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finlep [7]
4 years ago
14

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)
Chemistry
1 answer:
Pavel [41]4 years ago
8 0

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

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