[MelgaShop]

Hi all, I've been thinking in a way how to solve this problem, but I can't fine it. Here it goes:

There are 7 mol of NO2 closed in a bottle at 25ºC (298 Kelvin). After some time, it will react this way:

2 NO2 <-> N2O4 Kc= 175

Calculate the volume of the bottle, knowing that the fraction of mols of the NO2, in the equilibrium, is 15,8%.



What do I have to do? I know that at the final, there will be 1.11 mol of NO2 and 0,55 of N2O4.
Then how will I get it? From the Kc equation? Or from the PV=nRT one?

I'll be glad to get your tips. Thanks for your time!

[Donaldson Tan]

what is the unit of Kc? I assume it is dm³/mol

using PV = nRT suggest that your gaseous mixture is a perfect gas but it may not be true. However, we need not use the perfect gas assumption because we are given the mole fraction of NO2 at equilibrium.

let x be mole of NO2 dimerised

at equilibrium
no. of mole of NO2 = 7 - x
no. of mole of N2O4 = x/2
total number of moles = 7 - x + x/2 = 7 - x/2

(7-x)/(7-x/2)= 0.158
(14-2x)(14-x) = 0.158
14 - 2x = 2.212 - 0.158x
x = 6.4moles

no. of moles of NO2 = 7 - x = 0.6moles
no. of moles of N2O4 = x/2 = 3.2moles

let V be volume of system
[NO2] = 0.6/V
[N2O4] = 3.2/V
K = [N2O4]/[NO2]²
solve for V