl • as reactive intermediates
Every overall chemical reaction can be divided into a sequence of elementary reaction.
Every reaction has a mechanism, defined as the sequence of elementary reactions that
occur at appreciable rates when the reactants come together and react to form
products
CH3CH2HC=CH2
H

 CH3HC=CHCH3
Mechanism:
+
H
CH3CH2HC=CH2 + H+
+
H
CH3CH2HC=CH2 + H+
CH2
+
H
CH3CH2HC
CH3HC=CHCH3 + H+
CH2
CH3CH2HC
CH2
+
H
CH3CH2HC
CH3CH2HC
CH2
CH2 =CHCH2CH3 + H+
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Kinetic of Elementary Reactions
A + B —2 P +Q
r2 = k2 [A] [B]
-rA = k2 [A] [B]
-rB = k2 [A] [B]
-rA = -rB = +rP = +rQ
2 A —4 P +Q
r4 = k4 [A] [A] = k4 [A]2
+rP = k4 [A]2
+rQ = k4 [A]2
-
+rP = k2 [A] [B]
+rQ = k2 [A] [B]
rA/2 =
-rA = 2k4 [A]2
+rP/1 = +rQ/1 = k4 [A]2
Collosion Partner
Incorrect:
A —1  P
-rA = k1 [A]
Correct:
A + X—1 P + X
-rA = k1 [A] [X]
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Rates of Overall Reaction
AP
1
3
A  H  I  P  H


2
For each reaction:
r1  k1  [ A ]  [H ]
r2  k 2  [I]
r3  k 3  [I]
For each species:
rA  r1  r2
r   r1  r2  r3
H
rI  r1  r2  r3
rP  r3
In a constant volume batch reactor:
d[ A ]
 k1  [ A ]  [H ]  k 2  [I]
dt
d[H ]
 k1  [ A ]  [H ]  k 2  [I]  k 3  [I]
dt
d[I]
 k1  [ A ]  [H ]  k 2  [I]  k 3  [I]
dt
d[P]
 k 3  [I]
dt
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Pseudo-Steady-State Hypothesis
AP
1
3
A  H  I  P  H


2
According to pseudo-steady-state approximation, one can compute accurate values of
the concentrations of all of the intermediates in a reaction by assuming that the net
rate of the intermediates is negligible.
d[H ]
 k1  [ A ]  [H ]  k 2  [I]  k 3  [I]  0
dt
d[I]
 k1  [ A ]  [H ]  k 2  [I]  k 3  [I]  0
dt
k k
d[P]
 k 3  [I]  1 3  [ A ]  [H ] 
dt
k2  k3
According to stoichiometry:
 [I] 
k1
 [ A ]  [H ]
k2  k3
rp  ko  [A]  [H ]
 rA  rp  k o  [ A]  [H ]
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Another Example:
Rates of Overall Reaction
(CH3)2N2  C2H6 + N2
AZO  C2H6 + N2
At low pressures below 50 mmHg
At high pressures greater than 1 atm
:
:
-rN2 ~ CAZO2
-rN2 ~ CAZO
Reaction mechanism [F.A. Lindemann,Trans. Faraday Soc., 17, 598 (1922)]
(CH3)2N2 + (CH3)2N2 —k1 (CH3)2N2 + [(CH3)2N2]*
rAZO*= k1.CAZO2
(CH3)2N2* + (CH3)2N2 —k2 (CH3)2N2 + (CH3)2N2
rAZO*= -k2.CAZOCAZO*
(CH3)2N2* —k3 C2H6 + N2
rAZO*= -k3.CAZO*
PSSH: rAZO*= k1.CAZO - k2.CAZOCAZO* -k3.CAZO* 0
2
Then,
rN2  k 3  CAZO*
k1  k 3  C AZO2

k 2  C AZO  k 3
 C AZO *
k1  C AZO 2

k 2  C AZO  k 3
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H4.1.3
Find Rate Expression of Overall Reaction….
2 N2O5  4 NO2 + O2
Mechanism:
k
1


X  N2O5 
 NO2  NO3  X
k2
k
3
NO2  NO3 

 NO  O 2  NO2
k
4
NO  NO3 

 2 NO2
 What rate expression is consistent with this mechanism?
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Two Proposed Mechanism
can give rise to the same rate expression
2 NO + 2 H2  N2 + 2 H2O
Mechanism A
k
1
: 2 NO  H2 

N2  H2O2
k
2 2 H O
H2O2  H2 

2
k
Mechanism B
3



: 2 NO 
 N2O2
k4
k
5
N2O2  H2 

 N2  2 H2O2
k
2 2H O
H2O2  H2 

2
 What rate expression is consistent with these mechanism?
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H4.2.1
Example of Chain Reaction:
Free Radical as Active Intermediate
H2+ Br2  2 HBr
Mechanism:
Initiation
X + Br2 —1 2 Br• + X
Propagation
Br • + H2 —2 HBr + H •
H • + Br2 —3 HBr + Br •
Terminatiion
X + 2 Br • —4 Br2 + X
H • + HBr —5 H2 + Br •
 What rate expression is consistent with this mechanism?
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Chain Reactions
H-Il4.3
Mekanisme berantai di bawah ini diusulkan untuk reaksi dekomposisi ozon:
Inisiasi
:
k1
Cl 2  O3 
ClO ClO2 
Propagasi
:
k2

 ClO3  O 2
 ClO2  O 3 

k3

ClO


O


ClO2  2O 2
3
3

Terminasi
:
k4

 Cl 2  3O 2
ClO3  ClO3  

k5

ClO


ClO



Cl 2  O 2

 Bagaimana persamaan laju reaksi dekomposisi ozon menurut mekanisme
ini?
 Hasil percobaan pada suhu rendah menunjukkan bahwa persamaan laju
dekomposisi ozon mengikuti persamaan:
3
1
d[O3 ]
2

 k  [Cl 2 ]  [O3 ] 2
dt
Apakah mekanisme yang diusulkan konsisten dengan hasil percobaan ini? 77
Chain Reactions
H4.1
Houser & Lee [J. Phys. Chem., 71 (3422), 1967] have studied the pyrolysis of
ethyl nitrate using a stirred flow reactor. They have proposed the following
mechanism for the reaction.
Initiation
:
1
C2H5ONO2 

C2H5O  NO2
Propagation
:
2
C2H5O  

 CH3  CH2O
k
k
k
3
CH3  C2H5ONO2 

 CH3NO2  C2H5O 
Termination
:
k
4  CH CHO C H OH
2C2H5O  

3
2 5
 What rate expression is consistent with this mechanism?
78
Chain Reactions:
Thermal Cracking of Ethane
Ex.7-2
The thermal decomposition of ethane to ethylene, methane, butane, and hydrogen
is believed to proceed in the following sequence:
Initiation:
k
1
C2H 6 
2CH 3 
Propagation:
k
2
CH 3  C2H 6 
CH 4  C2H5 
k
3
C2H5  
C2H 4  H 
k
4
H  C2H 6 
C2H5   H 2
T ermin ation:
k
5
2C2H5  
C4H10
Use PSSH to derive a rate law for the formation of ethylene
79
Chain Reactions: Flame Retardants
P7-3B
Hydrogen radicals are important to sustaining combustion reactions. Consequently,
if chemical compounds that can scavenge the hidrogen radicals are introduced, the
flame can be extinguished. While many reactions occur during the combustion
process, we shall choose CO flames as a model system to ilustrate the process [S.
Senkan et al., Combustion and Flame, 69, p. 113 (1987)] . In the absence of
inhibitors:
O2 
 O  O 
The last two reactions are rapid compared to
the firs two. When HCl is introduced to the
flame, the following additional reactions
occur:
H 2O  O  
 2OH 
CO  OH  
 CO 2  H 
H  O2 
 OH  O 
H   HCl 
 H 2  Cl 
H  Cl  
 HCl
Derive a rate law for consumption of CO for both when no retardant present and
80
when HCl is introduced
Chain Reactions:
The Pyrolysis of Acetaldehyde
P7-4A
The pyrolysis of acetaldehyde is believed to take place according to the
following sequence:
k
1
CH3CHO 
CH3  CHO 
k
2
CH3  CH3CHO 
CH3  CO  CH 4
k
3
CHO  CH3CHO 
CH 3  2 CO  H 2
k
4
2 CH3  
C2H 6
Derive the rate expression for the rate of disappearance of acetaldehyde
81
Chain Reactions in Tribology
Engine Oil Degradation
P7-7C
One of the major reasons for engine oil degradation is the oxidation of the motor oil. To
retard the degradation process, most oils contain an antioxidant [see Ind. Eng. Chem.
26, 902 (1987)].
Without an inhibitor to oxidation present, the
suggested mechanism at low temperature is:
k
0
I 2 
2I 
k
i
I   RH 
R   HI
k
Where I2 is an initiator and RH is the hydrocarbon in
the oil.
When the temperature is raised to 100oC, the
following additional reaction occurs as a result of
the decomposition of the unstable ROOH:
p1
R  O 2 
 ROO 
k
p2
ROO   RH 

ROOH  R 
k
t
2 ROO  
inactive
k p3
ROOH 
 RO   HO 
k p4
RO   RH 
 ROH  R 
k p5
HO   RH 
 H 2O  R 
Derive the rate expression for the degradation of
the uninhibited motor oil:
a. At low temperature (25oC)
82
o
b. At high temperature (100 C)
Engine Oil Degradation:
The Role of Antioxidant
P7-7C (cont’)
When an antioxidant is added to retard degradation at low temperatures, the following
additional termination step occur:
k
A1
ROO   AH 

 ROOH  A 
k
A4
A   ROO  

 inact ive
Derive the rate expression for the degradation of the uninhibited motor oil:
a. At low temperature (25oC)
b. At high temperature (100oC)
83
Free Radical Polymerization
1. The Reaction
INITIATION
This reaction produces the formation of the Primary Radical
PROPAGATION
TERMINATION
Transfer
To solvent
To chain transfer agent
To monomer
To initiator
Addition
Disproportionation
84
Rate-determining (-limiting) Step
When one of the steps is much slower than all of the other steps in the
mechanism, the rate of this step is fully control the overall rate, thus considerable
1
simplification can be gained:


AX


A#  X
2
3
A# 
 B
Using PSSH:
rB 
k 3  k1  [A]  [X]
k 2  [X]  k 3
If it is known that reaction (3) is much slower than (1) & (2) reactions, it is easily derived
that:
k 3  k1
rB 
 [A]
k3
85
Rate-determining (-limiting) Step
When one of the steps is much slower than all of the other steps in the
mechanism, the rate of this step is fully control the overall rate, one can often
derive a suitable rate equation for the reaction using somewhat less algebra
2 N2O5  4 NO2 + O2
Mechanism:
k
1


X  N2O5 
 NO2  NO3  X
k2
k
3
NO2  NO3 

 NO  O 2  NO2
k
4
NO  NO3 

 2 NO2
 Fast
 Slow
Fast
 What rate expression is consistent with this mechanism?
86
Example : P7-8A
Consider the application of the PSSH to epidemology. We shall treat each of the following
steps as elementary in that the rate will be proportional to the number of people in a particular
state of health. A healthy person, H, can become ill, I, spontaneously,
k1
(P7-11.1)
H
I
Or he may become ill through contact with another ill person
k2
I+H
2I
The ill person may become healty
I
k3
H
(P7-11.2)
(P7-11.3)
Or he may expire
k4
(P7-11.4)
I
D
The reaction given in equation (P7-11.4) is normally considered completely ireversible, although
the reverse reaction has been reported to occur :
(a) Derive an equation for death rate.
(b) At what concentration of healty people does the death rate become critical?
(c) Comment the validity of the PSSH under the condition of part (b).
87