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Deriving Rate Laws Using the SteadyState Approximation  part I
Skills to develop

Explain steady state and steadystate approximation.

To derive a rate law when a mechanism is given but the rate determining
step is not identified

Derive a general expression of the rate law using the steadystate
approximation.

Make appropriate assumptions so that the derived
rate law agrees with the observed rate law
Deriving Rate Laws Using the SteadyState Approximation  part I
When a reaction mechanism has several steps of comparable rates, the
ratedetermining step is often not obvious. However, there is an
intermediate in some of the steps. An intermediate is a species that
is neither one of the reactants, nor one of the products.
The steadystate approximation is a method used to derive a
rate law. The method is based on the assumption that one intermediate in
the reaction mechanism is consumed as quickly as it is generated.
Its concentration remains the same in a duration of the reaction.
The Steady State
When a reaction involves one or more intermediates, the concentration
of one of the intermediates remains constant at some stage of the reaction.
Thus, the system has reached a steadystate, hence the name of the
technique is called steady state approximation.
The concentration of one of the intermediate, [Int] varies with time
as shown on the right. At the start and end of the reaction, [Int] does vary
with time.
When a reaction mechanism has several steps with comparable rates, the
ratedetermining step is not obvious. However, there is an intermediate
in some of the steps. The steadystate approximation implies that you
select an intermediate in the reaction mechanism, and calculate its
concentration by assuming that it is consumed as quickly as it is generated.
In the following, an example is given to show how the steadystate
approximation method works.
Example 1
If the reaction
2 N_{2}O_{5} > 4 NO_{2} + O_{2}
follows the following mechanism,
i. let k_{f} and k_{b} be forward and backward rate constants
k_{f}
N_{2}O_{5} ==== NO_{2} + NO_{3}
k_{b}
ii. NO_{3} + NO_{2} > NO + NO_{2} + O_{2}
(rate constant k_{2})
iii. NO_{3} + NO > 2 NO_{2}
(rate constant k_{3})
use the steadystate approximation to derive the rate law.
Solution
In these steps, NO and NO_{3} are intermediates. You have
production rate of NO = k_{2} [NO_{3}] [NO_{2}]
consumption rate of NO = k_{3} [NO_{3}] [NO]
A steadystate approach makes use of the assumption that the rate of
production of an intermediate is equal to the rate of its consumption.
Thus, we have
k_{2} [NO_{3}] [NO_{2}] = k_{3} [NO_{3}] [NO]
and solving for [NO] gives the result,
[NO] = k_{2} [NO_{3}] [NO_{2}] / (k_{3} [NO_{3}]) . . . (1)
For the other intermediate NO_{3},
production rate of NO_{3} = k_{f} [N_{2}O_{5}]
consumption rate of NO_{3} = k_{2}[NO_{3}] [NO_{2}] + k_{3}[NO_{3}] [NO] + k_{b}[NO_{3}] [NO_{2}]
Applying the steadystate assumption gives:
k_{f} [N_{2}O_{5}] = k_{2}[NO_{3}] [NO_{2}] + k_{3}[NO_{3}] [NO] + k_{b}[NO_{3}] [NO_{2}]
Thus,
k_{f} [N_{2}O_{5}]
[NO_{3}] =  . . . . (2)
k_{2}[NO_{2}] + k_{3}[NO] + k_{b}[NO_{2}]
Let's review the three equations (steps) in the mechanism:
step i is a equilibrium and thus can not give a rate expression.
Step ii leads to the production of some products, and the active species
NO causes further reaction in step iii. This consideration led to a rate
expression from step ii as:
d[O_{2}]
 = k_{2} [NO_{3}] [NO_{2}] . . . . . (3)
dt
Substituting (1) in (2) and then in (3) gives
d[O_{2}] k_{f} k_{2} [N_{2}O_{5}]
 =  = k [N_{2}O_{5}]
dt k_{b} + 2 k_{2}
where k = k_{f} k_{2}/(k_{b} + 2 k_{2}).
Carry out the above manipulation yourself on a piece of paper.
Simply reading the above will not lead to solid learning yet.
This is the differential rate law, and it agrees with the experimental
results.
More examples are given in the link
Steady State Approximation  Part II
Confidence Building Questions

In the steadystate approximation, the assumption is
(a) the flow of reactants occurs at constant speed
(b) a gasphase reaction
(c) the concentration of intermediates is constant
(d) concentrations of intermediates are zero.
Skill 
Explain steady state and steadystate approximation.
 The reaction between H_{2} and Br_{2}
in the presence of light is consistent with the following chain mechanism:
k_{1}
i. Br_{2} === 2Br
k_{2}
k_{3}
ii. Br* + H_{2} === HBr + H*
k_{4}
k_{5}
iii. H* + Br_{2} > HBr + Br*
What is the rate of generating H*?
(a) k_{3}[Br*][H_{2}]
(b) k_{5}[H*][Br_{2}]
(c) k_{4}[HBr][H*]
Skill 
Recognize the step that generates H*
 The reaction between H_{2} and Br_{2} in the presence of light is consistent
with the following chain mechanism:
k_{1}
i. Br_{2} === 2Br
k_{2}
k_{3}
ii. Br* + H_{2} === HBr + H*
k_{4}
k_{5}
iii. H* + Br_{2} > HBr + Br*
The rate of consuming H* is
(a) k_{5}[H*][Br_{2}] + k_{4}[HBr][H*]
(b)  k_{5}[H*][Br_{2}]  k_{4}[HBr][H*]
Skill 
Work out the rate law.
 The reaction between H_{2} and Br_{2} in the presence of light is consistent
with the following chain mechanism:
k_{1}
i. Br_{2} === 2Br
k_{2}
k_{3}
ii. Br* + H_{2} === HBr + H*
k_{4}
k_{5}
iii. H* + Br_{2} > HBr + Br*
If we use K to represent some expression of the rate constants
k_{1}, k_{2}, ... k_{5},
then the rate law is
a. rate = K [Br_{2}] [H_{2}]
b. rate = K [Br_{2}]^{1/2} [H_{2}]
c. rate = K [Br_{2}]^{2} [H_{2}]
d. rate = K [Br_{2}]^{3/2} [H_{2}]
e. rate = K [Br_{2}]^{3/2}
Skills 
The skills to derive the steady state approximation has been
broken down to to smaller steps in these questions.
Review them and get the whole picture
More examples are given in
Steady State Approximation  Part II
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