2N2O5(g) → 4NO2(g) + O2(g),at a certain time, the rate of formation of NO2 is 0. 04 Ms^(-1).

Question

In the reaction,

2N₂O_5(g) → 4NO_2(g) + O_2(g),at a certain time, the rate of formation of NO₂ is 0. 04 Ms^-1.

Find the rate of consumption of N₂O₅, rate of formation of O and the rate of the reaction.

Answer 1

Given :

2N₂O_5(g) → 4NO_2(g) + O_2(g)

Rate of formation of NO₂ = \(\frac{d[NO_2]}{dt}\) = 0.04 Ms^-1

From the reaction,

Rate of consumption of N₂O₅ is half the rate of formation of NO₂ since when 2 moles of N₂O₅ are consumed, 4 moles of NO₂ are formed.

Rate of formation of O₂ is one-fourth rate of formation of NO₂.

Rate of reaction,

∴ (i) Rate of consumption of N₂O₅ = 0.02 Ms^-1

(ii) Rate of formation of O₂ = 0.01 Ms^-1

(iii) Rate of reaction = 0.01 Ms^-1