Question

Dinitrogen pentoxide (N2O5) decomposes in chloroform as a solvent to yield NO2 and O2. The decomposition is first order with a rate constant at 45 ∘C of 1.0×10−5s−1. Calculate the partial pressure of O2 produced from 1.63 L of 0.659 M N2O5 solution at 45 ∘C over a period of 24.2 h if the gas is collected in a 12.4-L container. (Assume that the products do not dissolve in chloroform.)

Answer 1

Solution-

The reaction is as follow
2 N₂O₅ → 4 NO₂ + O₂

Reaction is first order with respect to dinitrogen pentoxide, i.e.
d[N₂O₅]/dt = - k∙[N₂O₅]

Solving this differential equation with initial concentration [N₂O₅]₀ leads
to integrated rate law:
ln([N₂O₅]) = ln([N₂O₅]₀) - k∙t
<=>
[N₂O₅] = [N₂O₅]₀ ∙e^(-k∙t)

So the concentration after 24.2 hours is:
[N₂O₅] = 0.659mol/L ∙ e^(-1.0×10⁻⁵s⁻¹ ∙ 24.2∙3600s) = 0.276mol/L

The amount N₂O₅, which has reacted away, is equals to the change of concentration times the volume of the solution:
∆n(N₂O₅) = ∆[N₂O₅] ∙ V
= (0.659mol/L - 0.276mol/L) ∙ V
= 0.383mol

According to reaction equation one mole of oxygen is formed per two moles of N₂O₅ decomposed,
therefore the amount of oxygen formed in the 24.2 hours is:
n(O₂) = (1/2)∙∆n(N₂O₅) = 0.1915mol

Partial pressure in collection container can be found from ideal gas law:
p(O₂)∙V = n(O₂)∙R∙T
=>
p(O₂) = n(O₂)∙R∙T / V
= 0.1915mol ∙ 8.314472Pam³/molK ∙ (273.15 + 45)K / 0.0124m³
= 40852.03Pa
= 0.40atm