Transcript Document 7397981

Self-test 22.8 Derive the rate law for the decomposition of ozone in the
reaction 2O3(g) → 3O2(g) on the basis of the following mechanism
O3 → O2 + O
k1
O2 + O → O3
k1’
O + O3 → O 2 + O 2
k2
Solution:
First write the rate law for the reactant O3 and the intermediate product O
d [O3 ]
  k1 [O3 ]  k1' [O2 ][O ]  k 2 [O ][O3 ]
dt
d [O ]
dt
 k1[O3 ]  k1' [O2 ][O]  k2 [O][O3 ]
Applying the steady state approximation to [O]
0  k1 [O3 ]  k1' [O2 ][O ]  k 2 [O ][O3 ]
 [O] 
k1[O3 ]
k1' [O2 ]  k 2 [O3 ]
Plug the above relationship back to the rate law of [O3]
d [O3 ]
k1[O3 ]
  k1 [O3 ]  (k1' [O2 ]  k 2 [O3 ]) '
dt
k1[O2 ]  k 2 [O3 ]
Rate determining step
Simplifications with the rate determining step
•
Suppose that k2 >> k1,
then k2 – k1 ≈ k2
e  k2t  e  k1t
therefore concentration C

k1e  k2t  k 2e  k1t 
[C ]  1 
[ A]0
k 2  k1


can be reorganized as
[C] ≈ (1 - e  k t)[A]0
1
•
The above result is the same as obtained with steady state approximation
Kinetic and thermodynamic control
of reactions
• Consider the following two reactions
A + B → P1 rate of formation of P1 = k1[A][B]
A + B → P2 rate of formation of P2 = k2[A][B]
• [P1]/[P2] = k1/k2 represents the kinetic
control over the proportions of products.
•
Problems 22.6 The gas phase decomposition
of acetic acid at 1189 K proceeds by way of
two parallel reactions:
(1) CH3COOH → CH4 + CO2, k1 = 3.74 s-1
(2) CH3COOH → H2C=C=O + H2O, k2 = 4.65 s-1
What is the maximum percentage yield of the
ketene CH2CO obtainable at this temperature.
Pre-equilibrium
• Consider the reaction:
A + B ↔ I → P
when k1’ >> k2, the intermediate product, I, could reach an
equilibrium with the reactants A and B.
• Knowing that A, B, and I are in equilibrium, one gets:
and K  k1
[I ]
K
[ A][ B]
k1'
• When expressing the rate of formation of the product P in terms of
the reactants, we get
d[ P ]
 k 2 [ I ]  k 2 K [ A][ B]
dt
Self-test 22.9: Show that the pre-equilibrium mechanism in which 2A
↔ I followed by I + B → P results in an overall third-order reaction.
Solution: write the rate law for the product P
d[ P ]
 k 2 [ I ][ B]
dt
Because I, and A are in pre-equilibrium
K
so
[I ]
[ A]2
[I] = K [A]2

d[ P ]
 k 2 K [ A]2 [ B]
dt
Therefore, the overall reaction order is 3.
Kinetic isotope effect
• Kinetic isotope effect: the decrease in the rate of a chemical
reaction upon replacement of one atom in a reactant by a heavier
isotope.
• Primary kinetic isotope effect: the kinetic isotope effect observed
when the rate-determining step requires the scission of a bond
involving the isotope:
~
k (C  D)
 CH 1 / 2 

hc
v
(C  H ) 
with
e

) 
1  (
k (C  H )
2kT
 CD


• Secondary kinetic isotope effect: the variation in reaction rate
even though the bond involving the isotope is not broken to form
product:
k (C  D)
 e 
k (C  H )
~
with
~
 CH 1 / 2 
hc {v  (C  H )  v (C  H )} 

) 
1  (
2kT

CD


Kinetic isotope effect
22.8 Unimolecular reactions
•
The Lindemann-Hinshelwood mechanism
A reactant molecule A becomes energetically excited by collision with
another A molecule:
A + A → A* + A
d [ A*]
 k1 [ A]2
dt
The energized molecule may lose its excess energy by collision with
another molecule:
A + A* → A + A
d [ A*]
  k1' [ A][ A*]
dt
The excited molecule might shake itself apart to form products P
A* → P
d [ A*]
  k 2 [ A*]
dt
The net rate of the formation of A* is
d [ A*]
 k1 [ A] 2  k1' [ A][ A*]  k 2 [ A*]
dt
• If the reaction step, A + A → A* + A, is slow enough to be
the rate determining step, one can apply the steady-state
approximation to A*, so [A*] can be calculated as
d [ A*]
 k1[ A]2  k1' [ A][ A*]  k 2 [ A*]  0
dt
Then
[ A*] 
k1[ A]2
k1' [ A]  k2
The rate law for the formation of P could be reformulated as
d [ P]
k1k 2 [ A]2
 k2 [ A*]  '
dt
k1[ A]  k 2
Further simplification could be obtained if the deactivation of A* is
much faster than A*  P, i.e., k1' [ A ][ A*]  k2 [ A*] then k1' [ A]  k2
d [ P ] k1 k 2
 ' [ A]
dt
k1
'
in case k1[ A]  k2
d[ P ]
 k1 [ A]2
dt
• The equation
reorganized into
d [ P]
k1k 2 [ A]2
 k2 [ A*]  '
dt
k1[ A]  k 2
can be
d [ P]
k1k 2 [ A]
( '
)[ A]
dt
k1[ A]  k 2
• Using the effective rate constant k to represent
k
k1k 2 [ A]
k1' [ A]  k 2
• Then one has
'
1
k1

k
k1k 2
1

k1[ A]
The Rice-Ramsperger-Kassel
(RRK) model
• Reactions will occur only when enough of required energy
has migrated into a particular location in the molecule.

E 

P  1 
E 

*
s 1
s 1

E 
 kb
kb ( E )  1 
E 

*
• s is the number of modes of motion over which the energy
may be dissipated, kb corresponds to k2
The activation energy of combined reactions
• Consider that each of the rate constants of the following reactions
A + A → A* + A
A + A* → A + A
A* → P
has an Arrhenius-like temperature dependence, one gets
k1 k 2 A1e  E 1/ RT A2 e  E2 / RT

'
k1'
A1' e  E1 / RT

A1 A2
A1'
e ( E1  E2  E1 ) / RT
'
Thus the composite rate constant also has an Arrhenius-like form
with activation energy,
E = E1 + E2 – E1’
Whether or not the composite rate constant will increase with
temperature depends on the value of E,
if E > 0, k will increase with the increase of temperature
Combined activation energy
• Theoretical problem 22.20
The reaction mechanism
A2 ↔ A + A (fast)
A + B → P (slow)
Involves an intermediate A. Deduce the
rate law for the reaction.
• Solution: