Question

For the reaction, 2N₂O₅(g) → 4NO₂ (g) + O₂(g), the concentration of NO₂ increases by 2.4 × 10^–2 Mol lit.^–1 in 6 second. What will be the rate of appearance of NO₂ and the rate of disappearance of N₂O₅ -

(a)2 × 10^–3 mol lit.^–1 sec^–1, 4 × 10^–3 mol lit^.–1 sec^–1

(b) 4 × 10^–3 mol lit.^–1 sec^–1, 2 × 10^–3 mol lit^.–1 sec^–1

(C) 2 × 10^–3 mol lit.^–1 sec^–1,2 × 10^–3 mol lit.^–1 sec^–1

(D) None of these

Answer 1

Correct Option (b) 4 × 10^–3 mol lit.^–1 sec^–1, 2 × 10^–3 mol lit^.–1sec^–1

Explanation: 

Rate of reaction  = -1/2Δ[N₂O₅]/Δt = 1/4Δ[NO₂]/Δt

Since NO₂ is the product, therefore, its concentration when t = 0 is zero.

Rate of appearance of NO₂ i.e. Δ[N₂O₅]/Δt  = 2.4 x 10^-2/6

= 4 × 10^–3 mol lit.^–1 sec^–1

Thus, rate of reaction = -1Δ[NO₂]/4Δt

4 x 10^-3/4  mol lit.^–1 sec^–1

1 x 10^-3  mol lit.^–1 sec^–1

Rate of disappearance of N₂O₅ i.e^_.  Δ[N₂O₅]/Δt = 2 x Rate of reaction

2 × 10^–3 mol lit.^–1 sec^–1