Posted by **kriti** on Sunday, May 12, 2013 at 11:41am.

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.

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

- physical chemistry -
**DrBob222**, Monday, May 13, 2013 at 5:13pm
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.

- oops--physical chemistry -
**DrBob222**, Monday, May 13, 2013 at 5:28pm
I believe that is 15 atm for Ptotal and not 1.5.

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