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 oC of 1.0 x 10–5 s–1. Calculate the partial pressure (in atm) of O2 produced from 1.00 L of 0.600 M N2O5 solution at 45 oC over a period of 20.0 h if the gas is collected in a 10.0-L container. (assume that the products do not dissolve in chloroform.)

Answer 1

The reaction of decomposition of Dinitrogen pentoxide (N₂O₅) in chloroform as a solvent to yield NO₂ and O₂ as follows:

2 N₂O₅ → 4 NO₂ + O₂

Reaction is first order with respect to dinitrogen pentoxide, i.e.

Rate = - k [concentration of N₂O₅]
d[N₂O₅]/dt = - k∙[N₂O₅]

The integrated rate law for this reaction is as follows:
ln([N₂O₅]) = ln([N₂O₅]₀) - k∙t
or we can write as follows:
[N₂O₅] = [N₂O₅]₀ ∙e^(-k∙t)

Here t= 20.0 h= 72000s and k = 1.0 x 10⁻⁵s⁻¹

[N₂O₅] = 0.600 /1.0 = 0.600 M
Now calculate the concentration of N₂O₅ after 20.0 h is:
[N₂O₅] = 0.600mol/L ∙ e^(-1.0×10⁻⁵s⁻¹ ∙ 72000 s)

= 0.29 mol/L

now determine the change of concentration of N₂O₅ with times :

∆n(N₂O₅) = ∆[N₂O₅] ∙ V
= (0.600mol/L - 0.292mol/L) ∙ V
= 0.308 mol

2 N₂O₅ → 4 NO₂ + O₂

According to reaction equation one mole of oxygen is formed per two moles of N₂O₅ decomposed, so calculate the moles of oxygen formed in the 20 h is as follows:

Moles of (O₂) =0.308 mol N₂O₅ *1 mol O₂ / 2 mol N₂O₅

= 0.154 mol O₂

Now calculate the Partial pressure of O₂ with the use of ideal gas law:
p(O₂)∙V = n(O₂)∙R∙T or

p(O₂) = n(O₂)∙R∙T / V

Here n= 0.154 mol O₂ T= 45 C= 45+273=318 K, V= 10.0L

R = 0.0821 L atm / K mol

= 0.154 mol ∙ 0.0821 L atm / K mol∙ (318)K / 10.0 L

= 0.402 atm