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 0.589 M N2O5 solution at 45 ∘C over a period of 18.4 h if the gas is collected in a 11.2-L container. (Assume that the products do not dissolve in chloroform.)

I came up with .05757 atm but its telling me I'm wrong.

Answer 1

N₂O5 ---> N₂O +2O₂

Moles of N2O5 in 1.63 L of 0.589 M = 1.63 x 0.589 = 0.96007

Since the decomposition is first order

% conversion X_A at 18.4 hrs = 18.4 x 60 x 60 = 66240 sec

-ln(1-X_A) = K x t

-ln(1-X_A) = (1 x 10-5) x 66240 = 0.6624

(1-X_A) = e^-0.6624

1-X_A = 0.5156

X_A = 0.4844

So moles of N2O5 converted= 0.96007 x 0.4844 = 0.4650

Moles of oxygen formed = twice the moles of N2O5 decomposed= 2*0.4650 = 0.2325

Moles of N2O formed= 0.4650

Total moles = 0.96007 + 0.4650= 1.42512

The total pressure of product gases is calculated by using, PV= nRT

R= 0.08206 atm.L/mole.K

T= 45 deg C = 318.15 K

n = 1.42512

P= nRT/V

P = (1.42512 x 0.08206 x 318.15) 11.2

= 3.321 atm

Partial pressure of O2 = Mole fraction * Total pressure of oxygen

= (0.96007 / 1.42512) x 3.321

= 2.237 atm