Question

  use the steady state approximation

2N2O5 (g) → 4NO2 (g) + O2 (g)

The mechanism is based on:

  N2O5 → NO2 + NO3    ka

  NO2 + NO3 → N2O5    k'a

  NO2 + NO3 → NO2 + O2 +NO   kb

  NO + N2O5 → NO2 + NO2 + NO2   kc

Answer 1

The mechanism is based on:

step 1 N2O5 → NO2 + NO3    ka

step 2 NO2 + NO3 → N2O5    k'a

step 3 NO2 + NO3 → NO2 + O2 +NO   kb

step 4 NO + N2O5 → 3 NO2   kc

In these steps, NO and NO₃ are intermediates.

For the other intermediate NO,

Formation rate of NO = kb [NO3] [NO2]
Consumption rate of NO = kc [NO] [N2O5]

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

kb [NO3] [NO2] = kc [NO] [N2O5]

and solving for [NO] gives the result,

[NO] = kb [NO3] [NO2] / (kc [N2O5]) . . . (1)

For the other intermediate NO₃,

Formation rate of NO3 = ka [N2O5]
Consumption rate of NO3 = k'a[NO3] [NO2] + kb[NO3] [NO2]

Applying the steady-state assumption gives:

ka [N2O5] = k'a[NO3] [NO2] + kb[NO3] [NO2]

Thus,

ka [N2O5]
[NO3] = -------------------------------- . . . . (2)
k'a[NO2] + kb[NO2]

Let's review the three equations (steps) in the mechanism:
Step 3 leads to the production of some products, and the active species NO causes further reaction in step 4. This consideration led to a rate expression from step 3 as:

      d[O2]
      ----- = kb [NO3] [NO2] . . . . . (3)
       dt

Substituting (2) in (3) gives

                            ka [N2O5]
rate = k[N2O5] = -------------------------------- [NO2]
                        k'a[NO2] + kb[NO2]