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.

Part A Calculate the partial pressure of O2 produced from 1.44 L of 0.543 M N2O5 solution at 45 ∘C over a period of 15.0 h if the gas is collected in a 12.6-L container. (Assume that the products do not dissolve in chloroform.)

Answer 1

N2O5---à 2NO2 +1/2O2

-d[N2O5]/dt= K[N2O5] = 1*10-5[N2O5]

When integrated   [N2O5] = [N2O5]0* exp(-1*10-5*t)

[N2O5]0 = concentration of N2O5 at zero time and [N2O5] is concentration of N2O4 at time t

After 16 hrs,= 15*60*60 second = 54000

[N2O5] = 0.543* exp^{(-54000* 1*10-5)} =0.32 M

Moles of N2O5 reacted= (0.543-0.32)*1.44 =0.32 moles

From the reaction, moles of oxygen formed = 0.32/2= 0.16 moles

From PV= nRT, P= nRT/V= 0.16*0.0821*(45+273)/12.6 =0.3315 atm, partial pressure of oxygen