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 1.15 L of 0.592 M N2O5 solution at 45 ∘C over a period of 24.3 h if the gas is collected in a 12.1-L container. (Assume that the products do not dissolve in chloroform.)

Answer 1

K= 1x10^-5 /s

t= 23.4x3600s=84240

N2O5----->2NO2 + O2

0.6808            0       0      I

-x                 2x           x   C

this is 1st oder reaction

-Kt = ln[N2O5] f / [N2O5]initial

mole of N2O5 = .592x1.15=0.6808

ln N2O5 = ln [N2O5]initial - Kt= ln .6808 - 1x10^-5 x 84240=- 0.45791

[N2O5] = 0.2933

total mole of gases=0.2933+0.3875+.775=1.4558

P2= n2RT/ V2= 1.4558 x 318 x0.082 / 12.1= 3.137 atm

mole of O2 = mole of N2O5 decompose= x = 0.6808- .2933 = 0.3875

mole fraction of O2 = 0.3875 /1.4558= 0.266

partial pressure = total pressurex mole fraction = 3.137x0.266=0.8349atm