Question

For the reaction,

\[\ce{N2O5_{(g)} -> 2NO2_{(g)} + \frac{1}{2} O2_{(g)}}\]

the value of rate of disappearance of N₂O₅ is given as 6.25 × 10⁻³ mol L⁻¹ s⁻¹. The rate of formation of NO₂ and O₂ is given respectively as:

Options

• 6.25 × 10⁻³ mol L⁻¹ s⁻¹ and 6.25 × 10⁻³ mol L⁻¹ s⁻¹

• 1.25 × 10⁻² mol L⁻¹ s⁻¹ and 3.125 × 10⁻³ mol L⁻¹ s⁻¹

• 6.25 × 10⁻³ mol L⁻¹ s⁻¹ and 3.125 × 10⁻³ mol L⁻¹ s⁻¹

• 1.25 × 10⁻² mol L⁻¹ s⁻¹ and 6.25 × 10⁻³ mol L⁻¹ s⁻¹

Answer 1

1.25 × 10⁻² mol L⁻¹ s⁻¹ and 3.125 × 10⁻³ mol L⁻¹ s⁻¹

Explanation:

The given reaction is 

\[\ce{N2O5_{(g)} -> 2NO2_{(g)} + \frac{1}{2} O2_{(g)}}\]

Rate of disappearance of N₂O₅:

\[\ce{\frac{-d[N2O5]}{dt}}\] = 6.25 × 10⁻³ mol L⁻¹ s⁻¹

From the balanced equation:

1 mole of N₂O₅ produces 2 moles of NO₂

1 mole of N₂O₅ produces 0.5 moles of O₂

So,

\[\ce{\frac{d[NO2]}{dt} = 2 \times \frac{d[N2O5]}{dt}}\] = 2 × 6.25 × 10⁻³ = 1.25 × 10⁻² mol L⁻¹ s⁻¹

\[\ce{\frac{d[O2]}{dt} = \frac{1}{2} \times \frac{d[N2O5]}{dt}}\] = 0.5 × 6.25 × 10⁻³ = 3.125 × 10⁻² mol L⁻¹ s⁻¹