Question

Nitrogen dioxide exists in the gas phase as a mixture of NO2 and N2O4. At 25 °C and 0.98 atm, the density of the gas mixture is 2.7 g/L. What is the partial pressure of each gas?

Answer 1

let us assume we had 1 liter volume

then mass of gas mixture = 2.7 g  

hence mNO2 + mN2O4 = 2.7 .........................................(1)

we   use formula PV = nRT to get moles of mixture n

0.98 x 1 = n x 0.08206 x 298

n = 0.04 = moles of NO2 + moles of N2O4

now we have formula moles = mass / molar mass

hence moles of NO2 = mass of NO2 / 46   = 0.021739 mNO2

moles of N2O4 = masss of N2O4 / 92   = 0.0108696 mN2O4

now 0.4 = 0.021739mNO2 + 0.0108696mN2O4 .....................(2)

we solve by (1) and ( 2)

0.4 = 0.021739 mNO2 + 0.0108696( 2.7 - mNO2)

mNO2 = 0.98   , mN2O4 = 1.72

moles of NO2 = 0.98 /46 = 0.0213

moles of N2O4 = 1.72 / 92 = 0.018696

mol fraction of NO2 = moles ofNO2 / ( total gas moles) = ( 0.0213 /0.0213+0.018696) = 0.53256

mol fraction of N2O4 = 1- mol fc of NO2 = 1-0.53256 = 0.46744

partial P of NO2 = mol fc of nO2 x total P = 0.53256 x 0.98 = 0.522 atm

partial P of N2O4 = 0.98 - 0.522 = 0.458 atm