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:
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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