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Posted by 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 - , 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 - , Monday, May 13, 2013 at 5:28pm

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

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