The system N2O4 <--->2NO2 maintained in a closed vessel at 60º C & a pressure of 5 atm has an average (i.e. observed) molecular weight of 69, calculate Kp.

Question

At what pressure at the same temperature would the observed molecular weight be (230/3)

Answer 1

I would do this.
Let x = fraction of gas that is NO2
The 1-x = fraction that is N2O4
molar mass NO2 = 46
molar mass N2O4 = 92

46(x) + (92(1-x) = 69
Solve for x and I get 0.5 which is the mole fraction of these gases.
Then pNO2 = XNO2*Ptotal
and pN2O4 = XN2O4*Ptotal

Finally, Kp = p^2NO2/pN2O4
I think Kp = 2.5 but you need to confirm that.

Part 2 is similar.
46(x) + 92(1-x) = 230/3 = 76.67
x = about 0.333 = fraction NO2 = XNO2
1-x = about 0.666 = XN2O4
Then pNO2 = XNO2*Ptotal
and pN2O4 = XN2O4*Ptotal

Substitute pNO2 and pN2O4 into Kp and solve for Ptotal
Kp = p^2NO2/pN2O4 = (0.333P)^2/(0.667P)
and I solved for Ptotal as 1.5 atm. You should confirm that.

Answer 2

I believe that is 15 atm for Ptotal and not 1.5.

Answer 3

Answer is 15 atm lol