Question

The reaction, \[\ce{2N2O5_{(g)} -> 4NO2_{(g)} + O2_{(g)}}\] is carried out in a closed vessel. The concentration of N₂O_5(g) is found to decrease by 2 × 10⁻² mol L⁻¹ in 10 seconds. Calculate the rate of reaction and the rate of appearance of NO_2(g).

Answer 1

Given:

Decrease in concentration of N₂O₅ = 2 × 10⁻² mol L⁻¹

Time (t) = 10 seconds

Rate of disappearance of N₂O₅ = \[\ce{-\frac{d[N_2O_5]}{dt}}\]

= \[\ce{\frac{2 \times 10^{-2}}{10}}\]

= 2 × 10⁻³ mol L⁻¹ s⁻¹

Rate of reaction = \[\ce{\frac{1}{2} -\frac{d[N_2O_5]}{dt}}\]

= \[\ce{\frac{1}{2} \times 2 \times 10^{-3}}\]

= 1 × 10⁻³ mol L⁻¹ s⁻¹

From the stoichiometry

\[\ce{2N2O5 -> 4NO2_{(g)} + O2_{(g)}}\]

So, the rate of appearance of NO₂ is

\[\ce{\frac{4}{2} \times 2 \times 10^{-3}}\]

= 4 × 10⁻³ mol L⁻¹ s⁻¹