# che452lect09

advertisement ```ChE 452 Lecture 09
Mechanisms & Rate Equations
1
Objective

Describe relationship between rate
equation and mechanism (Quasi steady
state approximation)



Why does it arise
When does it work
When does it fail
2
Previous Lectures: Measurement Of
Rate Laws For Reactions
Rate laws are hard to determine experimentally.
•
•
•
Accurate rate laws only from direct rate measurements
Even then, more than one rate law will fit a given set
of data. (see example 3.C)
The optimal parameters and the quality of the fit
depend on the way the data is fit as well as the rate
law.
•
An accurate rate law fit by a poor procedure can look worse
than the wrong rate law fit by a better procedure.
3
Today: How Else Can We
Get A Rate Law?
•
•
•
Start with mechanisms
Derive fundamental equations
Fit rate data
4
Method To Find A Rate Law

Find the mechanism of the reaction


Computationally
Experimentally
Use the quasi-steady state approximation
to derive a rate equation
Generally more accurate-but we need a
mechanism

5
Idea That Reactions Follow Mechanisms
First Proposed By Van’t Hoff
Experimentally no
correlation between
reaction and rate law
• Thermo, MD
suggested simple
relationship between
rate law and mechanism
• Van’t Hoff suggested
there must be extra
reactions
•
6
Examples Considered By
Van’t Hoff
Reaction
4PH 3  P4  6H 2
2AsH 3  As 2  3H 2
2PH 3  4O 2  P2 O 5 + 3H 2 O
Rate equation
(2.T.1)
r AsH 3 = - k 4
(2.T.5)
r PH 3 = - k 5  PH 3
C12 H 22 O 11  H 2 O  C 6 H 12 O 6  C 5 H 9 O 5 CH 2 O H
(2.T.7)
H+
(2.T.9)
H+
CH 3 COOH  ROH  CH 3 COOR  H 2 O
 AsH 3
(2.T.3)
H+
CH 3 COOR  H 2 O  CH 3 COOH  ROH
rPH3   k 3  PH 3 
(2 .T.11)
ClCH 2 COOH  H 2 O  HOCH 2 COOH  HCl
 O 21/ 2
rS  - k 6 [sucrose] [ H + ]
rAc   k 7 CH 3COOR  H  
(2.T.2)
(2.T.4)
(2.T.6)
(2.T.8)
(2.T.10)
rAc   k 8  CH 3 COOH  ROH   H  
rC 2 H 3ClO 2   k 9  C 2 H 3 ClO 2 
(2.T.12)
(2.T.14)
(2.T.13)
7
All Reactions Actually Occur By A Series of
Chemical Transformations Called Elementary
Reactions
Elementary reaction – a reaction which goes from reactants to
products without going through any stable intermediates
Mechanism – the sequence of elementary steps which occur
when the reactants come together to form products.



H
CH 3CH 2 HC = CH 2  H  CH 3CH 2 HC `CH 2 





H
+
CH 3CH 2 HC `CH 2   CH 3 HC = CHCH 3  H





H
+
CH 3CH 2 HC `CH 2   CH 3CH 2 HC = CH 2  H





H
+
CH 3CH 2 HC `CH 2   H 2 C = CHCH 2 CH 3  H


+
(a)
(b)
(c)
(d)
(4.13)
8
Other Key Definitions
Continued
•
•
•
•
•
•
•
Reactive intermediate
Molecularity
Unimolecular
Bimolecular
Termolecular
Overall reaction
Stoichiometric reaction
9
All Elementary Reactions Have At Least
Two Reactants And Two Products
H + HBr  H2 + Br
(4.11)
X + H2  2H + X
(4.12)
10
Kinetics Of Elementary
Reactions
2
A+BP+Q
4
2A  P  Q
(4.19)
r2 = k2[A][B]
r4  k 4[A]
rA = -k2[A][B]
rA  2k4 [A]2
(4.20)
(4.21)
2
(4.23)
(4.24)
11
Multiple Reactions
5
rA   ri A ,i
i =1
(4.25)
Memorize this equation
12
Next: Example Illustrating Rates Of Overall
Reactions In terms Of Elementary Rates;



H
CH3CH 2 HC = CH 2  H +  CH3CH 2 HC `CH 2 





H
+

CH
CH
H
C
C
H

CH
HC
=
CHCH

H
`
3
2
2
3
3







H
+
CH3CH 2 HC `CH 2   CH3CH 2 HC = CH 2  H





H
+

CH
CH
H
C
C
H

H
C
=
CHCH
CH

H
`
3
2
2
2
2
3




(a)
(b)
(c)
(d)
(4.13)
1
+
A + H+ 
I

P
+
H
2
(4.44)
13
Differential Equations For
Each Specie
5
rA   riA,i
d[ I ]
 r1  r2  r3
dt
(4.45)
i 1
(4.25)
d[A]
 r2 - r1
dt
(4.46)
14
Next Substitute Rate Laws
A
1 3
+
+
+ H I  P + H
2
r2  k 2 [I]
(4.48)
(4.44)
r3  k 3[I]
(4.49)
15
Resultant Differential Equations
d[A]
+
 k 2[ I ] - k1[A][H ]
dt
(4.50)
d[ I ]
+
 k1[A][H ] - (k 2 + k3 )[ I ]
dt
(4.51)
16
Integration Of The Rate
Equation
Three approaches:
• Analytical integration of the differential
equations
• Numerical integration of the differential
equations
• Approximate integration of the rate
equation
17
Analytical Solution:
d[A]
+
 k 2[ I ] - k1[A][H ]
dt
(4.50)
d[ I ]
+
 k1[A][H ] - (k 2 + k3 )[ I ]
dt
(4.51)
2 equations and 2 unknowns
18
Analytical Solution
0  (k 4
 k 6 )exp(-k 5t) - (k 5 - k 6 )exp(-k 4 t) 

(k 4  k 5 )

(4.53)
 k 6 )(k 6 - k 5 ) 
exp(-k 5t) - exp(-k 4 t)
k 2 (k 4  k 5 ) 
(4.54)
[A] = [A] 

0  (k 4
[I] = [A] 

1
k 4   k 6  k 2  k3 
2
k6  k 2  k3 2  4k6k3 
(4.55)
1
k5   k 6  k 2  k3 
2
k6  k 2  k3 2  4k6k3 
(4.56)


 
k 6  k1 H 
(4.57)
Derivation
21
Plot Of Result
1
Concentration, Mol/Lit
[A]
0.8
[I]
S
0.6
0.4
[P]
0.2
[I]
0
0
1
2
3
4
Time, Min
22
Next: The Pseudo-Steady State
Approximation
According to the 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
formation of the intermediates is negligible (i.e.,
the derivatives with respect to time of the
concentrations of all intermediates are negligible
compared to other terms in the equation.)
23
For Our Example
d[I]

0
 k1[A][H ]  (k 2  k3 )[I]
dt
(4.61)
Or
 

k
[
H
]
X
1

[I]  [A ]
k k 
3
 2
(4.60)
24
7E-09
7E-09
6E-09
6E-09
Concentration, Mol/Lit
Concentration, Mol/Lit
Plot Of [I]X Calculated By
Steady State To Exact
5E-09
4E-09
Key
3E-09
Exact
2E-09
Steady
State
Approx
1E-09
0
5E-09
4E-09
Key
3E-09
Exact
2E-09
Steady
State
Approx
1E-09
0
5E-08
1E-07
1.5E-07
Time, Mins
2E-07
0
0
1
2
3
4
Time, Mins
25
(4.61)
0.25
0.25
0.2
0.2
Rate, Mole/Lit-Min
d[ I ]
 k1[A][H + ] - (k 2 + k3 )[ I ]
dt
Rate, Mole/Lit-Min
Why Does The Steady State
Approximation Work?
0.15
Key
0.1
k1 [A][H +]
0.15
Key
0.1
k1 [A][H +]
(k 2+ k 3)[I]
(k 2+ k 3)[I]
0.05
0.05
Deriv
0
0
5E-08
1E-07
1.5E-07
Time, Mins
Deriv
2E-07
0
0
1
2
3
4
Time, Mins
Derivative much smaller than other terms in the equation.
26
Steady-State Approximation
Does Not Always Work
• Assumes derivatives small compared to other
terms in equation
• Generally holds only when the concentration of
intermediate small compared to reactants and
products
• Need rate constants for destruction of intermediates to
be more than a factor of 100 larger than the rate
constants for formation of intermediates
• Common examples of failures include:
• Cyclic reactions, biological reactions
• Most reactions in plasma reactors
• Explosions, combustion
27
Next: Example
H 2  Br2  2HBr
(4.67)
1
X + Br2 
 2Br + X
2
Br + H 2 
 HBr + H
3
H + Br2 
 HBr + Br
4
X + 2Br 
 Br2 + X
5
H + HBr 
 H 2 + Br
Calculate as rate equation.
28
Solution
rBr 2   r1  r3  r4
(4.69)
r1  k1[X][Br2 ]
r3  k3[H][Br2 ]
r4  k1[X][Br]2
1
X + Br2 
 2Br + X
2
Br + H 2 
 HBr + H
3
H + Br2 
 HBr + Br
4
X + 2Br 
 Br2 + X
5
H + HBr 
 H 2 + Br
rBr2  k1[X][Br2 ]  k 3[H][Br2 ]  k 4[X][Br] 2
29
Solution Continued
d[Br]
 2 r1  r2  r3  2 r4  r5
dt
(4.74)
d[Br]
 rBr  2 k1[X][Br2 ] - k 2 [H 2 ][Br] + k 3[H][Br2 ]
dt
 2 k 4 [ Br]2 [X] + k 5[H][HBr]
(4.75)
d[H]
 rH = k 2 [H 2 ][Br] - k 3[H][HBr2 ] - k5[H][HBr]
dt
(4.76)
30
Next Apply S.S. Approximation
0  2k1[X][Br 2 ] - k 2 [H 2 ][Br] + k 3[H][Br 2 ]
 2k 4 [Br] [X] + k 5 [H][HBr]
2
(4.77)
0   k 2[H 2 ][Br] - k 3[H][Br2 ] - k 5[H][HBr]
(4.78)
Adding the two equations together
0  2k1[X][Br2 ]  2k 4[Br] [X]
2
(4.79)
31
Solution Continued
Repeating the equation
0  2k1[X][Br2 ]  2k 4[Br]2[X]
(4.79)
Solving for [Br]
1/ 2
k 
[Br] =  1 
 k4 
[Br2 ]1/ 2
(4.80)
32
Solving for [H]
0=k 2[H2 ][Br]-k3[H][HBr2 ]-k5[H][HBr]
(4.78)
Moving the k3 and k4 terms to the other side
1/ 2
 k1 
k 3[H][Br2 ]  k 5 [H][HBr]  k 2 [H 2 ][Br]=k 2 [H 2 ]   [Br2 ]1/2
 k4 
Substituting or [Br]
1/ 2
 k1 
k 3[H][Br2 ]  k 5 [H][HBr]  k 2 [H 2 ][Br]=k 2 [H 2 ]   [Br2 ]1/2
 k4 
1/ 2
 k1 
Answer:
k 2   [H 2 ][Br2 ]1/2
k4 

[H] =
k 3 [Br2 ] + k 5 [HBr]
(4.82)
33
We Can Extend This Procedure To Any
Reaction.
1) Set up the differential equation for the
species of interest in terms of rate of all of
the elementary reactions using equation
(4.25) to keep track of the coefficients.
2) Substitute the expression for the rate of
each of the elementary reactions using
equations from section 4.3.
3) Set the derivatives of the intermediate
concentrations to zero.
34
General Steps Continued
4) Eliminate terms in the expression in (1)
which contain the concentrations of
unstable intermediates other than the
species of interest.
(Usually done by adding equations
together)
5) Solve the reactant expression for the
concentration of the species of interest.
35
Discussion Problem
Derive an equation for [H]
1
 2OH 
H 2 + O2 
2
 H 2O + H 
OH   H 2 
3
 OH  +O :
H  +O 2 
4
 OH  +H 
O : +H 2 
5
 wall
H
6
 HO 2  +X
H  O2 + X 
7
 wall
HO 2  +X 
8
 H 2O 2 + X
2HO + X 
(4.85)
36
After Considerable Algebra
2k1[H2 ][O2 ]
[H]=
k 5 +(k 6 [X]-2k 3 ) O2 
(4.98)
Derivation
Notice that denominator goes to zero at a
critical value of O2  [H] grows to  ! (Not
physical)
37
The Key Thing To Remember When
Simplifying Rate Expressions
•
•
•
•
Write down the differential equations in
terms of the rates and then substitute in
the rate equations.
Keep track of what terms you want to
eliminate and eliminate them.
Adding equations together helps.
Steady-state approximation sometimes
fails.
40
Class Question

What did you learn new today?
41
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