Question

In the gas phase, nitrogen dioxide is actually a mixture of nitrogen dioxide (NO2) and dinitrogen...

In the gas phase, nitrogen dioxide is actually a mixture of nitrogen dioxide (NO2) and dinitrogen tetroxide (N2O4). If the density of such a mixture is 6.76 g/L at 74°C and 3.50 atm, calculate the partial pressures of the gases and KP for the dissociation of N2O4.

Pressure of NO2:

Pressure of N2O4:

KP:

Homework Answers

Answer #1

Apply:

Davg = P*MWavg/(RT)

MWavg = MW*x1 + MW*x2

solve for MWavg

MWavg = (D*RT)/P = (6.76*0.082*(74+273))/3.5

MW avg = 54.956 g/mol

MW of N2O4 = 92.011 g/mol

MW of NO2 = 46.00550 g/mol

so

54.956 = (x-N2O4 )(92.011 ) + (x-NO2)*46.00550

since 1 = x-NO2+x-N2O4

let

x-NO2 = 1-N2O4

54.956 = (x-N2O4 )(92.011 ) + (1-N2O4)*46.00550

54.956 = 92.011 *x + 46 - 46x

54.956 - 46 = (92.01 - 46)x

x = (54.956 - 46 ) /(92.01 - 46) = 0.1946

1-x = 0.8054

NO2 = 0.8054

N2O4 =0.1946

P-NO2 = 0.8054*3.5 = 2.8189 atm

P-N2O4 =0.1946*3.5 = 0.6811 atm

so

N2O4 <--> 2NO2

Kp = (P-NO2)^2 / (P-N2O4)

Kp = (2.8189 ^2)/(0.6811) = 11.666

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