Nitrogen dioxide dimerizes to give dinitrogen tetroxide: 2NO2 (g) → N2O4 (g). At 298 K, 12.55 g of an NO2/N2O4 mixture exerts a pressure of 0.963 atm in a volume of 6.92 L. What are the mole fractions of the two gases in the mixture?

To answer this item, we assume that the gases are ideal for us to be able to use the ideal gas law.

PV = nRT

where P is pressure, V is volume, n is the number of moles, R is the universal gas constant and T is the temperature (in K)

Substituting the known values,

(0.963 atm) (6.92 L) = n (0.0821 L. atm/mol. K) (298 K)

The value of n from the equation is n = 0.27237 moles

We let x and y be the number of moles of NO2 and N2O4, respectively. Given the calculated total moles above and the total mass,

x + y = 0.27237

30x + 60y = 12.55

The values of x and y are:

x = 0.126 moles

y = 0.146 moles

The mole fractions of each gases are therefore:

mole fraction of NO2 = 0.126 / (0.126 + 0.146) = 0.46

mole fraction of N2O4 = 0.146 / (0.126 + 0.146) = 0.54