Question

Nitrogen dioxide (NO2) cannot be obtained in a pure form in the gas phase because it exists as a mixture of NO2 and N2O4. At 16°C and 0.94 atm, the density of this gas mixture is 2.7 g·L−1. What is the partial pressure of each gas? (2 sig fig)

Answer 1

PNO₂ = 0.49 atm

PN₂O₄ = 0.45 atm

Step-by-step explanation:

Let's begin with the equation of ideal gas, and derivate from it an equation that involves the density (ρ = m/V).

PV = nRT

n = m/M (m is the mass, and M the molar mass)

`PV = (m)/(M)RT`

`PxM = (m)/(V)RT`

PxM = ρRT

ρ = PxM/RT

With the density of the gas mixture, we can calculate the average of molar mass (Mavg), with the constant of the gases R = 0.082 atm.L/mol.K, and T = 16 + 273 = 289 K

`2.7 = (0.94xMavg)/(0.082x289)`

0.94Mavg = 63.9846

Mavg = 68.0687 g/mol

The molar mass of N is 14 g/mol and of O is 16 g/mol, than
`M_(NO2) = 46` g/mol and
`M_(N2O4) = 96` g/mol. Calling y the molar fraction:

`Mavg = M_(NO2)y_(NO2) + M_(N2O4)y_(N2O4)`

And,

`y_(NO2) + y_(N2O4) = 1`

`y_(N2O4) = 1 - y_(NO2)`

So,

`68.0687 = 46y_(NO2) + 92x(1 - y_(NO2))`

`68.0687 - 92 = 46y_(NO2) - 92y_(NO2)`

`46y_(NO2) = 23.9313`

`y_(NO2) = 0.52`

`y_(N2O4) = 0.48`

The partial pressure is the molar fraction multiplied by the total pressure so:

PNO₂ = 0.52x0.94 = 0.49 atm

PN₂O₄ = 0.48x0.94 = 0.45 atm