Question

For the reaction, \[\ce{N2O4_{(g)} <=> 2NO2_{(g)}}\], it has been found that the pressure of N₂O₄ falls from 0.64 atm to 0.38 atm in 28 minutes. Calculate the rate of reaction and the rate of appearance of NO_2(g).

Answer 1

Given:

\[\ce{N2O4_{(g)} <=> 2NO2_{(g)}}\]

Initial pressure of N₂O₄ = 0.64 atm

Final pressure of N₂O₄ = 0.38 atm

Time = 28 minutes

Change in pressure N₂O₄:

\[\ce{\Delta P_{N_2O_4}}\] = 0.64 − 0.38

= 0.26 atm

Since the rate is based on N₂O₄ disappearance and its stoichiometric coefficient is 1:

Rate of reaction = \[\ce{\frac{\Delta P_{N_2O_4}}{\Delta t}}\]

= \[\ce{\frac{0.26}{28}}\]

= 9.28 × 10⁻³ atm min⁻¹

Rate of appearance of NO₂:

From the balanced reaction:

\[\ce{1N2O4 −> 2NO2}\]

⇒ Rate of appearance of NO₂

= 2 × Rate of reaction

= 2 × 9.28 × 10⁻³

= 1.86 × 10⁻² atm min⁻¹