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Deriving Rate Laws Using the Steady-State Approximation - part I

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Deriving Rate Laws Using the Steady-State Approximation - part I

When a reaction mechanism has several steps of comparable rates, the rate-determining 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 steady-state 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 steady-state, 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 rate-determining step is not obvious. However, there is an intermediate in some of the steps. The steady-state 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 steady-state approximation method works.

Example 1

If the reaction 2 N2O5 -> 4 NO2 + O2 follows the following mechanism, i. let kf and kb be forward and backward rate constants               kf
N2O5 ==== NO2 + NO3
              kb
ii. NO3 + NO2 --> NO + NO2 + O2       (rate constant k2)
iii. NO3 + NO --> 2 NO2           (rate constant k3)
use the steady-state approximation to derive the rate law.

Solution
In these steps, NO and NO3 are intermediates. You have

production rate of NO = k2 [NO3] [NO2]
consumption rate of NO = k3 [NO3] [NO]
A steady-state 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 k2 [NO3] [NO2] = k3 [NO3] [NO]
and solving for [NO] gives the result, [NO] = k2 [NO3] [NO2] / (k3 [NO3]) . . . (1)
For the other intermediate NO3, production rate of NO3 = kf [N2O5]
consumption rate of NO3 = k2[NO3] [NO2] + k3[NO3] [NO] + kb[NO3] [NO2]
Applying the steady-state assumption gives: kf [N2O5] = k2[NO3] [NO2] + k3[NO3] [NO] + kb[NO3] [NO2]
Thus,                   kf [N2O5]
[NO3] = -------------------------------- . . . . (2)
            k2[NO2] + k3[NO] + kb[NO2]
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[O2]
      ----- = k2 [NO3] [NO2] . . . . . (3)
       dt
Substituting (1) in (2) and then in (3) gives
      d[O2]    kf k2 [N2O5]
      ----- = -------------  =  k [N2O5]
       dt      kb + 2 k2
where k = kf k2/(kb + 2 k2).

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

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