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.

Answer 1

Answer:

The rate of appearance of NO2 and the rate of disappearance of N2O5 would be 4 × 10–3 mol lit.–1 sec–1, 2 × 10–3 mol lit.–1 sec–1 . The Rate of reaction = -1/2Δ[N2O5]/Δt = 1/4Δ[NO2]/Δt. As NO2 is product, therefore, the concentration is zero when t = 0. 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 or, 4 x 10-3/4  mol lit.–1 sec–1or, 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.