Question

The mechanism for the reaction described by

2N₂O₅(g) ---> 4NO₂(g) + O₂(g)
is suggested to be

(1) N₂O₅(g) (k1)--->(K-1) NO₂(g) + NO₃(g)
(2) NO₂(g) + NO₃(g) --->(K2) NO₂(g) + O₂(g) + NO(g)
(3) NO(g) + N₂O₅(g) --->(K₃) 3NO₂(g)

Assuming that [NO₃] is governed by steady-state conditions, derive the rate law for the production of O2(g) and enter it in the space below.
Rate of reaction= ?[O₂]/?t = ?

Answer 1

The rate law for the production of O₂(g) is

Rate = (1/2)k₁k₂[N₂O₅] / (1 + k₃[NO] / k₂)

How can you derive the rate law for the production of O₂(g)?

the rate expressions for each step

1: N₂O₅(g) → NO₂(g) + NO₃(g)

Rate = k₁[N₂O₅]

2: NO₂(g) + NO₃(g) → NO₂(g) + O₂(g) + NO(g)

Rate = k₂[NO₂][NO₃]

3: NO(g) + N₂O₅(g) → 3NO₂(g)

Rate = k₃[NO][N₂O₅]

The steady-state approximation states that the rate of formation of an intermediate species is equal to the rate of its consumption. In this case, the intermediate species is NO₃. Therefore, we can write:

k₁[N₂O₅] = k₂[NO₂][NO₃] + k₃[NO][N₂O₅]

[NO₃] = (k₁[N₂O₅]) / (k₂[NO₂] + k₃[NO])

Rate = k₂NO₂: (k₁[N₂O₅] / (k₂[NO₂] + k₃[NO]))

Rate = (k₁k₂[NO₂][N₂O₅]) / (k₂[NO₂] + k₃[NO])

Simplifying the rate expression

Rate = k₁k₂[N₂O₅] / (1 + k₃[NO] / k₂)

Since one molecule of O₂ is produced for every two molecules of NO₂, we can express the rate in terms of [O₂] as follows:

Rate = (1/2)k₁k₂[N₂O₅] / (1 + k₃[NO] / k₂)

Therefore, the rate law for the production of O₂(g) is:

Rate = (1/2)k₁k₂[N₂O₅] / (1 + k₃[NO] / k₂)