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?

Question

Kristen Oconnor

Chemistry

17 June, 00:10

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?

+5

Answers (1)

Know the Answer?

Kingston Mosley · Answer 1

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