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

Answer:

PNO₂ = 0.49 atm

PN₂O₄ = 0.45 atm

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)

$PV = \frac{m}{M}RT$

$PxM = \frac{m}{V}RT$

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

$2.7 = \frac{0.94xMavg}{0.082x289}$

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